mm-458 The usual method is to define sgn(x) in terms of the heaviside functionH(x) = { 0, x < 0; 1, x > 0; 0.5, x = 0 }, and write formallydH/dx = delta(x) where delta(x) = { 0, x != 0 } is the Dirac delta'function' (integral of delta(x) over any interval containing 0 is 1 bydefinition, and 1/2 if 0 is an endpoint).> This is a joke, right?No. The OP requested a formal solution, so I provided a formal solution.You may be misreading my compact notation for H(x) = 0 if x < 0, 1 ifx > 0 or 0.5 if x = 0, in which case I have just clarified it for you, oryou may not fully understand the properties of the Dirac delta.H(0) = 1/2 is a definition made to ensure that the formal statementdH/dx = delta(x) is consistent with the definition of the integralof delta. d/dx might not denote the classical analytic derivative, butit had still better be true that it is the inverse of indefiniteintegration and that it satisfies the Liebnitz product rule and the chainrule. If it didn't, the formalism would be useless.The knock-on effect that sgn(0) = 0 seems fairly logical: there is noreason why sgn(0) should be +1 or -1.The fact that H(0) = int_-oo^0 delta(x) dx = 1/2 follows trivially fromthe observations (i) delta(x) is an even function and (ii) the integral ofdelta over the whole real line is 1 (by definition).What is wrong is my statement that d/dx of delta(x) is undefined; it canbe multiplied by any differentiable f(x) and integrated by parts to yield-f'(0) if the domain of integration includes 0, and 0 if 0 lies outsidethe domain, so in this sense it is a well-defined distribution. Having 0on the boundary is fatal in this case (the boundary contribution isf(0)*delta(0), which is undefined). In fact, all derivatives of delta(x)are well-defined distributions:(d^n/dx^n delta(x)): f(x) |-> (-1)^n*d^n/dx^n(f)(0).However, whereas delta(x) has some meaning outside of integrals, the sameis not true of its derivatives, which is why I didn't bother quoting termsinvolving delta'(x) or higher derivatives.See http://en.wikipedia.org/wiki/Dirac_delta_function andhttp://en.wikipedia.org/wiki/Distribution-- 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: How big can a manifold be?>what about using whitney's embedding theorem? a jump>in dimension will have no effect on the cardinality.I suggest that you check the hypotheses of Whitney's embeddingtheorem (I'd even do it for you but I am a bit busy); I thinkyou'll find that paracompact (or something essentiallyequivalent) is necessary. (For instance, the long lineisn't paracompact, and I am pretty sure it doesn't embedin R^3, either...isn't the set of joining points inits construction both discrete with the relative topology, and of too big a cardinality to embed in R^3, never mindthe stuff in between?)Lee Rudolph === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> <8765e4q2fe.fsf@phiwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@phiwumbda.org> <87vfm4o85i.fsf@phiwumbda.org> <87n07fo9qr.fsf@phiwumbda.org> <87k72jnxtp.fsf@phiwumbda.org> Discussion, linux)> This TM will find a representation not on your tape:> It is a three state machine and I can provide a state> transition table if you like.>> 1) Find a blank> 2) Find a second blank> 3) Backup and write a 1 on the previous blank> repeat steps (1) through (3)>> This TM will produce a tape that contains exactly one blank.> The contiguous string of 1's preceding this blank will be> a representation not on your tape.> It will produce no such tape. It will produce a tape consisting of> all 1's. This is obvious.> Suppose that it contains a blank. Then that blank must occur at some> square on the tape, say square n. It is trivial to see that, by the> nth iteration of your three steps, square n is no longer blank.> (After the first iteration, square 1 is not blank, square two was> never blank, square three is not blank after the second iteration and> hence is also not blank after the third, and so on.)> This TM always checks to see if there is another blank on the tape> before overwriting the previous blank. That is what step 2 does.> It is easy to prove there is a blank on the output tape.Wrong.It is easy to prove that at each step n, there is another blank on thetape.Unfortunately, you want to talk about the output after an infinitenumber of steps. You have proved that at each finite step n, there isa blank on the tape. That is not sufficient to prove that after aninfinite number of steps, there is still a blank.Moreover, there is a very simple proof that there is no blank. Lookup at the paragraph that begins, Suppose that it contains a blank.> Your proof of this claim is simply another confusion.> It is a three state TM.> How confusing can it be?The TM is not confusing and yet you are confused. Huh.I didn't say your TM was likely to confuse a man of averageintelligence. I merely said you were confused. Evidently, youconfused these two statements also.>> No, no, no. A TM cannot extend the representation of natural numbers>> for two reasons: (1) It's a ing Turing machine, isn't it? I don't>> want to argue about requirements for intentionality or whether TMs are>> capable of intelligence, but this TM is not part of the negotiations>> regarding our conventions.>> No intelligence required.> My TM just adds all the numbers> together (plus 1 for each addition).> A TM is too stupid to know that> the sum is supposed to be infinity.> If your tape contains a single contiguous block of an infinite number> of 1's, then your tape does not contain a set of natural numbers.> A single contiguous block of an infinite number of 1's followed by> a blank in a finite position? I never said the output of my TM> contained the set of all natural numbers. I said it would contain> a representation not on the initial input tape. The output contains> the representation of exactly one natural number.And of course you're simply wrong.You repeat the same basic mistake repeatedly hereafter. There's nopoint in my writing that you're mistaken three more times. I'vesnipped the rest.My above explanation is sufficient.-- Jesse F. HughesI thought it relevant to inform that I notified the FBI a couple ofmonths ago about some of the math issues I've brought up here. -- James S. Harris gives Special Agent Fox a new assignment. === Subject: Re: covering compact set w/squares>Oh, I think I did misread. How about this:>Given a compact set K in the plane s.t. each pt x is the center of a square>Q_x, prove that you can find a subsequence Q_x_i of squares s.t. K is>covered by the Q_x_i and>sum(over i) Char(Q_x_i) <= 4 Char (union(over i) Q_x_i),>where Char(X) is the characteristic (indicator) function of X.>Are you assuming the squares are all oriented with sides parallel to the >axes? I don't know whether _he's_ been assuming that, but I have been.>If not, there's a counterexample with K consisting of 7 points at >the vertices of a regular heptagon; take a square centered at each of>these points, oriented so the centre of the heptagon is on a diagonal>of the square, and large enough so that the square contains a>neighbourhood of the centre of the heptagon but each point of K is>outside the squares centred at other points of K. It makes a>rather pretty picture: .>Robert Israel israel@math.ubc.ca>Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2************************ === Subject: Re: Axioms defining a finite field === >> Subject: Axioms defining a finite field> Let (F, +, *) be a finite set with two operations> and constants 0, 1 such that the following rules hold:> (1) a + (b + c) = (a + b) + c> (2) a + 0 = a> (3) for every a there's a b so that a + b = 0> (4) a*(b*c) = (a*b)*c> (5) a*1 = a> (6) 1 is distinct from 0> (7) a*(b + c) = a*b + a*c> (8) (a + b)*c = a*c + b*c> (9) a*b = 0 => a=0 or b=0> Show that F is a field. Can one of the rules be omitted> so that F still has to be a field?>> let a /= 0, b0 = 0, b1 = 1>> A = { a bj | 0 <= j <= |F| }> A| = |F| because>> if a bj = a bk: a(bj - bk) = 0; bj - bk = 0; bj = bk>> As F is finite, 1 in A>> which shows a has right multiplicative inverse a_r.>> This with a1 = a, shows F0 is a group under *.>> Did you forget a+b = b+a, ab = ba?> I don't know about the a+b = b+a, but if we add that then> ab = ba follows: it's a theorem that every finite division> ring is a field. (It's not quite trivial, as I recall. Possibly> it actually is trivial and just didn't seem trivial to me> at the time; that was an algebra class when I was an> undergraduate...)>No, not trivial . This is Wedderburn's theorem; Proofs from the Book gives>a nice (what else?) four page proof (chapter 5) due to Witt (and mention 7>or 8 more, using quite different ideas). Witt begins by proving, using>linear algebra, that the centraliser of s , is of dimension q^(n_s) , where>q is the characteristic of F (q.1=0) and that n_s divides |F|. Then, he>imbeds F in the roots of unity in C, and conclude by a clever argument on>the cyclotomic polynomials...You have any idea whether a + b = b + a follows from the conditionsabove?> ************************> ************************ === Subject: Re: JSH: Making it personal <3c65f87.0402190553.cdc1fc0@posting.google.com> Discussion, linux)> You're an asshole Dik Winter. No doubt about it.> Look who's talking! The arch-asshole of all.Virgil, no offense, but when you feel tempted to call someone anarch-asshole, maybe you should just wait for the muse to return.-- It has been shown that no man can sit down to write without a very profounddesign. Thus to authors in general trouble is spared. A novelist, for example,need have no care of his moral. It is there -- that is to say, it is somewhere-- and the moral and the critics can take care of themselves. -.A. Poe === Subject: Re: I got low score on math test, please advise me and take a look> David C. Ullrich schrieb:>problem 9: 4.5 out of 5. I would have given you fewer points>than that, because there's a bit of explanation missing>(you need to point out that the function is not even>_defined_ at the point in question, because that>denominator vanishes).>Am I missing something here? How can you say that a function is>discontinuous at a point, where it's not defined? That function is>continuous at each point of its domain - Unless of course you define>f(2) = 3.01 or something like that.Well yes. The meaning of continuous seems to shift a littlebetween calculus and mathematics on exactly this point...Regardless, _my_ point was just that he didn't give anadequate explanation of why that point is not includedin the set where f is continuous.>Thomas************************ === Subject: Sequences Involving Binary & Prime-FactorizationFirst, we have the sequence formed by simply lettinga(m) = 1, if the sum of all exponents of the prime-factorization of mhas no carries when summed in in base-2.a(m) = 0, if there are any carries in the summing of the exponents ofthe prime-factorization of m.So, for example, a(12) = 1 because 12 = 2^2 *3^1,and, in base-2, 2 = '10', 1 = '1',and '10' and '1' have their ones in different positions.But, a(24) = 0, because 24 = 2^3 *3^1,and, in base-2, 3 = '11', 1 = '1',which both share a rightmost one.Unless I made an error, this sequence is not yet in the EncyclopediaOf Integer Sequences.a(k) -> 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1,...(Defining a(1) = 1 makes b()-recursion below work.)Now, let b(m) = sum{k=1 to m} a(k).b(m) gives the number of positive integers <= m and with an a() of 1,obviously.My main math result related to these sequences:(a somewhat interesting method of calculating {b()} using recursion):b(1) = 1;b(m) =b(floor(sqrt(m)) + sum{p=primes<=m} b(floor(sqrt(m/p))),(or, if we choose to define 1 as a prime:b(m) =sum{p=primes<=m} b(floor(sqrt(m/p))) )And, unless I made an error, this sequence is not yet in theEncyclopedia Of Integer Sequences either.b(k) -> 1, 2, 3, 4, 5, 5, 6, 7, 8, 8, 9, 10, 11, 11, 11, 12, 13,...Leroy Quet === Subject: Re: equivalent expressions>I'm sure this is just a print error, but a Foundations book I have>asks the following:>[Q] Give an example of two equivalent [propositional] expressions that>do not have the same truth table.>Is there something deep I'm missing here, because what I see is a>contradiction. Perhaps my misunderstanding lies in what they mean by>truth table.Seems like nonsense to me...Hmm, unless they meant something silly like this:P & ~P and Q & ~Q are equivalent, but they havedifferent truth tables because there's a P columnin one table and no P column in the other?>Any help is greatly appreciated.>R************************ === Subject: Re: the anticlassicalist }{ : mitchism for Daleks <40325A2C.21936CE9@hate.spam.net> <103513r3gafocff@corp.supernews.com> <40328829.1E5C341D@hate.spam.net> <1035esdt30h4d38@corp.supernews.com> <1039dr5kpjloif1@corp.supernews.com> <+No$awZ0HMNAFw0n@baesystems.com>> In message <1039dr5kpjloif1@corp.supernews.com>, galathaea>: >You should have realised that anything full of -ism and -ist which is>spewed>: >to sci.lang and sci.logic is not going to contain any physics.>: >>: (Nor language; I can't speak for sci.logic ;-)>:>: I'm well aware of that. What I was wondering was whether the _OP_>: realises that there's a difference between these isms and physics, and>: what it consists of.>Installment vi> >>Before you lies the Void...>details the properties of the closed linear subspace lattice>of a Hilbert space, detailing the relationship with observables and>prediction in quantum systems. This builds off of the mathematical theory>You do understand the physical content, do you not?> Haven't seen any.> Dirac compressed quantum mechanics into ~300 pages of terse elegance,> and didn't use ontology once.Obviously, Dirac knew the tastes of the audience to whom he was writing. Theman had to get published, didn't he? Taxes needed to be paid. The wife andchildren needed to eat. Little things like that have often gotten in the wayof principle.But one must not to say Dirac did not have ontology in mind....Therefore, there is plenty of room for improvementand for seeking to discover a new and logically superiorfoundation for the theory. I suspect that it is simpleAristotelian logic (in the context of Platonic ontology)that is missing from the theory, and this is due to thepainful divorce of physics from philosophy.Fragmented people think fragmented thoughts thatnever add up and ultimately don't make sense, whateverillicit success may initially obtain. The deepest urge inmy being is to understand the principles of physicsfrom the first principles of logic. Without that ourclaim to knowledge is nought but a pretense, and oursouls are divided against themselves, leaving usculturally schizophrenic and socially insane. http://www.geocities.com/saint7peter/ DiracslectureonQFT.htmlWhat a great legacy you physics clowns leave everyone. Since Aristotlecomplained of fatalists in his own work, I doubt your kind of 'ism' offers muchhope for change. Moreover, Hobbes observed that the utility of mathematicsresides in building weapons of war. Surely, one cannot deny that physics--thepristine science of non-isms--has its ideological protectors.Methinks the Daleks doth protest too loudly.In any case, whatever Dirac did or did not do has little in common with thosepeople who advertise physics in the modern information age. Ontology flashesbetween every word like the old subliminal buy popcorn frames.:-)mitch----If one were actually paying any attention to anything but the choir and itslaw-giving preachers, one might actually have a clue to how mathematicalphysics is linked into philosophy by Heyting algebras.In Distributive Lattices Balbes and Dwinger define a closure algebra asfollows:A Boolean algebra L with an additive closureoperator ^c in which 0^c=0 is called a closurealgebra. An element (a e L) is called open if(a-bar e L^c). The set of open elements isdenoted by L^oThey make that definition in order to establish a representation theorem forHeyting algebras:If L is a closure algebra, then L^o is a Heytingalgebra under the partial ordering of L and is alsoa D01-subalgebra of LMore remarkable is the fact that every Heytingalgebra can be represented this wayNow, what one needs to recognize immediately is that the Heyting algebra hereis entangled relative to consequence relations in the closure algebra. InProtoalgebraic Logics by Janusz Czelakowski you will find the remark,Thus any consequence is a closure operator ona sentential language. Following logical tradition,the separate term 'consequence operation' isreserved to denote such closure operations.Since Mr. Herring is personally being such a pissant, perhaps he can explainwhat is meant by physical content anyway? Does that mean using words andphrases that depend on Cauchy's (?) epsilon-delta justification for thederivative or Robinson's non-standard analysis justifying naive use ofdifferentials? His apparent interest in Dirac extends to I let other peopledo my 'isms' so I can act the fool... just like the well-received FranzHeymannThe development of the limit statement and derivative in terms of open setsintimately links truth in physics with Heyting algebras. Cech's association ofa nerve with open sets links truth in physics with simplicial homology. Andthe work in foundations with topos have associated the logic of open sets withconsistent logic. In other words, the mathematicians have been justifyingphysicists' crap while they play at predicting how the machines they build willbehave in the experiments in which they imagine them to be used.The problem is understanding the complexity between the Heyting algebras andthe Boolean algebras. We get some sense of that with Kuratowski 14 setproblem,Consider the collection of all subsets A of thetopological space X. The operations of closureA ---> A-barand complementationA ---> X-Aare functions from this collection to itself.(a) Show that starting with a given set A, onecan form no more than 14 distinct sets by applyingthese two operations successively.Unfortunately, while that problem helps us a little bit with respect to closureoperations and complementation between closed sets and open sets, the number 14here is significant.For example, in tense logic we have Benthem reporting on the Hamblin FourteenTenses Theorem:Of course, questions of correspondence andaxiomatization are not the only queries arisingin this semantics. For instance, an obvious questionis, on any given temporal structure, what happensto the potentially infinite number of 'tenses' that is,sequences of operators F, P, G, H in our language.A good exercise to become familiar with thepeculiarities of our formalism is to prove Hamblin'sFourteen Tenses Theorem: On the real time axis, there are only 14 logically distinct sequences of operators.We cannot simply draw pictures if we are to understand what is happening here.I could go through a great deal of complexity, but no one would understand thateither. The fact of the matter is that problems must be reduced to what Fregecalled 'judgeable content.' The appropriate algebra for this is a de Morganalgebra,[Balbes and Dwinger]The reader should note that the operation ~ is, ingeneral, not set theoretic complementation.... there is a one-to-one correspondence betweenall complete de Morgan fields of subsets of a setX and the involutions on X.In other words, one first needs a restriction to statements capable ofrefutation and confirmation before embarking on an empirical science groundedin an envelope of refutations. So, where we understand logic with respect toconsequence relations, the modern Boolean logic and Heyting logic decompose thesituation to reflect that empirical sciences operate according to datasemantics. These are very close to intuitionistic logics.In any case, the next technical concept of interest the free Boolean algebras.This is because they satisfy the countable chain condition,[Balbes and Dwinger]Definition 20Let a be an element in a lattice L. A nonmptysubset (S subset L) is a-disjoint provided thatxy=awhenever x, y are distinct elements of S. S iscalled disjoint if L has a 0 and S is 0-disjoint. Alattice which has no uncountable disjoint subsetsis said to satisfy the countable chain condition.Then,The countable chain condition derives its namefrom the fact that in a complete Boolean algebra L,the countable chain condition holds if and only ifevery well ordered chain in L is countable. Theseconditions are not, in general, equivalent. However,in the course of this section, we will show that bothconditions hold in F_B(X) and F_D(X)F_B(X) and F_D(X) are the free Boolean algebras and the free distributivelattices.Here are two of the exercises that follow the remarks quoted above,Let X be a countable infinite set. Show that 2^Xsatisfies the countable chain condition but containsa chain isomorphic with the reals.Let Y be a set such that |Y|=2^(Aleph_0) and let Lbe the Boolean algebra of finite and cofinite subsetsof Y. Show that every chain in L is countable but thatL does not satisfy the countable chain condition.The first question should get us back to physics somewhat. I do believe theystill think in terms of real spaces, except perhaps with loop quantum gravity(they think in terms of tetrhedral simplexes and are wondering why some oftheir objects seem to require recursive definition. Go figure.)In fact, it tells us a little more. One of the questions in the metaphysics ofphysics asks why the mathematics reduces to probabilistic models built onHilbert spaces. Well, if we simply look at the discussion of descriptive settheory in Set Theory by Jech, we discover that there is a minor problem withLebesgue measure,[Jech]Although both 'null' and 'meager' mean in a sense'negligible,' the following exercise shows that thereal line can be decomposed into a null set and ameager set:Exercise 39.12. There is a null set of reals whosecomplement is meager[Let q_1, q_2, ... be an enumeration of the rationals.For each n>=1 and k>=1, let I_nk be the openinterval with center q_n and length 1/(k*2^n). LetD_k=bigcup_(n=1 to oo) I_nk andA = bigcap_(k=1 to oo) D_kEach D_k is open and dense, and m(D_k)<=1/k.Hence A is null and R-A is meager]With respect to the second problem from Balbes and Dwinger, we can concludethat the Boolean algebra of finite and cofinite subsets is not a free Booleanalgebra. Indeed, they go on to write,Every free Boolean algebra satisfies the countablechain condition.Every chain in a free Boolean algebra is countable.Recalling from above that the Heyting algebra representations are associatedwith closure spaces, we might get some sense of what is involved here by citingSchmidt's theorem,[Czelakowski]For a closure system C the following conditions areequivalent: (i) C is finitary (ii) C is inductive(iii) The union of every nonmpty directed subset of C belongs to CIf C is a finitary closure system, then (C, subset) is analgebraic lattice and the sets J(X), X-finite are compactelements of this lattice.Well, J(X) refers to the closure operation, and, the difference betweenCzelakowski's closure systems and the closure algebras of Balbes and Dwinger isan additivity condition. Czelakowski's definition of finitary and inductiveused in the Schmidt's theorem is straightforward,[Czelakowski]The notions of a closure operator and of a closuresystem are thus coextensive. A closure operator Jon a set A is called finitary, if for any (X subset A)and (a e A)if (a e J(X)), then (a e J(X_f) for some finiteset (X_f subset X)A closure system is finitary if the correspondingclosure operator is finitary....A family C of subsets of A is said to be inductiveif the *set-theoretic union* of every nonmptychain in C belongs to C (A chain is being understoodhere as a chain with respect to set-theoretic inclusion).In case anyone missed my long discourses on Tore Langholm's Partiality, Truth,and Persistence, the notion of a neighborhood assignment with which heanalyzes truth definitions is contrary to the description of finitary closureoperators,For a given model M, a neighborhood assignmentis a function eta from the finite subsets of |M| tosubsets of |M| satisfying the following,1. (D subset eta(D))2. if (D_0 subset D_1) then (eta(D_0) subset eta(D_1))Since Langholm is investigating first-order logic relative to strong Kleenetruth, however, this does make sense. In the current context, closure systemsrefer to loci separated from the loci in which Langholm is interested.Apparently, it takes no more than seven alternations of closure andcomplementation to achieve the separation needed to satisfy the compactnesscondition for (X-finite) in a finitary closure system.If we simply ignore the closure/complementation relationship, we are reflectedelsewhere.As per the statement above, the representation for a Heyting algebra admits itsinterpretation as a D01-subalgebra. According to Balbes and Dwinger, there areno projectives in D and the only projective in D01 is 2 ({0,{0}}). Withrespect to world semantics (whatever) the 2-universal models in http://www.illc.uva.nl/Publications/ResearchReports/MoL-2001- 09.text.pdfare asssociated with 7 graphs generating 26 Heyting algebras. So, there isthe magic (Dalek) number of 26 again.Do the Daleks see what comes of running around telling people one knows what istrue and false in the universe? Refuting other people's belief systems is notthe same as truth. Ahh,... but such GREAT BOMBS!For my part, I am guessing that the concept of a question for datasemantics/strong Kleene truth seems to reside with the free de Morgan algebraon one generator, 1 | | * / / a ~a / / * | | 0The proof concerning projectives in D and D01 involves a map from 3 into 2x2, 2 -------> (1,1) | / | / 1 ---> (1,0) (0,1) | / | / 0 -------> (0,0)0 ---> (0,0)1 ---> (1,0)2 ---> (1,1)But, when 3 is taken as a de Morgan algebra, 1=(~1). So, a comparable diagramin the class of de Morgan algebras to the free algebra on one generator wouldsuggest dissociation of a Shannon bit into yes and no distinguishablefrom 0 and 1. It is a simple and meaningful construct.I mentioned Hamblin's tense logic above in comparison with Kuratowski's 14 settheorem. Malinowski specifically notes that tense logic is a subclassicallogic to which presupposition via negation applies, http://www.uni.torun.pl/~jacekm/bimatrix.pdfMoreover, that process of discerning a presumed consequence only works where deMorgan laws hold. Otherwise, there is simply no logic known (to Malinowski)that enables one to investigate presupposition. That is why I believe aquestion in quantum mechanics (per Mackey's seventh axiom) must be understoodwith respect to idempotents in the sense of de Morgan.This kind of thing is already appearing in quantum topology with theformalization of idempotent Jones-Wentzel projectors.Are the Daleks confused yet?Welcome to Cantor's paradise where ALL means ALL.lol:-)mitch === Subject: Re: equivalent expressions> I'm sure this is just a print error, but a Foundations book I have> asks the following:> [Q] Give an example of two equivalent [propositional] expressions that> do not have the same truth table.> Is there something deep I'm missing here, because what I see is a> contradiction. Perhaps my misunderstanding lies in what they mean by> truth table.> Any help is greatly appreciated.> RHow about two expressions which are both always true (or both always false) but which have different numbers of variables? === Subject: Probability Instinct - Sufficiently Discrete to Survive in the JungleDiscrete enough to outrun a predator or the slowest member of your ownspecies.THE PROBABILITY INSTINCTIt looks as if Kant, who thought our minds structure our perceptions, wasright. Probability was built into our minds. Our minds, the electrochemicalsymphony that our narrowly evolved neural ganglia play, impose aninfrastructure on our thinking. The mind imposes a background of time andspace and causal connectedness. Scientists have never seen a causality inthe wild. They have seen, and they predict, only space-time events thatfollow space-time events. Apples on the tree, then apples in the air, thenapples on the ground. Equations and correlations have replaced causes, justas science has largely replaced philosophy and religion as a theory ofthings. No causal germ in one event unfolds into another event. But themind, as eighteenth-century philosopher David Hume observed, makes it seemso and inserts the causal links in the event chain.Probability seems to be part of the same mental infrastructure. It formspart of our mental background or viewing screen along with time and spaceand causality and similarity and the topological notions of continuity andconnectedness. We see probability everywhere because it lies in our glasses.I believe that probability or randomness is a psychic instinct or Jungianarchetype or mental trend that helps us organize our perceptions andmemories and most of all our expectations. Probability gives structure toour competing causal predictions about how the future will unfold in thenext instant or day or season or millennium.Probability ranks or weights the future alternatives. Our expectations thenblend or average these future alternatives into a singleprobability-weighted average. The probability weights do not exist outsideour minds. They have no physical reality but have a powerful psychologicalreality rooted in our neural mi-crostructure. Hume also thought that we makeup probability as we go and use it to fill in gaps in our mind schemes orworld views: Though there be no such thing as chance in the world, ourignorance of the real cause of any event has the same influence on theunderstanding and begets a like species of belief.This probability instinct seems to cut across cultures and may cut acrossspecies. Besides the probability-laden psychology of scientists and mostnonscientists, the widespread gambling and games of chance in primitive andmodern cultures suggest that probability reasoning may be a culturalconstant like hero worship or fertility rituals or incest and adulterytaboos. A cultural constant suggests a biological substrate, and thatrequires an evolutionary history.Ranking future alternatives can help pass on genes. Those who could so rankmay have eaten those who could not. It allows us to bet before we act andimprove the outcome of acting. That forward-looking ability has supremesurvival value in biological evolution, the genetic variation and selectionin the last few million years that has finely sculpted our brains and minds,and in the prior evolution that sculpted the brains and minds of ourmammalian ancestors in the last 220 million years. Natural selection filtersout organisms as they cross the fuzzy line from the present to the future.Natural selection favors brain mechanisms that help an organism make itsnext move in a changing and dangerous world. These forward-looking brainmechanisms may run deep in the structure of mammalian and even reptilianbrains. Future studies may find that the brains of chimps and apes andlesser-brained mammals house a forward-looking probability instinct. At theother extreme we should not be surprised that scientists have exaltedprobability ranking into their grand organizing principle of maximumprobability. Scientists follow their probability instincts as their hominidforefathers followed theirs. Scientists just know more math.Fuzzy Thinking - The New Science of Fuzzy Logic Koskohttp://www.amazon.com/exec/obidos/tg/detail/-/078688021X/ === Subject: Re: Model theory puzzle> Find a finite system of axioms (feel free to introduce operations,> relations, constants) so that all of its finite models would have a> prime number of elements, and for every prime p there's a model to> that system of size p.> Posted Via Usenet.com Premium Usenet Newsgroup Services> ----------------------------------------------------------> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **> ---------------------------------------------------------- > http://www.usenet.comFinite fields? === Subject: Re: Coprime Grid: Filling Infinite Quadrant> >.....>>Fun perhaps: Show if my algorithm is foolproof...or just foolish.>>Well, what I know about number theory and 50 cents will get me a cup>of coffee, but it seems to me that the way to attack this solution>is to show that once the peninsula length exceeds some number N,>you will always have a noncoprime pair when it loops its way back.>>It seems to me that this ought to be true. >>Of course, that stil wouldn't prove that your algorithm can't>work, because you might never have to buld a peninsula out that big,>but it would be a start.>>(I know I do not give the peninsula's length. Also fun perhaps: try to>>determine the shortest length needed for each peninsula, given that>>the algorithm never leads to a problem.)>Leroy Quet>>George>I do not believe you are necessarily right about the filling-algorithm>working for all peninsula-lengths above N(k), for the k_th peninsula.> That's not what I said. I'm not trying to make your algorithm work,> I'm trying to show it won't work, I find that to be more fun :-)> My thought was that, in general, the longer, the peninsula,> the more likely there will be at least 1 pair of noncoprime> numbers when it loops back. So maybe beyond a certain length> you will always have at least one such pair.> For example, if I start at one, and try various peninsula lengths,> 2 and 3 are ok, but 4 doesn't work because 3 matches with 6> 1 2 3 4> 8 7 6 5> further, any penisula with lenngth 3n+1 will also have a problem with> multiples of 3 matching> 1 2 3 4 5 6 7> 14 13 12 11 10 9 8> similarly, if the peninsula is of length 7 then multiples of 5 will> match, and there will be a match for any peninsula length 5n+2.> similarly, for a peninsula of length 10 then 7 will mathch 14, and> you'll get a match for every peninsula lentgh of 7n+ 3> So the density of possible peninsula lengths decreases in a manner> analogous to the way that the density of primes themselves.> But although there are an infinite number of primes, it's not clear to> me right now if there are an infinite number of possible peninsula> lengths. To those of you who know lots of number theory,> it may be really obvious, but it isn't to me.> George>Leroy QuetSorry...Let us ignore the issue of the path going around bends (for now,anyway).If we were to let the peninsula be of length r, and r works, thenevery peninsula of lengthr + m *product p,where p = primes dividing terms of previous outside-section of path,should work too.(Because every integer in outer path-section would remain the same(mod p), for each p dividing terms in previous path-section,whatever integer m is.)So, if there are NO solutions for peninsula lengths > some N,then there are no solutions at all using my algorithm.Does anyone feel like programming a computer using my 'algorithm' tofind how far it IS successful, if it eventually fails?(I know it goes as high as 99, as figured by hand, unless I made anerror.)(thanks)Leroy Quet === Subject: Re: Axioms defining a finite field Adjunct Assistant Professor at the University of Montana.>> (1) a + (b + c) = (a + b) + c>> (2) a + 0 = a>> (3) for every a there's a b so that a + b = 0>> (4) a*(b*c) = (a*b)*c>> (5) a*1 = a>> (6) 1 is distinct from 0>> (7) a*(b + c) = a*b + a*c>> (8) (a + b)*c = a*c + b*c>> (9) a*b = 0 => a=0 or b=0>You have any idea whether a + b = b + a follows from the conditions>above?First, from (1,2,3) we can derive that 0+a=a+0=a for all a, and thatfor all a there exists b such that a+b=b+a=0, by the usual methodsvalid for groups. Then, by 1, addition is associative. Now considera + a + b + b = a*1 + a*1 + b*1 + b*1 (by 5) = a*(1+1) + b*(1+1) (by 7) = (a+b)*(1+1) (by 8) = (a+b)*1 + (a+b)*1 (by 7) = a + b + a + b (by 8)c+(a+a+b+b) + d = c+(a+b+a+b)+dhence(c+a) + (a+b) + (b+d) = (c+a)+(b+a)+(b+d) 0 + (a+b) + 0 = 0+(b+a)+0 a + b = b + a.Unless I've missed something...-- === ==Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: Definition of Separable Space (basic topology question)>Let A be a subset of (X,T). Then A is dense in X iff for every nonmpty>open subset U of X, A / U != {}.>I have seen two definitions of 'separable topological space':>a) (X,T) is separable if there exists A X, where A is countableand>dense in X>b) (X,T) is separable if there exists A X, where A is>countable and dense in X>Given def (a), its simple to prove that all countable spaces X are>automatically separable since the subset X is countable and non-trivially>intersects every nonmpty open subset. However, this won't work givendef>(b). I first became suspicious when I saw a proof that all countablespaces>were separable that seemed ridiculously complex in comparison to the>seemingly obvious 1 step proof above. I later found definitions of>separable like def (b).> The integers with the usual topology is separable.Right. This would follow from the fact that any topology on a countablespace is separable.> Definition (a) is the correct one. BTW, finite> spaces are separable.I consider finite spaces to be a subset of the class of countable spaces.However, from what I recall, this isn't a universal definition.http://en.wikipedia.org/wiki/CountableA countable (or denumerable) set is a set which is either finite orcountably infinite.Is there something else I'm missing here?l8r, Mike N. Christoff === Subject: Re: No Set Contains Every Computable Natural (was Church-Turing compared to Zuse-Fredkin thesis)> I realize this is gross over-posting, but I just had to add this quote from> Mathworld:> In fact, all theorems of calculus remain true if the field of real numbers> is replaced by the field of definable numbers, sequences are replaced by> definable sequences, sets are replaced by definable sets and functions by> definable functions.> Quite interesting.I find it quite natural!!! === Subject: Re: How many long primes are there?My apologies for topposting. I didn't want to cut any of the data, and I didn't want to require scrolling through it all to get to my response. I gather what you are on about is, you want to know stuff about the length L of the period when you expand 1/N to base b. I think that by a long prime to a given base b you mean a prime p such that the period of 1/p base b is p - 1. If b and N are relatively prime and you look at the numbers b, b^2, b^3, ..., and look at their remainders on division by N you will eventually see remainder 1. When that happens, the exponent on b is the L you are looking for. That is, saying L is the period length is the same as saying 1) b^L has remainder 1 when you divide by N, and 2) if 0 < r < L then b^r doesn't leave remainder 1 when you divide by N. You mention the Totient rule so I take it you're familiar with Euler's totient function, more commonly known as the phi-function. L is at most phi(N), and is always a divisor of phi(N). If N is a prime then phi(N) = N - 1. If b is a k-th power residue (mod N) [and not an m-th power residue for any m > k dividing N - 1] then L is (N - 1) / k. Artin's conjecture states that for any b there are infinitely many primes N such that b is a primitive root (mod N); that is, b is not a k-th power residue for any k > 1 dividing N - 1; that is, N is what you are calling a long prime to base b. Hooley proved that the Generalized Riemann Hypothesis (GRH) implies Artin's conjecture. GRH is a bit of a long story, and Hooley's proof is way beyond anything I would attempt to summarize on Usenet. More recent work has shown that there are at most a few exceptions to Artin's conjecture. This is also very advanced stuff. > N can represent any integer> L represents the length of the period of the expansion of 1/N> The residues at the same level have the same period lengths> For each subsequent residue the period length halves until the length is> odd. At that level the residues form a cycle.> Can anyone tell me how many long primes there are to various bases?> Can anyone tell me what was the deep work that Christopher Hooley has done> that John Conway and Richard Guy alludes to in The Book of Numbers> > Why didn't they mention the correlation of period length to the quadratic> graphs below? They mentioned the Totient rule and Wilsons theorem. Is this> too basic?> Could someone enlighten me please> N L Graph of quadratic residues b where b also represents the base> in which the expansion of 1/n occurs> 2 1 1 > > 3 2 2 > 1 1 > 4 2 3 > 1 1 > > > 5 4 |3|2| > 2 4> 1 1 > > 7 6 |3|5| > 3 | |2=4| > > 11 10 |7|6|8|2|> 5 |5=3=9=4|> > 2 11 > 1 1 > > 13 12 |2|11|6|7|> 6 | 4 | 10|> 3 | 3 = 9 |> > 4 |8|5| > 2 |12 | > 1 | 1 | > > 17 16 |6|11| 7|10|5|12|3|14|> 8 | 2 | 15 | 8 | 9 |> 4 | 4 | 13 |> 2 | 16 |> 1 | 1 |> > 19 18 |14|13|2|15|3|10|> 9 | 6=17=4=16=9= 5|> > 6 |12|8| > 3 |11=7| > > 2 |18| > 1 |1| > > 23 22 |5|21|19|7|20|14|11|17|10|15|> 11 |2= 4=16=3= 9=12= 6=13= 8=18|> > 2 |22| > 1 | 1| > > > 29 28 |2|27|10|19|11|18|> 14 | 4 | 13 | 5 |> 7 |16 = 24 =25 |> > 28 |3|26|8|21||14|15|> 14 | 9 | 6 | 22 |> 7 |23 = 7 = 20 |> > 4 |2|17| > 2 | 28 |> 1 | 1 | > > 31 30 |3|22|12|11|> 15 |9=19=20=28|> > 30 |21|24|13|17|> 15 |7=18=14= 10|> > 10 |23|29|27|15|> 5 |2= 4=16= 8|> > 6 |6|26| > 3 |5=25| > > 2 |30| > 1 | 1| > -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: equivalent expressions>I'm sure this is just a print error, but a Foundations book I have>asks the following:>[Q] Give an example of two equivalent [propositional] expressions that>do not have the same truth table.>Is there something deep I'm missing here, because what I see is a>contradiction. Perhaps my misunderstanding lies in what they mean bytruth table.> Seems like nonsense to me...> Hmm, unless they meant something silly like this:> P & ~P and Q & ~Q are equivalent, but they have> different truth tables because there's a P column> in one table and no P column in the other?Or maybe something silly like P -> P and (P V Q) -> (P V Q) are equivalent, but they have different truth tables because one has two rows and the other, four?-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: No Set Contains Every Computable Natural>> This TM will find a representation not on your tape:>> It is a three state machine and I can provide a state>> transition table if you like.> 1) Find a blank>> 2) Find a second blank>> 3) Backup and write a 1 on the previous blank>> repeat steps (1) through (3)> This TM will produce a tape that contains exactly one blank.>> The contiguous string of 1's preceding this blank will be>> a representation not on your tape.>It will produce no such tape. It will produce a tape consisting of>all 1's. This is obvious.>Suppose that it contains a blank. Then that blank must occur at some>square on the tape, say square n. It is trivial to see that, by the>nth iteration of your three steps, square n is no longer blank.>(After the first iteration, square 1 is not blank, square two was>never blank, square three is not blank after the second iteration and>hence is also not blank after the third, and so on.)This TM always checks to see if there is another blank on the tapebefore overwriting the previous blank. That is what step 2 does.It is easy to prove there is a blank on the output tape.> Wrong.> It is easy to prove that at each step n, there is another blank on the> tape.> Unfortunately, you want to talk about the output after an infinite> number of steps. You have proved that at each finite step n, there is> a blank on the tape. That is not sufficient to prove that after an> infinite number of steps, there is still a blank.Yes it is.> Moreover, there is a very simple proof that there is no blank. Look> up at the paragraph that begins, Suppose that it contains a blank.Look at the definition of my TM.Suppose that it contains a blank. Step (2) will then search foranother blank. The blank on the tape will not be overwrittenunless a second blank is found. The output tape will ALWAYScontain one blank. Even after an infinite number of steps.Russell- 2 many 2 count === Subject: Re: How many ways to put 5 balls into 500 ordered cups? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1JN2H932606;Yes, I took that into account in my step (2). Unfortunatley, my method is ineffiecient.I now remember deriving the efficient way to this problem a few years ago, but I had forgotten about that, and gave an answer in haste.Have a good day.>> SUMMARY:> (1) Find the ways to partition the number 5:> 5> 4 1> 3 2> 3 1 1> 2 2 1> 2 1 1 1> 1 1 1 1 1 <<>The gimmick is that the cups are ordered, so (2,3) is not the same as>(3,2) fo the two cup case and so forth upt to 500. === Subject: Re: Axioms defining a finite field by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1JN2Gl32579;You might get somewhere with(1+a)*(1+b) = 1*(1+b)+a*(1+b) = 1+b+a+ab(1+a)*(1+b) = (1+a)*1+(1+a)*b = 1+a+b+abbut I don't know if all the necessary cancellations can be done.Don Coppersmith...>You have any idea whether a + b = b + a follows from the conditions>above?> === Subject: When is a finite field a cyclic field? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1JN2Gk32583;I don't intend to introduce new notion,but for the moment let's call a finite field F(+,*) a cyclic field ifthe abelian group (F,+) is cyclic.(We know that (F0,*) is always cyclic when F is finite).When is (F,+) also cyclic?May I ask Derek Holt who gave such a nice and clear solutionto a recent problem for finite fields,to give answer to this?Thank you! === Subject: Re: infinitely many palindromic primes?> On page 228 of Albert Beiler's _Recreations in the Theory of Numbers_> (Dover, 1964) one finds the raw assertion that there are infinitely> many palindromic primes, primes whose base 10 representation reads the> same forwards and backwards.> The result is certainly believable, e.g., as a standard heuristic> predicts infinitely many primes consisting solely of 1's, but I was> surprised to see it claimed as a theorem. Is this really the case? An> old (1984) sci.math post from Bob Silverman suggests that it is and> that the proof is connected with results on automorphs of binary> quadratic forms. (The post in question discusses classes of> automorphic binary quadratic forms, but the references to Dickson> seem to be to theorems about automorphs of binary forms and their> connection with the Pell equation.) Can anyone elucidate for me the> connection of this theory to palindromic primes?I hesitate to contradict Bob Silverman, who knows vastly more about primes than I do, and I haven't had a look at Dickson, but to the best of my knowledge the existence of infinitely many palindromic primes is unproved.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Coprime Grid / Increasingly-Sized Steps> Here is a question which combines 2 earlier puzzles of mine into one.> As in:> http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off& threadm=b4be2fdf.0402151952.6a0480a%40posting.google.com&rnum= 12&prev=> we attempt to: > Start with an infinite (in every direction) rectangular grid/lattice.> Put 1 in a square of the grid (or at a vertex of the lattice).> Place 2, 3, 4, 5, ....infinity, by some algorithm, into the> grid/lattice so that each integer (1+m) is exactly m squares (in only> directions of either up,down,left,or right) from the square/vertex> with the integer m, for all m's.> And only one, no more/ no fewer, integer per vertex/square.> But...> as in:> http://groups.google.com/groups?dq=&hl=en&lr=&ie=UTF-8&threadm =266426e1.0401280041.5b6ea997%40posting.google.com&prev=> the integers are to be placed so that EVERY pair> of adjacent squares contains two integers which are COPRIME with each> other...> (Adjacent is as defined as immediately next to in the directions of> either up, down, left, or right.)> A few things:> I am wondering if it can be proved it is possible to eventually place> all the integers, using the restictions above, into the infinite> unbounded grid with exactly one integer per grid-square.> My instinct is that there are ways to fill the infinite grid> successfully.> And if so, what would the 'most compact' filling be?> By 'most compact', I mean:> which placement of integers would minimize:> limit{n -> oo}> (area of convex hull of first n integers' positions)/n> ? > (I could also ask this question about the less specific infinite> variations of the 2 puzzles from which this one was derived.)> Leroy QuetHere is why I sense a solutionis possible.Let m be odd.We can move, if all positions are previously available and are not aproblem as far as adjacent-uncoprimes,(Use fixed-width font to view.)1) (m+2)(m) ---------> (m+1) <-------------2)(m+4) ^ (m)----->(m+1) ! ! ! ! ! V(m+3)<----------(m+2) ((m+4) and m are at difference of 2 positions in both horizontal andvertical directions.)3) (m+4) ^(m+2)(m) ---------> (m+1) ! ! <------------- ! ! ! ! ! V (m+3)etc..And we can combine these.(If, say, we take (2), we can extend it as far as we want, so(m+4n) is at a distance of (2n) difference in both the vertical andhorizontal from m, simply by having n 'rectangles', if this is allowedby the other integers already on the grid.) So, perhaps, we can concentrate the primes near the 'center' of thegrid (center is nearer to 1), if possible, and arbitrarily choose ourmoves and their orientations so as to avoid undesirable already-placedintegers.Anyone have any luck with filling the grid to a significantly highinteger?Leroy Quet === Subject: differentiationCould anyone help me differentiate the following in order to gain avalue for for 'r double dot':r dot = (-0.5sin theta)*theta dotwhere: theta = pi/4 and theta dot = 0.6James Tunic === Subject: Re: Pascals Triangle> There was an interesting puzzle in the Sunday Times in the UK recentlythat> set me thinking about the issue that the puzzle raised.> The idea is that of a set of points arranged in the familiar Pascal's> triangle format (rows 1, 2, 3, ... etc. containing 1, 2, 3, ... points ina> triangular format) after which the issue is that of the total number of> equilateral triangles that can be found in this set of points.For N in 1..10, I get 0, 1, 5, 13, 27, 48, 78, 118, 170, 235, which can beexpressed asfloor((N - 1) (N + 1) (2 N - 1) / 8)and appears in Sloane's database:http://www.research.att.com/cgi-bin/access.cgi/as/ njas/sequences/eisA.cgi?Anum=A002717> This itself is not hard but the subsequent question seems more difficult -> for such an arrangement with N rows, what is the minimum number of points> that need to be blocked in order to reduce this total number ofequilateral> triangles to zero?> I wondered if anyone here might have come across this or have someinsights> about a general solution.> GladmanThe second problem can be formulated as a set covering problem and solvedusing integer programming:minimize sum [i in 1..N, j in 1..i] x[i,j]subject tox[i1,j1] + x[i2,j2] + x[i3,j3] >= 1 for each equilateral triangle ((i1,j1),(i2,j2), (i3,j3))x[i,j] in {0,1} for all (i,j)Solving this IP for N in 1..11 yields 0, 1, 2, 4, 7, 9, 14, 18, 23, 29, 36.This sequence does not appear in Sloane's database.Rob Pratt === Subject: Re: two other sides of triangle> How we can find two sides of triangle if the third side, the radius of> inscribed circle and the angle opposite to the third side is given?>Let c be the known side, C the corresponding angle and r the inradius.