mm-173 >Conjecture (TPC) are true, if the Generalized Riemann Hypothesis (GRH) is true. GRH implies twin primes? News to me. http://members.tripod.com/~american_almanac http://larouchepub.com/radio/index.html === Subject: Re: Factorial/Exponential Identity, In'nity ThatÍs a good point. Maybe what I need to have in the restriction on the interchange rules is that if you change a zero to a one then you have to change a one to a zero that is within a given 'nite distance of the one changed, for example q or q-1 or one for the set with p/q ones and (p-q)/q zeros. That might be too restrictive, I am to convert the sequence (01)... to 00001111(01).... 0101010101... 0011010101... 0010110101... 0001110101... 0001101101... 0001011101... 0000111101... Converting (01).. to (10)... is as follows: 01010101... 10101010... Each zero element can be changed to a one and each one within one element of that previous zero element is changed to a zero, and vice versa. Oh good, I have this idea within a few minutes of reading your post. Yet, that might only slow the process and then still I must show if that would not allow the conversion of any sequence (001)... through (1)... to all other sequences with in'nitely many ones and zeros. In your example you set for the sequence (01)... indexing from 1 each odd element to one and for each of those each multiple of fourth element to zero, getting (1110).... According to this rule, then, after changing an element from zero to one then youÍd have to change an element that is a one to a zero that is next to it. For example you could interchange each 4x-1 and 4xÍth elements, (0110).... Another problem with this restriction is that I need to show that it would allow any sequence of given contrived density to convert eventually to a sequence with the same known contrived density, as well that it would not convert to one of a different density. With this restricted sequence element interchange then 10000(0)... isnÍt convertible to 00001(0)..., instead it is to 01000(0)... which is to 00100(0)... which is to 00010(0)... which is to 00001(0).... How about (011) to (110)...? The list of intermediate sequences between them is in'nite. A rule to convert them all at once for each triplet is to convert (011) to (101) to (110). For the previous example the elements are interchanged in a two element sliding window. How about that? Basically what I want here is a description of a method to convert a sequence of a contrived rational density of ones and zeros to any other sequence with that same rational density but not any other, restricted sequence element interchange. Ross === Subject: Normal to a plan In 2-dimensions.. if I have a line such as the one below: Will the normal to this plane always be (0, 0)? How do I calculate the ignorance) Rick === Subject: Re: Contractible Spaces > If a topological space S is contractible to some point p in S, > is S contractible to every point in S? > Yes. Suppose S is contractible to a particular point s0 in S, with F: S x [0,1] --> S giving a contraction and s1 is any other point in S. De'ne G: S x [0,1] --> S by F(s,2t) if t in [0,1/2] G(s,t) = . F(s1,2-2t) if t in [1/2,1] Then G gives a contraction of S to s1. > If a topological space S is strongly contractible to some point > p in S, is S strongly contractible to every point in S? > Probably not in general, although I donÍt know offhand of any relevant example. There are theorems that say things like If A is a closed subset of X and both A, X are ANRs[*], then A is a deformation retract of X if and only if A is a strong deformation retract of X. Your question corresponds to the case A = {point} -- but itÍs not clear (to me, anyhow) just how to apply that theorem ... ThereÍs probably also an obstruction theory that applies to your question ... [*] ANR = absolute neighborhood retract -- I believe the above theorem actually applies to spaces that are ANRs for the class of compact metric spaces -- X is such a thing if any imbedding of X as a closed subspace of a compact metric space is a retract of some open neighborhood of the image ... > WhatÍs an example of a space that is > contractible but not strongly contractible? S is contractible to a when thereÍs some continuous h:Sx[0,1] -> S with > for all x in S, h(x,0) = x, h(x,1) = a, > and strongly contractible when in addition > for all t, h(a,t) = a === Subject: 4th order Bspline I have a stupid question about the 4th order Bspline. In the paper( see link), they use summation to denote a B-spline. But why they use a vector (4 j)~T. Can anyone help me to understand it? http://www.ri.cmu.edu/pub_'les/pub2/wu_yu_te_1998_1/wu_yu_te_ 1998_1.pdf === Subject: Re: Normal to a plan > In 2-dimensions.. if I have a line such as the one below: Will the normal to this plane always be (0, 0)? How do I calculate the ignorance) Rick > You need 3d to get a normal to a plane. Any normal to a plane has to go at right angles to every line in the plane, but in 2d, the plane and all its lines are the whole thing and there is nowhere else to go. === Subject: Linear Algebra question IÍm not sure if/why this would be true or false. IÍm having a hard time understanding how I could explain that it was true. I would appreciate any help. Given a linear transformation T:F^n-->F^m. Is it true that there exists an m x n matrix A such that T(x)=Ax (where we think of x as an n x 1 matrix? === Subject: Re: Factorial/Exponential Identity, In'nity > ThatÍs a good point. Maybe what I need to have in the restriction on > the interchange rules is that if you change a zero to a one then you > have to change a one to a zero that is within a given 'nite distance > of the one changed, for example q or q-1 or one for the set with p/q > ones and (p-q)/q zeros. What is the point of all this? It will not provide next s in densely ordered sets, because an essential property of such dense ordering is that there are no next s. You are trying to create something like a 4-sided triangle, i.e., something inherently impossible because it contradicts its own de'nition. === Subject: BlissÍs Theorem Does anyone know of a statement of BlissÍs Theorem on a website? If so, will you please share the link with me? In case there is more than one BlissÍs Theorem, itÍs the theorem that it used to justify the arc length formula for parametric equations being a Riemann integral, when the derivation doesnÍt lead to a Riemann sum. John === Subject: Re: Boolean Algebra - Arithmetic Relationship IÍd like to recommend some reading. For logic generally, see Tarski An >Introduction to Logic and the Methodology of the Deductive Sciences >OUP. (Dreadful title, best book.) For recursion theory IÍm at a bit of >a loss, how about Boolos & Jeffrey Computability and Logic CUP? IÍll look for both of these at B&N. Are they fairly accessible to the > Layman? Tarski de'nitely. Boolos & Jeffrey is a bit harder. For recursion theory there is also a book by Cutland Computability: An Introduction to Recursive Function Theory CUP. Amazon says What can computers do in principle? What are their inherent theoretical limitations? These are questions to which computer scientists must address themselves. The theoretical framework which enables such questions to be answered has been developed over the last 'fty years from the idea of a computable function: intuitively a function whose values can be calculated in an effective or automatic way. This book is an introduction to computability theory (or recursion theory as it is traditionally known to mathematicians). Dr Cutland begins with a mathematical characterisation of computable functions using a simple idealised computer (a register machine); after some comparison with other characterisations, he develops the mathematical theory, including a full discussion of non-computability and undecidability, and the theory of recursive and recursively enumerable sets. The later chapters provide an introduction to more advanced topics such as GildelÍs incompleteness theorem, degrees of unsolvability, the Recursion theorems and the theory of complexity of computation. Computability is thus a branch of mathematics which is of relevance also to computer scientists and philosophers. Mathematics students with no prior knowledge of the subject and computer science students who wish to supplement their practical expertise with some theoretical background will 'nd this book of use and interest. I have a rather interesting hypothesis for you: The role of mathematics, as I understand it, is to create various > logical structures that help predict/explain the nature of empirical > phenonmenon. We link these structures to empirical reality by making > assumptions. Maybe youÍd like reading Popper on the relationship ïtwixt logic and experiment. > Assuming an atom behaves like the Bohr model, that > assumptions are, they are always just approximations. The further you > develop any one logical model less its behavior matches that of > reality. Why should that be? I take it that reality is logical, so it is an exceedingly re'ned model of itself that is logical. > Would it therefore be good argument, that focus of > mathematical development should be on enhancing the breadth (Variety > of Symbolic Systems / Logical Perspectives) and not the depth > (Complexity of any one Symbolic System) of mathematics? I donÍt think that mathematics will be limited either in breadth or width. Both expand rapidly. > Because no > matter how great our assumptions are in the beginning, if we carry the > logic far enough they wonÍt match reality. Abstract mathematics gets applied to physics. Do you read John BaezÍs This WeekÍs Finds in Mathematical Physics at http://math.ucr.edu/home/baez/TWF.html ? In the end there seems to be no a priori reason why mathematics is useful in physics, but it seems to be so. ThatÍs a mystery. -Steve -- G.C. e-mailing me. === Subject: Re: Linear Algebra question > Given a linear transformation T:F^n-->F^m. Is it true that there exists an > m x n matrix A such that T(x)=Ax (where we think of x as an n x 1 matrix? Yes, there exists such a matrix, because L(F^n, F^m) is isomorphic to M_m,n(K), K being the 'eld over which is constructed F. Sam -- If sharing a thing in no way diminishes it, it is not rightly owned if it is not shared. - St Augustine === Subject: Re: Normal to a plan > You need 3d to get a normal to a plane. Any normal to a plane has to go at right angles to every line in the > plane, but in 2d, the plane and all its lines are the whole thing > and there is nowhere else to go. algorithm and canÍt 'gure out how IÍd be able to do this in 2D. What would you Rick === Subject: Re: Core error, FEAR is a natural response : The de'nition of algebraic integers as roots of monic polynomials >: with integer coef'cients gives the ability to give two supposed >: proofs that contradict each other. I donÍt get it. I understand the part about two arguments contradicting >each other, and one of them being wrong. But what does this have to >do with the de'nition of the algebraic integers? DonÍt the arguments >have to use the same de'nitions in order to be contradictory? And if >one of the arguments wrong, isnÍt the problem with the argument, not >the de'nition? The de'nition of algebraic integers excludes ALL roots of non-monic polynomials with integer coef'cients, when the polynomial is irreducible over Q. ItÍs that arbitrary exclusion that leads to a way to *appear* to prove two different and contradictory things. ItÍs not even a subtle proof to show it, but mathematicians have been running from the shock. (Neat, eh? Proving an error in core with a proof. Great fun.) It seems an entire math area--algebraic number theory--has to be re-worked. LOTS of textbooks to change. But then again, change can be fun. Apparently itÍs no longer necessary to assume anything in order to > arrive at a contradiction. JH states that the mere act of de'ning the > concept of algebraic integers is something that leads to contradicting > mathematical results, which must mean that mathematics itself is as > pointless as a very blunt object as without de'nitions there is no > mathematics. But how can we use mathematics to prove that mathematics > cannot be used to prove anything? Well that position requires that you not actually pay attention to what IÍm saying, which is the *easy* way out, eh? Social forces are powerful. Most of you need to accept that as human beings you are naked apes. You canÍt eat a math proof, so for many of you ideas are nothing, unless society TELLS you theyÍre ok. > The other option is that JHÍs proof is *gasp* wrong. The fact that > he accepts the contradicting proof about the algebraic integers as > being valid actually directly implies that JH accepts his proof must > be wrong. Now thatÍs bizarre as in fact, I note that thereÍs the *appearance* of dueling proofs. But proofs donÍt, canÍt duel. Social forces are so powerful, but you people get annoying because youÍre so damn irrational! Now why donÍt you try *actually looking at my work* instead of tossing out what you think the group expects, but then, you probably donÍt believe in ideas, do you? > What really scares me is, how can a physics major be so ignorant about > mathematical formulation and the logical requirements for a proof? I > have 'rst-hand experience that a lot of math is just regurgitated at > students without adequate explanation of the proofs and theories > underneath so that you can learn a lot of advanced mathematical > methods without understanding what the hell youÍre actually doing, but > physics of all subjects should require a solid understanding of math. What scares you is actually thinking for yourself, rather than letting society TELL you what is the truth. IÍve talked with a mathematician at my alma mater Vanderbilt University, and explained the entire argument to him...then he basically ran away. People, thereÍs the easy way, and thereÍs the hard way. The hard way is for you to come out and chatter nonsense like you just left the trees this morning, or you can actually pay attention and follow the logical argument. If you donÍt want to stand with intellectuals, and instead wish to be another cow in the herd, youÍll be treated accordingly, if I notice you and care to have some fun. James Harris === Subject: Re: Assignment Question (Cam6996) >If anyone can help with the following, please post a reply at >http://www.assignmenthelp.com/viewtopic.php?t=8 I have the answers to this problem but I donÍt know how to get there. >A child in danger of drowning in a river is being carried downstream >by a current that has a speed of 2.50km/h. The child is 0.600km form >shore and 0.800 km upstream of a boat landing when a rescue boat sets >out. >a) If the boat proceeds at its maximum speed of 20.0km/h relative to >the water what heading relative to the shore should the pilot take? >b)What angle does the boat velocity make with the shore? c)How long >does it take the boat to reach the child? the answers are a)39.6 b)41.6 and c) 3.00 minutes Please post a reply at http://www.assignmenthelp.com/viewtopic.php?t=8 Without loss of generality, letÍs assume the river §ows due south and the boat starts on the left bank. LetÍs de'ne the starting position of the boat as the origin of cartesian coordinates where x is east and y is north. Let the angle the boat makes with the shore be a, where a = 0 means directly upstream (north), a = 90 means directly across the river (east), and a = 180 means directly downstream (south), then the velocity vector of the boat can be broken into two perpendicular components: the upstream component v_u = 20 * cos(a) - 2.5 the across stream component v_a = 20 * sin(a) At time t (expressed in hours), the coordinates of the boat are (v_a * t, v_u * t) At time t, the coordinates of the child are (.6, .8 - 2.5 * t) If the boat rescues the child at time T, then the coordinates of the child at time T must match the coordinates of the boat at time T. v_a * T = .6 v_u * T = 8 - 2.5 * T Two equations in two unknowns (T and a). You should be able to take it from there. <> === Subject: Re: Use of variable independence, core error >... do the algebraic >integers indeed form a ring? IÍve never said that algebraic integers donÍt form a ring. What they form is a §awed ring, which doesnÍt include all the numbers > it must to prevent the possibility of appearing to prove two different > but opposite things. Well, that seems clear enough... but... is it possible that the §awed ring of algebraic integers includes too many numbers? Is that a possibility? -- Clive Tooth http://www.clivetooth.dk === Subject: Re: multiplication negs In sci.math, Dunphy <3ffhb.5153$G_.420768@news20.bellglobal.com>: Hopefully, IÍm in the right place here.... IÍve been making my way throught the Principia, taking my time, working > along. One day I was talking to my young neice, helping her out with some > simple algebra. We were discussing mulitiplication and division of negative > numbers and my neice brought up a question and for some reason I have been > unable so far to reason my way through the answer: exactly why is it that > when one multiplies 2 negative numbers one ends up with a positive? ItÍs > disturbing me that I cannot come up with a solid answer, and certainly > ïWeÍve been taught that that is the case. will not suf'ce. > Any help would be greatly appreciated. > TIA > Attempt computation of the value (1 - 1) [Times] (1 - 1), or (1 + (-1) ) [Times] (1 + (-1) ), and remember the distributive law. Method A: 1 + (-1) = 0; therefore 0 [Times] 0 or 0. Method B: Distribute it out, yielding 1 [Times]1 + 1 [Times](-1) + (-1) [Times]1 + (-1) [Times](-1). I may have to back up here and start with the following. First, we start with PeanoÍs axioms or something equivalent and de'ne + and [Times] over N [Times]N=>N, in the usual fashion. One can prove commutativity, associativity, and the distributive law for both operations as well. We then de'ne 0 and a magical -a as the arithmetic inverse of a, or a + (-a) = 0, for any a in the natural numbers N; we de'ne thereby J, J = N union {0} union {-a: a in N}. (Note that 0 = -0.) We also de'ne W = N union {0}, for convenience. [*] We de'ne subtraction in a fairly straightforward way over N [Times]N=>N and extend addition and subtraction to N [Times]N=>J, then J [Times]W=>J and W [Times]J=>J, and ultimately J [Times]J=>J. We extend multiplication to W [Times]W=>W trivially: 0 [Times] 0 = 0 [Times] a = a [Times] 0 = 0. (Such is required if we want to keep the distributive laws.) [Times]: W [Times]J=>J and [Times]: J [Times]W=>J has to be de'ned such that a [Times](-b) = -(a [Times]b) and (-a) [Times]b = -(a [Times]b) as well. This may look a bit circular (and probably is!), but if we de'ne 0 [Times] a et al to be anything else things can get nastily inconsistent. Now we attempt (-a) [Times] (-b). To prove that this equals a [Times]b, one has to show the following: [1] We know the distributive law (a+b)c = ac+bc works in N [Times]N [Times]N. ItÍs trivial to show that it works for W [Times]W [Times]W; one then has to work it out for the union of J [Times]J [Times]W and W [Times]W [Times]J, given the de'nition (-a) [Times]b = -(a [Times]b), and a [Times](-b) = -(a [Times]b). The leap to J [Times]J [Times]J requires a little work, although weÍre helped considerably by the observation that (a+b) [Times](-c) = -((a+b) [Times] c) and a [Times](-c) + b [Times](-c) = -(a [Times]c) + (-(b [Times]c)). [2] If we donÍt want to muck around with commutativity we need to prove a(b+c) = ab+ac as well, using similar methods. [3] We now can evaluate the expression (a + (-a) ) [Times] (b + (-b)). By using [1] we get a [Times] (b + (-b)) + (-a) [Times] (b + (-b)). By using [2] twice we get (a [Times] b + a [Times] (-b)) + ((-a) [Times] b + (-a) [Times](-b)). By using associativity and the de'nition of [Times] over J [Times]W union W [Times]J, we get a [Times]b - a [Times]b - a [Times]b + (-a) [Times] (-b) = 0 or -(a [Times]b) + (-a) [Times](-b) = 0 or (-a) [Times](-b) = ab. QED. IÍm not sure if this is quite rigorous enough but itÍs clear that It Must Be That Way(tm). :-) IÍll admit itÍs probably more appropriate for a college professor than a 12-year-old (?), although the underlying concepts donÍt appear all that dif'cult to grasp. (Of course, itÍs been a long time since *I* was 12 years old...) [*] Minor controversy; in some textbooks N = {0,1,2,...}. IÍm assuming here N = {1,2,3,...}. -- #191, ewill3@earthlink.net ItÍs still legal to go .sigless. === Subject: Re: Core error, FEAR is a natural response Nntp-Posting-Host: apps.cwi.nl ... >>I donÍt get it. I understand the part about two arguments contradicting >>each other, and one of them being wrong. But what does this have to >>do with the de'nition of the algebraic integers? DonÍt the arguments >>have to use the same de'nitions in order to be contradictory? And if >>one of the arguments wrong, isnÍt the problem with the argument, not >>the de'nition? The de'nition of algebraic integers excludes ALL roots of non-monic > polynomials with integer coef'cients, when the polynomial is > irreducible over Q. Yes, so what? ItÍs that arbitrary exclusion that leads to a way to *appear* to > prove two different and contradictory things. It is *not* an arbitrary exclusion. A rational number is *not* an integer when it is the root of a non-monic polynomial with integer coef'cients, when the polynomial is irreducible over Q. See? The same kind of exclusion. How can a *de'nition* lead to a contradiction? You simply attempt to prove stuff about the objects you have just de'ned. Either they lead to a contradiction, which does *not* mean that the de'nition is wrong, but which means that either the proof is wrong, or that our method of proof is wrong. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: oh...i am sorry..problem reloaded f(0) = 0 integral (0 ~ 1) f(x) dx = 1 fÍ(x) is continuous 'nd interval of maximum value M of fÍ(x) on the interval [0,1] ----------------------------- answer : M >= 2 how do you solve it? i have the illusion. Forgive. === Subject: Re: Square root modulo a power of two Use HenselÍs lemma or Newton iteration or the binomial theorem, but you need to start with a solution modulo 8 for it to work. If a is even, youÍll need to mess around a bit 'rst, but IÍll just do the case that a is odd and use NewtonÍs method, which in practice is extremely fast. So, 'rst solve x^2 = a (mod 8), call the solution x1. Then Newton iteration (i.e., to 'nd a root of x^2-a) says to let x2 = (x1^2 + a)/(2*x1) (mod 16). Note this makes sense, because you know that x1^2 = a + 8*y1, so what it really means is to set x2 = (a + 4*y1)/x1 (mod 16). Since a is assumed odd, it follows that x1 will also be odd, so it has an inverse modulo 16. To see that x2 is really a square root of a modulo 16, we can compute x2^2 = (a^2 + 8*a*y1 + 16*y1^2)/x1^2 = (a^2 + 8*a*y1 + 16*y1^2)/(a + 8*y1) since x1^2 = a + 8*y1 = a + (16*y1^2/(a + 8*y1)) = a (mod 16). Next write x2^2 = a + 16*y2 and take x3 = (x2^2 + a)/(2*x2) = (a + 8*y2)/x2 (mod 32) and you can check that x3^2 = a (mod 32). Etc. Actually, the convergence is even faster than this, so you can actually compute x3 mod 64 and get a solution to x^2 = a (mod 64). Each time, the exponent on the power of 2 will almost double, re§ecting the fact that NewtonÍs method converges quadratically exponentially, once youÍre close enough to a root. JHS > How does one solve the quadratic congruence x^2=a(mod 2^n) ? === Subject: Re: JSH: About time > Ok, I 'gured it out. If your logic were correct it would apply >to other functions. Yup. Let Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3. Say Q(m) is factored in the form Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f). Now note Q(0)/f^2 = x + f. That is, Q(0) = f * f * (x + f). Therefore when m = 0, we can say a1 = 0, a2 = 0, >and a3 = 1. Note that when m = 0, a1 and a2 are divisible by f. Yup. Now by your logic, for values of m other than 0, >a1 and a2 must be divisible by f and a3 is relatively >prime to f. Do you agree with this? This is a test of your >method. We need to know before we go to the next step. Nora B. The answer is that yes, but with the quali'cation that m not equal 1, > but I have a better explanation for why than the wacky reply when I > initially freaked out. Look again at your Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f). Here the constant terms for the factors are revealed to be f, f and x + f where it just so happens that for your a3 you have something > like--yup, I know some will probably not like this, but itÍs the > reality--introducing h for the functions h(m), a3 = h_1(m) - mx h_2(m) + x + f where h_(1) has f^{2/3} as a factor and h_2(1)=1. And yes, the same thing can happen with my argument, but it requires mf^2 = 1 but both m and f are integers in that argument, so that condition > doesnÍt occur. > You have suggested elsewhere that I need to solve for the aÍs with expression. LetÍs do that. Recall that Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3. Let m = 1: [1] Q(1) = f^2*x^3 + f^3 = f^2*(x^3 + f). Assume this is factored in the form Q(1) = (a1*x + f)*(a2*x + f)*(a3*x + f). The roots of this polynomial are [2] r1 = -f/a1, r2 = -f/a2, and r3 = -f/a3. Of course from [1] I can compute the roots explicitly. They are: r1 = -f^{1/3} r2 = -f^{1/3}*(-1 + sqrt(3))/2 and r3 = -f^{1/3}*(-1 - sqrt(3))/2. and a3, since ai = -f/ri : a1 = f^{2/3}, a2 = f^{2/3} * (-1 - sqrt(3))/2, and a3 = f^{2/3} * (-1 + sqrt(3))/2. Note that a1, a2, and a3 are all algebraic integers. Note that each of them has a factor of f^{2/3}. Note that none of them have f as a factor. No surprises here. What has this got to do with your wanting m*f^2 to equal 1 ? > (If I didnÍt say they were before, well they are now.) For the more adventurous, check Q(m) with m NOT equal to 1, and yes, > you will 'nd that *two* of the aÍs have a factor that is f. > See my other post on this. Nora B. PS: Oh yes, you also asked if I would be convinced if a computer veri'es your proof. I would say *no*. I know exactly what is wrong with your proof already. If someone said a computer veri'ed your proof, I would have to have a guarantee of at least three things: (1) That the computer program and the computing hardware were infallible. (2) That the translation of your proof into the programÍs language was done correctly. (3) That the person reporting the results could be trusted. All of these three conditions will be harder to ascertain than what I already know about your proof . I cannot see how (1) can be guaranteed at all. As for (2), checking that would probably considerably more tedious than what I have already done. And for (3), the most likely person to report positive results would be the least trustworthy. So no. But donÍt let me stop you. Go ahead and carry it out. Maybe you will convince somebody else. > It might be a fun exercise, assuming I didnÍt miss something. If any of you out there think you can 'nd fault with my conclusion > here, please try. It is math after all. And itÍs also a lot of fun. > James Harris === Subject: Re: identical property count >Since they now are identical-with-somethings, if they shared >all the same properties they would be one cake rather than >two. How can something be identical with some things? > And if they shared all their properties, how could you presume to count > ïtwoÍ? > You present an either/or relationship as a relational one. === Subject: Re: chisquare test question > I calculated Chi-square for a set of data..... > but, then im asked to compare the calculated value of Chi-squared with > (n-1) - a.k.a degrees of freedom. Why would I compare my calculated value of Chi-Squared with n-1? I donÍt > get the purpose.... The purpose of this is that, in general, the value of chi-square should be about equal to the number of degrees of freedom you have. If you compute a value of chi-squre which is signi'cantly larger than the number of degrees of freedom, then either (i) your particular experiment represents a *very* low probability statistical §uke or (ii) your data do not 't the theoretical relationships you are trying to impose upon them. If you compute a value of chi-square signi'cantly smaller than the number of degrees of freedom, then you either have (i) a statistical §uke or (ii) you have overestimated the uncertainties at the data points. To be somewhat more concrete, suppose you have a 'xed amount of gas held in a tube at a 'xed pressure, and you are watching the volume expand as a function of temperature. Let us suppose at temperatures T_1,...,T_n you have measured the volume to be V_1,...,V_n. Now, let us suppose that you believe (that is, you have _a_priori_ reasons to conclude) that the temperature measurements are so good that the uncertainty in the temperature measurements can be ignored. Let us suppose that your errors are in sigma_1, ..., sigma_n. Now, let us suppose that you know the pressure exactly , that you know the number of moles of gas exactly (i.e., you have decided that the uncertainties in this are negligible). The, using PV=nRT you could compute theoretical volumes V(T_i), where V() indicates V as a function of temperature. Since you know n, R, and P, there are no parameters to 't, so for each data point you expect (V_i - V(T_i))^2/sigma_i^2 to be about 1, where sigma_i is you estimate in the uncertainty of the i-th volume measurement. So you expect chi^2 to be about equal to n. Again, if the observed chi-squre is larger than this, you would conclude that the something is wrong--either the ideal gas law is wrong, or you are looking in a range of volumes and temperatures where the gas behaves in a non-ideal manner, or something else is wrong. Now, suppose you didnÍt know ïnÍ or ïRÍ, but you never-the-less wanted to test whether V is proportional to T. So now you 'nd the best constant alpha such that *for your data* V_i = V_alpha(T_i) = alpha * T_i. [Here we write V_alpha to indicate how the expected value of volume depends upon temperature, and that one adjustable parameter is involved]. Here, best means youÍve minimized chi-square. Now, again, you would expect alpha to be approximately nR/P and chi-square to be about equal to n. But note--since alpha is an adjustable parameter, you expect the 't now to be just a bit better, because while alpha should be about nR/P, it can wiggle a little to better 't your particular data set. So, according to the statistiticians, if you have 't to a one paramter family of curves (ie, you have adjust one free parameter to minimize chi-square) it turns out that you expect chi-square to be about n-1. The chi-square has gone down because you can adjust alpha to 't your data better than if you had compared to V=(nR/P)T directly. Again, suppose you were 'tting to V=alpha*T+beta. Perhaps you know the termpature in the Celcius scale and the experiment aims at 'xing absolute zero in the Celcius scale. Now, you have more wiggle room when you 't, and you would expect chi-square to be about n-2. Note that for a reasonable experiment the number of measurement points should be signi'cantly greater than the number of parameters you are 'tting. For example, if you are 'tting V=alpha*T+beta and you have made measurents at two temperatures, so n=2, indeed you will 'nd that chi-squre=n-2=0, because you should be able to 'nd a line which passes exactly through the two data points. This, of course, tells you nothing about whether the ideal-gas law holds for your data. So that is a general summary of how this works. I have left out some details. For example, the reason why things are added squared (ie., you raise things to the second power) is related to the assumption that the uncertainties are approximated by Gaussian distributions. There are some other assumptions that go into this. Again, note that if you make enough measurements, then n,n-1, and n-2 are approximately equal, and you donÍt have to worry about this (expect perhaps to prove that you are aware of this). The real problem comes when the number of adjustable parameters m gets close to the number of data-points n , so that the number of degrees of freedom n-m is signi'cantly less that the number of measurement n . Sometimes the amount of data one can get is limited. Generally, this can result is multiple papers reanlyzing the same data set :-). There really is not much that can be done. People will use fancier and fancier statistical methods. Note, by the way, that lots of assumptions go into this, such as knowing the TÍs exactly, etc. In general, oneÍs estimates of oneÍs uncertainties are not very precise, you should not need to sweat this too much. The rules of thumb are: Make sure that the number of measurements n is larger than the number of parameters m . Chi-square should be about n-m -- you can always make the 't look better by adding more parameters, ie., you add parameters and chi-square goes down. If chi-square is much above n-m then either this is evidence against the theoretical model which you are using or something went wrong. If chi-square is much less than n-m, you have probably overestimated your uncertainties. DonÍt sweat this--if the difference between n and n-1 *really* matters, then either you really need to take more measurements or you need to 'nd an expert in statistics and really analyze the particular experiment at hand, and event then you really wonÍt get much more. Hope this helped. By the way, what experiment were you doing? Best wishes, Mike === Subject: Re: Question on Hilbert & Godel > What did Hilbert ask and claim concerning Foundations of Mathematics >> (sets, predicate calculus), metamathematics, Logic, Incompleteness, >> etc? As a mathematician, he did not *claim* things he could not prove. DidnÍt he claim that a certain mathematical problem must have a > solution when in fact it doesnÍt? What particular mathematical problem do you have in mind? >he did not reach >this goal which was shown by G .9adel to be unreachable. One cannot say >he was contradicted ; every serious mathematician has goals that >prove unreachable - only claiming to have reached such goals can be >contradicted. What about claiming that a particular goal is reachable only to have > it proven that the goal is not reachable? Is that a contradiction? What particular goal is that? > Charlie Volkstorf > Cambridge, MA Helmut Richter -- Alan Smaill School of Informatics tel: 44-131-650-2710 University of Edinburgh === Subject: Re: Question on Hilbert & Godel What did Hilbert ask and claim concerning Foundations of Mathematics > (sets, predicate calculus), metamathematics, Logic, Incompleteness, > etc? > Just a side note: >> Bourbaki was cited as the best example so far of mathematics >> organized into a coherent framework. According to Andre Weil, >> Perhaps