mm-263 Can anyone recall the transcendental number which, if crudely approximated,>> appears as an integer? It's something like e^(sqrt159) or e^(pi*sqrt159.>> >> 262537412640768743.9999999999992Also: (Pi*Sqrt(163))^Exp(1)= === 22806.9992386Furthermore:quoting:Subject: Re: pinewsgroups: rec.org.mensa,sci.mathRefer to: A Classical Introduction to Modern Number Theory byIreland & Rosen, and note the following:Theorem. Let R be the ring of integers of an imaginary quadraticnumber field; then R is an elliptic lattice, so it makes sense tospeak of its j-invariant. Statement: that j-invariant is an algebraicinteger whose degree is equal to the class number of R.Apply that theorem to the ring of integers of Q(sqrt(-163)): it isgenerated by the element (1+sqrt(-163))/2, so j((1+sqrt(-163))/2)is an algebraic integer, and since the ring in question is a PID,that is, its class number is 1, the j-invariant in question is in Z.In fact, it is -640320^3. So if q=exp(i pi(1+sqrt(-163))/2)=-exp(-pi sqrt(163)), we have j((1+sqrt(-163))/2) = 1/q + 744 + higher-order terms in q -640320^3 = -exp(pi sqrt(163)) + 744 + small terms exp(pi sqrt(163)) = 640320^3 + 744 + small termsEnd quoteAnd lastly: http://www.cacr.caltech.edu/~roy/upi/episqrtn.htmlPiece of cake...er pie.John Baileyhttp://home.rochester.rr.com/jbxroads/mailto.html === Subject: e^(pi*sqrt(163)) and all that (was: Re: Almost an Integer - e^?)|> |> Can anyone recall the transcendental number which, if crudely|> approximated, appears as an integer? It's something like|> e^(sqrt159) or Len||e^(pi*sqrt(163)) = 262537412640768743.9999999999992||Google for the thread About the property of 163 in sci.math|July 2, .95Jul5163105%40mogul.osf.orgi have some questions about some messages in that thread:||>According to Maple, exp(Pi*sqrt(37))=199148647.999978 correct to 15|>figures. This looks close to an integer to me even though 37 isn't|>in the list above. [of discriminants of quadratic imaginary fields|>with class number 1] Is there a good reason why it is so close?||Yes: it's trying to approximate not j(sqrt(-37)) but H2(sqrt(-37))|where H2 parametrizes pairs of 2-isogenous elliptic curves.|Q(sqrt(-37)) does not have unique factorization, but is idoneal ---|the class group is entirely accounted for by genus theory. An even|better example of this is|exp(Pi*sqrt(58))=24591257751.999999822... which I think was known|empirically to Ramanujan.from browsing around trying to find out what some of those words likegenus and idoneal mean, i have some wild guesses about the gist ofwhat elkies is saying here, but it would be helpful if someone couldsay how close to the mark my guesses are.what i think's going on is something like this:apparently, there's something called the genus field of an abelianextension k of the rationals, and it's the maximal unramifiedextension of k that's abelian over the rationals. so apparently thegenus field is smaller than the hilbert class field which (i think)is just the maximal unramified extension. (apparently the genus fieldis the part of the hilbert class field that fits inside a cyclotomicextension of the rationals, and that makes it rather easier to dealwith than the hilbert class field?) so if the terminology works theway i'm guessing it does, it seems then that the relationship betweenclass and genus here is anti-linnaean; that is, genus seems tobe a coarser grouping than class. that is, the ideal class group isthe galois group of the hilbert class field, and there's some subgroupof the ideal class group that corresponds to the genus field in thegalois correspondence between sub-groups of the ideal class group andsub-fields of the hilbert class field; and the genuses of ideals arethe cosets of that sub-group.is that anything like the way it actually is?and is there some more conceptual description of what a genus ofideals (or of quadratic forms or whatever, if that's the main specialcase) is (or of the equivalence relation on ideals for which thegenuses are the equivalence classes)?actually i was going to ask some more questions here about idonealnumbers and so forth, but i don't think i'm quite ready for that yetso instead i'll ask === some more basic questions:|Subject: Re: Ramanujan, e, pi and 163||>a deep reason why e^{pisqrt{163}} is so close to an integer.||The reason has to do with complex multiplication and imaginary|quadratic fields of class number 1. If K = Q(sqrt(d)) is an imaginary|quadratic field, and (1,w) is a basis for the integers of K, then the|theory of complex multiplication tells us that j(w) has degree h over|K (and in fact over Q), where h is the class number of K and j() is|the elliptic modular function. Now for d = -163, w = (1 + sqrt(d))/2,|and since (as Gauss was aware) the class number of K in this case is|1, this says j((1+sqrt(-163))/2) is a rational integer. But j has a|q-expansion|| j(z) = 1/q + 744 + 196884q + 21493760q^2 + ...||where q = exp(2*pi*i*z). When we replace z by (1+sqrt(-163))/2, the|1/q term becomes -e^(pi*sqrt(163)). The second term is an integer,|the remaining terms are _quite_ small, and the whole sum must be an|integer, so e^(pi*sqrt(163)) must differ from an integer by something|small. In fact, e^(pi*sqrt(163)) =|262537412640768743.9999999999992... and j((1+sqrt(-163))/2) =|-262537412640768000 (which, incidentally, is -640320^3).||Notes in Mathematics #320, Modular Functions of One Variable I.it says above that j(w) has degree h over k and over the rationalswhere h is the class number of k. this provokes me to ask the obviousdumb question: are the values j(x) where x ranges over latticesrepresenting all of the ideal classes of k all the conjugate roots ofj(w), and is there a nice way to explicitly describe the minimalpolynomial of j(w)? (there's probably a bunch of other dumb questionsthat ought to be asked following from this, but i'll leave it at thatfor === now.)-- [e-mail address jdolan@math.ucr.edu]Subject: Re: Almost an Integer - e^?e^(Pi*sqrt(163))This is I.N.Herstein, Topics in Algebra (2nd edition), chapter 2(Groups), section 2.9 (Automorphisms), problem 19:Let G be a group and T an automorphism of G. If N is a normal subgroupof G such that (N)T is contained in N, show how you could use T todefine an automorphism of G/N.I think I'm probably missing something obvious here. I defined T' onG/N by (Ng)T' = N(gT). When defining a function on a quotient group,we have to take care that the function is well-defined. With this inmind, my T' seemed the natural choice, since the condition that (N)Tis contained in N is precisely the one we need for T' to bewell-defined. Certainly T' is a homomorphism, and since T is anautomorphism, T' is surjective.But T' need not be injective: Let G be the group of real numbers underaddition and let N be the subgroup of integers. Define T:G -> G by xT= 2x. Then G, T and N satisfy the conditions in the question but T' isnot an automorphism. For example, (N + 1/4)T' = (N + 3/4)T' = N + 1/2.My T' works whenever (N)T = N. In particular, it works whenever G isfinite. But it doesn't work in the general case. So what am I missing?John Harrison> This is I.N.Herstein, Topics in Algebra (2nd edition), chapter 2> (Groups), section 2.9 (Automorphisms), problem 19: Let G be a group> and T an automorphism of G. If N is a normal subgroup of G such that> (N)T is contained in N, show how you could use T to define an> automorphism of G/N.> > I think I'm probably missing something obvious here. I defined T' on> G/N by (Ng)T' = N(gT). When defining a function on a quotient group,> we have to take care that the function is well-defined. With this in> mind, my T' seemed the natural choice, since the condition that (N)T> is contained in N is precisely the one we need for T' to be> well-defined. Certainly T' is a homomorphism, and since T is an> automorphism, T' is surjective.> > But T' need not be injective: Let G be the group of real numbers> under addition and let N be the subgroup of integers. Define T:G ->> G by xT = 2x. Then G, T and N satisfy the conditions in the question> but T' is not an automorphism. For example, (N + 1/4)T' = (N +> 3/4)T' = N + 1/2.> > My T' works whenever (N)T = N. In particular, it works whenever G is> finite. But it doesn't work in the general case. So what am I> missing?Maybe I'm missing something, too, but I think you're right: Thecondition we need is that the inverse image of N w.r.t. T is containedin N, rather than the way the problem is stated.-- Nils G.9ascheDon't ask for whom the tolls.PGP key ID #xEEFBA4AF>> This is I.N.Herstein, Topics in Algebra (2nd edition), chapter 2>> (Groups), section 2.9 (Automorphisms), problem 19: Let G be a group>> and T an automorphism of G. If N is a normal subgroup of G such that>> (N)T is contained in N, show how you could use T to define an>> automorphism of G/N.>> >> I think I'm probably missing something obvious here. I defined T' on>> G/N by (Ng)T' = N(gT). When defining a function on a quotient group,>> we have to take care that the function is well-defined. With this in>> mind, my T' seemed the natural choice, since the condition that (N)T>> is contained in N is precisely the one we need for T' to be>> well-defined. Certainly T' is a homomorphism, and since T is an>> automorphism, T' is surjective.>> >> But T' need not be injective: Let G be the group of real numbers>> under addition and let N be the subgroup of integers. Define T:G ->>> G by xT = 2x. Then G, T and N satisfy the conditions in the question>> but T' is not an automorphism. For example, (N + 1/4)T' = (N +>> 3/4)T' = N + 1/2.>> >> My T' works whenever (N)T = N. In particular, it works whenever G is>> finite. But it doesn't work in the general case. So what am I>> missing?>Maybe I'm missing something, too, but I think you're right: The>condition we need is that the inverse image of N w.r.t. T is contained>in N, rather than the way the problem is stated.But then T' would not be well-defined. I think you need the conditionT(N) = N for this question to make much sense.Derek Holt.>Maybe I'm missing something, too, but I think you're right: The>condition we need is that the inverse image of N w.r.t. T is contained>in N, rather than the way the problem is stated.> > But then T' would not be well-defined. I think you need the> condition T(N) = N for this question to make much sense.Ah, quite right.-- Nils G.9ascheDon't ask for whom the === tolls.PGP key ID #xEEFBA4AFSubject: mathcadI am turning to you with a problem that i have whith Mathcad 2001.I woud like to type something like that (after defining a=2 ,b=3 )when i will type g=a+b and then ,g= i woud like to get somthing like g=2+3=6not g=6.I have heard that it is posible and i woud like to know how to do it .I looked === everware and i could't find it .Subject: mathcad 2001I am turning to you with a problem that i have whith Mathcad 2001.I woud like to type something like that (after defining a=2 ,b=3 )when i will type g=a+b and then ,g= i woud like to get somthing like g=2+3=6not g=6.I have heard that it is posible and i woud like to know how to do it .I looked === everware and i could't find it .Subject: Topology QuestionI'm trying to prove that for any topological space (Y,T), with a subset X ofY, that the relative topology T_X (the collection of all sets of the form Vintersect X, where V an element of T) is a topology for X.However, I'm trying to prove it by assuming that every topological spaceobeys 3 axioms, then proving those axioms.The one I'm having trouble with is1) If T_X is a topology for X, then the null set and X are open sets in T_X.Null set is trivial.However, if X is closed, I run into a wall. Could someone give me a hint howto deal with this case?(This is an assignment === question, so only hints please.)GREGSubject: Re: Topology Question> The one I'm having trouble with is> 1) If T_X is a topology for X, then the null set and X are open sets in T_X.> However, if X is closed, I run into a wall. Could someone give me a hint how> to deal with this case?Hint: Open Set is just another name for a set being an element of atopology. Example: [0,1] is closed in R, but it is open and closedwhen understood as a topological subspace of R:R intersect [0,1] = [0,1]As R is an element of the standard topology on R, [0,1] must be anelement of the induced topology on [0,1].Another example: [0,1) is neither open or closed in R, but it is anopen subset if understood as a topological subspace. It is also closedthen.Rene.-- Ren.8e MeyerStudent of Physics & === MathematicsZhejiang University, Hangzhou, ChinaSubject: Re: Topology Question>I'm trying to prove that for any topological space (Y,T), with a subset X of>Y, that the relative topology T_X (the collection of all sets of the form V>intersect X, where V an element of T) is a topology for X.>However, I'm trying to prove it by assuming that every topological space>obeys 3 axioms, then proving those axioms.>The one I'm having trouble with is>1) If T_X is a topology for X, then the null set and X are open sets in T_X.>Null set is trivial.>However, if X is closed, I run into a wall. Could someone give me a hint how>to deal with this case?There will (typically; not always, but essentially always) bemore than one subset Z of Y whose intersection with X is X.All you need is for *one* such Z to be an open set V in T.>(This is an assignment question, so only hints please.)Typically (same disclaimer as before), X itself is a badcandidate for Z, but then again, X is pretty small.Think === big!Lee RudolphSubject: Map as graph : unreachable regionI've learnt how to represent a map as the set of pairs of connectedvertices. Accordingly, the object {(A, B)} represents both (1) and(2): (1) (2) ___________ _____| | | | | A | | A || _____ | |_____|| | | | | || | B | | | B || |_____| | |_____||___________|But clearly (1) and (2) differ with respect to the reachability ofregion B from the outside [(2) reachable, (1) not]. How is this kindof fact formally expressed usually? (I can think of a couple of waysbut I'm sure this problem has been dealt before so no need to reinventthe wheel and/or use a 'strange' === mathematician playing with map coloring)Subject: Re: Map as graph : unreachable regionM?rio Amado Alves> I've learnt how to represent a map as the set of pairs of connected> vertices. Accordingly, the object {(A, B)} represents both (1) and> (2):> (1) (2)> ___________ _____> | | | |> | A | | A |> | _____ | |_____|> | | | | | |> | | B | | | B |> | |_____| | |_____|> |___________|> But clearly (1) and (2) differ with respect to the reachability of> region B from the outside [(2) reachable, (1) not]. How is this kind> of fact formally expressed usually? (I can think of a couple of ways> but I'm sure this problem has been dealt before so no need to reinvent> the wheel and/or use a 'strange' structure.)> mathematician playing with map coloring)Country A is simply connected, as they say, in (2) but not in (1). If werepresent countries by vertices rather than by faces, people speak of abridge (resp. cutpoint) as an edge (resp. vertex) which, if omitted, causesthe graph to become disconnected. So, if there are other countries outsideA, A would be a cutpoint. But the terminology of graph theory is === not verywell standardized.LHSubject: Re: Map as graph : unreachable region> M?rio Amado Alves> I've learnt how to represent a map as the set of pairs of connected> vertices. Accordingly, the object {(A, B)} represents both (1) and> (2):>> (1) (2)> ___________ _____> | | | |> | A | | A |> | _____ | |_____|> | | | | | |> > | | B | | | B |> | |_____| | |_____|> |___________|>> But clearly (1) and (2) differ with respect to the reachability of> region B from the outside [(2) reachable, (1) not]. How is this kind> of fact formally expressed usually? (I can think of a couple of ways> but I'm sure this problem has been dealt before so no need to reinvent> the wheel and/or use a 'strange' Alves> (amateur mathematician playing with map coloring)> Country A is simply connected, as they say, in (2) but not in (1). If we> represent countries by vertices rather than by faces, people speak of a> bridge (resp. cutpoint) as an edge (resp. vertex) which, if omitted, causes> the graph to become disconnected. So, if there are other countries outside> A, A would be a cutpoint....Ok. But with this I still cannot represent the fact explicitly.Consider maps (3), (4) which are (1), (2) resp. plus a region Cconnected (only) to A. The symbolic represention of *both* (3) and (4)is {(A, B), (A, C)}.I'm playing with constructing (colored) maps by adding one region at atime (and exchanging colors). For that I want to represent theunreachability condition, for example expressing the fact that I canadd pair (B, C) to (4) but not to (3).Maybe I should say a litle more. I intuit that is possible to keep thenumber of colors of the periphery of any map, below 4, where peripheryis the set of vertices available for connection to a newly addedvertex. I intuit this is possible by exchanging colors in the newlycreated map. This is of course a quest for a different (?) proof ofthe four color theorem. Maybe I'm on a beaten track. I don't know muchabout four color theorem proving, other then the 'fact' that existingproofs are not elegant enough for my === transcendence of these numbers. How to find out?> |> |>> |> Yes, the numbers you're thinking of are called Liouville numbeafter> |> Liouville who showed that if an irrational number can be approximated> |> extremely well by rationals then it must be transcendental. If the> |> 1-digits are spaced apart by factorial then the terminating decimals> |> stopping at each 1 are excellent rational approximations.> |> > |> > |> Uhm, in what sense do you mean extremely well here? It's obviously> |> not is a limit of a sequence of rationals, since the rationals are> |> dense in R. (Just curious)> |> |x is a Liouville number if, for every positive integer k, there exist> |integers m,n so that |x - m/n| < 1/n^k.> |> |That is approximated extremely well!> > You were discussing irrational numbers. Note that Liouville> numbers are required to be irrational. There are various> ways to rephrase the definition to incorporate that requirement.> There's a sense in which rational numbers are badly> approximated by other rational numbers.> > A number can be approximated by rationals to order k if there> exists a constant C>0 and infinitely many pairs of integers (m,n)> such that 0<|x-m/n| to order 1, and no higher order. All irrational numbers can be> approximated to order 2, with C=1/sqrt(5). Liouville numbers are> those numbers which can be appoximated to all orders. It's not> too hard to prove that they are transcendental, by proving that> algebraic numbers of degree d can't be approximated to order> greater than d:> > If r is a root of a polynomial P(r)=0 of degree d, then there exists> a C>0 such that |P(x)|=|P(x)-P(r)| mean value theorem. But when m and n are integen<>0,> n^k P(m/n) is an integer, and if m/n itself is not a root, and> |r-m/n|>1, then |P(m/n)| > 1/n^k, which implies |r-m/n| > 1/Cn^k.> > This is why Liouville numbers are transcendental. More is true,> however. By the Thue-Siegel-Roth theorem, any number which> can be approximated to a degree >2 is a quick follow-up question: is there anextension of this concept which applies to say, continuity ordifferentiability? (Putting constraints on the error term, I mean). Also, are there any good references for === into science classesAgain, my response to this did not come onto the newsgroup the first time, soI try again.parker@wt.net askes this:>Have any of you had particular success at a certain approach to>integrating math skills into your science courses (e.g. through Excel>exercizes, etc.)???Mathematics skills ARE integrated into most or even ALL of the physicalsciences, and at least a little bit in the other sciences. Methods of datatreatment range from using a (sometimes teacher-created) computer program, toas low level stuff at graphing onto graph paper using pencil and templatedevices. Expressing conventionally with symbolism is always included wheneverand however appropriate. If mathematics were not integrated with the learningof sciences, we would have a hell of a bad time === studying the sciences.G CSubject: Re: Integrating math into science classes>Have any of you had particular success at a certain approach to>integrating math skills into your science courses (e.g. through Excel>exercizes, etc.)???It is more important to integrate concepts than computations.Excel does not do mathematics; it does computation. Using its limited features tends to get students to do only thosetypes of problems which the computer package has been programmedto carry out; a computer is a super-fast sub-imbecile.