>Then you>have the equations (a, b being the unknown sides) :>ab sinC = r (a + b + c) --> a + b = ab/r sin C - c (1) (Formula for>inradius)>c^2 = (a + b)^2 - (2 + 2cosC) ab (2) (Cosine rule)>Substitute in (2) the formula for a+b given in (1) and you have a>(remarkably simple !) quadratic in ab. Solve, evaluate a+b using (1)>and then solve x^2 - (a+b)x + ab = 0 whose two roots will give you the>sides.Unfortunately, since the method above needs to solve 2 quadratics, weget four values, not two. Here is another method, similar in idea, butonly requiring the solution of one quadratic, giving the two sides thatwere requested.The area of a triangle is both ab/2 sin(C) and r(a+b+c)/2, therefore 2ab sin(C) = 2r(a+b+c) [1]The law of cosines says c^2 = a^2 + b^2 - 2ab cos(C), therefore 2ab cos(C) = a^2 + b^2 - c^2 [2]Thus, 2ab(1+cos(C)) = (a+b)^2 - c^2 = (a+b+c)(a+b-c) [3]Dividing [1] by [3], we get tan(C/2) = 2r/(a+b-c) [4]We can easily solve [4] to get a+b = c + 2r cot(C/2) [5]Dividing [3] by 2+2cos(C) yields ab = (a+b+c)(a+b-c)/(2+2cos(C)) [6]Plug a+b from [5] into [6], we get both a+b and ab. Now use thequadratic formula to solve x^2 - (a+b)x + ab = 0 to get the values ofa and b. === Subject: Re: . The hardest of all hard facts .> Hi Eleaticus, Re: How you think,> Michelson-Morley and Kennedy-Thorndike do indeed fit> Galilean ( c + v ) physics ,>> All throughout the annals of history ...> No premise has been better tested than this premise:> The speed of light is the same for all observers.>> That makes it: The hardest of all hard facts.No experiment ever showed that the speed of light isthe same for all observers. Indeed any determinationof the one way speed of light can be used to demonstratethe speed of light is NOT the same for all observers.... A light----> B <-you < ----------- L --------------> v m/sUse syncronised clocks at A amd B to time how long it takeslight to travel a distance of L meters across the laboratory..Speed of light relative to the laboratory = L/ (tB - tA) = c where 'tA' is the time at which the light left A and 'tB' is the at which the light arrived at BNow repeat the experiment while running towards B at v m/sNote that 'in your frame of reference' the point B is moving ,so that the light must travel an extra distance = v * (tB - tA)which is the distance B has moved as the light travels fromA to B.Therefore:Speed of light relative to you = (L+ v * (tB - tA)) / (tB - tA) = c + vkeith stein> You had measured the relative velocity between a light ray front and a> moving you with v velocity. Notice that the velocities of your two> moving entities are c and v, measured in the Laboratory inertial> frame. As c and v refer to the same inertial frame, you have the right> to use ordinary vector algebra obtaining c+v, because Relative> velocity can be >c and is NOT invariant (see my post with that> title). If you measured the relative velocity in the inertial frame> you is at rest you obtain c as the result,Perhaps you could show HOW you obtain c as the result, Mr. Hidalgo-Gato> compatible with the> second postulate of Special Relativity. ANY light continue having the> same vacuum speed c for ANY observer AT REST in ANY inertial frame,Note that the distance the light travelled relative in the inertial frame inwhich you is AT REST is L+ v * (tB - tA) as shown above, and yourhardest of all hard facts is in fact a totally unjustified assumption.keith stein> The hardest of all hard facts .> RVHG === Subject: Help with a proofHello all I need a little with this proof: Let G be a group and let g E G be an element of finite order m. Ifm=kn for some k,n E N, then prove that the order of g^n is k.Here is what I think is right:(g^n)^k=1I just can't get my mind to wrap around anything that will make anysense can someone please help me. === Subject: re:Axioms defining a finite fieldThe set of rules you have give a ring. In order to get a field, youneed a multiplicative inverse for all non-zero elements. For example,integers mod 4, 2 does not have a multiplicative inverse. SeeTable.2*0=02*1=22*2=02*3=2 Posted Via Usenet.com Premium Usenet Newsgroup Services------------------------------------------------------ ---- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **---------------------------------------------------------- http://www.usenet.com === Subject: Re: No Set Contains Every Computable Natural <8765e4q2fe.fsf@phiwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@phiwumbda.org> <87vfm4o85i.fsf@phiwumbda.org> <87n07fo9qr.fsf@phiwumbda.org> <87k72jnxtp.fsf@phiwumbda.org> <87ptcau3xj.fsf@phiwumbda.org> Moreover, there is a very simple proof that there is no blank. Lookup at the paragraph that begins, Suppose that it contains a blank.> Look at the definition of my TM.> Suppose that it contains a blank. Step (2) will then search for> another blank. The blank on the tape will not be overwritten> unless a second blank is found. The output tape will ALWAYS> contain one blank. Even after an infinite number of steps. A TM never takes an infinite number of steps. After any stepit takes, it was some finite time. And if we are to homour you and pretend that it does take aninfinite amount of steps, then there are no 0s. Since assume thereis some 0; there must have been a 0 after it (since the string of1s after it is always finitely long) therefore that 0 was turned toa 1, contradiction. Take an analogy to the natural numbers. Your removal of 0s is like this:If, for any odd number 2k+1, there is an odd number coming after it, thenremove 2k+1 from the set. Your argument is that the remaining set would still have an odd number,but it's not hard to see that the set {1,2,...,N} up to ANY N at all wouldnever have an odd number.J === Subject: Re: I'd like to join this group!girls = time * moneytime = moneygirls = money * money = money^2money = root(evil)girls = root(evil)^2girls = evil === Subject: Re: smallest eigenvalue of Laplacian> there is a theorem that says that an eigenfunction corresponding to> the smallest eigenvalue of the Laplacian on some bounded domain has no> zeros inside that domain.> Is the reverse implication also true? Meaning: Does every> eigenfunction without zeros inside the domain belong to the smallest> eigenvalue?> Thank you in advance for any helpful comments.> Yours sincerely,> Tobias N.8ahring> What a great question.Robert Israel's answer was much better than mine! === Subject: Can something be undeterminable?In his new book A new kind of science, Stephen Wolfram refers to a problemas potentially undeterminable, and which is linked to goedel incompletenesstheorem.But is it logically possible for something, a mathematical statement, to beabsolutely undeterminable? === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs> My initial guess is 1 - pi*sin(sqrt(6)) / sqrt(6). Based on a> heuristic argument.Did you REALLY mean:1 - sin(sqrt(6)) / sqrt(6),with no 'pi*'?I and Rob Johnson got the no-pi solution.(I am not saying you are wrong or dumb....I mis-type/mis-calculate too sometimes...)(EVEN I!...);)Leroy Quet === Subject: Re: One Of Them Sequence-Puzzles AgainOkay,solution below:> One more clue below original post.> I will post answer in a couple/few days.Here is the beginning of the sequence:1, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36,38, 40, 42, 44, 3, 52, 54, 56, 58, 60, 62, ...It, as a whole, has these 2 characteristics:1) It is a permutation of the positive integers.2) It relates somehow to a post I have made to sci.math/rec.puzzles inthe last few weeks.(And, yeah, I know, the sequence has an infinite number ofdefinitions.But this puzzle is for FUN. And what fun is it reminding us AGAIN ofthe fact that there is no specific answer?)Leroy Quet> As with the beginning of the sequence, the even integers are always> far more numerous than the odd integers among the first n terms, n >=> 3.> And the pattern of many strings of consecutive even integers continues> indefinitely as well.> Leroy QuetI just took my puzzle-solution from:http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&selm= b4be2fdf.0402021719.6fc8a5df%40posting.google.comAnd spiralled outward from the center (in a Ulam-spiral kind ofway...).I felt originally this was justified as being natural because thesolution itself forms a quasi-spiral, as far as the even integers areconcerned.Oh, well....Leroy Quet === Subject: Re: Pascals TriangleThis itself is not hard but the subsequent question seems moredifficult -for such an arrangement with N rows, what is the minimum number ofpointsthat need to be blocked in order to reduce this total number ofequilateral triangles to zero?I wondered if anyone here might have come across this or have someinsights about a general solution. Gladman> The second problem can be formulated as a set covering problem and solved> using integer programming:> minimize sum [i in 1..N, j in 1..i] x[i,j]> subject to> x[i1,j1] + x[i2,j2] + x[i3,j3] >= 1 for each equilateral triangle((i1,j1),> (i2,j2), (i3,j3))> x[i,j] in {0,1} for all (i,j)> Solving this IP for N in 1..11 yields 0, 1, 2, 4, 7, 9, 14, 18, 23, 29,36.> This sequence does not appear in Sloane's database.> Rob Prattprogram ran out of steam at about N = 15 and I hence wondered whether therewas a more efficient technique.Is there a standard reference describing this technique? Gladman === Subject: Re: Matrix math -- verify please?> explicitly stated, so I wanted to verify:> - The product of two rectangular matrices (e.g. A = 4x3 and B = 3x4) cannot> be invertible, since each matrix has at most rank 3.> - The product AB has at most rank 3. (But can be less) Yes to both of those.> - Even if A has rank 3 and B has rank 3, AB may still have rank < 3. When they're 4x3 and 3x4? How? Can you give an example? Think about the two successive linear mappings and the dimension of the image at each stage. Ken Pledger. === Subject: Re: Model theory puzzleFind a finite system of axioms (feel free to introduce operations,relations, constants) so that all of its finite models would have aprime number of elements, and for every prime p there's a model tothat system of size p.> Finite fields?No, because of the stipulation all of its finite models would have aprime number of elements. Prime fields would do the trick, exceptthat the original poster probably wanted *first-order* axioms. You canaccomplish that by adding a relation to the usual field set-up.E.g., structures (F,+,*,0,1,<) where (F,+,*,0,1) is a field, < is alinear ordering, and SPOILER. === Subject: Re: No Set Contains Every Computable Natural> Moreover, there is a very simple proof that there is no blank. Look> up at the paragraph that begins, Suppose that it contains a blank.Look at the definition of my TM.Suppose that it contains a blank. Step (2) will then search foranother blank. The blank on the tape will not be overwrittenunless a second blank is found. The output tape will ALWAYScontain one blank. Even after an infinite number of steps.> A TM never takes an infinite number of steps. After any step> it takes, it was some finite time.A number of people have stated that time is not an issuefor an idealized TM. I have shown this is equivalent toassuming that a TM can perform an infinite number ofoperations in a finite amount of time.There is no difference between assuming that a TMwill still be computing long after the stars have turnedto dust and assuming a TM can perform an infinite numberof operations.If we assume a TM requires a fixed, non-zero amount of timeto perform an operation, I can come up with a natural numberthat will takes billions of years for that TM to process.> And if we are to homour you and pretend that it does take an> infinite amount of steps,You are not humouring me. The definition of algorithmimplicitly assumes a TM can perform any number ofoperations in a finite amount of time.> then there are no 0s. Since assume there> is some 0; there must have been a 0 after it (since the string of> 1s after it is always finitely long) therefore that 0 was turned to> a 1, contradiction.Why is it a contradiction?You just said there a 0 (or a blank) following the 0that is overwritten. The output tape will always contain ablank after any number of operations.> Take an analogy to the natural numbers. Your removal of 0s is like this:> If, for any odd number 2k+1, there is an odd number coming after it, then> remove 2k+1 from the set.> Your argument is that the remaining set would still have an odd number,> but it's not hard to see that the set {1,2,...,N} up to ANY N at all would> never have an odd number.Huh?Consider a finite set (1,2,3).Using your algorithm I get (2,3).There is no odd number that comes after 3 in the set (1,2,3).Russell- 2 many 2 count === Subject: Minimally simple finite groups?Which of the finite simple groups are minimally simple, i.e.,have all of their proper subgroups solvable? Obviously the onlyalternating group that qualifies is A_5 =~ L(2,4) =~ L(2,5), andI know the list also includes L(2,7) =~ L(3,2), L(2,8), L(2,13),... On the other hand, I also know it *doesn't* include L(2,9) =~A_6 or L(2,11), both of which contain subgroups isomorphic toA_5.-- Jim Heckman === Subject: Re: No Set Contains Every Computable NaturalIt is obvious that your machine does not produce a (tape representinga) set of natural numbers even though at each finite step the tape themachine is working on represents a set of natural numbers.> My TM produces a tape with one representation of a natural number.> This representation is not on the initial input tape.What is the makeup of your initial tape? Is it finite or infinite? If finite, does it end with a 0 or a 1?Only in the case of an input tape starting with a zero and ending with a zero, does your TM work. And such never contains ALL naturals.> This TM produces a tape with a contiguous string of 1's> followed by a blank. The output tape contains the representation> of exactly one natural number.Only if your input tape ends with a finite string of 1's followed by a single 0. Which means that the input tape does not contain representations of all naturals.> Russell> - 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural>> This TM will find a representation not on your tape:>> It is a three state machine and I can provide a state>> transition table if you like.> 1) Find a blank>> 2) Find a second blank>> 3) Backup and write a 1 on the previous blank>> repeat steps (1) through (3)> This TM will produce a tape that contains exactly one blank.>> The contiguous string of 1's preceding this blank will be>> a representation not on your tape.> It will produce no such tape. It will produce a tape consisting of>> all 1's. This is obvious.> Suppose that it contains a blank. Then that blank must occur at some>> square on the tape, say square n. It is trivial to see that, by the>> nth iteration of your three steps, square n is no longer blank.>> (After the first iteration, square 1 is not blank, square two was>> never blank, square three is not blank after the second iteration and>> hence is also not blank after the third, and so on.)>> This TM always checks to see if there is another blank on the tape> before overwriting the previous blank. That is what step 2 does.> It is easy to prove there is a blank on the output tape.Wrong.It is easy to prove that at each step n, there is another blank on thetape.Unfortunately, you want to talk about the output after an infinitenumber of steps. You have proved that at each finite step n, there isa blank on the tape. That is not sufficient to prove that after aninfinite number of steps, there is still a blank.> Yes it is.Thus again proving that Russell does not understand what is going on here.Moreover, there is a very simple proof that there is no blank. Lookup at the paragraph that begins, Suppose that it contains a blank.> Look at the definition of my TM.> Suppose that it contains a blank. Step (2) will then search for> another blank. The blank on the tape will not be overwritten> unless a second blank is found. The output tape will ALWAYS> contain one blank. Even after an infinite number of steps.So you claim, but you also claim that the blank is in the last position, which makes the tape finite, as infinite tapes do not have lastpositions. That comes with the definition of infinite.> Russell> - 2 zany 2 count === Subject: Re: Sets That Resemble Derivatives Somewhat by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1K0Bfi06959;> 2) [(A-->B)(A'-->B')]' = AB' U A'B> 3) [(A'-->B)(A-->B')]' = A'B' U AB> >2) is just symmetric difference, assuming the existence of some universeThe symmetric difference is indeed AB' U A'B. You can also find itin Hewitt and Stromberg (1965) p. 4. It obeys the associative andcommutative laws and even a version of the distributive law underintersection (p. 6 ibid). However, Hewitt and Stromberg never gotas far as A-->B and so on since they weren't looking for it. Itdefinitely helps to not be looking for something else even whenwalking.See my replies to the other posts if I get that far today.Osher === Subject: Re: When is a finite field a cyclic field?Cron> I don't intend to introduce new notion,but for the moment> let's call a finite field F(+,*) a cyclic field if> the abelian group (F,+) is cyclic.(We know that (F0,*) is always cyclicwhen F is finite).> When is (F,+) also cyclic?There is a prime number p such that 1+1+...+1 (p summands) is zero, andtherefore x+x+...+x=0 by distributivity. Hence F cannot be cyclic unless ithas only p elements, in which case it is cyclic and any nonzero element is agenerator.LH === Subject: Re: If We Replaced Each Prime With -1...Let c(k) = the sum of the exponents in the prime factorization of k.So, if we exchanged each prime in the prime factorization of k with(-1),we would get (-1)^c(k).What I am wondering, however, iswhat is C(m) =sum{k=1 to m} (-1)^c(k)asymptotical towards?> The function (-1)^c(k) is usually denoted lambda(k) and is known as> Liouville's lambda function. Its partial sums (up to m) are o(m); this> is equivalent to the prime number theorem. The stronger statement that> its partial sums are O(m^{1/2+epsilon}) for each fixed epsilon > 0> (and m->oo) is equivalent to the Riemann hypothesis.> You may have seen the same results stated for the Mobius function. One> can relate the partial sums of lambda to those of mu and thereby go> back and forth between these results as follows. As you note,C(m) also equals:sum{k=1 to m} floor(sqrt(m/k)) *mu(k),> and this can be put in the alternate form> C(m) = sum_{k <= sqrt(m)} M(m/k^2),> where M(x):=sum(mu(n), n<=x); one can also verify the related identity> M(m) = sum_{k <= sqrt(m)} mu(k) C(m/k^2). > and similarly for the estimate mentioned in connection with the> Riemann hypothesis. On the other hand, neither C(m) = o(m^{1/2}) or> M(m) = o(m^{1/2}) can hold (as m->oo); this is because either would> imply zeta had no zeros on Re(s) = 1/2.> You may be interested in Chapter 4 of my online notes which presents> an elementary proof of the PNT (in the form M(x) = o(x) as x->oo) and> which discusses some of these points. See> http://www.princeton.edu/~ppollack/notes/> Hope this helps,> PaulAfter I posted, I entered the first few terms of the C-sequence into:http://www.research.att.com/~njas/sequences/index.html# Lwhich sent me to:http://mathworld.wolfram.com/LiouvilleFunction.htmlBut thanks for the reply!Leroy Quet === Subject: Re: Can something be undeterminable?> In his new book A new kind of science, Stephen Wolfram refers to a problem> as potentially undeterminable, and which is linked to goedel incompleteness> theorem.> But is it logically possible for something, a mathematical statement, to be> absolutely undeterminable?Given any mathematical system powerful enough to do ordinary arithmetic, it is possible to construct a well-formed mathematical question which that system is incapable of answering. You can then devise another, more powerful mathematical system, within which you can answer aforesaid question - but then there will be some other question which even the new system can't answer.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Genetics and Math-AbilityJust curious:What is the current scientific opinion regarding how math-ability isinherited?Is it, in general, a recessive trait or is it dominate?(I say recessive.)I am only half-serious, since math ability is most probably a resultof many traits and experiences and education.But I was wondering today because, even though there are some in myfamily who were math-teachers, almost everyone else had/has littleability with mathematics...even very little ability as relative tome...('relative to me'...snicker, snicker...)Leroy Quet === Subject: Re: Induced completeness to stronger metric spaces> Which Cauchy sequence in (X,d2) does not converge in (X,d2)?The sequence 1, 1/2, 1/3, ... === Subject: Re: Genetics and Math-Ability> Just curious:> What is the current scientific opinion regarding how math-ability is> inherited?> Is it, in general, a recessive trait or is it dominate?> (I say recessive.)> I am only half-serious, since math ability is most probably a result> of many traits and experiences and education.> But I was wondering today because, even though there are some in my> family who were math-teachers, almost everyone else had/has little> ability with mathematics...even very little ability as relative to> me...> ('relative to me'...snicker, snicker...)> Leroy QuetHere's my experience. My father was really good at math, but grew up in a time when trivial pursuits like math for math's sake weren't as important as learning a trade.My sister (the oldest in the family) had to go to summer school in high school because of math, twice. My oldest brother is a chemist of some sort and did well enough in math, but nothing spectacular. My older brother has a master degree in math and teaches high school math. And there is me, the youngest, who is getting a masters degree in math and planning on startting a Ph.D. program in August for math.So it seems the math ability in my family increased with each child. Maybe it is just a product of my environment, but we all went to the same grade school and high school and even had the same teachers.I know that is not conclusive, or probably doesn't even have a point either way, but that's just how it is in my family. - Tim-- Timothy M. BrauchGraduate StudentDepartment of MathematicsWake Forest Universityemail is:news (dot) post (at) tbrauch (dot) com === Subject: Re: Huge Perturbations>Given a matrix A + b B, where A and B are of the>same order, b >> 1, and B has a null space>dimensionality > 1,>I'm interested in both the eigenvalues of A + b B, and>roots of (A + b B) x = c (c is of the same order as A>and B) : the asymptotic case b -> inf and how it's>approached.Well, the thing to do would be to consider b^(-1) A as a perturbation ofB rather than b B as a perturbation of A.Of course the eigenvalues of A + b B are b times the eigenvalues of B + b^(-1) A, and (A + b B) x = c iff (B + b^(-1) A) (b x) = c. === Subject: Re: Model theory puzzle> > Find a finite system of axioms (feel free to introduce operations,> relations, constants) so that all of its finite models would have a> prime number of elements, and for every prime p there's a model to> that system of size p.>Hint: forget about model theory as such and ask yourself if you can>think of an algebraic structure that always has a prime number of>elements (or a particular kind of well-known algebraic structure...).That seems like the obvious thing to do. I can think of at least onesuch algebraic structure, group generated by any non-identityelement but I can't figure out a first-order way to _say_ that.And of course field almost works... probably there's somethingreally obvious.>Hthab************************ === Subject: Re: Help with a proofJordan> Hello all I need a little with this proof:> Let G be a group and let g E G be an element of finite order m. If> m=kn for some k,n E N, then prove that the order of g^n is k.> Here is what I think is right:> (g^n)^k=1True, and all that remains is to show that no smaller power of g^n is 1.There's a tip below.Argue by contradiction.LH === Subject: Re: Model theory puzzle> Find a finite system of axioms (feel free to introduce operations,> relations, constants) so that all of its finite models would have a> prime number of elements, and for every prime p there's a model to> that system of size p.> Finite fields?>No, because of the stipulation all of its finite models would have a>prime number of elements. Prime fields would do the trick, except>that the original poster probably wanted *first-order* axioms. You can>accomplish that by adding a relation to the usual field set-up.>E.g., structures (F,+,*,0,1,<) where (F,+,*,0,1) is a field, < is a>linear ordering, and SPOILER.and for every x, x + 1 > x unless x + 1 = 0.************************ === Subject: mathcad 11I have plugged the expression Sum(n=1->m)(10^n)/n, and asked for asymbolic solution. The answer that came back involved, among otherthings, the natural log of -9, a red m superimposed on aparenthesis, and something illegible (black)superimposed on a plussign. Did I do something wrong? I have to say that I am completelyat sea with computers. Any help will be appreciated, including theproper answer, if it exists, for my sum.C. F. Connie Eaton === Subject: Notation QuestionI'm wondering if there is a standard name (i.e. somethinglike the Kronecker Delta Function) for the following:(read d subscript i for any integer i)d_i = 1 ( i >= 0 ) = 0 ( i <= -1 )I realize that this is basically a discrete form ofthe Heaviside function, but somehow I doubt thatthis function goes by that name. Is there a standardname for this thing in the math/physics world?JB === Subject: Re: Church-Turing compared to Zuse-Fredkin thesis (two new papers)Distribution: inet <4030e88c.32841454@netnews.att.net> <40311933.36090852@netnews.att.net> <40311cb1.36985181@netnews.att.net> <40313eb6.37522811@netnews.att.net> <5jeYb.26213$Ov3.20124@newssvr25.news.prodigy.com> <%etYb.24944$2c3.7452@newssvr27.news.prodigy.com> [Followups set to sci.math, who may be able to explain what analgebraic number is better than I can.]The web page states, by implication, * All algebraic numbers are roots of Diophantine equations of degree 1 or 2. Cube root of 2 is an algebraic number which is not the root of anyDiophantine equation of degree 1 or 2. Hence, the web page's statementis disproved by this example. As I understand it, we may say that all algebraic numbers are rootsof Diophantine equations; but that's all.> I understand this differently: A Diophantine equation is an equation in> which only integer solutions are allowed.> So I don't see how bringing up Diophantine equations is relevant. Whoops! I misspoke: read polynomial equation with integercoefficients, not Diophantine equation. Sorry about that. Still,the web page is wrong.> That website focuses on numbers: rational, irrational and constructible.> ---------|----------------Real Numbers-----------|------------> Rational numbers Irrational numbers> are all algebraic are algebraic or> transcendental Correct, after re-formatting. :)All algebraic numbers (including algebraic irrationals)of the 1st or 2nd degree, or those having a power of two,e.g., x2, x4, x8, x16, x32, x64, x128, ..., etc., if given a linesegment of unit length, are numbers that can be constructed by theclassical Greek geometric method of straightedge and compass.> SH: The author is pointing out that some algebraic irrationals> are constructible. For instance the square root of 2. Right.> The cube root of 2 is not constructible and it well known that> in general cube roots (trisecting etc.) are not constructible. Right.> His explanation disallows including cube roots, which is> the type of counterexample you provided and thought injured> the correctness of his description. Right. His explanation disallows cube roots from being calledalgebraic; however, cube roots *are* called algebraic; thus,his explanation is flawed.> I don't know what that page means by algebraic equation.> You can find numerous definitions on Google. No, actually, I can't. If I had, you can be assured I wouldn'tbe posting otherwise! MathWorld doesn't know the term, and neitherdoes anyone on the first page of Google results. Post a URL tothe definition if you want to use it. However, I have made the assumption above that when you writealgebraic equation, you mean zero = some polynomial with integercoefficients. Is that right?-Arthur === Subject: Re: Nonlinear PDE HelpRobert Israel says...>We want to find f(x,y) subject to the following two conditions> 1. f(x,0) = g(x)> 2. (df/dx)^2 + (df/dy)^2 = 1>where g(x) is some known differentiable function of x>and where d/dx and d/dy are supposed to be partial derivatives.>Geometrically, your differential equation says that gradient(f) has length >1 everywhere. In principle, you should be able to get solutions by>starting with a more-or-less arbitrary curve C (either closed, or >going to infinity at both ends), on which you take f = c for some >constant c, and defining >f(x,y) = c + dist((x,y), C) on one side of the curve, > = c - dist((x,y), C) on the other side of the curvethe problem comes up in Special Relativity. If you have arigid ruler (or as rigid as possible in SR) and one endtravels in a straight-line path in the direction the ruleris oriented according to position at time t = g(t)then a point on the ruler that is initially a distance x fromthe end of the ruler will travel a path given by position at time t = f(x,t)where f is the solution to f(0,t) = g(t)and (df/dt)^2 + (df/dx)^2 = 1--Daryl McCulloughIthaca, NY === Subject: Re: No Set Contains Every Computable Natural> It is obvious that your machine does not produce a (tape representing> a) set of natural numbers even though at each finite step the tape the> machine is working on represents a set of natural numbers.My TM produces a tape with one representation of a natural number.This representation is not on the initial input tape.> What is the makeup of your initial tape? Is it finite or infinite?The initial input tape is 01011011101111...The assumption is that this tape has a representationfor every natural number. This would seem to indicatethat the input tape in infinite.> If finite, does it end with a 0 or a 1?How would it end if it were infinite?Does it end with an infinite string of 1's?It might end with 00.This would kind of like an end of file mark.I guess most people would say it never ends.> Only in the case of an input tape starting with a zero and ending with a> zero, does your TM work. And such never contains ALL naturals.No, there could be an infinite string of 1's at the end of the output tape.This would be the case if the input tape ended with an infinite string of1's.The output would be a finite string of 1's followed by a 0 (or blank)followed by an infinite string of 1's.This TM produces a tape with a contiguous string of 1'sfollowed by a blank. The output tape contains the representationof exactly one natural number.> Only if your input tape ends with a finite string of 1's followed by a> single 0. Which means that the input tape does not contain> representations of all naturals.Actually, I am trying to prove the input tape doesn't contain arepresentation of every natural. So, you are right, the inputtape does not contain a representation of every natural number.Russell- Zeno was right. Motion is impossible. === Subject: Re: I got low score on math test, please advise me and take a lookthe following argument did not work with the dean of student affairsthis school is really horrible. i can't believe the screwed up procedures at it.this is the arugment:Dear Dr. Johnson Jr., I am requesting that I be excused this one time to be withdrawn This was one day after the date of withdraw without a W. The class is Math 170, ticket # 3098, and meets on Monday and Wednesday from 6:00PM-8:00PM. Again, I did not have enough information to withdraw by February 17th since there was absolutely no graded work or assignments that were returned back by February 17th, one day before the exam results were made available. Please, I ask that I be excused this one time to be withdrawn without a W === Subject: Re: covering compact set w/squares>>Given a compact set K in the plane s.t. each pt x is the center of a square>>Q_x, prove that you can find a subsequence Q_x_i of squares s.t. K is>>covered by the Q_x_i and>>sum(over i) Char(Q_x_i) <= 4 Char (union(over i) Q_x_i),>>where Char(X) is the characteristic (indicator) function of X.>Are you assuming the squares are all oriented with sides parallel to the >axes? >I don't know whether _he's_ been assuming that, but I have been.Suppose for simplicity we want the point [0,0] to be in the interior of a region covered by more than 4 squares, such that noneof the centres of the squares is contained in a different square.Then at least two of the centres will be in one of the four quadrants;WLOG the first quadrant. And by translating slightly, we can assumethose centres are in the interior of the first quadrant, at[x_1,y_1] and [x_2,y_2]. If 2 r_1 and 2 r_2 are the sides of the two squares, then 0 < x_1, y_1 < r_1 0 < x_2, y_2 < r_2Since [x_1,y_1] is not in the square centred at [x_2,y_2], we needx_1 >= x_2 + r_2 or y_1 >= y_2 + r_2; let's say x_1 >= x_2 + r_2.Then r_2 <= x_1 - x_2 < x_1 < r_1.But the same argument, interchanging subscripts 1 and 2, shows r_1 < r_2, contradiction. === Subject: Re: For what purpose ?> The Michelson-Morley experiment was good enough> to suggest that the speed of light in a vacuum> is the same for all observers.Bull Mr. Relf.The MMX showed that the velocity of the light was relativeto the air it was passing through, as predicted by Maxwell eh!keith stein === Subject: Re: No Set Contains Every Computable Natural> Moreover, there is a very simple proof that there is no blank. Look> up at the paragraph that begins, Suppose that it contains a blank.>> Look at the definition of my TM.> Suppose that it contains a blank. Step (2) will then search for> another blank. The blank on the tape will not be overwritten> unless a second blank is found. The output tape will ALWAYS> contain one blank. Even after an infinite number of steps.> A TM never takes an infinite number of steps. After any step> it takes, it was some finite time.>A number of people have stated that time is not an issue>for an idealized TM. I have shown this is equivalent to>assuming that a TM can perform an infinite number of>operations in a finite amount of time.No. You have stated that these are equivalent. You have failedto show it.>There is no difference between assuming that a TM>will still be computing long after the stars have turned>to dust and assuming a TM can perform an infinite number>of operations.Yes there is. Just as there is a difference between an arbitrarilylarge finite number and infinity.Alan-- Defendit numerus === Subject: Re: combitorics charset=Windows-1252> Torrey loves Matt on some days, but Torrey hate Matt on other days.> Torrey also loves Paul on some days, but Torrey hates Paul on other> days.> Torrey also loves Sandy on some days, but she never hates Sandy.> Sandy hates Torrey on some days, but loves her on other days> Matt loves Sandy on some some days, but hates her on other other days> Paul Loves Matt on some days, but Paul hates other days.> paul hates Sandy on some days. > how can I figure out how many possible combinations there are that> desribes the love and hate relationship between Torrey, Matt, Paul,> and Sandy?> I have 32. but I'm not sure if i'm right. and i'm doing it brute force> way.A description of the love/hate relationships in the group would be given by a table as follows, with24 True/False values, but the given information isonly enough to fix the values shown: How A Feels About B Person A Person B Love Hate -------- -------- -------------------Torrey Matt T TTorrey Paul T T Torrey Sandy T FMatt Torrey Matt PaulMatt Sandy T TPaul TorreyPaul Matt T TPaul Sandy ? TSandy Torrey T TSandy MattSandy PaulIt's not quite clear how Paul feels about Sandy, but if the ? is known to be T, then that leaves 10 values undetermined; i.e, there are 2**10 = 1,024 differenttables consistent with the given information, and each one corresponds to a different set of love/hate relationships in the group. --r.e.s. === Subject: Re: Collatz Conjecture : Symmetry question.I have a program that can build each level from the previous one.> I also have written this program. I am using an arbitrary precision> lib to allow large numbers. I have two applications, one calculates> just end points, and the other traverses the tree. We should compare> notes Sure, I'll see if I can dig up my program. The final version used textfiles to hold the level data. I stopped at Level 84 because the textfile, at 3.3GB was getting too big to fit on a single CD when zipped.> and maybe work together and release a simple tool for others to> use. Just a thought.Simple, yes. Usefull? That remains to be seen.But I've got some other stuff that might be interesting. I finallysolved my Big Problem (multi-generation sequence vectors) and am working on a web page to document it. I can post some of it herealong with the programs if you're interested. === Subject: malta-new discoveryCan you believe that the oldest stone building in the world has amathematical perfection? Check:www.star-mysteries.com === Subject: malta-new discoveryCan you believe that the oldest stone building in the world has amathematical perfection? Check:www.star-mysteries.com === Subject: Re: Collatz Conjecture : Symmetry question.> http://arxiv.org/abs/math.NT/0312309> I think it's clear that it will never be proven, because there is just> not enough time to prove it.> CraigDo you mean proven true? It could be proven false with a single counterxample. === Subject: Re: To users of GMP>The FFT code in GMP currently contains a flaw that leads to>inaccurate results.Can you please give some order of magnitude for the number of bitsbelow which it can be safely assumed that the bug will not occur(maybe just because the FFT code is not even called).> The file gmp-mparam.h contains thresholds concerning which algorithms will> be used. Look for MUL_FFT_THRESHOLD, which will be in the thousands, and> multiply by the number of bits per limb, which will be 32 or 64 depending on> your machine word size. That will be the number of bits below which the FFT> code will not be called.>More information can be found at http://www.swox.com.Could not find more thanA bug in the FFT multiply code that can cause miscomputationhas been found. Until we can provide a fix, and until we haveperformed extensive further testing of the code, all users areurged to recompile GMP using the configure option --disable-fft.TIA,Francois Grieu> There is a new random number generator developed which can be found at> andomb.obj> which makes the span lengths proportional to the size of the output.> More information about the random code can be found at> .> The new random code triggered the bug in the FFT code almost immediately> after testing began.So what versions of GMP have this bug? I'm using the Windows gmpy module in Python which appears to be based on version 4.0. Does that have the bug? === Subject: Re: Nonlinear PDE Help>Here's a partial differential equation that I'm almost positive>has a unique solution, although I don't have any idea how to go>about finding it. Does anyone have suggestions?>We want to find f(x,y) subject to the following two conditions> 1. f(x,0) = g(x)> 2. (df/dx)^2 + (df/dy)^2 = 1>where g(x) is some known differentiable function of x>and where d/dx and d/dy are supposed to be partial derivatives.>Geometrically, your differential equation says that gradient(f) has length >1 everywhere. In principle, you should be able to get solutions by>starting with a more-or-less arbitrary curve C (either closed, or >going to infinity at both ends), on which you take f = c for some >constant c, and defining >f(x,y) = c + dist((x,y), C) on one side of the curve, > = c - dist((x,y), C) on the other side of the curve>where dist((x,y), C) is the minimum distance from (x,y) to C. Of course,>depending on the curve, this may be non-differentiable at some points>(which have more than one closest point on C). Two easy special cases>are where C is a circle (as in your example) or a straight line.>By the way, the latter example shows that in general f(x,0) = g(x) >does not uniquely determine the solution: consider f(x,y) = y and >f(x,y) = -y, both with g(x) = 0.>I don't know if there are other solutions (besides circles and >straight lines) with nice simple formulas: for most curves, >calculating the distance from a point to the curve is not very easy.On second thought, there's a geometric construction to get the curve Cgiven g. Of course we need to assume |g'(x)| <= 1 everywhere. Consider the family of circles centred at (x,0) with radius |g(x)|. For g(x) >= 0 take the envelope of these in the upper half plane, and for g(x) <= 0take the envelope in the lower half plane. Of course you could reflectit across the x axis and get another solution. Since the family of circles has implicit equation (x-s)^2 + y^2 = g(s)^2, an equation ofthe envelope is obtained (in principle) by eliminating s from the system (x-s)^2 + y^2 - g(s)^2 = 0 d/ds((x-s)^2 + y^2 - g(s)^2) = 2 (s-x) - 2 g'(s) g(s) = 0 === Subject: 1=ma+nbm,n belongs to Za and b is realtively prime.Is there any proof of this? or just an axiom?m,n belongs to Za and b is realtively prime.ma+nb=1 does not hold in {Z+}? If so prove it. === Subject: Re: Definition of Separable Space (basic topology question)Content-transferncoding: 8bitLet A be a subset of (X,T). Then A is dense in X iff for every nonmptyopen subset U of X, A / U != {}.I have seen two definitions of 'separable topological space':a) (X,T) is separable if there exists A X, where A is countable anddense in Xb) (X,T) is separable if there exists A X, where A iscountable and dense in XGiven def (a), its simple to prove that all countable spaces X areautomatically separable since the subset X is countable and non-triviallyintersects every nonmpty open subset. However, this won't work given def(b). I first became suspicious when I saw a proof that all countable spaceswere separable that seemed ridiculously complex in comparison to theseemingly obvious 1 step proof above. I later found definitions ofseparable like def (b).l8r, Mike N. Christoff> I have never seen definition (b). I have seen the symbol $subset$> used to mean subset, not proper subset, so maybe that is what you saw.> Certainly a one-point space with the discrete topology is to be> considered separable, even though it has no dense proper subsets.I've read the other follow-ups, and I think you've hit the nail on thehead: the OP is confused about the meaning of subset, and is takingA subset B to mean A is a proper subset of B.I have never seen definition (b) either. It would be very weird forfinite sets NOT to be separable. (And it would make the useful fact,A subspace of a separable metric space is also separable false.--Ron Bruck === Subject: Re: Can something be undeterminable? charset=Windows-1252In his new book A new kind of science, Stephen Wolfram refers to aproblemas potentially undeterminable, and which is linked to goedelincompletenesstheorem.But is it logically possible for something, a mathematical statement, tobeabsolutely undeterminable?> Given any mathematical system powerful enough to do ordinary arithmetic,> it is possible to construct a well-formed mathematical question which> that system is incapable of answering. You can then devise another,> more powerful mathematical system, within which you can answer aforesaid> question - but then there will be some other question which even the new> system can't answer.Has it been (or can it be) proved whether there exist questionsthat are *absolutely* unanswerable in the sense of there beingno sufficiently powerful mathematical system to answer them?--r.e.s. === Subject: Re: Can something be undeterminable?> In his new book A new kind of science, Stephen Wolfram refers to a problem> as potentially undeterminable, and which is linked to goedel incompleteness> theorem.> But is it logically possible for something, a mathematical statement, to be> absolutely undeterminable?What about.All true statments can be proven to be true.Can you prove that statement.look at the sitehttp://www.mtnmath.com/whatrh/node37.html === Subject: Re: 1=ma+nb>m,n belongs to Z>a and b is realtively prime.>Is there any proof of this? or just an axiom?>m,n belongs to Z>a and b is realtively prime.>ma+nb=1 does not hold in {Z+}? If so prove it.This is called Bezout's Identity. I have a page about it at === Subject: Re: No Set Contains Every Computable Natural>There is no difference between assuming that a TM>will still be computing long after the stars have turned>to dust and assuming a TM can perform an infinite number>of operations.> Yes there is. Just as there is a difference between an arbitrarily> large finite number and infinity.What is the difference between an arbitrarily large finite number andinfinity?Can you define an algorithm that will determine if a string of 1'srepresents a natural number or is infinitely long?Russell- Solution to halting problem: Ctrl Alt Del === Subject: Re: Sets That Resemble Derivatives Somewhat by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1K2cFc18878;>[Snip: in Doctorowese A' is the completement of A, A+B is the union>of A and B and AB is the intersection of A and B]>> >> 2) (A')' = A>> >> This doesn't correspond to differentiation,>Yet another thing that doesn't correspond to differentiation>is that >AB' + A'B =/= (AB)'.>Indeed I am at a loss reading these posts to find any analogy between>complementation and differentiation (save that Doctor Osherow uses>the same notation for both).>Evidently you've forgotten that derivatives satisfy> (f + g)' = (f')(g').Contrary to appearance, David C. Ullrich actually added the lasttwo lines above as a joke expanding on Robin Chapman's claim that I used A + B to be the union of A and B, or else Ullrich himself wovethis claim into Robin's post (it's hard to figure it out where thisquote came from, so I'm inclined to believe that Ullrich inventedthis claim based on the nature of Ingenious Imitation as it pervadesworld colleges at the faculty level). He then asserts that (f + g)'= (f')(g'). This also is his invention, the purpose of which beyondemotionality escapes me. I think that he is asserting a contradictionand projecting it upon the external world in the manner of some with whom we are very familiar from recent experiences. Osher Doctorow === Subject: Re: Sets That Resemble Derivatives Somewhat by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1K2cFf18885;>there's a general theme that some people have explored that says that>the fact that certain sorts of differential operators obey>leibniz-like identities is conceptually subordinate to the fact that>certain sorts of boundary operators in geometry and topology obey>leibniz-like identities, roughly (or sometimes precisely, depending on>the particular concept of boundary being used) boundary(x X y) =>(boundary(x) X y) + (x X boundary(y)).This looks interesting, although I think that themes of conceptualsubordination may be a bit premature here, but I am very encouragedby the lack of hostility in your message and by your ability todetect relationships. How did a nice guy like you get in the samethread as Robin and Ullrich?Osher Doctorow === Subject: Re: Sets That Resemble Derivatives Somewhat by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1K2cFZ18867;>in what?>No. The right side looks similar, 'cos of your denoting complement>by the same notation as derivative, but the right side is completely>different!>One idea? Not the idea of it being a limit of difference>quotients methinks.> Derivations are a topic in the literature on Lie Algebras and other> structures, but (2) and (3) have a rather different flavor.>I.e., by not resembling derivation at all?>applause!>Like er yeah! I mean yeah if x + h is different to x then er yeah>then x + h is outside x. Wow, groovy!Aha! The universe itslef. Next you'll be giving us the answer>to life the universe and everything.>Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html>Francis Wheen, _How Mumbo-Jumbo Conquered the WorldI omit Robin Chapman's quotation of Lacan on penises, and I omit mycomments to which these are replies (except that Lacan was Robin'sown quotation) for brevity.We don't have to really worry about compliments in what (Did Iwrite compliment? Perhaps a Freudian slip). Robin is in chargeof that.The next comment concerns the right side looking similar becauseof my denoting complement by the same notation as derivative....Let's stop there for a moment. So far, nothing's wrong beyondbeyond being novel in use of symbols. I mean, if a student ofyours claimed A is similar to B because the symbols used to expressA and B, perhaps he hasn't given the answer to a pile of homework,but it might even be deeper! Here I use the ordinary notation forcomplements of sets, prime ('). I point out that something = AB' + A'B in equation (2), and I leave it for readers to noticethat this is a similar form to the right side of (fg)' = fg' + f'g.(No, I'm not defining f and g here - take a guess!) Now, Matt Grime who I doubt wants to get involved in this argumentdid point out that AB' + A'B is the symmetric difference, whichRobin didn't, so Matt wasn't as upset as Robin at that point. If astudent in one's Math class said: Look at (fg)' = fg' + f'g. Isn'tthe right hand side similar in symbols to (AB)' = AB' + A'B, onemight well think of giving that student an A for noticing unusualrelationships, though admittedly nothing yet has been done to apply it to one's pet projects. However, Robin apparently took meto be a college teacher (right on!), and I can understand his disappointment that I only seemed to be at the level of an A student up to that point. Or perhaps Robin grades on the basis ofhomework exclusively, in which case slightly different problemsarise so to speak. Next, Robin used the word 'cos to mean because. Robin, cos withan argument is an abbreviation for cosine, and we discouragestudents for using it without an argument unless explicitly stated,but it's O.K. this time. Perhaps you mean some relationship between'cos and something else which escapes me, and if so, do explain.