the most important contribution of Bourbaki was to >> carry out a famous proposal made by the great German mathe- >> matician David Hilbert in 1900 that mathematics be placed on >> a more secure foundation. He noted: Hilbert just said so, >> and Bourbaki did it And just how did Bourbaki do as Weil claims? But didnÍt Hilbert actually claim that a lot more than that is >possible? DidnÍt he ask for (and claim that it must exist) a decision >procedure to determine if an arbitrary predicate calculus wff is valid More than that, an arbitrary wff of mathematics. Where did he ever claim that there was a decision procedure? -- Alan Smaill School of Informatics tel: 44-131-650-2710 University of Edinburgh === Subject: Re: Minimal Graph, Four Color Theorem Visiting Assistant Professor at the University of Montana. It seems to me that the reason you seem to think that people gave you bad data is that you never explained what was going on in your head, and that you did not read what you were given very carefully; combined with some preconception on the part of your repliers (myself included) about what you ->could<- be thinking (which apparently you were not). So let me try to summarize what I think has been going on. The 4 color theorem states: If a graph G is planar, then chi(G)<=4. This is equivalent to any number of things. For example, to If G is a graph and chi(G)=5, then G is not planar ; and If G is a graph, then it cannot be that G is planar and chi(G)=5. Apparently, you prefer to think of the 4 color theorem via the second version: If G is a graph and chi(G)=5, then G is not planar. So you think of graphs which have chi(G)=5, and try to see why they would not be planar. The usual approach, that described in Saaty and Kainen (which you quoted to begin with) is to use the 'rst version: If G is planar, then chi(G)<=4; in particular, one usually thinks of a graph which is ->planar<-, has the property that any proper subgraph is 4-colorable, and tries to prove that G must therefore also be 4-colorable. You started this thread by quoting from Saaty and Kainen, as follows: Suppose the 4CC is false, Then there is some planar graph G with chi(G)=5. Among all such 've-chromatic planar graphs one with a minimum number of vertices is called a ïminimalÍ graph. Thus planar graph G is minimal if chi(G)=5 but chi(H)<=4 whenever H is planar and has fewer vertices than G. Let H be any subgraph of G, where G has n vertices and H has n-1 vertices. Then, the description of H seems to imply that the deletion of ïanyÍ vertex from G will make chi(H)<=4. But this interpretation is generally false and is valid only for n=5!!! Apparently, instead of thinking G is the (hypothetical) smallest PLANAR graph with chi(G)=5 , you were thinking G Is the smallest graph with chi(G)=5 . Which would mean you did not read the statement carefully, which states that G is assumed to be PLANAR. That is, one usually thinks about a planar graph such that every proper subgraph is planar, and 4 colorable; assuming it is not 5-colorable and attempting to reach a contradiction. You, from your comments, prefer to think of a graph which has chi(G)=5, and try to show it is not planar. That is 'ne, but the problem is that the minimal counterexample argument does not work well with that situation; the minimal graph with the property that chi(G)=5 and such that every proper subgraph is planar and 4 colorable is not hypothetical, itÍs K5. But staring at K5 does not get you any closer to the 4 color theorem. The replies addressed the statement as written: assume G is PLANAR; apparently, you kept reading the replies as if it they were assume G has chi(G)=5 . Naturally, miscommunication arose. You kept thinking people were arguing that the 4 color theorem was false for some reason, when people were trying to explain what Saaty and Kainen were trying to say. Examples were given to explain certain facts about planar graphs. You thought they were being given to explain certain facts about graphs with chi(G)=5; again, this mostly arose from you not saying what you were thinking, and from people thinking you understood that they were arguing about planar graphs to begin with. Now, by far the most common false proof of the 4 color theorem consists of people proving that K5 is not planar, and thinking that this proves that a planar graph must be 4 colorable. Your continuous mention of K5 led many of us (or at least, me) to believe you might be thinking along those lines. So, for example, the cycle of length 5 example was given as an example of a graph which does NOT contain any K3, but which is nonetheless NOT 3-colorable; likewise, there are examples of graphs that do NOT contain any K4, and are NOT 3 colorable. This is by way of explaining why does not contain K5 is not enough to imply is 4 colorable , since in general, does not contain Kn is not enough to imply is (n-1)-colorable . My example was an example of why you cannot assume that if G is planar, then any 4-coloring of a G-{v} can be extended to a 4-coloring of G; you can assume that if v has fewer than 4 adjacencies, but I gave a graph in which every vertex had 4 or more neighbors. I did not intend it to be a graph with chi(G)=5, and said so explicitly several times; but you were 'xed on your approach, of starting with a graph with chi(G)=5 and showing non-planar, rather than with the usual approach of starting with a graph G which is planar, and showing that chi(G)<5. To my mind (and to the mind of several, judging from the replies), this confusion arose because you were not paying attention to what you were being told; your posts followed up, but did not address, the text written in the post you were following up. You were as careless reading the replies as you were reading the book you quoted. Having such a cavalier attitude with other peopleÍs time (by ignoring, not paying attention to, or simply misrepresenting what they to annoy people. Arturo Magidin magidin@math.berkeley.edu === Subject: Re: oh...i am sorry..problem reloaded > f(0) = 0 integral (0 ~ 1) f(x) dx = 1 fÍ(x) is continuous 'nd interval of maximum value M of fÍ(x) on the interval [0,1] ----------------------------- answer : M >= 2 how do you solve it? i have the illusion. Forgive. For f(x) = 2*x, one has M = 2, but for every positive M, there are functions satisfying the conditions for which fÍ(0) >= M. E.g., let f(x) = 2*M*x + o(x) as x -> 0+ Consider, for example, f(x) = (n+1)*(n+2)*x*(1-x)^n, which satis'es all the conditions and has fÍ(0) = (n+1)*(n+2) === Subject: Series Derivations Hi All, How would I derive a simple formula f(n) for the series: f(n) = (n-0)(n-1) + (n-1)(n-2) + (n-2)(n-3) + ... ? Does is have anything to do with partial sums? === Subject: Re: Boolean Algebra - Arithmetic Relationship >Assuming an atom behaves like the Bohr model, that >assumptions are, they are always just approximations. The further you >develop any one logical model less its behavior matches that of >reality. Why should that be? I take it that reality is logical, so it is an > exceedingly re'ned model of itself that is logical. > Well, I guess what IÍm trying to say here is best explained in the comparison of the Newtonian vs. Relativity models of physics. NewtonÍs model explains much of what we observe in day to day empirical interactions. However at a certain point (either the cosmological or quantum scale) this logical model developed by Newton falls apart, in other words fails to explain those portions of reality. EinsteinÍs model of relativity is superior in that it explains more phenomenon, or matches reality to greater degree. It seems that we keep Newton along just for pragmatic reasons like itÍs easier to teach and the math is simple. Even if there is some ultimate logical construct that lies behind physical reality, we really only grow closer to approximating it with each advance. (I personally believe the struggle is an asymtotic one governed by the ïlaw of diminishing returns. --But what I believe is neither here nor there, just a non-empirically based opinion.) > Abstract mathematics gets applied to physics. Do you read John BaezÍs > This WeekÍs Finds in Mathematical Physics at > http://math.ucr.edu/home/baez/TWF.html ? > I guess what IÍm trying to say by developing the breadth and not the depth of mathematics is this. I 'nd certain things easy to do in some programming languages and very dif'cult to do in others, scripting in Perl, and Object-Orientated programming in Java, etc. Similarly, I assume, there are certain proofs which are easier done geometry than their equivalents in another mathematical symbology (that geometry can be reduced to of course). My point: WhatÍs abstract in one form of mathematics may be readily apparent in another. Rather than seeking more and more abstract notions and proofs of a particular symbology, perhaps we have to look for a different symbology. Some symbology could be developed to convert the rather abstract nature of mathematics involved in relativity to be fairly apparent or easy to work with. Of course this may be nothing more than a hypothetical pipedream. My ignorance of the actual topic leaves open the possiblity that something such as this already have been developed and is widely used. Of course if it has its not yet simple enough for the layman to grasp. IÍm even told that Newton had to invent calculus to properly explain his theories. (Of course, some say Liebniz beat him to it!) Well anyway I look forward to your post. Humbly as always, -Steve === Subject: Re: oh...i am sorry..problem reloaded >f(0) = 0 integral (0 ~ 1) f(x) dx = 1 fÍ(x) is continuous 'nd interval of maximum value M of fÍ(x) on the interval [0,1] ----------------------------- answer : M >= 2 how do you solve it? i have the illusion. Forgive. For f(x) = 2*x, one has M = 2, but for every positive M, there are > functions satisfying the conditions for which fÍ(0) >= M. E.g., let f(x) = 2*M*x + o(x) as x -> 0+ Consider, for example, f(x) = (n+1)*(n+2)*x*(1-x)^n, which satis'es > all the conditions and has fÍ(0) = (n+1)*(n+2) I think sheÍs trying to show that if f is C^1 on [0,1], f(0) = 0, and int_[0,1] f(x) dx = 1, then fÍ(x) >= 2 for some x in [0,1]. Suppose to the contrary fÍ(x) < 2 for all x in [0,1]. Integrating by parts shows 1 = int_[0,1] f(x) dx = int_[0,1] fÍ(x)(1-x) dx, which then leads to a contradiction. === Subject: Re: Max. Non-Adjacent Vertices on 120-cell Hello ... d> I have reason to suspect that the [independence] number [of the > 120-cell] is an integer square I should point out now that after further investigation, (and, of course, in agreement with the results here http://www.csr.uvic.ca/~wendym/courses/582/120cell.ps ) I no longer believe the number is an integer square. (I said I had reason for my suspicion ... not that I had _valid_ reason.) Still, knowing the number is about 30 greater than where IÍd been 'shing is a help for the underlying goal of this investigation. Don === Subject: Re: Square root modulo a power of two >How does one solve the quadratic congruence x^2=a(mod 2^n) ? If we write a = 2^m * b where m is an integer >=0 and b is odd, then of course m has to be even, and (x/2^m)^2 = b (mod 2^{n-2m}), which reduces it to the case when a is odd. Note that if x is odd, then a=1 mod 8, so for n>2 you must have a=1 mod 8. Usually the simplest quick method of solving a congruence modulo a power of a prime is to solve it modulo successive powers of the prime. So you would solve it mod 8 (1^2 = 3^2 = 5^2 = 7^2 = 1 mod 8) and then use the solutions mod 2^k to get the solutions mod 2^{k+1}. The thing that makes this congruence different from the congruence mod p^n for an odd prime p, is that when x=y mod 2^k, for k>0, we get x^2=y^2 mod 2^{k+1}, because x^2-y^2 = (x-y)(x+y), and x-y is divisible by 2^k while x+y is even because x and y have the same parity. Generally, then, for n>2 (and a=1 mod 8) there will be four solutions, x, -x, x+2^k, and -x+2^k. But itÍs a waste of time to keep track of all four. Instead you can consider that the values of x are really de'ned mod 2^{k-1} instead of 2^k. Or to put it another way, once youÍve gotten a solution x^2=a mod 2^k, in order to get a solution mod 2^{k+1}, either x^2=a mod 2^{k+1}, or if not, then (x+2^{k-1})^2 = a mod 2^{k+1}. NewtonÍs method is faster, however, for large values of n. Start with x an odd number, so that x^2 = a mod 8. Then apply the Newton iteration x --> xÍ = x - (x^2-a)/2x where 2x is the derivative of x^2-a. We can simplify (x^2+a)/2x. The tricky part is the division: you have to know how to do a division mod 2^n, i.e., 'nd the solution to 2*x*xÍ = x^2-a mod 2^n. Convergence is quick, though; xÍ^2-a = (x^2-a)^2/2x, and since x is odd, if x is a solution to the congruence mod 2^k, then xÍ is a solution to the congruence modulo 2^kÍ where kÍ=2k-1. You can speed things up a bit more by observing that you donÍt need to compute the division modulo the original power of 2; you can compute it modulo powers of 2 which grow with each step, and the value of x at each step will still solve the congruence modulo the same power of 2. So for instance suppose I want to compute the square root of 17 modulo powers of 2. Start with the congruence class x=1 mod 8. The 'rst method proceeds one power at a time: 1^2 = 17 mod 16 1^2 != 17 mod 32, so replace it with 1+8=9, where 9^2=17 mod 32. 9^2 = 17 mod 64 9^2 != 17 mod 128, so replace 9 with 9+32=41. 41^2 != 17 mod 256, so replace 41 with 41+64=105. 105^2 != 17 mod 512, so replace 105 with 105+128=233. The second method takes longer for each step, but picks up more powers as it goes along, computing x mod 2^3, 2^5, 2^9, 2^17,.... x=1 mod 8. (x^2+17)/2 = 9. x=9 mod 32. (9^2+17)/(2*9) = 49/9 = 233 mod 512. x=233 mod 512. (233^2+17)/(2*233) = 27153/233 = 75495 mod 131072. x=75495 is a solution to x^2=17 mod 2^18. To illustrate doing the division mod 2^17: 233*x = 27153 mod 131072. The smallest multiple of 233 exceeding 131072 is 233*563: 563*233*x = 563*27153 mod 131072. 107*x = 82787 mod 131072. The smallest multiple of 107 exceeding 131072 is 107*1225: 1225*107*x = 1225*82787 mod 131072. 3*x = 1225*82787-773*131072 = 95419 mod 131072. The smallest multiple of 3 exceeding 131072 is 3*43691: 43691*3*x = 43691*95419 mod 131072. x = 75497 mod 131072=2^17. And in fact 75497^2 = 17 mod 2^18. Presumably this is still not the most ef'cient method for large problems, but I believe the best methods are related. IÍll admit that I made a mistake in the division mod 2^17 which took me a while to 'nd. One can, alternatively, use a second Newton iteration to do the division: x --> x*(2-xy) which is the Newton iteration for f(x) = 1/x - y. ThatÍs how I corrected my mistake. It may be that this is not so good for doing the calculation by hand, but by computer it works well. You also may or may not know that a sequence of compatible congruence classes modulo powers of a given prime p is known as a p-adic number. So in this example, I was in effect computing the 2-adic square root of 17: sqrt(17) = 1 + 2^3 + 2^5 + 2^6 + 2^7 + .... Keith Ramsay === Subject: Re: books - selling - books ... >Davies, Paul >The Fifth Element >Hard bound, great condition. Has withdrawn stamped on top side. >$5 > WhatÍs this about then? My 'rst thought was, Um ... boron. On second thought, I looke for the title on Best Book Buys and didnÍt 'nd it; by Davies did write a book _The Fifth Miracle: The Search for the Origin and Meaning of Life_. -- Chris Henrich Nanotechnology could be huge. -- Lord Sainsbury, Science and Innovation Minister (UK) === Subject: Re: Joe Uptaught (Was Re: David Ullrich on Identity) >> A propos, here are cites from Pierre BourdieuÍs >> _Language & Symbolic Power_ (the titles are mine). >> Enjoy! >> Censorship >> There Oughta Be A Law >> Cool-Hand Luke >The Social Conditions for the Effectiveness of Ritual Discourse Heretical Discourse The *Skeptron* > Symbolic Power & the Symbolism of Power Collusion in the ïHood Expertise as a Problem > Camaraderie of the Experts Knowledge Is Created in a Cultural Context We are proposing that we think of both physics and history as discrete examples of what Pierre Bourdieu calls a 'eld of cultural production (1993: 31 and passim). We hope that by thinking of knowledge production as a cultural activity we can recognize the ultimate cultural context within which knowledge is produced and evaluated, the sociological mechanisms that motivate and limit the actions of individuals in a group, and the legitimate demand for autonomy that is so powerfully expressed and experienced within these 'elds. We hope also to be able to suspend the questions about the objective truths produced within these 'elds by recognizing that these sociological mechanisms provide a powerful structure for creating knowledge but also that this structure may not be as effective outside the 'eld creating the knowledge. A 'eld of cultural production is an organized social group that establishes its own organization by agreeing more or less about the basic purposes of their collective endeavors and the basic rules of what counts as a legitimate claim or statement within the 'eld (in our case, an academic discipline). It can be helpful to think of cultural production as a game, albeit a very serious game, but also a game whose rules are under constant revision. BourdieuÍs conceptualization of cultural production thus escapes some of the problems associated with KuhnÍs paradigms (1962). In KuhnÍs discussion of scienti'c revolutions there seems to be no mechanism whereby individual actors (that is, scientists) come to change their minds. If they are not changing their minds because the new paradigm is better than the old one--that is, it provides a better understanding of the physical reality, for instance (an interpretation that Kuhn speci'cally dismissed)--then what could possibly be causing them to change their minds (Weinberg 1998)? BourdieuÍs understanding of 'elds provides a sociological explanation. Not only are the rules of the game under constant revision, the evaluation of the effectiveness of the rules is under constant revision. And what is at stake in these revisions is of absolute sociological signi'cance: the power and prestige of the participants as well as the very legitimacy of their claims to be real participants in this 'eld (to be a physicist or a historian, which means to be recognized as such by other physicists or historians). In order to be as clear as we can about how this works, we will begin by quoting Bourdieu, and then we will try to go on to explain what this means for our argument. We are afraid that we must begin with an abstract description: to facilitate understanding we are going to avoid introducing speci'cs about the ostensible object being produced--whether it be knowledge of the historical past or of physical reality. But we will be introducing examples from time to time in order to aid clarity. According to Bourdieu, any 'eld of cultural production is made up solely and totally by the positions taken by actors in the 'eld.[2] Every position [in the 'eld], even the dominant one, depends for its very existence, and for the determinations it imposes on its occupants, on the other positions constituting the 'eld; and. . . the structure of the 'eld . . . is nothing other than the structure of the distribution of the capital of speci'c properties which govern success in the 'eld and the winning of the external or speci'c pro'ts . . . which are at stake. (Pierre Bourdieu, _The Field of Cultural Production_, 1993: p. 30) This is BourdieuÍs description of the sociological operations within the 'eld, which he understands to be relations of power and determination. Dominant positions in the 'eld, in any 'eld, exert their power through their in§uence on what count as meaningful and signi'cant statements. For example, the 'eld of psycholinguistics is currently dominated by two main poles of thought, with successive versions of Chomskyan rule-based grammar competing with connectionist network-based models. Theorists and experimentalists base their arguments so that they can be recognized and positioned with relationship to these dominant theoretical perspectives. This is not simply KuhnÍs normal science, as it acknowledges competing paradigms operating simultaneously and provides a sociological mechanism that shapes the actions of individual participants. A psycholinguist risks the possibility of being completely misunderstood if she refuses to state her arguments in terms that one pole or the other can recognize. If she desires her work to be signi'cant and meaningful, it will be less risky to investigate some of the hypothetical consequences of one dominant position or the other (or perhaps to bring together predictive consequences of both). Each instance of such position-taking participates in the rati'cation of the dominance of the current poles. What happens if our renegade psycholinguist persists in stating her position in isolation from the dominant poles? While it is more likely that her work would not be recognized as psycho- linguistics, it is possible that this new position would gain adherents and would come to challenge the dominance of the other two poles. If this does occur, then the relationships of power and determination within the 'eld would be completely reorganized. If, for example, she 'nds herself suddenly championed by actors occupying already powerful positions (say, Steven Pinker or David Rummelhart), then this will legitimize the relative value and signi'cance of her position. And the much greater risk assumed by her rather radical position-taking will have paid off quite handsomely. With the coming to dominance of this new position, other actors in the 'eld will 'nd it necessary to position their work in relation to hers. What is important about this simpli'ed fable for our purposes is that the entire system, the speci'c 'eld of cultural (or knowledge) production, operates by virtue of the value accorded to the positions taken by the actors in the 'eld by other actors in the 'eld. Yet each actorÍs ability to evaluate is determined by the relative dominance of other positions in the 'eld. This is a sociological model that values the internal autonomy of the participants in a 'eld without making each of them into atomistic disconnected individuals. Science operates like a 'eld of cultural production, but recognizing the cultural basis of scienti'c practice does not strip science of its autonomy, rather it indicates the conditions of knowledge production that create that autonomy. Note 2. In BourdieuÍs sociology, position is a multivalent word. Its primary meaning involves an attempt to spatialize the relationships among differing legitimate intellectual statements in a speci'c 'eld. But the metaphoric resonances of the word are also meant to suggest both point of view and social status . One of the most important aspects of BourdieuÍs theoretical system is that point of view and social status are not extricable from each other. In addition, the value or signi'cance of an intellectual position within a particular 'eld is both a re§ection of and re§ects back on the social status and point of view of the actor holding the intellectual position. Finally, it is important to emphasize that the terms social status and point of view are simpli'ed versions of the more nuanced concepts that Bourdieu uses in his analysis. (Barry Shank et al, Pure Objects and Useful Knowledges (in _After the Science Wars_, Keith M. Ashman and Philip S. Baringer, eds., pp. 69-79): 73-75) === Subject: re: The Shape of Space PS They seem not be aware the the very relevant papers on gravitational instantons by Peter Kronheimer (1989, 1990) -- although the spherical spaces they consider are all SU(2)/g, where g is a 'nite subgroup of SU(2). The classi'cation of these spaces via the A-D-E Coxeter graphs, provides the big guns of Lie algebra theory and the many other objects classi'ed by these same Coxeter graphs, such as catastrophe bundles and 2-d conformal 'eld theories (which live on the 2-d string world-sheet). Saul Paul the d 2-d conformal 'eld theories (which live on the 2-d string world-sheet) are the world holograms whose projections or images are 4D spacetimes in Lp*^2 = hG*/c^3 = Lp^4/3(c/Ho)^2/3 = 10^40Lp^2 which numerically 'ts Abdus SalamÍs f-gravity idea of ~ 1973 hadronic string tension ~ c^4/G* ---> alphaÍ ~ 1/(1 Gev)^2 for universal geometrodynamical Regge trajectories of micro-geons of Kerr-Newmann type. === Subject: re: The Shape of Space I am still trying to get a Mickey Mouse heuristic picture for Pedestrians, a sculpture, of what Ellis et-al are suggesting. Rather than trying to picture it in 3D letÍs go back to 2D. OK, if Omega zero = 1.013 i.e. k = +1 in FRW metric in standard convention. Then our total Universe (sans one space dimension so we can visualize it) is topologically equivalent a 2D spherical surface without boundary (a cycle bc= 0 where b is the boundary operator dual to the exterior derivative d on Cartan forms f where c is a chain or space such that c = c1 + c2 + c3 + .... where bci =/= 0 ). We then tile or partition this spherical surface using great circle geodesic lines forming some kind of regular polygons that in actual 3D case correspond to EllisÍs the spatial sections of the Universe are dodecahedral sections of a space of positive curvature, 'tted together to make a 'nite three dimensional spaces. [without boundary] The questions is: ( The Question is: What is The Question? :-) What happens if a star ship attempts to cross the line, i.e. cross the 1D edges of the polygon in the 2D toy model case, or pass through the 2D walls separating the dodechahedral sections in the actual 3D case? Is it like the Pong Video Game or not? I mean as you exit right you enter left , or was that simply an example Ellis used that had no direct Creon, and printing them out. You have given me my 'rst look at the 9 October *Nature* paper by Luminet et al. The papers by Gausmann (et al.) Topological lensing in spherical spaces and Eigenmodes of three-dimensional spherical spaces and their application to cosmology are especially interesting to me. They seem not be aware the the very relevant papers on gravitational instantons by Peter Kronheimer (1989, 1990) -- although the spherical spaces they consider are all SU(2)/g, where g is a 'nite subgroup of SU(2). The classi'cation of these spaces via the A-D-E Coxeter graphs, provides the big guns of Lie algebra theory and the many other objects classi'ed by these same Coxeter graphs, such as catastrophe bundles and 2-d conformal 'eld theories (which live on the 2-d string world-sheet). It will take a while for me to digest all this new material -- but a lot of it looks very familiar! Saul-Paul ---------- === Subject: Papers coming on cosmological topology Saul Paul - I am sending you several papers in subsequent messages. pdf 'les. You may also be interested in: http://xxx.lanl.gov/abs/astro-ph/0212223 http://xxx.lanl.gov/abs/astro-ph/0303580 http://xxx.lanl.gov/abs/astro-ph/0310253 === Subject: re: Shape of the Universe === Subject: Re: WMD found! New York Times reports. Part III The Pong Video Game in The Sky with 2D Wall periodic boundaries may not be what Ellis et-al are suggesting. Ellis cites Omega zero = 1.02 +-0.02. meetings the Pundits like Mike Turner all said k = 0, in§ation is right, i.e spatially §at with Omega zero = 1 on the button. Ellis is saying no to that that there will be a Big Crunch with k = +1 and Omega zero > 1. That allows a 'nite 3D space without 2D wall boundaries. So, spatial sections of the Universe are dodecahedral sections of space of positive curvature, 'tted together to make 'nite three-dimensional spaces. Earlier Ellis does make the Pong Video Game analogy implicitly for the §at toroidal space, as you exit right you enter left and space is 'nite Does Ellis mean that there are sectors of 3D space with 2D walls between them which 't together like some 'nite Platonic solid? What happens when you walk through a wall so to speak? Is there a periodic boundary condition (i.e. a Star Gate in effect) or not? That is the total 'nite space of positive curvature has no boundary, but is partitioned into disjoint subspaces or sectors separated by 2D boundary walls? Is that the idea? Are these walls weird or not? Can we pass through them? What happens when we do? Or are they not there at all? Whatever the picture is here, Ellis is clear that this alternative is not compatible with chaotic in§ation that demands Omega zero = 1 with k = 0. Luminet has Omega zero = 1.013 Max TegmarkÍs picture of Level I Hubble universes in May Scienti'c American is not consistent with what Ellis et-al is suggesting here. That is Ellis rejects that spatial homogeneity extends outside our visual horizon forever .... and we are in one expanding bubble in the middle of innumerable other similar ones. But if Luminet et-al are correct, chaotic in§ation is ruled out: there is only one expanding bubble, and we can see almost all the way round it .... The WMAP data as interpreted by Luminet et-al... suggest that we might live in such a small closed universe. This is qualitatively globally different from what Mike Turner, Max Tegmark et-al are suggesting. So we have a schism between the k = 0 Camp and the k = +1 Camp. We need better precision than the current 2% to decide. === Subject: Re: Goldbach Computations In 1742, historian and mathematician Christian Goldbach (1690-1764) > effect, that every integer greater than 5 is the sum of three prime > numbers. A prime number is evenly divisible only by itself and 1. Nowadays, GoldbachÍs conjecture is expressed in the following > equivalent form: Every even number larger than 2 is the sum of two > prime numbers. Despite centuries of effort, no one has yet been able to prove GoldbachÍs > conjecture. Progress has been slow. > Vinogradov proved in 1937 that every suf'ciently large (more than a certain > number called ïVinogradov constantÍ) odd number is a sum of three primes. > This constant is large (> 2^65536), but anyway the problem is not very > interesting, since the quantity of numbers left to check is 'nite. I cannot immediately see the equivalence of the various conditions. If all even numbers can be expressed as the sum of two primes, then the following odd number can be expressed as the sum two primes, plus one. But one is not allowed as prime. So what does this gain us? I suppose the proof is by induction, with some attention to cases ... Either one of the prime decomposing an even integer larger than two is itself two, or it isnÍt. If one of the summands is two, than the other is still even, therefore composed of two primes, so the total sum is of three primes -- unless both summands are two, in which case we donÍt get two primes out of the remaining summand, but then the number in question was four, which wasnÍt part of the conjecture. Ok. Suppose neither of the summands is two. Then, both must be odd numbers greater than one. Suppose both are three ... then the number is six and the sum is two + two + two. Suppose one summand is three, and the other 've: than the target may be expressed as three + three + two. Suppose both summands are odd, and at least one is greater than 've. Then what? Then we may decompose the at least one summand into three primes, thereby decomposing the original target into a sum of four primes ... which buys us what, since we were looking for three primes? And we have yet to consider the odd numbers, provided of course they are larger than 've. The number of cases to consider is 'nite, but annoying! Which may explain my non-existence as a number theorist. IÍm not sure why a 'nite number of cases makes the thing uninteresting: investigators found the groups interesting, though AFAIK it turns out in cataloguing them that the cardinality of some phyla of groups are 'nite. Why should the detailed structure of all numbers less than 2^65536 be uninteresting to the number connoisseur? Maybe thereÍs a single counterexample to the Goldbach conjecture lurking in there, waiting to give its discoverer a lasting footnote! === Subject: Combinatorics question Say we have the number 4. There are 5 unique ways (up to ordering) of writing it as sums of integers lower than it. They are: 4 2 + 2 1 + 3 1+1+1+1 1+1+2 I believe these are called partitions of integers. Does anybody know how to 'nd the number of such partitons (with a proof) ?? Is this a famous combinatorial problem? If someone knows the answer, has some hints, or knows of a reference that contains an answer, please tell me. Bill === Subject: Re: Square root modulo a power of two > How does one solve the quadratic congruence x^2=a(mod 2^n) ? > By hand, one can solve it by asking what squared equals k*2^n+a. For example for n=1,a=1, then x=1. For n=2,a=1, then x=1,3,5,7 and for a=4, then x=2,6. If n is even, then it is only neccessay to consider the 'rst 2^(n/2) numbers and then extend to k*2^(n/2) plus minus a reference number. > I am sure that a general algorithm or rule exists, but I canÍt think of one. Maybe IÍll have one come morning, in which case youÍll be the 'rst to know. === Subject: Re: Boolean Algebra - Arithmetic Relationship |I guess what IÍm trying to say by developing the breadth and not the |depth of mathematics is this. I 'nd certain things easy to do in |some programming languages and very dif'cult to do in others, |scripting in Perl, and Object-Orientated programming in Java, etc. |Similarly, I assume, there are certain proofs which are easier done |geometry than their equivalents in another mathematical symbology |(that geometry can be reduced to of course). My point: WhatÍs |abstract in one form of mathematics may be readily apparent in |another. Rather than seeking more and more abstract notions and proofs |of a particular symbology, perhaps we have to look for a different |symbology. ItÍs nice to see someone thinking about this kind of issue. I would say that what to the interested layperson looks like more and more abstract notions of a particular symbology is in fact often much more about trying to develop a framework within which lots of things become much easier to do, in the way that using a more suitable programming language might do. Not that this is always successful (itÍs harder to do than elaborating on old ideas, because it requires new ideas). ItÍs the kind of mathematics that mathematicians like to see, however. Some mathematicians are very good at wielding elaborate technical apparatus to prove results, and thereÍs respect for that, but in the end people would rather have a proof which is less technical and more conceptual, if there is one. If all a 'eld does is pile on technical details higher and higher, it eventually goes dormant. Another thing to keep in mind is that thereÍs more of an emphasis on interapplicability than there is in a lot of sciences, like physics. developed at length, then be rather completely abandoned, so much so that almost nothing is salvaged for later use, because thereÍs no use for results about how the old theory worked in the development of the new one. Or at least so IÍm told. In mathematics, it seems that ideas have a greater lifespan; a genuinely good result is less apt to be discarded because it was only valid in an abandoned theory . Even inside of mathematics, there are occasions when something like a paradigm shift occurs, but a lot of the overlying mathematics acts like the table settings in the old parlor trick, where a tablecloth is whisked quickly away while the tablewear, plates, and glasses stay put. After working in software for awhile, the way that mathematics manages to achieve the kind of reusability we only wish we had in software, and keep ideas around for a long time, elaborating on them, without gradually drowning in legacy proofs , seems all the more amazing. Keith Ramsay === Subject: Re: Factorial/Exponential Identity, In'nity I 'nd an obvious §aw in my statement. Consider the sequence (01)..., interchange each 1 with the adjacent 0 of the next subsequence, 01010101010101... 