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: === (765)494-0558Subject: Re: Integrating math into science classesparker1@wt.net asks this:>Have any of you had particular success at a certain approach to>integrating math skills into your science courses (e.g. through Excel>exercizes, etc.)???Unnecessary inquiry. Most sciences, especially physical sciences absolutely require mathematics fortheir study. Computerization is used also. Think of the various laboratoryexercises which by design, a student produces graphable data. Decisions basedon this data can often be made by using a computer program, if not at least === byplotting points onto graph paper. G CSubject: re: Making Star Trek RealThis is most interesting!There is also a published paper (Phys. Essays) by Clark Jeffries who argues that the exponential metric handles the missing mass problem that looks interesting, though I have not really dug into the details.Hal PuthoffJS: There is an interesting exponential dependence in string theory likeG* ~ G(Newton)e^<(Dilaton)>if that is what you mean? I am studying all that now.That would give a double exponentialK = e^2G*M/c^2r = e^2G(Newton)e^(M/c^rI am wondering if the ~ Lp^2/zpf ?There is a basic string vertex strength gs for closed spin 2 graviton strings that move in the total hyperspace and gs^1/2 for open strings of spin 1/2 lepto-quarks & spin 1 internal gauge force bosons whose ends are anchored to 3Dim brane parallel universes floating in hyperspace.parallel brane worlds with ends of open strings attached to different parallel brane worlds because the spin gauge force fluxes form closed loops through the ends of the open string exactly like Wheeler's Charge without charge in his 1950's book Geometrodynamics. I do not know if anyone else has yet noticed this simple explanation for broken charge conjugation symmetry?Part II Discussion with ZielinskiThe Freud identity is supposed to hold for any theory of gravitation that isbased on a spacetime curvature model. It was cited by no less an authority thanrelativity.JS: True.PZ: In the context of GR, Freud's results imply(1) that the Einstein curvature G_uv can automatically be linearly decomposedinto the difference of two pseudotensors (quantities that act like tensors underlinear, but not under general, coordinate transformations),JS: By that I suppose is meant that they are tensors under the Poincare group i.e.globally flat background but not under the curved local Diff(4) group?Locally gauge the globally flat Poincare group to get the curved spacetimeguv(curved) - nuv(global flat) = huv(curved)as the compensating field.In no sense is huv small in this local gauging, which is not perturbativerelative to flat background. Unfortunately the same notation is usedin perturbation theory.A Poincare group local stress-energy density tensor for the first order perturbativegravity where h'uv is small can be defined of course, so that may be at least oneof the pseudotensors in the Freud identity?Again trivially even in the full non-perturbative caseTuv(Gravity) = (String Tension)Guv(Geometry)and in the non-exotic vacuum where the Diff(4) divergence-free Tuv(Source Mass-Energy) = 0Tuv(Gravity) = 0because the local Einstein field equation in that case isTuv(Gravity) + Tuv(Source Mass-Energy) = 0PZ: U_uv and u_uv:1/2 G_uv = U_uv - u_uv,where u_uv is specifically identified as the Einstein vacuum stress-energypseudotensor and U_uv is called the Freud pseudotensor (Freud decomposition)JS: Sure so what's the big deal? u_uv is Tuv(Vacuum Geometry relative to Poincare Group), U_uv is Tuv(Vacuum Geometry extra curvature terms). Only the total sum is the very same Diff(4) tensor I proposedTuv(Geometry) = (String Tension)Guv(Einstein)which = 0 in NON-EXOTIC VACUUM i.e. ALL residual micro-quantum zero point energy from strings and things in large scale GR low energy limit is completely damped down to ZERO COSMOLOGICAL CONSTANT by my VACUUM COHERENCE FIELD. Like Ben Franklin I have poured oil to calm the turbulent waters. This has nothing to do with fact that in string theory completely different problem, which is precisely why Ed Witten said he was very worried. String theory does not solvethe mystery of why Einstein's cosmological constant is so small.PZ: (2) the density divergence of the Freud pseudotensor U_uv vanishes identically:partial d_v(SQRT[-g] U_uv) = 0(using the Einstein summation convention) (Freud identity)JS: Yes, but probably assuming:i. metricityii zero torsioniii non-exotic vacuum, i.e. /zpf = 0Note in string theoryLp*^2 = Lp^2(Newton)e^(Dilaton)Where Dilaton is the compensating gauge field from the Dilation charge of the Conformal Group I think?PZ: According to Yilmaz's arguments inNuovo Cimento 107B, 941 (1992)it can be shown that in view of the above, the Freud and Einstein pseudotensors caneach be split into tensorial and non-tensorial parts:U_uv = tau_uv + z_uvu_uv = -t_uv + z_uv,where z_uv is a common *non-tensor* component that always vanishes identically in*some* coordinate system.JS: By tensor I assume you still mean with respect to the full Diff(4) group of local general coordinatetransformations?PZ: Yilmaz shows that if the non-tensor quantity z_uv is mathematically eliminated from theRHS (from both U_uv and u_uv), we get1/2 G_uv = tau_uv + t_uv (vac)where the Yilmaz vacuum stress-energy term t_uv (vac) is a true tensor quantity (worldtensor) that is interpreted by Yilmaz as a localizable vacuum gravitational stress-energythat itself locally induces Riemann curvature in the general case.JS: It would be good to keep dimensions explicit[(string tension)/2] G_uv = tau_uv + t_uv (vac)Why the formal split who ordered that? How can one physically distinguish this formal split?PZ: OK. Personally, I have been focusing on fundamental aspects of EinsteinGR vs. Yilmaz, and it has been some time since I looked at the detailsof PV vs. Einstein, so I'll leave this to you and Hal.To be fair, I think Jack's position is that if the arguments cited byMTW arecorrect, it is simply not possible to define a localized gravitationalfieldstress-energy density in an absolute (i.e. world tensor-covariant)mannerwithout generating conflicts with the so-called EEP under some widelyaccepted canonical interpretation of that principle. And if the priceof Yilmazis the EEP, then Jack says: trivially i.e.Tuv(Geometry) = (String Tension)GuvIn the non-exotic vacuum the geometrodynamic stress-energy densitycurrents is completely insulated from the corresponding matter currents.That is, the covariant 4-divergences of both Tuv(Geometry) andTuv(Matter) vanish separately. This makes metric engineeringimpossible because there is then no way to directly convert matter intogeometry and vice versa in a practical cost effective way IMHO.Also in this non-exotic vacuum degenerate limiting case, Tuv(Geometry)is trivially zero!BecauseTuv(Geometry) + Tuv(Matter) = 0You can also get this from the Classical Action Principle (no quantumeffects like zero point energy in this limiting case).PZ: OK.JS: Tuv(Matter) = (-2/g^1/2)Functional Derivative of the Matter Actionrelative to the metric guvSimilarlyTuv(Geometry) ~ Functional Derivative of the Einstein-HilbertGeometrodynamic action relative to the classical matter fields whichmust vanish in non-exotic classical vacuum.PZ: But there is also a published derivation of the Yilmaz field equations from analternate action principle for a tensor gravitational potential phi_ab which yields1/2 G_uv = tau_uv + t_uv (vac),i. e., the Yilmaz modification of the GR field equations, with localizabletensor t_uv (vac).JS: This is too vague for me to understand. Why should we take that action seriously?The physical picture is completely obscure and un-motivated to my mind.PZ: And it is claimed that this has a precise n-body Newtonian limit with Newtonianstatic gravitational interaction -- whereas standard GR may not.JS: Now there is an excellent purely mathematical argument for the EEP wellknown to Einstein. It goes like this.For a D dimensional metric manifold, EEP means finding a special classof Diff(4) local maps called tetrad transformations at point P such thatguv(curved) -> nab(flat)PZ: OK, let's pause right there.Here once again we see these familiar old weasel terms curved and flat.One favorite definition of guv(curved) is anything other than the *fixedflat level* Minkowski metric n_ab.JS: Yes, it's a good one. No reason to change it. If it ain't broke, don't fix itwith less with more rather than more with less.PZ: I would suggest the following definitions:Fixed means the tensor components of n_ab are constant everywhere in spacetime;Meaningless to my mind. Who needs it? This is less with more IMHO.Intrinsically unstable GLOBALLY FLAT space-time already hastensor components of n_ab are constant everywhere in spacetimeso why your redundant addition above?PZ: Flat means that the *second* derivatives of the components of g_uv with respectto the coordinates vanish at a point in spacetime;JS: Ditto. Redundant. Locally flat in EXACT sense already means that the local Riemann curvature tensorof 4th rank Ruvwl vanishes at P. EEP means ONLY that, within the two limits i) far from singularity and ii) not near Planck scale,that the geodesic deviation relative tidal acceleration between two of epsilon - delta like defining continuity of f(x) in calculus, were epsilon is a given sensitivity of a tidal force curvaturedetector placed at P.PZ: Level means that the *first* derivatives of the components of g_uv vanish ata point;JS: Ditto. Vanishing g-force defines local free-float timelike geodesic frame i.e. MTW's LIF.Not all first derivatives need vanish BTW only those combinations making the torsion-freeconnection field for parallel transport.PZ: Fixed flat level then means that all derivatives of g_uv = n_ab vanish everywheresince the components of n_ab are constants.Ditto. Who needs it?PZ: Another, more specific, definition of curved in this context is that the Riemanncurvature R_uvwl does not vanish everywhere -- which implies that certain secondderivatives of the variable components g_uv do not vanish everywhere in spacetime.Already in orthodox GR. Einstein's field equation is a LOCAL equation.Physics is simple when local. (Wheeler)PZ: If we pick a point P in spacetime, then we can say that the metric field is level thereif the gradients g_uv, w with respect to the x_u all vanish at P; but at the same time thethe Riemann curvature R_uvwl could still be finite at P, if certain second derivatives ofthe g_uv do not vanish there.But there is according to these definitions no such thing asguv(curved) --> nab(flat)inside GR,JS: Red Herring. The precise statement of EEP isguv(curved) ---> nab(APPROXIMATELY FLAT)operational definition is free float weightlessnesswith negligible stretch-squeeze tidal curvatureforces - a very excellent approximation for NASAEarth orbital operations, for Mars expedition.Voyager etc - not good for probing a black holeof course.Physics is not legal theory it is an empirical subjectand it is not mathematics subject to strict rigorin its informal descriptions. Of course it mustnot make formal errors to be an acceptabletheory. No serious physicist in the field wouldaccept your modifications here and withgood reason IMHO. There is simply nomarket demand for your alleged innovation.This is not where the boundary of thecutting edge is located in Physics TodayIMHO.PZ: since if g_uv is curved (in the strict sense of non-vanishing Riemann curvature)at a spacetime point P then it is curved at P *in every physical frame of reference*, andmore generally in every spacetime coordinate system.Strictly speaking, there is no nab(flat) in GR in the presence of gravitating matter *in anyframe of reference*. Strictly speaking, nab only exists in classic special relativity (Minkowskispacetime) since in classic SR a variable g_uv is *not even defined*.If we extend SR to allow a field-variable g_uv to accommodate accelerated frames, then itshould be clear that all the gradients g_uv, w (and thus all non-tidal g-forces) can vanishat a spacetime point *while the tensor R_uvwl may not*. However, in SR R_uvwl doesvanish everywhere because in SR there is no metric theory of physical gravitation, andaccelerated kinematical frames in and of themselves never result in non-vanishing R_uvwl.So we do not really have formal kinematic correspondence with classic SR by the EEP.We can only make an argument that if we voluntarily restrict ourselves to a certain class ofphysical measurements, then locally the metric tensor *might as well* be nab (unless thegravitational powers of Einstein's ponderable bodies (Einstein 1905) are all mysteriouslyswitched off).In other words, in GR a level g_uv can be *locally modeled as if it were* the SR n_ab,but then only with respect to a limited class of physical measurements.IMO that is the *entire objectively validated content* of the EEP.In my view this is really no different in essence from pointing out that within a small area ofa perfect sphere's surface, for all practical purposes the earth *might as well be* flat --unless we exceed the bounds of practicality (e.g by performing highly sensitive geodeticmeasurements using laser devices) -- even though we know theoretically that thesurface is nevertheless round everywhere (non-vanishing Gaussian curvature at every pointon the spherical surface).That is not to say that it is flat at a point -- which is sheer nonsense IMHO. It is simplyapproximately flat within a limited neighborhood -- which is a relative statement.Going to an infinitesimal neighborhood does not fundamentally alter this reality.Also, we have no compelling physical reason -- other than EEP -- for supposing thatall phenomena that result from non-vanishing R_uvwl can be made empirically negligiblein this way simply by diminishing the spacetime volume of the neighborhood. It's just thatthe equivalence principle tells us not to look for those other phenomena (negativeheuristic).JS: Essentially Paul your proposal is falling on deaf ears as far as the communityof top working theorists is concerned as you can clearly see from the onlinepre-prints. A small number of philosophers may find it of some interest. Ido not see you asking interesting questions here or supplying significantnew insights here in an already well ploughed field. All the movers andshakers are in an entirely different distant region of the conceptuallandscape and until you can come up with something amazing dealingwith the issues of the day like dark energy or metric engineering etc you will be a lonewolf howling in the wilderness. As someone commented reading your stuff herewas Scholastic like counting Angels on the head of a pin.JS: Make an infinitesimal coordinate transformation about point P expressedas a Taylor series.Go out to the cubic term. That's all you need to do to see the error ofyour ways Paul.PZ: I don't think so.JS: The first order terms at the metric tensor level have D^2 coefficientsto adjust to make the local metric flat but only at point P.PZ: Yes, this is what I call level -- all g_uv, w locally vanish.JS: No they do not locally vanish. You just made an obvious math error.The diagonal nuv are -1, +1, +1, +1JS: But the metric tensor is symmetric with D(D + 1)/2 independentcomponents.PZ: OK.JS: Therefore, the residualgauge freedom at first order in the Taylor series insD^2 - D(D + 1)/2 = D(D - 1)/2free local LIF -> LIF' transformations at P corresponding to thetangent vector space at Pwith local Lorentz group symmetry SO(D - 1, 1) of rotations and boosts.PZ: Fine.JS: But this is all in MTW and embodies the essence of the orthodox understanding of EEP.There is no interesting issue here IMHO. This worksin general like in Kaluza-Klein hyperspace with D = 5 as well asEinstein's D = 4.PZ: OK.JS: Next go to the second order, i.e. at the connection field for paralleltransport level, wherethere are D^2(D + 1)/2 connection field coefficients to adjust to makethe connection field g-force slamming you against the seat of your space fighter, with warpdrive, on disappear whilst dogfighting by locally shaping the geometry with zero pointenergy density macro-quantum wave interference modulation as I begin to show in thesediscussions.PZ: Yes, let's do. :-)JS: Since God is subtle and not malicious the math shows that D^2(D + 1)/2is precisely thesame number first derivatives of the metric at P that we need to adjustto zero to make thetorsion-free connection field vanish at P.PZ: OK. But I nevertheless fear He may have played a trick or two in this case.JS: I do not see that. I think your system point is in a false attractor on your mental landscapeseeing patterns that are not really there in this case.This is the hard core derivation of EEP that Yilmaz and whoever cannotabrogate.PZ: As far as I am aware no one is trying to. They are simply distinguishing and separatingthe gravitational and inertial contributions to the g_uv, w. At least from a mathematicalstandpoint, that is unassailable.JS: No that is NOT GOOD PHYSICS IMHO. It is not enough to make the formal distinctionUNLESS there is an OPERATIONAL WAY to distinguish the formal split! I mean atleast in principle as a Bohr-Einstein gedankenexperiment. Now this is a matterof culture which is why some theoretical physicists are disturbed by the mathematiciansin physicists clothing not unlike old hands at CIA and State are disturbed by Paul Wolfowitz'sand Richard Perl's Neo Cons Coup D' Etat at The Pentagon bungling the events in post-war Iraq.I am told 50,000 Iraqi dead most non-combatants, more than 2000 US Soldiers wounded inaddition to the hundreds killed. Also the oil fields are in such bad shape that the abilityto pay for the Nation building will fall mainly on the American taxpayer. These are seriousintelligence failures that can be laid on the doorstep of the Neo Con Cabal I am afraid.No doubt serious investigations in 2005 if the Demos do manage to get in?PZ: Whether this distinction has any *physical* meaning depends on how the GR formalismis to be physically interpreted -- which is of course the $64K question.JS: The burden of proof is on you. Right now I would say 64 old Iraqi dinars with Saddam's face on itplus a few wooden nickels and some Weimar marks and Confederate dollars. :-)Finally at the 3rd order curvature level (i.e. second order derivativesof the metric) you have D^2(D + 1)(D + 2)/6 curvature coefficients of the Taylorseries to try to make zero, but there are a larger number D^2(D + 1)^2/4second order partial derivatives of the metric tensor, so you have left over D^2(D^2 - 1)/12 secondorder partial second order partial derivatives that are undetermined.PZ: Exactly.JS: So where is the beef since I am only rehashing Einstein's original arguments? It's all part of his 1915 - 18 theory.PZ: This is the number ofindependent componentsof the 4th rank Riemann tensor for the relative tidal accelerationsbetween closelyPZ: Right. And this splits into a Ricci tensor R_uv and a Weyl tensor, and in the Einstein gravitationalvacuum, R_uv = 0.So you have simply confirmed my point above. We do not needg_uv(curved) ---> n_ab(flat),literally construed, for purposes of SR correspondence. We only needg_uv, w ---> 0 at a spacetime point (in an LIF)andRELATIVE MAGNITUDE of tidal effects ---> 0 as SCALE of spacetime neighborhood --> 0,where relative means relative to a certain class of physical measurements of the typical accuracyof those that historically -- and from the standpoint of GR, falsely -- provided empiricalconfirmation of classic SR.If it turns out there are measurable effects that result from non-vanishing R_uvwl that do not scaledown in this manner, then the EEP will have to be regarded as having only restricted empiricalvalidity within GR, since we could then *in principle* distinguish between a permanent gravity fieldand an inertial field even within an infinitesimal neighborhood.Although the argument cited by Jack (to be found on 467 et seq. of MTW)based on EEP is still a point of contention between us.This is a fundamental objection to a Yilmaz-type stress energy term inthe GR field equations which IMO would have to be fully resolved beforewe can know whether to take Yilmaz's theory too seriously.At the same time, I believe I am now in a position to resolve thisentire questionwith a fundamental argument based on an internal critique of thecanonicalinterpretation of the GR formalism and a mathematical proof thatprovides themissing link in Yilmaz's development.AL: I said it was error ridden because there wasenough obvious dissonance between the questions I'dbeen asking and the responses I'd been getting toallow the (correct) prediction that he hadn't reallyread the sources.PZ: I think he glanced at Yilmaz's papebut decided they were a wasteof time-- at least as far as he is concerned.Jack has an ambitious program, which is based on the emergence of*canonical*that and it isprobably expecting too much to ask that he invest a substantial amountof effortinto unraveling the details of Yilmaz's arguments.I suppose he is following his nose (or betting on horses).JS: Exactly.I will comment on Freud Identity another time.PZ: OK. I think you should take this Freud thing seriously since, based on a purelymathematical argument, it is apparently valid for all *internally consistent* theoriesof gravitation based on curved spacetime manifolds -- just like the Bianchiidentity.In other words -- it's not wood. It's marble.Z.JS: No I do not take it seriously. Assume for the moment, I do not know for sure, that the formal argument given by Yilmaz is the same that is given by Pauli, and that it is correct. That is not enough to make it GOOD PHYSICS. You also need a PHYSICAL ARGUMENT i.e. agedanken experiment showing how to === distinguish the two terms in the formal split.Subject: re: Making Star Trek Real> > > This is most interesting!