(Is that a current keyword, 'cos'?) But Robin continues: 'cosof your denoting complement by the same notation as derivative, butthe right side is completely different! So Robin doesn't detectthe relationship between AB' + A'B and fg' + f'g, so we know thatthis isn't an I.Q. test. I have also been chastised for using thesame symbol for sets like A, B and functions like f, g, in orderto point out a similarity in pattern - not bad for an I.Q. test, butthis isn't an I.Q. test (well, not by definition, anyway). The next comment of Robin is: One idea? Not the idea of it beinga limit of difference quotients methinks. Here Robin's Spirit ofScience-Math as opposed to Letter of Science-Math involves theidea that the limit is found in derivatives but not in set comple-ments. If this were stated as a simple fact, it would be interest-ing but still as a non-similarity one suspects that life wouldbe much more Creative looking for Similarities, and despite thewarning Robin has not yet eliminated Similarities. I won't quibbleabout Letters of Science-Math such as for example methinks,which are presumably not intended to conjure up confusion betweendrama and Mathematics, unless of course it is in the same codeas 'cos, one an expanding code and the other a contracting code (thismight in fact be the relationship that we're looking for!). I could go on and on like this, but I think that most readers get theidea. Osher Doctorow === Subject: Fresnel integrals1) int[-inf to inf] cos x^2 dx = sqrt(pi/2)2) int[-inf to inf] exp(-x^2) dx = sqrt(pi)Eqn (2) is easy to show by looking at the square of the left side(a double integral) and switching to polar coordinates. I tried thesame approach with (1). (There are other ways to get (1), I know.)The double integrand will be(cos x^2)(cos y^2) = (cos(x^2 - y^2) + cos(x^2 + y^2)) /2The first term on the right integrates to zero okay, but the secondterm, in polar coordinates, runs into a convergence problem.Moreover the conversion to polars requires some justification,because the Fresnel integrals are not absolutely convergent.Anybody know any quick fix for this one? I get the feeling I'mmissing something obvious...TIA,Larry === Subject: Re: Can something be undeterminable?> Has it been (or can it be) proved whether there exist questions> that are *absolutely* unanswerable in the sense of there being> no sufficiently powerful mathematical system to answer them?For that, you would have to rule out taking the question as an axiom.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: JSH: Non-uniqueness of factorizationDavid C. Ullrich> J. Edgar Harris>so you can factor *every* integer as 2j(j+1) or j(2j+1) with j an>integer in the ring of integers.> Huh????Bizarre, isn't it?>The guy is SCUM. And he's fooled you repeatedly, and that's why IAgain Harris is talking about himself. I wonder if he can keep it up muchlonger, before something really breaks.> Hint: When you call a person SCUM for making mathematical> comments you sound like a ing moron.He sounds like one for the same reason that ducks sound like ducks.packet driver for Windows, with anti-crank capabilities. Oughta be worth afortune.LH === Subject: Re: Definition of Separable Space (basic topology question)>Let A be a subset of (X,T). Then A is dense in X iff for everynonmpty>open subset U of X, A / U != {}.>I have seen two definitions of 'separable topological space':>a) (X,T) is separable if there exists A X, where A iscountable> and>dense in X>b) (X,T) is separable if there exists A X, where A is>countable and dense in X>Given def (a), its simple to prove that all countable spaces X are>automatically separable since the subset X is countable andnon-trivially>intersects every nonmpty open subset. However, this won't work given> def>(b). I first became suspicious when I saw a proof that all countable> spaces>were separable that seemed ridiculously complex in comparison to the>seemingly obvious 1 step proof above. I later found definitions of>separable like def (b).The integers with the usual topology is separable.> Right. This would follow from the fact that any topology on a countable> space is separable.Oops! I get you. ie: no proper subset's closure equals the integers....duhDefinition (a) is the correct one. BTW, finitespaces are separable.I was thinking you may have meant a similar thing about finite sets (ie: nodense proper subset), but:(X,T):X = {1,2,3,4}T = { X,{},{1,2},{3,4} }A = {1,3}l8r, Mike N. Christoff === Subject: Re: How many long primes are there?: Artin's conjecture states that for any b (other than -1 and perfect squares)there are infinitely many : primes N such that b is a primitive root (mod N); that is, b is : not a k-th power residue for any k > 1 dividing N - 1; that is, N : is what you are calling a long prime to base b. Ted === Subject: Re: Definition of Separable Space (basic topology question)> Let A be a subset of (X,T). Then A is dense in X iff for everynonmpty> open subset U of X, A / U != {}.>> I have seen two definitions of 'separable topological space':>> a) (X,T) is separable if there exists A X, where A iscountable and> dense in X> b) (X,T) is separable if there exists A X, where A is> countable and dense in X>> Given def (a), its simple to prove that all countable spaces X are> automatically separable since the subset X is countable andnon-trivially> intersects every nonmpty open subset. However, this won't workgiven def> (b). I first became suspicious when I saw a proof that all countablespaces> were separable that seemed ridiculously complex in comparison to the> seemingly obvious 1 step proof above. I later found definitions of> separable like def (b).>> l8r, Mike N. ChristoffI have never seen definition (b). I have seen the symbol $subset$used to mean subset, not proper subset, so maybe that is what you saw.Certainly a one-point space with the discrete topology is to beconsidered separable, even though it has no dense proper subsets.> I've read the other follow-ups, and I think you've hit the nail on the> head: the OP is confused about the meaning of subset, and is taking> A subset B to mean A is a proper subset of B.> I have never seen definition (b) either. It would be very weird for> finite sets NOT to be separable. (And it would make the useful fact,> A subspace of a separable metric space is also separable false.Ok. I'm tired and don't immediately see anything wrong with this:(X,T):X = {1,2,3,4}T = { X,{},{1,2},{3,4} }A = {1,3}Finite space with dense proper subset...l8r, Mike N. Christoff === Subject: Re: . The hardest of all hard facts .Dear Peter:>They found none. Miller's results could not be duplicated by anyone but>Miller with his apparatus.> The following paper describes a number of experiments that seem to> have detected motion relative to space or (something in space).> http://arxiv.org/pdf/physics/0312082> The following abstract is from:> The motion of the Solar System and the Michelson-Morley experiment> M. Consoli and E. Costanzo> Istituto Nazionale di Fisica Nucleare, Sezione di Catania> http://arxiv.org/pdf/astro-ph/0311576> ---QUOTE---> Historically, the Michelson-Morley experiment has played a crucial> role for abandoning the idea of a preferred reference frame, the> ether, and for replacing Lorentzian Relativity with Einsteins Special> Relativity. However, our re-analysis of the Michelson-Morley original> data, consistently with the point of view already expressed by other> authors, shows that the experimental observations have been> misinterpreted. Namely, the fringe shifts point to a non-zero> observable Earths velocity Vobs = 8 4 10 5 km s. Assuming the> existence of a preferred reference frame, and using Lorentz> transformations to extract the kinematical Earths velocity> that corresponds to this Vobs , we obtain a real velocity, in the> plane of the interferometer,v earth = 201 112 km s. This value is in> excellent agreement with Millers calculated value Vearth = 203 18> km/s and suggests that the magnitude of the fringe shifts is> determined by the typical velocity of the Solar System within our> galaxy. This conclusion, which is also consistent with the results of> all other classical experiments, leads to an alternative interpre-> tation of the Michelson-Morley type of experiments. Contrary to the> generally accepted ideas of last century, they provide experimental> evidence for the existence of a preferred reference frame. This point> of view is also consistent with the most recent data for the> anisotropy of the two-way speed of light in the vacuum.> --ND QUOTE---> The following paper proposes that motion relative to space (or> something in space) should be detectable, because space is full of> Vacuum condensates and ether-drift experiments> M. Consoli, A. Pagano and L. Pappalardo> Istituto Nazionale di Fisica Nucleare, Sezione di Catania> http://arxiv.org/pdf/physics/0306094> The following is also interesting.> Modern Michelson-Morley experiments and gravitationally-induced> anisotropy of c> M. Consoli> Istituto Nazionale di Fisica Nucleare, Sezione di Catania> http://arxiv.org/pdf/gr-qc/0306105> The abstract states:> The recent, precise Michelson-Morley experiment performed by Muller et> al. suggests a tiny anisotropy of the speed of light. I propose a> quantitative explanation of the observed effect based on the> interpretation of gravity as a density fluctuation of the Higgs> condensate.So your analysis of the literature has found that the purported anisotropyis always less-than-orqual-to the smallest resolution of the recordeddata, or that the anisotropy is a function of time.Surely the anisotropy decreased over the 100 years from your first citationto your last. Obviously you have discovered a new thorn for theaetherists. Good work! A secular change in the anisotropy of the aether!Amazing. How will they explain it. Surely they will not claimexperimental error, since they bitch whenever the establishment doesso.Us SRians are proud of your efforts.David A. Smith === Subject: Re: malta-new discoveryX-No-Archive: yesWhile still snuggled in a 'spider hole', ivanmohoric@volja.net (temacnik)scribbled: >Can you believe that the oldest stone building in the world has a>mathematical perfection? I believe someone could PRETEND that it does.To reply by email, remove the XYZ.Lumber Cartel (tinlc) #2063. Spam this account at your own risk.This sig censored by the Office of Home and Land Insecurity.... === Subject: Re: x^2 + y^4 = z^4>Would you explain why?A squarefree positive integer n is a congruent number if and only if n isthe area of a right triangle with rational sides. As examples, 1, 2, 3 are notcongruent numbers while 5, 6, 7 are congruent numbers.n - sides of rational Pythagorean triangle5 - (3/2, 20/3, 41/6)6 - (3, 4, 5)7 - (35/12, 24/5, 337/60)A number n is a congruent number if and only if n(m^2) is the area of arational right triangle. In particular, that 1 is not a congruent number isequivalent to there being no (rational) right triangle whose area is the squareof an integer.See Koblitz's book Introduction to Elliptic Curves and Modular Forms for moreinfo.John Robertson === Subject: Re: Can something be undeterminable? > Has it been (or can it be) proved whether there exist questions> that are *absolutely* unanswerable in the sense of there being> no sufficiently powerful mathematical system to answer them? Any statement only really has meaning within a system, does itnot? A system has to be fixed first, then the statement is madewithin that system. The property mentioned is that any (sufficiently strong) system will have unprovable statements.J === Subject: Re: Can something be undeterminable?> For that, you would have to rule out taking the question as an axiom.I kind of like the idea of a question being an axiom. === Subject: Re: . The hardest of all hard facts .as in mister Peter's quote,there just haven't *been* any duplications of M-M-M, becauseof the massive say-so about their results;there are certainly anomalies in both results.for this. D.C.Miller's refinement highlighted the anomoliesthat MM had found> They found none. Miller's results could not be duplicated by anyone but> Miller with his apparatus.> It was dismissed because it could not be repeated. It was dismissed> because it was not verfiable.of course, a lot of this is moot, if we allow Cheenyto grind us into Tony's McCrusade (Usama's MacJihad). see my sig.--Give the World a Trickier Dick Cheeny -- out of office after GIGA years.http://www.benfranklinbooks.com/http://www.rand.org/ publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanac === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> <8765e4q2fe.fsf@phiwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@phiwumbda.org> <87vfm4o85i.fsf@phiwumbda.org> <87n07fo9qr.fsf@phiwumbda.org> <87k72jnxtp.fsf@phiwumbda.org> <87ptcau3xj.fsf@phiwumbda.org> *sigh*... I don't know why I bother with this discussion...Again, I won't contribute after this.> There is no difference between assuming that a TM> will still be computing long after the stars have turned> to dust and assuming a TM can perform an infinite number> of operations. It's a huge difference. One is saying that after a huge amountof time T, it is computing. This is normal and fine, since T ishuge but finite. But saying that a turing machine performs an infinite numbersof steps is nonsense and goes against the standard definitions.Create your own difinitions is you want that property. In another post, you asked what the difference was in doingsomething for an arbitrarily large amount of time and doing somethinginfinitely long. This is the source of all your problems. Once you understand thatthere is a difference between these two, most of your problems willclear up. The first one is a finite computation (but for any largeamount of steps) and the second one is simply just not a computationby the definition of turing machine computations.> If we assume a TM requires a fixed, non-zero amount of time> to perform an operation, I can come up with a natural number> that will takes billions of years for that TM to process. That's totally fine. I give you some time T and you can createa number that takes longer than T steps to compute (or output orread or anything.) Our point is that for any computation you dotaking longer than this T-steps, its computation time will (and must)be bounded by some other number B.> You are not humouring me. The definition of algorithm> implicitly assumes a TM can perform any number of> operations in a finite amount of time. As long as any number is any finite number. Arbitrarily large,yes, but still finite.then there are no 0s. Since assume thereis some 0; there must have been a 0 after it (since the string of1s after it is always finitely long) therefore that 0 was turned toa 1, contradiction.> Why is it a contradiction?> You just said there a 0 (or a blank) following the 0> that is overwritten. The output tape will always contain a> blank after any number of operations. You need to learn what a proof by contradiction is. Assumption: There exists a 0 at the end of computation. We use this statement and show that it contradicts itself. Say there exists a '0' at the end of computation. If it existson the tape, it MUST exist at some finitely-indexed tape cell. But then if the tape originally had all the natural numbers,then this 0 must have been followed by a string of 1s, eventuallyanother 0, and another string of 1s. Since there is this other 0,the first 0 (that we assumed exists) must have been changed to a '1,'contradicting the fact that the 0 existed. Take an analogy to the natural numbers. Your removal of 0s is like this:If, for any odd number 2k+1, there is an odd number coming after it, thenremove 2k+1 from the set. Your argument is that the remaining set would still have an odd number,but it's not hard to see that the set {1,2,...,N} up to ANY N at all wouldnever have an odd number.> Huh?> Consider a finite set (1,2,3).> Using your algorithm I get (2,3).> There is no odd number that comes after 3 in the set (1,2,3). You missed the fact in elementary school that if n is some oddnumber, then so is n+2. I'm not talking about a finite set, I'm talking about takingthe natural numbers and applying that process to it. Your hypotheticaltape supposedly had all the natural numbers on it. Once the processis complete, then any set you look at {1..N} will have all its oddnumbers removed. Again, make sure you learn about the difference between 'artibrarilylarge' and 'infinite.' The set {1, 11, 111, 1111, 11111, ... } will contain strings that arearbitrarily long but will not contain the infinite string 1111....J === Subject: Re: Can something be undeterminable?>Has it been (or can it be) proved whether there exist questions>that are *absolutely* unanswerable in the sense of there being>no sufficiently powerful mathematical system to answer them?Suppose S is your current system, and Q is a question that itcan't answer (i.e. there is no proof in S that the answer is yes,and there is no proof in S that the answer is no). Then you canproduce two more powerful systems: S1, by adding to S the axiom the answer to Q is yes;S2, by adding to S the axiom the answer to Q is no.Both of these systems can answer the question. And, if it'sa meaningful yes-or-no question, one of these answers will be correct.So not only is there a system that answers the question, there's a system that answers the question correctly. The trouble is, wedon't happen to know which one! === Subject: Re: infinitely many palindromic primes?there are quite a few in base-1;00, for instance (that's two in decimal .-)The result is certainly believable, e.g., as a standard heuristicpredicts infinitely many primes consisting solely of 1's, but I was> but to the best of my knowledge the existence of infinitely > many palindromic primes is unproved.of course, a lot of this is moot, if we allow Cheenyto grind us into Tony's McCrusade (Usama's MacJihad). see my sig.--Give the World a Trickier Dick Cheeny -- out of office after GIGA years.http://www.benfranklinbooks.com/http://www.rand.org/ publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanac === Subject: Re: Can something be undeterminable?,Has it been (or can it be) proved whether there exist questionsthat are *absolutely* unanswerable in the sense of there beingno sufficiently powerful mathematical system to answer them?> Any statement only really has meaning within a system, does it> not? A system has to be fixed first, then the statement is made> within that system. The property mentioned is that any > (sufficiently strong) system will have unprovable statements.Spot on, but perhaps the questioner will be satisfied with The Halting Problem as absolutely unanswerable. No computer can be programmed to tell, correctly, whether an arbitrary computer will halt when presented with an arbitrary program.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Letter to Prof Ullrich and othersI am just curious: why are people like youinterested in even acknowleding James Harris and his ilk ?This guy has been oopsing on sci.math for ages (at least severalyears) I see. Is it not worthwhile for everyone to just ignore him at this point ? There is real danger of course that someone would take him seriously,but I feel thats remote.It seems that he has taken to calling universities to complain. Lets not giving him any more attention. Without attention, he willshrivel up and disappear.Is there a history or an obvious reason that I am missing ? === Subject: Re: How many long primes are there?> : Artin's conjecture states that for any b > (other than -1 and perfect squares)> there are infinitely many > : primes N such that b is a primitive root (mod N); that is, b is > : not a k-th power residue for any k > 1 dividing N - 1; that is, N > : is what you are calling a long prime to base b. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Can something be undeterminable? charset=Windows-1252Has it been (or can it be) proved whether there exist questionsthat are *absolutely* unanswerable in the sense of there beingno sufficiently powerful mathematical system to answer them?> Any statement only really has meaning within a system, does it> not? That seems reasonable.> A system has to be fixed first, then the statement is made> within that system. ^^^^OK; but, is it possible to prove the (non/)existence of statements having a certain property, without explicitly making them? IOW, show that a certain kind of statementexists, without exhibiting one of them? (Sorry if it's an ignorant question -- I don't know.)> The property mentioned is that any > (sufficiently strong) system will have unprovable statements.OK. But as I read Gerry's reply, the unprovability is only relative to the mathematical system in which it is made.--r.e.s. === Subject: functional analysis--separable, I found the following proposition in a book without a proof. Ihave difficulty in proving it. Can anyone help? Suppose B is a separable Banach space. Let {x_n} be a countabledense subset of the unit circle C={x in B : ||x||=1}. How can I provethat: every element x in B can be written as:x= summation[from 1 to infinity] {a_n x_n} where a_n is a absolutesummable sequence. That's:summation[from 1 to infinity] {|a_n|} < infinity. === Subject: Re: Silly question for someone with a big calculator....>4/3 = 1.333333...>24/17 = 1.411764...>816/577 = 1.414211...>941664/665857 = 1.414213...>If we let k represent that ratio for one of the exponents, the ratio for>the next in this list is (4k)/(k^2+2); it can be shown that the ratios>will converge to sqrt(2)....> I thought your following line was neat:>1 = 3^2-2^2-2^2 = 17^2-12^2-12^2 = 577^2-408^2-408^2 = ...> [and] 1 = 665857^2 - 470832^2 - 470832^2...I missed the beginning of this thread so don't know what the question might be, but for approximation of sqrt(2), a termof a Taylor series will reduce the error quadratically. If a^2-2b^2=x and f(x)=(2b^2+x)^(1/2), f'(x)=1/(2(2b^2+x)^(1/2)),so a ~ s*(b+x/(4b)) [where s=sqrt(2)]. a b s-a/b s-a/(b+x/(4b)) 3 2 -.086 .0024 17 12 -.0025 .0000021577 408 -2E-6 1.6E-12665857 470832 1.6E-11 9E-25-jiw === Subject: Re: Can something be undeterminable? charset=Windows-1252> ,> Has it been (or can it be) proved whether there exist questions> that are *absolutely* unanswerable in the sense of there being> no sufficiently powerful mathematical system to answer them? Any statement only really has meaning within a system, does itnot? A system has to be fixed first, then the statement is madewithin that system. The property mentioned is that any (sufficiently strong) system will have unprovable statements.> Spot on, but perhaps the questioner will be satisfied with > The Halting Problem as absolutely unanswerable. No computer > can be programmed to tell, correctly, whether an arbitrary computer > will halt when presented with an arbitrary program.That's actually the first example that occurred to me, but perhaps I was then misled by a previous poster's link to a page describing a hierarchy of unsolvable problems, claiming that Every Halting Problem can be solved at some level in this hierarchy...--r.e.s. === Subject: Re: Definition of Separable Space (basic topology question)Content-transferncoding: 8bit>> Let A be a subset of (X,T). Then A is dense in X iff for every> nonmpty> open subset U of X, A / U != {}.>> I have seen two definitions of 'separable topological space':>> a) (X,T) is separable if there exists A X, where A is> countable and> dense in X> b) (X,T) is separable if there exists A X, where A is> countable and dense in X>> Given def (a), its simple to prove that all countable spaces X are> automatically separable since the subset X is countable and> non-trivially> intersects every nonmpty open subset. However, this won't work> given def> (b). I first became suspicious when I saw a proof that all countable> spaces> were separable that seemed ridiculously complex in comparison to the> seemingly obvious 1 step proof above. I later found definitions of> separable like def (b).>> l8r, Mike N. Christoff>> I have never seen definition (b). I have seen the symbol $subset$> used to mean subset, not proper subset, so maybe that is what you saw.>> Certainly a one-point space with the discrete topology is to be> considered separable, even though it has no dense proper subsets.I've read the other follow-ups, and I think you've hit the nail on thehead: the OP is confused about the meaning of subset, and is takingA subset B to mean A is a proper subset of B.I have never seen definition (b) either. It would be very weird forfinite sets NOT to be separable. (And it would make the useful fact,A subspace of a separable metric space is also separable false.> Ok. I'm tired and don't immediately see anything wrong with this:> (X,T):> X = {1,2,3,4}> T = { X,{},{1,2},{3,4} }> A = {1,3}> Finite space with dense proper subset...Presumably re: my remark It would be very weird for finite sets NOT tobe separable. I didn't mean that ALL finite sets wouldn't beseparable. (Although, if the topology is Hausdorff, it means exactlythat.) It suffices for a SINGLE finite set not to be separable to setoff my weirdness alarm.But I've lost the thread of what I was thinking when I said it wouldfalsify subspace of separable metric space is separable. I'm sure Ihad something in mind...--Ron Bruck === Subject: Re: No Set Contains Every Computable Natural> Moreover, there is a very simple proof that there is no blank. Look> up at the paragraph that begins, Suppose that it contains a blank.>> Look at the definition of my TM.> Suppose that it contains a blank. Step (2) will then search for> another blank. The blank on the tape will not be overwritten> unless a second blank is found. The output tape will ALWAYS> contain one blank. Even after an infinite number of steps. A TM never takes an infinite number of steps. After any stepit takes, it was some finite time.> A number of people have stated that time is not an issue> for an idealized TM. I have shown this is equivalent to> assuming that a TM can perform an infinite number of> operations in a finite amount of time.You may have argued it, but that does not necessarily constitute a proof.> There is no difference between assuming that a TM> will still be computing long after the stars have turned> to dust and assuming a TM can perform an infinite number> of operations.In the first case, the TM may still only be able to perform a finite, though large, number of steps. Their may be no practical difference, in that no one will be around then, but there is a theoretical difference which is critically important.> If we assume a TM requires a fixed, non-zero amount of time> to perform an operation, I can come up with a natural number> that will takes billions of years for that TM to process.So? And if we are to homour you and pretend that it does take aninfinite amount of steps,> You are not humouring me. The definition of algorithm> implicitly assumes a TM can perform any number of> operations in a finite amount of time.What definition is that? The definitions of algorithms that I am aware of all require that the algorithm be completed in a finite number of steps. Perhaps you can give a reference for someone who agrees with your definition?then there are no 0s. Since assume thereis some 0; there must have been a 0 after it (since the string of1s after it is always finitely long) therefore that 0 was turned toa 1, contradiction.> Why is it a contradiction?> You just said there a 0 (or a blank) following the 0> that is overwritten. The output tape will always contain a> blank after any number of operations.If it starts with infinitely many zeros, as it must to represent ALL natural numbers, then there will always be infinitely many zeros left. Take an analogy to the natural numbers. Your removal of 0s is like this:If, for any odd number 2k+1, there is an odd number coming after it, thenremove 2k+1 from the set. Your argument is that the remaining set would still have an odd number,but it's not hard to see that the set {1,2,...,N} up to ANY N at all wouldnever have an odd number.> Huh?> Consider a finite set (1,2,3).> Using your algorithm I get (2,3).> There is no odd number that comes after 3 in the set (1,2,3).That is the point. In finite sets, there is a point that nothing comes after, but in an infinite set that is not the case.> Russell> - 2 zany 2 count === Subject: Re: Goldberg dual> A Goldberg polyhedron has all hexagonal faces except for 12 pentagons> or 6 fourgons or 4 trigons, and are multi-symmetrical. A soccor ball> is a Goldberg polyhedron. Goldberg-like is the same but asymmetrical.> Is there a name for an asymmetric Goldberg-like> polyhedron? What is the name of the dual of the Goldberg-like> polyhedron?> My guess is this. For 3(N-2) edge elements whose lengths are variable,> and where N is is an integer greater that 2, ther is a maximum> diameter polyhedron that can be contructed and a minimum diameter> polyhedron that can be constructed. The maximim diameter polyhedron> that can be built with 3(N-2) edges is a Goldberg Polyhedron like> structure. The minimum diameter sphere will be an omnitriangulated> icosahedral arrangement of the edge elements.> Additionally, even smaller spheres can be built by doubling up edges,> tripling up edges, etc. === Subject: LogicI'm stuck on these 2 problems.Represent the following compound propositions, each involving one or more ofthese three propositons, as symbolic expressions in the variables d,e,f.1)e does not f unless d2)d and e does not imply fOf coure I am told what these propositions are.I'm sorry but I can't even show you what I've done so far--I'm completelylost.Steven === Subject: Re: Goldberg dualin other words, the fullerenes. your guess is ill-posed, since the fullerenes are allof 3-way vertices, their duals will be trigonated --but the lengths for duals is not the same, althoughit's the same *number* of edges.it's interesting that the fullerenes can thus be modelledwith trigona, and that Bucky never noticed that!> A Goldberg polyhedron has all hexagonal faces except for 12 pentagons> or 6 fourgons or 4 trigons, and are multi-symmetrical. A soccor ball> is a Goldberg polyhedron. Goldberg-like is the same but asymmetrical.> My guess is this. For 3(N-2) edge elements whose lengths are variable,> and where N is is an integer greater that 2, ther is a maximum> diameter polyhedron that can be contructed and a minimum diameter> polyhedron that can be constructed. The maximim diameter polyhedron> that can be built with 3(N-2) edges is a Goldberg Polyhedron like> structure. The minimum diameter sphere will be an omnitriangulated> icosahedral arrangement of the edge elements.of course, a lot of this is moot, if we allow Cheenyto grind us into Tony's McCrusade (Usama's MacJihad). see my sig.--Give the World a Trickier Dick Cheeny -- out of office after GIGA years.http://www.benfranklinbooks.com/http://www.rand.org/ publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanac === Subject: Re: No Set Contains Every Computable Natural>> It is obvious that your machine does not produce a (tape representing> a) set of natural numbers even though at each finite step the tape the> machine is working on represents a set of natural numbers.>> My TM produces a tape with one representation of a natural number.> This representation is not on the initial input tape.What is the makeup of your initial tape? Is it finite or infinite?> The initial input tape is 01011011101111...> The assumption is that this tape has a representation> for every natural number. This would seem to indicate> that the input tape in infinite.If finite, does it end with a 0 or a 1?> How would it end if it were infinite?Are you too illiterate to understand IF?> Does it end with an infinite string of 1's?> It might end with 00.> This would kind of like an end of file mark.> I guess most people would say it never ends.Only in the case of an input tape starting with a zero and ending with azero, does your TM work. And such never contains ALL naturals.> No, there could be an infinite string of 1's at the end of the output tape.> This would be the case if the input tape ended with an infinite string of> 1's.> The output would be a finite string of 1's followed by a 0 (or blank)> followed by an infinite string of 1's.The output could not exist, as the TM cold never finish.>> This TM produces a tape with a contiguous string of 1's> followed by a blank. The output tape contains the representation> of exactly one natural number.Only if your input tape ends with a finite string of 1's followed by asingle 0. Which means that the input tape does not containrepresentations of all naturals.> Actually, I am trying to prove the input tape doesn't contain a> representation of every natural. So, you are right, the input> tape does not contain a representation of every natural number.Not if it is finite. But if it is infinite it can have all naturals represented, but the TM never finishes. === Subject: Suggestions requested for notation of functions such as sincI mentioned the sine cardinal function (1 if x = 0 sinc(x) = ( ( sin(x)/x otherwisein a thread here not long ago. Since then it has appeared in other threads.Example 1. Rob Johnson, in the thread puzzle: GCDs of Infinite Set ofInteger Pairs, showed a certain probability to be 1 - sinc(sqrt(6)).Example 2. In the Mathematica newsgroup thread how to explain this weirdeffect? Integrate, I noted that the integral of sin(m x) sin(n x) from 0to 2 pi can be given nicely as pi(sinc(2 pi (m - n)) - sinc(2 pi (m + n))),which is valid for all m and n. OTOH, the expression given byMathematica for the integral is not, literally, valid when m = n or m = -n.Besides sin(x)/x, there are other commonly occurring functions of the formf(x)/x having removable singularities at x = 0. I think it would be nice tohave a consistent notation for the functions obtained once thesingularities have been removed. Of course, some people will say that nospecial notation is needed since we can always just say something likeSingularities are supposed to be understood as having been removedwhenever possible. But that doesn't work very well with my computeralgebra systems, for example. Another possibility which avoids giving a newname to such functions is to use limit notation. For example, g(x) = lim_{t -> x} sin(t)/tis the same as sinc(x). But I'm not fond of that alternative since it's notin closed form.The only special notation I've seen so far for such functions is based onanalogy with sinc. The notation sinhc(x) is already in use, at least tosome extent, for the function which is 1 if x = 0, sinh(x)/x otherwise. Andat MathWorld, Eric Weisstein has introduced the notation tanc(x) for thefunction which is 1 if x = 0, tan(x)/x otherwise. But in those twoinstances, I feel that there is no real need for new notation since each ofthose functions can be expressed nicely in terms of sinc(x). Namely,instead of sinhc(x), we could always write sinc(i x), and instead oftanc(x), we could always write sinc(x)/cos(x). In other cases, however,when functions cannot be expressed handily in terms of sinc, I think that aspecial notation would be nice. Here's an example. For a spheroid havingpolar radius a and equatorial radius b, what's the surface area? Perhapsyou'd say it's A = 2 pi b ( a Asin(d)/d + b )where d = Sqrt(1 - (b/a)^2) and Asin denotes the principal-valued inversesine function.But for the sphere itself (which I certainly consider to be a spheroid),that formula does not work, if taken literally, since 0/0 is encountered.However, if we have the function f(x) = 1 if x = 0, Asin(x)/x otherwise,then we may say, for any spheroid (prolate, sphere, or oblate), that thearea is given by A = 2 pi b ( a f(d) + b ).What's a nice name for function f? Should we, by anaolgy with sinc, call itAsinc? (But then might not some misguided fellow misconstrue it as beingthe inverse of sinc instead? And BTW, how should we denote that inverse?)Am I correct in thinking that the name sine cardinal function and theabbreviation sinc are due to E. T. Whittaker? Why did he choose that name?Would names such as Asinc be based on false analogies with sinc?I would appreciate any thoughtful comments, suggestions for notation, etc.David Cantrell === Subject: Re: No Set Contains Every Computable Natural>There is no difference between assuming that a TM>will still be computing long after the stars have turned>to dust and assuming a TM can perform an infinite number>of operations.Yes there is. Just as there is a difference between an arbitrarilylarge finite number and infinity.> What is the difference between an arbitrarily large finite number and> infinity?The difference is, itself, infinite.> Can you define an algorithm that will determine if a string of 1's> represents a natural number or is infinitely long?No. Has anyone claimed to be able to? === Subject: Re: . The hardest of all hard facts .Dear Quincy Hutchings:> as in mister Peter's quote,> there just haven't *been* any duplications of M-M-M, because> of the massive say-so about their results;> there are certainly anomalies in both results.There are *dozens* of documented MMX experiments with different apparati.Nothing *repeatable* has been found within their ability to resolve.Except for Peter's secular variation in the aether anisotropy, there hadbeen nothing extraordinary.David A. Smith === Subject: Re: 1=ma+nb> m,n belongs to Z> a and b is realtively prime.> Is there any proof of this? or just an axiom?> m,n belongs to Z> a and b is realtively prime.> ma+nb=1 does not hold in {Z+}? If so prove it.If a and b are members of Z and are relatively prime (have no non-unit common factors, then there are m and n in Z such that a M + b N = 1.If a and b are of the same sign, then m and n must be of opposite signs.If a and b are of opposite signs then m and n must be of the same sign.One method of finding suitable m and n for given a and b is through the extended Euclidean algorithm.See, for example: http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm === Subject: Drawing subgroup lattices for research papersI'm working on a thesis (using MikTex and Winedt) that requires me toput in a lot of subgroup lattices. This results in several problems. For an early version of the paper, I made .bmps and cut and paste themin the appropriate spots, but now I want to put them in the actualcode of the paper itself. My method was to import into Mathematicaand export as .eps. First of all, the .eps files are huge. I triedto import a 3k grayscale gif and got a 400k eps file. Then, usinggraphicx I was able to get the picture to show up after Latexing. However, the graphic looks terrible. It is clearly something that ishappening when I Latex the paper because I checked the eps file inGhostview and though it was degraded some from the original bmp thesubgroups were still recognizable.I've also tried downloading xypic and followed the instructions, butcan't figure out how to get MikTex or WinEdt to recognize the newfiles.Does anyone either have a suggestion on getting what I have to work ora more tex-friendly way to draw lattices? === Subject: Re: Logic> I'm stuck on these 2 problems.> Represent the following compound propositions, each involving one or moreof> these three propositons, as symbolic expressions in the variables d,e,f.> 1)e does not f unless dDoes 1 this mean If d and e the f????> 2)d and e does not imply fDoes 2 mean not d or not e does imply f???> Of coure I am told what these propositions are.> I'm sorry but I can't even show you what I've done so far--I'm completely> lost.> Steven === Subject: Re: LogicI'm stuck on these 2 problems.Represent the following compound propositions, each involving one ormore> ofthese three propositons, as symbolic expressions in the variables d,e,f.1)e does not f unless d> Does 1 this mean If d and e the f????Typo: Does 1 mean If d and e the f????2)d and e does not imply f> Does 2 mean not d or not e does imply f???Of coure I am told what these propositions are.I'm sorry but I can't even show you what I've done so far--I'mcompletelylost.Steven === Subject: Re: Genetics and Math-AbilityLeroy Quet> Just curious:> What is the current scientific opinion regarding how math-ability is> inherited?> Is it, in general, a recessive trait or is it dominate?> (I say recessive.)I don't have any formal stats about it, but it seems not to run in familiesvery much. Among the famous names, there were several Bernoulli's and acouple of Jacobi's. Today there are the Chudnovsky's and the Voevodsky's,but that's all I can think of, not that I know much about it.But I'll bet a lot of mathematicians have near relatives in other sciences.I'd be interested in any stats about birth order, because temperament seemsto matter in this game. Consider two brothers, one a doctor and the other amath professer. I'll bet the doctor is the older of the two.LH === Subject: Re: No Set Contains Every Computable Natural>There is no difference between assuming that a TM>will still be computing long after the stars have turned>to dust and assuming a TM can perform an infinite number>of operations.> Yes there is. Just as there is a difference between an arbitrarily> large finite number and infinity.>What is the difference between an arbitrarily large finite number and>infinity?One is finite and the other isn't.>Can you define an algorithm that will determine if a string of 1's>represents a natural number or is infinitely long?I think a sufficiently rigorous definition of algorithm requires thatthe input (or problem) be finite in size, so I suspect that what you areasking for is not within the scope of an algorithm. Alan-- Defendit numerus === Subject: Re: Can something be undeterminable? A system has to be fixed first, then the statement is madewithin that system. ^^^^> OK; but, is it possible to prove the (non/)existence of > statements having a certain property, without explicitly > making them? IOW, show that a certain kind of statement> exists, without exhibiting one of them? (Sorry if it's an > ignorant question -- I don't know.) The best way to show that there exist unprovable statementsis it construct it, much like how Goedel's proof works.> OK. But as I read Gerry's reply, the unprovability is only > relative to the mathematical system in which it is made. Actually, Robert Israel made a good follow-up that you couldextend any system to have that constructed unprovable theoremtrue or false (whichever you like.) A common example of thatis the axiom of choice; you can choose to accept it or not, butthere is a host of problems whose only known proofs depend on theaxiom of choice, and usually a proof that doesn't use the extrapower of the axiom of choice is preferred.J === Subject: Re: Suggestions requested for notation of functions such as sinc> I mentioned the sine cardinal function> (1 if x = 0> sinc(x) = (> ( sin(x)/x otherwise> in a thread here not long ago. Since then it has appeared in otherthreads.> Example 1. Rob Johnson, in the thread puzzle: GCDs of Infinite Set of> Integer Pairs, showed a certain probability to be 1 - sinc(sqrt(6)).> Example 2. In the Mathematica newsgroup thread how to explain this weird> effect? Integrate, I noted that the integral of sin(m x) sin(n x) from 0> to 2 pi can be given nicely as pi(sinc(2 pi (m - n)) - sinc(2 pi (m +n))),> which is valid for all m and n. OTOH, the expression given by> Mathematica for the integral is not, literally, valid when m = n or m= -n.> Besides sin(x)/x, there are other commonly occurring functions of the form> f(x)/x having removable singularities at x = 0. I think it would be nicetoDavid, here's my take on this.Although a computer programming function to evaluate sin(x)/x, for example,would use the the definition you provided above since it would otherwisecalculate 0/0, actually there is no singularity in sin(x)/x at x=0 becaused[sin(x)]dx / (dx/dx) = cos(x)/1 which evaluated at x=0 is 1.0/1.0 = 1.0 andthis is by definition how one resolves 0/0.KeithK> have a consistent notation for the functions obtained once the> singularities have been removed. Of course, some people will say that no> special notation is needed since we can always just say something like> Singularities are supposed to be understood as having been removed> whenever possible. But that doesn't work very well with my computer> algebra systems, for example. Another possibility which avoids giving anew> name to such functions is to use limit notation. For example,> g(x) = lim_{t -> x} sin(t)/t> is the same as sinc(x). But I'm not fond of that alternative since it'snot> in closed form.> The only special notation I've seen so far for such functions is based on> analogy with sinc. The notation sinhc(x) is already in use, at least to> some extent, for the function which is 1 if x = 0, sinh(x)/x otherwise.And> at MathWorld, Eric Weisstein has introduced the notation tanc(x) for the> function which is 1 if x = 0, tan(x)/x otherwise. But in those two> instances, I feel that there is no real need for new notation since eachof> those functions can be expressed nicely in terms of sinc(x). Namely,> instead of sinhc(x), we could always write sinc(i x), and instead of> tanc(x), we could always write sinc(x)/cos(x). In other cases, however,> when functions cannot be expressed handily in terms of sinc, I think thata> special notation would be nice. Here's an example. For a spheroid having> polar radius a and equatorial radius b, what's the surface area? Perhaps> you'd say it's> A = 2 pi b ( a Asin(d)/d + b )> where d = Sqrt(1 - (b/a)^2) and Asin denotes the principal-valued inverse> sine function.> But for the sphere itself (which I certainly consider to be a spheroid),> that formula does not work, if taken literally, since 0/0 is encountered.> However, if we have the function f(x) = 1 if x = 0, Asin(x)/x otherwise,> then we may say, for any spheroid (prolate, sphere, or oblate), that the> area is given by> A = 2 pi b ( a f(d) + b ).> What's a nice name for function f? Should we, by anaolgy with sinc, callit> Asinc? (But then might not some misguided fellow misconstrue it as being> the inverse of sinc instead? And BTW, how should we denote that inverse?)> Am I correct in thinking that the name sine cardinal function and the> abbreviation sinc are due to E. T. Whittaker? Why did he choose that name?> Would names such as Asinc be based on false analogies with sinc?> I would appreciate any thoughtful comments, suggestions for notation, etc.> David Cantrell === Subject: help is needed!!!i want play whit MAPLE but i can not find sourse for it if any body know some sourses for it or any one can help and supportme please say more thank you pad === Subject: Re: Genetics and Math-Ability > I don't have any formal stats about it, but it seems not to run in> families very much. Among the famous names, there were several> Bernoulli's and a couple of Jacobi's. Today there are the Chudnovsky's> and the Voevodsky's, but that's all I can think of, not that I know much> about it. And apparently Euler had something like 11 or 12 children, but we didn't even see a 2 or 3 generation reign of Eulers like the Bernoulli family had. (At least not to my knowledge.)J === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> <8765e4q2fe.fsf@phiwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@phiwumbda.org> <87vfm4o85i.fsf@phiwumbda.org> <87n07fo9qr.fsf@phiwumbda.org> <87k72jnxtp.fsf@phiwumbda.org> <87ptcau3xj.fsf@phiwumbda.org> Discussion, linux)> A TM never takes an infinite number of steps. After any step> it takes, it was some finite time.There is no reason why we can't allow a definition that a TM convergesafter an infinite number of steps if each square on the TM's taperemains unchanged after a certain step. That is, if for all squares x, there exists a step n in thecomputation such that for all steps m > n, the value written in x at mis the same as the value written in x at n. -- Jesse F. HughesWhat I represent is the unknowable future--the power of change. Inthat sense I'm a force of Nature, a force of the Universe, a livingemodiment of change itself. --James Harris and his sense of humility === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> <8765e4q2fe.fsf@phiwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@phiwumbda.org> <87vfm4o85i.fsf@phiwumbda.org> <87n07fo9qr.fsf@phiwumbda.org> <87k72jnxtp.fsf@phiwumbda.org> <87ptcau3xj.fsf@phiwumbda.org> Discussion, linux)>> This TM will find a representation not on your tape:>> It is a three state machine and I can provide a state>> transition table if you like.> 1) Find a blank>> 2) Find a second blank>> 3) Backup and write a 1 on the previous blank>> repeat steps (1) through (3)> This TM will produce a tape that contains exactly one blank.>> The contiguous string of 1's preceding this blank will be>> a representation not on your tape.> It will produce no such tape. It will produce a tape consisting of>> all 1's. This is obvious.> Suppose that it contains a blank. Then that blank must occur at some>> square on the tape, say square n. It is trivial to see that, by the>> nth iteration of your three steps, square n is no longer blank.>> (After the first iteration, square 1 is not blank, square two was>> never blank, square three is not blank after the second iteration and>> hence is also not blank after the third, and so on.)>> This TM always checks to see if there is another blank on the tape> before overwriting the previous blank. That is what step 2 does.> It is easy to prove there is a blank on the output tape.> Wrong.> It is easy to prove that at each step n, there is another blank on the> tape.> Unfortunately, you want to talk about the output after an infinite> number of steps. You have proved that at each finite step n, there is> a blank on the tape. That is not sufficient to prove that after an> infinite number of steps, there is still a blank.> Yes it is.A withering retort! I am stunned by the force of logic.And yet, you're simply wrong.> Moreover, there is a very simple proof that there is no blank. Look> up at the paragraph that begins, Suppose that it contains a blank.> Look at the definition of my TM.> Suppose that it contains a blank. Step (2) will then search for> another blank. The blank on the tape will not be overwritten> unless a second blank is found. The output tape will ALWAYS> contain one blank. Even after an infinite number of steps.You have proven by induction that for all n, at step n in thecomputation, there is a blank space. This does not prove that afteromega steps, there is a blank space. Your proof by bald assertiondoesn't cut it.Your proof allows for this silly argument. I can prove that, for alln, there exists a number m such that n < m. *Therefore*, there mustbe a number m bigger than every number n. (I suppose that thisreductio won't be persuasive to you, since you have put forwardsomething like this argument a million times so far.)Now, if you'll read my argument again, you will see that, on thecontrary, each blank space is overwritten within a finite number ofsteps of the computation and therefore, after omega steps, there areno blank spaces on the tape at all. For, if there was a blank space,that space must occur at some finite position, say n. However, weknow that by the n+1st iteration of your algorithm, the nth space is*not* blank. Therefore, there is no blank space on your tape after infinitely manysteps of computation. (Here is the point where you wittily rejoinder,yes, there is, so that I may admit defeat in the face of suchunassailable arguments.)