00101010101010... Then interchange each 'fth and later 1 with the following 0: 00100101010101... Repeat for each 3x-1Íth one, and the output sequence results in (001).... I consider this, then. Divide the sequence with the known density p/q into subsequences of 'nite length xq for any positive integer x. Then, for each of those 'nite subsequences its elements can be permuted. The resultant sequence is to have the same density as the input sequence and shall not be convertible to a sequence known to be of any other density. That appears to be an adequate method for that, but IÍm concerned that it would not be able to convert into each other sequence of the same density, that is, some of the irrational sequences. It would appear to allow converting (01)... to (10)... but not to (001).... I think it would allow the conversion of (01)... to normal sequences. Almost all real numbers are absolutely normal. Ross === Subject: generating a random discrete vector given pair-correlations Suppose we have an unknown probability distribution on the set of vectors in (Z/4Z)^40 . We explicitly donÍt assume that the module structure is related to the probability distribution. What we do know are the C(40,2) coordinate-pair correlations. In other words, if V denotes the random variable with unknown distribution, we have: Prob( V_i = a and V_j = b ) = c_{a,b,i,j} for all 1<=iCompared with the near $100 for the 3rd. I am just looking for a gentle >introduction to Calculus/Analysis that I can use as self-teaching text. For >my background, I 'nd Rudin too terse. Whoo-hoo. Gentle is not the word IÍd have chosen 'rst, or tenth, > to describe SpivakÍs _Calculus_. The problems, in particular, tend > to be brutal. Have you looked at ApostolÍs _Mathematical Analysis_? Or, for that > matter, CourantÍs calculus textbook? I have never seen RudinÍs > calculus text, just (many many years ago) his _Real and Complex > Analysis_, and slightly more recently a monograph or two on some > topics in several complex variables. But I donÍt think Spivak > is notably less terse than Rudin, and I do think that both > Apostol and Courant are probably less terse while being quite > as rigorous as necessary. Lee Rudolph Lee, are you refering to SpivakÍs Calculus , or to his Calculus on manifolds ? Lurch === Subject: Re: Linear Algebra question > IÍm not sure if/why this would be true or false. IÍm having a hard time > understanding how I could explain that it was true. I would appreciate any > help. Given a linear transformation T:F^n-->F^m. Is it true that there exists an m > x n matrix A such that T(x)=Ax (where we think of x as an n x 1 matrix? If your notation means what I think it means, this is pretty standard and will be in almost any Linear Algebra text. The idea is to take the usual basis of F^n, { e_1, ... , e_n }. Let the usual basis of F^m be { v_1, ...., v_m }. Write T(e_j) = a_1*v_1 + a_2*v_2 + ... + a_m*v_m. Then the transpose of (a_1, ... , a_m) is column j of A. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Goldbach Computations |I cannot immediately see the equivalence of the various conditions. I had to stop a bit too. |If all even numbers can be expressed as the sum of two primes, then |the following odd number can be expressed as the sum two primes, plus |one. But one is not allowed as prime. So what does this gain us? Use 3 instead! Let me attempt a little summary (unnecessary as it must be now that everyone has stopped a bit and 'gured it out): G: Goldbach in the modern form. Every even number >=4 can be written as a sum of two primes. G2: Goldbach in roughly the original form. Every number >5 is a sum of three primes. G3: Goldbach in a pecular form. Even even number >5 can be written as a sum of three primes. V: What Vinogradov wanted to prove. Every odd number >5 is a sum of three primes. Obviously a sum of two primes has to be at least 4 and a sum of three has to be at least 6. G2 is just G3 and V . G is equivalent to G3 because when n is even, n+2 can be written as a sum of three primes only as 2+p+q, which is equivalent to being able to write n as a sum of two primes p+q. The even numbers >5 are just two more than the even numbers >=4. Also, G implies V, because if n is even and can be written as a sum of two primes, then n+3 can be written as a sum of three primes. G starts with n=4, V starts with n=7. ThereÍs no obvious way to use V to prove G, so in that subjective sense itÍs weaker. The version Vinogradov actually proved, that every odd number above some big bound can be written as a sum of three primes, is weaker still. So since G2 is just G3 and V where G3 is equivalent to G and V is implied by G, we get that G2 is equivalent to G also. Actually, for some reason some people like to state Goldbach as every even number n>4 can be written as a sum of two odd primes . I donÍt know why. Maybe theyÍre trying to do even more of our thinking for us, by sparing us having to realize that 2+2=4 and 2 is the only even prime and so on. Keith Ramsay === Subject: Re: multiplication negs Hopefully, IÍm in the right place here.... IÍve been making my way throught the Principia, taking my time, working > along. One day I was talking to my young neice, helping her out with some > simple algebra. We were discussing mulitiplication and division of negative > numbers and my neice brought up a question and for some reason I have been > unable so far to reason my way through the answer: exactly why is it that > when one multiplies 2 negative numbers one ends up with a positive? ItÍs > disturbing me that I cannot come up with a solid answer, and certainly > ïWeÍve been taught that that is the case. will not suf'ce. > Any help would be greatly appreciated. > TIA > I will offer a geometric arguement I saw some where. Suppose you have a line or some other 2d geometric 'gure. Now, if you take the re§ection of a point on a line, or the line('gure) itself, one gets the negative of that point. Take the re§ection of the re§ection and you get the point back again. Or, similarly, think of §ipping a line once, then twice, to get back to where you started. Lurch === Subject: Re: Question on Hilbert & Godel |> But didnÍt Hilbert actually claim that a lot more than that is |> possible? DidnÍt he ask for (and claim that it must exist) a decision |> procedure to determine if an arbitrary predicate calculus wff is valid |> |> More than that, an arbitrary wff of mathematics. | |Where did he ever claim that there was a decision procedure? In 1931 heÍs supposed to have said, There are absolutely no unsolvable problems. Instead of the foolish ignorabimus [we donÍt know], our answer is on the contrary: We must know, We shall know. I donÍt know that he stated proving the existence of a decision procedure as a goal. I donÍt know for certain whether (a) he claimed his formalization of mathematics would demonstrate this, and (b) whether he realized that if a formal system is complete, then thereÍs a decision procedure (i.e., search until a proof or disproof is found). I tend to assume (b) just because he was smart enough, and the relationship between formalism and algorithms seems obvious enough. As for (a), did he go through all that drama in order to produce a formalization of mathematics that would be mostly pretty good ? I sure got the impression he meant for it to be a formalization which would cover all the reasoning people do. IÍm not sure why we should fault someone for occasionally setting an unsatis'able goal. I seem to remember that Goedel had been working on the Hilbert program when he hit upon the ideas for his most famous work. Probably it was a bene'cial failure all told. Keith Ramsay === Subject: Re: Combinatorics question [...] |I believe these are called partitions of integers. Does anybody know how to |'nd the number of such partitons (with a proof) ?? Is this a famous |combinatorial problem? Yes, assorted methods exist, and yes, this is very famous. http://mathworld.wolfram.com/PartitionFunctionP.html Keith Ramsay === Subject: Re: hello...my question is ?? f(0) = 0 >integral (0 ~ 1) f(x) dx = 1 >fÍ(x) is continuous 'nd M such that fÍ(x) <= M Does this mean we need to 'nd an M such > that fÍ(x) <= M for all x, or for at least one x? > Consider f(x) = 6x - 6x^2. Thus problem is to 'nd at least one. Answer, some x in [0,1] with fÍ(x) <= 2. Proof: Assume otherwise. for all x in [0,1], 2 < fÍ(x) Integrating both sides from 0 to x 2x < integeral(0,x) fÍ(x) dx = f(x) As fÍ(x) is continuous at 0 thereÍs some small eps and interval about 0 with 2 + eps < fÍ(x) for all x in the interval. Because of this, itÍs assured that 2x < integral... instead of 2x <= integral... Now integrating both sides from 0 to 1 1^2 - 0^2 < integral(0,1) f(x) dx = 1 which is a contradiction. (Again < instead of <= because f is continuous) === Subject: Re: Goldbach Computations > In 1742, historian and mathematician Christian Goldbach (1690-1764) > effect, that every integer greater than 5 is the sum of three prime > numbers. A prime number is evenly divisible only by itself and 1. Nowadays, GoldbachÍs conjecture is expressed in the following > equivalent form: Every even number larger than 2 is the sum of two > prime numbers. Despite centuries of effort, no one has yet been able to prove GoldbachÍs > conjecture. Progress has been slow. >Vinogradov proved in 1937 that every suf'ciently large (more than a certain >number called ïVinogradov constantÍ) odd number is a sum of three primes. >This constant is large (> 2^65536), but anyway the problem is not very >interesting, since the quantity of numbers left to check is 'nite. I cannot immediately see the equivalence of the various conditions. > If all even numbers can be expressed as the sum of two primes, then > the following odd number can be expressed as the sum two primes, plus > one. But one is not allowed as prime. So what does this gain us? Three is prime. The following -after- the following odd number can be expressed as the sum of two primes plus three. Minor Crank === Subject: Dice There are impossibility arguments concerning probabilities of certain rolls of two dice, invoking a cyclotomic polynomial and given here some time ago. Does anyone have a reference? === Subject: Re: Goldbach Computations > I cannot immediately see the equivalence of the various conditions. > If all even numbers can be expressed as the sum of two primes, then > the following odd number can be expressed as the sum two primes, plus > one. But one is not allowed as prime. So what does this gain us? But then every suf'ciently large odd can then be expressed as the sum of 3 primes including a 3 as one of them. === Subject: Re: Combinatorics question > Say we have the number 4. There are 5 unique ways (up to ordering) of > writing it as sums of integers lower than it. They are: 4 > 2 + 2 > 1 + 3 > 1+1+1+1 > 1+1+2 I believe these are called partitions of integers. Does anybody know how to > 'nd the number of such partitons (with a proof) ?? Is this a famous > combinatorial problem? If someone knows the answer, has some hints, or knows of a reference that > contains an answer, please tell me. Bill See http://www.mathpages.com/home/kmath383.htm or do a Google search for partitions of n === Subject: Re: Michael Gordon Nagus - February 18th 1978 In alt.math.undergrad Probababbilities > IÍm not the slightest bit scared of you nor your threats. And Michael, > you keep in mind that you can beat the crap out of me and itÍs very > unlikely that the Saskatoon City Police will ever charge you with > assault, for I only have the right to be assaulted and brutalized. If I thought that were true, IÍd look up MichaelÍs address and mail him a copy of this myself. In fact, I think I will anyway. -- Wayne Brown (HPCC #1104) | When your tailÍs in a crack, you improvise fwbrown@bellsouth.net | if youÍre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Core error, proof checking |The nice thing about a really short proof, like whatÍs necessary to |show the de'nition error in core, is that it should be checkable by |an automated proof checker. Why not try Isabelle? People seem to like it, and you can download yourself a copy: http://www.cl.cam.ac.uk/Research/HVG/Isabelle/ Some work has gone into a number theory package: http://www.cl.cam.ac.uk/Research/HVG/Isabelle/library/HOL/ NumberTheory/ind ex.html They have a proof of quadratic reciprocity. The format is a little cryptic, but humanly readable. For example, FermatÍs little theorem that if a is relatively prime to n, then a^phi(n)=1 mod n appears in this form: theorem Euler_Fermat: 0 < m ==> zgcd (x, m) = 1 ==> [x^(phi m) = 1] (mod m) given the previously de'ned terms in the library. Then the proof is a couple dozen lines. It takes a lot of detail, but getting down to such a level of detail would presumably break this deadlock. Keith Ramsay === Subject: recurrence relation IÍve turned up the following recurrence: x_0=x_1=0 and x_n=(n+1)x_{n-1}+[(n-1)/3](n+1)! for n>=2 I donÍt know a recipe for solving a recurrence of this form (ie: with a nonconstant coef'cient on the x_{n-1}), and there doesnÍt seem to be any inspiration on the way. Can anybody give me a hint? === Subject: Re: Laurent series ('nding the principal part) > I have those two examples: > f(z) = e^(-1/(z^4)) > and > f(z) = (e^z - 1)/(e^z + 1) > and iÍd like to 'nd the principal part of their laurent > expansions (the 'rst one about a=0, the second a=i*pi). f(z) = e^(-1/(z^4)) = 1 + -1/(z^4) + (-1/(z^4))^2/2 + (-1/(z^4))^3/3! + ... = 1 - z^(-4) + (1/2).z^(-8) - (1/3!).z^(-12) - ... Therefore, the principal part is - z^(-4) + (1/2).z^(-8) - (1/3!).z^(-12) - ... On the other hand, if f(z) = (e^z - 1)/(e^z + 1) = (-e^(z - i.pi) - 1)/(-e^(z - i.pi) + 1) = (-e^(z - i.pi) - 1)/(z - i.pi) . (z - i.pi)/(-e^(z - i.pi) + 1), the principal part of the Laurent series of the left member is -2/(z - i.pi). The right member has no principal part and the constant member of the Laurent series (which is, therefore, a power series) is -1. So, the principal part of f(z) at i.pi is 2/(z - i.pi). Jose Carlos Santos === Subject: Re: recurrence relation >IÍve turned up the following recurrence: >x_0=x_1=0 and x_n=(n+1)x_{n-1}+[(n-1)/3](n+1)! for n>=2 >I donÍt know a recipe for solving a recurrence of this form (ie: with a >nonconstant coef'cient on the x_{n-1}), and there doesnÍt seem to be any >inspiration on the way. Can anybody give me a hint? If I havenÍt made a mistake then I think I found an equivalent form x_0=0, x_1=0, x_2=2, x_n= n(n+1)/(n-2)x_{n-1} for n>2 That gets rid of the additive term in your expression. And it puts it in a form where generating functions apply. === Subject: I still donÍt understand by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9C49Kq18266; I even asked the ta last week & he spent the tutorial trying to understand how to do the qn & got nowhere :( He didnt understand hints :( For the added hint that was given - what are my I_nÍs speci'cally? Just closed intervals where f_n > 1/2? A grad student told us teh rationals are not countable intersection of open sets (somthing about Baire Category thm - I dontÍ know this, have heard of it htough). Then said de'ne A_N = union(n >= N to in'nity) of inverse images of f_n of open balls centered at 1 radius 1/2. Then said rationals Q = intersection(N=1 to in'nity)A_N .. & the said that was the contradiction. The grad student said that is what the hint was saying - is it really? === Subject: Re: Goldbach Computations >I cannot immediately see the equivalence of the various conditions. >If all even numbers can be expressed as the sum of two primes, then >the following odd number can be expressed as the sum two primes, plus >one. But one is not allowed as prime. So what does this gain us? Three is prime. The following -after- the following odd number can be > expressed as the sum of two primes plus three. Oh. Good point. :-) Now about the evens ... any similar way of seeing how the sum-of-two-primes condition becomes the sum-of-three primes? Oh again. Same idea, but using two . Never mind :-). === Subject: Re: 4th order Bspline > I have a stupid question about the 4th order Bspline. In the > paper( see link), they use summation to denote a B-spline. But > why they use a vector (4 j)~T. Can anyone help me to understand it? http://www.ri.cmu.edu/pub_'les/pub2/wu_yu_te_1998_1/wu_yu_te_ 199 > 8_1.pdf It is not a vector but a binomial coef'cient 4 choose j or choose j from 4 . === Subject: Re: Core error, FEAR is a natural response >The de'nition of algebraic integers excludes ALL roots of non-monic >polynomials with integer coef'cients, when the polynomial is >irreducible over Q. ItÍs that arbitrary exclusion that leads to a way to *appear* to >prove two different and contradictory things. Only if you can prove there exists a root of a monic polynomial which is also the root of a non-monic polynomial irreducible over Q. The proof is not logically complete, nor have you been able to construct such an example. That does not mean the result is necessarily false, but given what it would imply it almost certainly is false. >ItÍs not even a subtle proof to show it, but mathematicians have been >running from the shock. (Neat, eh? Proving an error in core with a proof. Great fun.) Often thinking about accomplishing something can be more fun than actually accomplishing anything. >It seems an entire math area--algebraic number theory--has to be >re-worked. So where are your corrections to all the supposedly false theorems of algebraic number theory? If algebraic number theory is internally §awed, then itÍs almost certain one or more theorems are false. Which one(s)? >LOTS of textbooks to change. But then again, change can be fun. According to your social theory ( mathematicians are afraid of discovering new things ), if it was up to mathematicians weÍd still be stuck with EuclidÍs textbooks as our only source of knowledge. Why does this sound unlikely to me? >Now thatÍs bizarre as in fact, I note that thereÍs the *appearance* of >dueling proofs. But proofs donÍt, canÍt duel. They can if mathematics is inconsistent. Your proof is not convincing enough to suggest this to be the case. >Social forces are so powerful, but you people get annoying because >youÍre so damn irrational! Have you got something against irrationals then? Are they not equal members of your object ring? >Now why donÍt you try *actually looking at my work* instead of tossing >out what you think the group expects, but then, you probably donÍt >believe in ideas, do you? I think youÍre deeply confused. Were your proof correct, it would indicate towards inconsistency in mathematics. Something IÍm sure mathematicians would jump at instead of sweeping under the rug. By consistently claiming mathematicians are running away from mathematics to avoid having to rethink their positions is absurd and simply shows you donÍt know what youÍre saying. To claim that mathematics is internally inconsistent requires more than jumbling some algebra together, proclaiming yourself a genius and berating anyone who can see through the holes in your proof as a liar and a coward. >IÍve talked with a mathematician at my alma mater Vanderbilt >University, and explained the entire argument to him...then he >basically ran away. Maybe he didnÍt have time to dwell into your theory because he was busy working on his own. How would you like it if somebody forced themselves on you with a theory that did not concern your 'eld and, after taking an hour of their time to graciously look through that personÍs argument, be labelled a coward on a public forum? >naked apes[...]left the trees this morning[...]another cow in the herd WhatÍs with the biology references? IÍd stay out of the 'eld of social studies and stick with algebra, at least you seem to know your way around numbers. === Subject: Re: Question on Hilbert & Godel > What did Hilbert ask and claim concerning Foundations of Mathematics > (sets, predicate calculus), metamathematics, Logic, Incompleteness, > etc? > As a mathematician, he did not *claim* things he could not prove. DidnÍt he claim that a certain mathematical problem must have a >solution when in fact it doesnÍt? What particular mathematical problem do you have in mind? Any that he made this claim about. Are there several? How many are you aware of? >> he did not reach >> this goal which was shown by G .9adel to be unreachable. One cannot say >> he was contradicted ; every serious mathematician has goals that >> prove unreachable - only claiming to have reached such goals can be >> contradicted. What about claiming that a particular goal is reachable only to have >it proven that the goal is not reachable? Is that a contradiction? What particular goal is that? Any mathematical goals. If it is claimed that it is reachable then proven to not be reachable, isnÍt that a contradiction for any mathematical goals? >Charlie Volkstorf >Cambridge, MA > Helmut Richter === Subject: preregular/preopen sets I chanced upon the following terms: preopen , preregular sets but what did i wrong? martin === Subject: Re: Goldbach Computations >> I cannot immediately see the equivalence of the various conditions. >> If all even numbers can be expressed as the sum of two primes, then >> the following odd number can be expressed as the sum two primes, plus >> one. But one is not allowed as prime. So what does this gain us? Three is prime. The following -after- the following odd number can be >expressed as the sum of two primes plus three. Oh. Good point. :-) Now about the evens ... any similar way of seeing how the > sum-of-two-primes condition becomes the sum-of-three primes? Oh again. Same idea, but using two . Never mind :-). Minor Crank === Subject: Re: Question on Hilbert & Godel What did Hilbert ask and claim concerning Foundations of Mathematics >(sets, predicate calculus), metamathematics, Logic, Incompleteness, >etc? Just a side note: Bourbaki was cited as the best example so far of mathematics > organized into a coherent framework. According to Andre Weil, > Perhaps the most important contribution of Bourbaki was to > carry out a famous proposal made by the great German mathe- > matician David Hilbert in 1900 that mathematics be placed on > a more secure foundation. He noted: Hilbert just said so, > and Bourbaki did it >> And just how did Bourbaki do as Weil claims? >> But didnÍt Hilbert actually claim that a lot more than that is >> possible? DidnÍt he ask for (and claim that it must exist) a decision >> procedure to determine if an arbitrary predicate calculus wff is valid More than that, an arbitrary wff of mathematics. > (true of all interpretations) - was it? >> conventional >> wisdom was wrong and mathematicians gained new insight into the >> nature of mathematics. >> Or are you one of those diehards who clings to the failed ideas and >> theories of the past? >> (Actually, the most secure foundation possile is software that carries Software is 'ne if youÍre a strict 'nitist. Some mathematicians like >a set theory with _huge_ axioms of in'nity in which to do their >category theory. ??? Do you know of any examples of proofs which take in'nitely much paper to write? ThereÍs no problem using software to prove things involving _huge_ axioms of in'nity... (because actually the axioms are _'nite_, not huge - theyÍre talking about huge sets.) >> out both logic and metamathematics, as I have implemented and descibed >> in my papers below.) >> Charlie Volkstorf >> Cambridge, MA >> http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/ 20021008.1/1 >> http://www.arxiv.org/html/cs.lo/0003071 > (Quote from the QED Manifesto) F. ************************ David C. Ullrich === Subject: Re: Core error, proof checking linux) > WhatÍs important here is that my work IS correct, and luckily for me, > since I have mathematicians running in terror all over the place, itÍs > short enough that it should be machine checkable. Whether your proof is short or not, it is not easily checkable. First, it must be translated to a purely formal exposition, which is tedious but not too dif'cult in principle. Second, all of the prior results on which your proof depends must be also translated to a formal exposition (the statements of the theorems at the least, if not the proofs as well). This is again tedious but not too dif'cult in principle. The in principle easy but tedious parts become fairly more dif'cult, given that no one else on the newsgroup really seems to understand *how* you justify each of your conclusions. Trying to formalize and check your proof is not a simple task. Moreover, after anyone attempts to formalize it and gets to a bit where they simply canÍt see how the conclusion is supposed to follow from the antecedents, they must simply give up or guess what you mean or ask you and hope that they understand your response. While some of us may know a proof checker or two, I donÍt know anyone here (aside from you) that claims to understand your argument. This understanding is necessary in order to formalize the proof and seems rather harder than the technical know-how required to formalize your argument. (ThatÍs not to say that formalizing all of the algebra required as background for your proof is easy -- it is not.) So, donÍt hold your breath waiting for someone else to formalize your work. Sit down with an introduction to mathematical logic and also a tutorial and reference manual for an automated proving environment and do it yourself, if you think the exercise is valuable. If itÍs not valuable enough to spend your own time on it, no one else will volunteer theirs. -- [T]hereÍs no point in telling any of you what mathematicians IÍm in email contact with, just like thereÍs no point in going into detail about my contacts in a major news organization. [...] [P]olite disinterest is what IÍve found. --JSH on his important contacts. === Subject: Re: calculus problem > Does anyone can help me with this problem? Show that do not exist a sequence of which set of > accumulation points is Q. (a in R is *accumulation point* of sequence x_n, if there exist > subsequence x_{n_k} such that lim x_{n_k} = a ) In fact, I donÍt know why this is true. > We know that set of rational numbers is countable. > Let q_1, q_2, ... be the sequence of ratinal numbers. > So we can build a new sequence like that a_n = (q_1, q_1,q_2, q_1,q_2,q_3, q_1,q_2,q_3,q_4, ...) and in this sequence every rational number occur > in'nitly many times. So for every q in Q there > exist subsequence a_{n_k} such that lim a_{n_k} = q. what did i wrong? > martin I think the problem means: Q and nothing else. Your example has other accumulation points, too. === Subject: Re: recurrence relation >> IÍve turned up the following recurrence: >> x_0=x_1=0 and x_n=(n+1)x_{n-1}+[(n-1)/3](n+1)! for n>=2 >> I donÍt know a recipe for solving a recurrence of this form (ie: with a >> nonconstant coef'cient on the x_{n-1}), and there doesnÍt seem to be any >> inspiration on the way. Can anybody give me a hint? Elementary, Watson! x_n=n(n-1)(n+1)!/6 Holmes === Subject: Re: calculus problem >Does anyone can help me with this problem? Show that do not exist a sequence of which set of >accumulation points is Q. (a in R is *accumulation point* of sequence x_n, if there exist > subsequence x_{n_k} such that lim x_{n_k} = a ) In fact, I donÍt know why this is true. >We know that set of rational numbers is countable. >Let q_1, q_2, ... be the sequence of ratinal numbers. >So we can build a new sequence like that a_n = (q_1, q_1,q_2, q_1,q_2,q_3, q_1,q_2,q_3,q_4, ...) and in this sequence every rational number occur >in'nitly many times. So for every q in Q there >exist subsequence a_{n_k} such that lim a_{n_k} = q. what did i wrong? Saying that the set of accumulation points is Q says that the elements of Q are the _only_ accumulation points - in your example you didnÍt show that there are no irrational accumulation points (which is impossible, because there _are_ some). Hint for doing the original problem: Show that any limit point of the set of accumulation points is an accumulation point. >martin ************************ David C. Ullrich === Subject: help me...my problem?? sequence {An} A1 = 1 A2 = 8 An = root (An-1 * An-2) (n=3,4,5....) 'nd lim An (n->in'nite) --------------------------------- i want your warm advice. help me...please === Subject: Random Numbers I was going through a C++ book when I encountered an problem asking me to de'ne some classes for providing random numbers of certain distributions(like uniform and exponential). The problem is that IÍm not exactly sure on what an exponential or a uniform distribution exactly is. :-( Is there a certain algorithm or formula which I can use for generating a random number for a given distribution? Can someone point me to some basic material on random numbers for uniform and exponential distribution? Sumit. === Subject: Re: help me...my problem?? A2 = 8 >An = root (An-1 * An-2) (n=3,4,5....) 'nd lim An (n->in'nite) --------------------------------- i want your warm advice. help me...please Since you are always taking the geometric mean of the last two numbers, it sure looks to me right off that it will converge somewhere between 1 and 8. Probably havenÍt rigorously proved that, but it seems pretty clear to me. Use brute force. Calculate 10 or 20 terms, see if you can see where you are headed. Gene Nygaard Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ === Subject: Re: help me...my problem?? A2 = 8 >An = root (An-1 * An-2) (n=3,4,5....) 'nd lim An (n->in'nite) --------------------------------- i want your warm advice. help me...please warm advice for a hot girl! Hmmm-mm-m === Subject: Re: help me...my problem?? >sequence {An} A1 = 1 >A2 = 8 >An = root (An-1 * An-2) (n=3,4,5....) 