> > There is also a published paper (Phys. Essays) by Clark Jeffries who > argues that the exponential metric handles the missing mass problem that > looks interesting, though I have not really dug into the details.> > Hal Puthoff> > > > JS: There is an interesting exponential dependence in string theory like> > G* ~ G(Newton)e^<(Dilaton)>> > if that is what you mean? I am studying all that now.> > That would give a double exponential> > K = e^2G*M/c^2r = e^2G(Newton)e^(M/c^r> > I am wondering if the ~ Lp^2/zpf ?> > There is a basic string vertex strength gs for closed spin 2 graviton > strings that move in the total hyperspace and gs^1/2 for open strings of > spin 1/2 lepto-quarks & spin 1 internal gauge force bosons whose ends > are anchored to 3Dim brane parallel universes floating in hyperspace.> parallel brane worlds with ends of open strings attached to different > parallel brane worlds because the spin gauge force fluxes form closed > loops through the ends of the open string exactly like Wheeler's Charge > without charge in his 1950's book Geometrodynamics. I do not know if > anyone else has yet noticed this simple explanation for broken charge > conjugation symmetry?> > Part II Discussion with Zielinski> > > The Freud identity is supposed to hold for any theory of gravitation that is> based on a spacetime curvature model. It was cited by no less an > authority than> relativity.> > JS: True.> > > PZ: In the context of GR, Freud's results imply> > (1) that the Einstein curvature G_uv can automatically be linearly > decomposed> into the difference of two pseudotensors (quantities that act like > tensors under> linear, but not under general, coordinate transformations),> > JS: By that I suppose is meant that they are tensors under the Poincare > group i.e.> globally flat background but not under the curved local Diff(4) group?> > Locally gauge the globally flat Poincare group to get the curved spacetime> > guv(curved) - nuv(global flat) = huv(curved)> > as the compensating field.> > In no sense is huv small in this local gauging, which is not perturbative> relative to flat background. Unfortunately the same notation is used> in perturbation theory.> > A Poincare group local stress-energy density tensor for the first order > perturbative> gravity where h'uv is small can be defined of course, so that may be at > least one> of the pseudotensors in the Freud identity?> > Again trivially even in the full non-perturbative case> > Tuv(Gravity) = (String Tension)Guv(Geometry)> > and in the non-exotic vacuum where the Diff(4) divergence-free > Tuv(Source Mass-Energy) = 0> > Tuv(Gravity) = 0> > because the local Einstein field equation in that case is> > Tuv(Gravity) + Tuv(Source Mass-Energy) = 0> > > PZ: U_uv and u_uv:> > 1/2 G_uv = U_uv - u_uv,> > where u_uv is specifically identified as the Einstein vacuum stress-energy> pseudotensor and U_uv is called the Freud pseudotensor (Freud decomposition)> > JS: Sure so what's the big deal? u_uv is Tuv(Vacuum Geometry relative > to Poincare Group), U_uv is Tuv(Vacuum Geometry extra curvature terms). > Only the total sum is the very same Diff(4) tensor I proposed> > Tuv(Geometry) = (String Tension)Guv(Einstein)> > which = 0 in NON-EXOTIC VACUUM i.e. ALL residual micro-quantum zero > point energy from strings and things in large scale GR low energy limit > is completely damped down to ZERO COSMOLOGICAL CONSTANT by my VACUUM > COHERENCE FIELD. Like Ben Franklin I have poured oil to calm the > turbulent waters. This has nothing to do with fact that in string theory > completely different problem, which is precisely why Ed Witten said he > was very worried. String theory does not solve> the mystery of why Einstein's cosmological constant is so small.> > > PZ: (2) the density divergence of the Freud pseudotensor U_uv vanishes > identically:> > partial d_v(SQRT[-g] U_uv) = 0> > (using the Einstein summation convention) (Freud identity)> > JS: Yes, but probably assuming:> > i. metricity> > ii zero torsion> > iii non-exotic vacuum, i.e. /zpf = 0> > Note in string theory> > Lp*^2 = Lp^2(Newton)e^(Dilaton)> > Where Dilaton is the compensating gauge field from the Dilation charge > of the Conformal Group I think?> > PZ: According to Yilmaz's arguments in> > Nuovo Cimento 107B, 941 (1992)> > it can be shown that in view of the above, the Freud and Einstein > pseudotensors can> each be split into tensorial and non-tensorial parts:> > U_uv = tau_uv + z_uv> > u_uv = -t_uv + z_uv,> > where z_uv is a common *non-tensor* component that always vanishes > identically in> *some* coordinate system.> > JS: By tensor I assume you still mean with respect to the full Diff(4) > group of local general coordinate> transformations?> > PZ: Yilmaz shows that if the non-tensor quantity z_uv is mathematically > eliminated from the> RHS (from both U_uv and u_uv), we get> > 1/2 G_uv = tau_uv + t_uv (vac)> > where the Yilmaz vacuum stress-energy term t_uv (vac) is a true tensor > quantity (world> tensor) that is interpreted by Yilmaz as a localizable vacuum > gravitational stress-energy> that itself locally induces Riemann curvature in the general case.> > JS: It would be good to keep dimensions explicit> > [(string tension)/2] G_uv = tau_uv + t_uv (vac)> > Why the formal split who ordered that? How can one physically > distinguish this formal split?> > > > > PZ: OK. Personally, I have been focusing on fundamental aspects of Einstein> GR vs. Yilmaz, and it has been some time since I looked at the details> of PV vs. Einstein, so I'll leave this to you and Hal.> > To be fair, I think Jack's position is that if the arguments cited by> MTW are> correct, it is simply not possible to define a localized gravitational> field> stress-energy density in an absolute (i.e. world tensor-covariant)> manner> without generating conflicts with the so-called EEP under some widely> accepted canonical interpretation of that principle. And if the price> of Yilmaz> is the EEP, then Jack can define one trivially i.e.> > Tuv(Geometry) = (String Tension)Guv> > In the non-exotic vacuum the geometrodynamic stress-energy density> currents is completely insulated from the corresponding matter currents.> That is, the covariant 4-divergences of both Tuv(Geometry) and> Tuv(Matter) vanish separately. This makes metric engineering> impossible because there is then no way to directly convert matter into> geometry and vice versa in a practical cost effective way IMHO.> > Also in this non-exotic vacuum degenerate limiting case, Tuv(Geometry)> is trivially zero!> > Because> > Tuv(Geometry) + Tuv(Matter) = 0> > You can also get this from the Classical Action Principle (no quantum> effects like zero point energy in this limiting case).> > PZ: OK.> > JS: Tuv(Matter) = (-2/g^1/2)Functional Derivative of the Matter Action> relative to the metric guv> > Similarly> > Tuv(Geometry) ~ Functional Derivative of the Einstein-Hilbert> Geometrodynamic action relative to the classical matter fields which> must vanish in non-exotic classical vacuum.> > PZ: But there is also a published derivation of the Yilmaz field > equations from an> alternate action principle for a tensor gravitational potential phi_ab > which yields> > 1/2 G_uv = tau_uv + t_uv (vac),> > i. e., the Yilmaz modification of the GR field equations, with localizable> tensor t_uv (vac).> > JS: This is too vague for me to understand. Why should we take that > action seriously?> The physical picture is completely obscure and un-motivated to my mind.> > > PZ: And it is claimed that this has a precise n-body Newtonian limit > with Newtonian> static gravitational interaction -- whereas standard GR may not.> > JS: Now there is an excellent purely mathematical argument for the EEP well> known to Einstein. It goes like this.> > For a D dimensional metric manifold, EEP means finding a special class> of Diff(4) local maps called tetrad transformations at point P such that> > guv(curved) -> nab(flat)> > PZ: OK, let's pause right there.> > Here once again we see these familiar old weasel terms curved and flat.> > One favorite definition of guv(curved) is anything other than the *fixed> flat level* Minkowski metric n_ab.> > JS: Yes, it's a good one. No reason to change it. If it ain't broke, > don't fix it> with less with more rather than more with less.> > > PZ: I would suggest the following definitions:> > Fixed means the tensor components of n_ab are constant everywhere in > spacetime;> > Meaningless to my mind. Who needs it? This is less with more IMHO.> Intrinsically unstable GLOBALLY FLAT space-time already has> tensor components of n_ab are constant everywhere in spacetime> so why your redundant addition above?> > PZ: Flat means that the *second* derivatives of the components of g_uv > with respect> to the coordinates vanish at a point in spacetime;> > JS: Ditto. Redundant. Locally flat in EXACT sense already means that > the local Riemann curvature tensor> of 4th rank Ruvwl vanishes at P. EEP means ONLY that, within the two > limits i) far from singularity and ii) not near Planck scale,> that the geodesic deviation relative tidal acceleration between two > of epsilon - delta like defining continuity of f(x) in calculus, were > epsilon is a given sensitivity of a tidal force curvature> detector placed at P.> > PZ: Level means that the *first* derivatives of the components of g_uv > vanish at> a point;> > JS: Ditto. Vanishing g-force defines local free-float timelike geodesic > frame i.e. MTW's LIF.> Not all first derivatives need vanish BTW only those combinations making > the torsion-free> connection field for parallel transport.> > PZ: Fixed flat level then means that all derivatives of g_uv = n_ab > vanish everywhere> since the components of n_ab are constants.> > Ditto. Who needs it?> > PZ: Another, more specific, definition of curved in this context is > that the Riemann> curvature R_uvwl does not vanish everywhere -- which implies that > certain second> derivatives of the variable components g_uv do not vanish everywhere in > spacetime.> > Already in orthodox GR. Einstein's field equation is a LOCAL equation.> Physics is simple when local. (Wheeler)> > PZ: If we pick a point P in spacetime, then we can say that the metric > field is level there> if the gradients g_uv, w with respect to the x_u all vanish at P; but > at the same time the> the Riemann curvature R_uvwl could still be finite at P, if certain > second derivatives of> the g_uv do not vanish there.> > But there is according to these definitions no such thing as> > guv(curved) --> nab(flat)> > inside GR,> > JS: Red Herring. The precise statement of EEP is> > guv(curved) ---> nab(APPROXIMATELY FLAT)> > operational definition is free float weightlessness> with negligible stretch-squeeze tidal curvature> forces - a very excellent approximation for NASA> Earth orbital operations, for Mars expedition.> Voyager etc - not good for probing a black hole> of course.> > Physics is not legal theory it is an empirical subject> and it is not mathematics subject to strict rigor> in its informal descriptions. Of course it must> not make formal errors to be an acceptable> theory. No serious physicist in the field would> accept your modifications here and with> good reason IMHO. There is simply no> market demand for your alleged innovation.> This is not where the boundary of the> cutting edge is located in Physics Today> IMHO.> > PZ: since if g_uv is curved (in the strict sense of non-vanishing > Riemann curvature)> at a spacetime point P then it is curved at P *in every physical frame > of reference*, and> more generally in every spacetime coordinate system.> > Strictly speaking, there is no nab(flat) in GR in the presence of > gravitating matter *in any> frame of reference*. Strictly speaking, nab only exists in classic > special relativity (Minkowski> spacetime) since in classic SR a variable g_uv is *not even defined*.> > If we extend SR to allow a field-variable g_uv to accommodate > accelerated frames, then it> should be clear that all the gradients g_uv, w (and thus all non-tidal > g-forces) can vanish> at a spacetime point *while the tensor R_uvwl may not*. However, in SR > R_uvwl does> vanish everywhere because in SR there is no metric theory of physical > gravitation, and> accelerated kinematical frames in and of themselves never result in > non-vanishing R_uvwl.> > So we do not really have formal kinematic correspondence with classic SR > by the EEP.> We can only make an argument that if we voluntarily restrict ourselves > to a certain class of> physical measurements, then locally the metric tensor *might as well* > be nab (unless the> gravitational powers of Einstein's ponderable bodies (Einstein 1905) > are all mysteriously> switched off).> > In other words, in GR a level g_uv can be *locally modeled as if it > were* the SR n_ab,> but then only with respect to a limited class of physical measurements.> > IMO that is the *entire objectively validated content* of the EEP.> > In my view this is really no different in essence from pointing out that > within a small area of> a perfect sphere's surface, for all practical purposes the earth > *might as well be* flat --> unless we exceed the bounds of practicality (e.g by performing highly > sensitive geodetic> measurements using laser devices) -- even though we know theoretically > that the> surface is nevertheless round everywhere (non-vanishing Gaussian > curvature at every point> on the spherical surface).> > That is not to say that it is flat at a point -- which is sheer > nonsense IMHO. It is simply> approximately flat within a limited neighborhood -- which is a > relative statement.> > Going to an infinitesimal neighborhood does not fundamentally alter > this reality.> > Also, we have no compelling physical reason -- other than EEP -- for > supposing that> all phenomena that result from non-vanishing R_uvwl can be made > empirically negligible> in this way simply by diminishing the spacetime volume of the > neighborhood. It's just that> the equivalence principle tells us not to look for those other > phenomena (negative> heuristic).> > JS: Essentially Paul your proposal is falling on deaf ears as far as the > community> of top working theorists is concerned as you can clearly see from the online> pre-prints. A small number of philosophers may find it of some interest. I> do not see you asking interesting questions here or supplying significant> new insights here in an already well ploughed field. All the movers and> shakers are in an entirely different distant region of the conceptual> landscape and until you can come up with something amazing dealing> with the issues of the day like dark energy or metric engineering > etc you will be a lone> wolf howling in the wilderness. As someone commented reading your stuff here> was Scholastic like counting Angels on the head of a pin.> > JS: Make an infinitesimal coordinate transformation about point P expressed> as a Taylor series.> Go out to the cubic term. That's all you need to do to see the error of> your ways Paul.> > PZ: I don't think so.> > JS: The first order terms at the metric tensor level have D^2 coefficients> to adjust to make the local metric flat but only at point P.> > PZ: Yes, this is what I call level -- all g_uv, w locally vanish.> > JS: No they do not locally vanish. You just made an obvious math error.> The diagonal nuv are -1, +1, +1, +1> > > JS: But the metric tensor is symmetric with D(D + 1)/2 independent> components.> > PZ: OK.> > JS: Therefore, the residual> gauge freedom at first order in the Taylor series ins> > D^2 - D(D + 1)/2 = D(D - 1)/2> > free local LIF -> LIF' transformations at P corresponding to the> tangent vector space at P> with local Lorentz group symmetry SO(D - 1, 1) of rotations and boosts.> > PZ: Fine.> > JS: But this is all in MTW and embodies the essence of the orthodox > understanding of EEP.> There is no interesting issue here IMHO.> > > This works> in general like in Kaluza-Klein hyperspace with D = 5 as well as> Einstein's D = 4.> > PZ: OK.> > JS: Next go to the second order, i.e. at the connection field for parallel> transport level, where> there are D^2(D + 1)/2 connection field coefficients to adjust to make> the connection field g-force slamming you against the seat of your space > fighter, with warp> drive, on disappear whilst dogfighting by locally shaping the geometry > with zero point> energy density macro-quantum wave interference modulation as I begin to > show in these> discussions.> > PZ: Yes, let's do. :-)> > JS: Since God is subtle and not malicious the math shows that D^2(D + 1)/2> is precisely the> same number first derivatives of the metric at P that we need to adjust> to zero to make the> torsion-free connection field vanish at P.> > PZ: OK. But I nevertheless fear He may have played a trick or two in > this case.> > JS: I do not see that. I think your system point is in a false attractor > on your mental landscape> seeing patterns that are not really there in this case.> > This is the hard core derivation of EEP that Yilmaz and whoever cannot> abrogate.> > PZ: As far as I am aware no one is trying to. They are simply > distinguishing and separating> the gravitational and inertial contributions to the g_uv, w. At least > from a mathematical> standpoint, that is unassailable.> > JS: No that is NOT GOOD PHYSICS IMHO. It is not enough to make the > formal distinction> UNLESS there is an OPERATIONAL WAY to distinguish the formal split! I > mean at> least in principle as a Bohr-Einstein gedankenexperiment. Now this is > a matter> of culture which is why some theoretical physicists are disturbed by > the mathematicians> in physicists clothing not unlike old hands at CIA and State are > disturbed by Paul Wolfowitz's> and Richard Perl's Neo Cons Coup D' Etat at The Pentagon bungling the > events in post-war Iraq.> I am told 50,000 Iraqi dead most non-combatants, more than 2000 US > Soldiers wounded in> addition to the hundreds killed. Also the oil fields are in such bad > shape that the ability> to pay for the Nation building will fall mainly on the American > taxpayer. These are serious> intelligence failures that can be laid on the doorstep of the Neo Con > Cabal I am afraid.> No doubt serious investigations in 2005 if the Demos do manage to get in?> > PZ: Whether this distinction has any *physical* meaning depends on how > the GR formalism> is to be physically interpreted -- which is of course the $64K question.> > JS: The burden of proof is on you. Right now I would say 64 old Iraqi > dinars with Saddam's face on it> plus a few wooden nickels and some Weimar marks and Confederate dollars. :-)> > > Finally at the 3rd order curvature level (i.e. second order derivatives> of the metric) you have D^2(D + 1)(D + 2)/6 curvature coefficients of > the Taylor> series to try to make zero, but there are a larger number D^2(D + 1)^2/4> second order partial derivatives of the metric tensor, so you have left > over D^2(D^2 - 1)/12 second> order partial second order partial derivatives that are undetermined.> > PZ: Exactly.> > JS: So where is the beef since I am only rehashing Einstein's original > arguments? It's all part of his 1915 - 18 theory.