-- If you see math knowledge as a tool--as a hammer--with whichyou can attack other people then ... you defeat rational discourse.I get to call my proof the Hammer. It's more powerful than *any*physical object. It is overwhelming force. -- Two JSH quotes === Subject: Re: the anticlassicalist }{ vi: into the quantum> I want to address ontology.then get real. ;-)> There _is_ a problem there. Should it be Mr Ontology, Mrs> Ontology...? I was about to suggest Comrade Ontology but... ah, the> English speaker's lot is not a happy one! If only we spoke Japanese:> Ontorogii-san. Done! (Why Ontorogii and not Ontorojii? Dunno. Sounds> better).Riddle of the day: are ontologists for real? === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> <8765e4q2fe.fsf@phiwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@phiwumbda.org> <87vfm4o85i.fsf@phiwumbda.org> <87n07fo9qr.fsf@phiwumbda.org> <87k72jnxtp.fsf@phiwumbda.org> <87ptcau3xj.fsf@phiwumbda.org> Discussion, linux)> A number of people have stated that time is not an issue> for an idealized TM. I have shown this is equivalent to> assuming that a TM can perform an infinite number of> operations in a finite amount of time.This is wrong. I don't have anything against extending the commondefinition of convergence to allow for convergence after omega steps(for *some*, but not *all* TMs and inputs), but it *is* an extension.You have not shown equivalence. I don't recall your argument, but Iwill show my psychic powers: Your argument involves swapping the orderof quantification at an essential step.Well? Was I right? Huh? Was I?-- Even if [...] a communistic regime should come [to China], the oldtradition [...] will break Communism and change it beyond recognition,rather than Communism [...] break the old tradition. It must be so. -- Lin Yutang on Socialism with Chinese characteristics in 1935 === Subject: Re: No Set Contains Every Computable Natural Discussion, linux)> I think a sufficiently rigorous definition of algorithm requires> that the input (or problem) be finite in size, so I suspect that> what you are asking for is not within the scope of an algorithm.That depends on context. In algorithmic information theory as developed by Chaitin, forinstance, the input tape is filled with 0's and 1's from the get-go,with no restriction on the number of 1's. (On the other hand, I don't want to press the claim that Chaitin isthe model of rigor in mathematics.)-- I've been thinking about my problems with getting any kind ofadmission that my math arguments showing the core error in mathematicsare correct, so I've gone to marketing books. -- James S. Harris, on when mathematics isn't enough === Subject: Re: Collatz Conjecture : Symmetry question. === >Subject: Re: Collatz Conjecture : Symmetry question.>Message-id: > I have a program that can build each level from the previous one.> > I also have written this program. I am using an arbitrary precision> lib to allow large numbers. I have two applications, one calculates> just end points, and the other traverses the tree. We should compare> notes >Sure, I'll see if I can dig up my program. The final version used text>files to hold the level data. I stopped at Level 84 because the text>file, at 3.3GB was getting too big to fit on a single CD when zipped.> and maybe work together and release a simple tool for others to> use. Just a thought.>Simple, yes. Usefull? That remains to be seen.>But I've got some other stuff that might be interesting. I finally>solved my Big Problem (multi-generation sequence vectors) and am >working on a web page to document it. I can post some of it here>along with the programs if you're interested.I originally did this using a list in memory, but I ran out of memory.Here's the text file based version written in Python (which also does Big Arithmetic). The program reads the previous level out of a file,doubles every number and if a number is both == 1 (mod 3) AND== 0 (mod 2), it spawns a new branch by using the inverse 3x+1rule.# usage: python lread.py n# where n is level to process# reads file named Ln.txt# and outputs next level#import syslevel = sys.argv[1]filein = 'L' + level + '.txt'f = open(filein,'r')s = 'begin'while s != '': if s != 'begin': n = long(s) print n*2 p3 = divmod(n,3) if (p3[1]==1): p2 = divmod(n,2) if (p2[1]==0): print (n-1)/3 s = f.readline()f.closeStart by creating a text file named L5.txt that contains just16Running the program using L5.txtpython lread.py 5produces the following output:325To build up successive levels, redirect the output to a filepython lread.py 5 > L6.txtOf course, you can batch file the low levels since they go quick.It gets slow by the time you get to Level 70. And it starts eating up disk space: 65 File(s) 448,531,354 bytesThe one good thing is that you only need to keep the last levelon hand. All the previous ones can be archived. I originally wanted to go all the way to Level 100, but since Level 84took 3.3 GB, I lost interest at that point. === Subject: Re: Can something be undeterminable? charset=Windows-1252>Has it been (or can it be) proved whether there exist questions>that are *absolutely* unanswerable in the sense of there being>no sufficiently powerful mathematical system to answer them?> Suppose S is your current system, and Q is a question that it> can't answer (i.e. there is no proof in S that the answer is yes,> and there is no proof in S that the answer is no). Then you can> produce two more powerful systems: > S1, by adding to S the axiom the answer to Q is yes;> S2, by adding to S the axiom the answer to Q is no.> Both of these systems can answer the question. And, if it's> a meaningful yes-or-no question, one of these answers will be correct.> So not only is there a system that answers the question, there's a > system that answers the question correctly. The trouble is, we> don't happen to know which one!Ah. Now I see what Daniel Johnson must have meant about not taking the [answer to the] question as an axiom. Your explanation is very enlightening to me. Apparentlyboth S1 and S2 qualify for what Gerry Myerson called systemsmore powerful than the one we started with (S). When I readhis comments about there being a more powerful system that can answer what is unanswerable in a less powerful one, I didn't appreciate the axiomatic nature of the situation --specifically, that more powerful does not imply correct. --r.e.s. === Subject: Re: Letter to Prof Ullrich and othersBob> I am just curious: why are people like you> interested in even acknowleding James Harris and his ilk ?...> Is it not worthwhile for everyone to just ignore him at this point ?Some responders have kicked the Harris habit, but others pick it upmeanwhile. I'm been on and off the wagon for a couple of years, alas. Andif, by a fluke of statistics, one of his threads gets no responses even forone day, he barrages the board with spew.> There is real danger of course that someone would take him seriously,> but I feel that's remote.I agree.> It seems that he has taken to calling universities to complain.> Lets not giving him any more attention. Without attention, he will> shrivel up and disappear.But Harris is good at getting attention, because he has given it years ofpractice, and cares about nothing else, as far as I can see.LH === Subject: Re: Church-Turing compared to Zuse-Fredkin thesis (two new papers)> [Followups set to sci.math, who may be able to explain what an> algebraic number is better than I can.]I and the website author http://members.ispwest.com/r-logan/narrative.htmlunderstand that an algebraic number is the solution to a algebraic equation.Your criticism is unfounded and I think is caused by not realizing thatan algebraic equation and polynomial equation with integer coefficients,mean just about the same thing. Knowing that is not so uncommon for somebodywho feels qualified to point out errors in math websites.>> The web page states, by implication,>> * All algebraic numbers are roots of Diophantine equations of> degree 1 or 2.>> Cube root of 2 is an algebraic number which is not the root of any> Diophantine equation of degree 1 or 2. Hence, the web page'sstatement> is disproved by this example.>> As I understand it, we may say that all algebraic numbers are roots> of Diophantine equations; but that's all.I understand this differently: A Diophantine equation is an equation inwhich only integer solutions are allowed.So I don't see how bringing up Diophantine equations is relevant.> Whoops! I misspoke: read polynomial equation with integer> coefficients, not Diophantine equation. Sorry about that. Still,> the web page is wrong.polynomial equation with integer coefficients,That looks to me like the definition of algebraic equation:http://www.cut-the-knot.org/do_you_know/numbers.shtml #algebraic ...Real roots of such equations are said to be algebraic. In other words, anumber a is called algebraic if it satisfies an algebraic equation: Pn(a)=0for some polynomial Pn(x) with integer coefficients.Which is P_n(a)=0 for some polynomial P_n(x) with integer coefficientsin case the formatting is lost.That website focuses on numbers: rational, irrational and constructible.---------|----------------Real Numbers-----------|------------Rational numbers Irrational numbersare all algebraic are algebraic or transcendental> Correct, after re-formatting. :)> All algebraic numbers (including algebraic irrationals)> of the 1st or 2nd degree, or those having a power of two,> e.g., x2, x4, x8, x16, x32, x64, x128, ..., etc., if given a line> segment of unit length, are numbers that can be constructed by the> classical Greek geometric method of straightedge and compass.SH: The author is pointing out that some algebraic irrationalsare constructible. For instance the square root of 2.> Right.The cube root of 2 is not constructible and it well known thatin general cube roots (trisecting etc.) are not constructible.> Right.His explanation disallows including cube roots, which isthe type of counterexample you provided and thought injuredthe correctness of his description.> Right. His explanation disallows cube roots from being called> algebraic; however, cube roots *are* called algebraic; thus,> his explanation is flawed.Wrong. Read this again. All algebraic numbers (including algebraicirrationals)...[SH: then follows a rule excluding cube roots as algebraicsolutions to algebraic equations] can be *constructed*His rule is explaining that irrational cube roots are not solutionsto algebraic equations which are at the same time constructible.He is defining _constuctible_ algebraic irrational solutions whichof course are some but not all solutions to algebraic equationshaving irrational solutions.Of course he calls cube roots algebraic (or more preciselyalgebraic equations which have algebraic irrational solutions).being called algebraic; satisfy algebraic equations such as x^ 2 - 2 = 0 and x^3 - 11 = 0.I previously quoted this from his webpage.x^3 -11 = 0 will have the solution: x = the cube root of 11The author is stating that this cube root solution is not constructible.Not, that the cube root is not an algebraic solution; because he givesan example of algebraic irrational cube root equation in his definition.I don't know what that page means by algebraic equation.You can find numerous definitions on Google.> No, actually, I can't. If I had, you can be assured I wouldn't> be posting otherwise! MathWorld doesn't know the term, and neither> does anyone on the first page of Google results. Post a URL to> the definition if you want to use it.> However, I have made the assumption above that when you write> algebraic equation, you mean zero = some polynomial with integer> coefficients. Is that right?> -Arthurhttp://members.ispwest.com/r-logan/glossary.html#glo same authorYes. The author of the website you find flawed also providesthe polynomial with integer coefficients definition as ageneralized equation.Algebraic number - A real number that is the root of an equationin the form ... [ see webpage for equation with correct formatting]http://mathforum.org/dr.math/faq/ faq.impossible.construct.htmlThree geometric construction problems from antiquity puzzledmathematicians for centuries: the trisection of an angle, squaringthe circle, and duplicating the cube. Are these constructionsimpossible?The impossibility proofs depend on the fact that the only quantities you canobtain by doing straightedge-and-compass constructions are those you can getfrom the given quantities by using addition, subtraction, multiplication,division, and by taking square roots. These numbers are called Euclideannumbers, and you can think of them as the numbers that can be obtained byrepeatedly solving the quadratic equation.These three problems require either taking a cube root or constructing pi. Acube root is not a Euclidean number, and Lindemann showed that pi is atranscendental number, which means that it is not the root of an *algebraicequation with integer coefficients*, making it too nonuclidean.The author of the website to which you object is not stating that cube rootsolutions are not the roots of algebraic equations. He is saying they arenot constructible [see above] quote. Transcendental numbers like Pi, are thenumbers which are neither constructible nor the root of an algebraicequation (polynomial equation with integer coefficients). Algebraicequations have integer coefficients. What Pi and cube roots have in commonis their inability to be constructed-- not that they both are non-algebraicequations-- and the website author doesn't claim they are both non-algebraicas you impute, but that they are both non-constructible.SH: As to finding results on google it was easier for me since I knew thatan algebraic number was a solution generated by an algebraic equation.Actually what I thought was strange about your post was that you had acmu.edu email address and were imprecise/limited with your mathdefinitions.This was a google result:http://www.mayer.dial.pipex.com/samples/pi/tex4ht/ pisample.htmlA complex number is algebraic over Q if it is a root of a polynomialequation with rational coefficients.http://groups.google.com/groups?q=%22algebraic+ equation%22+root+definition&hl=en&lr=&ie=UTF-8&oe=UTF-8&selm= 2540%40tecsun1.tec.army.mil&rnum=12. :TRANSCENDENT 1a 3. a. incapable of being the root of an algebraic equation with rational coefficients (Pi is transcendental number).An AltaVista search for algebraic equation returned AltaVista found 8,739 resultsStephen === Subject: Re: I got low score on math test, please advise me and take a lookYou are really making this a bigger deal then what it is.The idea of withdraw dates as far as i know is to see if you like thesubject content and if you can understand it. Its not about how well youwill go in the subject. You will never graduate with a good degree if youare always droping subjects as you are not getting the highest possablegrade.instead about worring about this spend the time doing some study.I dont understand why you are doing a degree where you dont want to do thesubjects to get it.stephen> the following argument did not work with the dean of student affairs> this school is really horrible. i can't believe the screwed up procedures> at it.> this is the arugment:> Dear Dr. Johnson Jr.,> I am requesting that I be excused this one time to be withdrawn> available until one week and two days, Wednesday the 18th of February> This was one day after the date of withdraw without a W. The class is Math> 170, ticket # 3098, and meets on Monday and Wednesday from 6:00PM-8:00PM.> Again, I did not have enough information to withdraw by February 17thsince> there was absolutely no graded work or assignments that were returned back> by February 17th, one day before the exam results were made available.> Please, I ask that I be excused this one time to be withdrawn without a W === Subject: Re: differentiation> Could anyone help me differentiate the following in order to gain a> value for for 'r double dot':> r dot = (-0.5sin theta)*theta dot> where: theta = pi/4 and theta dot = 0.6r' = -(sin t) t'/2r = -(1/2)(cos t) (t')^2 - (1/2)(sin t) twhat's t at the location in question? === Subject: Re: No Set Contains Every Computable Natural> *sigh*... I don't know why I bother with this discussion...> Again, I won't contribute after this.There is no difference between assuming that a TMwill still be computing long after the stars have turnedto dust and assuming a TM can perform an infinite numberof operations.> It's a huge difference. One is saying that after a huge amount> of time T, it is computing. This is normal and fine, since T is> huge but finite.> But saying that a turing machine performs an infinite numbers> of steps is nonsense and goes against the standard definitions.There are lots of ways to define a Turing machine.In Turing's original paper, he said a number was computable only ifthere exists a TM that computes that number's binary representationand DOESN'T halt.You have chosen to use a definition that you think supports your position.It doesn't matter. My argument still holds.> Create your own difinitions is you want that property.OKAn accelerated Turing machine will perform an infinite number ofoperations in a finite amount of time.> In another post, you asked what the difference was in doing> something for an arbitrarily large amount of time and doing something> infinitely long.> This is the source of all your problems. Once you understand that> there is a difference between these two, most of your problems will> clear up. The first one is a finite computation (but for any large> amount of steps) and the second one is simply just not a computation> by the definition of turing machine computations.If we assume a TM requires a fixed, non-zero amount of timeto perform an operation, I can come up with a natural numberthat will takes billions of years for that TM to process.> That's totally fine. I give you some time T and you can create> a number that takes longer than T steps to compute (or output or> read or anything.) Our point is that for any computation you do> taking longer than this T-steps, its computation time will (and must)> be bounded by some other number B.Then I will give you another number that takes longer than B.We could do this for a really long time.Which one us will come up with the biggest number?I bet I do.There are 31,556,926 seconds in a year.This is about 3 * 10^7.Assume we have have computer that performs an operationevery 10^(-43) sec. This is as fast as we can expectany physical process to occur. Let's say the universe endsin about 3 * 10^10 years.3*10^7 * 10^43 * 3*10^10 = 9 * 10^60Nearly all natural numbers are larger than 10^61.I win. I found a number a TM can't read beforethe universe freezes over.You are not humouring me. The definition of algorithmimplicitly assumes a TM can perform any number ofoperations in a finite amount of time.> As long as any number is any finite number. Arbitrarily large,> yes, but still finite.Which one of us gets to decide what is finite?> then there are no 0s. Since assume there> is some 0; there must have been a 0 after it (since the string of> 1s after it is always finitely long) therefore that 0 was turned to> a 1, contradiction.Why is it a contradiction?You just said there a 0 (or a blank) following the 0that is overwritten. The output tape will always contain ablank after any number of operations.> You need to learn what a proof by contradiction is.A bad proof?> Assumption: There exists a 0 at the end of computation.Of course there is. How can there not be?My TM will never overwrite the last 0.> We use this statement and show that it contradicts itself.> Say there exists a '0' at the end of computation. If it exists> on the tape, it MUST exist at some finitely-indexed tape cell.Yep.> But then if the tape originally had all the natural numbers,That's a big if, isn't it?> then this 0 must have been followed by a string of 1s, eventually> another 0, and another string of 1s. Since there is this other 0,What other 0? There is only one 0 on the final tape.> the first 0 (that we assumed exists) must have been changed to a '1,'> contradicting the fact that the 0 existed.You just said there was this other 0.You must be talking about some intermediate version of the tape.So, the intermediate tape still contains a 0 at some finite position.> Take an analogy to the natural numbers. Your removal of 0s is likethis:> If, for any odd number 2k+1, there is an odd number coming after it,then> remove 2k+1 from the set.> Your argument is that the remaining set would still have an oddnumber,> but it's not hard to see that the set {1,2,...,N} up to ANY N at allwould> never have an odd number.Huh?Consider a finite set (1,2,3).Using your algorithm I get (2,3).There is no odd number that comes after 3 in the set (1,2,3).> You missed the fact in elementary school that if n is some odd> number, then so is n+2.I only see two odd numbers in the set (1,2,3).3+2 = a number not in my set.3+2 must not be finite.> I'm not talking about a finite set, I'm talking about taking> the natural numbers and applying that process to it. Your hypothetical> tape supposedly had all the natural numbers on it. Once the process> is complete, then any set you look at {1..N} will have all its odd> numbers removed.Supposedly is the key word here.There will be such an N. N could be quite large (it might even equal 3).Obviously, the original tape didn't contain every natural number.> Again, make sure you learn about the difference between 'artibrarily> large' and 'infinite.'Why don't you give me a TM that decides if a string is'arbitrarily large' or 'infinite.'> The set {1, 11, 111, 1111, 11111, ... } will contain strings that are> arbitrarily long but will not contain the infinite string 1111....And you have an algorithm that proves this?How do you know the input tape doesn't contain an infinite string of 1's?How do prove there is such a thing as an infinite string of 1's?I have given a proof that every string is finite as far as a TM isconcerned,even for TM's that perform an infinite number of operations.Russell- 2 many 2 count === Subject: What is the complexity of Gradient optimation method?Dear All,I am trying to figure out the complexity level of the gradient methodfor minimizing an objective function.Maybe it is a stupid and confusing assumption of my problem.Suppose the data points number is N, and dimension is d,what is the complexity level? It is O(dN) or O(dN^2)?Fred === Subject: Re: Huge Perturbations>Given a matrix A + b B, where A and B are of the>same order, b >> 1, and B has a null space>dimensionality > 1,> >I'm interested in both the eigenvalues of A + b B, and>roots of (A + b B) x = c (c is of the same order as A>and B) : the asymptotic case b -> inf and how it's>approached.> Well, the thing to do would be to consider b^(-1) A as a perturbation of> B rather than b B as a perturbation of A.> Of course the eigenvalues of A + b B are b times the eigenvalues of > B + b^(-1) A, and (A + b B) x = c iff (B + b^(-1) A) (b x) = c.I'm afraid I can not do that. B has a null space withdimensionality > 1 (see above) i.e. not only is it singular(I can't use it as the unperturbed part in the linearequations problem), but its eigenvalues aremultiple (so I can't use it as the unperturbed part inthe eigenproblem either. And the perturbation of that zeroeigenvalue is the one I would be interested in. === Subject: Re: 1=ma+nbVirgilm,n belongs to Za and b is realtively prime.Is there any proof of this? or just an axiom?...> See, for example:> http://en.wikipedia.org/wiki/Extended_Euclidean_algorithmA similar procedure, called Blankinship's algo, goes like this, for a=39 andb=23:39 1 023 0 116 1 -17 -1 22 3 -51 -10 17 ;and now 1 = -10*39 + 17*23The first two rows area 1 0b 0 1and each subsequent row is a linear combination of the two previous rows, asin Euclid's algo. When the leftmost column reaches gcd(a,b), you can readoff m and n.LH === Subject: Re: HELL :-) (Was: Re: the anticlassicalist }{ ii: the spectre continues)> [...]> You even picked up Conway and Sloan... That was the first book I ever> checked out of a university library (I didn't understand it then, but I kept> coming back to try again).Now that I know that, I can make a better picture.I want to bring what follows back to the discussion of hexacodes and F_4. F_4is listed as{0, 1, w, w-bar}with specific code shapes based on a quadratic form.I found the picture for a free de Morgan algebra on one generator Balbes andDwinger, 1 | | * / / a ~a / / * | | 0So, structurally I am looking at (w) and (w-bar) as being in (vague)correspondence with (a) and (~a). These are de Morgan idempotents now. So,with regard to classical truth table semantics, think of the fact that there aretwo complete connectives. That is, there are three concepts here, and, the(vague) correspondence in intended to extend to NAND and NOR as well.In Malinowski's paper on the lattice of orthomodular logics, http://www.uni.torun.pl/~jacekm/latoml.pdfyou will find a lattice diagram labeled 'c.' The diagram above can be embeddedinto that lattice so that (a) and (~a) correspond to the two nodes in theisolated paths on the right. The asterisks in the diagram above coincide withTOP and BOTTOM on Malinowski's lattice. You may place the 0 and 1 in thediagram above on any of the remaining six nodes provided that the 0 and 1 arenot connected by any path not passing through an asterisk.So, now we have embedded a de Morgan idempotent into Malinowski's diagram.Some time ago, I did a post using a seven element Steiner triple system. It hasa 1-factorization given by{{#,0},{1,3},{2,6},{4,5}}{{#,1},{0,3},{2,4},{6,5}}{{#,2},{ 0,6},{1,4},{3,5}}{{#,3},{0,1},{2,5},{4,6}}{{#,4},{0,5},{1,2},{ 3,6}}{{#,5},{0,4},{2,3},{1,6}}{{#,6},{0,2},{3,4},{1,5}}For our purposes, the 0 and 1 in our diagram map to # and 0 in the Steinerquasigroup. If we take the first factorization where {#,0} appears asreflecting the listing of Boolean idempotents, any of the remainingfactorizations is a combination that can be interpreted in our diagram.In terms of truth tables, the de Morgan isomorphism being reflected here can beunderstood with respect to A B |------------ T T | T F | F T | F F |maps to ~B ~A |------------ F F | T F | F T | T T |It is not that one can match the numbers up. Rather, it is that NAND and NORbecome OR and AND, respectively. In like fashion, IFF and XOR become XOR andIFF, respectively. That is the exchange captured in the pairings with # and 0in factorizations like,{{#,1},{0,3},{2,4},{6,5}}{{#,3},{0,1},{2,5},{4,6}}I can repost the Steiner quasigroup and the Steiner idempotent quasigroup toshow you what syntactic associations are being held constant under the algebraicproducts. But, the end result was a labeling of the incident matrix, a/1 b/5 c/3 d/4 e/2 f/6{1,2,4} x x x{1,6,5} x x x{2,3,5} x x x{3,4,6} x x xthat contained no reference to 0/1 or 1/#.All of these forms coincide properly without specification for logicalequivalence or exclusive disjunction. This intrinsic ambiguity appears in Lemma4 of http://citeseer.nj.nec.com/feigelson97forbidden.htmlSo, what I have done here is consistent with the treatment oftruth-functionality in the combinatorial framework of threshold logic.Also, before returning to F_4, I note that the incidence matrix abovecorresponds to a tetrahedral simplex. So, there is the notion of a relativemetric in the spatial sense. But, the quadratic associated with code shapes isinformational.Now, Conway and Sloane list three additive products which are the first matterof interest,1 + w = w-bar1 + w-bar = ww + w-bar = 1This usage of 1 corresponds with the 0 in the Steiner quasigroup product. Forexample, 0*5=4 | | | 1 | | 0 | | N | | | | | | | | 0 | | 1 | | O | * | | = | | | | | 0 | | 1 | | T | | | | | | | | 1 | | 0 | 4*5=0 | 0 | | 1 | | | | | | | | N | | 1 | | 0 | | | | | * | | = | O | | 1 | | 0 | | | | | | | | T | | 0 | | 1 | | | 4*0=5 | 0 | | | | 1 | | | | N | | | | 1 | | | | 0 | | | * | O | = | | | 1 | | | | 0 | | | | T | | | | 0 | | | | 1 | 5*0=4 | 1 | | | | 0 | | | | N | | | | 0 | | | | 1 | | | * | O | = | | | 0 | | | | 1 | | | | T | | | | 1 | | | | 0 | 5*4=0 | 1 | | 0 | | | | | | | | N | | 0 | | 1 | | | | | * | | = | O | | 0 | | 1 | | | | | | | | T | | 1 | | 0 | | | 0*4=5 | | | 0 | | 1 | | N | | | | | | | | 1 | | 0 | | O | * | | = | | | | | 1 | | 0 | | T | | | | | | | | 0 | | 1 |With respect to NOT, remember that XOR and IFF are de Morgan conjugates as wellas Boolean complements. I had been thinking of Boolean conjugates when I firstfigured out the association with the triple system. But, when I realized thecorrespondence with F_4, the only thing that mattered was de Morgan conjugation.The hexacode is described as a 3-dimensional code of length 6 over F_4.According to Conway and Sloane, it has a word,W(phi) = ab cd effor each quadratic functionphi(x) = ax^2 + bx + cdefined over F_4. The first three digits (a,b,c) specify the function phi, andthe last four (c,d,e,f) give the values for 0, 1, w, w-bar:c = phi(0)d = phi(1)e = phi(w)f = phi(w-bar)Now, remember that the context here is no longer the Steiner quasigroups. NANDand NOR are in (vague) correspondence with (w) and (w-bar). 0 and 1 do have(vague) correspond to TOP and BOTTOM Malinowski's diagram, but that (vague)correspondence is relative to the association with logical equivalence andexclusive disjunction as per the choices associated with the Steiner quasigroup.Thus, Malinowski's diagram is interpreted relative to a classical truth tablesemantics. It simply has not been given a definite labeling.Each word of the hexacode has a slope, (s), and the concatenation of couplesab cd efis a codeword if and only if it satisfies the following,1-rule:a + b = c + d = e + f = 1(w)-rule:a + c + e = a + d + f = b + c + f = b + d + e = (w)(s)(w-bar)-rule:b + d + f = b + c + e = a + d + e = a + c + f = (w-bar)(s)Hopefully, you can see how these equations would lead to a diagram likeMalinowski's relative to a free de Morgan algebra on one generator. The realproblem is visualizing information in terms of de Morgan isomorphisms on deMorgan idempotents.I could go on... But, no one (except, perhaps, Galathaea) has any sense of whatI am trying to talk about.You are all so used to thinking in terms of Boolean complements and linearlogic. How does one prove anything with de Morgan logic??? The de Morganisomorphisms are not order isomorphisms.All of the symmetries here are appearing in physical cosmology because we arenot differentiating between thermodynamic entropy and information-theoreticnegentropy. It is not clear that we can. But, I would really like to know....:-)mitch === Subject: Re: Collatz Conjecture : Symmetry question.First: Won't there be duplicates in the large files?Second: The two divmod's can be combined into one.-Michael. === Subject: Re: functional analysis--separable> I found the following proposition in a book without a proof. I> have difficulty in proving it. Can anyone help?> Suppose B is a separable Banach space. Let {x_n} be a countable> dense subset of the unit circle C={x in B : ||x||=1}. How can I prove> that: every element x in B can be written as:> x= summation[from 1 to infinity] {a_n x_n} where a_n is a absolute> summable sequence. That's:> summation[from 1 to infinity] {|a_n|} < infinity.Hint: Let x be in B. Suppose you've approximated x nicely with a finite sum s of the sort under discussion. So x = s + (x-s) and |x-s| is small. Now approximate (x-s)/|x-s| by y in {x_n} as closely as you like. Then (x-s) - |x-s|*y will be even smaller. Continue ... === Subject: Re: How big can a manifold be?> what about using whitney's embedding theorem? a jump> in dimension will have no effect on the cardinality.> I suggest that you check the hypotheses of Whitney's embedding> theorem (I'd even do it for you but I am a bit busy); I think> you'll find that paracompact (or something essentially> equivalent) is necessary. (For instance, the long line> isn't paracompact, and I am pretty sure it doesn't embed> in R^3, either..The exact result is : any continuous map from the long line to R^n iseventually constant (i.e there exist a in R^n and x in L such that for ally>x , f(y)=a).isn't the set of joining points in> its construction both discrete with the relative topology,> and of too big a cardinality to embed in R^3, never mind> the stuff in between?)> Lee Rudolph === Subject: Re: Axioms defining a finite field> The set of rules you have give a ring. In order to get a field, you> need a multiplicative inverse for all non-zero elements. For example,> integers mod 4, 2 does not have a multiplicative inverse. See> Table.> 2*0=0> 2*1=2> 2*2=0> 2*3=2One of his axioms was that a*b=0 implies a=0 or b=0, which does not holdfor the integers mod 4.That axiom, combined with a*0=0 (easily proved from his axioms), leads to acancellation theorem, a*b=a*c => b=c for a!=0. Combine that with F beingfinite (specified in the original post), and it is a short step to everya!=0 having a multiplicative inverse.-- --Tim Smith === Subject: Re: Can something be undeterminable?Content-transferncoding: 8bitFor that, you would have to rule out taking the question as an axiom.> I kind of like the idea of a question being an axiom.Better than the idea of an axiom being questionable...--Ron Bruck === Subject: Re: Can something be undeterminable?|But is it logically possible for something, a mathematical statement, to be|absolutely undeterminable?Goedel once expressed an interest in developing a concept of absoluteprovability, but so far as I know nobody has done it, so the question issomewhat undefined.I think it's plausible that certain questions in mathematical logic arein fact undeterminable in a strong sense.Chaitin has defined an uncomputable real number called Omega.Omega is uncompressable in the sense that for each programminglanguage there exists a constant C such that for any N, a programwhich correctly computes the first N bits of Omega is at least N-C bitslong itself.One way to construct a program for computing bits of Omegais to have it search for proofs in a given formal system whoseconclusion is that a given bit of Omega is 0 or that it is 1. Theprogram can then output them in order. So any formal systemcapable of proving what the first N bits of Omega are has to beclose to N bits long as well.I think this provides a pretty convincing argument that all but thefirst few bits of Omega are beyond any reasonable determination.There may be longish formal systems that we believe to becorrect as far as their arithmetic consequences go, but presumablybecause of more fundamental grounds. It seems unlikely that wehave a large amount of innate knowledge of mathematics, in thesense of knowing facts of arithmetic that absolutely cannot bejustified except by assuming a lengthy set of axioms. I've seenZF condensed to a relatively short (smaller) set of axioms. Higheraxioms of infinity don't require much more.The theorem mentioned above leaves open the possibility, of course,of an axiom system being short and deciding the n-th bit of Omegafor some large n, without deciding all the previous ones. Omega isdefined as the probability of a Turing machine halting, given acertain way of constructing one at random. It seems implausible tome that a natural axiom system would decide bits very muchbeyond the first one it couldn't decide.This is just a plausibility argument, of course, and without an actualdefinition of absolute provability there's no way to make such anargument rigorous.Keith Ramsay === Subject: Re: Teaching philosophy> teaching philosophy. I think that gives me a good idea of the genre.> What about the length of the statement?> Teaching Philosophy: Les fleurs qui pousse de la merde> by Allan Adler> Evil is the root of all money. Hence if funding has been found> for a position, it necessarily cometh of evil. Yet, though mushrooms> nourish themselves on excrement, they in turn nourish people and, in> like manner, our fate as educators is to be a kind of mushroom. Let me> therefore describe some recipes in which this mushroom has been used.> [details omitted for the purpose of this posting]> (I know, cut the first paragraph...) Use the first as a cover sheet. Since Philosophy is known to be orthogonal to money. And state plainly in the second paragraph that whether or not it's orthogonal to intelligence depends on crucial ongoing experiments in Bejing. Hence the education commitee needs to organize a Buddhist revival movement in order to effectively prove that mathemathematics is truly a logic system, rather than simply a faint remnent of the after-glow of the exploding super-galaxy, PI-ZETA-4* in the Constellation of SETI-Alpha. > 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: Re: Pascals TriangleRob Pratt escribi.97 en el mensaje> There was an interesting puzzle in the Sunday Times in the UK> recently that set me thinking about the issue that the puzzle raised.> The idea is that of a set of points arranged in the familiar Pascal's> triangle format (rows 1, 2, 3, ... etc. containing 1, 2, 3, ...> points in a triangular format) after which the issue is that of the> total number of equilateral triangles that can be found in this set> of points.> For N in 1..10, I get 0, 1, 5, 13, 27, 48, 78, 118, 170, 235, which> can be expressed as> floor((N - 1) (N + 1) (2 N - 1) / 8)> and appears in Sloane's database:http://www.research.att.com/cgi-bin/access.cgi/as/ njas/sequences/eisA.cgi?An> um=A002717It seems that you only count triangles with sides paralel to array sides.But there are other equilateral triangles with vertices in the points of theequilateral triangular array with its sides in another orientation.Exactly, there areComb(n, 4) = n(n + 1)(n - 1)(n + 2)/24in a triangular array of order n. For n = 1 to 10,0, 1, 5, 15, 35, 70, 126, 210, 330, 495http://math.smsu.edu/~les/POW03_01.htmlhttp:// www.research.att.com/projects/OEIS?Anum=A000332-- === Subject: Re: Genetics and Math-Ability> Just curious:> What is the current scientific opinion regarding how math-ability is> inherited? Math ability depends entirely on what is currently fashionable in math. If cloning is currently fashionable, and you can add, you are a genuis. If not, well there's always chemistry.> Is it, in general, a recessive trait or is it dominate?> (I say recessive.)> I am only half-serious, since math ability is most probably a result> of many traits and experiences and education.> But I was wondering today because, even though there are some in my> family who were math-teachers, almost everyone else had/has little> ability with mathematics...even very little ability as relative to> me...> ('relative to me'...snicker, snicker...)> Leroy Quet === Subject: re:Axioms defining a finite fieldYour example violates the last rule. Pay attention. Posted Via Usenet.com Premium Usenet Newsgroup Services------------------------------------------------------ ---- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **---------------------------------------------------------- http://www.usenet.com === Subject: Re: Definition of Separable Space (basic topology question)>N.>> Let A be a subset of (X,T). Then A is dense in X iff for everynonmpty> open subset U of X, A / U != {}.>> I have seen two definitions of 'separable topological space':>> a) (X,T) is separable if there exists A X, where A iscountable and> dense in X> b) (X,T) is separable if there exists A X, where Ais> countable and dense in X>> Given def (a), its simple to prove that all countable spaces X are> automatically separable since the subset X is countable andnon-trivially> intersects every nonmpty open subset. However, this won't workgiven def> (b). I first became suspicious when I saw a proof that allcountablespaces> were separable that seemed ridiculously complex in comparison tothe> seemingly obvious 1 step proof above. I later found definitionsof> separable like def (b).>> l8r, Mike N. Christoff>> I have never seen definition (b). I have seen the symbol $subset$> used to mean subset, not proper subset, so maybe that is what yousaw.>> Certainly a one-point space with the discrete topology is to be> considered separable, even though it has no dense proper subsets.>> I've read the other follow-ups, and I think you've hit the nail on the> head: the OP is confused about the meaning of subset, and is taking> A subset B to mean A is a proper subset of B.>> I have never seen definition (b) either. It would be very weird for> finite sets NOT to be separable. (And it would make the useful fact,> A subspace of a separable metric space is also separable false.>Ok. I'm tired and don't immediately see anything wrong with this:(X,T):X = {1,2,3,4}T = { X,{},{1,2},{3,4} }A = {1,3}Finite space with dense proper subset...> Presumably re: my remark It would be very weird for finite sets NOT to> be separable. I didn't mean that ALL finite sets wouldn't be> separable. (Although, if the topology is Hausdorff, it means exactly> that.) It suffices for a SINGLE finite set not to be separable to set> off my weirdness alarm.> But I've lost the thread of what I was thinking when I said it would> falsify subspace of separable metric space is separable. I'm sure I> had something in mind...The following assumes topological spaces, and subspaces must be nonmpty.Let X be a separable metric space, and S a single element subspace of X.Assume A is a proper dense subset of S under the induced topology. Then Amust equal {}, but it must also non-trivially intersect S, which isimpossible.l8r, Mike N. Christoff === Subject: Re: Pascals TriangleIgnacio Larrosa Ca.96estro escribi.97 enel mensaje> It seems that you only count triangles with sides paralel to array> sides. But there are other equilateral triangles with vertices in the> points of the equilateral triangular array with its sides in another> orientation.> Exactly, there are> Comb(n, 4) = n(n + 1)(n - 1)(n + 2)/24Obviously, it must beComb(n + 2, 4) = n(n + 1)(n - 1)(n + 2)/24> in a triangular array of order n. For n = 1 to 10,> 0, 1, 5, 15, 35, 70, 126, 210, 330, 495> http://math.smsu.edu/~les/POW03_01.html> http://www.research.att.com/projects/OEIS?Anum=A000332-- === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> <8765e4q2fe.fsf@phiwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@phiwumbda.org> <87vfm4o85i.fsf@phiwumbda.org> <87n07fo9qr.fsf@phiwumbda.org> <87k72jnxtp.fsf@phiwumbda.org> <87ptcau3xj.fsf@phiwumbda.org> <87fzd6ti0j.fsf@phiwumbda.org> Discussion, linux)> Look at the definition of my TM.> Suppose that it contains a blank. Step (2) will then search for> another blank. The blank on the tape will not be overwritten> unless a second blank is found. The output tape will ALWAYS> contain one blank. Even after an infinite number of steps.> You have proven by induction that for all n, at step n in the> computation, there is a blank space. This does not prove that after> omega steps, there is a blank space. Your proof by bald assertion> doesn't cut it.In fact, once again you seem to make the error of commutingquantifiers.You have proved[1] that, for all steps n, there exists a square x suchthat x is blank.You have asserted that there exists a square x such that for all stepsn, x is blank.This is simply false. This sort of reasoning[2] would yield thefollowing argument.Start with a blank tape. Execute the algorithm:(1) Write a 1 in the current square, and move right one square. Go tostate 1. Now let square 0 be the start square, 1 the square to the right of it,2 the square to the right of square 1 and so on.Clearly, for every step of the computation, there is a positive n suchthat n is blank.Nonetheless, it is obviously not the case that after an infinitenumber of steps, there is an n such that n is blank.Obviously here means obvious to anyone capable of simple logicalarguments. Since I won't presume that's true of all parties here,I'll give the argument.For any n, after the n+1st step, square n contains a 1. Thereforethere is no n such that, after an infinite number of steps, square nis blank.Footnotes: [1] I'm being charitable here.[2] I'm being facetiously charitable here.-- It has been shown that no man can sit down to write without a very profounddesign. Thus to authors in general trouble is spared. A novelist, for example,need have no care of his moral. It is there -- that is to say, it is somewhere-- and the moral and the critics can take care of themselves. -.A. Poe === Subject: Re: Huge PerturbationsHere's a suggestion. Write A + b B = b [ (1-b)/b * A + (B+A) ]. Now considerB+A as the unperturbed part, and (1-b)/b * A as the perturbation.Here's another more general suggestion. You should try to separate B'snullspace, i.e. diagonalize B and write x = (x_b, x_b') where x_b has thesame dimensionality as B's nullspace. Because in the limit b->infinity, thedimensionality of the problem is reduced, and x_b becomes undetermined.-Michael. === Subject: Re: Are the derivatives of abs[(x-a)^3] different for x>a and x The usual method is to define sgn(x) in terms of the heaviside function> H(x) = { 0, x < 0; 1, x > 0; 0.5, x = 0 }, and write formally> dH/dx = delta(x) where delta(x) = { 0, x != 0 } is the Dirac delta> 'function' (integral of delta(x) over any interval containing 0 is 1 by> definition, and 1/2 if 0 is an endpoint).> This is a joke, right?-- === Subject: Re: Sets That Resemble Derivatives Somewhat>>in what?>No. The right side looks similar, 'cos of your denoting complement>by the same notation as derivative, but the right side is completely>different!>One idea? Not the idea of it being a limit of difference>quotients methinks.>> Derivations are a topic in the literature on Lie Algebras and other>> structures, but (2) and (3) have a rather different flavor.>I.e., by not resembling derivation at all?>applause!>Like er yeah! I mean yeah if x + h is different to x then er yeah>then x + h is outside x. Wow, groovy!> Aha! The universe itslef. Next you'll be giving us the answer>to life the universe and everything.>Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html>Francis Wheen, _How Mumbo-Jumbo Conquered the World> I omit Robin Chapman's quotation of Lacan on penises, and I omit my> comments to which these are replies (except that Lacan was Robin's> own quotation) for brevity.For brevity, or for deliberate obscurity?-- === Subject: Re: Sets That Resemble Derivatives Somewhat>>[Snip: in Doctorowese A' is the completement of A, A+B is the union>>of A and B and AB is the intersection of A and B]> > 2) (A')' = A> > This doesn't correspond to differentiation,>>Yet another thing that doesn't correspond to differentiation>>is that>>AB' + A'B =/= (AB)'.>>Indeed I am at a loss reading these posts to find any analogy between>>complementation and differentiation (save that Doctor Osherow uses>>the same notation for both).>Evidently you've forgotten that derivatives satisfy> (f + g)' = (f')(g').> Contrary to appearance, David C. Ullrich actually added the last> two lines above as a joke expanding on Robin Chapman's claim that I used A> + B to be the union of A and B, or else Ullrich himself wove this claim> into Robin's post (it's hard to figure it out where this quote came from,> so I'm inclined to believe that Ullrich invented this claim based on the> nature of Ingenious Imitation as it pervades> world colleges at the faculty level). He then asserts that (f + g)'> = (f')(g'). This also is his invention, the purpose of which beyond> emotionality escapes me. I think that he is asserting a contradiction> and projecting it upon the external world in the manner of some> with whom we are very familiar from recent experiences.Why don't you talk proper?-- === Subject: Re: Sets That Resemble Derivatives Somewhat>there's a general theme that some people have explored that says that>the fact that certain sorts of differential operators obey>leibniz-like identities is conceptually subordinate to the fact that>certain sorts of boundary operators in geometry and topology obey>leibniz-like identities, roughly (or sometimes precisely, depending on>the particular concept of boundary being used) boundary(x X y) =>(boundary(x) X y) + (x X boundary(y)).> This looks interesting, although I think that themes of conceptual> subordination may be a bit premature here, but I am very encouraged> by the lack of hostility in your message and by your ability to> detect relationships. How did a nice guy like you get in the same> thread as Robin and Ullrich?Detect what relationship. As I pointed out there is no analogy betweencomplementation and boundary/derivative.-- === Subject: Re: two other sides of triangle> Unfortunately, since the method above needs to solve 2 quadratics, we> get four values, not two. Here is another method, similar in idea, but> only requiring the solution of one quadratic, giving the two sides that> were requested.I said that the first quadratic was remarkably simple. If you actuallywork it out, you will see that one of its roots corresponds to ab = 0,which is obviously not the one we are looking for. Therefore the firstquadratic is not really a 'quadratic' but a linear equation, and thereis no ambiguity.Dimitris === Subject: Re: When is a finite field a cyclic field?>I don't intend to introduce new notion,but for the moment >let's call a finite field F(+,*) a cyclic field if>the abelian group (F,+) is cyclic.(We know that (F0,*) is always cyclic when F is finite).When is (F,+) also cyclic?>May I ask Derek Holt who gave such a nice and clear solution>to a recent problem for finite fields,to give answer to this?>Thank you! Yes you may ask! The answer is when |F| is prime.the smallest positive integer p such that p.1 = 0 is prime, but thenit follows that p.a = 0 for all a in F, and so (F,+) cannot be cyclicunless |F| = p.Derek Holt. === Subject: Re: Pascals Triangle> Rob Pratt escribi.97 en el mensaje>There was an interesting puzzle in the Sunday Times in the UK>recently that set me thinking about the issue that the puzzle raised.>The idea is that of a set of points arranged in the familiar Pascal's>triangle format (rows 1, 2, 3, ... etc. containing 1, 2, 3, ...>points in a triangular format) after which the issue is that of the>total number of equilateral triangles that can be found in this set>of points.For N in 1..10, I get 0, 1, 5, 13, 27, 48, 78, 118, 170, 235, whichcan be expressed asfloor((N - 1) (N + 1) (2 N - 1) / 8)and appears in Sloane's database:http://www.research.att.com/cgi-bin/access.cgi/as/ njas/sequences/eisA.cgi?Anum=A002717> It seems that you only count triangles with sides paralel to array sides.> But there are other equilateral triangles with vertices in the points ofthe> equilateral triangular array with its sides in another orientation.> Exactly, there are> Comb(n, 4) = n(n + 1)(n - 1)(n + 2)/24> in a triangular array of order n. For n = 1 to 10,> 0, 1, 5, 15, 35, 70, 126, 210, 330, 495> http://math.smsu.edu/~les/POW03_01.html> http://www.research.att.com/projects/OEIS?Anum=A000332Hi Ignacio,Thank you for your input - I must admit that I also missed the rotatedtriangles in my search algorithm. Do you have any insights into the minimumnumber of grid points that need to be removed in order to eliminate allequilateral triangles?It seems possible that this might actually be easier to solve with therotated triangles since all the points (except one) on one side of allnon-rotated sub-triangles always need to be removed. Gladman === Subject: Re: Axioms defining a finite field> The set of rules you have give a ring. In order to get a field, you> need a multiplicative inverse for all non-zero elements. For example,> integers mod 4, 2 does not have a multiplicative inverse. See> Table.