'nd lim An (n->in'nite) --------------------------------- i want your warm advice. help me...please > I suggest thinking about using logarithms to base 2... David Bernier === Subject: Re: Groups with 16 elements > I found an interesting program for doing things > with groups of order < 32 at http//math.ucsd.edu/~jwavrick . > program (or using telnet); > This program will do groups of order 32 also. Van Jacques === Subject: Re: Groups with 16 elements >I found an interesting program for doing things >with groups of order < 32 at http//math.ucsd.edu/~jwavrick . >program (or using telnet); I want to look at the 9 non-Abelian groups that the groups32 program calls #34-42. Recall that GAP said about groups of order 16; There are 14 groups of order 16. >> There are sorted by their ranks. >> 1 is cyclic. >> 2 - 9 have rank 2. >> 10 - 13 have rank 3. >> 14 is elementary abelian. I will assume rank = # of generators. (GAP gives the response for every group of order 16, so I must be doing something wrong.) After removing the Abelian groups, this leaves 3 with 3 generators, and 6 with 2 generators. Z_2 x D_4 and Z_2 x Q have 3, leaving only one other with 3 generators. D_8 has 2 generators, leaving 5 more with 2 generators. I donÍt know if this agrees with the info in my previous post, and below, from the groups32 program. In a previous post I copied from the groups32 program; ORDERS for Group Number 34 Group number 34 of Order 16 1 elements of order 1: A 11 elements of order 2: C E G I J K L M N O P 4 elements of order 4: B D F H 0 elements of order 8: 0 elements of order 16: This is the only group with these orders, so it is Z_2 x D_4. = ORDERS for Groups Number 35 and 38 Group number 35 of Order 16 1 elements of order 1: A 3 elements of order 2: C E G 12 elements of order 4: B D F H I J K L M N O P 0 elements of order 8: 0 elements of order 16: Both 35 and 38 have the same center; Z = {A E C G}. This is the same distribution of orders as in Z_2 x Q, so 35 or 38 is Z_2 x Q. For 35 and 38, there are 6 = 12/2 elements of order 4, and if we call these x_i; i=1,6, then let x_1^2 = x_2^2, x_3^2 = x_4^2, x_5^2 = x_6^2, which gives 3 elements of order 2, all of which are in the center. == I would like to look 'rst at the 5 groups 34 thru 38 groups more closely. (Note that none of these 5 have an element of order 8.) Also, #36 and 37 have the same distribution of orders. ORDERS for Group Number 36 Group number 36 of Order 16 1 elements of order 1: A 7 elements of order 2: C E G I K N P 8 elements of order 4: B D F H J L M O 0 elements of order 8: 0 elements of order 16: ORDERS for Group Number 37 Group number 37 of Order 16 1 elements of order 1: A 7 elements of order 2: C E G I K M O 8 elements of order 4: B D F H J L N P 0 elements of order 8: 0 elements of order 16: CENTER of Group Number 36 { A B C D } (cyclic--Z_4) CENTER of Group Number 37 { A C E G } (Z_2 x Z_2) TABLE for Group Number 36 _A_B_C_D_E_F_G_H_I_J_K_L_M_N_O_P_ A |A B C D E F G H I J K L M N O P B |B C D A F G H E J K L I N O P M C |C D A B G H E F K L I J O P M N D |D A B C H E F G L I J K P M N O E |E F G H A B C D O P M N K L I J F |F G H E B C D A P M N O L I J K G |G H E F C D A B M N O P I J K L H |H E F G D A B C N O P M J K L I I |I J K L M N O P A B C D E F G H J |J K L I N O P M B C D A F G H E K |K L I J O P M N C D A B G H E F L |L I J K P M N O D A B C H E F G M |M N O P I J K L G H E F C D A B N |N O P M J K L I H E F G D A B C O |O P M N K L I J E F G H A B C D P |P M N O L I J K F G H E B C D A SUBGROUPS of Group Number 36 * = Normal subgroup Generators Subgroup 0 { } *{ A } 1 { C } *{ A C } 2 { E } { A E } 3 { G } { A G } 4 { I } { A I } 5 { K } { A K } 6 { N } { A N } 7 { P } { A P } 8 { B } *{ A B C D } 9 { C E } *{ A C E G } 10 { F } *{ A C F H } 11 { C I } *{ A C I K } 12 { J } *{ A C J L } 13 { M } *{ A C M O } 14 { C N } *{ A C N P } 15 { B E } *{ A B C D E F G H } 16 { B I } *{ A B C D I J K L } 17 { E I } *{ A C E G I K M O } 18 { F J } *{ A C F H J L M O } 19 { F I } *{ A C F H I K N P } 20 { E J } *{ A C E G J L N P } 21 { B N } *{ A B C D M N O P } 22 { B E I } *{ A B C D E F G H I J K L M N O P } B,F,J,M are the 8/2 = 4 elements of order 4, and all generate normal subgroups. So 36 has 3 generators, and all the rest have 2 generators (if GAP is right and rank = # generators). CENTER { A B C D } EVALUATE FJF^(-1) = L = J^3 = J^(-1) EVALUATE JFJ^(-1) = H POWERS for element B ; A B C D POWERS for element C ; A C POWERS for element F ; A F C H POWERS for element J ; A J C L POWERS for element M ; A M C O EVALUATE EBE= B EVALUATE ECE= C EVALUATE EFE= F EVALUATE EJE= L EVALUATE EME= O EVALUATE EOE= M EVALUATE GBG= B EVALUATE GFG= F EVALUATE GJG= L EVALUATE GMG= O EVALUATE IBI= B EVALUATE IFI= H EVALUATE IJI= J EVALUATE IMI= O EVALUATE KBK= B EVALUATE KFK= H EVALUATE KJK= J EVALUATE KMK= O EVALUATE NBN= B EVALUATE NFN= H EVALUATE NJN= L EVALUATE NMN= M EVALUATE PBP= B EVALUATE PFP= H EVALUATE PJP= L EVALUATE PMP= M EVALUATE EBE= B EVALUATE ECE= C EVALUATE EFE= F EVALUATE EJE= L EVALUATE EME= O EVALUATE EOE= M EVALUATE GBG= B EVALUATE GFG= F EVALUATE GJG= L EVALUATE GMG= O EVALUATE IBI= B EVALUATE IFI= H EVALUATE IJI= J EVALUATE IMI= O EVALUATE KBK= B EVALUATE KFK= H EVALUATE KJK= J EVALUATE KMK= O EVALUATE NBN= B EVALUATE NFN= H EVALUATE NJN= L EVALUATE NMN= M EVALUATE PBP= B EVALUATE PFP= H EVALUATE PJP= L EVALUATE PMP= M This shows the structure of 36, but I havenÍt 'gured out how to write it concisely, as one does for D_8, e.g. I plan to look at the rest of the groups. I donÍt yet see how one decides how many groups there are with each distribution of orders. It didnÍt turn out quite like I thought--there are more groups with no element of order 8 than I would have guessed. There must be a systematic way of doing this--I know there are papers, and I think some books, on groups of order 2^n, but I am not at a University, so I donÍt have the easy access that I used to have. Van Jacques === Subject: Re: Groups with 16 elements I found an interesting program for doing things > with groups of order < 32 at http//math.ucsd.edu/~jwavrick . > program (or using telnet); I want to look at the 9 non-Abelian groups that the groups32 program >calls #34-42. >Recall that GAP said about groups of order 16; There are 14 groups of order 16. > There are sorted by their ranks. > 1 is cyclic. > 2 - 9 have rank 2. > 10 - 13 have rank 3. > 14 is elementary abelian. I will assume rank = # of generators. (GAP gives the response > for every group of order 16, so I >must be doing something wrong.) After removing the Abelian groups, this leaves 3 with 3 generators, >and >6 with 2 generators. Z_2 x D_4 and Z_2 x Q have 3, leaving only one >other with 3 generators. ThatÍs right. The other one with three generators is the central product of Z_4 and D_4, which is de'ned as the direct product of Z_4 and D_4 but with their central elements of order 2 amalgamated. This turns out to be isomorphic to the central product of Z_4 and Q and is group number 13 in the GAP list. >D_8 has 2 generators, leaving 5 more with 2 generators. Yes. Of these, 3 have elements of order 8, and two do not. I can answer any speci'c queries about these groups. ... >I plan to look at the rest of the groups. I donÍt yet see >how one decides how many groups there are with each distribution of >orders. It didnÍt turn out quite like I thought--there are more >groups with no element of order 8 than I would have guessed. > Here is a GAP loop and printout which provides this sort of information. gap> for i in [1..14] do > G := SmallGroup(16,i); > Print( Group number ,i, n ); > Print( Group abelian: ,IsAbelian(G), n ); > Print( Rank = ,Rank(G), n ); > Print( Orders of elements: ); > l:=List(G,Order); Sort(l); Print(l, n ); > Print( n ); > od; Group number1 Group abelian:true Rank = 1 Orders of elements: [ 1, 2, 4, 4, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 16, 16 ] Group number2 Group abelian:true Rank = 2 Orders of elements: [ 1, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 ] Group number3 Group abelian:false Rank = 2 Orders of elements: [ 1, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4 ] Group number4 Group abelian:false Rank = 2 Orders of elements: [ 1, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 ] Group number5 Group abelian:true Rank = 2 Orders of elements: [ 1, 2, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8 ] Group number6 Group abelian:false Rank = 2 Orders of elements: [ 1, 2, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8 ] Group number7 Group abelian:false Rank = 2 Orders of elements: [ 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 8, 8, 8, 8 ] Group number8 Group abelian:false Rank = 2 Orders of elements: [ 1, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8 ] Group number9 Group abelian:false Rank = 2 Orders of elements: [ 1, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8 ] Group number10 Group abelian:true Rank = 3 Orders of elements: [ 1, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4 ] Group number11 Group abelian:false Rank = 3 Orders of elements: [ 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4 ] Group number12 Group abelian:false Rank = 3 Orders of elements: [ 1, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 ] Group number13 Group abelian:false Rank = 3 Orders of elements: [ 1, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4 ] Group number14 Group abelian:true Rank = 4 Orders of elements: [ 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2 ] Derek Holt. === Subject: Re: Groups with 16 elements I found an interesting program for doing things with groups of order < 32 at http//math.ucsd.edu/~jwavrick . program (or using telnet); It doesnÍt require the experience of GAP, and gives all the properties below--see the commands. This includes order of all elements, and a group multiplication table. Here are some of the results. Connected to euclid.ucsd.edu. Escape character is ï^]'. SunOS 5.8 login: groups32 ; Password: Groups32 Ver 6.4g - September 24, 1999 copyright 1990-1999 John J Wavrik Dept of Math - UCSD Running on: GForth - 0.4.0 Written by Anton Ertl and Bernd Paysan copyright Free Software Foundation CENTER CENTRALIZER CHART CONJ-CLS GROUP HELP INFO ISOMORPHISM KEYFIX LEFT NORMALIZER ORDERS PERMGRPS POWERS QUIT RIGHT SEARCH STOP SUBGROUPS TABLE WEBPAGE X CHART Order of Groups (1-32 or 0) Number 16 29 30 31 32 33 34* 35* 36* 37* 38* 39* 40* 41* 42* There are 14 Groups of order 16 5 abelian and 9 non-abelian (*). ORDERS for Group Number 34 Group number 34 of Order 16 1 elements of order 1: A 11 elements of order 2: C E G I J K L M N O P 4 elements of order 4: B D F H 0 elements of order 8: 0 elements of order 16: ORDERS for Group Number 35 Group number 35 of Order 16 1 elements of order 1: A 3 elements of order 2: C E G 12 elements of order 4: B D F H I J K L M N O P 0 elements of order 8: 0 elements of order 16: ORDERS for Group Number 36 Group number 36 of Order 16 1 elements of order 1: A 7 elements of order 2: C E G I K N P 8 elements of order 4: B D F H J L M O 0 elements of order 8: 0 elements of order 16: ORDERS for Group Number 37 Group number 37 of Order 16 1 elements of order 1: A 7 elements of order 2: C E G I K M O 8 elements of order 4: B D F H J L N P 0 elements of order 8: 0 elements of order 16: ORDERS for Group Number 38 Group number 38 of Order 16 1 elements of order 1: A 3 elements of order 2: C E G 12 elements of order 4: B D F H I J K L M N O P 0 elements of order 8: 0 elements of order 16: ORDERS for Group Number 39 Group number 39 of Order 16 1 elements of order 1: A 3 elements of order 2: E I M 4 elements of order 4: C G K O 8 elements of order 8: B D F H J L N P 0 elements of order 16: ORDERS for Group Number 40 Group number 40 of Order 16 1 elements of order 1: A 9 elements of order 2: E I J K L M N O P 2 elements of order 4: C G 4 elements of order 8: B D F H 0 elements of order 16: ORDERS for Group Number 41 Group number 41 of Order 16 1 elements of order 1: A 5 elements of order 2: E I K M O 6 elements of order 4: C G J L N P 4 elements of order 8: B D F H 0 elements of order 16: ORDERS for Group Number 42 Group number 42 of Order 16 1 elements of order 1: A 1 elements of order 2: E 10 elements of order 4: C G I J K L M N O P 4 elements of order 8: B D F H 0 elements of order 16: = CENTER of Group Number 34 { A C E G } CENTER of Group Number 35 { A C E G } CENTER of Group Number 36 { A B C D } CENTER of Group Number 37 { A C E G } CENTER of Group Number 38 { A C E G } CENTER of Group Number 39 { A C E G } CENTER of Group Number 40 { A E } CENTER of Group Number 41 { A E } CENTER of Group Number 42 { A E } TABLE for Group Number 34 _A_B_C_D_E_F_G_H_I_J_K_L_M_N_O_P_ A |A B C D E F G H I J K L M N O P B |B C D A F G H E L I J K P M N O C |C D A B G H E F K L I J O P M N D |D A B C H E F G J K L I N O P M E |E F G H A B C D M N O P I J K L F |F G H E B C D A P M N O L I J K G |G H E F C D A B O P M N K L I J H |H E F G D A B C N O P M J K L I I |I J K L M N O P A B C D E F G H J |J K L I N O P M D A B C H E F G K |K L I J O P M N C D A B G H E F L |L I J K P M N O B C D A F G H E M |M N O P I J K L E F G H A B C D N |N O P M J K L I H E F G D A B C O |O P M N K L I J G H E F C D A B P |P M N O L I J K F G H E B C D A TABLE for Group Number 35 _A_B_C_D_E_F_G_H_I_J_K_L_M_N_O_P_ A |A B C D E F G H I J K L M N O P B |B C D A F G H E L I J K P M N O C |C D A B G H E F K L I J O P M N D |D A B C H E F G J K L I N O P M E |E F G H A B C D M N O P I J K L F |F G H E B C D A P M N O L I J K G |G H E F C D A B O P M N K L I J H |H E F G D A B C N O P M J K L I I |I J K L M N O P C D A B G H E F J |J K L I N O P M B C D A F G H E K |K L I J O P M N A B C D E F G H L |L I J K P M N O D A B C H E F G M |M N O P I J K L G H E F C D A B N |N O P M J K L I F G H E B C D A O |O P M N K L I J E F G H A B C D P |P M N O L I J K H E F G D A B C TABLE for Group Number 42 _A_B_C_D_E_F_G_H_I_J_K_L_M_N_O_P_ A |A B C D E F G H I J K L M N O P B |B C D E F G H A P I J K L M N O C |C D E F G H A B O P I J K L M N D |D E F G H A B C N O P I J K L M E |E F G H A B C D M N O P I J K L F |F G H A B C D E L M N O P I J K G |G H A B C D E F K L M N O P I J H |H A B C D E F G J K L M N O P I I |I J K L M N O P E F G H A B C D J |J K L M N O P I D E F G H A B C K |K L M N O P I J C D E F G H A B L |L M N O P I J K B C D E F G H A M |M N O P I J K L A B C D E F G H N |N O P I J K L M H A B C D E F G O |O P I J K L M N G H A B C D E F P |P I J K L M N O F G H A B C D E To use GAP effectively, you will need to devote several hours to reading > the documentation (on-line) and playing with it. Here is sample session > concerning groups of order 16. Note that the generators of a group G are > typed in as G.1, G.2,..., but (confusingly) they are printed by GAP as > f1, f2,... gap> NumberSmallGroups(16); > 14 > gap> G:=SmallGroup(16,11); > IsAbelian(G); > false > gap> Size(Centre(G)); > 4 > gap> IsCyclic(Centre(G)); > false > gap> NilpotencyClassOfGroup(G); > 2 > gap> Size(CommutatorSubgroup(G,G)); > 2 > gap> Order(G.3); > 2 > gap> G.1 in Centre(G); > false > gap> G.4*G.3*G.2*G.1; > f1*f2*f3 > gap> Size(AutomorphismGroup(G)); > 64 > gap> IsomorphismGroups(G,DihedralGroup(16)); > fail > for i in [1..14] do >Print(i, ,IsomorphismGroups(SmallGroup(16,i), DihedralGroup(16)), n ); >od; > 1 fail > 2 fail > 3 fail > 4 fail > 5 fail > 6 fail > 7 Pcgs([ f1, f2, f3, f4 ]) -> [ f1*f2, f1*f4, f3, f4 ] > 8 fail > 9 fail > 10 fail > 11 fail > 12 fail > 13 fail > 14 fail === Subject: Re: formula for roots of quadratic matrix equation? Content-Length: 602 Originator: rusin@vesuvius Is there a closed form formula for >the roots of the quadratic matrix equation QAQ - BQ + C = 0 where A, B, C and Q are all symmetric nxn matrices, n>1? If not, how about for the special case B=I (identity matrix)? Let sqrt(X) be a symmetric matrix satisfying sqrt(X)^t sqrt(X) = X. It is not completely trivial that there is always such a matrix. In particular, one may need to work in an algebraically closed 'eld. Arnold Neumaier === Subject: Re: formula for roots of quadratic matrix equation? Content-Length: 695 Originator: rusin@vesuvius > Is there a closed form formula for > the roots of the quadratic matrix equation QAQ - BQ + C = 0 where A, B, C and Q are all symmetric nxn matrices, n>1? If not, how about for the special case B=I (identity matrix)? Let sqrt(X) be a symmetric matrix satisfying sqrt(X)^t sqrt(X) = X. Then Q = sqrt(A)^{-1} [ sqrt( F^t F - C ) - F ] where F = -sqrt(A)^{-t} [ I + B ] If B=I, we get F = -2 sqrt(A)^{-t} and Q = sqrt(A)^{-1} sqrt( 4 A - C ) + 2 A I got this from matching terms on ( sqrt(A) Q + F )^t ( sqrt(A) Q + F ) = F^t F - C so it may not give all the roots. -Thomas C === Subject: Re: Does the world have an edge? > You ainÍt in Kansas anymore. > No, but you sure seem to be in Oz. Watch out for those §ying monkeys Jack. Perion === Subject: Does the world have an edge? You ainÍt in Kansas anymore. Our Topsy Turvy Universe - A Hall of Funny Mirrors in The Cosmic Coney Island of a Lunatic Mind? ;-) This make you feel like a gold'sh in a bowl with really weird star gate portal 2D faces in fact 12 of them shaped as pentagons - positively Medieval. The regular dodecahedron is the Platonic solid composed of 20 polyhedron vertices , 30 polyhedron edges , and 12 pentagonal faces ,. It is also uniform polyhedron and Wenninger model . It is given by the Schl .8a§i symbol and the Wythoff symbol . http://mathworld.wolfram.com/Dodecahedron.html with many cheerful facts about the square of the hypotenuse Major General, Pirates of Penzance , Gilbert and Sullivan Jack, Your question about the geometry of the dodecahedral space is answered Do you mean this starting on p. 593: The Poincare dodecahedral space is a dodecahedral block of space with opposite faces abstractly glued together, so objects passing out of the dodecahedron across any face return from the opposite face. Now this is trivial for a topologist but not for a physicist like Richard Feynman for example, who would immediately want to know how such a think could be physically implemented in our Universe. Exotic vacua / zpf(x,L) =/= 0 will do the trick. The astounding stranger than Sci Fi paper Dodecahedral space topology as an explanation for weak wide-angle temperature correlations in the cosmic microwave background from a Franco-American team in Paris, NYC, Saclay and Orsay (JP Luminet, JR Weeks, A Riazuelo, R. Lehoucq & JP Uzan respectively continues: Light travels across the faces in the same way, so if we sit inside the dodecahedron and look outward across a face, our line of sight re-enters the dodecahedron from the opposite face. We have the illusion of looking into an adjacent copy of the dodecahedron. OK light rays are null geodesics ds^2 = 0. Suppose instead we take an alleged retrieved alien ET §ying saucer with / zpf(x,L) zero point energy exotic vacuum timelike geodesic warp drive and send some of our space-jockeys to one of the 12 pentagon 2D faces, of the single 3D spherical dodecahedron we are inside of, to its literal Star Gate edge far way in space billions of light years from Earth in the blink of an eye on a self-generated timelike geodesic with ds^2 very close to zero at each point, but timelike mind you inside the local light cone. What happens when the saucer enters one of the 12 pentagonal faces. Obviously it must exit at the opposite face. This mean it must be a Star Gate, this means it must have an interesting / zpf(x,L) 'eld con'guration at the edge of the world. If you are beginning to feel like Jason and the Argonauts in Search of the Golden Fleece, or a Viking or Chris Columbus thatÍs understandable. ;-) If we take the original dodecahedral block of space not as a euclidean dodecahedron (with edge angles ~ 117 degrees) but as a spherical dodecahedron (with edge angles exactly 120 degrees), then adjacent images of the dodecahedron 't together snugly to tile the hypersphere (Fig 3b) ... 120 spherical dodecahedra tile the surface of the hypersphere. A hypersphere is the three-dimensional surface of a four-dimensional ball. See http://www.geometrygames.org/(CurvedSpaces). Note that the 4th dimension here is not time as in space-time, but is, rather, an extra space dimension that we can actually do without and just do intrinsic Riemannian geometry on the 3D hypersphere that is expanding and accelerating in time from / > 0 in the large scale where Omega(/ > 0) is somewhere between 0.66 and 0.73 depending on which Pundit one reads these days. The entire 3D hypersphere snugly tiled with the 120 spherical dodecahedra has no boundary, i.e. bc = 0. However, my question here is: Are we trapped inside only one of the 120 spherical dodecahedra? That is, every time we send a star ship through one of its 12 pentagonal faces it instantly reappears at the opposite face c/H(t) light years away? That is the CMB temperature is the same - or, is this a Cosmic Time Travel Machine? Can we go both to past and future through The Pentagonal Faces at the 12 Edges of The World? I mean the sub-world of the 120 that make the total hypersphere with no boundary? This is beginning to sound like the Captain Marvel comics I read in the 1940Ís with Shazam and The Black Rock of Tauhid like the one in Mecca? Which way to Mecca? :-) Let us ponder. To be continued. Also you should re-read the description of dodecahedral space from Lou Kauffman I forwarded to you this morning. Especially relevant to your question is his statement: [And keep your Pong game picture in mind]: Yeah, thatÍs Star Gate metric engineering. That is the sort of description that the dodecahedral ïidentify by 1/5 twist the opposite facesÍ description uses. In that description a ray of light heading for one face §ies right through and comes in again via the opposite face. But the ïfacesÍ are not there. That is just a way to describe the periodicity, and it is traces of this periodicity that are supposed to be inherent in the astronomical data. Here is a 2-d analog that makes this clear. [This can be done as a gedanken experiment.] Take a rectangular sheet of paper. Glue together two opposite sides (1-d faces ). Now you have a cylinder. Now connect up the two ends of the cylinder. You now have a 2-d torus. Clearly the edges that you glue together are in arbitrary position on the torus. We could cut the torus at some arbitrary place and then cut the resulting cylinder lenghthwise in another arbitrary place. So from the point of view of a creature trapped on the 2-d torus there are no edges to go through -- although these 2-d creatures could still describe their toroidal space as a rectangle with opposite edges identi'ed (provided that their mathematical imagination was suf'ciently advanced). Saul-Paul ---------- === Subject: Re: Vigier V: EscherÍs Star Gate shape of the universe Part III The Pong Video Game in The Sky with 2D Wall periodic boundaries may not be what Ellis et-al are suggesting. Ellis cites Omega zero = 1.02 +-0.02. meetings the Pundits like Mike Turner all said k = 0, in§ation is right, i.e spatially §at with Omega zero = 1 on the button. Ellis is saying no to that that there will be a Big Crunch with k = +1 and Omega zero > 1. That allows a 'nite 3D space without 2D wall boundaries. spatial sections of the Universe are dodecahedral sections of space of positive curvature, 'tted together to make 'nite three-dimensional spaces. Earlier Ellis does make the Pong Video Game analogy implicitly for the §at toroidal space, as you exit right you enter left and space is 'nite Does Ellis mean that there are sectors of 3D space with 2D walls between them which 't together like some 'nite Platonic solid? What happens when you walk through a wall so to speak? Is there a periodic boundary condition (i.e. a Star Gate in effect) or not? That is the total 'nite space of positive curvature has no boundary, but is partitioned into disjoint subspaces or sectors separated by 2D boundary walls? Is that the idea? Are these walls weird or not? Can we pass through them? What happens when we do? Or are they not there at all? Saul-Paul: There are no walls. See my description above. When I look at Fig 3 and read look outward across a face, our line of sight re-enters the dodecahedron from the opposite face that sounds like a wall or face to me. Physically, it is a Star Gate portal. It may not be a wall to the topologist in the strict formal sense but it looks like a wall I mean a 2d surface to this physicist in the physical informal sense or operational sense. I mean is the ship transported instantly billions of light years across space, or perhaps forward or backward in the global cosmic time of the FRW metric? Whatever the picture is here Ellis is clear that this alternative is not compatible with chaotic in§ation that demands Omega zero = 1 with k = 0. Luminet has Omega zero = 1.013 Max TegmarkÍs picture of Level I Hubble universes in May Scienti'c American is not consistent with what Ellis et-al is suggesting here. That is Ellis rejects that spatial homogeneity extends outside our visual horizon forever .... and we are in one expanding bubble in the middle of innumerable other similar ones. But if Luminet et-al are correct, chaotic in§ation is ruled out: there is only one expanding bubble, and we can see almost all the way round it .... The WMAP data as interpreted by Luminet et-al... suggest that we might live in such a small closed universe. This is qualitatively globally different from what Mike Turner, Max Tegmark et-al are suggesting. So we have a schism between the k = 0 Camp and the k = +1 Camp. We need better precision than the current 2% to decide. Jack, Post Script: Looking again at the picture of the dodecahedral space in *Nature* (9 October, page 594), I think I see what your conceptual problem is--i.e., why you think of walls. First they show the 12 spherical pentagons covering the 2-sphere (soccer ball). Then they show some of the 120 spherical dodecahedra tesellating the 3-sphere. Yes, I am staring at Fig 3. What they DONÍT explain is that the cosmological space is NOT supposed to be the 3-sphere but the orbit space created by the action of the Icosahedral Double group on the three sphere. This group is usually called the binary Icosahedral group, but I prefer the nomenclature of the crystallographers (since binary has the more common connotation of binary numbers). Since SU(2) is geometrically a three sphere, this orbit space can be written as SU(2)/ID. OK so you say that the proper total 3D physical space is not the hypersphere with no boundary, but this orbit space that is a quotient space mod an equivalence relation in this instance ID the Icosahedral Double Group. Now ID is a 120 element group, and the orbits are all cosets of ID in SU(2). In other words ID acts on any point in one of the spherical dodecahedra and jumps that point in a discrete orbit through the analogous points--one point in each of the 120 dodecahedra. The set of orbits (each of 120 points) constitute a new 3-d space of very different topology -- i.e., the topology equivalent that created by identifying the opposite faces (with a 1/5 twist) of an ordinary (non-curved) dodecahedron. Incidentally, since ID is a 'nite subgroup of SU(2), the 120 elements of ID are located at the mid-points of the 120 spherical dodecahedra. These constitute an orbit of ID (called the identity coset) within the SU(2) Lie group. Saul-Paul OK, but I need to recast this in more operational physical terms like what happens when we send a star ship with zero point energy exotic vacuum warp drive on a self-generated timelike ds^2 ~ 0 (but not exactly 0) to the edge of the world does it come back at the opposite face or not? Or does that no longer have meaning? But it must according to the intuitive operational picture for global physical properties of SU(2)/ID. You have one point event of the orbit space SU(2)/ID with an internal structure of 120 points of the original SU(2). Each 'nite coset with 120 elements is a single point event of SU(2)/ID correct? There must be some physics here in addition to this pure formal images of the same structure but in different directions in the sky . Now that is almost an operational de'nition of entering and leaving a Star Gate - and that must happen in SU(2)/ID in your above interpretation! Now imagine we are not tracking light signals, but actual ships with warp drive. DonÍt they see multiple images of the UFOs at times? I mean we cannot get rid of the weirdness even if there are no edges, though I am not sure what that means. Operationally all I mean by an edge or a wall is a sudden transport or teleportation of a material object across a large distance including possible shifts in cosmic time either way. That seems to be an objective feature of this weird topology model that is inescapable and that gives more credibility to the otherwise incomprehensible UFO allegations that led Jacques Vallee to prematurely embrace the irrational point of view like Bohr did in quantum physics. === Subject: product of arithmetic sequence terms Which is the fastest algorithm for calculating the product of the n terms of the following arithmetic sequence: s(n) = 1, 3, 5, ..., 2n-1 === Subject: Re: product of arithmetic sequence terms > Which is the fastest algorithm for calculating the product of the n terms of > the following arithmetic sequence: > s(n) = 1, 3, 5, ..., 2n-1 You want a exact result or an aproximate one? In the 'rst case, you must multiply all the numbers. If you only need a aproximate result, you can do P(n) = 1*3*5*...*(2n-1) = 1*3*5*...*(2n-1)*(2*4*...*2n-2)/(2^(n-1)*(n-1)!) = (2n-1)!/(2^(n-1)*(n-1)!) And then use the Stirling aproximation: n! ~= (n/e)^n*sqrt(2*pi*n) Then P(n) ~= ((2n - 1)/e)^(2n - 1)sqrt(2pi(2n - 1))/(2^(n-1)((n - 1)/e)^(n - 1)sqrt(2pi(n - 1))) = (2n - 1)^((4n - 1)/2)/(2^(n-1)*e^n*(n-1)^((2n-1)/2)) Or perhaps better, Ln(P(n)) ~= (2n-1/2)Ln(2n-1) - (n-1/2)Ln(n-1) - (n-1)Ln(2) - n -- Ignacio Larrosa Ca .96estro A Coru .96a (Espa .96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Minimal Graph, Four Color Theorem Visiting Assistant Professor at the University of Montana. >I judge you to be malicious and hypocritical. Your prerogative. >Malicious in that you >chose to deliberately insult me. Hypocritcial by dissing my manners >and my personality. I stated that ->if<- what you were doing was something speci'c, then I found you to have no manners. You chose to take this as an insult, which I assume means that what you were doing ->was<- what I put in the premise of that conditional statement. As for your reading of malice and hypocrisy, all I can say is that your read on this has been about as accurate as your read on pretty much everything you have posted in this thread: a lot of what was in your head before you began, and little about what was actually written. >Finally, you are ïpig-headedÍ if you do not offer an apology; even if >I do not deserve one. Oh, excellent! That one really takes the cake. >I hope that in the future we may be civil to each other. We will, for the simple reason that I will not read anything else you have to say. <*plonk*> Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Minimal Graph, Four Color Theorem I did NOT claim it was 5-chroma. There was no incorrect data, except > in your imagination. > I apologize. How could G(7) be 5-chroma and also planar? I guess I have 5-chroma on the brain? You have no idea what you are talking about. You are either not > reading what you are replying to, or not understanding it, or being > purposely dishonest. Which is it? > understand, I do not always see the relevance. Of course if I do not understand something, then it is irrelevant to me. I have a very good idea of what I am talking about. Its just that my 'ngers seem to be tongue-tied. The problem, from my point of view, is that you donÍt know what I am talking about. I judge you to be malicious and hypocritical. Malicious in that you chose to deliberately insult me. Hypocritcial by dissing my manners and my personality. I might add conceited, but I will assume that if you take a square with one diagonal added, and then you join the other two opposite vertices by an edge lying outside the square, you do get a minimal graph with chi(G)=3 , to be merely a typo, rather than ïinaccurate dataÍ. Finally, you are ïpig-headedÍ if you do not offer an apology; even if I do not deserve one. I hope that in the future we may be civil to each other. Bill J. === Subject: Re: That Hilbert guy is giving me a lot of trouble |> 3) Does having a complex linear space instead of a real one improves > |> the situation > | > |Yes. Complex numbers are algebraically complete. This helps. The property is termed algebraically closed . -- === Subject: Re: That Hilbert guy is giving me a lot of trouble |> 3) Does having a complex linear space instead of a real one improves |> the situation | |Yes. Complex numbers are algebraically complete. This helps. The property is termed algebraically closed . Keith Ramsay === Subject: Re: Boolean Algebra - Arithmetic Relationship Is there some master reducibilty graph that mathematicians use to show when one symbology is reducible to another? IÍve seen something like that, that relates NP-Completeness problems to one another. If I pursue mathematics IÍd like to be a big picture sort of guy. === Subject: Re: Boolean Algebra - Arithmetic Relationship Is there some master reducibilty graph that mathematicians use to > show when one symbology is reducible to another? Do mathematicians reduce or do they build up ? Sometimes they uncover the commonality between two apparently different branches of mathematics--but thatÍs not reducing itÍs 'nding a common foundation. Category theory is a good example. David RusinÍs Mathematical Atlas http://www.math-atlas.org/welcome.html shows relations between different 'elds of mathematics. IÍve seen something like that, that relates NP-Completeness problems > to one another. If I pursue mathematics IÍd like to be a big picture sort of guy. Good, IÍm sure that narrow specialization is not good if it can be avoided (maybe the professionals have no choice) but nor is attention to detail avoidable. -- G.C. e-mailing me. === Subject: Re: Boolean Algebra - Arithmetic Relationship Mathematics seems like an entirely chaotic enterprise at the macrosopic level, it hard to believe that anyone could exercise leadership to in§uence it in a particular direction. >Is there some master reducibilty graph that mathematicians use to >show when one symbology is reducible to another? Do mathematicians reduce or do they build up ? Sometimes they > uncover the commonality between two apparently different branches of > mathematics--but thatÍs not reducing itÍs 'nding a common foundation. > Category theory is a good example. This is rather disheartening, I was hoping that reducibility might lead to a hierarchical organization of mathematic symbology. I was under the impression that some symbologies where, in fact, conceptual subsets of others. Or in other words, everything that can be proven in symbology X can be proven in symbology Y, but Y can prove even more. If this was true, it might be possible to propagate mathematical proofs up the chain of symbologies. Like say I prove something in Boolean Algebra which is reducible to Propositional Logic. I could then translate that proof to Propositional Logic. If Propositional Logic was reducible to 'rst order logic, I could then translate it again to 'rst order logic... and so on. You could move it all the way up the chain, to say some symbology in Set Theory, that acts as a giant repository for proofs. Possibly you could propogate these down again too. Of course, if some maths are not truly subsets of others this would run into problems. > David RusinÍs Mathematical Atlas http://www.math-atlas.org/welcome.html > shows relations between different 'elds of mathematics. Interesting website. Although from what I see I think an undirected graph of relationships might be a better map for the mathematical landscape than the 2D picture he uses. === Subject: Re: Boolean Algebra - Arithmetic Relationship .. > Or at least so IÍm told. In mathematics, it seems that ideas have a greater > lifespan; a genuinely good result is less apt to be discarded because it > was only valid in an abandoned theory . Still, mathematicians can be the victims of fashion like the rest of us. And old fashions can come round again: special functions went out of fashion and then de Branges used them to prove the Bieberbach conjecture. -- G.C. e-mailing me. === Subject: why is Hamel dimension well-de'ned? Would someone please point me to a proof of this theorem? Any two Hamel bases of a vector space have the space have the same cardinality. This is driving me crazy. Have a tolerable existence. Eli === Subject: Re: why is Hamel dimension well-de'ned? >Would someone please point me to a proof of this theorem? Any two Hamel >bases of a vector space have the space have the same cardinality. This is >driving me crazy. Say B1 and B2 are in'nite bases for the same vector space. Let S be the set of 'nite subsets of B2. Then S has the same cardinality as B2 (right?). De'ne f : B1 -> S by letting f(b1) be the set of all elements of B2 that appear when we write b1 as a linear combination of elements of B2. If f were 1-1 then this would show card(S) >= card(B1). But f _is_ 'nite-to-one, meaning that an element of S has only 'nitely many inverse images; surely(?) this shows that card(S) >= card(B1) as well. I bet, maybe. Yeah: At least with AC, we could partition B1 into countable many subsets such that f is 1-1 on each subset. Hence each subset has cardinality <= card(S), and hence card(B1) <= card(S). >Have a tolerable existence. Eli ************************ David C. Ullrich === Subject: Re: why is Hamel dimension well-de'ned? > Would someone please point me to a proof of this theorem? Any two Hamel > bases of a vector space have the space have the same cardinality. This is > driving me crazy. > Let S be a Hamel Basis of smaller cardinality that the Hamel Basis T. Consider pairs (SÍ,TÍ,fÍ) where SÍ is a subset of S, TÍ is a subset of T, SÍ_UNION_TÍ is linearly independent and TÍ is not empty and SÍ_UNION_TÍ has cardinality |T| where the identity map on TÍ and fÍ on SÍ give a 1-1 map from SÍ_UNION_TÍ to T (T fÍ(SÍ)=TÍ) (fÍ(SÍ) is the set of elements in T we replace by elements of S so that SÍ_UNION_TÍ remains linearly independent) Order them by: (SÍ,TÍ) > (S ,T ) iff SÍ contains S T contains TÍ fÍ extends f (basically