> > PZ: This is the number of> independent components> of the 4th rank Riemann tensor for the relative tidal accelerations> between closely> > PZ: Right. And this splits into a Ricci tensor R_uv and a Weyl tensor, > and in the Einstein gravitational> vacuum, R_uv = 0.> > So you have simply confirmed my point above. We do not need> > g_uv(curved) ---> n_ab(flat),> > literally construed, for purposes of SR correspondence. We only need> > g_uv, w ---> 0 at a spacetime point (in an LIF)> > and> > RELATIVE MAGNITUDE of tidal effects ---> 0 as SCALE of spacetime > neighborhood --> 0,> > where relative means relative to a certain class of physical > measurements of the typical accuracy> of those that historically -- and from the standpoint of GR, falsely -- > provided empirical> confirmation of classic SR.> > If it turns out there are measurable effects that result from > non-vanishing R_uvwl that do not scale> down in this manner, then the EEP will have to be regarded as having > only restricted empirical> validity within GR, since we could then *in principle* distinguish > between a permanent gravity field> and an inertial field even within an infinitesimal neighborhood.> > Although the argument cited by Jack (to be found on 467 et seq. of MTW)> based on EEP is still a point of contention between us.> > This is a fundamental objection to a Yilmaz-type stress energy term in> the GR field equations which IMO would have to be fully resolved before> we can know whether to take Yilmaz's theory too seriously.> > At the same time, I believe I am now in a position to resolve this> entire question> with a fundamental argument based on an internal critique of the> canonical> interpretation of the GR formalism and a mathematical proof that> provides the> missing link in Yilmaz's development.> > AL: I said it was error ridden because there was> enough obvious dissonance between the questions I'd> been asking and the responses I'd been getting to> allow the (correct) prediction that he hadn't really> read the sources.> > PZ: I think he glanced at Yilmaz's papebut decided they were a waste> of time> -- at least as far as he is concerned.> > Jack has an ambitious program, which is based on the emergence of> *canonical*> that and it is> probably expecting too much to ask that he invest a substantial amount> of effort> into unraveling the details of Yilmaz's arguments.> > I suppose he is following his nose (or betting on horses).> > JS: Exactly.> > I will comment on Freud Identity another time.> > > PZ: OK. I think you should take this Freud thing seriously since, based > on a purely> mathematical argument, it is apparently valid for all *internally > consistent* theories> of gravitation based on curved spacetime manifolds -- just like the > Bianchi> identity.> > In other words -- it's not wood. It's marble.> > Z.> > JS: No I do not take it seriously. Assume for the moment, I do not know > for sure, that the formal argument given by Yilmaz is the same that is > given by Pauli, and that it is correct. That is not enough to make it > GOOD PHYSICS. You also need a PHYSICAL ARGUMENT i.e. a> gedanken experiment showing how to distinguish the two terms in the > formal split.When was the last time you === actually had sex, Jack? Like with a woman.Subject: Re: Making Star Trek Real> > When was the last time you actually had sex, Jack? Like with a woman.Never was there a white man who === so badly needed a blow-job.Bob Kolker> Subject: =?ISO-8859-1?Q?K=F6nnt_Ihr_mir_diese_Formel_aufl=F6sen=3F?= Da ich kein Mathegenie bin w.8are es nett wenn Ihr mir die Formelausschreiben w.9frdet. Besonders diese EXP1,65*0,000125^(AusgabeDichte-1)*(1-EXP(-0,04*kochzeit1te) === )/4,15*1tealpha*10/AuschlagmengeDankeMikeSubject: Re: =?iso-8859-1?Q?K=F6nnt?= Ihr mir diese Formel =?iso-8859-1?Q?aufl=F6sen=3F?=> Da ich kein Mathegenie bin w.8are es nett wenn Ihr mir die Formel> ausschreiben w.9frdet. Besonders diese EXPMike: This is an english newsgroup. For german help on maths seede.sci.mathematikIn German:Mike, dies ist eine englische Gruppe. Das deutsche Pendant istde.sci.mathematikRene.-- Ren.8e MeyerStudent of Physics & MathematicsZhejiang University, === Hangzhou, ChinaSubject: Re: =?iso-8859-1?Q?K=F6nnt?= Ihr mir diese Formel =?iso-8859-1?Q?aufl=F6sen=3F?=>> Da ich kein Mathegenie bin w.8are es nett wenn Ihr mir die Formel>> ausschreiben w.9frdet. Besonders diese EXP>Mike: This is an english newsgroup. De facto, perhaps, but not de jure. The group has no charter (it's old)and USENET protocols as a whole do not specify a language. Post in anylanguage you like as long as it's mathematics. Of course if you post inEsperanto you'll get many few responses than if you post in English(and you'll have to accept people's right to === respond in Volapuk).daveSubject: Re: =?ISO-8859-1?Q?K=F6nnt?= Ihr mir diese Formel =?ISO-8859-1?Q?aufl=F6sen=3F?=> .... > Mike: This is an english newsgroup. For german help on maths see> de.sci.mathematik> .... That was probably good advice; but I believe sci.math is international and should be open to any language. The preponderance of American contributors makes English its main language, but it does use others from time === to time. Ken Pledger.Subject: Jordan curves and n holesLet P and Q be differentiable functions continuous over a open set Rwhich is path connected with n-holes, if d(P)/dy= d(Q)/dx for all _R.The number of different values for the line integrals P dx + Q dy overregular Jordan curves is 2^(n+1) - 1. Anyone know a proof of thisusing combinatorics? I read oneusing mathematical induction. Any other way to === prove this? Feel freeto mail.Subject: existence of a probability measureIs it true that there exists a (uniform) probability measure on thespace of open or closed subsets of === [0,1] ?Subject: Re: existence of a probability measureContent-Length: 421Originator: rusin@vesuvius> Is it true that there exists a (uniform) probability measure on the> space of open or closed subsets of [0,1] ?At least two possibilities come to mind:1. Is there a uniform probability measure on [0,1] that assigns a measure to all open sets? Or one that does so to all closed sets?2. Is there a measure on the space of open subsets of [0,1] that === is uniform?DaleSubject: Re: existence of a probability measure>Is it true that there exists a (uniform) probability measure on the>space of open or closed subsets of [0,1] ?You need to clarify what the question means (because Lebesguemeasure seems to give an affirmative answer, but the idea thatyou're asking this question but you've never heard of Lebesguemeasure seems unlikely, so I conjecture you mean something else...)************************David C. === UllrichSubject: Re: existence of a probability measure>Is it true that there exists a (uniform) probability measure on the>space of open or closed subsets of [0,1] ?It's not at all clear what the question means. A probability measureis defined on a sigma-algebra; Lebesgue measure is a uniformprobability measure on the Borel algebra on [0,1], which containsthe open and closed sets. But you must know that, so that can'tbe what you mean.Next guess is you're wondering about a measure defined on somealgebra of subsets of X, where X is the set of open or closed subsetsof [0,1]. Is that what you're wondering about? If so it's not clearwhat algebra you have in mind, nor what uniform should mean...************************David C. === UllrichSubject: how to test whether data fits poisson distribution?I have about 10,000 integeand I would like to know whether theyconform to a poisson distribution. Any poisson distribution at all.Ideally there would exist a program somewhere which would take my listof integers as input (and nothing else), and spit out a probabilitythat the integers fit a poisson distribution.None of the 'statistics calculators' that I have found online do thisor anything like it as far as I can see. Does anyone know of any suchprogram, or how I could most closely approximate my === test whether data fits poisson distribution?>I have about 10,000 integeand I would like to know whether they>conform to a poisson distribution. Any poisson distribution at all.>Ideally there would exist a program somewhere which would take my list>of integers as input (and nothing else), and spit out a probability>that the integers fit a poisson distribution.The basic test is formulated like this: The idea is to test whether your data might follow a poissondistribution with parameter lambda. Since you don't seem to knowlambda, you'll have to estimate it. For poisson distributionlambda=mean, so you can simply use the arithmetic mean of your datafor this estimation.You then divide your data into a table of values containing:o the sample frequencyo the actual poisson distribution probability with estimated lambdao the probability times the sample set sizefor each integer within the range of integers you are seeing, and thenuse them to calculate the test value. You then compare it to the knownchi-squared distribution with n-2 degrees of freedom (I think you loseone when estimating lambda, somebody correct me if I'm wrong) where nis the sample set size, with the hypothesis that it is poissondistributed. Picking a suitable alpha value and comparing the twovalues will then tell you if it's === safe to assume the data to bepoisson distributed.Subject: Re: how to test whether data fits poisson distribution?Well, one very quick test is to calculate the mean and the variance of yoursample. For a Poission distribution these should be approximately equal.-Michael.> I have about 10,000 integeand I would like to know whether they> conform to a poisson distribution. Any poisson distribution at all.> Ideally there would exist a program somewhere which would take my list> of integers as input (and nothing else), and spit out a probability> that the integers fit a poisson distribution.> None of the 'statistics calculators' that I have found online do this> or anything like it as far as I can see. Does anyone know of any such> program, or how I could most closely approximate my ideal scenario> above? === poisson distribution?>I have about 10,000 integeand I would like to know whether they>conform to a poisson distribution. Any poisson distribution at all.>Ideally there would exist a program somewhere which would take my list>of integers as input (and nothing else), and spit out a probability>that the integers fit a poisson distribution.>None of the 'statistics calculators' that I have found online do this>or anything like it as far as I can see. Does anyone know of any such>program, or how I could most closely approximate my chi-squared goodness-of-fit test. This is very standard; you should find it in any basic statistics text (or on the web).-- Stephen J. Herschkorn === herschko@rutcor.rutgers.eduSubject: Re: Axiomatic Set Theory and FoundationLast time I read about it a few years ago, I learned that one of theobstacles to using Conway's theory of games to study the game of Go isthe difficulty of dealing with ko fights, since they screw up wellfoundedness. I wondered whether one might be able to do better by dealingwith non-well founded models of set theory to set up Conway's theory ofgames and maybe apply it to Go. I don't claim that it would let onedecide how to win a ko fight, but it might at least let one talk about it.Maybe someone has figured out another way to apply Conway's stuff to ko?Ignorantly,Allan Adlerara@zurich.ai.mit.edu******************************** ********************************************* ** Disclaimer: I am a guest and *not* a member of the MIT Artificial ** Intelligence Lab. My actions and comments do not reflect ** in any way on MIT. Moreover, I am nowhere near the Boston ** metropolitan area. ** *********************************************************** === ******************Subject: Re: Axiomatic Set Theory and Foundation> Last time I read about it a few years ago, I learned that one of the> obstacles to using Conway's theory of games to study the game of Go is> the difficulty of dealing with ko fights, since they screw up well> foundedness. I wondered whether one might be able to do better by> dealing with non-well founded models of set theory to set up Conway's> theory of games and maybe apply it to Go. I don't claim that it would> let one decide how to win a ko fight, but it might at least let one> talk about it.> Maybe someone has figured out another way to apply Conway's stuff to> ko?Yes, they did (see the work of Berlekamp and al). But set-theory has nothingto do with it. The problem is not of finding a good model, but of findingthe right theorems to make calculations...> Ignorantly,> Allan Adler> ara@zurich.ai.mit.edu*************************************** *************************************> *> * * Disclaimer: I am a guest and *not* a member of the MIT> Artificial * * Intelligence Lab. My actions and> comments do not reflect * * in any way on MIT.> Moreover, I am nowhere near the Boston * *> metropolitan area. * *> ************************************************************ === *****************Subject: Subgroups of S4? Uniqueness of Subgroups of Order Half of Group and other qnsHello everyone,I have been trying to work out all the subgroups of S4, the permutation group of 4 objects. And I have quite a few questions.(1) Is there a systematic manner to find out all the subgroups of a given group? Or, to make the question less general, of S_N?(2) I'm having difficulty trying to figure out whether I've found all subgroups of S4 of order 6 and 8. The groups I've found are:(a) For order 6, I note that by fixing one of the four numbers {1,2,3,4} and permuting the rest, I've creating 4 subgroups of S4 that is isomorphic to S3. But I am not sure if there are more, or if there are no more how to prove that it is the case.(b) For order 8, I note that there are three inequivalent ways of labeling the corners of a square with {1,2,3,4} - by inequivalent, I mean there is no way to bring the four numbers on one of these squares into the four numbers on another square under an operation of D4, the symmetry operations of a square (physically I think of prohibiting the square from being 'twisted' or 'torn up' - I wonder if such geometric ideas can be generalized in some way). And so I can find at least three subgroups of S4 isomorphic to D4. Again, however, I am not sure how to find more or how to prove there is no more subgroups of order 8.(3) I notice that there is only 1 order 12 subgroup. This can be proved using the fact that it must contain complete classes because it must be invariant. But this does make me wonder: do groups always have at most one === help!Yi-ZenSubject: Re: Subgroups of S4? Uniqueness of Subgroups of Order Half of Group and other qns>Hello everyone,>I have been trying to work out all the subgroups of S4, the permutation >group of 4 objects. And I have quite a few questions.>(1) Is there a systematic manner to find out all the subgroups of a >given group? Or, to make the question less general, of S_N?Well, provided that the group is finite, you could consider all subsets ofthe group, and check which of them are subgroups! For general (infinite)groups, probably not.For a more efficient approach for finite groups, there are results ofAschbacher, Scott, Kovacs, which effectively reduce the problem to almostsimple groups G; i.e. groups with S <= G <= Aut(S), for a nonabeliansimple group S. A lot is now known about maximal subgroups of (almost)simple groups, but there is no complete description, nor is there likelyto be one any time soon. But there are practical algorithms which will workfor groups of moderate order. For S_N, general purpose algorithms will workup to about N=11 or 12. There are compelte lists of trnasitive subgroupsup to N=31 and primitive subgroups up to N=1000.>(2) I'm having difficulty trying to figure out whether I've found all >subgroups of S4 of order 6 and 8. The groups I've found are:>(a) For order 6, I note that by fixing one of the four numbers >{1,2,3,4} and permuting the rest, I've creating 4 subgroups of S4 that >is isomorphic to S3. But I am not sure if there are more, or if there >are no more how to prove that it is the case.Yes, these are all subgroups of order 6. Any group of order 6 has a normalsubgroup of order 3, which must fix one point. The normalizer of a subgrouppermutes the fixed points of that subgroup, so the groups of order 6 must allfix a point. Hence there are just four of them.>(b) For order 8, I note that there are three inequivalent ways of >labeling the corners of a square with {1,2,3,4} - by inequivalent, I >mean there is no way to bring the four numbers on one of these squares >into the four numbers on another square under an operation of D4, the >symmetry operations of a square (physically I think of prohibiting the >square from being 'twisted' or 'torn up' - I wonder if such geometric >ideas can be generalized in some way). And so I can find at least three >subgroups of S4 isomorphic to D4. Again, however, I am not sure how to >find more or how to prove there is no more subgroups of order 8.Yes, there are just 3 subgroups of order 8. This is most easily proved fromSylow's Theorem. >(3) I notice that there is only 1 order 12 subgroup. This can be proved >using the fact that it must contain complete classes because it must be >invariant. But this does make me wonder: do groups always have at most >one subgroup of order half of its own order?No, for example the Klein 4-group, which is the direct product of two groupsof order 2 has 3 subgroups of order === 2.Derek Holt.Subject: How to solve logarithm simulataneous equationsI have a simultaneous equation of the form:m^h = a (m to the power of h = a)(m+1)^h = b ( [m+1] to the power of h = b)and I need to express m in terms of a and b only. In other words, m and hare variables, a and b are constants...is there a way to solve these type of === equations?Teng JunbinSubject: Re: How to solve logarithm have a simultaneous equation of the form:> > m^h = a (m to the power of h = a)> (m+1)^h = b ( [m+1] to the power of h = b)> > and I need to express m in terms of a and b only. In other words, m and h> are variables, a and b are constants...> > is there a way to solve these type of equations?Explicitly, in closed form? Unlikely.Assuming you mean for m, a, and b to be positive, h not necessarily so,your equations are equivalent to h log(m) = log(a), h log(m+1) = log(b).Hence unless a = 1 and b = 1 (hence h = 0) you will have log(m)/log(m+1) = log(a)/log(b), with log(b) ne 0.This one-variable equation can be solved e.g. by Newton's method. Thefunction f(x) = log(x)/log(1+x) seems to be strictly increasing on(0,+infty), with range (-infty, 1), (I haven't officially verifiedthis) so for any a and b for which log(a)/log(b) < 1 it will have aunique solution m, f(m) = log(a)/log(b).Once you have m you can use either the first equation to solve for h, h = log(a)/log(m),or if you are unlucky enough to have m = 1, the second, h = log(b)/log(m+1).If log(b)/log(a) is 1, 2, 3, or 4--in other words, if b = a, a^2, a^3or a^4 -- you can solve the equations explicitly, since you have log(m)/log(m+1) = 1, 1/2, 1/3 or 1/4,in other words (m+1) = m (no solution), m+1 = m^2 (one positivesolution), m+1 = m^3 (one positive solution), m+1 = m^4 (one positivesolution). But these seem to be the only cases.Hmmm. I suppose there's b = a+1 too. === Then you get m = a.--Ron BruckSubject: Re: How to solve logarithm simulataneous equationsI think I stated the problem too generally.. let me rephrase the problem.Basic definitions for the terms used after the problem statement.Given a full m-ary tree where each node has either m children or 0 children,and the height of the tree is h. The value of each non-terminal node is 1 /(m+1) times the value of any of its parent. The value of each terminal nodeis 1.Examples:m = 5h = 2therefore, there are 5^2 = 25 terminal nodesthe value of the root node is 6^2 = 36the value of each second level node is 6the value of each terminal node is 1 (by definition)Given the number of terminal nodes in such a tree, and given the value ofthe root node, find the number of internal vertices.Worked example:If the value of the root node is 216, and the number of terminal nodes is125,m^h = 125(m+1)^h = 216clearly m = 5, h = 3therefore, the number of internal vertices = 31 (1 root node, 5 level 1nodes, 25 level 2 nodes)Using basic tree analysis, we can determine that:the number of internal vertice = number of terminal nodes - 1 / m - 1Provided that m > 1.Therefore, our problem is simplified to simply finding out m given:m^h = a(m+1)^h = bwhere a and b are constants, m and h are variableswhich is the original question I posted.Basic definitions:full m-ary tree: A full m-ary tree is a tree where each node in the tree iseither a terminal node (ie: has 0 children nodes) or has exactly m childrennodesheight of a tree: The height of a tree is the maximum length of the pathfrom any terminal node to the root node. So a root node by itself is a treeof height 0. A root node with children which are all terminal nodes is atree of height 1.terminal node: A terminal node is also called a leave node, terminalvertice, etc. It is simply a node with no children.internal vertice: An internal vertice is a node/vertice in a tree which isnot a terminal node. A root node with at least one child node is therefore,an internal vertice.Teng Junbin> I have a simultaneous equation of the form:>> m^h = a (m to the power of h = a)> (m+1)^h = b ( [m+1] to the power of h = b)>> and I need to express m in terms of a and b only. In other words, m andh> are variables, a and b are constants...