> 2*0=0> 2*1=2> 2*2=0> 2*3=2>One of his axioms was that a*b=0 implies a=0 or b=0, which does not hold>for the integers mod 4.>That axiom, combined with a*0=0 (easily proved from his axioms), leads to a>cancellation theorem, a*b=a*c => b=c for a!=0. Combine that with F being>finite (specified in the original post), and it is a short step to every>a!=0 having a multiplicative inverse.But this example is important, because it shows that the axiom a*b = 0implies a=0 or b=0 is definitely not redundant. Part of the problem, whichnobody has answered yet, was to decide whether any of the axioms can beomitted. I am guessing no, but I could be wrong. The axiom 1 != 0 is notredundant, because F = { 0 } satisfies all other conditions.Two down, seven to go!Derek Holt. === Subject: Re: Fresnel integrals> 1) int[-inf to inf] cos x^2 dx = sqrt(pi/2)> 2) int[-inf to inf] exp(-x^2) dx = sqrt(pi)> Eqn (2) is easy to show by looking at the square of the left side> (a double integral) and switching to polar coordinates. I tried the> same approach with (1). (There are other ways to get (1), I know.)> The double integrand will be> (cos x^2)(cos y^2) = (cos(x^2 - y^2) + cos(x^2 + y^2)) /2> The first term on the right integrates to zero okay, but the second> term, in polar coordinates, runs into a convergence problem.> Moreover the conversion to polars requires some justification,> because the Fresnel integrals are not absolutely convergent.> Anybody know any quick fix for this one? I get the feeling I'm> missing something obvious...Dunno, but (1) is the wrong thing to consider --- you should considerintegral_{-infinity}^infinity exp(i x^2) dxof which (1) is the real part. Of course these are *improper* integrals).-- === Subject: ATTENTION: Open dispute with my college about Procedures. Please read.My web site www.johncho.us has all the information. I am seeking people who are familiar with procedures. I am also considering getting a lawyer. === Subject: Re: Drawing subgroup lattices for research papers> I'm working on a thesis (using MikTex and Winedt) that requires me to> put in a lot of subgroup lattices. This results in several problems. > For an early version of the paper, I made .bmps and cut and paste them> in the appropriate spots, but now I want to put them in the actual> code of the paper itself. My method was to import into Mathematica> and export as .eps. First of all, the .eps files are huge. I tried> to import a 3k grayscale gif and got a 400k eps file. Then, using> graphicx I was able to get the picture to show up after Latexing. > However, the graphic looks terrible. It is clearly something that is> happening when I Latex the paper because I checked the eps file in> Ghostview and though it was degraded some from the original bmp the> subgroups were still recognizable.> I've also tried downloading xypic and followed the instructions, but> can't figure out how to get MikTex or WinEdt to recognize the new> files.I suggest that you post the same question at the comp.text.texnewsgroup.Jose Carlos Santos === Subject: Re: Fresnel integralsGeometrically the curvature of this curve (Cornu spiral) has curvatureproportional to arc length, tangent slope is proportinal to square ofarc, it is so intrinsically . === Subject: Re: Solving Equation with ln> I was working on a math problem at work and ran into something I don't> know how to solve and was hoping someone could give me a clue on this> fairly simple problem.> Original Equation:> 227.11 = 5.8{ln(38/x)} + (38-x)Well, there is a solution with x positive, but *really really* close tozero. In that case you could replace the last term 38-x simply with 38.Then you can solve for x and very that it indeed is small.> Simplfied Equation:> 189.11 = 5.8*{ln(38/x)} - x> of course you can reduce this down to> 32.60 = ln(38/x) - x/5.8> and further (I think) to> 1.44 x 10^14 = 3.8/x - e^(x/5.8)This is incorrect. It should be 1.44 x 10^14 = 3.8/x * e^(-x/5.8)> but I am stuck after this. Is it possible to solve for X?Not using elementary functions.-Michael. === Subject: Re: No Set Contains Every Computable Natural>>There is no difference between assuming that a TM>>will still be computing long after the stars have turned>>to dust and assuming a TM can perform an infinite number>>of operations.>Yes there is. Just as there is a difference between an arbitrarily>large finite number and infinity.>What is the difference between an arbitrarily large finite number and>infinity?> One is finite and the other isn't.>Can you define an algorithm that will determine if a string of 1's>represents a natural number or is infinitely long?> I think a sufficiently rigorous definition of algorithm requires that> the input (or problem) be finite in size, so I suspect that what you are> asking for is not within the scope of an algorithm.So there is no way to determine if a string is infinite?Why do you believe some strings are finite and otherstrings are infinite if there is no possible way to tellthe difference?Russell- 2 many 2 count === Subject: Re: the anticlassicalist }{ vi: into the quantum> Riddle of the day: are ontologists for real?Ah.... ontology is about existence. L'.90tre et len.8eant (sounds much better, doesn't it?)Can something that does not exist be ontological?Does an ontologist need to exist in order to... er... be an ontologist? (Never mind what anontologist is, I have a hunch that you couldargue the same of a communist, a capitalist,a typist, a writer, a plumber, a feather). === Subject: Re: the anticlassicalist }{ : mitchism for Daleks <40325A2C.21936CE9@hate.spam.net> <103513r3gafocff@corp.supernews.com> <40328829.1E5C341D@hate.spam.net> <1035esdt30h4d38@corp.supernews.com> <1039dr5kpjloif1@corp.supernews.com>> In message <1039dr5kpjloif1@corp.supernews.com>, galathaea>: >You should have realised that anything full of -ism and -ist which is>spewed>: >to sci.lang and sci.logic is not going to contain any physics.>: >>: (Nor language; I can't speak for sci.logic ;-)>:>: I'm well aware of that. What I was wondering was whether the _OP_>: realises that there's a difference between these isms and physics, and>: what it consists of.>>Installment vi> >>Before you lies the Void...>details the properties of the closed linear subspace lattice>of a Hilbert space, detailing the relationship with observables and>prediction in quantum systems. This builds off of the mathematical theory>>You do understand the physical content, do you not?> Haven't seen any.> Dirac compressed quantum mechanics into ~300 pages of terse elegance,> and didn't use ontology once.>Obviously, Dirac knew the tastes of the audience to whom he was writing.The Nobel committee?> The>man had to get published, didn't he? Taxes needed to be paid. The wife and>children needed to eat. Little things like that have often gotten in the way>of principle.I think you may be confusing him with Hawking. I doubt if the royalties on PQM, which is hardly a dumbing-down for popular tastes, would have kept him in hot dinners.>But one must not to say Dirac did not have ontology in mind....>Therefore, there is plenty of room for improvement>and for seeking to discover a new and logically superior>foundation for the theory. I suspect that it is simple>Aristotelian logic (in the context of Platonic ontology)>that is missing from the theory, and this is due to the>painful divorce of physics from philosophy.>Fragmented people think fragmented thoughts that>never add up and ultimately don't make sense, whatever>illicit success may initially obtain. The deepest urge in>my being is to understand the principles of physics>from the first principles of logic. Without that our>claim to knowledge is nought but a pretense, and our>souls are divided against themselves, leaving us>culturally schizophrenic and socially insane.> http://www.geocities.com/saint7peter/ DiracslectureonQFT.htmlAre you sure all those words are Dirac and not Mutnick? The way it's laid out, the separation of quotes and comment isn't entirely clear.http://www.geocities.com/saint7peter/index.html>Since Mr. Herring is personally being such a pissant, perhaps he can explain>what is meant by physical content anyway? Does that mean using words and>phrases that depend on Cauchy's (?) epsilon-delta justification for the>derivative or Robinson's non-standard analysis justifying naive use of>differentials?No, it means that at some point you have to hook into what physicists *observe*. Will it tell us the energy levels of a hydrogen atom and predict the Lamb shift? If you're interested in developing a better mathematical formalism, that's fine, go ahead. Just don't tell us it's physics.> His apparent interest in Dirac extends to I let other people>do my 'isms' so I can act the fool... just like the well-received Franz>Heymann>The development of the limit statement and derivative in terms of open sets>intimately links truth in physics with Heyting algebras. Cech's association of>a nerve with open sets links truth in physics with simplicial homology. And>the work in foundations with topos have associated the logic of open sets with>consistent logic.... and it would help your case if you could make it read less like the hermeneutics of quantum gravity.[snip again ]-- Richard Herring === Subject: Re: Drawing subgroup lattices for research papers> I'm working on a thesis (using MikTex and Winedt) that requires me to> put in a lot of subgroup lattices. This results in several problems. > For an early version of the paper, I made .bmps and cut and paste them> in the appropriate spots, but now I want to put them in the actual> code of the paper itself. My method was to import into Mathematica> and export as .eps. First of all, the .eps files are huge. I tried> to import a 3k grayscale gif and got a 400k eps file. Then, using> graphicx I was able to get the picture to show up after Latexing. > However, the graphic looks terrible. It is clearly something that is> happening when I Latex the paper because I checked the eps file in> Ghostview and though it was degraded some from the original bmp the> subgroups were still recognizable.> I've also tried downloading xypic and followed the instructions, but> can't figure out how to get MikTex or WinEdt to recognize the new> files.> Does anyone either have a suggestion on getting what I have to work or> a more tex-friendly way to draw lattices?xypic will do most things you can imagine, and some you can't. simpler butless flexible is paul taylor's diagrams package. diagrams is available from ctan archive. I don't understand what you mean by can't get winedt or miktex torecognize the new files, which new files, what are the error messages? one simple suggestion: have you updated the miktex installation afterputting xypic in your path? === Subject: Re: Fresnel integrals>1) int[-inf to inf] cos x^2 dx = sqrt(pi/2)>2) int[-inf to inf] exp(-x^2) dx = sqrt(pi)>Eqn (2) is easy to show by looking at the square of the left side>(a double integral) and switching to polar coordinates. I tried the>same approach with (1). (There are other ways to get (1), I know.)>The double integrand will be>(cos x^2)(cos y^2) = (cos(x^2 - y^2) + cos(x^2 + y^2)) /2>The first term on the right integrates to zero okay, but the second>term, in polar coordinates, runs into a convergence problem.>Moreover the conversion to polars requires some justification,>because the Fresnel integrals are not absolutely convergent.>Anybody know any quick fix for this one? I get the feeling I'm>missing something obvious...This is usually done by contour integration, which we will do below.However, let us try fixing up your idea. Let |oo 2 2 A+iB = | (cos(x ) + i sin(x )) dx [1] |-ooThen we can try the same trick that worked for exp(-x^2): 2 (A+iB) |oo |oo 2 2 2 2 = | | (cos(x ) + i sin(x ))(cos(y ) + i sin(y )) dx dy |-oo |-oo |oo |oo 2 2 2 2 = | | (cos(x + y ) + i sin(x + y )) dx dy |-oo |-oo |2pi |oo 2 2 = | | (cos(r ) + i sin(r )) rdr d@ | 0 | 0 |oo = pi | (cos(r) + i sin(r)) dr [2] | 0However, this integral does not converge, so I don't think this methodwill work. Let's try contour integration.Let w = (1-i)/sqrt(2), so that w^2 = -i. Let z = wx. Then |oo 2 2 | (cos(x ) + i sin(x )) dx |-oo |oo 2 = | exp(ix ) dx |-oo 1 |oo w 2 = - | exp(-z ) dz [3] w |-oo wThe path of integration is a line through the origin with slope -1.For x > 0, along the vertical path from the path of integration in [3]to the x-axis, | |0 2 | | | exp(-(x+iy) ) i dy | | |-x | |0 2 2 <= | exp(-(x - y ) dy |-x |x 2 2 = | exp(-x + (y-x) ) dy | 0 |x <= | exp(-xy) dy | 0 <= 1/xA similar situation holds for x < 0. Since exp(-z^2) has no poles andthe integrals along the paths connecting the path of integration and thex-axis vanish at oo, we have [3] 1 |oo 2 = - | exp(-z ) dz w |-oo 1+i = ------- sqrt(pi) sqrt(2) = (1+i)sqrt(pi/2) [4]Thus, [4] tells us that |oo 2 |oo 2 | cos(x ) dx = | sin(x ) dx = sqrt(pi/2) |-oo |-oo === Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read.> My web site www.johncho.us has all the information. I am seeking people > who are familiar with procedures. I am also considering getting a lawyer.As you insist on thinking legally, you ought to find out the officialpolicy of your college and post all the relevant details here. Include allportions which refer to its expectations of you. If you just ask us for anuninformed opinion you will, IMHO, mainly get two kinds of answers, those of us who teach undergrad classes who think you should shut up and those who take undergrad classes and think you're hard done by. === Subject: Re: Axioms defining a finite field> No, not trivial . This is Wedderburn's theorem; Proofs from the Book gives> a nice (what else?) four page proof (chapter 5) due to Witt (and mention 7> or 8 more, using quite different ideas). Witt begins by proving, using> linear algebra, that the centraliser of s , is of dimension q^(n_s) , where> q is the characteristic of F (q.1=0) and that n_s divides |F|. Then, he> imbeds F in the roots of unity in C, and conclude by a clever argument on> the cyclotomic polynomials...There are a couple of proofs at PlanetMath.org: http://planetmath.org/?op=getobj&from=objects&id=3627 http://planetmath.org/?op=getobj&from=objects&id=4198Neither seem to be credited, but the second one looks like it could be theWitt proof you describe, or something related to it. -- --Tim Smith === Subject: Re: Quadratics and transformation> That is, if I have 3x^2 + 7x + 2, can I easily relate the coefficients> and constant to the dilations, reflections and translations that this> function represents compared to x^2 ?> Yes. 3(x^2 + 7/3*x) + 2, complete the square inside the parentheses. So> you end up with a(x-b)^2 + c. b is translation along the x-axis, c along> the y-axis, a is the dilation factor. If a is negative, that's a> reflection.> Jon,refreshed myself on completing the square that is :)Having solved one of lifes little mysteries I shall now wander off to contemplate polynomials in general :) ... this, I suspect, will be rather more difficult but the fun is in the hunt :)Ivan. === Subject: Re: Quadratics and transformation< Be careful when substituting A,B & C are all signed> CarlI appreciate the now snipped assistance.I think it's mostly just carelessness but whatever the reason I manage to mess up signage nearly every time! I suspect your advice was useful to me in the test I sat this morning :)Ivan. === Subject: Re: Axioms defining a finite field> Unless I've missed something...Doesn't look like it...so does this mean that a+b=b+a doesn't need to be inthe field axioms?It is annoying enough that the group axioms usually have a*e=e*a=a anda*a'=a'*a=e, when you only need a*e=a and a*a'=e, and I've just barelymanaged to surpress my rage at that...this field thing is going to pushme over the edge. :-)-- --Tim Smith === Subject: Re: covering compact set w/squares>Given a compact set K in the plane s.t. each pt x is the center of a square>Q_x, prove that you can find a subsequence Q_x_i of squares s.t. K is>covered by the Q_x_i and>sum(over i) Char(Q_x_i) <= 4 Char (union(over i) Q_x_i),>where Char(X) is the characteristic (indicator) function of X.>>Are you assuming the squares are all oriented with sides parallel to the >>axes? >I don't know whether _he's_ been assuming that, but I have been.>Suppose for simplicity we want the point [0,0] to be in the >interior of a region covered by more than 4 squares, such that none>of the centres of the squares is contained in a different square.>Then at least two of the centres will be in one of the four quadrants;Ah - there we go.>WLOG the first quadrant. And by translating slightly, we can assume>those centres are in the interior of the first quadrant, at>[x_1,y_1] and [x_2,y_2]. If 2 r_1 and 2 r_2 are the sides of the two >squares, then > 0 < x_1, y_1 < r_1 > 0 < x_2, y_2 < r_2>Since [x_1,y_1] is not in the square centred at [x_2,y_2], we need>x_1 >= x_2 + r_2 or y_1 >= y_2 + r_2; let's say x_1 >= x_2 + r_2.>Then r_2 <= x_1 - x_2 < x_1 < r_1.>But the same argument, interchanging subscripts 1 and 2, shows >r_1 < r_2, contradiction.So we just need to get a subsequence such that no center iscontained in one of the other squares, don't need to worry aboutdecreasing size as I was yesterday. Very good.>Robert Israel israel@math.ubc.ca>Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2************************ === Subject: Re: Minimally simple finite groups?>Which of the finite simple groups are minimally simple, i.e.,>have all of their proper subgroups solvable? Obviously the only>alternating group that qualifies is A_5 =~ L(2,4) =~ L(2,5), and>I know the list also includes L(2,7) =~ L(3,2), L(2,8), L(2,13),>... On the other hand, I also know it *doesn't* include L(2,9) =~>A_6 or L(2,11), both of which contain subgroups isomorphic to>A_5.The classification of minimal simple groups was a consequence ofJohn Thomspon's classification of nonsolvable groups in which nontrivialsolvable subgroups have solvable normlaizers, which was one of the bigclassification theorems that came before CFSG.This is in Bull. Amer. Math. Soc. 74, 1968, 383-437.The simple groups coming out of Thompson's Theorem are L_2(q), Sz(q),L_3(3), M_11, A_7, U_3(3). Of course, these are not all minimal simple.L_3(3) is, but M_11, A_7, U_3(3) are not.Sz(2^e) is minimal simple whenever it does not contain a smallerSz(2^f), which I guess is equivalent to e being prime.For L_2(p^e) to be minimal simple, its order must not be divisible by 60(otherwise it contains A_5). Also it must not contain any smallersimple L_2(p^f), so if p>=3, then we must have e=1. But L_2(2^e) andL_2(3^e) will be minimal simple for e prime provided their order isno divisible by 60.Something like that, anyway!Derek Holt. === Subject: Re: I got low score on math test, please advise me and take a look>the following argument did not work with the dean of student affairsWhy are you telling us this? You didn't notice that you got a large number of replies, of which essentially none expressed any sympathyfor the idea that you should be allowed to drop after the drop date?>this school is really horrible. i can't believe the screwed up procedures >at it.Yeah, sounds that way. They have a rule that you can't drop a classafter a certain date, and then they actually don't let you drop a class after that date! Simply incredible.What planet are you from, by the way?>this is the arugment:>Dear Dr. Johnson Jr.,> I am requesting that I be excused this one time to be withdrawn >This was one day after the date of withdraw without a W. The class is Math >170, ticket # 3098, and meets on Monday and Wednesday from 6:00PM-8:00PM. >Again, I did not have enough information to withdraw by February 17th since >there was absolutely no graded work or assignments that were returned back >by February 17th, one day before the exam results were made available. >Please, I ask that I be excused this one time to be withdrawn without a W ************************ === Subject: Re: Letter to Prof Ullrich and others>I am just curious: why are people like you>interested in even acknowleding James Harris and his ilk ?>This guy has been oopsing on sci.math for ages (at least several>years) I see. >Is it not worthwhile for everyone to just ignore him at this point ? >There is real danger of course that someone would take him seriously,>but I feel thats remote.>It seems that he has taken to calling universities to complain. He's called universities with totally bogus complaints in the past,later admitting here that the reason he did so was to get someone(me, actually) to stop replying to his posts. Seems to me that it'simportant that he be absolutely certain that that sort of intimidationwill not have the effect that he wants. (Because although thepeople here just laughed at the complaints someone elsewheremight not be so lucky.)>Lets not giving him any more attention. Without attention, he will>shrivel up and disappear.>Is there a history or an obvious reason that I am missing ?************************ === Subject: Re: Probability Instinct - Sufficiently Discrete to Survive in the JungleI'm not sure I understand where all of this is going. You assert thatscientists have never directly observed causality. Are youasserting that causation does not exist?I'm also not clear on where your essay on probability is leading. Ifthe outcome of events are not known beforehand, what difference doesit make whether chance is real or percevied? If an observer doesn'tknow ahead of time, he doesn't know. Even were one to grant thatthere is no real randomness in the universe, we are still saddledwith the fact that we possess limited sense faculties, which providenoisy, incomplete and even self-contradictory information. Probability gives us tools for dealing with such uncertainties.-Will Dwinnellhttp://will.dwinnell.com === Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read.>My web site www.johncho.us has all the information. I am seeking people >who are familiar with procedures. Did you bother to read any of the replies you got in the first thread you started on this topic? (If no: if you don't read replies why bother posting? If yes: If you read _those_ replies why wouldyou think you're going to get anything out of this second post?)>I am also considering getting a lawyer.Good luck with that. Since you've given us absolutely no evidencethat any school rules or procedures were violated I doubt you'regoing to find a lawyer interested in taking the case on a contingency basis. If you've got plenty of your own money I'mcertain you'll be able to find lots of interested lawyers...************************ === Subject: Re: 1=ma+nb>Virgil> m,n belongs to Z> a and b is realtively prime.>> Is there any proof of this? or just an axiom?>...> See, for example:> http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm>A similar procedure, called Blankinship's algo, goes like this, for a=39 and>b=23:>39 1 0>23 0 1>16 1 -1>7 -1 2>2 3 -5>1 -10 17 ;and now 1 = -10*39 + 17*23>The first two rows are>a 1 0>b 0 1>and each subsequent row is a linear combination of the two previous rows, as>in Euclid's algo. When the leftmost column reaches gcd(a,b), you can read>off m and n.Last April, I developed this algorithm and called it the Euclid-WallisAlgorithm:Bill Dubuque mentioned that he had previously posted this algorithm,which he called the Extended Euclidean Algorithm: .I see that Blankinship published this algorithm in 1963 and it has anentry at MathWorld: === Subject: Re: two other sides of triangle> Unfortunately, since the method above needs to solve 2 quadratics, we> get four values, not two. Here is another method, similar in idea, but> only requiring the solution of one quadratic, giving the two sides that> were requested.> >I said that the first quadratic was remarkably simple. If you actually>work it out, you will see that one of its roots corresponds to ab = 0,>which is obviously not the one we are looking for. Therefore the first>quadratic is not really a 'quadratic' but a linear equation, and there>is no ambiguity.Since the first quadratic could be factored so easily, it made sense todivide by the irrelevant factor leaving a linear equation. That is justwhat I did. === Subject: Re: the anticlassicalist }{ vi: into the quantumRiddle of the day: are ontologists for real?> Ah.... ontology is about existence. L'tre et le> nant (sounds much better, doesn't it?)How am I to know? What's that capital sigma after and the theta after ?> Can something that does not exist be ontological?Ya, I imagine so. How is real reality different than reality?> Does an ontologist need to exist in order to> ... er... be an ontologist?What if a thought was that no ontologist thought?> (Never mind what an ontologist is, I have a hunch that you could argue> the same of a communist, a capitalist, a typist, a writer, a plumber, a> feather).Does the corrupt president be need to be elected to be corrupt?Riddle of the day: what do ontologists imagine reality is? === Subject: Re: the anticlassicalist }{ vi: into the quantum> Riddle of the day: are ontologists for real?Ah.... ontology is about existence. L'.90tre et len.8eant (sounds much better, doesn't it?)> How am I to know? What's that capital sigma after > and the theta after ?A sign your newsreader isn't using the charset directions in theheaders correctly - the post was in iso-8859-1, and those were lettersfound in the Frenchy-French, into which Jacques suffers occasionalrelapses.[...]> Riddle of the day:> what do ontologists imagine reality is?An illusion. Lunch-reality doubly so.Deswill contemplate _l'.90tre de l'.8etant_ for food.-- [T]he structural trend in linguistics which took root with theInternational Congresses of the twenties and early thirties [...] hadclose and effective connections with phenomenology in its Husserlianand Hegelian versions. -- Roman Jakobson === Subject: Simple idea, mathematics and common-senseI've had a time explaining some *very* simple mathematical ideas thatlead to a few complexities, but it's been fruitful to explain, or tryto explain, as I work to figure out why these simple ideas eitherexcite derision, anger or confusion, and I think I have it figuredout.I'm sure many of you are put off by mathematics, so I assure you upfront that what I'll be talking about will mostly be *very* simple,and there will only be a few slightly complicated things at the end.First of all, I'm going to talk about a case where mathematicians gaveup because they couldn't see something, and assumed that because theycouldn't see something it didn't exist!You know how with simple quadratics like x^2 + 3x + 2, it's easyenough to see factors of 2 in the roots?I mean, it's just (x^2 + 3x + 2) = (x+2)(x+1), and there they are.However, if it's something like x^2 + 7x + 2, you can use thequadratic formula and get the roots to findx = (-7 +/- sqrt(41))/2and who can see factors of 2 in that thing?There's something else important here which is the ambiguity of thesquare root operator, which may sound complicated but it's easy todemonstrate with another root of a quadratic:(1+sqrt(9))/2and you may think, silly, why show sqrt(9) when sqrt(9) = 3, but yeah,that's *one* of its solutions, as sqrt(9) = -3 as well, so you have*two* numbers(1+sqrt(9))/2 = 2 or -1as either solution will work. I've had people argue with me that bydefinition (really by convention) you take the positive root. Butimagine the world of Contrary. On Contrary the mathematicians forsome odd reason *by definition* take the negative! Is mathematicsreally changing depending on such decisions.Imagine you've forgotten that you can resolve sqrt(9) to 3 or -3, soyou write these numbers like (1+sqrt(9))/2 and (1-sqrt(9))/2 andmercifully discover that you can get rid of that square root sign byadding them together, or multiplying them together.Like adding them gives 1, and multiplying them together gives (1 - 9)/4 = -2and your mathematicians scratched their heads and contemplated suchnumbers, and decided that there was *no way* to understand factors of2 of numbers like(1+sqrt(9))/2 and (1-sqrt(9))/2 except as to consider them to be unique factors of 2, in some kind ofmysterious way.But wait, that's not a problem here, of course, because you can justevaluate the square root, but look back now at x^2 + 7x + 2, wherex = (-7 +/- sqrt(41))/2and consider that you *cannot* resolve sqrt(41) in any way that willhelp you here with this question.Sure, you can write it out in decimal format with a lot of numbersafter the decimal place. My computer tells me that sqrt(41)approximately equals 6.4031242374328486864882176746218, but of course,you can keep going out to infinity trying just to see sqrt(41).If you drop some of those numbers (to make it easier) and move thosedecimal places, to get 64031242374, squaring gives4099999999957933155876, which looks VERY CLOSE to 41 * 10^20, and youcan see what our approximations actually are.We approximate irrational numbers like square roots by finding someREALLY BIG natural number, and moving the decimal place to the left.However, you STILL don't have a simple idea of where factors of 2 go,like you had with the easy and you now might think comforting exampleof(x^2 + 3x + 2) = (x+2)(x+1).Now then numbers like (-7 + sqrt(41))/2 defy our ability to analyzebecause we really, really like integers, but to get an integer youhave to use (-7 - sqrt(41))/2, as then you can add them together toget -7 and multiply them to get 2, but mathematicians could not figureout a way to handle such numbers in less than pairs!That's important. Mathematicians could never figure out a way tohandle such numbers in less than pairs.So some of them decided that what they couldn't see, wasn'tmeaningful.It'd be kind of like the weird mathematicians, who couldn't evaluatesqrt(9), and were looking at (1+sqrt(9))/2 and (1-sqrt(9))/2 decidingthat where factors of 2 resided was a mystery to them. But we *can*evaluate sqrt(9), but we *cannot* really evaluate sqrt(41) whichleaves a mystery with (-7 + sqrt(41))/2 and ((-7 - sqrt(41))/2, andsome mathematicians have decided that that's it.For them, that's all you need to know. Here are these numbers wherewe can't evaluate the square root. Sorry, no go there, they decided,you're stuck, isn't it obvious?You see they decided that the limit on what they could *see* was alimit on what could mathematically exist for those examples--irationalroots--where they could not see. As far as those mathematicians wereconcerned, end of story.But that's where my story begins.My mathematical research is about how the kind of patterns you seewith integers, like with(x^2 + 3x + 2) = (x+2)(x+1)where one root has all the meaningful factors of 2 *continues* intorealms where we can't see it directly because we can't get past thosedanged irrationals coming in at least pairs!I found an *indirect* way of looking, which involves puttingexpressions like these polynomials I've shown here, but morecomplicated into a special but VERY simple mathematical tool.For example,a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)can be peered into, by figuring out that its roots can be consideredin the following mathematical structure:(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where the a's are the roots of that complicated looking cubic that Istarted with, and I've managed to place them in a structure opposite apolynomial that has 49 as a factor.In one sense that's easy. You can take just about any expression likethat, focus on some factor of its last term, and build something likewhat I did. And it being so easy may explain some of the problems I'mhaving with people taking it seriously!What I figured out though is that what mathematicians couldn't seebefore can now be logically seen, and in fact, the way you divide offthat 49 is to get something like(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22but some vocal mathematicians don't like the idea as it contradictswhat they've been taught, and believed because by their thinking andtraining, you can't get around the barrier.Notice that I picked two roots arbitrarily and in fact even if yousolve for the roots you can't look and tell *which* roots should bedivided by 7. All you know is that two of them should be, but cannever know *which* two.Which makes my job of explaining harder, but it's just that ambiguitything coming back into place. We like integers. The math doesn'tcare as it just handles numbers.We can't look *directly* at the roots, but we can use logic toconsider them. What we can't see directly here *does* exist becausethe mathematics says it does.To some extent, my problems are with people and mathematicians whotrust logic less than what they've been taught and their own personalsense of what makes since.In a nutshell that's the basis for the arguments.I say that just because we can't see the factors of irrational rootslike we can with rational ones our limitation is not a mathematicalconstraint, while some people, including a lot of very obsessiveposters, refuse to acknowledge the possibility, fighting for their oldview.Is it important?Actually, it is important if mathematicians have an error in theirthinking.It may be the case that things thought to have been proven inmathematics, really haven't.It's quite possible that any number of arguments claimed to be proofsassumed that what they couldn't see, could not exist, as this issuehas been around for over a hundred years.Yes, I can talk it all out rigorously and in a heavily mathematicalformat, but I hope that giving you some idea what all the fighting isreally about, will help in understanding what's going on, as well aswhy I don't just capitulate.Mathematicians may think that seeing is believing, but I think inmathematics, logic is king. === Subject: Re: Can something be undeterminable?> Has it been (or can it be) proved whether there exist questions> that are *absolutely* unanswerable in the sense of there being> no sufficiently powerful mathematical system to answer them?> --r.e.s.So, you are a Platonist? === Subject: Re: I got low score on math test, please advise me and take a look> the following argument did not work with the dean of student affairs> this school is really horrible. i can't believe the screwed up procedures > at it.Oh you poor put-upon little rich spoiled brat. I simply can'tunderstand why there's an institution out there who won't cave in toyour every little stupid whim. They must be 'really horrible.' === Subject: Re: Can something be undeterminable?, Jim> Any statement only really has meaning within a system, does it> not? A system has to be fixed first, then the statement is made> within that system. The property mentioned is that any > (sufficiently strong) system will have unprovable statements.Perhaps it is a statement about natural numbers...Some folks (Platonists) believe that the natural numbersexist, independent of any system we use to describe them.And that such statements must be either true or false.Goedel showed, then, that no system completely capturesthe true and false statments, though.Now that I said this, I say that the answer to the originalquestion is NO...> Has it been (or can it be) proved whether there exist questions> that are *absolutely* unanswerable in the sense of there being> no sufficiently powerful mathematical system to answer them?Suppose P is some statement. I claim it is not absolutelyunanwerable in the sense given. In fact, I claim thereis a system that answers it. I can in fact list two systems,and one of them is the system we want...SYSTEM I, contains the axiom P.SYSTEM II, contains the axiom not-P.One of those is a system that correctly determines truth or falsity ofstatement P. But we don't know which system it is. === Subject: Re: Axioms defining a finite fieldSorry for intendedly top-posting: hereafter you find a full quote ofso... (see below!) === >Subject: re:Axioms defining a finite field>Your example violates the last rule. Pay attention.> Posted Via Usenet.com Premium Usenet Newsgroup Services>------------------------------------------------------ ----> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **>---------------------------------------------------------- > http://www.usenet.com...Huh?!?Whom/what does you comment apply to?Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc === Subject: Re: help is needed!!!> i want play whit MAPLE but i can not find sourse for it > if any body know some sourses for it or any one can help and support> me please say moreMaple is a commercial product. You may as well ask for the sources forMicrosoft Word.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Genetics and Math-AbilityThere is an old saying about inheritance of mathematical ability.Unlike most interitance, this goes from a man to his son-in-law.The explanation being, I guess, that when the professor's daughter isof the right age, she meets the professor's current student, andmarries him.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: I got low score on math test, please advise me and take a look> my website states my case and has jpg files of the four pages of the test> please take a look and advise me or give me opinions> http://www.johncho.usYou got a 77 on your first calc test. A brilliant woman, advanceddegrees in Divinity, self-taught in Greek, told me that the only thingthat kept her out of Phi Beta Kappa was a 69.5 average in first-yearGerman (instead of a 70). You are not badly off.If you want to remediate a W in Calc I, work with someone who is doingwell in the course and then take it over. Do exercises -- somethinglike Calculus on Web at www.math.temple.edu -- and you increase yourchances.David Ames === Subject: Re: 1=ma+nb> snip> A similar procedure, called Blankinship's algo, goes like this, for a=39 and> b=23:>snip> and each subsequent row is a linear combination of the two previous rows, as> in Euclid's algo. When the leftmost column reaches gcd(a,b), you can read> off m and n.> LHThis is great and much more suited to a spreadsheet than the algorithmI currently use.It is also obvious that it works, since lhs=a*rhs_1+b*rhs_2 is aninvariant on each line, and is guaranteed to reach lhs=gcd(a,b) byEuclid.Who or why or when or what is Blankinship? Good on him/her.JJ === Subject: Re: I got low score on math test, please advise me and take a lookTim,I had run into a few professors throughout my college years who were asarrogant and utterly unthoughtful (in the analytical way, not in the loveydovey way) as you are; luckily, I do not find quite so many in graduateschool. Allow me to explain: Your students are purchasing a service fromyou, if the course title is Calculus and the description in the handbook is'Introduction to Calculus. Topics include convergent sequences, limits,differentiation [... insert additional Calculus topics here ...]' then yourstudents are purchasing a course in Calculus. Your comments about markingstudents wrong when it takes additional time to read their answer and yourpersonal vendetta against students squeezing work in the margin when thereis a back page are fine for your own time - in fact, you can even explain toyour students that you, personally, like this and don't like that. But thesecond that you take points off of a student's grade for work that you don'tlike, you are commiting fraud and breaking the law in the worst way.You see, not only have you, mid course, decided that Calculus will no longerbe the topic of discussion, but (addressing the 'in the worst way' comment)your misleading of the students and change of topic may very well destroy astudent's whole future. Imagine the scenario where a student is geneticallypredispositioned to write in margines, but is Godel smart in mathematicallogic and wants to study at Princeton University. A grade of 'F' in the'Test taking, my way' class that the student inadvertently signed up for(entitled 'Calculus') could very well destroy these chances.I don't expect you to understand the above argument, but perhaps you'llunderstand this and copy the behavior. Whenever I ran into a professor whothought so highly of themselves that they dared to dictate the format that astudent could write their test answers, above and beyond the standard 'showyour work, write so that I can read it and circle your final answer' (butnever 'look to see if there is a back page, if so: do not write in marginif not: blah blah blah' etc.), they were always very untallentedtheoretically. That is, they would never be able to understand the'theoretical' scenario that I proposed above, they would never be able tounderstand the fact that they are 'theoretically' changing the course topicand commiting fraud (false advertising). This lack of tallent invariablycarried over into mathematics (or computer science, depending upon thecourse) as well. I never ran into a very smart professor who did this sortof thing. Perhaps you'll want to be thought of as smart, too, and get downfrom your high horse.Please note, I think that this 'Spockie Hendrick' guy is a pathetic whineyexcuse for a student who refuses to read his handbook which, almostcertainly, states that a student can't drop a corse past the drop date forany reason whatsoever. This post has nothing to do with 'Spockie Hendrick'... This is only addressed to Tim and his arrogant thought processes thattell him that his way of test taking is right and must be spread around theworld.my website states my case and has jpg files of the four pages of thetest please take a look and advise me or give me opinionshttp://www.johncho.us> Aside from the all the politics of whether the grade was fair and if the> test should be returned in time, I have a few comments.> First, for all the instructors out there, how many of you take of 1/4 of> a point? To me, if you make the problem worth 8 points, anything less> than perfection is 7. And, when I am assigning point values to> problems, I usually look over the problem and pick out the two or three> concepts being tested in that question and give each of those concepts a> value of 1 or 2 or even 3 points. I just can't igaine trying to grade> with fractional points. But, that is just my grading style and I'm not> really saying anything is wrong with giving fractional points,I just> find it uncommon.> Secondly, if I am grading and I cannot follow the work (perhaps because> it is not neatly written), I have a hard time giving full credit. That> might sound harsh, but I have often found that students write two or> three possible answers for the problem in their mess and never clearly> mark which one they want me to grade. Also, judging from your scans of> the page, it looks like you had the back of each page left blank (which> is pure speculation on my part). I despise it when students try to> squeeze in something in the margins, leaving the whole back side of a> page blank. You aren't going to use that page for anything else, why> squeeze things together? Now, if you really want to get your instructor> mad, you could ask him (or her) to please write the comments on your> test neater. But that would serve no purpose otehr than to point you> out as a snot-nosed brat.> Okay, about the grading. Personally, I don't know what topics were> stressed in class so I don't feel like I can make a judgement call. For> example, I gave a test in Trigonometry II on Tuesday (which I am in the> process of grading and won't be given back until Monday). For this test> the material I stressed was changing degrees to radians, law of sines,> and law of cosines. On one question I asked about the area of a> triangle. If a student accidentally used the formula area = base *> height, I will not take off many points probably one out of 10 points,> because they were not being tested on geometry (and had they asked me> the correct formula, I would have told them, since that wasn't the topic> they were being tested on). Or, maybe another example, if someone> computed 2+3=6 in part of a larger problem, multiplying instead of> adding, and they correctly carried that mistake through the problem> (that is if they had used 5 they would have gotten the correct answer) I> would probably only take off one point out of 10.> Those are my thoughts. Your grade, suck it up. If you ask the> instructor to look over the test again, he or she might find some places> when you should hae lost more points that you did. At least, if someone> quibles over two or three points I say I will look it over, and I do and> make sure I *took off* all the points I should have.> - Tim> -- > Timothy M. Brauch> Graduate Student> Department of Mathematics> Wake Forest University> email is:> news (dot) post (at) tbrauch (dot) com === Subject: Re: Simple idea, mathematics and common-sense>I've had a time explaining some *very* simple mathematical ideas that>lead to a few complexities, but it's been fruitful to explain, or try>to explain, as I work to figure out why these simple ideas either>excite derision, anger or confusion, and I think I have it figured>out.It's very simple. This is one of the many things that you neverseem to grasp, no matter how many times it's explained, nomatter how many people explain it:The reason people contradict you is because you're simply wrong.The reason people find you hilarious is because no matter howmany times you finally admit you were wrong, the next time youhave something to say you always insist that you understandthe math better than _every_ professional mathematician onthe planet. (Sometimes that's just implicit in the things yousay, sometimes you actually say so explicitly.)The reason people get angry with you is that you're sucha ing asshole.>I'm sure many of you are put off by mathematics, so I assure you up>front that what I'll be talking about will mostly be *very* simple,>and there will only be a few slightly complicated things at the end.And condescending to boot.>First of all, I'm going to talk about a case where mathematicians gave>up because they couldn't see something, and assumed that because they>couldn't see something it didn't exist!>You know how with simple quadratics like x^2 + 3x + 2, it's easy>enough to see factors of 2 in the roots?>I mean, it's just (x^2 + 3x + 2) = (x+2)(x+1), and there they are.>However, if it's something like x^2 + 7x + 2, you can use the>quadratic formula and get the roots to find>x = (-7 +/- sqrt(41))/2>and who can see factors of 2 in that thing?>[...]>Like adding them gives 1, and multiplying them together gives >(1 - 9)/4 = -2>and your mathematicians scratched their heads and contemplated such>numbers, and decided that there was *no way* to understand factors of>2 of numbers like>(1+sqrt(9))/2 and (1-sqrt(9))/2 >except as to consider them to be unique factors of 2, in some kind of>mysterious way.>[...]>Now then numbers like (-7 + sqrt(41))/2 defy our ability to analyze>because we really, really like integers, but to get an integer you>have to use (-7 - sqrt(41))/2, as then you can add them together to>get -7 and multiply them to get 2, but mathematicians could not figure>out a way to handle such numbers in less than pairs!>That's important. Mathematicians could never figure out a way to>handle such numbers in less than pairs.>So some of them decided that what they couldn't see, wasn't>meaningful.>It'd be kind of like the weird mathematicians, who couldn't evaluate>sqrt(9), and were looking at (1+sqrt(9))/2 and (1-sqrt(9))/2 deciding>that where factors of 2 resided was a mystery to them. But we *can*>evaluate sqrt(9), but we *cannot* really evaluate sqrt(41) which>leaves a mystery with (-7 + sqrt(41))/2 and ((-7 - sqrt(41))/2, and>some mathematicians have decided that that's it.>For them, that's all you need to know. Here are these numbers where>we can't evaluate the square root. Sorry, no go there, they decided,>you're stuck, isn't it obvious?>You see they decided that the limit on what they could *see* was a>limit on what could mathematically exist for those examples--irational>roots--where they could not see. As far as those mathematicians were>concerned, end of story.>[...]>What I figured out though is that what mathematicians couldn't see>before can now be logically seen, and in fact, the way you divide off>that 49 is to get something like>(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22>but some vocal mathematicians don't like the idea as it contradicts>what they've been taught, and believed because by their thinking and>training, you can't get around the barrier.>[...]>We can't look *directly* at the roots, but we can use logic to>consider them. What we can't see directly here *does* exist because>the mathematics says it does.>To some extent, my problems are with people and mathematicians who>trust logic less than what they've been taught and their own personal>sense of what makes since.>In a nutshell that's the basis for the arguments.>I say that just because we can't see the factors of irrational roots>like we can with rational ones our limitation is not a mathematical>constraint, while some people, including a lot of very obsessive>posters, refuse to acknowledge the possibility, fighting for their old>view.>Is it important?>Actually, it is important if mathematicians have an error in their>thinking.>It may be the case that things thought to have been proven in>mathematics, really haven't.>It's quite possible that any number of arguments claimed to be proofs>assumed that what they couldn't see, could not exist, as this issue>has been around for over a hundred years.>Yes, I can talk it all out rigorously and in a heavily mathematical>format, but I hope that giving you some idea what all the fighting is>really about, will help in understanding what's going on, as well as>why I don't just capitulate.>Mathematicians may think that seeing is believing, but I think in>mathematics, logic is king.What a truly awesome load of crap. _You_ are the person who'sbeen making ridiculous statements about how factors that youcan't see cannot exist.Hint: Just because you don't understand something does notimply that it's not understood by mathematicians.This really is a new twist. For newcomers, what's traditionalis something like this:James: Black is white.Someone: Uh, no, black is black and white is white - they'reactually different colors entirely.[someone else inserts quibble about whether they're really_colors_...]James: It can be hard for me to explain why black is white,because you think if you didn't learn it in school it can't be right.Someone: Huh? Black is white.James: That's what people thought until the Object Ring was invented.Someone: This is ridiculous. Why are you ignoring the proofs thatpeople have showed you that black is not white?James: OFF!!!!!!!! Your replies are NOT wanted - how manytimes do I have to say that?Someone: Here's an extremely simple proof that black is not white:James: I've contacted your university, informing them of youracademic fraud. I'm considering hiring a lawyer.Someone: Uh, right. That really doesn't change the fact thatblack is not white.James: Liar. Doesn't anyone here care about mathematical Truth?And so on for months or years, until finallyJames: Well, I was wrong, black is not white. No big deal.That's what usually happens - then a day or so later the cyclestarts again, this time a debate over whether red is blue.But this time we have a new twist:James: See, mathematicians have thought that black waswhite. It's kind of hard to see why black is not white - I'lltry to explain it without a lot of math. The reason mathematicianshave thought that black is white is just that that's the waythey were trained...