lookin at subset of S which we can shove into T and keep a linearly independent subset if we drop a corresponding number of elements out of T) For any 'nite subset SÍ of S, there exists a TÍ with this property (heck, we can shove each element of SÍ into T by dropping one element of T to reduce to TÍ as in the proof for 'nite dimensional spaces). Is the set of (SÍ,TÍ,fÍ) inductive? Well, for any totally ordered set of them (chain) let S~ be the union of the SÍ and T~ the intersection of TÍ and let f~ be de'ned on S~ in the natural way (since each of the fÍs extends the fÍ for smaller sets). S~_UNION_T~ is linearly independent (for linear dependency would require a 'nite subset to be dependent which would give a particular set SÍ_UNION_TÍ which is dependent). f~ on S~ and the identity on T~ is a 1-1 map onto T (since T fÍ(SÍ)=TÍ at each step, just take the intersection). T~ is not empty (of the cardinality of T is larger than that of S, since f~(S~) only has the cardinality of S). (S~,T~,f~) > (SÍ,TÍ,fÍ) for all (SÍ,TÍ,fÍ) in the chain. It looks like it is inductive. Let (A,B,F) be a maximal element (A a subset of S). Can A be smaller than S? If s is in S A, then as in the argument for 'nite dimensional spaces we could move s into A_UNION_B by dropping an element of B (if s is not in the span of A_UNION_B just add it, if s is in the span, it is not in the span of A, since they are both in S, so s=SUM[number_j*A_j]+SUM[numberÍ_k*B_k] and one, at least, of the numberÍ_kÍs is non zero and we can replace the corresponding element, B_k of B with s - this is the same as for 'nite sets) So, the maximal element must have A=S. But then we would have A_UNION_B being linearly independent (where B is not the empty set) and A=S. Then ... S does not span! (I have just done this while sitting here and have not checked for typos or extra details that may have to be included, but am sure this sort of argument will do it - it is the same sort of argument one uses to show that basis in a 'nite dimensional space have the same cardinality, by embedding each element of one basis, one after the other, into the other basis - if the second basis has larger cardinality, then one would embed the smaller basis into the larger and have a linearly independent set larger than the smaller basis! The only difference here is that one cannot do it one element after the other but ask for a largest emedding, using ZornÍs lemma to show that it exists, and then show that it works.) === Subject: Re: why is Hamel dimension well-de'ned? to do but I kept getting stuck at one step. In reading your proof, I got stuck on the corresponding step. > T~ is not empty (of the cardinality of T is larger > than that of S, since f~(S~) only has the cardinality > of S). (S~,T~,f~) > (SÍ,TÍ,fÍ) for all (SÍ,TÍ,fÍ) in the chain. The one step you omitted here is to prove that ( |S~ UNION T~| = |T| ) which you used to prove that T~ isnÍt empty and that (S~,T~,f~) is an object of the type you de'ned at the beginning. The set T~ is the intersection of a possibly in'nite chain. In'nite chains donÍt always behave well under intersection. For example, consider open intervals in the real line. In particular, INTERSECTION{(0,1/n) : n is a positive integer} is empty. Do you know how to deal with this? Have a tolerable existence. Eli === Subject: Re: why is Hamel dimension well-de'ned? trying to do but I > kept getting stuck at one step. In reading your proof, I got stuck on the > corresponding step. > T~ is not empty (of the cardinality of T is larger >> than that of S, since f~(S~) only has the cardinality >> of S). >> (S~,T~,f~) > (SÍ,TÍ,fÍ) for all (SÍ,TÍ,fÍ) in the chain. The one step you omitted here is to prove that ( |S~ UNION T~| = |T| ) which > you used to prove that T~ isnÍt empty and that (S~,T~,f~) is an object of > the type you de'ned at the beginning. The set T~ is the intersection of a possibly in'nite chain. In'nite > chains donÍt always behave well under intersection. For example, consider > open intervals in the real line. In particular, > INTERSECTION{(0,1/n) : n is a positive integer} > is empty. T f~(SÍ)=TÍ take inersections (the intersection of the complements of the f~(SÍ) is the complement of their union and their union is f~(S~)) Does that not work? === Subject: Re: why is Hamel dimension well-de'ned? trying to do but >> I >> kept getting stuck at one step. In reading your proof, I got stuck on >> the corresponding step. > T~ is not empty (of the cardinality of T is larger > than that of S, since f~(S~) only has the cardinality > of S). (S~,T~,f~) > (SÍ,TÍ,fÍ) for all (SÍ,TÍ,fÍ) in the chain. >> The one step you omitted here is to prove that ( |S~ UNION T~| = |T| ) >> which you used to prove that T~ isnÍt empty and that (S~,T~,f~) is an >> object of the type you de'ned at the beginning. >> The set T~ is the intersection of a possibly in'nite chain. In'nite >> chains donÍt always behave well under intersection. For example, >> consider >> open intervals in the real line. In particular, >> INTERSECTION{(0,1/n) : n is a positive integer} >> is empty. T f~(SÍ)=TÍ take inersections (the intersection of the complements > of the f~(SÍ) is the complement of their union and their > union is f~(S~)) Does that not work? looked through just state the result without even giving a reference to a proof. ItÍs damn frustrating, let me tell you. Have a tolerable existence. Eli === Subject: Re: why is Hamel dimension well-de'ned? <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h }Y L`!h_XXr5Q> _nGsY2_ [[ This message was both posted and mailed: see the To, Cc, and Newsgroups headers for details. ]] > Would someone please point me to a proof of this theorem? Any two Hamel > bases of a vector space have the space have the same cardinality. This is > driving me crazy. Have a tolerable existence. Eli > There is a proof in Real and Abstract Analysis by Hewitt & Stromberg. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: why is Hamel dimension well-de'ned? > Would someone please point me to a proof of this theorem? Any two Hamel > bases of a vector space have the space have the same cardinality. This is > driving me crazy. Have a tolerable existence. Eli Have you considered the group of automorphisms of the space. === Subject: Re: why is Hamel dimension well-de'ned? > Would someone please point me to a proof of this theorem? Any two Hamel >> bases of a vector space have the space have the same cardinality. This >> is driving me crazy. >> Have a tolerable existence. Eli Have you considered the group of automorphisms of the space. If I could 'nd an automorphism that mapped one basis to the other, that would solve the problem. But why should such an automorphism exist if the bases donÍt have the same cardinality? Have a tolerable existence. Eli === Subject: Re: why is Hamel dimension well-de'ned? >Would someone please point me to a proof of this theorem? Any two Hamel >bases of a vector space have the space have the same cardinality. This is >driving me crazy. Let V be a vectorspace. If V has a 'nite Hamel basis, then V is 'nite-dimensional, so by standard results (of matrix algebra, if you will) any two Hamel bases of V have the same ('nite) cardinality. Let V have an in'nite Hamel basis of cardinality k. Let the ground 'eld F of V have cardinality f. The cardinality of V is then ... some simple function of f and k which I am not immediately able to describe, and which (moreover) has the property that, for a 'xed f, it is an injective function of k. Some proof, eh? Maybe I should just not post this. But what the heck. Lee Rudolph === Subject: Re: why is Hamel dimension well-de'ned? <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h }Y L`!h_XXr5Q> _nGsY2_ > Let V have an in'nite Hamel basis of cardinality k. > Let the ground 'eld F of V have cardinality f. The > cardinality of V is then ... some simple function of > f and k which I am not immediately able to describe, > and which (moreover) has the property that, for a 'xed > f, it is an injective function of k. That wonÍt do. All vector spaces over R with Hamel dimension from aleph_0 up to c have the same cardinality. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: why is Hamel dimension well-de'ned? > Let V have an in'nite Hamel basis of cardinality k. >> Let the ground 'eld F of V have cardinality f. The >> cardinality of V is then ... some simple function of >> f and k which I am not immediately able to describe, >> and which (moreover) has the property that, for a 'xed >> f, it is an injective function of k. That wonÍt do. All vector spaces over R with Hamel dimension from >aleph_0 up to c have the same cardinality. (Namely c, I suppose?) Drat. ThatÍs what I get for posting with a head full of cottonwool. Lee Rudolph === Subject: Re: why is Hamel dimension well-de'ned? Let V have an in'nite Hamel basis of cardinality k. > Let the ground 'eld F of V have cardinality f. The > cardinality of V is then ... some simple function of > f and k which I am not immediately able to describe, > and which (moreover) has the property that, for a 'xed > f, it is an injective function of k. >>That wonÍt do. All vector spaces over R with Hamel dimension from >>aleph_0 up to c have the same cardinality. (Namely c, I suppose?) Drat. ThatÍs what I get for posting with a head full of >cottonwool. I coulda swore it was just a cardinality thing, as you said. (Maybe it says so in some book we both read in our youth or something...) >Lee Rudolph ************************ David C. Ullrich === Subject: Re: why is Hamel dimension well-de'ned? >Would someone please point me to a proof of this theorem? Any two Hamel >>bases of a vector space have the space have the same cardinality. This >>is driving me crazy. Let V be a vectorspace. If V has a 'nite Hamel basis, then V is 'nite-dimensional, > so by standard results (of matrix algebra, if you will) any > two Hamel bases of V have the same ('nite) cardinality. Let V have an in'nite Hamel basis of cardinality k. > Let the ground 'eld F of V have cardinality f. The > cardinality of V is then ... some simple function of > f and k which I am not immediately able to describe, > and which (moreover) has the property that, for a 'xed > f, it is an injective function of k. Some proof, eh? Maybe I should just not post this. But > what the heck. Lee Rudolph your argument? Have a tolerable existence. Eli === Subject: Re: Core error, proof checking Visiting Assistant Professor at the University of Montana. multiple copies of your messages. You might want to quit your browser and restart. === Subject: Re: Core error, proof checking > multiple >copies of your messages. You might want to quit your browser and >restart. Is it Google who is doing this? IÍve seen ISP stutters from [what appear to be] other newservers, too. Caveat: I havenÍt done a trace to see if thereÍs common hub. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Core error, proof checking linux) > In order to be checkable automatically, you must have a set of formal >> axioms, and EACH statement in your proof must be translated into a >> formal statement in a formal language. Then a proof becomes a sequence >> of statements. Then you check: each statement must either be in the >> list of axioms, or else it must be of the form B where there exist >> two prior lines, one of the form A->B and the other of the form >> A . If this is true, in turn, for each statement in your sequence, >> then your sequence is a correct proof. An interesting question - how to formalize his favorite term should ? The formal statements A is B and A is not B are quite common, and the > meaning is obvious, but I never heard of a predicate notation A > should be B . A should be B is commonly translated in terms of a deontic operator (a special form of modal operator). Of course, that only gives a formal means of expressing A should be B, without necessarily giving a real meaning for should here. I am *not* suggesting that if we move to a modal logic, then JamesÍs use of should would magically become coherent. -- But remember, as long as one human being follows the rules of mathematics, then mathematics as a human discipline survives. Right now IÍm that one human being, so mathematics survives. -- James S. Harris === Subject: Re: Is it mass; or is it weight > So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net > which stretches an additional 4.4-ft before thrusting the man back, would > the potential energy of the net at the instant it plunges the man back be U = mgh; > U = 220 * (36 + 4.4). >> You left off the factor of g . Which either means that you forgot about it >> or that you are using a system of units within which the factor of g >> out. >> For instance, if you measure m in pounds mass, h in feet and energy in = mh. >> from the jump site >> with standard g from the ratio of pounds force to pounds mass. >> Gene Nygaard would be correct to point out that there is no single standard >> g that is of'cially sanctioned for this purpose. While initially attempting this question, I converted pounds-mass to > kilograms-mass, and the height in feet to metres, and computed the potential > energy through U = mgh; > U = (97.99) * (9.8) * (12.73); which revealed the netÍs potential energy to be at the time it sent the man > back upwards almost twelve kilo-joules. My Physics teacher, however, disagreed, and prosposed that the given units of > the mass and height be maintained as is, and the potential energy be > calculated without the factor ïgÍ present anywhere in the computation. This > method outputs U = mh; > U = (220) (36 + 4.4); > U = 8888 lb-ft; >> Your physics teacher is apparently an idiot, taught by idiots out of textbooks >> written by idiots. Sadly, thatÍs quite typical >> For a coherent system of units, U = mgh. >> For example, the following situations involve coherent sets of units: >> U in foot-pounds(force), m in slugs, g in feet/sec^2, h in feet U in joules, m >> in kilograms, g in meters/sec^2, h in meters U in foot-poundals, m in >> pounds(mass), g in feet/sec^2, h in feet >> For systems of units that are not coherent, U = kmgh where k is a constant >> that depends on the system of units. >> For example >> If U is in foot-pounds(force), m is in pounds(mass), g is in feet/sec^2 and h >> is in feet, k will be equal to approximately 1/32.17 >> The precise value of k can be determined based upon if the precise de'nition >> of the pound force and the pound mass. Since one pound force is generally >> de'ned as the force required to support one pound mass under one standard >> gravity, this value will be 1/standard-g. >> If you are willing to overlook the difference between the standard g (the >> de'ning constant in the relationship between the standard pound force and the >> standard pound mass) and local g, the formula obviously simpli'es to >> U = kmgh = 1/g * m * g * h = mh >> John Briggs If I were given physical quantities in incoherent sets of units, and if I >calculated the answer by converting the physical quantities involved into one >of the coherent sets of units, would I then have to worry about the ïkÍ factor >in the original equation? ThatÍs what the k is, the factor you need to make that conversion. It doesnÍt matter if the 'nal result you end up with is an energy unit in _some_ coherent system, if the units you used for the other factors are not the ones which are coherent with it. You can always use U = mgh if you are willing to let the units of U be determined by the units of the other quantities. ItÍs only when you want to specify the units for all the quantities that you might need a k that is not equal to 1. A coherent system of units is suf'cient for k to be 1, but you donÍt actually need to be using a complete system of mechanical units. For example, if you measure m in pounds and g in gees and h in feet, you will get U in foot-pounds force, even though if you wanted to have a coherent system with those units, youÍd have a strange and unfamiliar unit of time. Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ === Subject: Re: Is it mass; or is it weight >> So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net >> which stretches an additional 4.4-ft before thrusting the man back, would >> the potential energy of the net at the instant it plunges the man back be >> U = mgh; >> U = 220 * (36 + 4.4). You left off the factor of g . Which either means that you forgot about it > or that you are using a system of units within which the factor of g > out. For instance, if you measure m in pounds mass, h in feet and energy in mh. > the jump site > with standard g from the ratio of pounds force to pounds mass. Gene Nygaard would be correct to point out that there is no single standard > g that is of'cially sanctioned for this purpose. >> While initially attempting this question, I converted pounds-mass to >> kilograms-mass, and the height in feet to metres, and computed the potential >> energy through >> U = mgh; >> U = (97.99) * (9.8) * (12.73); >> which revealed the netÍs potential energy to be at the time it sent the man >> back upwards almost twelve kilo-joules. >> My Physics teacher, however, disagreed, and prosposed that the given units of >> the mass and height be maintained as is, and the potential energy be >> calculated without the factor ïgÍ present anywhere in the computation. This >> method outputs >> U = mh; >> U = (220) (36 + 4.4); >> U = 8888 lb-ft; Your physics teacher is apparently an idiot, taught by idiots out of textbooks > written by idiots. Sadly, thatÍs quite typical For a coherent system of units, U = mgh. For example, the following situations involve coherent sets of units: U in foot-pounds(force), m in slugs, g in feet/sec^2, h in feet U in joules, m > in kilograms, g in meters/sec^2, h in meters U in foot-poundals, m in > pounds(mass), g in feet/sec^2, h in feet For systems of units that are not coherent, U = kmgh where k is a constant > that depends on the system of units. For example If U is in foot-pounds(force), m is in pounds(mass), g is in feet/sec^2 and h > is in feet, k will be equal to approximately 1/32.17 The precise value of k can be determined based upon if the precise de'nition > of the pound force and the pound mass. Since one pound force is generally > de'ned as the force required to support one pound mass under one standard > gravity, this value will be 1/standard-g. If you are willing to overlook the difference between the standard g (the > de'ning constant in the relationship between the standard pound force and the > standard pound mass) and local g, the formula obviously simpli'es to U = kmgh = 1/g * m * g * h = mh John Briggs If I were given physical quantities in incoherent sets of units, and if I calculated the answer by converting the physical quantities involved into one of the coherent sets of units, would I then have to worry about the ïkÍ factor in the original equation? -- Ayaz Ahmed Khan Yours Forever in, Cyberspace. === Subject: Topology of the Universe No walls from a topologist or differential geometerÍs point of view. No walls any more than a torus need have walls to be the shape it is. Yes, I understand in 2D it is like a bug crawling on the surface of a sphere with handles if a multiply-connected global topology like a torus. I understand that formally you can simple construct these spaces. What I am asking is how can the Universe physically have such a multiply-connected global topology? How is that topology implemented? For that we need the exotic vacuum zero point 'eld / (x,L)zpf in order to explain how such a topology can come into being and becoming. The handle in the 2D toy model becomes a 3D Star Gate that needs / (x,L)zpf =/= 0. This same 'eld in the limit that scale L -> c/H(t) explains Omega(Dark Energy) ~ 0.66 - 0.73 so with all that anti-gravitating exotic vacuum dark energy out there perhaps it is not so surprising that the Universe has this multiply-connected proposed? 120, or 12x120, or? I understand for example, that a simple orientable 2D torus or doughnut can formally be pictured as a rectangle with opposite edges identi'ed like in periodic boundary conditions for a quantum wave of a single here without encountering an actual edge ( wallÍ) in the case of the 2D doughnut. How do we implement the elegant formalism with the informal physical interpretation? How do we connect the conceptual dots with the physical lines? How do we §esh out the paper mache of formal math with the iron posts of observation (J.A. Wheeler)? On the other hand, it is not clear what they are really talking about from Fig 3 on p. 594 made an Ansatz that the physical 3D space is actually SU(2)/ID quotient space and that Fig 3 is misleading, especially 2D 'g 3a thinking of a bug crawling on the surface of the sphere that is partitioned in 12 spherical pentagonal sectors. It is clear there that the bug, or Flatlander in AbbottÍs now curved 2D space encounters a physical edge that is a Star Gate instantly teleporting the bug to a mirror point on a Luminet et-al illustrate this point well with the image of an insect crawling around the surface of a cylinder... It this is so, we may be able to identify multiple images of the same structure but in different regions in the sky. Now it seems to me that Ellis and Luminet et-al have not fully thought through that analogy. OK suppose there are no actual discrete physical walls that a Star Ship will encounter at the Edge of this neo-medieval cosmos of Platonic solids that Ptolemy and The Church Fathers of GalileoÍs time would 'nd familiar and congenial. Suppose instead that there is this continuous multiply connected 3D spatial topology of a hyper N torus. If the bug is simple crawling around the cylinder so to speak it will take time! But here the multiple images showing the temperature §uctuation correlations over ~ 60 degrees in the sky i.e., strange loss of power Bang when matter and radiation decouple. ThatÍs the problem here. That means a Star Gate effect of instant teleportation of the light rays or anything else in order to get SIMULTANEOUS multiple images of the same structure and that does require dark energy! In other words it is not enough for the theoretical physicist to say My, this formal topology works. . We must explain how it works and this requires more physics to show how the formal topology is implemented. Then the issue of whether the edges or walls are physically there as Star Gate portals is purely empirical and cannot be decided apriori with purely formal global topology constructions. Those constructions are needed of course but must be explained. In particular I do not as yet see any physical necessity for Saul-PaulÍs Ansatz that physical 3D expanding accelerating cosmic space is actually SU(2)/ID rather than the SU(2) shown in Fig 3 with the 1D edges in 2D Fig 3a and 2D walls in 3D in Fig 3b as physically detectable geometrodynamic structures i.e. Star Gates that physically implement the formal global multiply connected topology in which the 12 2D pentagon walls of the spherical dodecahedron have 3D handles connecting them into what I imagine are 6 pairs of opposite walls that are The Doors in and out either way so to speak. There are then, it appears, 120 of these 3D chunks that snugly tile a spherical 3D hypersurface with no boundaries. But if no tv game with walls and magik reentry then how §at. Two possible answers from mathematical structure point of view: 1. The manifold is one whole shape in a higher dimesional space from which it inherits its curvature. 2. The manifold is an abstract space in void with appropriate topology and metric. No walls needed anymore than grid pattern of coordinates on Euclidean space has any reality. Topology added to dust of in'nitely many discrete points makes the shape of the space. OR TAKE A HUGE AND FINITE COLLECTION OF TETRADEDRA AND GLUE THEM TOGETHER IN THE DESIRED PATTERN. NOW LOTS OF WALLS. EVERY TETRAHEDRAL FACE A STARGATE INTO THE NEXT ROOM. THIS IS ORDINARY REALITY. TO GET FROM MY LIVING ROOM TO THE KITCHEN I PASS THRU THE STAR GATE CALLED A doorway . But now keep the pattern and remove the grid and there is the space, 'nite bounded turning in on itself in the pattern of a three manifold. No walls, all stargate. Best, Lou K. Yes, all star gate with no detectable edges like, presumably Saul-PaulÍs Ansatz SU(2)/ID ? is 'ne with me. I see we agree on the necessity for something really gigantically weird - the Star Gate with its instant teleportation to account for the alleged fact of multiple images of the same object but in different directions in the sky . Ellis should have added at the same cosmic time t in sense of FRW metric. But he didnÍt. Had he, he would have seen that his analogy, or LuminetÍs et-alÍs analogy of the insect crawling around the surface of a cylinder is not a good one throwing the baby out with the bathwater. The insect does not crawl slowly taking time, rather, the evidence shows that it is for all practical purposed INSTANTLY TELEPORTED over 60 degrees of the celestial sphere of the past light cone reaching back almost 13.7 billion years to the Big Bang sans a mere 380,000 years or so from the initial singularity i.e. the micro-quantum to macro-quantum vacuum phase transition in the BCS pairing of virtual electrons and positrons at the edge of the negative -mc^2 branch or energy band Fermi surface of the Dirac electron vacuum in UNSTABLE globally §at Minkowski spacetime. Both EinsteinÍs gravity guv(curved spacetime) and the uni'ed dark energy/matter / zpf local 'eld emerge triumphant, as it were, like BotticelliÍs Venus Arising, from the modulations of the Goldstone phase and Higgs amplitude respectively of the LOCAL though long-range coherent vacuum order parameter. EinsteinÍs gravity is a Goldstone phase collective mode from SakharovÍs metric elasticity from cohering of the vacuum zero point §uctuations of ALL 'elds. SakahrovÍs metric elasticity of Hagen KleinertÍs world crystal lattice at renormalized Planck scale Lp* = Lp^2/3(c/Ho)^1/3 = 1 fermi at present epoch, is a special case of PW AndersonÍs generalized phase rigidity of the More is different principle of spontaneously self-organizing emergent complex order in the condensed matter approach to quantum gravity. Cc: sarfatti@pacbell.net === Subject: Re: FW: Vigier V: EscherÍs Star Gate shape of the universe I agree! Of course another wonderful SciFi way to view it is that the universe for the Dodec space IS the Three Sphere but there is a 120-fold periodicity of all events in the pattern of the dodec tiling of the three-sphere. Who would believe this? 120 copies of this email all being written together in an unimaginable simultaneity. I will vote for the is competing with Astounding SciFi these days) he suggests even crazier and in'nte parallelisms, so a mere 120 parallelism would be a mere nothing in his world of speculation. Best, Lou But it is speculation backed up by empirical data. Jack, Post Script: Looking again at the picture of the dodecahedral space in *Nature* (9 October, page 594), I think I see what your conceptual problem is--i.e., why you think of walls. First they show the 12 spherical pentagons covering the 2-sphere (soccer ball). Then they show some of the 120 spherical dodecahedra tesellating the 3-sphere. What they DONÍT explain is that the cosmological space is NOT supposed to be the 3-sphere but the orbit space created by the action of the Icosahedral Double group on the three sphere. This group is usually called the binary Icosahedral group, but I prefer the nomenclature of the crystallographers (since binary has the more common connotation of binary numbers). Since SU(2) is geometrically a three sphere, this orbit space can be written as SU(2)/ID. Now ID is a 120 element group, and the orbits are all cosets of ID in SU(2). In other words ID acts on any point in one of the spherical dodecahedra and jumps that point in a discrete orbit through the analogous points--one point in each of the 120 dodecahedra. The set of orbits (each of 120 points) constitute a new 3-d space of very different topology -- i.e., the topology equivalent that created by identifying the opposite faces (with a 1/5 twist) of an ordinary (non-curved) dodecahedron. Incidentally, since ID is a 'nite subgroup of SU(2), the 120 elements of ID are located at the mid-points of the 120 spherical dodecahedra. These constitute an orbit of ID (called the identity coset) within the SU(2) Lie group. Saul-Paul ---------- === Subject: Re: Vigier V: EscherÍs Star Gate shape of the universe Jack, Your question about the geometry of the dodecahedral space is answered on Also you should re-read the description of dodecahedral space from Lou Kauffman I forwarded to you this morning. Especially relevant to your question is his statement: [And keep your Pong game picture in mind]: That is the sort of description that the dodecahedral ïidentify by 1/5 twist the opposite facesÍ description uses. In that description a ray of light heading for one face §ies right through and comes in again via the opposite face. But the ïfacesÍ are not there. That is just a way to describe the periodicity, and it is traces of this periodicity that are supposed to be inherent in the astronomical data. Here is a 2-d analog that makes this clear. [This can be done as a gedanken experiment.] Take a rectangular sheet of paper. Glue together two opposite sides (1-d faces ). Now you have a cylinder. Now connect up the two ends of the cylinder. You now have a 2-d torus. Clearly the edges that you glue together are in arbitrary position on the torus. We could cut the torus at some arbitrary place and then cut the resulting cylinder lenghthwise in another arbitrary place. So from the point of view of a creature trapped on the 2-d torus there are no edges to go through -- although these 2-d creatures could still describe their toroidal space as a rectangle with opposite edges identi'ed (provided that their mathematical imagination was suf'ciently advanced). Saul-Paul ---------- === Subject: Re: Vigier V: EscherÍs Star Gate shape of the universe Part III The Pong Video Game in The Sky with 2D Wall periodic boundaries may not be what Ellis et-al are suggesting. Ellis cites Omega zero = 1.02 +-0.02. meetings the Pundits like Mike Turner all said k = 0, in§ation is right, i.e spatially §at with Omega zero = 1 on the button. Ellis is saying no to that that there will be a Big Crunch with k = +1 and Omega zero > 1. That allows a 'nite 3D space without 2D wall boundaries. spatial sections of the Universe are dodecahedral sections of space of positive curvature, 'tted together to make 'nite three-dimensional spaces. Earlier Ellis does make the Pong Video Game analogy implicitly for the §at toroidal space, as you exit right you enter left and space is 'nite Does Ellis mean that there are sectors of 3D space with 2D walls between them which 't together like some 'nite Platonic solid? What happens when you walk through a wall so to speak? Is there a periodic boundary condition (i.e. a Star Gate in effect) or not? That is the total 'nite space of positive curvature has no boundary, but is partitioned into disjoint subspaces or sectors separated by 2D boundary walls? Is that the idea? Are these walls weird or not? Can we pass through them? What happens when we do? Or are they not there at all? Saul-Paul: There are no walls. See my description above. Whatever the picture is here Ellis is clear that this alternative is not compatible with chaotic in§ation that demands Omega zero = 1 with k = 0. Luminet has Omega zero = 1.013 Max TegmarkÍs picture of Level I Hubble universes in May Scienti'c American is not consistent with what Ellis et-al is suggesting here. That is Ellis rejects that spatial homogeneity extends outside our visual horizon forever .... and we are in one expanding bubble in the middle of innumerable other similar ones. But if Luminet et-al are correct, chaotic in§ation is ruled out: there is only one expanding bubble, and we can see almost all the way round it .... The WMAP data as interpreted by Luminet et-al... suggest that we might live in such a small closed universe. This is qualitatively globally different from what Mike Turner, Max Tegmark et-al are suggesting. So we have a schism between the k = 0 Camp and the k = +1 Camp. We need better precision than the current 2% to decide. Part III The Pong Video Game in The Sky with 2D Wall periodic boundaries may not be what Ellis et-al are suggesting. Ellis cites Omega zero = 1.02 +-0.02. meetings the Pundits like Mike Turner all said k = 0, in§ation is right, i.e spatially §at with Omega zero = 1 on the button. Ellis is saying no to that that there will be a Big Crunch with k = +1 and Omega zero > 1. That allows a 'nite 3D space without 2D wall boundaries. spatial sections of the Universe are dodecahedral sections of space of positive curvature, 'tted together to make 'nite three-dimensional spaces. Earlier Ellis does make the Pong Video Game analogy implicitly for the §at toroidal space, as you exit right you enter left and space is 'nite Does Ellis mean that there are sectors of 3D space with 2D walls between them which 't together like some 'nite Platonic solid? What happens when you walk through a wall so to speak? Is there a periodic boundary condition (i.e. a Star Gate in effect) or not? That is the total 'nite space of positive curvature has no boundary, but is partitioned into disjoint subspaces or sectors separated by 2D boundary walls? Is that the idea? Are these walls weird or not? Can we pass through them? What happens when we do? Or are they not there at all? Whatever the picture is here Ellis is clear that this alternative is not compatible with chaotic in§ation that demands Omega zero = 1 with k = 0. Luminet has Omega zero = 1.013 Max TegmarkÍs picture of Level I Hubble universes in May Scienti'c American is not consistent with what Ellis et-al is suggesting here. That is Ellis rejects that spatial homogeneity extends outside our visual horizon forever .... and we are in one expanding bubble in the middle of innumerable other similar ones. But if Luminet et-al are correct, chaotic in§ation is ruled out: there is only one expanding bubble, and we can see almost all the way round it .... The WMAP data as interpreted by Luminet et-al... suggest that we might live in such a small closed universe. This is qualitatively globally different from what Mike Turner, Max Tegmark et-al are suggesting. So we have a schism between the k = 0 Camp and the k = +1 Camp. We need better precision than the current 2% to decide. === Subject: Re: Series Derivations > Hi All, How would I derive a simple formula f(n) for the series: f(n) = (n-0)(n-1) + (n-1)(n-2) + (n-2)(n-3) + ... ? Does is have > anything to do with partial sums? You meant