>> is there a way to solve these type of equations?> Explicitly, in closed form? Unlikely.> Assuming you mean for m, a, and b to be positive, h not necessarily so,> your equations are equivalent to> h log(m) = log(a),> h log(m+1) = log(b).> Hence unless a = 1 and b = 1 (hence h = 0) you will have> log(m)/log(m+1) = log(a)/log(b), with log(b) ne 0.> This one-variable equation can be solved e.g. by Newton's method. The> function f(x) = log(x)/log(1+x) seems to be strictly increasing on> (0,+infty), with range (-infty, 1), (I haven't officially verified> this) so for any a and b for which log(a)/log(b) < 1 it will have a> unique solution m,> f(m) = log(a)/log(b).> Once you have m you can use either the first equation to solve for h,> h = log(a)/log(m),> or if you are unlucky enough to have m = 1, the second,> h = log(b)/log(m+1).> If log(b)/log(a) is 1, 2, 3, or 4--in other words, if b = a, a^2, a^3> or a^4 -- you can solve the equations explicitly, since you have> log(m)/log(m+1) = 1, 1/2, 1/3 or 1/4,> in other words (m+1) = m (no solution), m+1 = m^2 (one positive> solution), m+1 = m^3 (one positive solution), m+1 = m^4 (one positive> solution). But these seem to be the only cases.> Hmmm. I suppose there's b = a+1 too. Then you get m = a.> --Ron === BruckSubject: Re: How to solve logarithm simulataneous equations>I think I stated the problem too generally.. let me rephrase the problem.>Given a full m-ary tree where each node has either m children or 0 children,>and the height of the tree is h. The value of each non-terminal node is 1 />(m+1) times the value of any of its parent. The value of each terminal node>is 1....>Given the number of terminal nodes in such a tree, and given the value of>the root node, find the number of internal vertices....>m^h = a>(m+1)^h = b>where a and b are constants, m and h are variables>which is the original question I posted.The big difference is that you're interested in m and h being positive integers. Note that m or m+1, and therefore a or b (but not both), is even.Find the greatest j such that 2^j divides a or b, and h is eitherthat j or a divisor of it. Unless a and b are enormous, there can't be very many possibilities to check. For each possibleh, check if a^(1/h) and b^(1/h) are integers differing by 1.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British === Columbia Vancouver, BC, Canada V6T 1Z2Subject: Re: How to solve logarithm simulataneous stated the problem too generally.. let me rephrase the problem.> >Given a full m-ary tree where each node has either m children or 0 children,>and the height of the tree is h. The value of each non-terminal node is 1 /> >(m+1) times the value of any of its parent. The value of each terminal node>is 1.> ...> >Given the number of terminal nodes in such a tree, and given the value of>the root node, find the number of internal vertices.> > ...> >m^h = a>(m+1)^h = b> >where a and b are constants, m and h are variables> >which is the original question I posted.> > The big difference is that you're interested in m and h being > positive integers. > > Note that m or m+1, and therefore a or b (but not both), is even.> Find the greatest j such that 2^j divides a or b, and h is either> that j or a divisor of it. Unless a and b are enormous, there > can't be very many possibilities to check. For each possible> h, check if a^(1/h) and b^(1/h) are integers differing by 1.I dunno. I think it might be easier to solve log(m)/ log(m+1) = log(a)/log(b)approximately, using Newton's method; if there's to be an integralsolution, this will have to be close to an integer. Purify m bysetting it to be this close integer. (The iteration is fast.)Once we know m we can solve for h from h = log(a)/log(m). This shouldbe an integer (if there really is an integral solution), but due toroundoff it might not be. Replace it by the integer which is veryclose, and we should have our solution (m,h). Check.Let's see, starting from his example m = 5, h = 3, and starting withinitial guess m = 2, it takes 7 iterations to get m = 5 to 11 decimals(9 iterations get it to 32 decimals). It takes virtually no time at allto divide Log[125] by Log[5] (of course, in this case one can see inone's head the quotient is 3).--Ron === BruckSubject: Re: How to solve logarithm simulataneous equations>I dunno. I think it might be easier to solve> log(m)/ log(m+1) = log(a)/log(b)>approximately, using Newton's method; if there's to be an integral>solution, this will have to be close to an integer. Purify m by>setting it to be this close integer. (The iteration is fast.)For large m, log(m)/log(m+1) = 1 - 1/(m log(m)) + o(1/m^2), so theinitial guess could be m_0 = 1/((1-t) W(1/(1-t)))where t = log(a)/log(b) and W is the Lambert W function (ProductLog in Mathematica). This satisfies 1 - 1/(m_0 log(m_0)) = t.According to Maple, m - m_0 = -1/2 + O(1/log(m)), so round(m_0-1/2) should be good for large m. In fact, it appears that m = round(m_0-1/2)for all positive integers m. Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British === Columbia Vancouver, BC, Canada V6T 1Z2Subject: Re: How to solve logarithm simulataneous equationsI would like to thank you for your valuable time and advice.Teng Junbin>I think I stated the problem too generally.. let me rephrase the problem.>Given a full m-ary tree where each node has either m children or 0children,>and the height of the tree is h. The value of each non-terminal node is1 /> >(m+1) times the value of any of its parent. The value of each terminalnode>is 1.> ...>Given the number of terminal nodes in such a tree, and given the value of>the root node, find the number of internal vertices.> ...>m^h = a>(m+1)^h = b>where a and b are constants, m and h are variables>which is the original question I posted.> The big difference is that you're interested in m and h being> positive integers.> Note that m or m+1, and therefore a or b (but not both), is even.> Find the greatest j such that 2^j divides a or b, and h is either> that j or a divisor of it. Unless a and b are enormous, there> can't be very many possibilities to check. For each possible> h, check if a^(1/h) and b^(1/h) are integers differing by 1.> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British === Columbia> Vancouver, BC, Canada V6T 1Z2Subject: Cycle Structures and Classes of SnHi everyone,Could someone explain why cycle structures in S_n determine its classes? Or is there some place I could read up on this? I have two introductory abstract algebra textbooks but they do not even === questionmy daughter had the following multiple choice question on her maths test. Idont agree with any of the answers.This is the question in fullQ Julie kept a record of the number of pairs of shoes she had.year no of pairs of shoes1996 111997 91998 121999 102000 13The mode isa. 1996b 1997c 1998d 1999e 2000the correct answer given was e 2000.I say there is no mode and if there was a mode , then the answer would notbe a year but a number of pairs of shoes. What do the statistics === my daughter had the following multiple choice question on her maths> test. I dont agree with any of the answers.> This is the question in full> Q Julie kept a record of the number of pairs of shoes she had.> year no of pairs of shoes> 1996 11> 1997 9> 1998 12> 1999 10> 2000 13> The mode is> a. 1996> b 1997> c 1998> d 1999> e 2000> the correct answer given was e 2000.> I say there is no mode and if there was a mode , then the answer> would not be a year but a number of pairs of shoes. What do the> statistics experts think?You are correct. What difference does it make when the five observationsare made? Instead of once a year, they could just as well be made once aday (in which case the mode would be Thursday?) or once an hour (in whichcase the mode would be 5:00 p.m.?), or in different cars (in which case themode would be Chevrolet?), or by different people (in which case the modewould be Sally?). Good grief. There are five observations. Each occursonce, so either there is no mode, === or the mode is 9, 10, 11, 12, and 13.Subject: Re: mode question>What do the statistics experts think?e is correct, because the year 2000 has a greater frequency than === the otheryears.DougSubject: Re: mode question>What do the statistics experts think?> e is correct, because the year 2000 has a greater frequency than the other> years.> Doug>>Doug I dont follow this at all. 13 is not the frequency , it is one ofthe possible values of the number of shoes.Actually on thinking some more I have changed my own answer. All theobservation have an equal frequency of 1 therefore they are all modes.But I am still not sure whether the answer should be the numbers or theyear. Every other example of mode question I have seen seems to be obviousto me.For example in antoher example 14 students get a mark for a test mark a b c d eno 3 4 2 3 2in this example the mode is b ( ie the score) because this score had thehighest frequencyexample 2a group of students eat a different no of lolliesstudent a b c d eno of lollies 2 5 3 8 5I would say the mode here was 5 because it is the most common choice of thenumber of lollies eaten. I dont say d because it has the highest numberof lollies.See how depending on the problem in one question I have chosen the thingnot the number and I have chosen the thing with the highest number and inthe second example I have chosen the number not the thing, and I have chosenthe most common number.Its seems to be a question of interpreting what is a frequency and what isan individual value and also determining what is the thing of interest ieis it the mark in the exam or the number of students that achieved it. Inall other examples of mode questions I have seen it has seemed perfectlyobvious. I thought the shoes were too but obviously others dont agree.No wonder my 13 year old daughter is === confused.TerrySubject: Re: mode question>For example in antoher example 14 students get a mark for a test> mark a b c d e>no 3 4 2 3 2>in this example the mode is b ( ie the score) because this score had the>highest frequencyThis is correct: Here, we are counting the number of each grade. E.g., 3/14 = 21.4% of the students got a's.>example 2>a group of students eat a different no of lollies>student a b c d e>no of lollies 2 5 3 8 5>I would say the mode here was 5 because it is the most common choice of the>number of lollies eaten. I dont say d because it has the highest number>of lollies.Here you are right: The number of lollies is not a frequency! The teacher's interpretation make no more sense than in the first one (with the shoes) you cited.It seems to me that the teacher does not understand the concept of mode; perhaps you should arrange a meeting with him/her to discuss the matter.>[...]>I thought the shoes were too but obviously others dont agree.I think only one person here agreed with the teacher; in my opinion, he is wrong. To repeat, it makes absolutely to use counts of shoes or lollies as frequencies in these examples.-- === Stephen J. Herschkorn herschko@rutcor.rutgers.eduSubject: Re: mode question> I think only one person here agreed with the teacher; in my opinion, > he is wrong. To repeat, it makes absolutely to use counts of shoes or > lollies as frequencies in these examples. I meant to type it makes absolutely no sense to use counts...-- Stephen J. === Herschkorn herschko@rutcor.rutgers.eduSubject: Re: mode question> >>What do the statistics experts think?>> >e is correct, because the year 2000 has a greater frequency than the other>years.How is the number of shoes a frequency? See my second post in this thread.-- Stephen J. Herschkorn === herschko@rutcor.rutgers.eduSubject: Re: mode question>my daughter had the following multiple choice question on her maths test. I>dont agree with any of the answers.>This is the question in full>Q Julie kept a record of the number of pairs of shoes she had.>year no of pairs of shoes>1996 11>1997 9>1998 12>1999 10>2000 13>The mode is>a. 1996>b 1997>c 1998>d 1999>e 2000>the correct answer given was e 2000.>I say there is no mode and if there was a mode , then the answer would not>be a year but a number of pairs of shoes. What do the statistics experts>think?The mode is 13, not the year. You are correct that the correct answer was not offered.-- Stephen J. Herschkorn === herschko@rutcor.rutgers.eduSubject: Re: mode question>> my daughter had the following multiple choice question on her maths >> test. I>> dont agree with any of the answers.>> This is the question in full>> Q Julie kept a record of the number of pairs of shoes she had.>> year no of pairs of shoes>> 1996 11>> 1997 9>> 1998 12>> 1999 10>> 2000 13>> The mode is>> a. 1996>> b 1997>> c 1998>> d 1999>> e 2000>> the correct answer given was e 2000.>> I say there is no mode and if there was a mode , then the answer >> would not>> be a year but a number of pairs of shoes. What do the statistics >> experts>> think?> The mode is 13, not the year. You are correct that the correct answer > was not offered.Oops. Typed without thinking. 10 sqrt(7) pi / 3 lashes with a wet noodle.As others have pointed out, 13 is not the mode. Or rather, I would agree with someone else, that each of the values 9 through 13 is a mode.In any case, the teacher is wrong. S/he may have had something else (i.e., 1996 appearing eleven times, etc.) in mind, but that would be a rather unusual (even nonsensical) data set for this situation.-- Stephen J. Herschkorn === herschko@rutcor.rutgers.eduSubject: Re: mode question >>> my daughter had the following multiple choice question on her maths>> test. I>> dont agree with any of the answers.>>>> This is the question in full>>>> Q Julie kept a record of the number of pairs of shoes she had.>>>> year no of pairs of shoes>>>> 1996 11>> 1997 9>> 1998 12>> 1999 10>> 2000 13> >>>> The mode is>> a. 1996>> b 1997>> c 1998>> d 1999>> e 2000>>>> the correct answer given was e 2000.>>>> I say there is no mode and if there was a mode , then the answer>> would not>> be a year but a number of pairs of shoes. What do the statistics> >> experts>> think?>>> The mode is 13, not the year. You are correct that the correct answer> was not offered.>> Oops. Typed without thinking. 10 sqrt(7) pi / 3 lashes with a wet noodle.> As others have pointed out, 13 is not the mode. Or rather, I would> agree with someone else, that each of the values 9 through 13 is a mode.> In any case, the teacher is wrong. S/he may have had something else> (i.e., 1996 appearing eleven times, etc.) in mind, but that would be a> rather unusual (even nonsensical) data set for this situation.Minor change in question produces a plausible situation:} Q Julie has marked each pair of shoes with the year in which she} purchased them, and now wants to know which mark occurs most} frequently.}} year no of pairs of shoes marked with that year}} 1996 11} 1997 9} 1998 12} 1999 10} 2000 13}} The mode is} a. 1996} b 1997} c 1998} d 1999} e 2000I think we can agree that (e) is the right answer to this question.-- P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will find these people and we === will bring them to justice.Subject: Re: mode question>As others have pointed out, 13 is not the mode. Or rather, I would >agree with someone else, that each of the values 9 through 13 is a mode.You've convinced me. There is no mode in this case; each total appearsexactly once.The problems I was having earlier were a result of the vagueness of theinitial question; the problem could certainly be formulated in such a waythat 2000 was the mode, but it clearly wasn't set up in that fashion. === Myapologies.DougSubject: Re: mode question>The mode is 13, not the year. You are correct that the correct answer >was not offered.The mode is *not* 13. The mode is 2000. === DougSubject: Re: mode question> > >The mode is 13, not the year. You are correct that the correct answer>was not offered.> > The mode is *not* 13. The mode is 2000.> > DougThe most common value obtained in a set of observations.from: http://mathworld.wolfram.com/Mode.htmlThe set of observations is here: {11,9,12,10,13}. In other wordsthe mode is not unique, and all five of them are the modes.So, you are totally correct:I say there is no modeYes. (or all five of them are the mode, it is nowhere statesin the definition it must be unique.)and if there was a mode , then the answer would notbe a year but a number of pairs of === shoes.Also correct.WilbertSubject: Re: mode question>>my daughter had the following multiple choice question on her maths test. I>>dont agree with any of the answers.>>This is the question in full>>Q Julie kept a record of the number of pairs of shoes she had.>>year no of pairs of shoes>>1996 11>>1997 9>>1998 12>>1999 10>>2000 13>>The mode is>>a. 1996>>b 1997>>c 1998>>d 1999>>e 2000>>the correct answer given was e 2000.>>I say there is no mode and if there was a mode , then the answer would not>>be a year but a number of pairs of shoes. What do the statistics experts>>think?> The mode is 13, not the year. You are correct that the correct answer > was not offered.Why is the mode 13, rather than 9, 10, 11, or 12? Each appears once.I think I see what the teacher had in mind, but I don't agree with it.If you plot each of the data points by considering a pair of shoes to bethe abscissa and the year to be the ordinate, then 2000 is indeed themode. But that is not the first interpretation that springs to mind onreading the question.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.Subject: Re: Vedic Mathematics --- Myth and Reality<>> At any rate, this proves that the book on Vedic maths is not a fraud. > Whether in 16 books or 1600, it took me but half a page of that book> in the English language to find out a new and better multiplication> method. Who knows what will happen to the field of maths when all then, if all mathematicians are> trained to be dogmatic morons, how can that ever happen?Do not worry, great brahmin mathematicians and scientists are publishing their inventions on pseudo natinal papers of India.By the by, we understood all modern sceince discoveries(like special theory of relativity) are explained in Vedasin detail. But could you enlight us discoveries yet to be happened(Like design of perpetual machine.) from Vedas?> > Arindam === Banerjee.Subject: Re: Vedic Mathematics --- Myth and Reality> <>> At any rate, this proves that the book on Vedic maths is not a fraud. > Whether in 16 books or 1600, it took me but half a page of that book> in the English language to find out a new and better multiplication> method. Who knows what will happen to the field of maths when all the> 16 books are fully mathematicians are> trained to be dogmatic morons, how can that ever happen?> > Do not worry, great brahmin mathematicians and scientists are publishing their > inventions on pseudo natinal papers of India.> > By the by, we understood all modern sceince discoveries> (like special theory of relativity) are explained in Vedas> in detail.No, that is wrong. The Indian philosophical thought - Sanatanadharma, or the way of life beyond the scope of time - is completelydifferent from the modern and dominant Jewish thinking, that isreflected in all that passes for modern physics. But could you enlight us discoveries yet to be happened> (Like design of perpetual machine.) from Vedas?Vimans - which should work upon that principle - are mentioned in theancient Indian epics. People seeking enlightenment should first gettheir grammar and spelling correct. Knowledge is wasted uponlow-minded and deliberate === fools.Arindam Banerjee.Subject: Re: Vedic Mathematics --- Myth and Reality> Tractenberg apparently did his work in 1910.