************************ === Subject: Re: . The hardest of all hard facts .> Hi Eleaticus, Re: How you think,> Michelson-Morley and Kennedy-Thorndike do indeed fit> Galilean ( c + v ) physics ,>> All throughout the annals of history ...> No premise has been better tested than this premise:> The speed of light is the same for all observers.>> That makes it: The hardest of all hard facts.>> No experiment ever showed that the speed of light is> the same for all observers. Indeed any determination> of the one way speed of light can be used to demonstrate> the speed of light is NOT the same for all observers....>> A light----> B <-you> < ----------- L --------------> v m/s>> Use syncronised clocks at A amd B to time how long it takes> light to travel a distance of L meters across the laboratory..>> Speed of light relative to the laboratory = L/ (tB - tA) = c> where 'tA' is the time at which the light left A> and 'tB' is the at which the light arrived at B>> Now repeat the experiment while running towards B at v m/s> Note that 'in your frame of reference' the point B is moving ,> so that the light must travel an extra distance = v * (tB - tA)> which is the distance B has moved as the light travels from> A to B.>> Therefore:> Speed of light relative to you = (L+ v * (tB - tA)) / (tB - tA)> = c + v>> keith steinYou had measured the relative velocity between a light ray front and amoving you with v velocity. Notice that the velocities of your twomoving entities are c and v, measured in the Laboratory inertialframe. As c and v refer to the same inertial frame, you have the rightto use ordinary vector algebra obtaining c+v, because Relativevelocity can be >c and is NOT invariant (see my post with thattitle). If you measured the relative velocity in the inertial frameyou is at rest you obtain c as the result,> Perhaps you could show HOW you obtain c as the result, Mr. Hidalgo-Gatocompatible with thesecond postulate of Special Relativity. ANY light continue having thesame vacuum speed c for ANY observer AT REST in ANY inertial frame,> Note that the distance the light travelled relative in the inertial frame in> which you is AT REST is L+ v * (tB - tA) as shown above, and your> hardest of all hard facts is in fact a totally unjustified assumption.> keith steinThe hardest of all hard facts .RVHGAll the distances and times that you use are referred to theLaboratory inertial frame. Respect de inertial frame where you is atrest are valid different ones, every inertial frame has a differenttime and space. I refer you to Einstein's original June 30-1905 paper.In it he DEFINE what simultaneity is for two events that occur indifferent A and B space points. The following are part of Einstein'swords: We have so far defined only an 'A time' and a 'B time'. Wehave not defined a common time for A and B, for the later cannot bedefined at all unless we establish BY DEFINITION that the 'time'required by light to travel from A to B equals the 'time' it requiresto travel from B to A. I recommend you to study in detail thisfundamental paper, near to reach a century from written. You have allthe right to reject Einstein's second postulate stating that vacuunlight speed is the same c for all inertial frames, making your owntheory to model Nature. But do not forget that there exist a hugequantity of experimental data that require explanation. The theorywith the best performace win (until a better one is created), that isthe rule in the scientific game.RVHG === Subject: Re: JSH: Splitting field, algebraic integer factorsPreviously I posted that if you can't *see* the factors betweenirrational algebraic integers then they're not there, but morecorrectly the situation is that two algebraic integers have to bemembers of the same splitting field to have a non-unit algebraicinteger in common.(I say more correctly as there may be some terminology issues herebecause what mathematicians currently call a splitting field is nota true field, but something close, like the field of rationals. Butthat's another issue for another time.)That's a nifty and powerful result. Why am I the one who had todiscover it?> Well I was wrong. I should have known in retrospect, but it was yet> another situation where I kind of decided that I wanted something and> thought I had it, only to find out later I was wrong.Bump the OOPS counter again.> There's no point to hanging on to wrong things though.Since when? What's changed?> Luckily, mathematics is supposed to be an arena where the truth> matters.> James HarrisDavid Moran === Subject: Re: the anticlassicalist }{ vi: into the quantumRiddle of the day: what do ontologists imagine reality is?> An illusion. Lunch-reality doubly so.> Des will contemplate _l'.90tre de l'.8etant_ for food.Before swallowing that check to see if existence of existence exists.> [T]he structural trend in linguistics which took root with the> International Congresses of the twenties and early thirties [...] had> close and effective connections with phenomenology in its Husserlian and> Hegelian versions. -- Roman JakobsonPhenomenally unphenomenal phenomena. === Subject: Re: How many different resistances with n resistors?Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow)>We had a thread in rec.puzzles called 'R = PI ohms' on this sort of topic a>while back. You'll still find it in the archives (e.g. google groups) II found it in the archives and enjoyed reading it. Are you able toanswer the Puzzle Corner question with your program? That is, howmany distinct resistances are achievable with n resistors? In order tocircumvent the subtleties involved in distinguishing between exactlyn resistors and at most n resistors, suppose we focus on the case ofat most n resistors.-- Tim Chow tchow-at-alum-dot-mit-dotduThe range of our projectiles--ven ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Simple idea, mathematics and common-sense> Yes, I can talk it all out rigorously and in a heavily mathematical> format, but I hope that giving you some idea what all the fighting is> really about, will help in understanding what's going on, as well as> why I don't just capitulate.You've never demonstrated any ability to think or write mathematics rigorously. In fact just the opposite seems to be true of your ramblings === Subject: Re: Collatz Conjecture : Symmetry question.> Do you mean proven true? It could be proven false with a single > counterxample.The Collatz conjecture is: The Collatz sequence is allways finite and ends in 1.How then, can it be a counterxample?Luis R. === Subject: JSH: Prime concept in Object RingAs a note to myself, one of the reasons I found myself continuallygoing back to the Decker quadratic was I realized I hadn't worked itall out for polynomials that weren't of a special class. That is, I'dfocused on one particular type of polynomial as that's where I cameinto this area, but the Decker example caused me to focus on otherareas.Apparently most polynomials are prime-like in that you can't separateoff non-unit object factors of their constant terms within the ring,but have to go into the operator space.In retrospect, it's kind of obvious.Now I can move forward to the story of my work versus actively workingon my own understanding, which hopefully will produce some progresshere.Doing active research while trying to explain it as you go along iskind of difficult. === Subject: Re: strain softening spring> >>i.e. the equation I am trying to solve is as follows:>>y'' + cy' + [ko*e^(-alpha*t)]y = 0>>Any help would be most appreciated.>>sincerely>Paul Joseph>(pjoseph@excite.com)>> Maple gives a solution in terms of Bessel functions. I posted a file> called DE.pdf showing the solution at:>> http://math.asu.edu/~kurtz/de/>> --LynnHi Lynn,This is in regard to the solution you kindly posted above.Can the first argument of the BesselY function be negative?> Now, having done the required manipulations by hand[1], I get> u^2 d^2f/du^2 + u df/du + (b^2 - u^2) = 0,> where> u = A(k,a,c)*exp(-at/2), b = c/a, f(t) = y(t)*exp(ct/2)> and A(k,a,c) is whatever it is, which has solutions J_b(u) and Y_b(u).> In any case, I'm not sure that exp(-ct/2)Y(A(k,a,c)*exp(-at/2)) is> actually bounded as t -> +oo for all positive k,a, and c, so that this> solution is therefore unphysical.I tried to implement this in Excel, but Excel's BesselY function appearsto require that the first argument be positive. I am not sure if thisis a quirk of Excel... or whether this is a mathematical requirement.> Conventionally[2], J_b(x) and Y_b(x) are indexed with positive b.> [1]: The original equation was a linear ODE in one unknown, with one> non-constant coefficient. It should not be necessary to throw this at a> computer to solve it, unless you want the actual numerical values.> [2]: In the UK, at least.Thank you very much for this useful information. Most important to meis that the solution may be unphysical - in which case, I am barkingup the wrong tree.Also useful to me was your identification of this as a linear ODE withone unknown and one non-constant coefficient.sincerelyPaul === Subject: Re: Simple idea, mathematics and common-sense[snip James' math lessons for those less magnificent than himself]> But wait, that's not a problem here, of course, because you can just> evaluate the square root, but look back now at x^2 + 7x + 2, where> x = (-7 +/- sqrt(41))/2> and consider that you *cannot* resolve sqrt(41) in any way that will> help you here with this question.> Sure, you can write it out in decimal format with a lot of numbers> after the decimal place. My computer tells me that sqrt(41)> approximately equals 6.4031242374328486864882176746218, but of course,> you can keep going out to infinity trying just to see sqrt(41).Whoa! What do you mean to see sqrt(41)? You are now talking about a*decimal expansion* of sqrt(41). Do you think you have to go out toinfinity with 0.333333333... to see 1/3? How about going out toinfinity with 1.0000000... to see unity?> If you drop some of those numbers (to make it easier) and move those> decimal places, to get 64031242374, squaring gives> 4099999999957933155876, which looks VERY CLOSE to 41 * 10^20, and you> can see what our approximations actually are.Yawn.[snip interminable, rambling which attempts to demonstrate the genius ofJSH, but which actually confirms his idiocy]> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.----http://www.crbond.com === Subject: Re: strain softening spring>Can the first argument of the BesselY function be negative? I tried>to implement this in Excel, but Excel's BesselY function appears to>require that the first argument be positive. I am not sure if this is>a quirk of Excel... or whether this is a mathematical requirement.>Since c and a are both real and positive numbers, -c/a will be>negative...no?>thanks!>Paul> Apparently this is a quirk of Excel (you did notice that Excel> requires the arguments in the reverse order?). You can read more about> properties of BesselY at> http://functions.wolfram.com/BesselAiryStruveFunctions/BesselY /> If you have particular values of a, c, and k, I could plug them in and> post a graph of the solution for you. Probably Maple or Mathematica> would be a more appropriate tool for you.> --LynnThank you very much Lynn for your kind offer. I have a niece who is agrad. student at a local university and I will ask her to use herstudent status to buy a copy of Maple or Mathematica!I am trying to curve fit some data to this equation, so I don't havevlues (yet) of a, c and k.Mr. Smith's comment below that the solution may be unphysical wouldmean that I am barking up the wrong tree - I am trying to solve a verypractical, physical problem - the shearing of soil, which I amconvinced can be described as using dynamical system theory.sincerelyPaul === Subject: Re: errors in an argument by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KEBHn07123;> The way it seems to me (and I'm not talking as an expert in aerodynamics> either) is that even for extremely small and light creatures, the ability> to develop a functional limb that will allow it even the smallest> maneuverability by mutation benefit is unlikely. Remember that it has to> be aerodynamic to some extent, and to be evolutionary beneficial. It's> very hard for me to imagine how such a thing would look like.>Imagine this:>http://www.animalnetwork.com/ critters/profiles/flyingsquirrel/default.asp>Keep in mind that embryos of many species typically have webbing during>development; it would not take much of a mutation for the webbing to>persist.>A little research also turns up an explanation of feathers that has>nothing to do with their eventual use in flight. >-- This looks like a flawed argument. Assuming evolution, we are already at a stage that allows the appearance of wings or semi wings, so this is circular. However, the existence of similar limbs in fish like someone else noted here combined with the huge advantage even the slightest ability for such maneuverability can sound plausible.My problem is more general. Too many times we get a very complicated and fine tunedmechanism that would require a major jump in order to have any effectiveness. Fish thatuse electricity to stun and kill opponents is a remarkable one. The amount of electricitygenerated, and the coordination involving it is very big. It looks like it would take a big number of mutations to achieve it, with any partial mutation or a small effect totallyuseless. A functional yet small weapon of this kind is even punishing given the amount ofenergy needed for it, and lack of effective results. === Subject: functional analysis--separable by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KEBEv07013;, I found the following proposition in a book without a proof. I have difficulty in proving it. Can anyone help? Suppose B is a separable Banach space. Let {x_n} be a countable dense subset of the unit circle C={x in B : ||x||=1}. How can I prove that: every element x in B can be written as: x= summation[from 1 to infinity] {a_n x_n} where a_n is a absolute summable sequence. That's: summation[from 1 to infinity] {|a_n|} < infinity. === Subject: Cavemen by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KEBHw07133;Where can I find pics of cavemens language e.g. A pic of a wall inprint === Subject: Re: Collatz Conjecture : Symmetry question.> > Do you mean proven true? It could be proven false with a single> counterxample.> The Collatz conjecture is: The Collatz sequence is allways finite and> ends in 1. How then, can it be a counterxample?A non-trivial cycle would be a (finitely checkable) counterexample.-- === Subject: Re: Collatz Conjecture : Symmetry question. === >Subject: Re: Collatz Conjecture : Symmetry question.>Message-id: <4035b84c$0$286$edfadb0f@dread12.news.tele.dk>>First: Won't there be duplicates in the large files?There shouldn't be- if the Collatz Conjecture is true- if my program is correctly generating the tree- but I haven't checkedThe program doesn't make any attempt to organize the numbersinto proper branches, just count them. But that can easily bedone by converting the numbers to binary.Every odd number is the start of a branch and every even numberhas an odd seed pattern in binary that is the same as the oddnumber of the start of its branch.The text file L6.txt has just two numbers 32 and 5. In binary, these are100000101Note the odd seed pattern of 16 is 1. At each subsequent level,these binary patterns get a 0 appended to them. At level 16they will have become10000000000000001010000000000All numbers that have the same odd seed pattern are on thesame branch. Since every level has a different number of 0s,there can't be any duplication across levels. And IF each number on a given level has a different odd seed pattern,there can't be any duplication within a level.The purpose of the mod 3, mod 2 requirement is to ensurethat the newly spawned numbers (those generated by (n-1)/3)create a new odd seed pattern that hasn't appeared before (in which case, neither it nor any of its descendants will bea duplicate of any other number). Any number == 1 (mod 3) will give you an integer when you apply (n-1)/3, but you cannot have two consecutive odd numbers in a Collatz Sequence, so odd numbers are not allowed to spawn new branches, otherwise you would get duplications.>Second: The two divmod's can be combined into one.How? My thinking was to do the divmod 3 first since it willfail the test two out of three times allowing the divmod 2test to be skipped.>-Michael.--MensanatorAce of Clubs === Subject: Re: Simple idea, mathematics and common-senseO> I found an *indirect* way of looking, which involves putting> expressions like these polynomials I've shown here, but more> complicated into a special but VERY simple mathematical tool.> For example,> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)> can be peered into, by figuring out that its roots can be considered> in the following mathematical structure:> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> where the a's are the roots of that complicated looking cubic that I> started with, and I've managed to place them in a structure opposite a> polynomial that has 49 as a factor.> In one sense that's easy. You can take just about any expression like> that, focus on some factor of its last term, and build something like> what I did. And it being so easy may explain some of the problems I'm> having with people taking it seriously!> What I figured out though is that what mathematicians couldn't see> before can now be logically seen, and in fact, the way you divide off> that 49 is to get something like> (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22Prove that is true. I think the many counter examples to this might be ahindrance. The only thing you are claiming is that if (f(x)+a)(g(x)+b) is divisible by a, it must be that f(x) is divisible by afor all x where f and g are just two functions from the underlying ring toitself, with no particular special propertes that ensure divisibility, andnothing special about the ring either. Complete and utter crap. It may well be that there is some function thatwill do what you want, but the one you've got ain't it.There is still the x(x+1) counter example in Z. You claim that's because Zis a UFD, well, let j be the root of any square free integer where Z[j] isnot a UFD, such exist. Then x(x+1)(x+j) is divisible by two again, but youcan't tell presume it will divide the same factor always. === Subject: Re: JSH: Prime concept in Object Ring> As a note to myself,Next time just use a PostIt on the refrigerator.> one of the reasons I found myself continually> going back to the Decker quadratic was I realized I hadn't worked it> all out for polynomials that weren't of a special class.The other reason is that you are too stupid to understand the errors inyour own work.> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.----http://www.crbond.com === Subject: Re: JSH: Prime concept in Object Ring> As a note to myself, one of the reasons I found myself continually> going back to the Decker quadratic was I realized I hadn't worked it> all out for polynomials that weren't of a special class. That is, I'd> focused on one particular type of polynomial as that's where I came> into this area, but the Decker example caused me to focus on other> areas.Hmm, so you were wrong for one type of polynomial, but dammit your rightfor the general one! > Apparently most polynomials are prime-like in that you can't separate> off non-unit object factors of their constant terms within the ring,> but have to go into the operator space.Eh? What the buggery does that mean?> In retrospect, it's kind of obvious.something is certainly> Now I can move forward to the story of my work versus actively working> on my own understanding, which hopefully will produce some progress> here.> Doing active research while trying to explain it as you go along is> kind of difficult.> James Harris === Subject: Re: Simple idea, mathematics and common-sense> O> > I found an *indirect* way of looking, which involves putting> expressions like these polynomials I've shown here, but more> complicated into a special but VERY simple mathematical tool.> > For example,> > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)> > can be peered into, by figuring out that its roots can be considered> in the following mathematical structure:> > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => > 49(300125 x^3 - 18375 x^2 - 360 x + 22)> > where the a's are the roots of that complicated looking cubic that I> started with, and I've managed to place them in a structure opposite a> polynomial that has 49 as a factor.> > In one sense that's easy. You can take just about any expression like> that, focus on some factor of its last term, and build something like> what I did. And it being so easy may explain some of the problems I'm> having with people taking it seriously!> > What I figured out though is that what mathematicians couldn't see> before can now be logically seen, and in fact, the way you divide off> that 49 is to get something like> > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) => > 300125 x^3 - 18375 x^2 - 360 x + 22> > Prove that is true. I think the many counter examples to this might be a> hindrance. > The only thing you are claiming is that if > (f(x)+a)(g(x)+b) is divisible by a, it must be that f(x) is divisible by a> for all x where f and g are just two functions from the underlying ring to> itself, with no particular special propertes that ensure divisibility, and> nothing special about the ring either.> Complete and utter crap. It may well be that there is some function that> will do what you want, but the one you've got ain't it.> There is still the x(x+1) counter example in Z. You claim that's because Z> is a UFD, well, let j be the root of any square free integer where Z[j] is> not a UFD, such exist. Then x(x+1)(x+j) is divisible by two again, but you> can't tell presume it will divide the same factor always.correction x(x+1)(x+j)(x+j+1) is divisible by two always. === Subject: Re: Logic by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KErjr10530;Hint for (1) - Unless means the same thing as if not. (2) you should be able to work out on your own.> I'm stuck on these 2 problems.> Represent the following compound propositions, each involving one or>more> of> these three propositons, as symbolic expressions in the variables d,e,f.> 1)e does not f unless d> Does 1 this mean If d and e the f????>Typo: Does 1 mean If d and e the f????> 2)d and e does not imply f> Does 2 mean not d or not e does imply f???> Of coure I am told what these propositions are.> I'm sorry but I can't even show you what I've done so far--I'm>completely> lost.> Steven> === Subject: Re: Simple question regarding frequency detection by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KErjR10534;>I am writing an application to detect a particular frequency(viz. 4.5>Khz) in an input audio signal. I have written the code to capture the>sound using a microphone; but I have to properly design a digital filter>which will return TRUE as soon as it detects sound of this frequency in>the input signal. Also my project demands that I have to detect the>signal by looking at a very small number of samples e.g. 44. So I will>apply the digital filter on a window of 44 samples of the input signal>repeatedly till I get a high response indicating that I have got the>reqd. frequency.>I know DFT/FFT is an option but are there any better algorithms I can>use for my project. Please suggest the best algorithm that will do my>work. I am writing the code in C++ so if anyone can point me to existing>code on the net that will be really helpful.>Atri.The problem is a bit more involved than you indicate.What is the desired probability of detection? What is the maximum allowable probability of false detection?What is the minimum signal-to-noise ratio at the input?etc. etc.phil === Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read.>My web site www.johncho.us has all the information. I am seeking people >who are familiar with procedures. > Did you bother to read any of the replies you got in the first thread > you started on this topic? (If no: if you don't read replies why > bother posting? If yes: If you read _those_ replies why would> you think you're going to get anything out of this second post?)Actually, I think it's becoming increasingly clear that he's just trolling. === Subject: Re: Letter to Prof Ullrich and others>I am just curious: why are people like you>interested in even acknowleding James Harris and his ilk ?>This guy has been oopsing on sci.math for ages (at least several>years) I see. >Is it not worthwhile for everyone to just ignore him at this point ? >There is real danger of course that someone would take him seriously,>but I feel thats remote.>It seems that he has taken to calling universities to complain. > He's called universities with totally bogus complaints in the past,> later admitting here that the reason he did so was to get someone> (me, actually) to stop replying to his posts. Seems to me that it's> important that he be absolutely certain that that sort of intimidation> will not have the effect that he wants. (Because although the> people here just laughed at the complaints someone elsewhere> might not be so lucky.)David Ullrich is a tenured math professor at Oklahoma StateUniversity, and he made *public* statements that I thought reflectedbadly on him and on that university.He has been defensive about those statement for years, so it's easierfor me to direct you to them. You can go to Google Groups, advancedsearch, athttp://www.google.com/advanced_group_search?hl=enand put David Ullrich as the author, and racial slur in the searchfield.Or you can click onhttp://groups.google.com/groups?as_q=racial%20slur&safe=off& ie=ISO-8859-1&as_ugroup=sci.math&as_uauthors=David%20Ullrich& lr=&hl=enOh yeah, in case you're wondering about what Ullrich finds sooffensive that the subject of racial slur came to his mind, I saidthat he'd acted as my lapdog in an instance.That was YEARS ago. === Subject: Re: Simple idea, mathematics and common-sense> You know how with simple quadratics like x^2 + 3x + 2, it's easy> enough to see factors of 2 in the roots?> I mean, it's just (x^2 + 3x + 2) = (x+2)(x+1), and there they are.> However, if it's something like x^2 + 7x + 2, you can use the> quadratic formula and get the roots to find> x = (-7 +/- sqrt(41))/2> and who can see factors of 2 in that thing?Anyone can, if they look at it in the right way. If you denotethe roots by x_1 and x_2, then since x_1 and x_2 satisfyx_1 * x_2 = 2, the roots are both factors of 2. I'll admitthat in one sense one can't see the factors of 2, butif you use the equation that the x's satisfy then you seethe factors of 2 immediately.The problem becomes trickier when you have a number like a = 1 + sqrt(-5)Now I know that that number divides 6, since (1 + sqrt(-5))(1 - sqrt(-5)) = 6so a will have some factor, w, that divides 6. Thus, we'relooking for some algebraic integer w that divides botha and 2. That would be easy if everything in sightwere an integer: we could simply use w = gcd(a, 2), andin the integers it's easy to calculate the gcd.Well, it turns out that gcds exist in the algebraic integersas well, but it's by no means easy to calculate them.Furthermore, it often happens that we can't expectto get an easy solution to, say gcd(a, 2); instead,we often have to look for gcd(a^k, 2^k), for some integer k.This is where the class number comes in, but that'llbe a matter for later discussion.Now we're lucky in this case. Trying k = 2 (which is theclass number of Q(sqrt(-5)), we see that a^2 = (1 + sqrt(-5))^2 = -4 + 2sqrt(-5)and, of course 2^2 = 4Now it's truly obvious in this case that 2 will divideboth a^2 and 4, so sqrt(2) will divide both a and 2,so the factor of 2 we're looking for in 1 + sqrt(-5)happens to be sqrt(2). That's not the only one, ofcourse: -sqrt(2), or sqrt(-2), or -sqrt(-2) willall work equally well, as will infinitely many others.The point here is that while there will always bea way to find the desired factors, it won't alwaysbe easy to do so. As an exercise, you might wantto show that a common factor of 3 and 1 + sqrt(-5)is sqrt(2 - sqrt(-5)), among others. > I found an *indirect* way of looking, which involves putting> expressions like these polynomials I've shown here, but more> complicated into a special but VERY simple mathematical tool.> For example,> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)> can be peered into, by figuring out that its roots can be considered> in the following mathematical structure:> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> where the a's are the roots of that complicated looking cubic that I> started with, and I've managed to place them in a structure opposite a> polynomial that has 49 as a factor.> In one sense that's easy. You can take just about any expression like> that, focus on some factor of its last term, and build something like> what I did. And it being so easy may explain some of the problems I'm> having with people taking it seriously!> What I figured out though is that what mathematicians couldn't see> before can now be logically seen, and in fact, the way you divide off> that 49 is to get something like> (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22Unfortunately, no. The whole point of my example was to demonstratethat in general a factor on the right will be split evenly (insome sense) among the a's. In this case, the 49 on the rightwill almost always turn out to be w_1 * w_2 * w_3, wherew_i are not units and each w_i will be a divisor of a_i and 49.Rick === Subject: Re: Letter to Prof Ullrich and others> I am just curious: why are people like you> interested in even acknowleding James Harris and his ilk ?> This guy has been oopsing on sci.math for ages (at least several> years) I see. > Is it not worthwhile for everyone to just ignore him at this point ? > There is real danger of course that someone would take him seriously,> but I feel thats remote.People like you have come and gone, as your agenda is controllingother people's behavior.Why try?It's Usenet you know. Usenet is known for having people around whojust post or reply, isn't that amazing!And in all that posting and replying there are people like yourselfwho try to control the process.> It seems that he has taken to calling universities to complain. > Lets not giving him any more attention. Without attention, he will> shrivel up and disappear.> Is there a history or an obvious reason that I am missing ?Yeah, the history of Usenet, and I guess you're a newbie, eh?Or you should have known enough not to have made the post you did. === Subject: Re: Collatz Conjecture : Symmetry question.http://arxiv.org/abs/math.NT/0312309I think it's clear that it will never be proven, because there is justnot enough time to prove it.Craig> Do you mean proven true? It could be proven false with a single > counterxample.Yes. === Subject: Re: Letter to Prof Ullrich and others> Oh yeah, in case you're wondering about what Ullrich finds so> offensive that the subject of racial slur came to his mind, I said> that he'd acted as my lapdog in an instance.> That was YEARS ago.Would you mind telling us what is the statute of limitations on your posting debacles?> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable -- and the obvious.----http://www.crbond.com === Subject: Re: Letter to Prof Ullrich and others> Or you should have known enough not to have made the post you did.Speak for yourself. Your record of posting and defending ridiculous errors is legion. You should know betterthan to post any math at all.POSTER ACCURACY INDEX----------------------------JSH 0%MATHEMATICIANS 99%> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable -- and the obvious.----http://www.crbond.com === Subject: Re: errors in an argument>My problem is more general. Too many times we get a very complicated and fine tuned>mechanism that would require a major jump in order to have any effectiveness. Fish that>use electricity to stun and kill opponents is a remarkable one. The amount of electricity>generated, and the coordination involving it is very big. It looks like it would take a big >number of mutations to achieve it, with any partial mutation or a small effect totally>useless. A functional yet small weapon of this kind is even punishing given the amount of>energy needed for it, and lack of effective results.I think you should ask these questions on the newsgroup talk.origins(or some other newsgroup that talks about evolution/biology). I'm surethere are several people there who can explain this to you (to someextend). === Subject: Re: errors in an argument> My problem is more general. Too many times we get a very complicated and> fine tuned mechanism that would require a major jump in order to have any> effectiveness. Fish that use electricity to stun and kill opponents is a> remarkable one. The amount of electricity generated, and the coordination> involving it is very big.The major jump is only in your mind.> It looks like it would take a big number of mutations to achieve it, with> any partial mutation or a small effect totally useless. A functional yet> small weapon of this kind is even punishing given the amount of energy> needed for it, and lack of effective results.Two words: active sensorhttp://www.flmnh.ufl.edu/fish/tropical/JSA/gymno.htmNext time, do your own research.http://www.google.com/search?q=evolution+%22electric+ eel%22-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: No Set Contains Every Computable Natural > So there is no way to determine if a string is infinite?Not according to the scope in this thread. Of course if you have a humanholding both ends of the tape....Actually there's another question, if you have a theoretical peice ofstring, and you have an end in each hand but the middle of it extends toan infinite distance away from you before coming back is it of fintelength? Does such a string even exist?> Why do you believe some strings are finite and other> strings are infinite if there is no possible way to tell> the difference?We can proove the exitence of an infinte set. We can't (by itdefinition) give you a peice of paper with every member listed on it asit is impossible to list the second to last member.-- If you can read this you've gone too far.Stephen Jones (Zombywuf) D5A4 5342 E7BD E524 710A E44F E997 4422 7FEC E44E === Subject: Re: I got low score on math test, please advise me and take a look> the following argument did not work with the dean of student affairs> this school is really horrible. i can't believe the screwed up procedures > at it.> this is the arugment:> Dear Dr. Johnson Jr.,> > I am requesting that I be excused this one time to be withdrawn > This was one day after the date of withdraw without a W. The class is Math > 170, ticket # 3098, and meets on Monday and Wednesday from 6:00PM-8:00PM. > Again, I did not have enough information to withdraw by February 17th since > there was absolutely no graded work or assignments that were returned back > by February 17th, one day before the exam results were made available. > Please, I ask that I be excused this one time to be withdrawn without a W assuming that your report about progress feedback is accurate, arejection of the request above is a good glimpse into an incompetentinstitution. what kind of justification were you offered? somebureaucratic non-sense? === Subject: Re: tensors for tots>How far can one get in understanding the structure of tensors at a>point on a manifold in terms of our old school friends column>vectors, row vectors and matrices?> You'll get bogged down fairly quickly.>Even in general relativity all we have at each point on the >manifold is a real vector space with four components,> No. You also have, e.g., g and R. You could write the components of g> as a matrix, but that would be horribly misleading. R is a lost cause;> you really need to forget about matrices and just think of it as a> tensor.> at 02:28 PM, nulldev00@aol.com (Edward Green) said:>Or for that matter, say we want v* to>be the dual vector corresponding to v> You need a metric in order to do that.>orthogonal transformations> Meaningless without a metric. But, since a metric is nothing but a linear vector space, it's also the reason that Einstein called philosophers idiots, more often than he called Heisenburg an idiot.>I also hope not to get hung up on any terminological problems>between math and physics, perhaps by saying things like we'd like>this object to be invariant, instead of using terms likecovariant and contravariant, to start.> First, you'd need to define what you mean by invariant. Second, you> won't get very far if you don't distinguish covariant from> contravariant, and both from mixed. Since invariant means only one thing: Einstein index convention, it's simple. It's simpler than Turing machines.>Anyway, to take one more baby step, if I have not exhausted my>credits, am I correct in thinking that a reasonable way to introduce>a metric tensor into babyland would be to define the inner product>of a row vector and a column vector, in a particular coordinate>system, to be a bilinear form with a matrix sandwiched in the>middle, that matrix our metric?> No, you don't need a metric for that. You need a metric tensor to> define the inner product of two row vectors, the inner product of two> column vector and the transpose operation that takes row vectors into> column vectors and vice versa. No you need you don't need a metric tensor. Since it's doesn't matter what the hell tensors are to *General Relavity*. Since GR *assumes* that gravitional metrics are equivalent to curved space-time. === Subject: Re: Collatz Conjecture : Symmetry question.>Do you mean proven true? It could be proven false with a single >counterxample.> The Collatz conjecture is: The Collatz sequence is allways finite and ends in 1.> How then, can it be a counterxample?By coming up with a number n that does not reach a power of two. One of two things will be true: The collatz sequence for this n will be unbounded, or it will cycle and never hit a power of 2. If such an n can be found, the Collatz conjecture is false. So the falsity of the conjecture might be proven in a finite number of steps.Bob Kolker === Subject: Re: Simple idea, mathematics and common-sense> James: See, mathematicians have thought that black was white. It's kind of> hard to see why black is not white - I'll try to explain it without a lot> of math. The reason mathematicians have thought that black is white is> just that that's the way they were trained... I have been following, on and off, the postings of Mr. Harris for years.Like you, I used to feel infuriated, for I couldn't believe that anyonecould be so thick. Now that I have a bit more experience on the behaviorof mentally disturbed individuals, I have come to the conclusion that Mr.Harris is a poor lunatic, for whom sci.math is the main means to live hismadness. I guess we should feel compassionate about his regrettable predicament.He is very annoying all right but, what can be expected from a derangedindividual? === Subject: Re: Can something be undeterminable? > The Halting Problem as absolutely unanswerable. No computer > can be programmed to tell, correctly, whether an arbitrary computer > will halt when presented with an arbitrary program.Is it possible to define a system whereby the halting problem is notunanswerable, i.e. starting with there exists a program which can tellif a given program will halt and then go on to create a (non trivial)definition of computer capable of satisying the axiom? Is it provablyimpossible to do this?-- If you can read this you've gone too far.Stephen Jones (Zombywuf) D5A4 5342 E7BD E524 710A E44F E997 4422 7FEC E44E === Subject: Re: Simple idea, mathematics and common-sense> I've had a time explaining some *very* simple mathematical ideas that> lead to a few complexities, Your ignorance, incompetence, and psychosis are not of interest to theworld at large. Quite the contrary. You are not even an interestinglaughingstock.Hey stooopid loud troll James Always in error, never in doubt!Harris, put up or shut up. James Harris, King of the Primes! Whereare your sceptor and crown, delusional James Harris, your regalclothes? Is a $10,000 prize no questions asked too small to justifyyour submission of two little prime numbers? Or are you a psychoticimpotent gelding?http://www.rsasecurity.com/rsalabs/challenges/ factoring/faq.htmlhttp://www.rsasecurity.com/rsalabs/ challenges/factoring/numbers.htmlhttp://www.crank.net/ harris.html It's not every braying jackass that gets a whole page at crank.net--Uncle Alhttp://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: JSH: Non-uniqueness of factorization Discussion, linux)> To be fair I think Outlook does that. Anything behind a : at the start of> a> subject line is treated as if it were Re: and is removed. I'd suggest> switching> to [JSH] which is the convention in other groups and seems to work with> all> newsreaders.> I'm using Outlook now, and it did not do that. It left the JSH: in thereThat's almost encouraging, but Microsoft is so abysmal when it comesto Usenet standards, I am still not persuaded that Outlook doessomething sensible.Maybe Outlook does this: If a subject starts with xxx: blah blah blah,then strip out xxx: and put Re:. Since you replied to Re: JSH: blahblah blah, it would have stripped out Re: and then put Re:,making no effective change.But if you had replied to JSH: blah blah, it would have chosen Re:blah blah as a subject line.I don't know if this is right. I'm just guessing, based onMicrosoft's bad track record (like using different replacements forRe: depending on the language of the user, RFC standards bedamned). So, can you check? Select the root of this thread andcompose a followup (without posting) and let us know what the subjectline says.Also, I wonder if it matters whether you use Outlook or OutlookExpress.-- My proof has been checked very thoroughly, both by me and others.Those others apparently decided that they would not believe the proofwas correct, but cannot support that position using mathematics. Buthey, they're just human beings. --JSH, prover of Fermat's Last Thm === Subject: Re: the anticlassicalist }{ : mitchism for Daleks <40325A2C.21936CE9@hate.spam.net> <103513r3gafocff@corp.supernews.com> <40328829.1E5C341D@hate.spam.net> <1035esdt30h4d38@corp.supernews.com> <1039dr5kpjloif1@corp.supernews.com>>Since Mr. Herring is personally being such a pissant, perhaps he can explain>what is meant by physical content anyway? Does that mean using words and>phrases that depend on Cauchy's (?) epsilon-delta justification for the>derivative or Robinson's non-standard analysis justifying naive use of>differentials?> No, it means that at some point you have to hook into what physicists> *observe*. Will it tell us the energy levels of a hydrogen atom and> predict the Lamb shift? If you're interested in developing a better> mathematical formalism, that's fine, go ahead. Just don't tell us it's> physics.It is not about the formalism. It is about whether assumptions in the formalismthat make it coherent are even being respected.Robert Hermann discusses a particular negative result that was ignored for sometime:Dirac and his successors among the physicists werealways vague (purposely and wisely so, it turned out)about which Lie algebras fo classical observables thiswas supposed to apply to. The question was onlytreated in the beginning of a reasonable mathematicalway by L. van Hove in 1951. He showed that onecould not completely 'quantize' a classical system bythe simple-minded way suggested by the work of thepioneers. [...]Of course, van Hove's negative general result did notconstrain the physicists from using Dirac's idea to setup quantum systems which seemed to be an adequateversion of the classical ones. What they did in practicewas to choose not *all* such classical observables,but a 'physically reasonable' set, and represent it byoperators, with Poisson bracket -- so far as it wasdefined within the set -- going over to operatorcommutator.Your claim translates into a definition for physics as nothing more than achaotically organized aggregate of partial results. If that is the case, you shouldnot be relying on mathematical formalism for justifications--as in the many times Ihave seen the statement it works. If mathematics is a symbolic tool with noinherent connection to the material and the natural language descriptions arenonsensical, then you are just idiots generating random information.:-)mitch === Subject: Re: Graph Theory TextbookOriginator: dwildstr@euclid.ucsd.edu (Jake Wildstrom)The Prophet Jack Huizenga known to the wise as huizenga@uchicago.edu, opened the Book of Words, and read unto the people:> I'm an undergraduate, and this summer I will be participating in an>REU for discrete math and combinatorics. I am looking for a good>graph theory textbook to learn the basics from. I have seen the books>that Dover (which of course are very cheap) has to offer on the>subject, but unfortunately they seem too elementary. Is the Springer>GTM Graph Theory by Reinhard Diestel any good? Any suggestions are>highly welcomed.I learned from Diestel and found it quite readable and useful. Theother books mentioned on this thread are by and large alsoexcellent, but I thought I'd throw my support to Diestel's work.+-------------------------------------------------------- -----+| D. Jacob Wildstrom -- Math monkey and freelance thinker || Graduate Student, University of California at San Diego || A mathematician is a device for turning coffee into || theorems. -Alfred Renyi |+------------------------------------------------------------ -+The opinions expressed herein are not necessarily endorsed by theUniversity of California or math department thereof. === Subject: Re: the anticlassicalist }{ vi: into the quantumalt.philosophy,sci.lang,sci.logic,sci.math,sci.physics: [...]> Des> will contemplate _l'.90tre de l'.8etant_ for food.Raisins, no doubt. === Subject: Re: Simple idea, mathematics and common-sense Discussion, linux)> First of all, I'm going to talk about a case where mathematicians gave> up because they couldn't see something, and assumed that because they> couldn't see something it didn't exist!You know, I've *seen* mathematicians say something like this. Let mesee if I can recall what it was.Oh yeah, it was this.,----[ <3c65f87.0402140856.593bc13f@posting.google.com> ]| What I've shown in a rather simple way is that only such simple and| direct results will work, so if you don't *see* the non-unit factors| between two irrational algebraic integers, then you don't have any in| the ring of algebraic integers!`----Wait. My mistake. That wasn't a mathematician. It was you.-- Jesse F. HughesWell, I don't claim to be an expert, in fact I am a fry cook with anational burger chain, but I have solved many differential and partialdifferential equations numerically. --C. Bond === Subject: Re: Letter to Prof Ullrich and others Discussion, linux)> People like you have come and gone, as your agenda is controlling> other people's behavior.> Why try?> It's Usenet you know. Usenet is known for having people around who> just post or reply, isn't that amazing!Does this advice possess any relevance given your recent tantrumsregarding Nora Baron's audacity to post in your threads?Nah, never mind. -- Jesse F. HughesEven I, who know beyond doubt that my death will be caused by a sillygirl, will not hesitate when that girl passes by. -- Merlin, asreported by John Steinbeck. === Subject: Re: JSH: Non-uniqueness of factorization> That's almost encouraging, but Microsoft is so abysmal when it comes> to Usenet standards, I am still not persuaded that Outlook does> something sensible.> Maybe Outlook does this: If a subject starts with xxx: blah blah blah,> then strip out xxx: and put Re:. Since you replied to Re: JSH: blah> blah blah, it would have stripped out Re: and then put Re:,> making no effective change.> But if you had replied to JSH: blah blah, it would have chosen Re:> blah blah as a subject line.> I don't know if this is right. I'm just guessing, based on> Microsoft's bad track record (like using different replacements for> Re: depending on the language of the user, RFC standards be> damned). So, can you check? Select the root of this thread and> compose a followup (without posting) and let us know what the subject> line says.That is correct, at least for OE6.The latest RFC (2822) in section 3.6.5 describing the subject field,mentions that Re: *may* (emphasis mine) be used but that is neitherrequired nor to be expected. The same section cautions about building longstrings of such add-on text - because the field is unstructured and meantfor human eyes, not programs. I have seen the strings 'Reply:,Forward:, FWD:, FW: and many others. I think MS is just erring on theside of caution by deleting a string ending with . One cannot fault themfor not knowing who JSH is because, after all, he's a fan of Java, not C#. === Subject: What's the computation time of this Gradient optimization method, O(kN) or O(kN^2)?Dear All,I am trying to figure out the complexity level of the gradient method forminimizing an objective function.Assume J is the scalar objection function of N vectors, each vector is of kdimension.Now we are trying to find a k-dimension vector x such that J is minimized,e.g. J is the sum of squares of the distances from N points in R^k space toa line segment,whichis connected by a known vector a and the unknown point x in R^k space.So given this definition of J, the way we did is to calculate dJ/dx = 0 andfind the vector x.However, the expression dJ/dx is not usually in explicit form, i.e, x cannot directly be obtained andhave to be done by some iterative approach.Then for this kind of optimization problem, how to define the computationalcomplexity?It is O(kN) or O(kN^2)?And how to verify this?Fred === Subject: Statistical independence test for continuous variablesI need to know what are the available statistical independence testsfor continuous variables. I know that the correlation is a test forlinear dependence for continuous variables, but I was wondering ifthere were others. === Subject: de Morgan isomorphism> [...]