f(n) = n(n-1) + (n-1)(n-2) + .... + 3*2 + 2*1 ? If is it, f(n) = Sum(k(k+1), k, 1, n-1) = Sum(k^2 + k, k, 1, n-1) = Sum(k^2, k, 1, n-1) + Sum(k, k, 1, n-1) = (n-1)n(2n-1)/6 + n(n-1)/2 = n(n-1)(2n - 1 + 3)/6 = n(n-1)(n+1)/3 = (n^3 - n)/3 -- Ignacio Larrosa Ca .96estro A Coru .96a (Espa .96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Series Derivations > How would I derive a simple formula f(n) for the series: f(n) = (n-0)(n-1) + (n-1)(n-2) + (n-2)(n-3) + ... ? Does is have > anything to do with partial sums? It doesnÍt look like f(n) exists for any n. For example, f(0) = (0)(-1)+(-1)(-2)+(-2)(-3)+ ... = 2+6+12+20+... which grows larger than any 'xed value. Perhaps you meant to ask what a formula is for summing the partial sums of f(n) (this is different than asking for f(n)). But note that summing f(n) is impossible, and the partial sums will always just grow to +in'nity. J P.S. to 'nd a simple formula for the partial sums, you can try a few example values and spot a simple pattern. For example, for f(0) above, the series is (2)+(6)+(12)+(20) or (2)+(2+4)+(2+4+6)+(2+4+6+8). This might suggest a line of attack for you. === Subject: Re: Use of variable independence, core error linux) [...] > But remember, you canÍt eat a math proof. [...] As the poster who quotes you more often than anyone else on the planet, yourself included, I hope you will take the following criticism under consideration. The whole eating math proofs thing just didnÍt work. It didnÍt work more spontaneous. I think that we have suf'cient evidence to conclude that it wonÍt work on the 538th time. Give it a rest, willya? YouÍre putting me off my favorite quotemeister. -- If you see math knowledge as a tool--as a hammer--with which you can attack other people then ... you defeat rational discourse. I get to call my proof the Hammer. ItÍs more powerful than *any* physical object. It is overwhelming force. -- Two JSH quotes === Subject: Re: Use of variable independence, core error Nntp-Posting-Host: apps.cwi.nl ... >Yes, and so what? Algebraic integers are also roots of both monic and >non-monic polynomials. The de'nition of algebraic integers only states >that an algebraic number is an algebraic integer only if it is a root of >at least one monic polynomial. However, as repeatedly shown by various posters on this board, > algebraic integers cannot be roots of non-monic polynomials with > integer coef'cients that are irreducible over Q. (That is now what was shown, you should also state ïprimitiveÍ.) Normal integers cannot be roots of non-monic polynomials with integer coef'cients that are primitive and irreducible over Q. >What asymmetry? Note that the distinction between integers and true >rationals is that rationals are *not* a root of a monic polynomial with >integer coef'cients. The same distinction we have between algebraic >integers and other algebraic numbers. However, as repeatedly shown by various posters on this board, > algebraic integers cannot be roots of non-monic polynomials with > integer coef'cients that are irreducible over Q. See above. >This is nonsense. The de'nition of algebraic integers only label a >set of numbers with the name algebraic integers. How can such naming >lead to a contradiction? The name could also have been numbers that >are roots of monic polynomials . Would that name lead to a >contradiction? ItÍs not the name; itÍs the exclusion. You have the exclusion wrong. >The only possibility is that in your contradiction you use a result >on algebraic integers that is wrong. Which is either a theorem you >use or your own logic. So you should identify the theorem you are >using that is wrong. The de'nition in and of itself is *not* wrong, >it just gives a labeling. But my proof is very short, which means it is machine checkable. I have not yet seen a version that is machine chackable. > If the factors were polynomials, thereÍd be no argument, as who out > there believes that you can have a polynomial P(x), a polynomial > factor f(x) of P(x), where P(x) has 7 as a factor, for all algebraic > integers x, and f(0)=7, that 7 can be divided off such that P(0) is > coprime to 7, without f(x) having 7 as a factor? It looks complicated, but *writing* it out with actual polynomials > makes it easy. However, with non-polynomial factors you have to trust the > mathematical logic. That is your problem. You attempt to apply what holds for polynomial factors also does apply for non-polynomial factors. That is false. P(x) = (x + 1)(x + 2) is (whenever x is an integer) *always* divisible by 2. But 2 is not a polynomial factor of P(x), so it is never the same factor of P(x) that is divisible by 2, it is either (x + 1) or (x + 2), depending on the value of x. >One such theorem you use is that the algebraic integers form a ring. >Yes, that *is* a theorem that must be proven. To prove it is a ring >is to show the following (a, b and c algebraic integers): > R1: given a and b, a + b is also an algebraic integer (closed under > addition). > R2: given a and b, a * b is also an algebraic integer (closed under > multiplication) >the remaining requirements are trivial (a + b = b + a, etc) and follow >because algebraic integers are complex numbers. So to show that the >algebraic integers do not form a ring you have to show that either >R1 or R2 is false. The simplest proof of R1 and R2 is one where >given monic polynomials Pa(x) and Pb(x) of which a resp. b are roots, >two new monic polynomials Pc(x) and Pd(x) are constructed of which >a+b resp. a*b are roots. So either the construction is wrong, or >indeed the algebraic integers do form a ring. So what do you think? Is the construction wrong or do the algebraic >integers indeed form a ring? IÍve never said that algebraic integers donÍt form a ring. What they form is a §awed ring, which doesnÍt include all the numbers > it must to prevent the possibility of appearing to prove two different > but opposite things. In what way can a ring be §awed ? A ring is something that satis'es some particular properties. If the something satis'es them, it must be a ring. Period. When you attempt to prove contradictionary things about this ring the possibilities are that either you are using a theorem about rings that is wrong, or that your own theorems are wrong. Or the ring properties that de'ne a ring are contradictionary. Which is it? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: a division Sorry for an easy question. A general formula for x^n - y^n --------- = ? x -y === Subject: Re: a division from autumn_1999_1@yahoo.com: >Sorry for an easy question. A general formula for x^n - y^n >--------- = ? >x -y > Here is an approach from the street: start with n=1 and perform the division; do this for each integer up to n=4 or n=5. Look for a pattern. The trouble could be, that some of the result might yield a remainder. This would be a good exercise. Notice that when n is even, you will at least come up with some kind of result with, initially, no remainder......... not sure until it is really attempted. G C === Subject: Re: a division > Sorry for an easy question. A general formula for x^n - y^n > --------- = ? > x -y Sum(x^(n-k)*y^(k-1),k,1,n) === Subject: Re: a division Visiting Assistant Professor at the University of Montana. >Sorry for an easy question. A general formula for x^n - y^n >--------- = ? >x -y x^n - y^n = (x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+...+xy^{n-2}+y^{n-1}). 'rst and last term!) Arturo Magidin magidin@math.berkeley.edu === Subject: ppcm Je voudrais avoir un .8equivalent (bon, m .90me sans la constante explicite) de PPCM(1,2,...,n). thankyouverymuch Gabriel === Subject: Re: ppcm > Je voudrais avoir un .8equivalent > (bon, m .90me sans la constante explicite) > de PPCM(1,2,...,n). thankyouverymuch Gabriel What is PPCM(1,2,...,n)? -- Ignacio Larrosa Ca .96estro A Coru .96a (Espa .96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: ppcm >Je voudrais avoir un .8equivalent >(bon, m .90me sans la constante explicite) >de PPCM(1,2,...,n).... What is PPCM(1,2,...,n)? The least common multiple of 1,2,...,n. Ken Pledger. === Subject: Re: ppcm >> Je voudrais avoir un .8equivalent >> (bon, m .90me sans la constante explicite) >> de PPCM(1,2,...,n).... What is PPCM(1,2,...,n)? > The least common multiple of 1,2,...,n. Ken Pledger. Then PPCM(1, 2, .., n) = Product(p^§oor(log(n)/log(p))| p prime, p < = n) -- === Subject: Re: DonÍt go to 100 >database-intensive example in VB I had something like this: If MyRecordset!MyField = 1 And MyRecordset!MyField = 2 Then > Print How could we ever print this one? >End If Yet the message did get printed! Explanation below......... >[explanation snipped to bring some sense of mystery into the group] Mr. Anonymous, a quo haec excavavistis? This goes back two and a half years. And I see another one going back to 1998. === Subject: Re: DonÍt go to 100 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9BC2RA22354; >database-intensive example in VB I had something like this: If MyRecordset!MyField = 1 And MyRecordset!MyField = 2 Then > Print How could we ever print this one? >End If Yet the message did get printed! Explanation below......... Explanation: The 'eld MyField **did not exist** in the database. The If test was >throwing a run-time error, but then the On Error Resume Next was invoked >so that control ran to the next line - the object of the apparently >contradictory If statement. Could something similar be happening in your example? If your te or tp >was actually NULL, or something else which could not legitimately be tested, >the On Error Resume Next is taking you to the next line regardless. Try >debugging the routine withare measurable, so is fg (product). OR: g is itself a characteristic function. Show that g^(-1) (1) is a mesuarable set. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: how to prove that this is a measurable function >How can I prove that the function g(x,y) = X_A(x+y) X_b(y) is lebesgue >measurable in R^2 where A, B subset R are measurable sets and X_A, X_B >are the charactersitic functions? Any help appreciated, nojb. > Show that > 1) (x,y) |-> x+y is measurable; > 2) if f: X -> Y, g: Y -> Z are measurable, then so is g op f > (composition); and > 3) if f, g: X -> R are measurable, so is fg (product). 2) is false. The composition of Lebesgue measurable (LM) functions need not be LM . However, compositions work for Borel measurable functions, so one could work with that. Another approach: X_B(y) is clearly LM on R^2 and X_A(x+y) is the characteristic function of T(R x A), where T is a linear transformation on R^2. === Subject: Re: how to prove that this is a measurable function How can I prove that the function g(x,y) = X_A(x+y) X_b(y) is lebesgue >measurable in R^2 where A, B subset R are measurable sets and X_A, X_B >are the charactersitic functions? Any help appreciated, nojb. >>Show that >>1) (x,y) |-> x+y is measurable; >>2) if f: X -> Y, g: Y -> Z are measurable, then so is g op f >>(composition); and >>3) if f, g: X -> R are measurable, so is fg (product). 2) is false. The composition of Lebesgue measurable (LM) functions n ************* > For j = 1 To ne > For i = 1 To te > Randomize > 112: If i = te + 1 Then > GoTo 111 > Else > veille = 0 > End If > x(i) = valeur(Int(tp * Rnd + 1)) > v = x(i) > For k = 1 To i - 1 > If v = x(k) Then > veille = veille + 1 > End If > Next k > If veille <> 0 Then > GoTo 112 > Else > Cells(lig + i - 1, col + j - 1).Value = x(i) > End If > Next i > Next j > 111: > End > End If > 100: > End Sub Private Sub CommandButton1_Click() > UserForm1.Hide > End Sub > proven to not be reachable, isnÍt that a contradiction for any mathematical goals? >Charlie Volkstorf >Cambridge, MA > Helmut Richter === Subject: preregular/p I chaTb upon the following terms: preopen , preregular sets but b O O O O OTb O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O Ep Doctor f Pro'ler Caps Stationery Xications Servers To... Stationery < KendLP everyone. I have some question about De Moivre formula. I have been simply accepted that formula as : (cos t + i*sin t)^n = cos(tn)+i*sin(tn) and I regard it as true for every t and n. but, in case t=Pi/6 and n=2, De Moivre formula doesnÍt hold. So, IÍm so confusing with the condition for holding De Moivre formula. If anyone have obvious answer about my question, please post reply. i need help.. === Subject: Re: Curve Inside Another Of Equal Perimeter Quet) >Are there any two CONVEX closed curves of equal perimeter, where one >can be placed completely inside the other? (It is allowed that the curves touch at a 'nite number of points at >most.) Leroy Quet No. The outer curve must be longer than the inner curve. This is easy to see using CroftonÍs formula (a.k.a. the Cauchy-Crofton formula). John Mitchell === Subject: Re: Curve Inside Another Of Equal Perimeter > Are there any two CONVEX closed curves of equal perimeter, where one > can be placed completely inside the other? (It is allowed that the curves touch at a 'nite number of points at > most.) > Leroy Quet I would guess not. Here is an outline of a possible proof. Choose a point inside both curves. Consider the formula for the arc length as a function of r(theta) and theta, where r(theta) is the distance from the central point to the curve at angle theta. Since r_1(theta) <= r_2(theta), where r_1 and r_2 are the respective distances to the inner and outer curves, the length of the outer curve is greater than the length of the inner curve. Martin Cohen === Subject: Re: De Moivre formula > I have some question about De Moivre formula. > I have...accepted: > (cos t + i*sin t)^n = cos(tn)+i*sin(tn) > and I regard it as true for every t and n. > but, in case t=Pi/6 and n=2, De Moivre > formula doesnÍt hold. It does hold. Re-check your algebra. === Subject: Re: De Moivre formula >everyone. I have some question about De Moivre formula. I have been simply accepted that formula as : (cos t + i*sin t)^n = > cos(tn)+i*sin(tn) > and I regard it as true for every t and n. but, in case t=Pi/6 and n=2, De Moivre formula doesnÍt hold. So, IÍm so confusing with the condition for holding De Moivre formula. If anyone have obvious answer about my question, please post reply. i need help.. For n = 2 and t = pi/6, (cos(t) + i*sin(t)^n = (sqrt(3)/2 + i/2)^2 = 1/2 + i*sqrt(3)/2 and cos(2*t) + i*sin(2*t) = 1/2 + i*sqrt(3)/2. So what part of the formula does not hold? === Subject: Re: Construct a bounded set of real numbers . Need help. > IÍm working out of Walter RudinÍs Principles of mathematical analysis . > Problem #5 from Chapter 2 is: Construct a bounded set of real numbers with exactly 3 limit points. Would you please help me have steps (an algorithm) to solve this problem? > Construct a bounded set, S, of reals with exactly one limit point and form its union with two translations of it. === Subject: Re: De Moivre formula >everyone. I have some question about De Moivre formula. I have been simply accepted that formula as : (cos t + i*sin t)^n = > cos(tn)+i*sin(tn) > and I regard it as true for every t and n. Excellent news (as long as youÍre assuming that n is an integer). > but, in case t=Pi/6 and n=2, De Moivre formula doesnÍt hold. But you just said you regarded it as true for all t and n. So you regard it as true for t = pi/6 and n = 2 but you also regard it as false for t = pi/6 and n = 2. ? :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: ARCSIN function, single precision §oating point. -- Example routine needed? > IÍm trying to write ATAN2 function for a small basic language that has > IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are > availible in the language. > IÍve tried a few methods IÍve found but the results are way off due to > low precision, rounding, etc. > Are there any repositorys of old fortran routines or algorithms that I > could use to get a good accuracy single precision routine. Speed or > space arenÍt as important as reasonably good accuracy. == Let us introduce following constants T(j)= tan(j*pi/24) , j= 1,2,..., 11 and PI2=pi/2 . More precisely PI2=pi/2=1.570796 326794 896619 231... T(1) =0.131652 497587 395853 472... T(2) =0.267949 192431 122706 473... T(3) =0.414213 562373 095048 802... T(4) =0.577350 269189 625764 509... T(5) =0.767326 987978 960342 923... T(6) =1.000000 000000 000000 000... T(7) =1.303225 372841 205755 868... T(8) =1.732050 807568 877293 527... T(9) =2.414213 562373 095048 802... T(10) =3.732050 807568 877293 527... T(11) =7.595754 112725 150440 526... . Further denote (*) F(z) = z*P(z)/Q(z) with P(z)= 1155 + 1190*z^2 +231*z^4 , Q(z)= 1155 + 1575*z^2 + 525*z^4 + 25*z^6 , and I=[0,T(1)] , I(j)=[T(j), T(j+1)] , j=1,2,...,11 . Note that (1) arctan(z)=approx.= F(z) when -T(1) =< z =< T(1) . Observe that both functions arctan(x) and F(x) are odd functions. In the following we shall consider x in [0,infty) . When x in I(j) , then (x-T(j))/(1+x*T(j)) is in I(1) . Using equality arctan(x)= arctan(z(j)) + j*PI/24 where z(j)= (x-T(j))/(1+x*T(j)) , according to (1) we make approximations (1.1) arctan(x) =approx.= F(z(j)) + j*Pi/24 when x in I(j) , and (1.2) arctan(x) =approx.= Pi/2 -F(1/x) if x in (T(11),infty) . Therefore, a possible routine may be written in FORTRAN in the following way : FUNCTION ATAN(X) DIMENSION T(1) PI2 =1.570796 326794 896619 231 T(1)=0.131652 497587 395853 472 T(2)=0.267949 192431 122706 473 T(3)=0.414213 562373 095048 802 T(4)=0.577350 269189 625764 509 T(5)=0.767326 987978 960342 923 T(6)=1.000000 000000 000000 000 T(7)=1.303225 372841 205755 868 T(8)=1.732050 807568 877293 527 T(9)=2.414213 562373 095048 802 T(10)=3.732050 807568 877293 527 T(11)=7.595754 112725 150440 526 C=1. IF(X) 1,2,3 2 ATAN=0. RETURN 1 X=-X C=-1. 3 IF(X-T(1)) 21,22,23 21 Z=X*X P= 1155. + 1190.*Z +231.*Z*Z Q= 1155. + 1575.*Z + 525.*Z*Z + 25.*Z*Z*Z ATAN= C*X*P/Q RETURN 22 ATAN=C*PI2/12 RETURN 23 DO 100 j=1,10 H=j*PI2/12. ZJ=(Y-T(j))*(Y-T(j+1)) IF(ZJ) 10,20,100 20 ATAN =C*H RETURN 10 Z=ZJ*ZJ P= 1155. + 1190.*Z +231.*Z*Z Q= 1155. + 1575.*Z + 525.*Z*Z + 25.*Z*Z*Z F=ZJ*P/Q ATAN= C*(F+H) RETURN 100 CONTINUE W=1./X V=W*W P= 1155. + 1190.*V +231.*V*V Q= 1155. + 1575.*V + 525.*V*V + 25.*V*V*V F=W*P/Q ATAN= C*(PI2-F) RETURN END = Note : instead of approximation (1) you may consider (1Í) arctan(z) =approx.= = F(z):=a(1)*z+a(3)*z^3 +a(5)*z^5 +a(7)*z^7 + a(9)*z^9 for -T(1) =< z =< T(1), where a(1)= 0.999999 99843 a(3)= -0.333332 89364 a(5)= 0.199965 34780 a(7)= -0.141734 60613 a(9)= 0.094919 54952 , and then to extend the approximation on whole real axis (see (1.1)-(1.2)) . Using (1Í) one has |arctan(z)-F(z)| =< 2^{-42} for z in [-T(1),T(1)]. ItÍs better to approximate by means of rational function from (*). === Subject: Re: De Moivre formula Ooops!!! I misunderstood sin <=> cos for solving following exercise. : What is sum of natural #s n s.t. (sin t + i*cos t)^n=sin(nt)+i*cos(nt) ? Now, I see my misunderstanding point. > I have some question about De Moivre formula. I have been simply accepted that formula as : (cos t + i*sin t)^n = >cos(tn)+i*sin(tn) >and I regard it as true for every t and n. Excellent news (as long as youÍre assuming that n is an integer). but, in case t=Pi/6 and n=2, De Moivre formula doesnÍt hold. But you just said you regarded it as true for all t and n. So you regard it as true for t = pi/6 and n = 2 but you also > regard it as false for t = pi/6 and n = 2. ? :-( -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Needless to say, I had the last laugh. > Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Partition of R I was thinking about the interesting (and apparently dif'cult) >problem of 'nding sets A and B that form a partition of R and are >such that every interval of R contains uncountably many points of A >and B. In other words, every element of R should be a condensation >point of A and B. You can do far better than that. > R is a maximally resolvable space, > i.e., R can be partitioned into c=2^w many disjoint dense subsets. Here is one way to do it. > P = {Q + x | x in R} partitions R into 2^w counatble sets. > Ok, each Q+x is countable and there are c of them. > Partition P into PP consisting of 2^w subsets each of size 2^w. > Not just countable but of size c. How do I do that? > Then { US | S in PP } satis'es the bill. > US is the union of S. S in PP, means that S is collection of Q+xÍs. > The jist is partitioning 2^w into 2^w sets of size 2^w. Getting dense is > easy. > === Subject: Re: Two similar things >Since they now are identical-with-somethings, if they shared >all the same properties they would be one cake rather than >two. THerefore, since they are two cakes, they donÍt share >all their properties, and I can distinguish them by means >of any property that one has and the other doesnÍt. No, thereÍs a foot wrong there. Why would they be one cake if they shared > all the same properties? Consider Hesperus and Phosphorus. The properties of Hesperus include identity-with-Hesperus, and those of Phosphorus include identity-with- Phosphorus. So, if Hesperus and Phosphorus share *all* their properties, the properties of Hesperus will include identity-with-Phosphorus, and those of Phosphorus will include identity-with-Hesperus. But if Hesperus and Phosphorus have these properties, then Hesperus and Phosphorus are one and the same: that is, numerically identical. > Why not 3 cakes? If these cakes have all the same properties, for the reasons I have indicated these three cakes will be one and the same. > Would you tell the difference by counting? Just by counting, no. By counting correctly, yes. In other words, before it was discovered that Hesperus was Phosphorus, it was assumed that these heavenly bodies were different heavenly bodies rather than one and the same heavenly body. But this assumption proved to be wrong. --John === Subject: Re: quantum echo > Immortalist: Stop crossposting this to sci.physics.relativity > Are you saying it is possible for something to exist at location A [*SNIP*] Paul Teller having proposed, as an explanation for the divergence between quantum statistics and classical statistics, that ïquantaÍ lack haecceities, I posted to sci.physics.relativity what I took to be the import of TellerÍs position for the logic of identity. On this topic a fair amount has been written by philosophers of science who, unlike myself, are knowledgeable about the formal details of quantum mechanics. What these people have had to say on the topic may be of interest to some on sci.physics.relativity... --John === Subject: Re: HowÍd they do it? >Cut< >> ThatÍs not even taking into consideration their value as the worldÍs >> standards. >> Of course, guarded. >> Gene Nygaard >> http://ourworld.compuserve.com/homepages/Gene_Nygaard/ Even though physics would be better off with_out_ ïem; just using a >mathematical ratio: Like 1 kg = 2.2# sec/32.2Í; using 2.2 pints of water at >its maximum density, per kilogram. Gentlemen of the jury, Chicolini here may look like an idiot, > and sound like an idiot, but donÍt let that fool you: He > really is an idiot. > Groucho Marx ïThough you donÍt admit it, the slugÍs doinÍ okay without an artifact; using the mathematical equation: 1 slug = f/a = 1# sec/foot = 32Í sec/32Í. In physics, the slug is a _unit_ of mass derived from the fundamental concepts of the foot-pound-second British Gravitational System of weights and measures. In trade, the gram and kilogram are _absolute_ units of mass; with the kilogram artifact being the head honcho. === Subject: Re: HowÍd they do it? >Cut< >> ThatÍs not even taking into consideration their value as the worldÍs >> standards. >> Of course, guarded. >> Gene Nygaard >> http://ourworld.compuserve.com/homepages/Gene_Nygaard/ Even though physics would be better off with_out_ ïem; just using a >mathematical ratio: Like 1 kg = 2.2# sec/32.2Í; using 2.2 pints of water >at >its maximum density, per kilogram. >> Gentlemen of the jury, Chicolini here may look like an idiot, >> and sound like an idiot, but donÍt let that fool you: He >> really is an idiot. >> Groucho Marx ÍThough you donÍt admit it, the slugÍs doinÍ okay without an artifact; using >the mathematical equation: 1 slug = f/a = 1# sec/foot = 32Í sec/32Í. In physics, the slug is a _unit_ of mass derived from the fundamental >concepts of the foot-pound-second British Gravitational System of weights >and measures. Gentlemen of the jury, Chicolini here may look like an idiot, and sound like an idiot, but donÍt let that fool you: He really is an idiot. Groucho Marx Slugs are a little used 20th century invention, which donÍt even have an of'cial de'nition. There is no of'cial standard for either a slug or a pound force. There are absolutely no pints in the only subsystem which includes slugs. Dishonest Don S*head already knows this. Pounds do have an of'cial de'nition: 1 lb = 0.453 592 37 kg, EXACTLY Dishonest Don KNOWS this also, of course. Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ === Subject: Q wrt a number-theoretic generating function IÍm reading some introductory material about DirichletÍs generating functions of arithmetical functions, but could not 'nd anything about the obvious generating function Sum_p p^{-x}, where the sum is extended over all primes. Has it been studied? Is it an independent function or can it be expressed in terms RiemannÍs zeta? Anything else interesting about it? References? 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: Force substance, and bodies thereof; due to their impenetrability : The fact that the exact same place: Where they resist each otherÍs passage with thrusts that are proportional to the amount of material substance they contain. Weight in physics is that particular force exerted between a planetÍs terra further passage toward their common center of mass is restrained by a thrust contains. The magnitude of the thrusts are commonly measured with pushes or pulls of weight-scales. === Subject: Variation on an impossible problem Allowing only the use of the + operator, is it possible to use each of the digits 0 1 2 3 4 5 6 7 8 9 once only to make a total of 1000? One can combine digits to create sums like 210 + 345 + 6 + 78 + 9 = 648 etc. (This is obviously similar to use only + and 1 2 3 4 5 6 7 8 9 to make 100 which is impossible) Mitch. === Subject: Re: Variation on an impossible problem > Allowing only the use of the + operator, is it possible to use each of > the digits 0 1 2 3 4 5 6 7 8 9 once only to make a total of 1000? One can combine digits to create sums like 210 + 345 + 6 + 78 + 9 = > 648 etc. No. sum(i,i=0..9) = 0 mod 9, whereas 1000 = 1 mod 9. -- Julien Santini, France. === Subject: Discontinuous Derivative IÍm looking for a real function f:R -> R which is differentiable over all of R yet with a derivative discontinuous at 0. === Subject: Re: Discontinuous Derivative > IÍm looking for a real function f:R -> R > which is differentiable over all of R > yet with a derivative discontinuous at 0. > f(x) = x^2*sin(1/x), f(0)=0 f ï(x) = 2*x*sin(1/x)-cos(1/x) f ï(0) = 0, but lim x->0 f ï(x) doesnt exist. === Subject: Re: Discontinuous Derivative > IÍm looking for a real function f:R -> R > which is differentiable over all of R > yet with a derivative discontinuous at 0. > Standard example: x^2 * sin(1/x) , extended to 0 as 0. Variants: like x^2 * sin(1/x^2) ; this one has derivative unbounded near 0. === Subject: Re: Chessboard knight metric? >Much to my surprise, you all seem correct about the metric part >that the proof was in the de'nition itself. >I also appreciate the reader pointing out the regular shapes. >I had seen these shapes before- and would caution that the >shapes are perhaps as not as regular as they may seem. The octogon >that comes out at n=3 does not seem to be a true octogon (mabie IÍm >missing something again...) and for n = 4 the situation appears even >worse, with spaces way out on the perepherie being reached in (a >minimum of) 4 moves as well as spaces close to the starting square. >This was already pointed out by another reader- take x=(1,1) and >y=(1,2), then d(x,y)=4 and y surely cannot lie on the outer edge of >the octogon. Of course, it seems correct to point out that the >octogons keep turning up as part of the picture again and again for >higher n. >About the even / odd properties I mentioned before... at least that >does appear to be regular for all n, doesnÍt it? >C.Dement > Can you nevertheless learn to post properly? Your entire post appears to be a quote, even though it comes from you. ItÍs hard to follow your writing. -- Ioannis http://users.forthnet.gr/ath/jgal/ ___________________________________________ Eventually, _everything_ is understandable. === Subject: Re: Fundamental Reason for High Achievements of Jews === >Subject: Re: Fundamental Reason for High Achievements of Jews >> :> : Most historians believe that Jews avoid pork, >> :> : because the ancient Jews associated pigs with leprosy, >> :> : and pigs and people with leprosy were unclean . >> :> :> Name one historian who believes that, and give a citation to the >> :> place where he says it. >> : >> : As I recall, this was in TacitusÍ Histories >> : which was written in the 'rst century A.D. >> If thatÍs the best you can do, I think that we can safely ignore your >theory. ItÍs soooo easy Richard. Someone in your camp >should have the information at his 'ngertips. GOOGLE: tacitus histories ~13,600 hits > tacitus histories jews ~3,950 hits > tacitus histories jews leprosy ~154 hits >> This is all irrelevant. >> Quoting from Tom Potter: >> 2. This is the computer age. >> Almost everyone has access to the 'rst hand historical accounts, >> and can do wild card searches on the source material. >> It is STUPID to provide detailed cites, as these focus >> ONLY on the POINTS trying to be emphasized by the writer. >> It also STUPID to use second hand, accounts which have a >> racial, religious, national, or personal spin on them, rather than >> using the FIRST HAND historical accounts. >> Anything written by Tacitus would NOT be a FIRST HAND account, >> so only a STUPID person would allow himself to be brainwashed >> by TacitusÍ racial, religious, national, personal spin on history. It is interesting to see that Mensanator, >like many people, has been brainwashed to think that >Tacitus, who was one of the most unbiased, rational, >and correct historians, put a racial, religious > spin on history. Liar. YOU are the one who says that second hand accounts put a > spin on history . Or are you going to deny that that was your quote? > I wouldnÍt try it, I located it via a Google search, others can too. This attitude obviously has itsÍ roots in >conditioning, as the works of Tacitus >are far more rational and correct, >than the bible, the Greek and Roman mythologies, etc. But they are not a FIRST HAND account. So your brainwashed opinion of > their validity is irrelevant. Or are you now admitting that the quotation > about second hand accounts was stupid. You canÍt have it both ways, > so which is it? In other words, people who have been brainwashed to a point of view, >have a great dif'culty in accepting data >that con§icts with their conditioning. In other words, Tom Potter is too stupid to realize that heÍs just been > hoisted by his own petard. It is interesting to see that Mensanator does not comprehend that Herodutus, Tacitus, Josephus, and the ancient Greek and Roman Historians, are the closest FIRST HAND accounts of ancient history, and that works of modern historians use these works as their starting reference points. If you want to know the facts about ancient history, you start with the works of the ancient historians, so you can detect the spin put on history by modern historians who impose their present day agendas on the works. writers from different religions and nations, put a personal spin on the history, and a lot of information is lost as it is played through fallible, tuned 'lters. No doubt, some original historians put a personal spin on their works, but this does not seem to be the case with Tacitus, Herodutus, Thucydides, Xenophon, and Polybius, although one can detect a little Delphi Oracle spin in HerodutusÍ works. -- Tom Potter === Subject: Re: Question about the LÍHospital Rule Maybe, I was not clear enough. My post was motivated by the following paper: Counterexamples to LÍHpitalÍs Rule R. P. Boas, Northwestern University, Evanston, IL 60201 The American Mathematical Monthly, October 1986, Volume 93, Number 8, pp. 644645. This paper can be found online on the address: === Subject: Re: Discontinuous Derivative IÍm looking for a real function f:R -> R >which is differentiable over all of R >yet with a derivative discontinuous at 0. Standard example: x^2 * sin(1/x) , extended to 0 as 0. > fÍ(0) = lim(h->0) h sin 1/h = 0 fÍ(x) = 2x sin 1/x - x^2 (cos 1/x) x^-2 = 2x sin 1/x - cos 1/x > Variants: like x^2 * sin(1/x^2) ; this one has derivative > unbounded near 0. > fÍ(0) = lim(h->0) h sin 1/h^2 = 0 fÍ(x) = 2x sin 1/x^2 - x^2 (cos 1/x^2) 2x^-3 = 2x sin 1/x^2 - (2/x) cos 1/x^2 Wild! === Subject: hello...my question is ?? f(0) = 0 integral (0 ~ 1) f(x) dx = 1 fÍ(x) is continuous 'nd M such that fÍ(x) <= M ----------------------------- answer : M >= 2 how do you solve it? === Subject: Re: Discontinuous Derivative > IÍm looking for a real function f:R -> R > which is differentiable over all of R > yet with a derivative discontinuous at 0. Integrate the Wierstrass Function which is everywhere continuous (so the Riemann integral exists) and nowhere differentiable. Bob Kolker === Subject: Re: Invariant Galilean Transformations (FAQ) On All Laws Totally bogus crap. Bob Kolker > !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?- --!---? === Subject: Re: Just what is an L-series ? >Ah so. I was trying to compose a de'nition of L-series for one of the >volunteer-written online websites (www.planetmath.org). Maybe in the end, >L-series are not a structure but only a formalism, like generating ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ >functions. I may need to resort to arm-waving or (what is much the same >thing) category theory:) IÍm not sure what you mean exactly, but isnÍt F(x)=Sum a_n u_n(x) for a given sequence {u_n} of functions a general enough de'nition of generating function? 