> > So that was (just) before the (re)discovery of> math in the Vedic texts, right? And it was long before> all of this was exported to the west (1965).> Or am i missing something?> > BTW, it is unlikely that Trachtenberg invented the> method of calculation himself. He designed a method of> _education_, but probably used older material.> (But i don't know enough about Trachtenberg to be sure> about that. Just a hunch.)> If that is true, the calculation method was known in> the west before 1910...I thought you said you did not want to continue the discussion. Asfar as I am concerned *no one* did one line multiplication before Idid, even though I gave them ample time to do so, thus to pretend youall knew about it before is nothing but lies. Of course, nothing morethan lies can be expected from dishonourable people.> > BTW too: the method is not very special. As said also> in this thread, there are also children who invent it> themselves, Heh-heh. Let them also invent one-line division and one-line squareroot method as well. Don't expect me to explain any more! Let themdo the explaining and working-out by example. Next thing I will hearis some Farter saying that he always knew it, it was nothing special,and he did it as an infant aged 4! :) :)No more from me on this thread. Arindam Banerjee.>independently of Trachtenberg or anything> > Looks like there is a communication glitch.>> > clarifying that Tractenberg got his stuff from 16 books on Vedic> maths>> It sure looks like there is a communication glitch. :-)> Who mentioned Trachtenberg, i don't know, but it wasn't me.> I mentioned a book about school arithmetic by Kruijtbosch.> > The year (1936) makes it unlikely that he was influenced by> the current Vedic math cult.>> === Mathematics --- Myth and Reality> Tractenberg apparently did his work in 1910.>> So that was (just) before the (re)discovery of> math in the Vedic texts, right? And it was long before> all of this was exported to the west (1965).> Or am i missing something?>> BTW, it is unlikely that Trachtenberg invented the> method of calculation himself. He designed a method of> _education_, but probably used older material.> (But i don't know enough about Trachtenberg to be sure> about that. Just a hunch.)> If that is true, the calculation method was known in> the west before 1910...> I thought you said you did not want to continue the discussion.I had no choice continuing, since you put me words in mymouth that i hadn't written.> As> far as I am concerned *no one* did one line multiplication before I> did, even though I gave them ample time to do so, thus to pretend you> all knew about it before is nothing but lies.As said, i don't know about Trachtenberg, but an old book(Kruijtbosch, 1936) had it. Did you do single-line multiplicationbefore 1936?>No more from me on === Vedic Mathematics --- Myth and Reality> Tractenberg apparently did his work in 1910.>> So that was (just) before the (re)discovery of> math in the Vedic texts, right? And it was long before> all of this was exported to the west (1965).> Or am i missing something?>> BTW, it is unlikely that Trachtenberg invented the> method of calculation himself. He designed a method of> _education_, but probably used older material.> (But i don't know enough about Trachtenberg to be sure> about that. Just a hunch.)> If that is true, the calculation method was known in> the west before 1910...>> I thought you said you did not want to continue the discussion.> > I had no choice continuing, since you put me words in my> mouth that i hadn't written.Hmm, I am trying to put words in your mouth, eh? This is somethingnew, and should have come up in your earlier post!What words? You said that Vedic learning was verbose, and tried toprove it by saying that 16 volumes of Vedic arithmetic were publishedin 1910. Someone else had said that quick multiplication wasdescribed in a book that referenced Tractenberg's work of 1910. Ipresumed that Tractenberg got his ideas from what you said, the 16volumes of Vedic arithmetic. Or else the translators of Vedicarithmetic in 1910 stole from Tractenberg in 1910, and most peoplewere more honest and sharing in those days. Your great contributioncomes from your saying that 16 volumes of Vedic arithmetic werepublished in 1910, and for that I thanked you. I don't see how I putwords into your mouth.In any case, the vedic method is not taught in any Western school, andfrom the reaction of many Westerners in this thread, they did not haveany clue about it till I explained it with a working-out. Theautochthonous quality of Vedic arithmetic is to me at least, wellestablished.I suspect Tractenberg was someone like Tesla - greatest of allinventors - who admired ancient Indian thought. As a result he wassimilarly ostracised by the Western racists and bigots, and his workignored.> As> far as I am concerned *no one* did one line multiplication before I> did, even though I gave them ample time to do so, thus to pretend you> all knew about it before is nothing but lies.> > As said, i don't know about Trachtenberg, but an old book> (Kruijtbosch, 1936) had it. Did you do single-line multiplication> before 1936?No, I started doing it after I read the method in the book labelled asa fraud by proven genuine frauds.>No more from me on this thread.> > Good.Sorry, but you were trying to prove me a liar. I had to reply, sinceI still retain some regard for you.Arindam Banerjee.> > === --- Myth and RealityYou missed one...> In any case> FFT is no method for long multiplication, it is only for abmod(n) arithmeticTo which you should have replied: Bull.Phil-- Unpatched IE vulnerability: Web Archive buffer overflowDescription: Possible automated code execution.Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0303/107.html === Subject: Re: Vedic Mathematics --- Myth and Reality>>It's a matter of personal rebelliousness that prevents me from copying>>the packehowever profitable that may be. In fact, after I became>>a manager in the semiconductor company, I decided to raise a ruckus>>about these disparities.> > > Exactly what happened?This is meannesses.Also, I don't know the details. Basically, organized leashing ofof those that have appreciation of both vedic and non-vedic mathby those that know neither. I am told they used to hold regularmeetings to maintain the === racism.Arjoe> Subject: Re: Vedic Mathematics --- Myth and Reality>>It's a matter of personal rebelliousness that prevents me from copying>>the packehowever profitable that may be. In fact, after I became>>a manager in the semiconductor company, I decided to raise a ruckus>>about these disparities.> > > Exactly what happened?> meannesses.Who do you thank, and why? I think we scb-ers have a right to knowabout all the meannesses you have endured. They make you so vocal.> Also, I don't know the details. Basically, organized leashing of> of those that have appreciation of both vedic and non-vedic math> by those that know neither. But that's rightwing Western society, for you. The status ofengineers has always been very low in anglosaxon society, where statusis in direct proportion to money, and talent without money is of noworth. Engineers could at best marry the maidservants of theirbosses. (Read the play Strife by Galsworthy.) Only leftwingershave given prestige to math people. In India engineers have a highstatus, but that came from a basically leftwing political policy. Now, engineers are losing out to the money people, statuswise, as thecountry is becoming more rightwing. It may be that the low status ofengineers in Western society comes as a shock to those from othersocieties.> I am told they used to hold regular> meetings to maintain the racism.Tribalism is always the core issue in all human matters. Engineersshould understand this, and form their own tribe, irrespective ofracial matters.Arindam Banerjee.> === ArjoeSubject: Re: Vedic Mathematics --- Myth and Reality> >>It's a matter of personal rebelliousness that prevents me from copying>>the packehowever profitable that may be. In fact, after I became>>a manager in the semiconductor company, I decided to raise a ruckus>>about these disparities.>Exactly what happened?>>This is > > Who do you thank, and why? I think we scb-ers have a right to know> about all the meannesses you have endured. They make you so vocal.To the staof course!The stars of the night, for helping keep sanity in theface of what appeared to be a concerted effort to destroy it.> > > === >>Arjoe> Subject: real world problemhi guys, I've a real life math/physics problem that I would love help with,and someone here might enjoy a real world problem...I have a fish tank that is 4'x2'x2' along the horizontal axis.the stand that the tank is on is on a slight lean, so that the water levelin the highest corner is 12mm lower than the water level at the oppositediagonal corner.can anyone tell me the difference in pressure exerted outwards at thedifferent parts of the tank?the tank is filled with === world problemX-AUTHid: mulkscpaulgoodhew@bigpond.com says...>hi guys, I've a real life math/physics problem that I would love help with,>and someone here might enjoy a real world problem...>I have a fish tank that is 4'x2'x2' along the horizontal axis.>the stand that the tank is on is on a slight lean, so that the water level>in the highest corner is 12mm lower than the water level at the opposite>diagonal corner.>can anyone tell me the difference in pressure exerted outwards at the>different parts of the tank?>the help.1 gallon of water occupies 231 cubic inches and weighs 8 pounds,so a column of water 231 high (or 19' 3) weighs 8 lbs/sq_in.The pressure at a depth of 2' (24) would be8 * (24/231) = 0.831169 psiat 24 - (12/25.4) = 23.527559 === it would be8 * (23.527559/231) = 0.814807 psiSubject: Re: real world problemThe pressure is proportional to the weight of water above, or in otherwords, directly proportional to depth.> hi guys, I've a real life math/physics problem that I would love helpwith,> and someone here might enjoy a real world problem...> I have a fish tank that is 4'x2'x2' along the horizontal axis.> the stand that the tank is on is on a slight lean, so that the water level> in the highest corner is 12mm lower than the water level at the opposite> diagonal corner.> can anyone tell me the difference in pressure exerted outwards at the> different parts of the === for any help.Subject: Re: real world problemso in other words I have an extra few kilos at one end.. doesn't seem likemuch to worry about..> The pressure is proportional to the weight of water above, or in other> words, directly proportional to depth.> hi guys, I've a real life math/physics problem that I would love help> with,> and someone here might enjoy a real world problem...>> I have a fish tank that is 4'x2'x2' along the horizontal axis.>> the stand that the tank is on is on a slight lean, so that the waterlevel> in the highest corner is 12mm lower than the water level at theopposite> diagonal corner.>> can anyone tell me the difference in pressure exerted outwards at the> different parts of the tank?> >> the tank is filled with water obviously!!>> === real world problem>so in other words I have an extra few kilos at one end.. doesn't seem like>much to worry about..In any case, the depth below the water's surface (and hence thepressure) at every point will be less than it would be if the tankwere upright and filled to the brim, so if it can tolerate beingfilled when upright, it can tolerate being filled at a slant (unlessit falls off the stand).John Mitchell>> The pressure is proportional to the weight of water above, or in other>> words, directly proportional to depth.>> hi guys, I've a real life math/physics problem that I would love help>> with,>> and someone here might enjoy a real world problem...>>>> I have a fish tank that is 4'x2'x2' along the horizontal axis.>>>> the stand that the tank is on is on a slight lean, so that the water>level>> in the highest corner is 12mm lower than the water level at the>opposite>> diagonal corner.>> >>> can anyone tell me the difference in pressure exerted outwards at the>> different parts of the tank?>>>> === for any help.>>>>>>>>>>Subject: Re: a nice e^n like integer sequence > I am looking for a nice integer sequence which > grows asymptotically like e^n, up to a multiplicative > constant. > > One obvious sequence is floor(e^n), but I am curious > about more natural definitions....>>The Fibonnacci sequences grow exponentially.> > This is true, but it grows like phi^n, where phi> is the golden ratio, (1+sqrt(5))/2. I'm looking> for something which grows _exactly_ like e^n, in other> words an integer sequence a_n for which a_n / e^n -> 1> as n -> oo.> > If I were to replace e with gamma where gamma was> any algebraic constant, then finding a (rational) sequence> asymptotic to gamma^n would be easy, just using a polynomial> of which gamma was a root to define a recurrence relation,> along with carefully chosen initial conditions (actually,> now that I think about it, I'm not 100% sure that the necessary> init cond's would be rational -- good thing to try to prove).> > Anyway, my point here is that the fact that e is> transcendental makes things a bit more complicated.So are you looking for some sort of -integer- procedure that computes floor(e^n) exactly? As you point out, phi is irrational but still algebraic (the root of a poly, associated with the linear recurrence). And e is transcendental (there is no poly of which it is a root, or equivalently there is no linear recurrence (ooops forgot this with -constant- coeffs)).So there is no hope of a linear recurrence. Gerry's idea about derangements D(n) = (n-1)(D(n-1)+D(n-2)) is promising (but lim prod k!/D(k) is Theta(e^n) but not = e^n because of initial term(s)).So D(n) grows too fast. Maybe there are other similar recurrences with poly coeffs? Like: a(n) = a(n-1) + (n-1)a(n-2) http://www.research.att.com/projects/OEIS?Anum=A000085 === Subject: Re: a nice e^n like integer sequence> I am looking for a nice integer sequence which> grows asymptotically like e^n, up to a multiplicative> constant.>So there is no hope of a linear recurrence. Gerry's idea about >derangements D(n) = (n-1)(D(n-1)+D(n-2)) is promising (but lim prod >k!/D(k) is Theta(e^n) but not = e^n because of initial term(s)).D(k)/k! = 1/e (1 + O(1/k!)) so prod_{k=1}^infinity k!/(e D(k)) converges to some nonzero constant C, and thenprod_{k=1}^n k!/D(k) = e^n (C + O(1/n!))Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of === British Columbia Vancouver, BC, Canada V6T 1Z2Subject: Re: a nice e^n like integer sequence> > So D(n) grows too fast. Maybe there are other similar recurrences with > poly coeffs? Like:> > a(n) = a(n-1) + (n-1)a(n-2)> > http://www.research.att.com/projects/OEIS?Anum=A000085The D(n)-inspired sequence is fine by me, as it is theta(e^n). I thinkasking for something asymptotic up to constant 1 is being too picky(perhaps not?)I am really just curious about how easy it is to arrive at somenatural yet nontrivial asymptotics via nice integer sequences. Of course,how easy and nice are ambiguous, but these are human ideas which,while difficult to describe mathematically, are still valid concepts.The sequence a(n) = a(n-1) + (n-1)a(n-2) -- how fast does it grow?I don't know at all how to analyze such a sequence. Any ideas?Based on some empirical checking, the sequence appears to actuallygrow _faster_ than === e^n.-TylerSubject: Re: a nice e^n like integer sequence> > > So D(n) grows too fast. Maybe there are other similar recurrences with > poly coeffs? Like:> > a(n) = a(n-1) + (n-1)a(n-2)> > http://www.research.att.com/projects/OEIS?Anum=A000085> > The D(n)-inspired sequence is fine by me, as it is theta(e^n). I think> asking for something asymptotic up to constant 1 is being too picky> (perhaps not?)> > I am really just curious about how easy it is to arrive at some> natural yet nontrivial asymptotics via nice integer sequences. Of course,> how easy and nice are ambiguous, but these are human ideas which,> while difficult to describe mathematically, are still valid concepts.> > The sequence a(n) = a(n-1) + (n-1)a(n-2) -- how fast does it grow?> I don't know at all how to analyze such a sequence. Any ideas?> Based on some empirical checking, the sequence appears to actually> grow _faster_ than e^n.> > -TylerNote that a(n) grows faster than b(n) = (n-1)*b(n-2),which grows like const*(n/e)^(n/2), of course much faster than e^n.For a more precise analysis of a(n) = a(n-1) + (n-1)*a(n-2) itself,you can try the generating function method...f(x) = sum a(n)*x^n/n! satisfies a certain linear differentialequation, if you can solve that, then you can try to get theasymptoticsfor the coefficients. Or examine some of the other representationsfor A000085 given in the URL.-- G. A. Edgar === http://www.math.ohio-state.edu/~edgar/Subject: Re: a nice e^n like integer sequence> I am looking for a nice integer sequence which> grows asymptotically like e^n, up to a multiplicative> constant.Do you know the derangement numbers? The number of permutations of {1, 2, ..., n} with no fixed point. D_1 = 0, D_2 = 1, D_3 = 2, D_4 = 9, .... There's a simple recurrence, something like D_n = n D_(n-1) + (-1)^n. Anyway, it's known that D_n / (n!) approaches 1/e as n increases. So (n!) / D_n approaches e. So if you let a_n be the product of (k!) / D_k, k from 2 to n, you get a sequence of rationals growing as e^n.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for === email)Subject: Re: a nice e^n like integer sequence> > [this is my second reply to this thread, not a repeat of my first reply.]> > I am looking for a nice integer sequence which> grows asymptotically like e^n, up to a multiplicative> constant.>> One obvious sequence is floor(e^n), but I am curious> about more natural definitions. I've looked on> Sloane's but all of the close sequences I could> find (besides e^n itself) did not have the right> asymptotics. One exception -- A035508, which is> described as 2nd row of Inverse Stolarsky array> > I have no idea how to find the asymptotic growth> of that guy (let alone an explicit or even nice> recursive formula > -Tyler> > If we let a(n) be such> > a(n) is maximum (or minimum) positive integer where:> > n = floor(1 + 1/2 + 1/3 +....+ 1/a(n)),> > then a(n) = O(e^n), I believe.> > Leroy> QuetIn the OEIS there arehttp://www.research.att.com/projects/OEIS?Anum=A081881 http://www.research.att.com/projects/OEIS?Anum=A038625( http://www.research.att.com/projects/OEIS?Anum=A075880a(n)~ === exp(n)/n^(1/2) )Hugo PfoertnerSubject: Re: a nice e^n like integer sequence> LQ:> > I am surprised that the EIS does not have many sequences of O(e^n), though. > I bet you meant you could not *find* many such sequences by a manual search, right?> > > Yes, what you say is a bit more accurate. The only sequence> which definitely had the right asymptotic which I could manage> to find using a manual search was [e^n] itself> (that is floor(e^n)).> > By the way, I very much like your suggestion for> n = floor(1 + 1/2 use consistently, say, the minimum such integer(because there are many harmonic numbers which round down to anyparticular integer, many more so as n === increases).LeroyQuetSubject: Re: a nice e^n like integer sequenceHello!IIRC is an acronym used on the internet meaning If I remember/recall correctlyhttp://dictionary.reference.com/search?q=iirc&r= 67Anthony> > The Fibonnacci sequences grow exponentially.> > This is true, but it grows like phi^n, where phi> is the golden ratio, (1+sqrt(5))/2. I'm looking> for something which grows _exactly_ like e^n, in other> words an integer sequence a_n for which a_n / e^n -> 1> as n -> oo.> > If I were to replace e with gamma where gamma was> any algebraic constant, then finding a (rational) sequence> asymptotic to gamma^n would be easy, just using a polynomial> of which gamma was a root to define a recurrence relation,> along with carefully chosen initial conditions (actually,> now that I think about it, I'm not 100% sure that the necessary> init cond's would be rational -- good thing to try to prove).> > Anyway, my point here is that the fact that e is> transcendental makes things a bit more complicated.> > > IIRC, each is of form a*(b^n +- c^n) for suitable a,b and c.> > What's === IIRC?> > -TylerSubject: Re: a nice e^n like integer sequence>> I am looking for a nice integer sequence which> grows asymptotically like e^n, up to a multiplicative> constant.