> Its nice watching people have fun with their patterns, play with them,> fit the pieces together. Its almost as fun as doing it myself.I am going to attempt a labeled diagrams to express the de Morgan isomorphismfor Malinowski's diagram.The vectors may be interpreted according to standard truth table formatting.All exchanges reflect de Morgan conjugation.Initial state:[begin fixed width] | T | | | / | T | / | | / / | T | / / | | / / / | T | / / / / / / / / / / / / / / / / / / / / / / / / | T | | T | | F | | | | | | | | T | | F | | T | | | | | | | | F | | F | | F | | | | | | | | F | | T | | T | | | | | | / / | F| | F | / / / || | | x x / | T| | F | x X x || | | / x x | T| | F | / / / || | | / / | | | T| | T | | | |/ / | F | | F | | T |/ / | | | | | |/ / | F | | T | | F |/ / | | | | | |/ / | T | | T | | T |/ / | | | | | | // | T | | F | | F | / / / / / / / / / / / / / / / / / / / / | F | / / | | / / | F | / | | / | F | / | | / | F |[end fixed width]Exchanged state:[begin fixed width] | F | | | / | F | / | | / / | F | / / | | / / / | F | / / / / / / / / / / / / / / / / / / / / / / / / | T | | F | | F | | | | | | | | T | | T | | T | | | | | | | | F | | T | | F | | | | | | | | F | | F | | T | | | | | | / / | F| | F | / / / || | | x x / | F| | T | x X x || | | / x x | F| | T | / / / || | | / / | | | T| | T | | | |/ / | F | | T | | T |/ / | | | | | |/ / | F | | F | | F |/ / | | | | | |/ / | T | | F | | T |/ / | | | | | | // | T | | T | | F | / / / / / / / / / / / / / / / / / / / / | T | / / | | / / | T | / | | / | T | / | | / | T |[end fixed width]If I made the exchange so that it also reflected A B |------------ T T | T F | F T | F F |changing to ~B ~A |------------ F F | T F | F T | T T |then the exchangesNAND ---> ANDNOR ---> ORXOR ---> IFFIFF ---> XORwould have reflected Boolean negation.But, in the exchange under the de Morgan isomorphism, both labelings reflectsound and complete truth-functional connectivity.:-)mitch === Subject: Re: Collatz Conjecture : Symmetry question. === >Subject: Re: Collatz Conjecture : Symmetry question.>Message-id: I have a program that can build each level from the previous one.>>I also have written this program. I am using an arbitrary precision>lib to allow large numbers. I have two applications, one calculates>just end points, and the other traverses the tree. We should compare>notes >Sure, I'll see if I can dig up my program. The final version used text>files to hold the level data. I stopped at Level 84 because the text>file, at 3.3GB was getting too big to fit on a single CD when zipped.>and maybe work together and release a simple tool for others to>use. Just a thought.>Simple, yes. Usefull? That remains to be seen.>But I've got some other stuff that might be interesting. I finally>solved my Big Problem (multi-generation sequence vectors) and am >working on a web page to document it. I can post some of it here>along with the programs if you're interested.> I originally did this using a list in memory, but I ran out of memory.> Here's the text file based version written in Python (which also does > Big Arithmetic). The program reads the previous level out of a file,> doubles every number and if a number is both == 1 (mod 3) AND> == 0 (mod 2), it spawns a new branch by using the inverse 3x+1> rule.> # usage: python lread.py n> # where n is level to process> # reads file named Ln.txt> # and outputs next level> #> import sys> level = sys.argv[1]> filein = 'L' + level + '.txt'> f = open(filein,'r')> s = 'begin'> while s != '':> if s != 'begin':> n = long(s)> print n*2> p3 = divmod(n,3)> if (p3[1]==1):> p2 = divmod(n,2)> if (p2[1]==0):> print (n-1)/3> s = f.readline()> f.close> Start by creating a text file named L5.txt that contains just> 16> Running the program using L5.txt> python lread.py 5> produces the following output:> 32> 5> To build up successive levels, redirect the output to a file> python lread.py 5 > L6.txt> Of course, you can batch file the low levels since they go quick.> It gets slow by the time you get to Level 70. And it starts eating > up disk space:> 65 File(s) 448,531,354 bytes> The one good thing is that you only need to keep the last level> on hand. All the previous ones can be archived. I originally > wanted to go all the way to Level 100, but since Level 84> took 3.3 GB, I lost interest at that point.Hi Mensanator,You originally calculated these starting seeds for higher levelswhich I could not do because of my algorithm which found the levelrequested and thus all preceeding levels. So it was very slow findingall seeds for levels >20. What is surprising here is, your level counts do not match your oldlevel counts as shown below. Number of starting seeds for each level starting @ level 6 in the Collatz tree. Level # of seedsLevel 6 2 Level 7 2 Level 8 4Level 9 4Level 10 6Level 11 6Level 12 8 Level 13 10 Level 14 14 Level 15 18 Level 16 24 Level 17 29 Level 18 36Level 19 44 Level 20 58 Level 21 72 Level 22 91 Level 23 113 Level 24 143 Level 25 179 Level 26 227 Level 27 287 Level 28 366 Level 29 460Level 30 578 Level 31 732 Level 32 926 Level 33 1174 Level 34 1489 Level 35 1879 Level 36 2365 Etc.Am I interpeting somthing wrong here?Dan === Subject: bound on partitions of subsets Given a set S of n elements, i need to choose a subset S' of these elements and then choose a partitioning of S'. Can someone tell me a compact asymptotic upper bound on the number of ways this can be done? Anything that is better than (2^n)*(B_n). {B_n is the nth bell number}. Essentially a bound on the series sigma (nC_i B_i) If not, could someone direct me to a nice compact bound onthe bell number B_n. Lovasz's result is too complicated for a humblecomputer scientist like me :-) ~Deepak === Subject: Re: Teaching philosophyNotwithstanding the draft I submitted in this thread, and others Imight improvise, I am serious about the question of how to writethis kind of statement. As I mentioned, I think I now have some ideaof the genre. What I need to know now is how many yards of paper to use.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: Can something be undeterminable? charset=Windows-1252Has it been (or can it be) proved whether there exist questionsthat are *absolutely* unanswerable in the sense of there beingno sufficiently powerful mathematical system to answer them?--r.e.s.> So, you are a Platonist?Rather than developing an allegiance, it seems preferrable to try to develop an ability to see things as a *-ist wouldsee them, whatever the * might be. As I explained in reply to Robert Israel's posting, when I asked the above question, I wasn't properly appreciating theaxiomatic nature of the situation -- that one mathematical system is regarded as more powerful than another simply because it has an additional axiom, not because it bringsus closer to knowing the *correct* answer. (It does seem useful to entertain the a point of view in which one of a complete set of mutually exclusive answers is correct, regardless of whether we happen to know *which* answer is the correct one.)If we took more powerful in the sense in which I first (mis-)understood it, then what would be your reply to theabove question? (Simply adding an axiom is not enough --we need to know which system gives the correct answer.)--r.e.s. === Subject: Re: JSH: Non-uniqueness of factorization <87ad3den4w.fsf@phiwumbda.org> <25rZb.64909$KV5.49312@nwrdny01.gnilink.net> Discussion, linux)> I think MS is just erring on the side of caution by deleting a> string ending with . One cannot fault them for not knowing who> JSH is because, after all, he's a fan of Java, not C#.I guess we disagree on what constitutes caution. Mangling subjectlines is not cautious. The fact is that, JSH aside, there are plentyof opportunities in which one wants to start a thread with a subjectof the form, Blah: blah blah blah.That the MS coders couldn't think of any isn't surprising. Theycouldn't consider that (on rare occasions), lines might begin like this. Consequently, users of some of (which?) products ofMS software cannot read this sentence. Admittedly, it is not oftenthat a line begins with begin , but it is still ridiculous that MScoders chose to parse every such occurrence as a mime enclosure,regardless of whether the headers indicated there was an enclosure ornot.I suppose they were being cautious and ensuring that theycompensated for any mime enclosures that were incorrectlyconstructed. Or something.-- We are happy that you agree that customers need to know that OpenSource is legal and stable, and we heartily agree with that sentenceof your letter. The others don't seem to make as much sense, but wefind the dialogue refreshing. -- Linus Torvalds to Darl McBride === Subject: Re: help is needed!!!> i want play whit MAPLE but i can not find sourse for it> if any body know some sourses for it or any one can help and support> me please say moreYou should contact the company; I guarantee that they'll be more thanhappy to help you out.Doug === Subject: Re: Model theory puzzleFind a finite system of axioms (feel free to introduce operations,relations, constants) so that all of its finite models would have aprime number of elements, and for every prime p there's a model tothat system of size p. Posted Via Usenet.com Premium Usenet Newsgroup Services------------------------------------------------------ ---- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **---------------------------------------------------------- http://www.usenet.com> Finite fields?Not quite, _all_ of its finite models would have a prime number ofelements, there are finite fields with p^n elements for all primes pand each n >= 1.-- G.C. === Subject: Ping Jenny Wagner.You here?-- G.C. === Subject: Re: I got low score on math test, please advise me and take a look> this school is really horrible.Sounds like you should drop out.Doug === Subject: Re: I got low score on math test, please advise me and take a look> assuming that your report about progress feedback is accurate, a> rejection of the request above is a good glimpse into an incompetent> institution. what kind of justification were you offered? some> bureaucratic non-sense?My guess is that justification offered was THE RULE THAT YOU CAN'T DROP ACOURSE AFTER A CERTAIN DATE.Doug === Subject: Re: Genetics and Math-Ability>I don't have any formal stats about it, but it seems not to run in families>very much. Among the famous names, there were several Bernoulli's and a>couple of Jacobi's. Today there are the Chudnovsky's and the Voevodsky's,>but that's all I can think of, not that I know much about it.Among those I know are the three Browder brothers (Felix, Andrew and William), the two Fefferman brothers (Charles and Robert), the two Hsiang brothers (Wu-chung and Wu-yi), and the two Borwein brothers(Peter and Jo). There are many parent-and-child pairs.>But I'll bet a lot of mathematicians have near relatives in other sciences.>I'd be interested in any stats about birth order, because temperament seems>to matter in this game. Consider two brothers, one a doctor and the other a>math professer. I'll bet the doctor is the older of the two.The way I heard it, mathematicians are most likely to be firstborn.I don't know about physicians. === Subject: Re: Teaching philosophy> Evil is the root of all money. No, this is a common misconception. The love of evil is the rootof all money. === Subject: Re: Probability Instinct - Sufficiently Discrete to Survive in the Jungle> I'm not sure I understand where all of this is going. You assert that> scientists have never directly observed causality. Are you> asserting that causation does not exist?Do you mean do we observe events that stimulate the sights, sounds, tastes,smeels, or feels of things or do we just experience such properties ofsensory apperatus which we reason are stimulated by such things?> I'm also not clear on where your essay on probability is leading. If> the outcome of events are not known beforehand, what difference does> it make whether chance is real or percevied? If an observer doesn't> know ahead of time, he doesn't know. Even were one to grant that> there is no real randomness in the universe, we are still saddled> with the fact that we possess limited sense faculties, which provide> noisy, incomplete and even self-contradictory information.> Probability gives us tools for dealing with such uncertainties.Very good and I agree. Kosko argues that truth and fact are synonomous withcoherence and correspondance. Like a musical peice played on one instrumentlike a guitar while practicing. Although the device is tuned right[coherence] to play the music it may not be tuned to other instruments inthe band [correspondance] later that day at practice.The correspondence level of animal sensory perception may have been tuned tothe needs of survival, which may or may not correspond to what we use mathand logic for.Kosko is arguing for multi-valance in the economy of the brainstransactions.The cause thing and Hume...There are no ideas, which occur in metaphysics, more obscure and uncertain,than those of power, force, energy or necessary connection. ... When we lookabout us towards external objects, and consider the operation of causes, weare never able, in a single instance, to discover any power or necessaryconnection; any quality, which binds the effect to the cause, and rendersthe one an infallible consequence of the other.... Consequently, there isnot, in any single, particular instance of cause and effect, any thing whichcan suggest the idea of power or necessary connection. ...but the power orforce, which actuates the whole machine [the universe], is entirelyconcealed from us, and never discovers itself in any of the sensiblequalities of body.... It is impossible, therefore, that the idea of powercan be derived from the contemplation of bodies, in single instances oftheir operation; because no bodies ever discover any power, which can be theoriginal of this idea.When we entertain, therefore, any suspicion that a philosophical term isemployed without any meaning or idea (as is but too frequent), we need toenquire, from what impression is that supposed idea derived? And if it beimpossible to assign any, this will serve to confirm our suspicion. Now ifwe produce an idea, like power or necessary connection, that we maintain isnot derived from an antecedent impression, it is not incumbent upon Hume toproduce the impression or abandon his empiricism. Instead. ...our idea iswithout any meaning or idea. And as we can have no idea of any thing whichnever appeared to our outward sense or inward sentiment, the necessaryconclusion seems to be that we have no idea of connection or power at all,and that these words are absolutely without any meaning, when employedeither in philosophical reasonings or common life.This connexion, therefore, which we feel in the mind, this customarytransition of the imagination from one object to its ususal attendant, isthe sentiment or impression from which we form the idea of power ornecessary connection. Nothing farther is in the case.Causes and effects are discovered, not by reason but through experience,when we find that particular objects are constantly conjoined with oneanother. We tend to overlook this because most ordinary causal judgments areso familiar; we've made them so many times that our judgment seemsimmediate. But when we consider the matter, we realize that an (absolutely)unexperienced reasoner could be no reasoner at all (EHU, 45n). Even inapplied mathematics, where we use abstract reasoning and geometrical methodsto apply principles we regard as laws to particular cases in order to derivefurther principles as consequences of these laws, the discovery of theoriginal law itself was due to experience and observation, not to a priorireasoning.The mental imagery and associations may reflect laws of motion and sequencebut to claim they internally cause each other doesn't mean my mentalactivites causations reflect the causations of the observed sequences ofchanging atomic configurations.The Copy Principle accounts for the origins of our ideas. But our ideas arealso regularly connected. As Hume put the point in his Abstract of theTreatise, there is a secret tie or union among particular ideas, whichcauses the mind to conjoin them more frequently together, and makes the one,upon its appearance, introduce the other.A science of human nature should account for these connections. Otherwise,we are stuck with an eidetic atomism -- a set of discrete, independentideas, unified only in that they are the contents of a particular mind.Eidetic atomism thus fails to explain how ideas are bound together, andits inadequacy in this regard encourages us, as Hume thought it encouragedLocke, to postulate theoretical notions -- power and substance being themost notorious -- to account for the connections we find among our ideas.Eidetic atomism is thus a prime source of the philosophical hypothesesHume aims to eliminate.The principles required for connecting our ideas aren't theoretical andrational; they are natural operations of the mind, associations weexperience in internal sensation. Hume's introduction of these principlesof association is the other distinctive feature of his empiricism, sodistinctive that in the Abstract he advertises it as his most originalcontribution: If any thing can intitle the author to so glorious a name asthat of an inventor, 'tis the use he makes of the principle of theassociation of ideas.Hume locates three principles of connexion or association: resemblance,contiguity, and cause and effect. Of the three, causation is the onlyprinciple that takes us beyond the evidence of our memory and senses. Itestablishes a link or connection between past and present experiences withevents that we predict or explain, so that all reasonings concerning matterof fact seem to be founded on the relation of cause and effect. Butcausation and the ideas closely related to it also raise seriousmetaphysical problems: there are no ideas, which occur in metaphysics, moreobscure and uncertain, than those of power, force, energy or necessaryconnexionHume wants to fix, if possible, the precise meaning of these terms, andthereby remove some part of that obscurity, which is so much complained ofin this species of philosophy. This project provides a crucial experimentfor Hume's metaphysical microscope, one designed to prove the worth of hismethod, to provide a paradigm for investigating problematic philosophicaland theological notions, and to supply valuable material for theseinquiries.http://www.friesian.com/hume.htmhttp:// www.wutsamada.com/alma/modern/humepid.htmI suppose that at that point in history a milestone had been reached andnothing was to be the same henceforth and while through Kant and isappropriators...neurophysiology continues to grow.> -Will Dwinnell> http://will.dwinnell.com === Subject: Re: How many ways to put 5 balls into 500 ordered cups?Originator: jeyadev@kaveri>You are given 500 numbered cups and five identical balls. Any cup can>hold up to five balls. How many ways can you put the five balls into>the 500 cups?>As a warm-up, let's try smaller sets of ordered cups:> ....>I have a brute force solution using an SQL query from hell (my>specialty), which is not quite the same thing as a formula and proof. > I cannot remember enough Combinatorics and Finite Differences to make>the next step. ARRRGH!>It is bothering me enough that I will offer a copy of my next book>(TREES & HIERARCHIES IN SQL) as a prize for the best solution.Have you considered the following:How many terms are there in (x1 + x2 + ...... + xn)^m ??-- Surendar Jeyadev jeyadev@wrc.xerox.bounceback.com Remove 'bounceback' for email address === Subject: Re: Genetics and Math-Ability>I don't have any formal stats about it, but it seems not to run in>families very much. Among the famous names, there were several Bernoulli's>and a couple of Jacobi's. Today there are the Chudnovsky's and the>Voevodsky's, but that's all I can think of, not that I know much about it.> Among those I know are the three Browder brothers (Felix, Andrew and> William), the two Fefferman brothers (Charles and Robert), the two> Hsiang brothers (Wu-chung and Wu-yi), and the two Borwein brothers> (Peter and Jo). There are many parent-and-child pairs.In Britain we have Mary Rees (Liverpool) and Sarah Rees (Newcastle),Daughters of David Rees (Exeter emertius).-- === Subject: Re: Drawing subgroup lattices for research papersThe actual message is that it can't find xypic.sty. I triedpointing it there, but nothing happened. I read somewhere that thereis a command in MikTex that allows you to update the file databasethat it will read, but I couldn't find anything. Do I need to just doa fresh install?I'm working on a thesis (using MikTex and Winedt) that requires me toput in a lot of subgroup lattices. This results in several problems. For an early version of the paper, I made .bmps and cut and paste themin the appropriate spots, but now I want to put them in the actualcode of the paper itself. My method was to import into Mathematicaand export as .eps. First of all, the .eps files are huge. I triedto import a 3k grayscale gif and got a 400k eps file. Then, usinggraphicx I was able to get the picture to show up after Latexing. However, the graphic looks terrible. It is clearly something that ishappening when I Latex the paper because I checked the eps file inGhostview and though it was degraded some from the original bmp thesubgroups were still recognizable.I've also tried downloading xypic and followed the instructions, butcan't figure out how to get MikTex or WinEdt to recognize the newfiles.Does anyone either have a suggestion on getting what I have to work ora more tex-friendly way to draw lattices?> xypic will do most things you can imagine, and some you can't. simpler but> less flexible is paul taylor's diagrams package. diagrams is available > from ctan archive. > I don't understand what you mean by can't get winedt or miktex to> recognize the new files, which new files, what are the error messages? > one simple suggestion: have you updated the miktex installation after> putting xypic in your path? === Subject: Re: mathcad 11> I have plugged the expression Sum(n=1->m)(10^n)/n, and asked for a> symbolic solution. The answer that came back involved, among other> things, the natural log of -9, a red m superimposed on a> parenthesis, and something illegible (black)superimposed on a plus> sign. Did I do something wrong? I have to say that I am completely> at sea with computers. Any help will be appreciated, including the> proper answer, if it exists, for my sum.Hmmm. The same bug (and it is a bug) occurs in Mathcad 2001. === Subject: Re: the anticlassicalist }{ : mitchism for Daleks: No, it means that at some point you have to hook into what physicists: *observe*. Will it tell us the energy levels of a hydrogen atom and: predict the Lamb shift? If you're interested in developing a better: mathematical formalism, that's fine, go ahead. Just don't tell us it's: physics.Actually, all I did was point to the connection between the foundationalobjects of the theory and the observational objects it predicts. I calledthem the ontology and the epistemology of the model in a rigorous way toaccord somewhat with common usage, but any other names would work. But theformalism is all about observational propositions (what is the likelihoodthat A and B happen?, if C happens, what does that imply for D?, wherethe letters stand for quantum events like spontaneous decay). Anywhere youhave time series of quantum events or concurrent systems, you implicitly orexplicitly use the logic I mention, often in an algebraic setting.It _is_ physics. I _am_ interested in developing a better mathematicalformalism, but alas cannot take credit for this one. This one has beenstudied by physicists now for 3 quarters of a century.-- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea: prankster, fablist, magician, liar === Subject: Re: the anticlassicalist }{ vi: into the quantumFor the term to exist to mean anything at all, there must besomething that doesn't exist.It does not matter how the one is distinguished from the other or whois doing the distinguishing; it does not matter where the line isdrawn between that which exists and that which doesn't exist, how itis drawn, who is drawing it, whether it is sharp or fuzzy.What matters is that for the phrase X exists to mean anything atall, there must be some entity Y for which it is valid to state Ydoes not exist.Otherwise the word exists in the phrase X exists would expressnothing whatsoever.> Ah.... ontology is about existence. Or (in light of the above): ontology is about that which isdistinguished from that which does not exist.> Can something that does not exist be ontological?It would have to - just like something that does. For a distinction to take place requires that there is more thanpossible outcome.> Does an ontologist need to exist in order to> ... er... be an ontologist? Q: What is red and invisible?A: No tomatoes.Q: But what if they're green and invisible? A: Then they aren't ripe yet. === Subject: Infinite Galois GroupsLet F be the field of lengths constuctible by compass andstraighthedge. What is G(F/Q)?Let K be the field generated by the square roots of the prime numbers.What isG(K/Q)? What is G(F/K)? === Subject: Re: Teaching philosophy> Notwithstanding the draft I submitted in this thread, and others I> might improvise, I am serious about the question of how to write> this kind of statement. As I mentioned, I think I now have some idea> of the genre. What I need to know now is how many yards of paper to> use. Whatever else might be said, at least some people who willread the statement will be concerned with how _well-written_it is. Such people come later in the loop, such as the deanand the people on the search who are outside the department.Nevertheless, such documents are mostly ways to _lose_interviews, so if the question is how to write, thenthe answer is very well. Not just with exacting grammar,but also engagingly. In some sense, this is a sample ofyour teaching style in that it shows your mastery of English,your communications skills, and how well you read your audience.So in one sense, the topic is moot, as if they had just givenyou an essay question so you could show off your writing ability.Given that, how many yards of paper does it take to put youraudience off? I'm guessing about 1.5 is the correct number ofpages and that 3 pages would choke the entire committee. === Subject: Re: malta-new discovery> Can you believe that the oldest stone building in the world has a> mathematical perfection? Check:> www.star-mysteries.comHey a real money making scam. They make stupid claims and you buy the-------------------------------------------Quote Ancient Architects On Malta I - NEWMalta archipelago can be proud of the oldest stone-build structures onour planet supposed to be ancient temples build by ignorant farmersfive millennia ago. Yet they are geometrically perfect. Who designedthem any why?BUY THIS ARTICLE--------------------------------------------- === Subject : Re: malta-new discovery> Can you believe that the oldest stone building in the world has a> mathematical perfection? Check:> www.star-mysteries.comDoug === Subject: Re: Infinite Galois Groups Adjunct Assistant Professor at the University of Montana.>Let F be the field of lengths constuctible by compass and>straighthedge. What is G(F/Q)?Well, formally, it is the inverse limit of the Galois Groups of thefinite Galois subextensions, with the connecting morphisms being thecorresponding quotient maps.It should be fairly complicated, since it will include thesubextensions given by the 2^n-th and 3^n-th roots of all integers.>Let K be the field generated by the square roots of the prime numbers.>What is G(K/Q)? What is G(F/K)?This is the increasing union ofQ subset Q(sqrt(2)) subset Q(sqrt(2),sqrt(3)) subsetQ(sqrt(2),sqrt(3),sqrt(5)) subset ...In each case, the relative Galois group is Z/2Z, and the Galois gropuof the extensionQ(sqrt(2),...,sqrt(pn)) over Qwhere pn is the n-th prime is equal to (Z/2Z)^n (you would need toprove this; see for example Section 6.7 in Lang's Algebra, 3rdedition). The maps going down are... -> Z_2 x Z_2 x Z_2 --> Z_2 x Z_2 --> Z_2 --> 1where each map chops off the last coordinate. The inverse limit is aninfinite product of copies of Z_2.-- === ==Arturo Magidinmagidin@math.berkeley.edu===Subject: Re: Collatz Conjecture : Symmetry question.>>Do you mean proven true? It could be proven false with a single>counterxample.The Collatz conjecture is: The Collatz sequence is allways finite andends in 1. How then, can it be a counterxample?> A non-trivial cycle would be a (finitely checkable) counterexample.You can also say that every Collatz sequence is infinite and cycles onthe trivial loop 1. Why is 1 a trivial loop? Because every 3x+C systemloops at C. Thus, 3x+3 loops at 3, 3x+5 loops at 5, etc. But whereas3x+1, 3x+3, 3x+9 (or any where C is a power of 3) have only the singletrivial loop (as far as anyone knows), non-power-of-3 Cs have multipleloops. So the general conjecture any sequence of a 3x+C system isinfinite and cycles on C is false when C=5.The 3x+5 trivial loop: 80 40 20 10 5 20 10 5 (loop at 5)A 3x+5 counterexample:44 22 11 38 19 62 31 98 49 152 76 38 19 (loop at 19) === Subject: Re: e is transcendental by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KJtax05002;> Since e^[iPi]=cosPi+isinPi> or , e^[iPi]=-1+i[0]> then there are two solutions here, to the given equatio:> > A) e^[ipi]=-1 the real part solution and > > B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio.>No, the conclusion from e^[iPi]=-1+i[0] is>Re(e^[iPi]) = Re(-1+i[0]) = -1 AND>Im(e^[iPi]) = Im(-1+i[0]) = 0>-- >Daniel W. Johnson>panoptes@iquest.net>http:// members.iquest.net/~panoptes/>039 53 36 N / 086 11 55 WDaniel,Just saw Your message.I, This is it.Your help is great, and resolves my question completely.I, thank You all.Panagiotis Stefanides === Subject: Re: Teaching philosophy> Notwithstanding the draft I submitted in this thread, and others I> might improvise, I am serious about the question of how to write> this kind of statement. As I mentioned, I think I now have some idea> of the genre. What I need to know now is how many yards of paper to use.Having been asked to perform this silly exercise several timesin past years, I've developed two documents: one, about two pagesin length, that points out in broad strokes why I'm so wonderfulin general, and one, about 6 pages in length that points outin particular why I'm so wonderful in introductory courses, incourses for a general audience, in courses for upper-levelstudents, and in supervising research.This is a somewhat light-hearted description, but it's nottoo far from the truth. After a number of requests, one getsa certain facility at responding. The important thing to keepin mind is that within a very broad range anything you say willbe acceptable. When we are trying to hire a new faculty member,our department asks for a statement on teaching, expecting(and usually receiving) the longer of the above-describedversions. All we're really looking for is some facilityin English and no evidence that the person will be an obviousmismatch, likeIf my students don't understand a concept after several attemptsat explanation, I have no qualms about smacking them around.I am speaking English perfectly free.I expect students to concentrate on my course to the exclusionof everything else.Avoid obvous gaffes like these (two of which I've actuallyread), write with sincerity and (if you can do it) verveand you'll have nothing to worry about on this score.Rick === Subject: Re: Letter to Prof Ullrich and othersI am just curious: why are people like youinterested in even acknowleding James Harris and his ilk ?This guy has been oopsing on sci.math for ages (at least severalyears) I see. Is it not worthwhile for everyone to just ignore him at this point ? There is real danger of course that someone would take him seriously,but I feel thats remote.> People like you have come and gone, as your agenda is controlling> other people's behavior.> Why try?> It's Usenet you know. Usenet is known for having people around who> just post or reply, isn't that amazing!Some do more than just post and reply. They build web sites documenting cranks!> And in all that posting and replying there are people like yourself> who try to control the process.> It seems that he has taken to calling universities to complain. Lets not giving him any more attention. Without attention, he willshrivel up and disappear.Is there a history or an obvious reason that I am missing ?> Yeah, the history of Usenet, and I guess you're a newbie, eh?> Or you should have known enough not to have made the post you did.> James Harris === Subject: Re: Drawing subgroup lattices for research papers> The actual message is that it can't find xypic.sty. I tried> pointing it there, but nothing happened. I read somewhere that there> is a command in MikTex that allows you to update the file database> that it will read, but I couldn't find anything. Do I need to just do> a fresh install?I don't have MikTex anymore, but as I recall that there is some documentation on using kpathsea (the library that miktex uses to find things). I hope that helps.>>I'm working on a thesis (using MikTex and Winedt) that requires me to>>put in a lot of subgroup lattices. This results in several problems. >>For an early version of the paper, I made .bmps and cut and paste them>>in the appropriate spots, but now I want to put them in the actual>>code of the paper itself. My method was to import into Mathematica>>and export as .eps. First of all, the .eps files are huge. I tried>>to import a 3k grayscale gif and got a 400k eps file. Then, using>>graphicx I was able to get the picture to show up after Latexing. >>However, the graphic looks terrible. It is clearly something that is>>happening when I Latex the paper because I checked the eps file in>>Ghostview and though it was degraded some from the original bmp the>>subgroups were still recognizable.>>I've also tried downloading xypic and followed the instructions, but>>can't figure out how to get MikTex or WinEdt to recognize the new>>files.>>Does anyone either have a suggestion on getting what I have to work or>>a more tex-friendly way to draw lattices?>xypic will do most things you can imagine, and some you can't. simpler but>less flexible is paul taylor's diagrams package. diagrams is available >from ctan archive. >I don't understand what you mean by can't get winedt or miktex to>recognize the new files, which new files, what are the error messages? >one simple suggestion: have you updated the miktex installation after>putting xypic in your path? === Subject: Re: Axioms defining a finite field Adjunct Assistant Professor at the University of Montana.> Unless I've missed something...>Doesn't look like it...so does this mean that a+b=b+a doesn't need to be in>the field axioms?It doesn't have to be in the ring axioms in the presence of a 1 anddistributivity (and on this, Jacobson's algebra book agrees with me atany rate); if you know that the group is a (not necessarily abelian)group under +, a semigroup with identity under *, and thata*(b+c)=(a*b)+(a*c), (a+b)*c = (a*c) + (b*c), then commutativity of +will follow.This sort of exercise often shows up in algebra books after theintroduction of rings, as a way of explaining why one does notconsider the more general structure where we don't requirecommutativity of addition...>It is annoying enough that the group axioms usually have a*e=e*a=a and>a*a'=a'*a=e, when you only need a*e=a and a*a'=e, and I've just barely>managed to surpress my rage at that...Yes, but on the other hand, if you have a*e = a for all a, and for allb there exists b' such that b'*b = e, then you don't necessarily havea group...-- === ==Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: Prime concept in Object Ring> Doing active research while trying to explain it as you go along is> kind of difficult.Why not arrive at a useful result first, *prove* it, and *then* explain it?Skip === Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read.i do not respond to many of the posts because my argument is with this school and not people of the message board. The ideas that the board gives www.versuslaw.com after that i will try to do another google search for statewide policies and procedures regarding academic progress === Subject: Re: Simple idea, mathematics and common-sense> There's something else important here which is the ambiguity of the> square root operator, Which is easily resolved by stating what one means by the sqrt symbol. If one states that sqrt(x) requires that both x and sqrt(x) be non-negative reals, there is no ambiguity. If one allows that for non-zero complex numbers x, sqrt(x) can mean either of the complex numbers whose square equals x, THEN it is ambiguous.It is customary to presume the first meaning whenever possible, i.e., whenever x is a non-negative real.> which may sound complicated but it's easy to> demonstrate with another root of a quadratic:> (1+sqrt(9))/2> and you may think, silly, why show sqrt(9) when sqrt(9) = 3, but yeah,> that's *one* of its solutions, as sqrt(9) = -3 as well, so you have> *two* numbers> (1+sqrt(9))/2 = 2 or -1Only if you reject the customary interpretation and refuse to resolve the ambiguity in a civilized way. But no one has ever accused JSH of civility.> as either solution will work.Only if, contrary to custom, one assumes the second definition. Since JSH is so fervently anti-math, no one expects him to follow customary usage, but equally, those who are willing to follow customary usage should be very cautious about following JSH anywhere. I've had people argue with me that by> definition (really by convention) you take the positive root. As it is a conventional definition, it is really by both. > But> imagine the world of Contrary.That is certainly JSH's world.> On Contrary the mathematicians for> some odd reason *by definition* take the negative! Is mathematics> really changing depending on such decisions.Then the standard symbol would become -sqrt(x), no problem.> Imagine you've forgotten that you can resolve sqrt(9) to 3 or -3, Forgetting elementary properties of arithmetic is a favorite ploy of JSH.If done creatively, it can produce interesting ideas, but JSH is singularly uncreative in any mathematical sense, and his variations on this theme seems always to lead him into idiotic error.> so> you write these numbers like (1+sqrt(9))/2 and (1-sqrt(9))/2 and> mercifully discover that you can get rid of that square root sign by> adding them together, or multiplying them together.> Like adding them gives 1, and multiplying them together gives > (1 - 9)/4 = -2> and your mathematicians scratched their heads and contemplated such> numbers, and decided that there was *no way* to understand factors of> 2 of numbers like> (1+sqrt(9))/2 and (1-sqrt(9))/2 > except as to consider them to be unique factors of 2, in some kind of> mysterious way.What seems like a mystery to JSH is simple arithmetic to everyone else.> But wait, that's not a problem here, of course, because you can just> evaluate the square root, but look back now at x^2 + 7x + 2, where> x = (-7 +/- sqrt(41))/2> and consider that you *cannot* resolve sqrt(41) in any way that will> help you here with this question.It is certainly possible to add (-7 + sqrt(41))/2 to (-7 - sqrt(41))/2geting -7 or to multiply them getting (49-41)/4 = 2, so no further resolution is required.> Sure, you can write it out in decimal format with a lot of numbers> after the decimal place. My computer tells me that sqrt(41)> approximately equals 6.4031242374328486864882176746218, but of course,> you can keep going out to infinity trying just to see sqrt(41).But sinice the expression 6.4031242374328486864882176746218 is just as artificial, and less accurate, than sqrt(41), why bother?Decimal notation is a convenience, not a necessity, for representation of numbers. The fact that some numbers are not so representable is merely an inconvenience, not a disaster.> But that's where my story begins.[fairy story snipped]> Mathematicians may think that seeing is believing, but I think in> mathematics, logic is king.What JSH thinks about how mathematicians think is irrelevant to anything in the real world.If logic is king, then JSH has shown himself to be the court fool.> James Harris === Subject: Re: JSH: Prime concept in Object Ring> As a note to myself, Notes to yourself should only be published posthumously. === Subject: walking on a grate..hello everybody,suppose I have a grate, N x M. how many paths are from node (0,0) to (N-1,M-1) eg. from one corner to the diagonal-opposite one?allowable paths are only with zero or positive change in node's coordinate.how can I compute this? I would appreciate any suggestion and help..cheers,n. === Subject: Re: JSH: Non-uniqueness of factorization>To be fair I think Outlook does that. Anything behind a : at the startofa>subject line is treated as if it were Re: and is removed. I'd suggestswitching>to [JSH] which is the convention in other groups and seems to work withall>newsreaders.I'm using Outlook now, and it did not do that. It left the JSH: in there> That's almost encouraging, but Microsoft is so abysmal when it comes> to Usenet standards, I am still not persuaded that Outlook does> something sensible.> Maybe Outlook does this: If a subject starts with xxx: blah blah blah,> then strip out xxx: and put Re:. Since you replied to Re: JSH: blah> blah blah, it would have stripped out Re: and then put Re:,> making no effective change.> But if you had replied to JSH: blah blah, it would have chosen Re:> blah blah as a subject line.> I don't know if this is right. I'm just guessing, based on> Microsoft's bad track record (like using different replacements for> Re: depending on the language of the user, RFC standards be> damned). So, can you check? Select the root of this thread and> compose a followup (without posting) and let us know what the subject> line says.You are right. I tried it. In this thread Re:JSH:blahblah does not change,but in another thread with subject JSH:blahblah, the JSH is removed andreplaced with RE, thus changing JSH:blahblah to RE:blahblahKeithK> Also, I wonder if it matters whether you use Outlook or Outlook> Express.> -- > My proof has been checked very thoroughly, both by me and others.> Those others apparently decided that they would not believe the proof> was correct, but cannot support that position using mathematics. But> hey, they're just human beings. --JSH, prover of Fermat's Last Thm === Subject: Re: Axioms defining a finite field>The set of rules you have give a ring. In order to get a field, you>need a multiplicative inverse for all non-zero elements. For example,>integers mod 4, 2 does not have a multiplicative inverse. See>Table.>2*0=0>2*1=2>2*2=0>2*3=2>One of his axioms was that a*b=0 implies a=0 or b=0, which does not hold>for the integers mod 4.>That axiom, combined with a*0=0 (easily proved from his axioms), leads to a>cancellation theorem, a*b=a*c => b=c for a!=0. Combine that with F being>finite (specified in the original post), and it is a short step to every>a!=0 having a multiplicative inverse.> But this example is important, because it shows that the axiom a*b = 0> implies a=0 or b=0 is definitely not redundant. Part of the problem, which> nobody has answered yet, was to decide whether any of the axioms can be> omitted. I am guessing no, but I could be wrong. The axiom 1 != 0 is not> redundant, because F = { 0 } satisfies all other conditions.> Two down, seven to go!> Derek Holt.Axiom 4, the associativity of multiplication, is also not redundant. Thereexist finite counterexamples called semifields.On the other hand, I'm suspicious of axiom 5: a * 1 = a. Maybe I'm missingsomething obvious, but can there possibly be a finite ring without identitythat has no zero divisors? === Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read.>My web site www.johncho.us has all the information. I am seeking people>who are familiar with procedures. I am also considering getting a lawyer.The college is within its rights to require issuance of a 'W', and, in fact,may have a duty to do so under California law. If they reported yourattendance for funding purposes, I'm pretty certain they are committed toreporting your attendance in other areas.What is the problem with the 'W'? It does not affect your GPA. Mydaughters were admitted to UCLA with several of them. There are lots ofreasons for a 'W' - I think the majority of transfer students have them.I don't think it's worth your time or money to fight this particular battle.It's probably more appropriate to concentrate on your other classes.Hank Murphyspeaking only for myself === Subject: finite measureLet m be a finite measure on B(R), the borel sigma field generated bythe real line. I have to prove that there are at most countably many xin R (atoms) such that m(x)>0.Obviously, I want to end the proof with something likeSUM m(x) --> oosince a sum of uncountably many non-zero terms always diverges. Butthe problem is, I don't know exactly what sort of contradiction to setup.An uncountable collection of points in R always forms some sort ofinterval (a,b), right? Let I be the (uncountable) indexing set ofthese atoms. ThenUNION (over I) x_i = (a,b) (say), som( UNION (over I) x_i ) = m(a,b) < oo,so I'd like to show that the left-hand side = oo, but I can't doanything with an uncountable union since a measure only has countableadditivity (for disjoint sets...points in this case). It seems like Ineed more tools than just the measure itself (since only countableoperations are defined for it)...how can I proceed? === Subject: Re: there is no such thing as infinityDarryl L. Pierce,,, wibbled:>And can be written in exponentially more obfuscated a manner than even C>or assembler... :)Assembler? Bah!! Raw machine language burnt into a ROM is the only wayto find the largest number. > Ah, I remember when I first started programming. All I had was a really hot> needle and had to write my code directly to the SIMMs. It was a major leap> forward when I upgraded and could just type copy con file.exe...When I started, SIMMs hadn't been invented. We had people who had seen so many punch cards and papertapes they could read the holes by eye. I set the bootstrap instructions using an array of toggle switches on one mainframe we had. === Subject: Re: Partial-Sum -> Some Primes> Let a(1) = 1;> Let a(m) be the lowest yet unpicked positive integer such that:> sum{k=1 to m} k* a(k) is a prime.... > a(k) : 1, 2, 4, 3, 6, 5, 12, 7,...> Question:> Is this a permutation of the positive integers?...That is, choose a(m) so that m*a(m) + sum{k=1 to m-1} k*a(k) is prime and a(m) not equal a(k) if k I guess we disagree on what constitutes caution. Mangling subject> lines is not cautious. The fact is that, JSH aside, there are plenty> of opportunities in which one wants to start a thread with a subject> of the form, Blah: blah blah blah.> That the MS coders couldn't think of any isn't surprising. They> couldn't consider that (on rare occasions), lines might> begin...Ouch, the mime parsing error is a really horrible thing. I can't fathomthat one. They must be working around some horrible kludge that requiresit.I'll end this OT thread by saying that we do not disagree on the meaning of'caution'. I was commenting on MS and I *did* say that they 'erred'!FWIW, I think user agents should leave the subject field alone.--Stan Gula === Subject: Re: finite measure> Let m be a finite measure on B(R), the borel sigma field generated by> the real line. I have to prove that there are at most countably many x> in R (atoms) such that m(x)>0.Use the classical argument: E = Union({x in X; m(x)>1/n}; n in N*), whereeach set of this union is measurable and contains finitely many elements(because m is totally finite). === Subject: Eulerian Polyheron QuestionHow many forms of n-hedra are there for arbitrary n? For example,there is only one possible tetrahedron, i.e. various collineations ofthe simplex. There are two possible pentahedra: the triangular prism(and various distortions thereof) and the square pyramid. There areat least (I'm not totally sure) four possible hexahedra: thepseudo-parallelepiped as it might be called, with quadrilateralfaces and triangular vertices; the pentagonal pyramid; the triangularbi-pyramid; and a fourth solid formed by dividing a cube [0,1]^3 inhalf by intersecting it with z >= x/2 + y/2 (where >= is greater thanor equal, in the event of ambiguity). (Is that all?) Euler'sformula was no help with the original question, as it suggested aninfinite number of possible tetrahedra, simply by adding k to thevertex count and k to the edge count, which comes by dividing an edgein two or more, which isn't a legal move.--OWD [20 Feb '04] === Subject: Re: Genetics and Math-Ability> There is an old saying about inheritance of mathematical ability.> Unlike most interitance, this goes from a man to his son-in-law.> The explanation being, I guess, that when the professor's daughter is> of the right age, she meets the professor's current student, and> marries him.I have known three cases of identical twins, one of whom was amathematician and the other one wasn't. In one case, the other twinwent into physics, but in the other two they did something unrelatedto math. To me that says that while genetics may play a role, thereis a lot more to it than that. In my case, I am the only one in arather large extended family who went into math. I did have adistanct cousin who was a physicist, but that is the nearest, AFAIK.Even if it were genetic, it wouldn't make sense to ask if it wasdominant or recessive unless it was a simple Mendelian trait(essentially, one gene). Even simple Mendelian traits aren't always. For example, the common blood groups are supposedly simple Mendelianand they mainly are in the sense that the blood type is mostlydetermined by one gene. But more than one gene is involved inproducing the A and B alleles and it is entirely possible (thoughunlikely) for two type O parents to have an A or B offspring, contraryto what you read. Mathematical ability _and interest_ are clearlydetermined, insofar as they are genetic, by many, many genes and theirinteraction with the environment. === Subject: Re: walking on a grate.. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KLYl713563;>hello everybody,>suppose I have a grate, N x M. how many paths are from node (0,0) to >(N-1,M-1) eg. from one corner to the diagonal-opposite one?>allowable paths are only with zero or positive change in node's coordinate.>how can I compute this? I would appreciate any suggestion and help..>cheers,>n.You should be able to calculate this. Take advantage of symmetry.Start 1 position away from the end and work backwards.Good luckphil === Subject: Re: Eulerian Polyheron Question> How many forms of n-hedra are there for arbitrary n? For example,> there is only one possible tetrahedron, i.e. various collineations of> the simplex. There are two possible pentahedra: the triangular prism> (and various distortions thereof) and the square pyramid. There are> at least (I'm not totally sure) four possible hexahedra: the> pseudo-parallelepiped as it might be called, with quadrilateral> faces and triangular vertices; the pentagonal pyramid; the triangular> bi-pyramid; and a fourth solid formed by dividing a cube [0,1]^3 in> half by intersecting it with z >= x/2 + y/2 (where >= is greater than> or equal, in the event of ambiguity). (Is that all?) Euler's> formula was no help with the original question, as it suggested an> infinite number of possible tetrahedra, simply by adding k to the> vertex count and k to the edge count, which comes by dividing an edge> in two or more, which isn't a legal move.