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: Just what is an L-series ? >Ah so. I was trying to compose a de'nition of L-series for one of the >>volunteer-written online websites (www.planetmath.org). Maybe in the end, >>L-series are not a structure but only a formalism, like generating > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ >>functions. I may need to resort to arm-waving or (what is much the same >>thing) category theory:) IÍm not sure what you mean exactly, but isnÍt F(x)=Sum a_n u_n(x) for a given sequence {u_n} of functions a general enough de'nition of >generating function? And isnÍt what youÍve just written a perfect example of a formalism ? WhereÍs the structure ? Lee Rudolph === Subject: Re: Max. Non-Adjacent Vertices on 120-cell > WhatÍs the size of a largest subset of vertices on the 120-cell > ({5,3,3}) such that no two vertices in the subset are joined by > an edge? I have reason to suspect that the number is an integer square, > and the computer searches IÍve done come tantalizingly close, > but I havenÍt yet been able to conclude that IÍve found a maximal > set. > found a few notes on the web, if I understood it correctly the indenpendence number lies in the interval [220, 224]. So it cannot be a square. www.csr.uvic.ca/~wendym/courses/582/120cell.ps === Subject: Re: Ridicule This Crackpot > My big question is how can I insure that I receive credit for this > formula and algorithm? (Assuming that mathematicians decide > that it is novel and concede the slightest bit of signi'cance.) Any > advice would be appreciated. > My advice: youÍre in the wrong 'eld. Why donÍt you try investment banking, or better yet marketing? === Subject: Re: Boolean Algebra - Arithmetic Relationship > 3)Can all known Mathematical Notation and Symbolic manipulation be >> modeled by a turing machine?? > If itÍs recursive it can be represented by a Turing Machine. WhatÍs the signi'cance of recursion? IÍm confused by this. (IÍve > written non-recursive programs before of course.) Are all known Just to confuse you; your non-recursive programs that compute functions _are_ computing _recursive_ functions as mathematicians/logicians understand the word. > mathematics recursive? What is a _computable_ function? Who knows? Mathematicians (notably Turing and Church) have _de'ned_ computable function to be recursive function , where recursive function has a precise meaning. The essential question forming in my mind is this: Is there a > fundamental difference between mathematical symbolic notation and Symbols can be encoded with positive integers, so can strings of symbols. > computer programming languages?? Both represent logical structures Computer programs are limit by technology and human patients. If we ignore those limitations, then every recursively de'nable manipulation of symbols can, in theory (:-), can be carried out by a computer. > and their interactions. Both follow certain rules of manipulation. > For me itÍs looks like theyÍre two sides of the same coin. Are their certain properties of Zermelo-Fraenkel set theory that canÍt > be modeled with a Turing-Machine? I donÍt know what a properties of Zermelo-Fraenkel set theory is. A Turing Machine cannot prove all ZF theorems but it can check the correctness of any purported proof. ZF canÍt settle all mathematical problems. No single formal system can. to your > response. -Steve IÍd like to recommend some reading. For logic generally, see Tarski An Introduction to Logic and the Methodology of the Deductive Sciences OUP. (Dreadful title, best book.) For recursion theory IÍm at a bit of a loss, how about Boolos & Jeffrey Computability and Logic CUP? -- G.C. e-mailing me. === Subject: Re: 3D SurveyorÍs Formula (for volume)? Is there a version/generalization of the SurveyorÍs Formula for three >dimensions? IÍm imagining inputting a bunch of (x,y,z) points, and receiving as >output the volume of the polyhedron thus delimited. Of course, there will be issues involving exactly _how_ the points are >given - presumably itÍs more complicated than the 2D version - but it >seems like it should be possible.... Anyone know of the 3D version? >cdj See subject 5.19 of http://www.faqs.org/faqs/graphics/algorithms-faq/ === Subject: Re: Question about the LÍHospital Rule >Maybe, I was not clear enough. My post was motivated by the following paper: Counterexamples to LÍHpitalÍs Rule >R. P. Boas, Northwestern University, Evanston, IL 60201 >The American Mathematical Monthly, October 1986, Volume 93, Number 8, pp. >644645. This paper can be found online on the address: > You were perfectly clear. I explained that there are conditions under which the rule is correct stated in any calculus book. ThatÍs true. The paper you cite does not contain any counterexamples to the theorem in decent calculus books. In particular I pointed out, in case you couldnÍt 'nd a calculus book, that for example, if f -> 0 and g -> 0, and if fÍ/gÍ -> L then f/g -> L. Also if f -> in'nity and g -> in'nity. That paper does not contain any counterexamples to what I asserted here. (It contains examples where itÍs _not true_ that fÍ/gÍ -> L, but where a student might nonetheless think the rule applied.) ************************ David C. Ullrich === Subject: Re: hello...my question is ?? ----------------------------- answer : M >= 2 how do you solve it? > ************************ David C. Ullrich === Subject: Science is a human activity (was: Python syntax in Lisp and Scheme) Originator: claird@lairds.com (Cameron Laird) >Alex Martelli: >> would you kindly set right the guys (such as your >> namesake) who (on c.l.lisp with copy to my mailbox but not to here) are >> currently attacking me because, and I quote, >> Software is a department of mathematics. And anyone who doesnÍt think mathematics has its own >culture with ideas and even mistaken preferences for what >is right and wrong should read The Mystery of the Aleph: Mathematics, the Kabbalah, and the Human Mind to see how CantorÍs ideas of trans'nite numbers (and other ideas, >as I recall, like showing there are functions which are everywhere >continuous and nowhere differentiable) were solidly rejected by >most other mathematicians of his time. Mathematicians are people as well. . . . And let no one assume that these are mere foibles of the past that we moderns have overcome; mathematics remains stunningly incoherent in whatÍs labeled foundations . ThereÍs a wide, wide divergence between the intuitionism working mathematicians practice, and the formalism they profess. ïGood thing, too; our age enjoys the blessing of superb mathematicians, and IÍm relieved that philosophical in- consistencies donÍt (appear to) slow them down. -- Cameron Laird Business: http://www.Phaseit.net === Subject: Re: Question on Hilbert & Godel What did Hilbert ask and claim concerning Foundations of Mathematics >(sets, predicate calculus), metamathematics, Logic, Incompleteness, >etc? > As a mathematician, he did not *claim* things he could not prove. DidnÍt he claim that a certain mathematical problem must have a solution when in fact it doesnÍt? > he did not reach > this goal which was shown by Gdel to be unreachable. One cannot say > he was contradicted ; every serious mathematician has goals that > prove unreachable - only claiming to have reached such goals can be > contradicted. What about claiming that a particular goal is reachable only to have it proven that the goal is not reachable? Is that a contradiction? Charlie Volkstorf Cambridge, MA > Helmut Richter === Subject: Re: Question on Hilbert & Godel What did Hilbert ask and claim concerning Foundations of Mathematics >(sets, predicate calculus), metamathematics, Logic, Incompleteness, >etc? Just a side note: Bourbaki was cited as the best example so far of mathematics > organized into a coherent framework. According to Andre Weil, > Perhaps the most important contribution of Bourbaki was to > carry out a famous proposal made by the great German mathe- > matician David Hilbert in 1900 that mathematics be placed on > a more secure foundation. He noted: Hilbert just said so, > and Bourbaki did it And just how did Bourbaki do as Weil claims? But didnÍt Hilbert actually claim that a lot more than that is possible? DidnÍt he ask for (and claim that it must exist) a decision procedure to determine if an arbitrary predicate calculus wff is valid (true of all interpretations) - was it? conventional wisdom was wrong and mathematicians gained new insight into the nature of mathematics. Or are you one of those diehards who clings to the failed ideas and theories of the past? (Actually, the most secure foundation possile is software that carries out both logic and metamathematics, as I have implemented and descibed in my papers below.) Charlie Volkstorf Cambridge, MA http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/ 20021008.1/1 http://www.arxiv.org/html/cs.lo/0003071 > (Quote from the QED Manifesto) F. === Subject: Re: Uses of complex numbers by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9B5aUT30405; >All I know from my lessons is that complex numbers have geometric >interpretations... any help is most appreciated... IÍm sure some >meaning in science has been found for it... > === Subject: Re: Dif'cult Inequality by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9BC2Rk22336; I did it in a simpler way, using arithmetic-geometric mean ineqality (1+x) + (1+y) + (1+z) --------------------- >= ((1+x)(1+y)(1+z)) ^ (1/3) 3 (4/3)^ 3 >= (1+x) (1+y) (1+z) 64/27 >= (1+x) (1+y) (1+z) 64 >= 27 (1+x) (1+y) (1+z) Also x+y+z ----- >= (xyz) ^ 1/3 3 xyz <= 1/27 using the two 64 xyz >= (1+x) (1+y) (1+z) 64 >= (1+1/x) (1+1/y) (1+1/z) >
gwoegi@fmatbds01.tu-graz.ac.REMOVE.at
>
===
>Subject: Dif'cult Inequality
>Here is a pretty dif'cult inequality that I 
wasnÍt able to
solve
>so far.
>x,y,z are positive real numbers such that x+y+z=1
>Prove that(1+1/x)(1+1/y)(1+1/z)>=64
>>Set p= (xyz)^1/3. Then the arithmetic-geometric mean
ineqality yields
>>(1) x+y+z >= 3p
>>(2) xy+xz+yz >= 3p^2
>>Because of (1) and x+y+z=1, we also get
>>(3) 1/p >= 3
>>The righthand side of the desired inequality equals
>> (1+ x+y+z + xy+xz+yz + xyz) / xyz >= (1+3p+3p^2+p^3) / p^3
= (1+1/p)^3
>= 64
>>___________________________________________________________
>>Gerhard J. Woeginger (gwoegi@opt.math.tu-graz.ac.at)
>I solved it using just plain algebra without any problems,
starting
>out with things like z + x = 1 - y and factoring from there.
But IÍve seen so many solutions using all sorts of advanced
math that
>I wonder - is there a link between them all?
>--
A copper blade
>In'nitely many strings.
>
=== Subject: Re: Question on Hilbert & Godel > What did Hilbert ask and claim concerning Foundations of Mathematics >> (sets, predicate calculus), metamathematics, Logic, Incompleteness, >> etc? >Just a side note: Bourbaki was cited as the best example so far of mathematics > organized into a coherent framework. According to Andre Weil, > Perhaps the most important contribution of Bourbaki was to > carry out a famous proposal made by the great German mathe- > matician David Hilbert in 1900 that mathematics be placed on > a more secure foundation. He noted: Hilbert just said so, > and Bourbaki did it And just how did Bourbaki do as Weil claims? But didnÍt Hilbert actually claim that a lot more than that is > possible? DidnÍt he ask for (and claim that it must exist) a decision > procedure to determine if an arbitrary predicate calculus wff is valid More than that, an arbitrary wff of mathematics. > (true of all interpretations) - was it? conventional > wisdom was wrong and mathematicians gained new insight into the > nature of mathematics. Or are you one of those diehards who clings to the failed ideas and > theories of the past? (Actually, the most secure foundation possile is software that carries Software is 'ne if youÍre a strict 'nitist. Some mathematicians like a set theory with _huge_ axioms of in'nity in which to do their category theory. > out both logic and metamathematics, as I have implemented and descibed > in my papers below.) Charlie Volkstorf > Cambridge, MA > http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/ 20021008.1/1 > http://www.arxiv.org/html/cs.lo/0003071 (Quote from the QED Manifesto) F. -- G.C. e-mailing me. === Subject: chisquare test question I calculated Chi-square for a set of data..... but, then im asked to compare the calculated value of Chi-squared with (n-1) - a.k.a degrees of freedom. Why would I compare my calculated value of Chi-Squared with n-1? I donÍt get the purpose.... === Subject: Re: Science is a human activity (was: Python syntax in Lisp and Scheme) Laird) Well, since you crossposted this to sci.math you must be hoping for replies from that direction: >>Alex Martelli: > would you kindly set right the guys (such as your > namesake) who (on c.l.lisp with copy to my mailbox but not to here) are > currently attacking me because, and I quote, Software is a department of mathematics. >And anyone who doesnÍt think mathematics has its own >>culture with ideas and even mistaken preferences for what >>is right and wrong should read >>The Mystery of the Aleph: Mathematics, the Kabbalah, and the Human Mind >>to see how CantorÍs ideas of trans'nite numbers (and other ideas, >>as I recall, like showing there are functions which are everywhere >>continuous and nowhere differentiable) were solidly rejected by >>most other mathematicians of his time. >>Mathematicians are people as well. > . > . > . >And let no one assume that these are mere foibles of the >past that we moderns have overcome; mathematics remains >stunningly incoherent in whatÍs labeled foundations . >ThereÍs a wide, wide divergence between the intuitionism >working mathematicians practice, Actually inuitionism has a certain technical meaning, and actual intuitionism is not what most mathematicians practice. But never mind, I believe I know what you meant. >and the formalism they >profess. Far be it from me to insist weÍve overcome the foibles of the past. But: ItÍs certainly true that mathematicians do not _write_ proofs in formal languages. But all the proofs that IÍm aware of _could_ be formalized quite easily. Are you aware of any counterexamples to this? Things that mathematicians accept as correct proofs which are not clearly formalizable in, say, ZFC? >ÍGood thing, too; our age enjoys the blessing of superb >mathematicians, and IÍm relieved that philosophical in- >consistencies donÍt (appear to) slow them down. WhatÍs an actual example of one of those philosophical inconsistencies that luckily doesnÍt slow us down? ************************ David C. Ullrich === Subject: Re: Non-Polynomial Time Algorithms > ItÍs thought that NP properly contains P, but this is hard. ... > If any of the NP-complete > problems are in P, then the reduction would show that all NP problems > are in P. Do I understand correctly that you are saying NP might contain P as well as that P might contain NP? If that were true, of course, it would mean that P and NP are the same. Did I get that right? === Subject: i * (3,4)=( - 4, 3)= - 4 +3 * i and the Manifesto di Bombelli The opportunity to post the Manifesto di Bombelli sci.math. The applet - and much more - You can 'nd now also on www.insel.heim.at/mainau/331839 So, another month added to the twohundred years of avoiding WesselÍs ideas( no contributions up to now). Some - secretly knowing itÍs true - donÍt want to loose the nimbus of one, who comprehends the imaginary . Usually producing clouds to disorientate others - but for the time being avoiding, not to arouse attention. For some it might be a nut, too hard to crack : it costs a lot of brain-fat to replace imaginations, you cuddle. And some eagerly want to declare this nuts, but they fear for their reputation, if it turns out otherwise. So, what about knowledge ? Hero . > Introducing the dot-multiplication in the vector-space ( R2,+,r.s.m ) > results in an Euklidian vectorspace, inside you can 'nd a commutative > 'eld (R2,+,* ). > Here * is the Bombelli-multiplication: i * i = - 1 , or in HmiltonÍs > notation:(a,b)*(c,d)=(ac-bd,ad+bc), and i rotates a vector 90 degrees to the > left by i*(a,b)=(-b,a). > In modern-talking math:A vector, an ordered tupel by de'nition, can > be expressed in the terms of a vectorspace-basis. In R(=R1) the vector > (a) with the standard-basis ( 1 ) gives: (a)=a*1 and thatÍs -of > course - equal to a. Adjungate a foreign element to R, you go 2D.(a,b) > can be notated as a linear combination of the basis-elements: > c*1+d*element. With the standard-basis > ( (1,0) , (0,1) )=( 1 , (0,1) ) or ( 1 , i ) you get a + b * i, > looking different from (a,b), but identical. > You can write and calculate vectors in mixed mode now. > I propose the name Bombelli vector-space. Real arrows showing winds on a weather-map. > Attach (add) to a number of points the difference in coordinates from > one central point,multiplied by (0,1)=i.Improve this 'rst model by > multiplying > (cos @,sin @)=cos @+i*sin@ - allowing for friction alpha from 10 > (over sea) to 35 (over land)(-alpha on the southern hemisphere), and > scale everything by a factor 1/10 (rsp. -1/10) and add a common > velocity to the east. It models > the winds of the inner part of a depression > - done solely with BombelliÍs operations - and ........... > we stayed plain real. Two questions arise, sensitive questions for some - and the answer is > left to them : > Is there any difference to (C,+,*)? - > can you tell this difference to your computer? > If you call the y-axis an imaginary axis , do you add any > mathematical properties, or do you just change the name? Is the > Gauss-plane, complex > plane or the Aragand-diagramm more than just a Fata-Motgana-Re§ection > of the real plane? > Why - twohundred years after Wessel - this question berhaupt ? > In a math-lesson you get sometimes the impression, an extraterrestical > element has been adjungated to R. My opinion: > i - is - no - longer - imaginary > Copy this Hero > As in sci.math you can not display applets,you might take a look at > www.i-z.eu.tt or > www.i-is-no-longer-imaginary.gmxhome.de > There You 'nd the Manifesto, i quoted above, in a printable version > for the > pinboard and a (free) mixed-mode calculator. > As the computer makes no difference - do You have to ? > Hero. === Subject: quadratic reciprocity basic question Suppose I have the following if and only if condition: Let p and q be 2 distinct odd primes, then p is a square (mod q) <=> ((-1)^(q-1)/2) * q is a square (mod p). How do I deduce the quadratic reciprocity law from this if and only if condition??? Alex === Subject: Re: Variation on an impossible problem >Allowing only the use of the + operator, is it possible to use each of >the digits 0 1 2 3 4 5 6 7 8 9 once only to make a total of 1000? One can combine digits to create sums like 210 + 345 + 6 + 78 + 9 = >648 etc. No. sum(i,i=0..9) = 0 mod 9, whereas 1000 = 1 mod 9. every sum I could possibly form must be divisible be nine? I can see that if all such sums must be divisible by nine then 1000 cannot be a possible sum. Moreover, I can see that sum(i,i=0..9) = 0 mod 9 but how does this generalise to all possible sums formed? Mitch. === Subject: Re: Non-Polynomial Time Algorithms that P might contain NP? If that were true, of course, it would mean that > P and NP are the same. Did I get that right? Yes, you have that right. All that is known is that P is a subset of NP, but it is still unknown whether or not they are equivalent. J === Subject: Re: Curve Inside Another Of Equal Perimeter > Are there any two CONVEX closed curves of equal perimeter, where one >> can be placed completely inside the other? >> (It is allowed that the curves touch at a 'nite number of points at >> most.) >> Leroy Quet >I would guess not. Here is an outline of a possible proof. Choose a point inside both curves. >Consider the formula for the arc length as a function >of r(theta) and theta, where r(theta) is the distance from >the central point to the curve at angle theta. Since r_1(theta) <= r_2(theta), where r_1 and r_2 are >the respective distances to the inner and outer curves, >the length of the outer curve is greater than the >length of the inner curve. Martin Cohen That doesnÍt quite work, since the perimeter L = int(sqrt(r^2 + (dr/dtheta)^2)) dtheta and you havenÍt said anything about dr/dtheta. One quick approach to proving that the inner curve must be shorter than the outer one (unless theyÍre coincident) uses CroftonÍs formula, as I noted in another response. You can also use the fact that projection onto a convex set is (weakly) length-decreasing (by projection I mean assigning to a point P the nearest point in the convex set). So if you project the outer curve onto the (domain bounded by the) inner curve, the length is not increased; a little extra work is needed to show that itÍs actually decreased). John Mitchell === Subject: Re: Two similar things > Since they now are identical-with-somethings, if they shared > all the same properties they would be one cake rather than > two. THerefore, since they are two cakes, they donÍt share > all their properties, and I can distinguish them by means > of any property that one has and the other doesnÍt. >> No, thereÍs a foot wrong there. Why would they be one cake if they shared >> all the same properties? Consider Hesperus and Phosphorus. The properties of Hesperus include >identity-with-Hesperus, and those of Phosphorus include identity-with- >Phosphorus. So, if Hesperus and Phosphorus share *all* their properties, >the properties of Hesperus will include identity-with-Phosphorus, and >those of Phosphorus will include identity-with-Hesperus. But if Hesperus >and Phosphorus have these properties, then Hesperus and Phosphorus are >one and the same: that is, numerically identical. > Why not 3 cakes? If these cakes have all the same properties, for the reasons I have indicated >these three cakes will be one and the same. > Would you tell the difference by counting? Just by counting, no. By counting correctly, yes. In other words, >before it was discovered that Hesperus was Phosphorus, it was assumed >that these heavenly bodies were different heavenly bodies rather than >one and the same heavenly body. But this assumption proved to be wrong. --John ArenÍt you mixing physical properties with logical properties? Where is the justi'cation for that? Logically twins are one §esh. Physically they are two. === Subject: Re: quadratic reciprocity basic question > Suppose I have the following if and only if condition: > Let p and q be 2 distinct odd primes, then p is a square (mod q) <= ((-1)^(q-1)/2) * q is a square (mod p). How do I deduce the quadratic reciprocity law from this if and only if > condition??? I think you can use the EulerÍs criterion : If p is an odd prime and if a not equal to 0 mod p, then (a/p) = a ^ ((p-1)/2) mod p. First, consider the case where p is a square (mod q). Then, consider the case where p is not a square (mod q). - Bill Hale === Subject: Re: Discontinuous Derivative |> IÍm looking for a real function f:R -> R |> which is differentiable over all of R |> yet with a derivative discontinuous at 0. | |Integrate the Wierstrass Function which is everywhere continuous (so the |Riemann integral exists) and nowhere differentiable. By the fundamental theorem of calculus, the result is a function whose derivative is the Weierstrass function, which is continuous. Keith Ramsay === Subject: Re: Just what is an L-series ? |Ah so. I was trying to compose a de'nition of L-series for one of the |volunteer-written online websites (www.planetmath.org). Maybe in the end, |L-series are not a structure but only a formalism, like generating |functions. I may need to resort to arm-waving or (what is much the same |thing) category theory:) It would be interesting if there were a de'nition of L-series rather than just de'nitions of individual kinds of L-series. As far as I know, there may be one. But I never did see one, in spite of having seen a number of talks and lectures about speci'c kinds of L-series. It seems plausible to me, then, that what has happened is simply that as each kind of L-series has been de'ned (Artin L-series, etc.), itÍs gotten termed another kind of L-series because of its family resemblance to the other kinds of L-series previously de'ned. As far as I know they are all naturally expressed as Dirichlet series, sum_{n} a_n/n^s. Keith Ramsay === Subject: Re: quadratic reciprocity basic question > Suppose I have the following if and only if condition: > Let p and q be 2 distinct odd primes, then p is a square (mod q) <= ((-1)^(q-1)/2) * q is a square (mod p). How do I deduce the quadratic reciprocity law from this if and only if > condition??? I think you can use the special case of EulerÍs criterion : If p is an odd prime, then (-1/p) = (-1)^((p-1)/2). First, consider the case where q = 1 mod 4. Next, consider the case where p = 1 mod 4. Finally, consider the case where q = 3 mod 4 and p = 3 mod 4. - Bill Hale === Subject: Re: Core error, FEAR is a natural response : The de'nition of algebraic integers as roots of monic polynomials >: with integer coef'cients gives the ability to give two supposed >: proofs that contradict each other. I donÍt get it. I understand the part about two arguments contradicting >each other, and one of them being wrong. But what does this have to >do with the de'nition of the algebraic integers? DonÍt the arguments >have to use the same de'nitions in order to be contradictory? And if >one of the arguments wrong, isnÍt the problem with the argument, not >the de'nition? Apparently itÍs no longer necessary to assume anything in order to arrive at a contradiction. JH states that the mere act of de'ning the concept of algebraic integers is something that leads to contradicting mathematical results, which must mean that mathematics itself is as pointless as a very blunt object as without de'nitions there is no mathematics. But how can we use mathematics to prove that mathematics cannot be used to prove anything? The other option is that JHÍs proof is *gasp* wrong. The fact that he accepts the contradicting proof about the algebraic integers as being valid actually directly implies that JH accepts his proof must be wrong. What really scares me is, how can a physics major be so ignorant about mathematical formulation and the logical requirements for a proof? I have 'rst-hand experience that a lot of math is just regurgitated at students without adequate explanation of the proofs and theories underneath so that you can learn a lot of advanced mathematical methods without understanding what the hell youÍre actually doing, but physics of all subjects should require a solid understanding of math. === Subject: Re: Quanta and Cakes > Nice to read this. > I do not understand almost nothing of this thread. However, what is the > question? Is it this: What is the probability of 'nding some of the > Take a look into the box and 'nd what is the actual > con'guration. Do it many times. After, calculate the probability of > given con'guration (= number of this particular con'gurations / number of > all measurements). > You will 'nd that > probability of ll is 1/4 > probability of rr is 1/4 > probability of lr is 1/2 > Is this the point of this thread, right? > Palo Partly right. The point of the thread is that although > probability of ll is 1/4 > probability of rr is 1/4 > probability of lr is 1/2 are the probabilities that physicists would have expected to 'nd, the probabilities that they have found are: > probability of ll is 1/3 > probability of rr is 1/3 > probability of lr is 1/3 The question then arises, what is it about quanta that is responsible for the difference between classical statistics and quantum statistics. Some physicists have attributed this difference to what they refer to as a ïlack of individualityÍ in quanta. What I am suggesting is that from the standpoint of logic, things that ïlack individualityÍ do not satisfy ïx=xÍ. If this suggestion is correct, then classical logic--in which everything in the domain of individual variables satis'es ïx=xÍ--and quantum logic do not coincide. Of course, there is nothing new about the suggestion that classical logic and quantum logic do not coincide. To my knowledge, however, those who claim there is a difference between the two have not called into question the re§exivity of identity. --John > I should have asked in what respect *correct reasoning* about >> quanta resembles such reasoning about cakes-to-be, but >> differs from such reasoning about (extant) ïmedium size >> dry goodsÍ. > My answer to this question would be: correct reasoning about >> medium size dry goods should take these as satisfying >> both the right and the left sides of the following bi- >> conditional; > AxAy(Az(z in x <-> z in y) -> x=y) <-> Ax(x=x) >> [Identity of Indiscernibles] [Re§exive Law of Equality] > while correct reasoning about quanta--and cakes-to-be--should >> take these to satisfy neither. Sorry, you have obfuscated your meaning beyond my ability to extract >any sense from it, or even to guarantee that the language in use is >still English, so I shall withdraw from the discussion. xanthian. === Subject: Re: Technique of planning an elementary mathematical proof meant: show that the second derivative is always positive for x > 0, hence the value of f(x) where f ï (x) is zero will be the absolute minimum etc. I see that this type of problem probably comes up pretty often while studying partial derivatives (which IÍve just dipped into a little.) === Subject: Re: Non-Polynomial Time Algorithms |> ItÍs thought that NP properly contains P, but this is hard. | |... | |> If any of the NP-complete |> problems are in P, then the reduction would show that all NP problems |> are in P. | |Do I understand correctly that you are saying NP might contain P as well as |that P might contain NP? NP de'nitely does contain P. The fact that P is a subset of NP follows simply from the de'nitions of P and NP. By properly contains I meant that in addition to containing it, it contains things not in P as well. It seems reasonably common for people reading popular accounts of the P versus NP question to get the impression that problems in NP are supposed to be dif'cult, probably because a lot of the interest in NP is in some of the presumably hard problems in it. The way NP is de'ned, however, does not put any kind of lower bound on the dif'culty of a problem. A yes/no problem is in NP if it has a suitable type of witness , whose length is bounded by a polynomial in the original problem. There has to be an auxiliary algorithm which runs in polynomial time, having the property that if it accepts a witness, then the answer to the problem was yes , and conversely if the answer to the problem was yes , then there exists a witness that will be accepted by the algorithm. As an easy example, whether a number N is composite is easily seen to be a problem in NP. One can use as a witness a divisor of the number greater than 1 and less than the number. Given N and d, determining whether d divides N and 1({5,3,3}) such that no two vertices in the subset are joined by >an edge? I have reason to suspect that the number is an integer square, >and the computer searches IÍve done come tantalizingly close, >but I havenÍt yet been able to conclude that IÍve found a maximal >set. Suggestions? Well, I doubt IÍve caught up with you, since I just made > a simple pass at it based on some programs I have lying around > re 600 and 120-cell. I arranged the 600 vertices of the 120 cell into sets equidistant > from an arbitrary point, then looked at distances among these > sets and between them. There are 54 points in the equator between > the given point and its antipode, and these are the 15th nearest > neighbors to those two points. There are 6 of these with no ( 1st ) n-n among the 54, and the > other 48 have one n-n each, in pairs. So we can pick 30 of these > ( 6 + 48/2 ) and then radiate out to the poles. I donÍt know if > it matters which of each pair is picked, but it probably matters > which order you select candidates as you radiate . Anyway, doing it arbitrarily I got 193 points, so I suppose youÍre > aiming at 196. Is this right? Well, never mind that ! I get 201 just by progressing through the n-n planes starting from a point, taking the points in each plane in arbitrary order. The equator gets 20 points this way. Curiously, when I start at each end and go up to the equator, then pick points in the equator, I get 201 also, even though the equator only gets 15 points in this case. These results are subject to arbitrary variation, though. This equator is an interesting 'gure. It is 3-d since it lies in a hyperplane of the 4-d 'gure, and the points are on the surface of a 2-sphere, i.e. a sphere in 3-d space. The six points with no n-n in the equator are in antipodal pairs and form a set of orthonormal axes. The 48 remaining points are generated from a single nonsymmetrically placed point from the group O_h ( octahedral group ) so the 24 n-n pairs comprise the vertices of 2 snub cubes which map into each other under inversion. Lew Mammel, Jr. === Subject: Square root modulo a power of two How does one solve the quadratic congruence x^2=a(mod 2^n) ? === Subject: Re: hello...my question is ?? > f(0) = 0 integral (0 ~ 1) f(x) dx = 1 fÍ(x) is continuous 'nd M such that fÍ(x) <= M ----------------------------- answer : M >= 2 how do you solve it? Are you asking for the smallest possible M for which the conditions on f can all be satis'ed? This is not clear from your statement of the problem. In that case, f(x) = 2*x is the function with the smallest deriviative over all of 0 <= x <= 1 satisfying all the stated conditions, giving M = 2. === Subject: Re: quantum echo Metaphysical Background , Thomas McTighe asserted that the quiddity > of a thing is nothing other than unity itself. Hence, by virtue > of its positive content, the sun differs not at all from the moon > or any other particular thing. The diversity which is exhibited > by the natural world is merely the product of accidental > differences; no object possesses any speci'c form which > interposes itself between a particular existing thing and the > source of their being e.g. the Absolute.15 All individual > entities are nothing more than differing contractions of the > whole devoid of any being of their own. ...because the > restricted quiddity of a thing is the thing itself. http://www.crvp.org/book/Series01/I-10/chapter_ii.htm I didnÍt understand this the 'rst time around. This time IÍll > make an effort, by placing it in its context: Metaphysical Background , Thomas McTighe asserted that the quiddity > of a thing is nothing other than unity itself. Hence, by virtue > of its positive content, the sun differs not at all from the moon > or any other particular thing.14 The diversity which is exhibited > by the natural world is merely the product of accidental differences; > no object possesses any speci'c form which interposes itself between > a particular existing thing and the source of their being e.g. the > Absolute.15 All individual entities are nothing more than differing > contractions of the whole devoid of any being of their own. > Putting it in context didnÍt work. I have no idea what McTighe is > talking about. Do you? > I was trying to proceed through the evolution of these particulars i guess. I think he is trying to re§ect upon reductionism and it seems a bit like the whole in the part idea in fuzzy logic. I liked it because it addressed the macro or context and how features from that scale are only sub-components of the context and how the pre-text text and context each have their own rules to emulate identical functions. Like rules of de'nition with words and rules of subject/predicat in sentences and rules of topic/support sentences in paragraphs. in ideal form it would be like music when the riff double times the chord progression with 2 of the chord progression one scale down. draws the chord with double chord patterns but with different key combinations required for textcontext/context. but who knows the mind of the middle ages, lots of stuff hidden there, shrowded in god talk >> I take a quiddity to be a thingÍs *suchness* and a haecceity to be its >> *thisness*. A *thisness* I take to be, as Robert Adams does, the >> property of self-identity, although I disagree with an assumption >> which might be read into Adams, that an individual can lack >> self-identity and yet possess the property of being some >> individual or other). >> In Primitive Thisness and Primitive Identity (_The Journal >> of Philosophy_, Vol. 76, No. 1. (Jan., 1979), pp. 5-26), Adams >> A thisness is the property of being a certain particular >> individual, not the property of being some individual or other, >> but my property of being identical with me, your property of >> being identical with you, etc. These properties have recently been >> called ïessencesÍ, but that is historically unfortunate, for essences >> have normally been understood to be constituted by logical properties, >> and we are entertaining the possibility of nonqualitative >> thisnesses. In de'ning ïthisnessÍ as I have, I do not mean >> to deny that universals have analogous properties--for example, >> the property of being identical with the quality red. But since >> we are concerned here principally with the question whether >> the identity and distinctness of individuals is purely >> qualitative or not, it is useful to reserve the term >> ïthisnessÍ for the identities of individuals. >> It may be controversial to speak of a property of being >> identical with me. I want the word ïpropertyÍ to carry as >> light a metaphysical load here as possible. ïThisnessÍ >> is intended to be a synonym or translation of the traditional >> term ïhaecceityÍ (in Latin, ïhaecceitasÍ), which so far as I >> know was invented by Duns Scotus. >> Like many medieval philosophers, Scotus regarded properties as >> components of the things that have them. He introduced >> haecceities (thisnesses), accordingly, as a special sort >> of metaphysical component of individuals.