>> One obvious sequence is floor(e^n), but I am curious> about more natural definitions. I've looked on> Sloane's but all of the close sequences I could> find (besides e^n itself) did not have the right> asymptotics. One exception -- A035508, which is> described as 2nd row of Inverse Stolarsky array> I have no idea how to find the asymptotic growth> of that guy (let alone an explicit or even nice> recursive formula for the nth term) ?> n! is asymptotic to sqrt(2*Pi*n)*(n^n/e^n), so maybe you can cook up> what you want using factorials.> I was messing around with the above formula and by adding to the left side> (n)+1/6 it gives a MUCH closer approximation of each n!> Also like in the original formula it will never exceed n!> For values slightly > 1/6 it eventually will exceed n!> So 1/6 added to the left side (n) I believe is the limit.> n! ~ sqrt(2*pi*(n+(1/6)))*(n^n/e^n)> Is there a better approximation of factorials in some other closed form> method?Yes; you can go through the asymptotic analysis for Stirling's formula andinclude some higher order terms.-- P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will find these === people and we will bring them to justice.Subject: Re: Peano arithmetic and multiplication>1. 0 in N (zero is a natural number)>2. x in N -> s(x) in N (the successor of any # is also a #)> these first two just construct the naturals>3. 0 != s(x) (0 is the smallest #)>4. s(x) = s(y) -> x = y (successor is a one-to-one function)>5. [P(0) and forall x, P(x) -> P(s(x)) ] -> forall x, P(x)|OK, after some reading (just the questionable web search), it seems that | what I stated above (rules 1 through 5) are definitely what is |traditionally called the Peano Axioms, but that by themselves they are |not enough to express arithmetic.Beware vague phrases like express arithmetic.It's really important in this setting to distinguish between formal systemsand other axiom systems. The above axiom 5 refers to arbitrary predicates Pon the domain. No rules for deducing consequences of such an axiom arepresented. That's because (while there are popular sets of rules for suchdeductions) there's no canonical such set of rules. It's not a formalsystem.It *is* enough to express arithmetic in a certain sense. Any collectionof elements satisfying those axioms is isomorphic to the natural numberswith its successor function. Addition, multiplication, exponentiation andso on can all be defined by induction. Everything which is true of them isa consequence of the axioms. But not necessarily by following anyparticular set of deduction rules!There are on the other hand various formal systems affiliated with thatset of axioms. When you have a formal system, you can in principle write acomputer program which takes two strings in the alphabet of the system,and tell you whether or not the first string is a proof in the system ofthe second string. Typically, a formal system consists of a recursivelyenumerable set of first order axioms, with the theorems being thosesentences which can be proven from the axioms using first order logic.Unlike second order logic, first order logic has a complete axiomatization,so every sentence which is a logical consequence of the axioms can bededuced from them by following the rules.Goedel's theorem pertains to formal systems. I don't think he referredto PA, but to the formal system of Principia Mathematica (and relatedsystems).|You -need- to define addition and |multiplication on top of the Peano axioms in order to get what is |referred to as Peano Arithmetic (oh and you also need the language and |semantics of first order logic) You don't need to define in addition |things like exponentiation or less than because you can express those |using combinations of +,*, and FOL. Peano Arithmetic is also referred to |as First Order Arithmetic.There's a formal system typically referred to as PA. It does have rules foraddition and multiplication built into it. We can get away with not havingsuch rules built into other systems (like the second order axioms) becausethey are rich enough to permit these functions to be defined by induction.The language of PA is more confined, because in it you refer only toindividual natural numbenot arbitrary sets of them.Note especially that the induction axiom in PA is restricted. One only hasinduction for predicates which can be expressed in the language of PA.As this reference|My main reference is :||http://linas.org/mirrors/www.ltn.lv/2001.03.27/~ podnieks/gt3.htmlappears to point out, if you leave out *, the language is weak enough thatthere's an algorithm for determining which sentences are true and whichare false. It's just not expressive enough a language, and we can cook upa complete set of (first order) axioms for it. So you do want * built-inin this context.|Now it's cleared up for me how + and * are involved language-wise in PA.||But I'm still curious as to how Diophantine equations (the subject of |Hilbert's 10th) are related logically/language wise. Diophantine eqs |involve only naturals, addition and multiplication. What have I left out |logically that makes this different from PA?Hilbert's 10th problem is not about axiom systems directly. The questionis whether one of the systems of equations has a solution. So if youformalize it, you get a sentence with only existential quantifiers. That'sa very restricted kind of sentence.The problem is whether there exists an algorithm for deciding the truth ofsuch sentences. It's much easier to show that there is no such algorithmfor all sentences in the language, since with arbitrary quantifieyou candefine complicated relationships much more easily. A lot of the trouble withHilbert's 10th was showing that properties related to Turing machines couldbe encoded in the Diophantine problem (using just + and *).The existence of an algorithm is actually related to the existence of acomplete axiom system, although that's not the way Hilbert phrased theproblem. If you have a formal system with the property that every sentenceis either a theorem or its negation is a theorem, then there is an algorithm(just search for that proof or disproof). The opposite is simpler; if thereis an algorithm, the algorithm itself can be taken as the formal system.|Also, separately, general recursive functions: there are these rules |(constant, successor, projection, composition, primitive recursion, |unbounded search/minimization). Didn't we just allow recursion in our |definition of + and *? How are these different (GRF and PA)?Well, PA is an axiom system, and GRF are just functions....There are lots of variations on the language and axioms which one can make,which lead to a system equivalent to PA. I fairly often have seen logiciansdecide for convenience to consider all *primitive* recursive predicates tobe built-in.Unbounded search only yields a function which is defined everywhere in somecases. There is no general algorithm for determining whether or not thesearch always succeeds. So really you are dealing with arbitrary partialrecursive functions. Incorporating features into the language to permitpartially defined functions is possible but involves a little more overhead.You can consider y=f(x) where f is partial recursive to be a relationshipbetween x and y. But it can always be expressed as (Ez) P(x,y,z) where Pis primitive recursive. I think it's probably considered somewhat morenatural to incorporate just primitive recursive predicates, and write suchrelations with the implicit existential quantifier sitting there visibly.A recursive function corresponds to a relation P(x,y) satisfying(x)(E!y) P(x,y), where (E!y) means there exists a unique y.Given a choice between PA, having just + and * built in, and a similarsystem with primitive recursive predicates build in, I would guess thechoice is made based on things like notational convenience.||>However, sometimes I hear PA spoken of in the same breath as the|>theory of naturals with addition and multiplication ||That's because you can define them and prove the standard|(associative, commutative, distributive) properties using only PA.||>10th (and so the reduction is nontrivial)? ||It doesn't matter whether you start from PA or whether you take some|things as axioms that would otherwise have been theorems. You wind up|with the same body of results.||>Mitch, horribly confused by the names of things and what they may or|>may not refer to.||PA is simply a minimalist definition of the Naturals.Shmuel Metz appears to be using PA to refer to the second order axioms,which is probably a dangerous thing, since PA is so often used for theformal system.Of course, he killfiled me for saying he didn't really mean to say vectorspaces were defined as ordered sets having additional properties, so maybeI should assume he means exactly what he says. In that case, PA does notneed to define addition and multiplication, because they are undefined termsin PA (built in). It's true that the commutativity, associativity, and soon of the operations are theorems of PA.speak of the theory consisting of all true sentences in it, but that'svery far from being a formal system. It's just the abstract collection oftruths in the language.The second order Peano axioms (not being formal) could indeed be thought ofas a minimalist characterization of the natural numbebut PA the formalsystem is not.Keith === RamsaySubject: Re: Peano arithmetic and multiplicationX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS>However, sometimes I hear PA spoken of in the same breath as the>theory of naturals with addition and multiplication That's because you can define them and prove the standard(associative, commutative, distributive) properties using only PA.>10th (and so the reduction is nontrivial)? It doesn't matter whether you start from PA or whether you take somethings as axioms that would otherwise have been theorems. You wind upwith the same body of results.>Mitch, horribly confused by the names of things and what they may or>may not refer to.PA is simply a minimalist definition of the Naturals.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited 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 === functions as postscript files Matt> Or maybe there is a free program that plots real to real Matt> functions, analytic or numerical. I don't believe gnuplot Matt> is up to this job, but I could be wrong.What's wrong with gnuplot? Type help data to learn how to do whatyou want in gnuplot. For analytic functions, just do a plot sin(x)or something like that. Last but not least, you can generate EPS fromgnuplot.-- Lee Sau Dan +Z05biGVm-(Big5) ~{@nJX6X~}(HZ) E-mail: danlee@informatik.uni-freiburg.deHome page: === http://www.informatik.uni-freiburg.de/~danleeSubject: Re: drawing math functions as postscript files> > Matt> Or maybe there is a free program that plots real to real> Matt> functions, analytic or numerical. I don't believe gnuplot> Matt> is up to this job, but I could be wrong.> > What's wrong with gnuplot? Type help data to learn how to do what> you want in gnuplot. For analytic functions, just do a plot sin(x)> or something like that. Last but not least, you can generate EPS from> gnuplot.Just to plot sin(x) requires:Definition of the origin e.g. in mm.Definition of the axes lengths in mm.Definition of scaleX and scaleY.Definition of xmin and xmax at axis minimum and maximum.Definition of ymin and ymax at axis minimum and maximum.Definition of dx for the function (number of steps).Definition of axes line widths and colors.Definition of graph line width and color.Definition of tic marks by divisions and heights/widths.Definition of tic mark line widths.Definition of tic mark colors.Definition of graph visible bounding box (coordinate system frame).Definition of graph visible bounding box line width.Definition of graph visible bounding box color. Definition of graph visible bounding box miter mode.Writing numbers at tic marks with fixed format (not overlapping).Writing numbers for the y-axis with appropriate eventuallly graph line-gap-line-gap pattern.Definition of end caps for axes.Definition of end caps for graph.Definition of clipping area for graph.Maybe thats not all ...GnuPlot handles this automatically ?I prefer direct programming by PS. There are still some problems left.Mainly the rounding which is handled differently by different interpre-ters. PSAlter is an excellent editor, but the results have to be testedby Photoshop, PageMaker and Acrobat Distiller. The EPS can be consideredas correct if all these applications show the same results. Just do a plot sin(x) is wishful thinking.Illustrations & Examples === HoffmannSubject: Re: drawing math functions as postscript files> >> Matt> Or maybe there is a free program that plots real to real> Matt> functions, analytic or numerical. I don't believe gnuplot> Matt> is up to this job, but I could be wrong.>> What's wrong with gnuplot? Type help data to learn how to do what> you want in gnuplot. For analytic functions, just do a plot sin(x)> or something like that. Last but not least, you can generate EPS from> gnuplot.> > Just to plot sin(x) requires:> > Definition of the origin e.g. in mm.> Definition of the axes lengths in mm.> Definition of scaleX and scaleY.> Definition of xmin and xmax at axis minimum and maximum.> Definition of ymin and ymax at axis minimum and maximum.> Definition of dx for the function (number of steps).> Definition of axes line widths and colors.> Definition of graph line width and color.> Definition of tic marks by divisions and heights/widths.> Definition of tic mark line widths.> Definition of tic mark colors.> Definition of graph visible bounding box (coordinate system frame).> Definition of graph visible bounding box line width.> Definition of graph visible bounding box color.> Definition of graph visible bounding box miter mode.> Writing numbers at tic marks with fixed format (not overlapping).> Writing numbers for the y-axis with appropriate x-offsets depending> on the fixed format.> Definition eventuallly graph line-gap-line-gap pattern.> Definition of end caps for axes.> Definition of end caps for graph.> Definition of clipping area for graph.> Maybe thats not all ...> > GnuPlot handles this automatically ?> > I prefer direct programming by PS. There are still some problems left.> Mainly the rounding which is handled differently by different interpre-> ters. PSAlter is an excellent editor, but the results have to be tested> by Photoshop, PageMaker and Acrobat Distiller. The EPS can be considered> as correct if all these applications show the same results.> > Just do a plot sin(x) is wishful thinking.> > Illustrations & Examples are much appreciated.While it is true that you can set all of these attributes, most have defaults. Lets see what minimum we can get away with: set terminal postscript color set output sin.ps plot [-pi/2:pi/2] sin(x)So, apart from saying that we want postscript and where it is to go, yes it is === just do a plot sin(x). :-)-GregSubject: Re: drawing math functions as postscript files>>Or maybe there is a free program that plots real to real functions, >>analytic or numerical. I don't believe gnuplot is up to this job, but I >>could be wrong.> > > Well, this is the main purpose of gnuplot.> You can use data points and descriptions via formulas.> Gnuplot can also generate Postscript output (among several other formats),> so printing should not be a problem.> > MarcOkay, maybe my brain has failed me. I was thinking that the gnuplot resolution was not good. Maybe I am thinking of interactive mode with X.I need to be able to print things in exact scale on the page, for instance precisely 1 inch per unit, both horizontally and === vertically.Subject: Re: drawing math functions as postscript files> >Or maybe there is a free program that plots real to real functions, >analytic or numerical. I don't believe gnuplot is up to this job, but I >could be wrong.>> >> >> Well, this is the main purpose of gnuplot.>> You can use data points and descriptions via formulas.>> Gnuplot can also generate Postscript output (among several other formats),>> so printing should not be a problem.>[...]> I need to be able to print things in exact scale on the page, for > instance precisely 1 inch per unit, both horizontally and vertically.I do not know if you can have this fine-control with gnuplot.In fact, I am not sure, if it is really needed, === but of course thisis for you to decide.MarcSubject: Suggestions for Applied MathsAs anyone who's read my posts and remembers them (all two of you :) will probably have noticed, I'm something of a pure mathematician.I consider this quite unfortunate; I actually started my degree doing maths with physics, but Cambridge doesn't allow that option past the first year so I had to choose between maths and physics; as I'm interested in both pure maths and science, and the cambridge maths degree has a reasonably large applied maths component, I chose to do maths.Unfortunately I forgot to take something into account when doing this; The applied maths courses at Cambridge (or at least all the ones so far, and I'm in my third year now) are really really bad. The lecturing is on average not very good (although there are some good lecturers), and most of the time when the lecturing is good the course is crippled by a bad syllabus; all physical interest has been removed, just leaving the maths, and then the maths is done badly. (Note: This isn't just my opinion, a lot of others agree with me. I'm sure there are those who don't as well, but I haven't talked to them about it if there are). I've managed to find two courses so far which are bearable, only one of which was genuinely good. In case the lecturers read this group I'd rather not go on record as saying which ones those are. :)Anyway, I've recently been suffering from applied maths withdrawal. Too much pure maths has got to my brain and I've begun to get urges to scream things like Yes, ok. It's a splitting field. It's very pretty and all, but what on earth is it good for!?. I can appreciate subjects for their mathematical elegance, but I'm really starting to crave some actual practical applications.So, I'm looking for suggestions for the following:Areas of physics I should pursue.Any other interesting applications of mathematics.Good books for the above.To give you some idea of my background:In pure maths:Strong backgrounds in analysis, including complex analysis, a fair bit of general measure theory and functional analysis. Decent knowledge of hilbert spaces, banach spaces, etc.Quite a strong background in topology. Basic knowledge of differentiable manifolds, etc. but not that much - I'll be attending a course on it next term though, and have books on it which I can read.Fairly strong backgrounds in a variety of algebraic topics - number theory, galois theory, linear maths, group theory, etc.Not as strong as I should be on probability, but working on it. Reasonably decent knowledge of statistics, but again working on it. Fairly half-assed knowledge of markov chains and information theory, but I'm planning to do some fairly heavy reading of both these subjects.An ok background in methods and calculus - fourier series, sturm-liouville equations, fourier transforms, a variety of ways of calculating integrals, etc. I'm a bit out of practice at some of this, but used to be pretty good at it.Not that it's as relevant, but also quite a strong background in set theory and logic (useful for some computer related applied maths I guess).As far as science is concerned, by background is unfortunately a lot weaker. I have A-levels in physics and chemistry, plus the result of a first year of physics and some applied maths courses.Basically this includes a reasonable general knowledge of physics and chemistry, followed by specifics in the following:Special relativity in a rather baby-version. The maths is done properly, but without mention of things like tensocovectoetc. A general (but rather non-specific) idea of the tensor formulation of it, minkowski space-times, etc. from a later course.Basic understanding of stastistical physics - ideal gas laws, boltzmann's law, etc. (although I'd need to refresh my memory on these), from a physicist's perspective rather than a mathematician's.Basic quantum mechanics; Some idea of the general physical background, a course which involved not that much more than solving the schrodinger equation and some basic theory of operatoetc.A good working knowledge of classical dynamics. Certainly Hamiltonian and Lagrangian dynamics, although I'd have to go over a lot of the ideas about canonical transformations, hamilton-jacobi equations, etc. before I was completely solid on them.A decent enough knowledge of electromagnetism, although because I wasn't supervised on either course I've done on it due to time constraints on my part there are holes in my memory of some of the ideas, but nothing that would be too hard to fill in.In general I seem to have a fairly good physical intuition, which has readily adapted itself to the subject even when it isn't 'classical'.I'd quite like to start with learning a proper theory of quantum mechanics, but I'm not entirely sure where to go. Given that I'm entirely free of a university syllabus, I'm not really sure which flavour of QM I should go for. Also I'm starting from a rather different perspective than most physics students, as I already have a strong background in the maths which most would develop alongside the subject (or at least that's how it seems to work here). I'd quite like an approach that was mathematically elegant, but that's relatively secondary to a good presentation and an interesting physical background.After that, I'm completely open to suggestions as suggestions you can give, and I'm sorry for the long and === in the obvious way)Subject: Re: Suggestions for Applied Maths> I consider this quite unfortunate; I actually started my degree doing > maths with physics, but Cambridge doesn't allow that option past the > first year so I had to choose between maths and physics; as I'm > interested in both pure maths and science, and the cambridge maths > degree has a reasonably large applied maths component, I chose to do maths.Should Cambridge be this shortsighted? I mean modern physics in everypart is a great deal of mathematics, especially geometry, functionalcalculus and topology.Rene.