> --> OWD [20 Feb '04]See Counting Polyhedra athttp://home.att.net/~numericana/data/polyhedra.htm andhttp://home.att.net/~numericana/data/polycount.htmHugo Pfoertner === Subject: Re: the anticlassicalist }{ vi: into the quantum> What matters is that for the phrase X exists to mean anything at> all, there must be some entity Y for which it is valid to state Y> does not exist.> Otherwise the word exists in the phrase X exists would express> nothing whatsoever.Replace exists by sings. Same story. By snores. Samestory. By lives. Same story. > Or (in light of the above): ontology is about that which is> distinguished from that which does not exist.Yes. Carminology is about that which is distinguished fromthat which does not sing. (I was going to write cantologyby I retracted that)> Q: What is red and invisible?The bottle caps of a six-pack of Cooper's Sparkling Ale in a closed fridge. If I hadstudied medicine I might have come up witha cuter example. === Subject: Re: Letter to Prof Ullrich and others> People like you have come and gone, as your agenda is controlling> other people's behavior.> Why try?> It's Usenet you know. Usenet is known for having people around who> just post or reply, isn't that amazing!>Does this advice possess any relevance given your recent tantrums>regarding Nora Baron's audacity to post in your threads?Hey, I know, I know! Call on me, come on, I know the answer!>Nah, never mind. ************************ === Subject: Re: the anticlassicalist }{ vi: into the quantumDeswill contemplate _l'.90tre de l'.8etant_ for food.> Raisins, no doubt.No. There's a typo there. That should be l'.8etang.L'.90tre de l'.8etang. Some fish. Obvious. Unless...the Lady of the Lake? Des, a cannibal? Ah, il faut de tout pour faire un monde. === Subject: Re: When is a finite field a cyclic field? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KLmoh14652;>I don't intend to introduce new notion,but for the moment >let's call a finite field F(+,*) a cyclic field if>the abelian group (F,+) is cyclic.(We know that (F0,*) is always cyclic when F is finite).When is (F,+) also cyclic?>May I ask Derek Holt who gave such a nice and clear solution>to a recent problem for finite fields,to give answer to this?>Thank you! >Yes you may ask! The answer is when |F| is prime.>the smallest positive integer p such that p.1 = 0 is prime, but then>it follows that p.a = 0 for all a in F, and so (F,+) cannot be cyclic>unless |F| = p.>Derek Holt.Thank you Derek,thank you Larry for your replies!They are very helpful!Best wishes,Cron === Subject: Re: Simple idea, mathematics and common-sense> James: See, mathematicians have thought that black was white. It's kind of> hard to see why black is not white - I'll try to explain it without a lot> of math. The reason mathematicians have thought that black is white is> just that that's the way they were trained...> I have been following, on and off, the postings of Mr. Harris for years.>Like you, I used to feel infuriated, for I couldn't believe that anyone>could be so thick. I haven't been infuriated by him in a long time - these days he's justentertainment.>Now that I have a bit more experience on the behavior>of mentally disturbed individuals, I have come to the conclusion that Mr.>Harris is a poor lunatic, for whom sci.math is the main means to live his>madness. Could be. > I guess we should feel compassionate about his regrettable predicament.I don't see why. Supposing for the sake of argument he's a poor lunatic, that doesn't give him the right to behave the way he does.>He is very annoying all right but, what can be expected from a deranged>individual? ************************ === Subject: Re: Axioms defining a finite field>...so does this mean that a+b=b+a doesn't need to be in>the field axioms?>It is annoying enough that the group axioms usually have a*e=e*a=a and>a*a'=a'*a=e, when you only need a*e=a and a*a'=e, and I've just barely>managed to surpress my rage at that...this field thing is going to push>me over the edge. :-)Easy there, big fella! Take a breath and try to see the big picture.There is a certain delight in checking for independence of axioms, andcertainly there is elegance in minimal presentations of axiom systems.But that isn't the only possible goal for a set of axioms. Indeed, onecan trim down the set of axioms for a group to 1, I think, but thatone axiom is difficult to digest and certainly fails to convey thegestalt of what a group is.So too with the axioms for fields (or vector spaces, or ...). The pointof the mathematics is to study some interesting and useful things. Thoseinteresting and useful things are given a name, as soon as we seewhat characteristics make them similar. The point of the axioms is toestablish that definition in an unambiguous way so that results can beproved which apply to all these objects. Where is the harm in presentingthe axioms in a redundant way, if it helps provide a better picture ofthe objects? Doesn't that make it easier on you, the mathematician, asyou attempt to detect the interesting features of these things?My recommendation is to keep your interest in the independence of theaxioms but to store it in the back of your mind for those times whenyou find yourself stranded on a desert island with nothing else to doto occupy your time. In the mean time, get to work understanding theobjects being described, and prove theorems about them which the rest of us can use.dave === Subject: Re: Pi gallons problem - any ideas?> [...]>>Perhaps a hint is needed. Here a well known mathematical formula,>>along with a lesser known one:>>V1 = Ah/3 (r^3 - 1) / (r - 1), and>>V2 = Ah/3 (r^(3/2) - 1) / (r - 1).>>V1 is the volume of a frustum (not frustrum -- oops) with height = h,>>area of bottom circle = A, and ratio of radii of top circle to bottom>>circle = r.>>V2 is the volume of that subset of the frustum below a plane tangent to>>the top and bottom circles at opposite points: i.e., the volume of water>>that would remain in a frustum if it were tipped over until the water>>just touched the bottom circle.No takers? Okay, here's the answer:> [...]> I didn't really search, once you gave the tipped over hint...> I searched instead how to proove the> V2 = Ah/3 (r^(3/2) - 1) / (r - 1)> formula ... I didn't succeed....> (V1 is easy, thanks)This is something that I derived myself once. I haven't seen it anywhereelse, but if it's not well known, I'm sure it's at least known.I've cross-posted this to sci.math, hoping that it will be found interestingthere. The issue here is proving the formula V = (pi a^2 h/3) (r^(3/2) - 1) / (r - 1),where V is the volume of the subset of a (right circular) frustum belowa plane tangent to the top and bottom circles at opposite points, r = b/ais the ratio of the top to bottom radii, and h is the height. Here's apicture for those of you using monospace fonts: radius of upper circle = b o--------------------_o | o _/o | o _/.o | Side view o _/..o | <- height = h of frustum o A /...o | o _/....o | o _/.....o | o/______o | _/ o.....o <-- radius of lower circle = a _/ _ o...o _/ H_o.o / oAn easy way to prove this formula is to compute the volume of the shadedregion. The shaded region forms a cone (not a right circular cone, butan oblique, elliptical cone), so its volume is (1/3) base x height.Here, base is just the area of the ellipse, and height is thedistance H from the apex of the cone to the cutting plane.The area of an ellipse can be written as (pi/4) A B, where A is thelength of the major axis, and B, the length of the minor axis. Thus,the volume of the shaded region is V(shaded) = (pi/12) A B HBut we can simplify this by noting that T = A H/2 is the area of theshaded *triangle* pictured above, so V(shaded) = (pi/6) T B.The area T is simply T = a (h + h'), where h' is the distance from theapex to the lower circle. By similarity, h':a = (h+h'):b, soh' = h a/(b-a), whence T = h ab/(b-a).The minor axis B has length 2 sqrt(ab). This can be seen by consideringthe midplane between the top and bottom circles. The midplane intersectsthe original frustum in a circle of radius (a+b)/2. The center of theellipse lies in the midplane, at a distance (b-a)/2 from the center of thiscircle. Thus we have a right triangle with hypotenuse (a+b)/2, and twoadjacent sides of length (b-a)/2 and B/2, which yields B = 2 sqrt(ab).Therefore V(shaded) = (pi/6) h ab/(b-a) 2 sqrt(ab) = (pi h/3) (ab)^(3/2) / (b-a).To find the desired area V, we now subract the volume of the little cone V(little cone) = (pi h'/3) a^2 = (pi h/3) a^3 / (b-a),to get V = (pi h/3) ((ab)^(3/2) - a^3) / (b-a), = (pi a^2 h/3) (r^(3/2) - 1) / (r - 1). QED.Note that this proof assumes b > a. The formula also holds for the a < bcase (which can be seen by subtracting the above from the volume of theentire frustum). For a = b, there is a removable singularity which, whenremoved, gives the obvious answer pi a^2 h/2 (i.e., half the cylinder).-Jim Ferry === Subject: Re: combitoricssorry. I did leave out some information, about paul and sandy and I left outgary.here's more complete information. Torrey loves Matt on some days, but Torrey hates Matt on other days.Torrey loves Paul on some days, but Torrey hates Paul on other days.Torrey loves Sandy on some days, but Torrey never hates Sandy.Sandy hates Torrey on some days, but Sandy loves Torrey on other days.Sandy loves Gary on some days but Sandy never hates Gary.Matt loves Sandy on some days, but Matt hates Sandy on other daysPaul Loves Matt on some days, but Paul hates other days.Paul hates Sandy on some days, but Paul never loves Sandy. Gary hates Sandy, but Gary never loves Sandy.Gary Loves Torrey on some days but Gary hates Torrey on other days.I have reduced the problem to just hate and love relationship. isthere anything wrong with doing that? because I'm ignoring thetime/day information. what about the time/love/hate relationships?is there a way to figure out what pattern they have? with theinformation above, taking only the love and hate relationships, thereare 32 possibles. what happens if i take into consideration thedays? or is there a way to do that? what other information will oneneed in order to do that?I think gary and paul should be nicer to sandy. thanks all in advance. sean Torrey loves Matt on some days, but Torrey hate Matt on other days.Torrey also loves Paul on some days, but Torrey hates Paul on otherdays.Torrey also loves Sandy on some days, but she never hates Sandy.Sandy hates Torrey on some days, but loves her on other daysMatt loves Sandy on some some days, but hates her on other other daysPaul Loves Matt on some days, but Paul hates other days.paul hates Sandy on some days. > how can I figure out how many possible combinations there are thatdesribes the love and hate relationship between Torrey, Matt, Paul,and Sandy?I have 32. but I'm not sure if i'm right. and i'm doing it brute forceway.> A description of the love/hate relationships in the > group would be given by a table as follows, with> 24 True/False values, but the given information is> only enough to fix the values shown:> How A Feels About B > Person A Person B Love Hate > -------- -------- -------------------> Torrey Matt T T> Torrey Paul T T > Torrey Sandy T F> Matt Torrey > Matt Paul> Matt Sandy T T> Paul Torrey> Paul Matt T T> Paul Sandy ? T> Sandy Torrey T T> Sandy Matt> Sandy Paul> It's not quite clear how Paul feels about Sandy, but > if the ? is known to be T, then that leaves 10 values > undetermined; i.e, there are 2**10 = 1,024 different> tables consistent with the given information, and > each one corresponds to a different set of love/hate > relationships in the group. > --r.e.s. === Subject: Re: How many ways to put 5 balls into 500 ordered cups?> You are given 500 numbered cups and five identical balls. Any cup can> hold up to five balls. How many ways can you put the five balls into> the 500 cups?> As a warm-up, let's try smaller sets of ordered cups:> Cups Arrangements> === =================> 1 1> 2 6> 3 21> 4 56> 5 126> 6 252> 7 462> 8 792> 9 1287> 10 2002For what its worth! I can't explain it.Arrangements = Binomial Cofficient(nCm), where n = number of cups +4; m = 5Example; for C=5, n = C+4 = 9; 9C5 = 126 A = 504C5 = 265,661,562,600 for 500 cups.regards B. === Subject: Re: finite measure> Let m be a finite measure on B(R), the borel sigma field generated by> the real line. I have to prove that there are at most countably many x> in R (atoms) such that m(x)>0.Isn't it the case that each atom x has an interior? If so there is a rational number in it. Since there are only a countable number of rationals there must be a countable number of atoms.Bob Kolker === Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read.> My web site www.johncho.us has all the information. I am seeking people > who are familiar with procedures. I am also considering getting a lawyer.Why don't you put all this creative energy into studying? === Subject: Help needed from a group theorist by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KMShs18156;An exponent E(G) of a group G is the lcm of all the orders o(g),g in G.We have that G is cyclic if and only if E(G)=|G|. Does anybody have an idea how can we find all n such thatthe exponent of Z_n*,the group of units of Z/Zn is 2? E(Z_n*)=2 n=?The solution is n=2,3,4,6,8,12,24 ,but I don't have a cluewhere did this come from. === Subject: Re: Cavemen> Where can I find pics of cavemens language e.g. A pic of a wall inprintNot on sci.math, I can tell you that. === Subject: Re: Axioms defining a finite field>Axiom 4, the associativity of multiplication, is also not redundant. There>exist finite counterexamples called semifields.>On the other hand, I'm suspicious of axiom 5: a * 1 = a. Maybe I'm missing>something obvious, but can there possibly be a finite ring without identity>that has no zero divisors?Z_3 with addition as usual, and this multiplication table: 0 1 2 --------0 |0 0 0 1 |0 2 12 |0 1 2fits all the requirements-- Wim Benthem === Subject: Re: Polynomial solutions to Pell eqn>The Pell equation and the obvious cubic generalisation>X^2 - DY^2 = 1 and>X^3 + DY^3 +(D^2)Z^3 - 3DXYZ = 1>have simple polynomial solutions>(n,1) for D = n^2 1>(n^2, n, 1) for D = n^31>In the square case, you can use continued>fractions to expand expression of the form, say,>SQRT(an^2 + bn + c), but in the cubic case>no such option is available.>Simple cases can be guessed>for example>D = n^3 3>X = n^6 3n^3 1>Y = n^5 2n^2>X = n^4 n>D = n^3 +2, n even>X = (9n^6)/4 + (9n^3)/2 +1>Y = (9n^5)/4 + 3n^2>Z = 3n(3n^3 +2)/4>I was wondering if there is any>systematic approach to obtaining>solutions ?Not real sure about a systematic approach, but looking at a few data points forspecific D may be useful. Consider the following:D=10^3+5*10 X=62591931727331611501 D=100^3+5*100X=4910017588770392230083131681058781350001 D=1000^3+5*1000X= 489788270129887199471510415004171147924480921387800135000001 D=10000^3+5*1000000000001 D=100000^3+5*100000X= 489776026824440065292464512748268928258240960054041644806667920 0004480920000138780000001350000000001 D=1000000^3+5*1000000X= 489776025612244400640129246451200748268928002582409600005404164 480006667920000004480920000001387800000000135000000000001 D=10000000^3+5*10000000X= 489776025600122444006400012924645120000748268928000025824096000 000540416448000006667920000000044809200000000138780000000000135 00000000000001 D=100000000^3+5*100000000X= 489776025600001224440064000001292464512000000748268928000000258 240960000000054041644800000006667920000000000448092000000000013 8780000000000001350000000000000001 There is a pattern here that seems to indicate that a subset of {n^3+5*n} has apolynomial solution to P3. I'm not sure what the exact answer is, but it couldprobably be worked out, given sufficient time and interest. There are lots ofother curiosities that can be found this way.Rich === Subject: Re: Polynomial solutions to Pell eqn>The Pell equation and the obvious cubic generalisation>X^2 - DY^2 = 1 and>X^3 + DY^3 +(D^2)Z^3 - 3DXYZ = 1>have simple polynomial solutions>(n,1) for D = n^2 1>(n^2, n, 1) for D = n^31>In the square case, you can use continued>fractions to expand expression of the form, say,>SQRT(an^2 + bn + c), but in the cubic case>no such option is available.>Simple cases can be guessed>for example>D = n^3 3>X = n^6 3n^3 1>Y = n^5 2n^2>X = n^4 n>D = n^3 +2, n even>X = (9n^6)/4 + (9n^3)/2 +1>Y = (9n^5)/4 + 3n^2>Z = 3n(3n^3 +2)/4>I was wondering if there is any>systematic approach to obtaining>solutions ?Not real sure about a systematic approach, but looking at a few data points forspecific D may be useful. Consider the following:D=10^3+5*10 X=62591931727331611501 D=100^3+5*100X=4910017588770392230083131681058781350001 D=1000^3+5*1000X= 489788270129887199471510415004171147924480921387800135000001 D=10000^3+5*1000000000001 D=100000^3+5*100000X= 489776026824440065292464512748268928258240960054041644806667920 0004480920000138780000001350000000001 D=1000000^3+5*1000000X= 489776025612244400640129246451200748268928002582409600005404164 480006667920000004480920000001387800000000135000000000001 D=10000000^3+5*10000000X= 489776025600122444006400012924645120000748268928000025824096000 000540416448000006667920000000044809200000000138780000000000135 00000000000001 D=100000000^3+5*100000000X= 489776025600001224440064000001292464512000000748268928000000258 240960000000054041644800000006667920000000000448092000000000013 8780000000000001350000000000000001 There is a pattern here that seems to indicate that a subset of {n^3+5*n} has apolynomial solution to P3. I'm not sure what the exact answer is, but it couldprobably be worked out, given sufficient time and interest. There are lots ofother curiosities that can be found this way.Rich === Subject: Re: finite measure> An uncountable collection of points in R always forms some sort of> interval (a,b), right? Certainly not. Look at the Cantor set, or the irrationals. === Subject: Re: finite measure> Isn't it the case that each atom x has an interior? No, consider the unit point mass at the origin. === Subject: Re: Genetics and Math-Ability>There is an old saying about inheritance of mathematical ability.>Unlike most interitance, this goes from a man to his son-in-law.>The explanation being, I guess, that when the professor's daughter is>of the right age, she meets the professor's current student, and>marries him.My wife, her sister and her first cousin are also married to mathematicians. Obviously there's an inherited tendency to marry mathematicians. === Subject: Re: Pascals TriangleRob Pratt escribi.97 en el mensaje>> There was an interesting puzzle in the Sunday Times in the UK>> recently that set me thinking about the issue that the puzzle raised.> The idea is that of a set of points arranged in the familiar Pascal's>> triangle format (rows 1, 2, 3, ... etc. containing 1, 2, 3, ...>> points in a triangular format) after which the issue is that of the>> total number of equilateral triangles that can be found in this set>> of points.> For N in 1..10, I get 0, 1, 5, 13, 27, 48, 78, 118, 170, 235, which> can be expressed as>> floor((N - 1) (N + 1) (2 N - 1) / 8)>> and appears in Sloane's database:>http://www.research.att.com/cgi-bin/access.cgi/as/ njas/sequences/eisA.cgi?An> um=A002717It seems that you only count triangles with sides paralel to arraysides.But there are other equilateral triangles with vertices in the points of> theequilateral triangular array with its sides in another orientation.Exactly, there areComb(n, 4) = n(n + 1)(n - 1)(n + 2)/24in a triangular array of order n. For n = 1 to 10,0, 1, 5, 15, 35, 70, 126, 210, 330, 495http://math.smsu.edu/~les/POW03_01.htmlhttp:// www.research.att.com/projects/OEIS?Anum=A000332> Hi Ignacio,> Thank you for your input - I must admit that I also missed the rotated> triangles in my search algorithm. Do you have any insights into theminimum> number of grid points that need to be removed in order to eliminate all> equilateral triangles?> It seems possible that this might actually be easier to solve with the> rotated triangles since all the points (except one) on one side of all> non-rotated sub-triangles always need to be removed.> > GladmanOkay, the problem can still be formulated as a set covering problem. Onceall equilateral triangles (even the rotated ones) have been considered, forN in 1..11 we have0, 1, 2, 4, 7, 11, 16, 22, 28, 35, 44as the minimum number of grid points to be removed. This sequence is not inSloane's database.Rob Pratt === Subject: Re: 'erf' function in C> A possible improvement is to note that the even part of R(x),> (R(x)+R(-x))/2, is equal to 1/phi(x), thus only the odd part of R(x),> (R(x)-R(-x))/2, needs to be computed:> double Phi(double x)> {long double s,t=0,b=1,pwr=x;> int i;> s=x;> for(i=2;s!=t;i+=2)> { b/=(i+1);> pwr=pwr*x*x;> t=s;> s+=pwr*b;> }> return .5+s*exp(-.5*x*x-.91893853320467274178L);> }Very nice, and a definite improvement*! It speeds up the convergenceconsiderably which, quite possibly, accounts for more accurate results.i.e. w/CVF on Wintel, x phi(x) 0.123 0.5489464510164368 1.200 0.8849303297782917 2.400 0.9918024640754040 6.100 0.9999999994696567-6.100 0.0000000005303433-1.100 0.1356660609463827 7.200 0.9999999999996979* Actually, it's an understatement considering the Syziphian effort atLANL some odd thirty years ago. See, netlibfn lib.--Dr.B.Voh------------------------------------------------ ------Applied Algorithms http://sdynamix.com === Subject: Re: Collatz Conjecture : Symmetry question. === >Subject: Re: Collatz Conjecture : Symmetry question.>Message-id: I have a program that can build each level from the previous one.>> >> I also have written this program. I am using an arbitrary precision>> lib to allow large numbers. I have two applications, one calculates>> just end points, and the other traverses the tree. We should compare>> notes >>Sure, I'll see if I can dig up my program. The final version used text>files to hold the level data. I stopped at Level 84 because the text>file, at 3.3GB was getting too big to fit on a single CD when zipped. and maybe work together and release a simple tool for others to>> use. Just a thought.>>Simple, yes. Usefull? That remains to be seen.>>But I've got some other stuff that might be interesting. I finally>solved my Big Problem (multi-generation sequence vectors) and am >working on a web page to document it. I can post some of it here>along with the programs if you're interested.I originally did this using a list in memory, but I ran out of memory.Here's the text file based version written in Python (which also does Big Arithmetic). The program reads the previous level out of a file,doubles every number and if a number is both == 1 (mod 3) AND== 0 (mod 2), it spawns a new branch by using the inverse 3x+1rule.# usage: python lread.py n# where n is level to process# reads file named Ln.txt# and outputs next level#import syslevel = sys.argv[1]filein = 'L' + level + '.txt'f = open(filein,'r')s = 'begin'while s != '': if s != 'begin': n = long(s) print n*2 p3 = divmod(n,3) if (p3[1]==1): p2 = divmod(n,2) if (p2[1]==0): print (n-1)/3 s = f.readline()f.closeStart by creating a text file named L5.txt that contains just16Running the program using L5.txtpython lread.py 5produces the following output:325To build up successive levels, redirect the output to a filepython lread.py 5 > L6.txtOf course, you can batch file the low levels since they go quick.It gets slow by the time you get to Level 70. And it starts eating up disk space: 65 File(s) 448,531,354 bytesThe one good thing is that you only need to keep the last levelon hand. All the previous ones can be archived. I originally wanted to go all the way to Level 100, but since Level 84took 3.3 GB, I lost interest at that point.> Hi Mensanator,> You originally calculated these starting seeds for higher levels> which I could not do because of my algorithm which found the level> requested and thus all preceeding levels. So it was very slow finding> all seeds for levels >20. > What is surprising here is, your level counts do not match your old> level counts as shown below.> > > Number of starting seeds for each level starting @ level 6 in the Collatz tree.> Level # of seeds> Level 6 2 > Level 7 2 > Level 8 4> Level 9 4> Level 10 6> Level 11 6> Level 12 8 > Level 13 10 > Level 14 14 > Level 15 18 > Level 16 24 > Level 17 29 > Level 18 36> Level 19 44 > Level 20 58 > Level 21 72 > Level 22 91 > Level 23 113 > Level 24 143 > Level 25 179 > Level 26 227 > Level 27 287 > Level 28 366 > Level 29 460> Level 30 578 > Level 31 732 > Level 32 926 > Level 33 1174 > Level 34 1489 > Level 35 1879 > Level 36 2365 > Etc.> Am I interpeting somthing wrong here?Yes. The list I just posted is the size of the text files in bytes.L6.txt still has two seeds but it uses 7 bytes of disk space.I was pointing that out in case anyone wants to try running mylittle program (don't say you weren't warned).You can't get the count directly from the file size. Unlike Unix/Linux,there's no text file line counting utility in Windows. You can easily track the number of seeds generated and print that, but the count willend up in the file when doing simple re-direct.If one was ambitious, one could add a file write routine in place of theprint statements. That way you could print the count without it ending up in the level file.Or you could write a program that counts lines in a text file.> Dan === Subject: Re: Simple idea, mathematics and common-senseI know I shouldn't try, but the light... it's so beautifu...zap!>There's something else important here which is the ambiguity of the>square root operatorIt is not ambiguous. Sqrt(x) (or x^(1/2) or the notation with thesquare root symbol) all define the principal square root (provided xis a positive real number). So Sqrt(9) = 3. It is not -3 even though-3 is a square root of 9. Just read about it on the following link (You could also go to thelibrary and read a decent math-book there):http://mathworld.wolfram.com/SquareRoot.html === Subject: Re: CavemenWhere can I find pics of cavemens language e.g. A pic of a wall inprint> Not on sci.math, I can tell you that.No binaries, but maybe an ASCII rendering.To bring it on topic, van der Waerden in _Geometry and Algebra in AncientCivilizations_ quotes a translation from something ancient (how's that fornot consulting primary sources?) as if you do it this way [ruler andcompass construction, with no measurements] you won't have to worry aboutthe gods rejecting your sacrifice. === Subject: Re: help is needed!!!> i want play whit MAPLE but i can not find sourse for it> if any body know some sourses for it or any one can help and support> me please say more> thank you> padAs far as I know, the only one that's free (to students) is MuPad. And itmay not be free any more.google.after a little search, i found dmoz open directory project, and thenfollowing its internal links to science:math:software discovered a wholepageload of relevant links.the big six are Derive, Maple, Mathcad, Mathematica, MATLAB, and MuPad.But there are links to apparently specialized software in lots of area, soif you're more interested in Algebraic Geometry or Algebra, there are linksto software in those areas. === Subject: Re: Cavemen> Where can I find pics of cavemens language e.g. A pic of a wall inprintThis one comes close:http://www.jmilne.org/bitted.html:D === Subject: Re: Simple idea, mathematics and common-sense> Yes, I can talk it all out rigorously and in a heavily mathematical> format,James,I have been following your threads for some time now and I have seen on more occasions than I can count a request for you to do just this.I seem to recall that when you have deigned to acknowledge these requests it wasn't ...[talked] out rigourously and in a heavily mathematical format it was more of a diatribe against mathematicians and the evil society that they control.I have seen you claim that clearly stated definitions are wrong, that (as you did in this post) there is some inherent ambiguity and that professional mathematicians (as well as the very capable amatuers) are too stupid to understand but I have *NEVER* seen you post anything that even I would consider to be rigourous.Having now said that you can do this, when can I expect to see it?I'm sure that if I have any difficulty then either you or somebody else will be able to clarify.Ivan. === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> <8765e4q2fe.fsf@phiwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@phiwumbda.org> <87vfm4o85i.fsf@phiwumbda.org> <87n07fo9qr.fsf@phiwumbda.org> <87k72jnxtp.fsf@phiwumbda.org> <87ptcau3xj.fsf@phiwumbda.org> Assumption: There exists a 0 at the end of computation.> Of course there is. How can there not be?> My TM will never overwrite the last 0. This is the source of your error (and just about all your errors.) Thereis no last 0 on the *infinite* list, since there will always be a 0after it. But then if the tape originally had all the natural numbers,> That's a big if, isn't it? That was your assumption and you construction. *IF* your premises held,then a lot of wrong things happen. You missed the fact in elementary school that if n is some oddnumber, then so is n+2.> I only see two odd numbers in the set (1,2,3).> 3+2 = a number not in my set.> 3+2 must not be finite. Again, don't try redefining already-used terms. I'm not talking about a finite set, I'm talking about takingthe natural numbers and applying that process to it. Your hypotheticaltape supposedly had all the natural numbers on it. Once the processis complete, then any set you look at {1..N} will have all its oddnumbers removed.> Supposedly is the key word here.> There will be such an N. N could be quite large (it might even equal 3).> Obviously, the original tape didn't contain every natural number. Yes, it is obvious, and this is what everyone has been telling you allalong. (In case you forgot, you were the one claiming the existence ofsuch a tape) How many times do you need to be told that you will not havea tape with all the natural numbers on it? Again, make sure you learn about the difference between 'artibrarilylarge' and 'infinite.'> Why don't you give me a TM that decides if a string is> 'arbitrarily large' or 'infinite.' TMs don't decide anything about infinite things. They only accept finite-sized input. I'm telling you this again for the fourth time. The set {1, 11, 111, 1111, 11111, ... } will contain strings that arearbitrarily long but will not contain the infinite string 1111....> And you have an algorithm that proves this? No, that is what the notation ... means. Carry on the process. The set contains {1^k, for any natural number k} Perhaps you are confused about the fact that infinity is nota natural number.> How do prove there is such a thing as an infinite string of 1's? There isn't. There is no infinity in natural numbers. There arearbitrarily large numbers, though.> I have given a proof that every string is finite as far as a TM is> concerned, even for TM's that perform an infinite number of> operations. Every string that you give a TM *should* be finite in the first place. You're the one trying to feed a TM an infinite string and then trying to treat an infinite string as a finite number (The difference between 10, 110, 1110, 11110, ... , and your infinite string is that the infinite string does NOT has a last 0, like you claim.) ( Do you think that 1 - 0.999999999 has a last digit which is 1? )J === Subject: Re: 1=ma+nbJohn Jones> This [Blankinship's algo] is great and much more suited to a> spreadsheet than the algorithm I currently use....> Who or why or when or what is Blankinship? Good on him/her.Nobody seems to have noticed this tidy routine before W. A. Blankinship in1963. I learned it only recently, from this:http://mathworld.wolfram.com/BlankinshipAlgorithm.htmlLH= ==Subject: Re: Infinite Galois Groups>Let F be the field of lengths constuctible by compass and>straighthedge. What is G(F/Q)?> Well, formally, it is the inverse limit of the Galois Groups of the> finite Galois subextensions, with the connecting morphisms being the> corresponding quotient maps.> It should be fairly complicated, since it will include the> subextensions given by the 2^n-th and 3^n-th roots of all integers.I think I am missing something here. Why the 3^n-th roots? It is notpossible to construct a segment the length of the cube root of a givensegment with compasses and straightedge.Darren>Let K be the field generated by the square roots of the prime numbers.>What is G(K/Q)? What is G(F/K)?> This is the increasing union of> Q subset Q(sqrt(2)) subset Q(sqrt(2),sqrt(3)) subset> Q(sqrt(2),sqrt(3),sqrt(5)) subset ...> In each case, the relative Galois group is Z/2Z, and the Galois gropu> of the extension> Q(sqrt(2),...,sqrt(pn)) over Q> where pn is the n-th prime is equal to (Z/2Z)^n (you would need to> prove this; see for example Section 6.7 in Lang's Algebra, 3rd> edition). The maps going down are> ... -> Z_2 x Z_2 x Z_2 --> Z_2 x Z_2 --> Z_2 --> 1> where each map chops off the last coordinate. The inverse limit is an> infinite product of copies of Z_2.> -- > === =========================================================== === =====> It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> === =========================================================== === =====> Arturo Magidin> magidin@math.berkeley.edu===Subject: Re: Cavemen> Where can I find pics of cavemens language e.g. A pic of a wall inprintIs it a mathematical language? If not, not here. === Subject: Ideas for course on great ideas in (theoretical) CS? I have been coerced into teaching a Honors course the Fall(mostly for non-CS freshman/sophomore Honors students). Myidea was to do some Great Ideas/Problems/Puzzles etc. fromcomputer science -- emphasis on theory/algorithms and related areas like graph theory/combinatorics.Of course, the honors college wants a syllabus in one week! I looked at the book, Great Ideas in CS and though a nicebook, seems a bit light on the theory side ... given thatI want to focus on theory to keep me interested. I saw the course/web site at CMU Great Ideas in Theoretical Computer Scienceand may use that as a guide for some of the course. Examplesof some things I might discuss (besides a couple weeks onbasics/definitions/history) include Towers of Hanoi, ByzantineGenerals, voting problems, maybe a gentle discussionof interactive proofs, prisoner's dilemma, game of life, primalitytesting, graph coloring ... anything that can be discussed in a day or so to folks with no CS background, yet which has some theory component to it ... stuff that is surprising or counter-intuitiveis all the better ;) Anyway, if anyone has any suggestions for material/topicsthat I might cover, I would most appreciate it. Any pointerswould be accessible to students would be great (I have a couple)would be great.(or a link to one) to comp.theory in a week or so.Chip Klostermeyer === Subject: Math notation question: '('I just ran across a math question in a review and I'm totally stumpedby the answer. Then I realized that I may have been misreading thenotation.The statement was:There exist x, y, and z that are nonzero. Such that 1 ( y>x and xy=z.Does 1 ( y>x mean that both y & x are less than 1? Looking at theanswer that is the only thing that makes sense to me. I just don'tunderstand the notation.Ben.. === Subject: integralHey, can anyone help me in integrating from 0 to infinity e^(-(x^2))dx?-Greg R. === Subject: Re: integral> Hey,> can anyone help me in integrating from 0 to infinity e^(-(x^2))dx?> -Greg R.I'd suggest trying a power series.David Moran === Subject: Re: integral> ...integrating from 0 to infinity e^(-(x^2))dx?Let I be the number we want.I^2= (integral from 0 to infinity e^(-(x^2))dx) times (integral from 0 to infinity e^(-(y^2))dy)= double integral (0 to infinity in x and y)[e^-(x^2+y^2)]dy*dx(change to polar coordinates)= double integral(0 <= theta <= pi/2)(0 <= r < infinity)of the function: e^(-r^2) times r*dr*d(theta) infinity= (-1/2)e^(-r^2)| times pi/2 0= pi/4.Since I^2 = pi/4,I = (1/2)sqrt(pi). === Subject: Re: Axioms defining a finite field> Z_3 with addition as usual, and this multiplication table:> 0 1 2> --------> 0 |0 0 0 > 1 |0 2 1> 2 |0 1 2> fits all the requirementsYour 2 looks very much like an identiy.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: complex problemI have to solve equetions with complex coefficients:T11= - 1.50005 + 5.93649i;G1 = 0.08562 + 0.76492i;T12 = -6.4481 - 2.79474i;G1^2 * S22 + 2*G1*S11 - G1^2 + T11*G1 - T11*S11 = 0andG1*S22 - T12*G1 + S11 + T12*S11 = 0I know the solution which is:S11 = 0.13272 + 0.80569 iS22 = -0.52264 - 0.13305 iBut from analytical I 've gotS11 = G1S22 = -1Where I made a mistake? How to get this solution (S11 = 0.13272 + 0.80569 i;S22 = -0.52264 - 0.13305 i) ?--Miro === Subject: Re: Infinite Galois Groups>Let F be the field of lengths constuctible by compass and>straighthedge. What is G(F/Q)?> Well, formally, it is the inverse limit of the Galois Groups of the> finite Galois subextensions, with the connecting morphisms being the> corresponding quotient maps.> It should be fairly complicated, since it will include the> subextensions given by the 2^n-th and 3^n-th roots of all integers.cube root of 3 is constructible? === Subject: Re: Infinite Galois Groups||>Let F be the field of lengths constuctible by compass and|>straighthedge. What is G(F/Q)?|> |> Well, formally, it is the inverse limit of the Galois Groups of the|> finite Galois subextensions, with the connecting morphisms being the|> corresponding quotient maps.|> |> It should be fairly complicated, since it will include the|> subextensions given by the 2^n-th and 3^n-th roots of all integers.||cube root of 3 is constructible?i guess so. also, it's nice to know that you can duplicate the cube(cube root of 2) and trisect a 60-degree angle (9th root of -1).-- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Genetics and Math-Ability> My wife, her sister and her first cousin are also married to > mathematicians. Obviously there's an inherited tendency to marry > mathematicians.Have you noticed that people whose parent's did not have children, also tend not to have children.Bob Kolker === Subject: Re: Letter to Prof Ullrich and others > I am just curious: why are people like you > interested in even acknowleding James Harris and his ilk ? > This guy has been oopsing on sci.math for ages (at least several > years) I see. come from the discussions. I have learned quite a bit myself. Alsointeresting results come from it, like the Magidin-McKinnon theorem.One of the things I learned was how to deal properly with quadraticfields. > Is there a history or an obvious reason that I am missing ?There is a history indeed. Even for mathematicians it becomesincreasingly difficult to refute James' arguments because of theway he veils them.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Infinite Galois Groups> cube root of 3 is constructible?Not with ruler and compass. Any extension field constructed by adjunction of quantities that can be made of ruler and compass either is the same as the previous field or of degree 2 over it. In the end one can only get to fields of degree 2^n over the ground field.That is why it is impossible to trisect the angle with ruler and compass. To do so whould require getting the root of an irreducible cubic equation. No can do with ruler and compass.Bob Kolker === Subject: Re: What's the computation time of this Gradient optimization method, O(kN) or O(kN^2)?Content-transferncoding: 8bit> Dear All,> I am trying to figure out the complexity level of the gradient method for> minimizing an objective function.> Assume J is the scalar objection function of N vectors, each vector is of k> dimension.> Now we are trying to find a k-dimension vector x such that J is minimized,> e.g. J is the sum of squares of the distances from N points in R^k space to> a line segment,which> is connected by a known vector a and the unknown point x in R^k space.> So given this definition of J, the way we did is to calculate dJ/dx = 0 and> find the vector x.> However, the expression dJ/dx is not usually in explicit form, i.e, x can> not directly be obtained and> have to be done by some iterative approach.> Then for this kind of optimization problem, how to define the computational> complexity?> It is O(kN) or O(kN^2)?> And how to verify this?It seems to me there are (at least) two questions here.The first is, What's the operations count for a single gradient descentiteration? Presumably function evaluations and partial derivatives area single operation, else the question is completely fatuous. If youcan't figure this out... I'll say no more.The second is the far more interesting question, How many iterations(hence, how many function evaluations/flops) are required to get aniterate within epsilon of a minimizer? And this surely has no answer: it's unbounded, since the convergence of gradient descent can be asslow as we wish, even for convex functions.A modification of this: how many iterates does it take to get f(x_k)within epsilon of the minimum? As I recall, the best possible bound(i.e. in the worst possible case) for convex functions is somethinglike |f(x_k) - f(xmin)| <= c/sqrt(k),so it can take a LONG time. (The constant c involves unknowables,AFAIK.)--Ron Bruck === Subject: Re: there is no such thing as infinity> Ah, I remember when I first started programming. All I had was a really> hot needle and had to write my code directly to the SIMMs. It was a major> leap forward when I upgraded and could just type copy con file.exe...> When I started, SIMMs hadn't been invented. We had people who had seen> so many punch cards and papertapes they could read the holes by eye.> I set the bootstrap instructions using an array of toggle switches on> one mainframe we had.Paper cards? Ah, you were lucky! When I first started investigatingprogramming, we had to chisel instruction sets into very large rocks andcarry them ourselves to the CPU (Chiseling Processors Union) to get thework done. And we were thankful...-- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people think, Mr. Feynman? === Subject: Re: Axioms defining a finite field>Axiom 4, the associativity of multiplication, is also not redundant. There>exist finite counterexamples called semifields.>On the other hand, I'm suspicious of axiom 5: a * 1 = a. Maybe I'm missing>something obvious, but can there possibly be a finite ring without identity>that has no zero divisors?> Z_3 with addition as usual, and this multiplication table:> 0 1 2> --------> 0 |0 0 0 > 1 |0 2 1> 2 |0 1 2> fits all the requirementsNo, it doesn't; your 2 is the 1 specified in the definition. === Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read.> i do not respond to many of the posts because my argument is with this > school and not people of the message board. The ideas that the board gives > www.versuslaw.com after that i will try to do another google search for > statewide policies and procedures regarding academic progressWell, if this is for real, I really want you to keep us updated on thehorrible lawyers and horrible judges you're going to face. Soundslike great fun. === Subject: Re: integral can anyone help me in integrating from 0 to infinity e^(-(x^2))dx?-Greg R.> I'd suggest trying a power series.How could that possibly help? === Subject: Re: Letter to Prof Ullrich and others > Or you can click onhttp://groups.google.com/groups?as_q=racial%20slur&safe=off& ie=ISO-8859-1&as_ugroup=sci.math&as_uauthors=David%20Ullrich& lr=&hl=en> Oh yeah, in case you're wondering about what Ullrich finds so> offensive that the subject of racial slur came to his mind, I said> that he'd acted as my lapdog in an instance.> That was YEARS ago.It was three days ago:http://mathquest.com/discuss/sci.math/a/m/580751/580751Try Google withmoron factorization bulland guess who comes up. === Subject: Re: walking on a grate..>hello everybody,>suppose I have a grate, N x M. how many paths are from node (0,0) to >(N-1,M-1) eg. from one corner to the diagonal-opposite one?>allowable paths are only with zero or positive change in node's coordinate.>how can I compute this? I would appreciate any suggestion and help..>cheers,>n.You are going to take n steps one way and m steps the other for n+mtotal steps. You just have to count how many ways there are to choosethe n horizontal moves from n+m moves, i.e.: C(n+m, n) = (n+m)!/(n!m!)--Lynn === Subject: Re: Fresnel integrals>References regarding Fresnel's Integralssee Articles 560, 1166, 1169,1172,1323 in the followingnice books : === =========================================================== === ======Joseph EDWARDS ,,A Treatise on the Integral Calculus with Applications, Examples and Problems vol.I-vol.II , Chelsea Publishing Company, (Library of Congress Catalogue Card Number 55-234) === ==================================================== === =============Other incomplete references are : Cauchy [Comptes Rendus, tom XV, 534,573] Fresnel [Oeuvres, tom I] Gilbert [Memoires couronnes de l'Acad.de Bruxelles, t.XXXI, 1863] Knockenhauer [,,Die Undulationstheorie des Lichts,...?] Preston[,,Theory of Light , Art.141,...?] Verdet [Oeuvres tom V] = Perhaps help,Alex === ==Subject: Re: Help needed from a group theoristMartin> An exponent E(G) of a group G is the lcm of all the orders o(g),> g in G.We have that G is cyclic if and only if E(G)=|G|.> Does anybody have an idea how can we find all n such that> the exponent of Z_n*,the group of units of Z/Zn is 2?> E(Z_n*)=2 n=?This just means that x^2=1 for any invertible x in Z_n. Thus n cannot haveany prime factors other than 2 and 3. Write n=(2^s)(3^t). Splitting Z_n as aproduct of two rings, the cases are-- E(Z_(2^s)*)=2 and E(Z_(3^t)*)=1 i.e. s>=1 and t=0 (n=2,4,8,16,32,...)-- E(Z_(2^s)*)=1 and E(Z_(3^t)*)=2 i.e. s=0 and t=1 (n=3 only)-- E(Z_(2^s)*)=2 and E(Z_(3^t)*)=2 i.e. s>=1 and t=1 (n=6,12,24,48,...)I hope I haven't blundered somewhere in here :)LH === Subject: Re: Infinite Galois Groups Adjunct Assistant Professor at the University of Montana.>Let F be the field of lengths constuctible by compass and>straighthedge. What is G(F/Q)?> > Well, formally, it is the inverse limit of the Galois Groups of the> finite Galois subextensions, with the connecting morphisms being the> corresponding quotient maps.> > It should be fairly complicated, since it will include the> subextensions given by the 2^n-th and 3^n-th roots of all integers.>I think I am missing something here. Why the 3^n-th roots? It is not>possible to construct a segment the length of the cube root of a given>segment with compasses and straightedge.It should be 2^n-th roots; don't know where the heck that 3 came from...-- === ==Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: Letter to Prof Ullrich and others> I am just curious: why are people like you> interested in even acknowleding James Harris and his ilk ?> This guy has been oopsing on sci.math for ages (at least several> years) I see. > Is it not worthwhile for everyone to just ignore him at this point ? > There is real danger of course that someone would take him seriously,> but I feel thats remote.> It seems that he has taken to calling universities to complain. > Lets not giving him any more attention. Without attention, he will> shrivel up and disappear.> Is there a history or an obvious reason that I am missing ?I made this same point a while ago. I think you are right. For thosewho feel a need to reply to JSH keep innocent readers from being takenin, I suggest simply replying to everything he posts with This isbull! If he wants to waste his own time writing long longpostings of incorrect mathematics, professional mathematiciansshouldn't have to waste their time replying in detail. On the otherhand, maybe people are compassionate, taking pity on him, and tryingto set him straight in the hopes that someday he might actually dosomething worthwhile. Nora B. seems to fit this mold. === Subject: Re: Fresnel integralsLarry Hammick> 1) int[-inf to inf] cos x^2 dx = sqrt(pi/2)> 2) int[-inf to inf] exp(-x^2) dx = sqrt(pi)> Eqn (2) is easy to show by looking at the square of the left side> (a double integral) and switching to polar coordinates. I tried the> same approach with (1). ...highlights from advanced calculus, of which the Fresnel integrals will beone.)LH === Subject: Re: combitoricsSorry again...I made even more mistakes in my corrections. I hate it when I do that.now here's original problems that's been really corrected. Torrey loves Matt on some days, but Torrey hates Matt on other days. Torrey loves Paul on some days, but Torrey hates Paul on other days. Torrey loves Sandy on some days, but Torrey never hates Sandy. Sandy hates Torrey on some days, but Sandy loves Torrey on otherdays. Sandy loves Gary on some days, but Sandy never hates Gary. Matt loves Sandy on some days, but Matt hates Sandy on other days Paul Loves Matt on some days, but Paul hates Matt other days. Paul hates Sandy on some days, but Paul never loves Sandy. Gary hates Sandy on somedays, but Gary never loves Sandy. Gary Loves Torrey on some days, but Gary hates Torrey on other days.I have previously reduced the problem to just hate and loverelationship. is there anything wrong with doing that? because I'mignoring the time/day information. what about the time/love/haterelationships?is there a way to figure out what pattern they have? with theinformation above, taking only the love and hate relationships, thereare 32 possibles. what happens if i take into consideration thedays? or is there a way to do that? what other information will oneneed in order to do that?I think gary and paul should be nicer to sandy. thanks all in advance. sean === Subject: Re: integral> can anyone help me in integrating from 0 to infinity e^(-(x^2))dx?>> -Greg R.>>I'd suggest trying a power series.> How could that possibly help?I was trying to say take the power series for e^x, compose it with -x^2 andthen integrate that; since it's improper, you'd probably have to take alimit. I've seen something like this done in my calculus book.David Moran === Subject: re:Axioms defining a finite fieldSorry, you're right. Using rule 9, it is easy to show that everyelement has a multiplicative inverse. Fix a, then a*b must bedifferent for all b (if a*b=a*c, then a*(b-c)=0 violating rule 9 if band c different). Therefore the set {a*b} must be a permutation of{b} (that's where finite comes in), so that there is an element d sothat a*d=1. The same idea works for left inverse as well Posted Via Usenet.com Premium Usenet Newsgroup Services------------------------------------------------------ ---- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **---------------------------------------------------------- http://www.usenet.com === Subject: Re: Big Rip