-- Ren.8e MeyerStudent of Physics & === MathematicsZhejiang University, Hangzhou, ChinaSubject: Re: Suggestions for Applied Maths> >>I consider this quite unfortunate; I actually started my degree doing >>maths with physics, but Cambridge doesn't allow that option past the >>first year so I had to choose between maths and physics; as I'm >>interested in both pure maths and science, and the cambridge maths >>degree has a reasonably large applied maths component, I chose to do maths.> > > Should Cambridge be this shortsighted? I mean modern physics in every> part is a great deal of mathematics, especially geometry, functional> calculus and topology.> > Rene.> I agree completely. Unfortunately the Cambridge undergraduate course isn't really all it's cracked up to be. It has a lot of points in its favour of course; some of it is genuinely very good, but it has some fairly major problems as well.In the Maths with Physics case, there is a lot of maths available to do in the Natural Sciences tripos, but done from an applied and not very rigorous at all perspective (or at least that's my impression of it). Pure maths is fairly lacking in it as far as I know. So it's not like choosing to do physics cuts you off entirely from the maths, just the pure maths.I believe there have at various points been attempts to allow a joint maths with physics degree, but === in the obvious way).Subject: Re: Suggestions for Applied Maths> >>I consider this quite unfortunate; I actually started my degree doing >>maths with physics, but Cambridge doesn't allow that option past the >>first year so I had to choose between maths and physics; as I'm >>interested in both pure maths and science, and the cambridge maths > >>degree has a reasonably large applied maths component, I chose to do maths.> > > Should Cambridge be this shortsighted? I mean modern physics in every> part is a great deal of mathematics, especially geometry, functional> > calculus and topology.> > Rene.> > > I agree completely. Unfortunately the Cambridge undergraduate course > isn't really all it's cracked up to be. It has a lot of points in its > favour of course; some of it is genuinely very good, but it has some > fairly major problems as well.> > In the Maths with Physics case, there is a lot of maths available to do > in the Natural Sciences tripos, but done from an applied and not very > rigorous at all perspective (or at least that's my impression of it). > Pure maths is fairly lacking in it as far as I know. So it's not like > choosing to do physics cuts you off entirely from the maths, just the > pure maths.> > I believe there have at various points been attempts to allow a joint > maths with physics degree, but it's never gone through.> > David> worse than looking into the area of fluid mechanics. Thereare interesting problems people are working on to do with thesingularities that form when droplets form and leave the bulk of thefluid. There are also interesting problems in the area ofviscoelasticity. You may have to get into numerical analysis andcomputer programming to follow this up. Problems in fluid mechanicsmay also have industrial relevance. The rheology of grease, paint,suspensions, polymeetc flowing through pipes etc may not beparticularly exciting but is quite challenging mathematically and ofrelevance to many industries.I did Natural Sciences at Cambridge, graduating in 1984, and chosePhysics & Theoretical Physics. There was a lot of maths in the courseat the time, although it wasn't taught in a rigourous fashion at all(more in the spirit of the book Mathematical Methods of Physics byMatthews and Walker - a good book if you can get hold of a copy).which is online. The link === is:Hope this helpsIan TaylorSubject: Re: Suggestions for Applied MathsQ. Are you considering doing part three next year?If so there are many applied courses which are really pure in disguise. Lie algebras (which appeared in representation theory in my third year) areincredibly useful, for instance. if you want to know why (and this isn'tmy area so i won't make any wild claims) look at the schedules for partthree quantum field theory (i think). i've tried writing the second part of this about 4 times now and i seem toget stuck on how much there is to say. sadly i'm an algebraist so i keephaving to delete what i write as i realize i'm increasingly talking aboutstuff i either don't know or don't remember properly.solid advice:0. simply put, topology and geometry underpins a lot of modernmathematical physics. to see it in action look at sci.physics.research1 read john baez's stuff. he's at UCR maths dept. should be easy enough tofind.2. if you want to hear some incoherent ideas email me on my properaddress:matt dot grime at bris dot ac dot ukfor stuff algebra and category theory in mathematical physicsbut === hopefully you'll get some better answers. Subject: Re: Suggestions for Applied Maths> > Q. Are you considering doing part three next year?I am indeed, yes.> > If so there are many applied courses which are really pure in disguise. Yeah, I thought that was probably the case although I haven't run into too many so far. How much do these courses focus on applications as well as theory?> Lie algebras (which appeared in representation theory in my third year) are> incredibly useful, for instance. if you want to know why (and this isn't> my area so i won't make any wild claims) look at the schedules for part2. Is there a measure on the space of open subsets of [0,1] that is uniform?> three quantum field theory (i think). Yeah, I'd heard representation theory was very useful for physics; I can't see exactly how, but I expect it will become clearer after I've attended the course as right now I know very little about it.I can't seem to find an online copy of the part 3 schedules though, so I can't look up the quantum field theory syllabus yet, but will bear it in mind.> > i've tried writing the second part of this about 4 times now and i seem to> get stuck on how much there is to say. sadly i'm an algebraist so i keep> having to delete what i write as i realize i'm increasingly talking about> stuff i either don't know or don't remember properly.> > solid advice:> 0. simply put, topology and geometry underpins a lot of modern> mathematical physics. to see it in action look at sci.physics.researchYep, I thought it would based on my general impressions of things. Happily it's also a subject which I quite like from a mathematical perspective. :)> 1 read john baez's stuff. he's at UCR maths dept. should be easy you want to hear some incoherent ideas email me on my proper> address:> > matt dot grime at bris dot ac dot uk> > for stuff algebra and category theory in mathematical physics> > > but hopefully you'll get some better answers. > > === Applied Maths > I can't seem to find an online copy of the part 3 schedules though, so I> can't look up the quantum field theory syllabus yet, but will bear it in> mind.There are no Part III Schedules[1]. There is instead a collection ofsyllabi for individual courses:http://www.maths.cam.ac.uk/CASM/courses/descriptions That's for this year; it changes every year and is generally not issueduntil after July[2].mouse clicks.Returning to your original points, the only major gripe I have with thePart I/II(B) applied courses is that seperable solutions of Laplace'sequation in cylindrical polars are not covered anywhere. Finding J0'(r) interms of J1(r) may have appeared on a IB Methods examples sheet, but otherthan that there are no Bessel functions whatsoever in Parts I or II(B)(and the connection with Laplace's equation was not made explicit); Ican't comment on II(A) because I didn't take that option. (And when I0and I1 appear in the middle of a Part III course, it therefore comes as abit of a shock.)[1]: In the strict sense of a list of topics approved by the relevantuniversity committees and which will be examinable, but to the lecturermay chose to add additional material if s/he so desires.[3][2]: Given that lecturers are pretty much free to set whatever syllabusthey want and are unconstrained by Schedules, none of the problems youmention in your original post should apply.[3]: Paradoxically, some parts of the Schedules are marked 'notexaminable'.-- P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will find these people and we === will bring them to justice.Subject: Conjecture regarding the number 12Has this conjecture been made before? Proven? A X). Tau (X) / X < Tau (12) / 12, for any natural number X > === 12.Subject: Re: Conjecture regarding the number 12> Has this conjecture been made before? Proven? A similar > Tau (X) / X < Tau (12) / 12, for any natural number X > 12.I thought I could answer this immediately, but all I have now is thefollowing related result:In the thread (posted about a month back) two number-theoreticallimits (& bonus safe=off&threadm=b4be2fdf.0310301645.38b9a167% number of positive divisors of m, was suchthatlimit{n-> infinity} (1/n) (sum{m=1 to n} tau(m)) - ln(n)= 2*c - 1, where c is Euler's constant (.5772...).So, the sequence {tau(m)}'s AVERAGES is asymptotical towards log(n).But this does not help you particularly, since you are interested intau(m)/m.As for your conjecture, I highly doubt it is true, but this is only myimmediate hunch.Leroy === QuetSubject: Re: Conjecture regarding the number 12> Has this conjecture been made before? Proven? A similar > Tau (X) / X < Tau (12) / 12, for any natural number X > 12.If p is a prime and e is an integer >= 1, then Tau(p^e) / (p^e) = (1 + e) / (p^e)With q = p^e, we have Tau(q) / q <= (1 + log_2 q) / qIt follows that Tau(q) / q < 1 / 2 for any prime power q >= 9.We also have: Tau(2) / 2 = 1 Tau(3) / 3 = 2 / 3 Tau(4) / 4 = 3 / 4 Tau(5) / 5 = 2 / 5 Tau(7) / 7 = 2 / 7 Tau(8) / 8 = 1 / 2So if Tau(X) / X >= 1 / 2, then any prime power dividingX must be in the set {2, 3, 4, 8} (Tau(x) / x is of coursemultiplicative). Since Tau(24) / 24 is smaller than 1/2,the result follows.---J === K Hauglandhttp://www.neutreeko.comSubject: Re: Conjecture regarding the number 12M?rio Amado Alves a .8ecrit dans le message de> Has this conjecture been made before? Proven? A similar Tau (X) / X < Tau (12) / 12, for any natural number X > 12.I dunno if that's precisely true, but somehow it's trivial: a number n hasat most -say- 2*sqrt(n) divisoand 2*sqrt(n)/n -> === 0 as n grows.Subject: Re: Linear algebra problem over Z_2> Hi all,> > Let V be a finite-dimensional vector space over the field Z_2, let> B be a symmetric bilinear form from V x V into Z_2 and suppose that> there's a basis B of V such that B(v,v) = 1 for every v in B.> Problem: to prove that there is some vector w in V such that> B(v,w) = 1 for every v in B. I know that this is true because I> proved it years ago, but I have been unable to remember how I did> it or to find another proof.In terms of matrices, what this question amounts to is the following.Let A be a symmetric square matrix over Z_2 with all diagonalentries equal to 1. Then the all-ones vector lies in the row(or column) space of A.Let R be the row space of A, and let K be the set ofcolumn vectors v over Z_2 with Av = 0. Then the spacesK and R are dual in the sense that w in R iff wK = 0and v in K iff Rv = 0.Let w be in K. Then wAw = 0. But if say, w = (1, 1, ..., 1, 0, ..., 0)^twhere there are s 1s, then wAw is the sum of the first s diagonalentries of A, viz., s modulo 2. So s is even, and this appliesfor all w in K, the number of 1 entries in w is even.Thus (1, ..., 1)K = 0 and so (1, ..., 1) is in R.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 === times)Subject: Re: Linear algebra problem over Z_2> Hi all,> > Let V be a finite-dimensional vector space over the field Z_2, let> B be a symmetric bilinear form from V x V into Z_2 and suppose that> there's a basis B of V such that B(v,v) = 1 for every v in B.> Problem: to prove that there is some vector w in V such that> B(v,w) = 1 for every v in B. I know that this is true because I> proved it years ago, but I have been unable to remember how I did> it or to find another proof.> > And, of course, I am not assuming that B is non-degenerate;> otherwise, the statement would be trivial (and the symmetry of> B would not be needed).> Neat problem, here's a solution:Induct on the dimension, n, of V.Base case n==1 left to reader.Assume true for n-1 prove true for n:Let B={b_1, ..., b_n} be the basis with B(b_i,b_i)=1 for i=1..n.Put B_i=B-{b_i} (i.e. omit the single vector b_i from the basis).The induct hypothesis implies there is some v_i such that B(b_j,v_i)=1 for all i not equal to j, i.e. apply the inductive hypotesis to the space spanned by B_i. If B(b_i,v_i)=1 for any i, we're done, so assume B(b_i,v_i)=0.Now consider the vectors v_i+v_j for i not equal to j.B(b_k,v_i+v_j)=0 for k not i or j,B(b_i,v_i+v_j)=1, andB(b_j,v_i+v_j)=1.so for any vector u in (Z/2Z)^n with an even number of 1's, we can find a v such that B(b_i,v)=u_i.So if n is even, we're done.If n is odd, we just need some vector w such that the vector with entries B(b_i,w) i=1..n has an odd number of ones.Claim: for some j, the vector with entries B(b_i,b_j) i=1..n has an odd number of ones.otherwise, the matrix A_(i,j) = B(b_i,b_j) must have an even number of ones, each column having an even number of ones. But by hypothesis, A is symmetric with ones down the diagonal of odd dimension.. an odd number of ones down the diagonal, and because of symmetry, an even number of ones off the diagonal.That completes the proof.I hope that's clear enough - gotta go to === work now!!> > > > Jason LeasureSubject: Re: Prime issue challenge> Well I claim that my prime formula is a great discovery, while others> keep posting that it's not important at all!> > However, I know exactly how my formula works, so it seems to me that a> good check of others is to see if they do as well.> > So I have a simple challenge. Posters assertions imply expertise, and> at a minimum that expertise should involve understanding *how* my> formula works.> > So I'll give them a couple of weeks to try and explain it in this> thread, then if all goes according to plan I'll explain to you how it> works, and you can see who is the real expert.> > > Here is the partial difference equation and instructions for summing.> > dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,> sqrt(y-1))],> > S(x,1) = 0.> > And p(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS> from dS(x,2) to dS(x,y).> Okay, here's how it works, stripped to essentials.Throughout this discussion, all sets will be comprised ofpositive integers and q will always refer to a (positive)prime integer. |A| will denote the cardinality of the set Aand for a positive integer n, nA will denote the set{n*a | a in A}. We will use the notation op [cond : term]to denote the iteration of the operator op, while somecondition _cond_ is true, over the terms _term_, so forexample sum [0 < i <= n : i] = 1 + 2 + ... + n.In what follows, I've interspersed the exposition withparenthetical examples, delimited HTML-style with.... These may be eliminated withoutaffecting the demonstartion.Define p(x, y) to be the cardinality of the (disjoint) unionof P(x, y) and C(x, y), where P(x, y) = {q | q <= min(x, y)} C(x, y) = {n | 1 < n <= x and n is not divisible by any q in P(x, y)}. P(50, 3) = {2, 3} C(50, 3) = {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49}so p(50, 3) = 18We see immediately that p(x, 1) = x - 1 andp(x, floor(sqrt(x)) = the number of primes <= x.Theorem. If x >= y, thenx - 1 - p(x, y) = sum [q <= y : p(floor(x/q), q-1) - p(q-1, floor(sqrt(q-1))] Let x = 50, y = 3, then 49 - p(50, 3) = 49 - 18 = 31 = (p(50/2, 1) - p(1, 1)) + (p(floor(50/3), 2) - p(2, 1)) = (p(25, 1) - p(1, 1)) + (p(16, 2) - p(2, 1)) = (25 - 1 - 0) + (|P(16, 2) union C(16, 2)| - 1) = 24 + |{2} union {3, 5, 7, 9, 11, 13, 15}| -1 = 24 + 1 + 7 - 1 = 31Proof of theorem. We may regard x - 1 - p(x, y) as thecardinality of the set {2, 3, 4, ..., x} - P(x, y) - C(x, y),which is the set of all numbers between 2 and x inclusivethat are not primes <= y and are divisible by some prime <= y.This set is equal to the disjoint union union [q <= y : D(x, q)] [1]where D(x, q) is the set of composites 1 < c <= x that aredivisible by the prime q and not divisible by any prime < q. For x = 50, y = 3, we have the set {2, ..., 50} - {2, 3} - C(50,3) (listed above) = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27, 28, 30, 32, 33, 34, 36, 38, 39, 40, 42, 44, 45, 46, 48, 50} = {2*2, 2*3, ..., 2*25} union {3*3, 3*5, ..., 3*15} = D(50,2) union D(50, 3)Now we observe that the union in [1] may be expressed as union [q <= y : qC(floor(x/q), q - 1)] [2] D(50, 2) = 2{2, 3, ..., 25} = 2C(50/2, 1) D(50, 3) = 3{3, 5, 7, 9, 11, 13, 15} = 3C(floor(50/3), 2)and finally that P(x, q-1) = P(q-1, floor(sqrt(q-1)), when x >= q,from which with [2], we conclude the equality in the theoremstatement.Thus we see that the proposed prime counting methodis indeed correct, though that wasn't in doubt.Is this Legendre's method? That depends on one's point of view.It's certainly not an exact copy, but it is not difficult tosee that Legendre's prime counting formula may be derived fromthis method and conversely, so in that sense the two techniquesare equivalent. If one were inclined to be unchartiable, onecould call this obfuscated Legendre; otherwise, it could bedesignated as streamlined Legendre.Is it better than Legendre's method? Well, it does not requireknowing the primes, since they computed on the fly. Unfortunately,the resulting method is several orders of magnitude slower thansome existing methods, at least when implemented as written.Does it have any applicability other than to produce a slowmethod of counting primes? I don't see any, though it istidy and compact. It might have some use as a pedagogicaldevice, perhaps as an exercise in a text, but I have to admitthat I don't see any use beyond that.Is it publishable? Perhaps as a classroom note, but the journalwould have to be selected with care. It would face the samedifficulty as an improvement on an algorithm like bubble sort--the response by a great number of reviewers would be, I suspect,This is mildly interesting, but it === hardly advances the stateof the art.RickSubject: Re: Prime issue challenge> Correction to previous post:> > Dr. Joseph Mengele, *not* Mengala.Attention please. I hereby invoke Godwin's Law and declare thisdiscussion dead.-- === mail1dotstofanetdotdkSubject: Re: Prime issue challengeof ourse, pure math is a crock. here's a reference:http://www.wlym.com/antidummies/part52.html. is this an experiment in markeeting, perhaps? of course, I don't care about counting primes, butI'll try to read an explanation, as it ocmes, andsubmit my filled-out questionaire to the boardof mathforprofit,com. > Well I claim that my prime formula is a great discovery, > That is, go ahead mathematicians, admit that pure math is a crock! > http://mathforprofit.blogspot.com/--ils duces de Enron!