[4] I am not proposing >> to revive this aspect of his conception of a haecceity, because >> I am not committed to regarding properties as components of >> individuals. To deny that thisnesses are purely qualitative >> is not necessarily to postulate ïbare particularsÍ, substrata >> without qualities of their own, which would be what was left >> of the individual when all its qualitative properties were >> subtracted. Conversely, to hold that thisnesses are purely >> qualitative is not to imply that individuals are nothing >> but bundles of qualities, for qualities may not be components >> of individuals at all. (pp. 6-7) >> Acceptance of haecceities is a distinctive feature of the thought of many >followers of Scotus, though there are some sixteenth-century scholastics who >accept haecceities without accepting many other distinctively Scotist >teachings. Having said this, some early followers of Scotus reject >haecceities and the theory of the common nature altogether, and of those who >accept haecceities, some found the correct understanding of the nature of >the distinction between an individualÍs nature and its haecceity a >troublesome matter. One of the earliest Scotists, Francis of Meyronnes, writing his commentary >on the Sentences around 1320, accepts the theory of the non-numerical unity >of common natures, and the claim that individuation is by haecceity. But he >holds that it is inappropriate to talk of a formal distinction in this >context. Formal distinction obtains only between things that have some sort >of quidditative content. IÍm not sure what he/you are getting at here. > Francis of Marchia was not a faithful Scotist Marchia was uncomfortable with ScotusÍs stress on and use of a strong distinction between the divine intellect and will, and this led Marchia to oppose Scotus on issues such as the procession of the Holy Spirit and the mechanism of divine foreknowledge. Nevertheless, Scotus forms much of the backdrop for MarchiaÍs theology. ïActionÍ can be taken in three ways: either it can be taken actually, namely when an agent is actually acting; or it can be taken virtually, when an agent can act although he is not acting; or it can be taken in a middle way, not purely actually nor purely virtually, but in a middle way as ïdispositionallyÍ or ïaptitudinallyÍ, namely when an agent is not acting but is determined toward acting, although in actuality he is not acting - and he not only can act, but is determined to be acting later. Similarly there is a threefold ïdeterminationÍ of the agent: one actual, by which an agent actually determinately puts one part of a contradiction into effect; a second is a potential determination by which an agent posits or can determine any part of a contradiction dividedly; the other is, as it were, a ïdispositionalÍ or ïaptitudinalÍ determination, by which an agent is determined with respect to the future to putting one part of a contradiction [into effect]. Each determination presupposes the action corresponding to it, because an actual determination follows the action in actuality; the dispositional determination follows the action dispositionally, although it precedes the actual action; the potential determination follows the potential action, although it precedes that actual and dispositional action. (d. 35: Marchia 1999, pp. 89-90) Thus when an agent is determined de inesse to doing something in the future, that determination is like a disposition, and neither actual, because the event has not yet occurred, nor potential, because the possibility to do otherwise is not removed. Such a determination is not ïactuallyÍ in the agentÍs power, Marchia grants, but it is in his power ïdispositionallyÍ, for although the agent cannot act before he acts, he can be disposed to act so that he will in fact act. http://plato.stanford.edu/entries/francis-marchia/ http://www.wikipedia.org/wiki/Scholastic_philosophy Perhaps the most interesting non-qualitative approach is the theory, often associated with Aquinas (but attacked by Scotus in the forms presented by Godfrey of Fontaines and Giles of Rome), that individuation is by extended matter: by, as we might say, chunks of matter. ScotusÍs way of understanding the problem of individuation becomes important in his rejection of this theory. For his fundamental strategy against this sort of material individuation is that such a theory, while it may be able to explain numerical distinction, certainly cannot explain indivisibility: Quantity is not the reason for divisibility into individuals. . . . For a universal whole, which is divided into individuals and into subjective parts, is predicated of each of those subjective parts in such a way that each subjective part is it. But the quantitative parts into which a continuous whole is divided never admit of the predication of the whole that is divided into them. (Scotus, Ordinatio II, d. 3, p. 1, q. 4, n. 106 [Scotus (1950-), 7:443; Spade (1994), 85]) http://setis.library.usyd.edu.au/stanford/entries/ medieval-haecceity/ By such a reasoning, Dante concludes that the secular monarchy §ows from the divine unity sine ullo medio. John of Paris, however, asserts the proper autonomy of the secular authority, not only using Dionysius but indeed the very regula or lex which is authoritative for Boniface, Giles and de Meyronnes. This is a remarkable feat because, as A.P. Monahan remarks, As long as unquali'ed acceptance was given to the principle that all things must be subordinated to one, there was no logical escape from the doctrine that the temporal must be subordinated to the spiritual, whose ultimate unitary embodiment was the papacy.47 Moreover, he agrees with Dionysius in ecclesiastica hierarchia that the laity, with their kings, are in the lowest rank, though perfectible, as compared to the various ranks of the perfected and perfecting clergy: in in'mo gradu sunt laici cum suis regibus quasi imperfecti, perfectibiles tamen, supra quos sunt perfecti et supra illos sunt perfectiores, ut viri ecclesiastici, et in supremo est summus monarcha omnium, scilicet dominus papa. http://www.dal.ca/~claswww/DIONDIX.htm Haecceities have no such quidditative content, and thus cannot be formally >distinct from their nature. Rather, a haecceity is modally distinct from its >nature. What does Cross (or whoever) mean by Haecceities > cannot be formally distinct from their nature. ? > not sure but probably the distinction between this-ness and that-ness and limitations imposed by being one or the other? >A modal distinction, according to Meyronnes, obtains between a thing >and an intrinsic mode of that thing, where an intrinsic mode is something >which when added to a thing does not vary its formal de'nition . . . since >it does not of itself imply any quiddity or formal de'nition. A haecceity >does not affect a thingÍs kind; it is thus an intrinsic mode of the thing. In order for this to mean something to anyone who is not sure what > an intrinsic mode, or the formal de'nition, of something (in the > sense of Meyronnes) are--and is therefore unsure what ïvarying > somethingÍs formal de'nitionÍ might involve, illustrative examples > should be provided. > minimal convergence upon emulation or maybe water in the radiator could mean water from the lake in the radiator or water from the middle of the lake or water from unkown portion of the lake, is suf'cient for engine uses water? i donÍt know but i like middle ages thinking anyway. It may look as though this is just a terminological shift, but it is not so >in at least the following way: a modal distinction is a lesser kind of >distinction than a formal distinction. Formal distinctions obtain between >genus and speci'c difference; thus, the difference between species/nature >and haecceity, for Meyronnes, is less than the difference between genus and >difference. Scotus, contrariwise, makes no such distinction between degrees >of difference in this context (for this contrast between the two thinkers, >see Dumont [1987], 18). Still, without some principled way of spelling out >degrees of difference, this contrast between Scotus and Meyronnes amounts to >nothing of any philosophical interest. To this extent the difference between >the two thinkers might as well be merely terminological, and Meyronnes needs >to do more work if he is to make any signi'cant philosophical point here. Without any explanation of the before-shift and after-shift > terminology, how is one supposed to know which side is up? > Supposing that a haecceity is something real, where does it 't into the range of things that exist? Is it, for example, a form, or something else? According to Scotus, it is something like a form, and sometimes, indeed, he calls it such (while elsewhere denying the same claim: on these insigni'cant terminological shifts, see Dumont [1995]). The reason is that a haecceity is clearly something like a property of a thing -- hence like a form -- but is at the same time wholly devoid of any correspondence to any conceptual contents. It is not at all a qualitative feature of a thing -- not at all a quidditative feature, in the technical vocabulary. As irreducibly particular, it shares no real feature in common with any other thing. (This does not mean that haecceities cannot fall under the extension of a concept. Being an individuating feature is not a real property of a haecceity [it cannot be, since any haecceity is wholly simple, and shares no real features with any other thing]; but any concept of what a haecceity is certainly includes among its components being an individuating feature. A concept a haecceity includes representations merely of logical, not real, features of any haecceity.) >http://setis.library.usyd.edu.au/stanford/entries/ medieval-haecceity/ HereÍs the concluding paragraph from that Stanford Encyclopedia Note, of course, that ScotusÍs account of the common nature entails > something stronger than Adams is proposing: indeed, it entails > precisely the sort of minimal hypostatization that Scotus proposes. > And the reason for this, of course, is ScotusÍs view that individual > substances cannot themselves be primarily diverse -- a fact that is > explained by his claim that common natures have some sort of unity in > their instantiations: the nature in Socrates is (non-numerically) the > same as the nature in Plato. Natures, for Scotus, cannot be primarily > diverse; substances must include more than natures. But individual > natures in OckhamÍs view can indeed be primarily diverse, and this > surely amounts to a form of haecceitism -- nothing other than an > individual natureÍs own self-identity explains its distinction from > all other such natures. Maintaining that individual natures are > primarily diverse amounts not to having no theory of individuation, > but to accepting a form of haecceitism that, like AdamsÍs, does not > involve ontological commitment to the existence of real haecceities as > distinct real constituents of things. > Is the author of that one any relation to Mr. McTighe? > donÍt know but why does that matter [quidditical joking?] >http://plato.stanford.edu/entries/francis-marchia/ > --John === Subject: Re: quantum echo >Immortalist: Stop crossposting this to sci.physics.relativity > Are you saying it is possible for something to exist at location A [*SNIP*] Paul Teller having proposed, as an explanation for the divergence > between quantum statistics and classical statistics, that ïquantaÍ > lack haecceities, I posted to sci.physics.relativity what I took > to be the import of TellerÍs position for the logic of identity. > On this topic a fair amount has been written by philosophers of > science who, unlike myself, are knowledgeable about the formal > details of quantum mechanics. What these people have had to > say on the topic may be of interest to some on > sci.physics.relativity... --John the history of an idea is like the evolution of a species === Subject: Re: Max. Non-Adjacent Vertices on 120-cell don> WhatÍs the size of a largest subset of vertices on the 120-cell > ({5,3,3}) such that no two vertices in the subset are joined by > an edge? I have reason to suspect that the number is an integer square lew> I arranged the 600 vertices of the 120 cell into sets equidistant > from an arbitrary point, then looked at distances among these > sets and between them. There are 54 points in the equator between > the given point and its antipode, and these are the 15th nearest > neighbors to those two points. I can see the same arrangement in my data as well. lew> Anyway, doing it arbitrarily I got 193 points, so I suppose youÍre > aiming at 196. Is this right? 196 is indeed the target, and IÍve gotten about as close as you. (But IÍve had to adjust faulty logic in my algorithms a few times, so IÍve never quite trusted my results.) My latest approach is to start with a collection of 8 symmetrically- arranged 2-neighbors on a single dodecahedral cell, then to work my way out in shells of adjacent cells. An attempt at automating the random selection of 2-neighbors gave me worse results than starting at a single point (again, possibly faulty logic); manual selection seems to have shown me that there are good choices and bad choices to make at each stage, but after a couple of shells, I confuse myself :) so I havenÍt been able to codify what makes a good choice. Don === Subject: Re: Core error, FEAR is a natural response linux) > The other option is that JHÍs proof is *gasp* wrong. The fact that > he accepts the contradicting proof about the algebraic integers as > being valid actually directly implies that JH accepts his proof must > be wrong. Nonsense. There are such things as inconsistent theories. Mind you, James doesnÍt seem to grasp that (1) an incorrect de'nition cannot yield inconsistency[1] and (b) inconsistency is rather more troubling than 100 years of mathematics down the crapper. But the existence of two contradictory theorems does not imply that one of the proofs is invalid. Footnotes: [1] I am not considering a de'nition without referent to be incorrect. Rather, the error lies not in the de'nition but in the assumption that it is satis'ed. -- But in our enthusiasm, we could not resist a radical overhaul of the system, in which all of its major weaknesses have been exposed, analyzed, and replaced with new weaknesses. -- Bruce Leverett (presumably with apologies to Ambrose Bierce) === Subject: two envelope problem - root cause analysis Please share your remarks/opinions on the following root cause analysis of the two-envelope problem. (note: advanced detailed knowledge of the two-envelope problem is assumed in this post) As a starter two de'nitions: * De'nition 1 (game) A game G is a quadruple {A, S, payoff, p} in which A is a set of (player) actions, S is a set of (world) states, payoff is a function A x S -> IR, p is a probability measure on S. Lets assume for now (without loss of generality) that both A and S are countable (i.e. 'nite or countably in'nite). * De'nition 2 (proper game) A game G is proper in case the expected value of each action a in A, denoted by E(a), is well-de'ned. Note: - E(a) = sum [p(s)*payoff(a,s)] // here the sum is taken over the states s in S - E(a) is well-de'ned if and only if S is 'nite or the sum is absolute convergent - Absolute convergence is a stronger requirement than convergence (this is often overlooked when calculating expected values) According to probability theory (i.e. the law of large numbers) we know that the average payoff of an action a in a proper game G converges to E(a) in case G is repeated many times. Thats why current axiomatic decision theory (CADT) states that a (risk neutral) player of a proper game G should always select the action having the largest E(a) value. In fact CADT has nothing more to offer than this basic conclusion. Since the scope of CADT is limited to proper games we must always reason within the context of a proper game when dealing with the two-envelope problem. If we apply decision theoretic expected value (DTEV) calculations beyond the scope of proper games we are on illegal ground from the point of view of CADT. Well, the reason why (some versions of) the two-envelope problem seems paradoxal is simply because one of the following situations occur. 1. DTEV calculations are applied illegally, i.e. in a way such that an underlying proper game G cannot be de'ned. The author operates outside the CADT scope, 2. correct DTEV calculations are applied. However these calculations hold for another proper game than the game intended and/or suggested by the author. Most two-envelope papers run into (one of) these situations. The conclusion is that there are no probability nor decision theoretic paradoxal versions of the two-envelope problem. The only thing that is actually happening is that we are confronted with our currently too limited axiomatic decision theory (CADT) that is only able to deal with situations for which a proper game G can be de'ned. So the main goal should be to extend the scope of CADT by including decision theoretic what to do? theorems that are applicable beyond the context of proper games. Various approaches in this area have already been undertaken in the scienti'c community. Unfortunately a lot of decision theoretic situations cannot be mapped on a proper game G. Apart from (speci'c versions of) the two-envelope problem we can mention the famous St. Petersburg paradox. In this paradox the underlying game is not proper because E(a) is not convergent and thus not well-de'ned. Therefore, lets do for CADT what people before us did for naive set theory ( developing ZFC axiomatic set theory to extend the scope of naive set theory and to avoid misinterpretations such as the well-known Russell paradox ). === Subject: Re: Boolean Algebra - Arithmetic Relationship > IÍd like to recommend some reading. For logic generally, see Tarski An > Introduction to Logic and the Methodology of the Deductive Sciences > OUP. (Dreadful title, best book.) For recursion theory IÍm at a bit of > a loss, how about Boolos & Jeffrey Computability and Logic CUP? IÍll look for both of these at B&N. Are they fairly accessible to the Layman? I have a rather interesting hypothesis for you: The role of mathematics, as I understand it, is to create various logical structures that help predict/explain the nature of empirical phenonmenon. We link these structures to empirical reality by making assumptions. Assuming an atom behaves like the Bohr model, that assumptions are, they are always just approximations. The further you develop any one logical model less its behavior matches that of reality. Would it therefore be good argument, that focus of mathematical development should be on enhancing the breadth (Variety of Symbolic Systems / Logical Perspectives) and not the depth (Complexity of any one Symbolic System) of mathematics? Because no matter how great our assumptions are in the beginning, if we carry the logic far enough they wonÍt match reality. -Steve === Subject: Re: Max. Non-Adjacent Vertices on 120-cell %L;%tM$D+%zkQ$zp8f/vAx*mr6T79jgxh,SC!$,8.r%HBe}KZ)iMb$tB.Z,30 3QLpj-NoP*NzsIC,boYU]bQ]HÍy<#4ga3$21: > WhatÍs the size of a largest subset of vertices on the 120-cell > ({5,3,3}) such that no two vertices in the subset are joined by > an edge? Has nobody tried a search for independent-set 120-cell ? When I tried it I found http://www.csr.uvic.ca/~wendym/courses/582/120cell.ps Apparently the number of vertices is at least 220 and at most 224. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: circle with two centers In sci.math, riskbert a continuous, closed curve, C, such that both A & B are equidistant > from every point on the curve C. Erm, before we go any further: what embedding-dimension is the space? For example, I could postulate 3-D, and two points A and B. All points C in the space with distances AC = BC have to lie on the plane whose normal is along the vector AB and which contains the point M = (A+B)/2 -- the midpoint of the segment AB. If one stipulates a radius r > dist(AB)/2 one gets a circle on that plane, with center M and radius rÍ = sqrt(r^2 - dist(AB)^2/4). If youÍre using a distance-metric over the surface, that is of course a different problem (and I suspect thatÍs what youÍre really asking anyway). 1. Does such a surface exist? If so, under what conditions? As a trivial example, one could postulate two diametrically opposite points A and B on a spherical shell embedded in Euclidean 3-space and a family of circles equidistant from both points. > 2. If it does, can there be countably in'nite such centers? > Uncountably in'ntely many? If I understand your question correctly, the answer is yes. Take an arbitrary point A = (r,theta,phi) and its antipode (r, theta+pi, -phi), on a spherical shell of radius r. Since theta can range from [0,pi) and phi can range from (-pi, pi] (introducing thetas >= pi here will lead to doublecounting, although thatÍs not a major problem here), we get an uncountable in'nity of such double-center curves, even if we 'x the centers -- as one can parameterize the circles by specifying the distance d in the 3-D space from one of the points along the diameter of the sphere, leadig to a point X thereon, and then cutting the sphere with a plane perpendicular thereto, passing through X. I donÍt know whether you were looking for something more complicated, or what, but this works. :-) -riskbert -- #191, ewill3@earthlink.net ItÍs still legal to go .sigless. === Subject: Assignment Question If anyone can help with the following, please post a reply at http://www.assignmenthelp.com/viewtopic.php?t=8 I have the answers to this problem but I donÍt know how to get there. A child in danger of drowning in a river is being carried downstream by a current that has a speed of 2.50km/h. The child is 0.600km form shore and 0.800 km upstream of a boat landing when a rescue boat sets out. a) If the boat proceeds at its maximum speed of 20.0km/h relative to the water what heading relative to the shore should the pilot take? b)What angle does the boat velocity make with the shore? c)How long does it take the boat to reach the child? the answers are a)39.6 b)41.6 and c) 3.00 minutes Please post a reply at http://www.assignmenthelp.com/viewtopic.php?t=8 === Subject: RKutta order 2 or 4. How-to ? Do you kwow where I can found a smooth introduction to Runge Kutta methods (internet site). about a site for the non mathematician layman. === Subject: Re: Max. Non-Adjacent Vertices on 120-cell Hello ... s> if I understood it correctly the independence number [of the > 120-cell] lies in the interval [220, 224]. So it cannot be a > square. www.csr.uvic.ca/~wendym/courses/582/120cell.ps That does indeed seem to be the conclusion. However, the notes do include a need to formalize a proof of this result, so perhaps the upper bound really is 225. (Probably not.) === Subject: Re: quantum echo John: >Paul Teller having proposed, as an explanation for the divergence >between quantum statistics and classical statistics, that ïquantaÍ >lack haecceities, I posted to sci.physics.relativity what I took >to be the import of TellerÍs position for the logic of identity. >On this topic a fair amount has been written by philosophers of >science who, unlike myself, are knowledgeable about the formal >details of quantum mechanics. What these people have had to >say on the topic may be of interest to some on >sci.physics.relativity... The content of your thread contains no physics. My replies to one of those (who, I canÍt recall) also posting on this thread, indicates that whatever logic is being argued, totally neglects any physical meaning to the quantum mechanics. === Subject: Re: ab... = (a*b*...)^n ? | | >You get a lot further by examining only those | >numbers whose prime factors are less than B. | | >Still nothing in base 10 up to 10^200. | | > Would you describe further how you did that? ... | | The basic idea is to generate all numbers whose prime factors | are less than B. I initialize a set S={1}. I repeatedly remove | the smallest element e from S and for each prime p < B I | insert the product e p into S. This is especially fast if S | is implemented as a priority queue (see Knuth v3). | | An optimization to avoid inserting any element twice is to | insert only products e p such that p is the smallest prime | factor of e p. | | A further optimization for composite B is to avoid inserting any | product e p that is a multiple of B, since that number and its | multiples have zeroes in their base-B expansions. | | > Obviously, you didnÍt test each number to see | > whether its prime factors were less than 10, .... | | If there are k primes less than B, there are only O(n^k) | elements to process below B^n. If B is composite, the | second optimization reduces this to O(n^(k-1)). Either | of these is vastly smaller than B^n. Programming a priority queue (of bignums no less) is out of my reach, but it wouldÍve been fun to look beyond 10^200 ;o) === Subject: Laurent series ('nding the principal part) I have those two examples: f(z) = e^(-1/(z^4)) and f(z) = (e^z - 1)/(e^z + 1) and iÍd like to 'nd the principal part of their laurent expansions (the 'rst one about a=0, the second a=i*pi). I have tried all the metods (which is just one, sadly) i used on the other tasks but to no avail. Anybody willing to give a hint or (partial) solution? I tried to rewrite f(z) as a taylor series and then recognize the bad part and the holomorphic one. Then iÍd write the later as a power series looking like the part that iÍve extracted. But in the examples here, thereÍs no holomorphic part at all. Also, the bad part is not very nice either. Help! -- === Subject: Re: Core error, FEAR is a natural response > Given the people IÍve managed to contact about this particular > argument, who havenÍt found anything wrong or even claimed to 'nd > anything wrong, But they are not making any statements that you are right, right? > itÍs odd that to me that this Victor Eijkhout would > have the nerve to make his post. YouÍre right. ItÍs more in line for the other Victor Eijkhouts. I do come in six-packs, you know. > YouÍd think the poster chats casually with Barry Mazur, and Andrew > Granville on a daily basis!!! Nope. Not my 'eld. V. -- email: lastname at cs utk edu homepage: cs utk edu tilde lastname === Subject: Re: Does Goldbach imply Reimann thereÍs a proof of Twin Primes proving Goldbach, or vise versa, but I didnÍt actually read the thing in *Mathematics Mag.* It is possible prove the Ternary Goldbach Conjecture (TGC) and the Twin Prime >Conjecture (TPC) are true, if the Generalized Riemann Hypothesis (GRH) is >true. GRH implies twin primes? News to me. http://members.tripod.com/~american_almanac http://larouchepub.com/radio/index.html === Subject: Re: Factorial/Exponential Identity, In'nity ThatÍs a good point. Maybe what I need to have in the restriction on the interchange rules is that if you change a zero to a one then you have to change a one to a zero that is within a given 'nite distance of the one changed, for example q or q-1 or one for the set with p/q ones and (p-q)/q zeros. That might be too restrictive, I am to convert the sequence (01)... to 00001111(01).... 0101010101... 0011010101... 0010110101... 0001110101... 0001101101... 0001011101... 0000111101... Converting (01).. to (10)... is as follows: 01010101... 10101010... Each zero element can be changed to a one and each one within one element of that previous zero element is changed to a zero, and vice versa. Oh good, I have this idea within a few minutes of reading your post. Yet, that might only slow the process and then still I must show if that would not allow the conversion of any sequence (001)... through (1)... to all other sequences with in'nitely many ones and zeros. In your example you set for the sequence (01)... indexing from 1 each odd element to one and for each of those each multiple of fourth element to zero, getting (1110).... According to this rule, then, after changing an element from zero to one then youÍd have to change an element that is a one to a zero that is next to it. For example you could interchange each 4x-1 and 4xÍth elements, (0110).... Another problem with this restriction is that I need to show that it would allow any sequence of given contrived density to convert eventually to a sequence with the same known contrived density, as well that it would not convert to one of a different density. With this restricted sequence element interchange then 10000(0)... isnÍt convertible to 00001(0)..., instead it is to 01000(0)... which is to 00100(0)... which is to 00010(0)... which is to 00001(0).... How about (011) to (110)...? The list of intermediate sequences between them is in'nite. A rule to convert them all at once for each triplet is to convert (011) to (101) to (110). For the previous example the elements are interchanged in a two element sliding window. How about that? Basically what I want here is a description of a method to convert a sequence of a contrived rational density of ones and zeros to any other sequence with that same rational density but not any other, restricted sequence element interchange. Ross === Subject: Normal to a plan In 2-dimensions.. if I have a line such as the one below: Will the normal to this plane always be (0, 0)? How do I calculate the ignorance) Rick === Subject: Re: Contractible Spaces > If a topological space S is contractible to some point p in S, > is S contractible to every point in S? > Yes. Suppose S is contractible to a particular point s0 in S, with F: S x [0,1] --> S giving a contraction and s1 is any other point in S. De'ne G: S x [0,1] --> S by F(s,2t) if t in [0,1/2] G(s,t) = . F(s1,2-2t) if t in [1/2,1] Then G gives a contraction of S to s1. > If a topological space S is strongly contractible to some point > p in S, is S strongly contractible to every point in S? > Probably not in general, although I donÍt know offhand of any relevant example. There are theorems that say things like If A is a closed subset of X and both A, X are ANRs[*], then A is a deformation retract of X if and only if A is a strong deformation retract of X. Your question corresponds to the case A = {point} -- but itÍs not clear (to me, anyhow) just how to apply that theorem ... ThereÍs probably also an obstruction theory that applies to your question ... [*] ANR = absolute neighborhood retract -- I believe the above theorem actually applies to spaces that are ANRs for the class of compact metric spaces -- X is such a thing if any imbedding of X as a closed subspace of a compact metric space is a retract of some open neighborhood of the image ... > WhatÍs an example of a space that is > contractible but not strongly contractible? S is contractible to a when thereÍs some continuous h:Sx[0,1] -> S with > for all x in S, h(x,0) = x, h(x,1) = a, > and strongly contractible when in addition > for all t, h(a,t) = a ---- === Subject: 4th order Bspline I have a stupid question about the 4th order Bspline. In the paper( see link), they use summation to denote a B-spline. But why they use a vector (4 j)~T. Can anyone help me to understand it? http://www.ri.cmu.edu/pub_'les/pub2/wu_yu_te_1998_1/wu_yu_te_ 1998_1.pdf === Subject: Re: Normal to a plan > In 2-dimensions.. if I have a line such as the one below: Will the normal to this plane always be (0, 0)? How do I calculate the ignorance) Rick > You need 3d to get a normal to a plane. Any normal to a plane has to go at right angles to every line in the plane, but in 2d, the plane and all its lines are the whole thing and there is nowhere else to go. === Subject: Linear Algebra question IÍm not sure if/why this would be true or false. IÍm having a hard time understanding how I could explain that it was true. I would appreciate any help. Given a linear transformation T:F^n-->F^m. Is it true that there exists an m x n matrix A such that T(x)=Ax (where we think of x as an n x 1 matrix? === Subject: Re: Factorial/Exponential Identity, In'nity > ThatÍs a good point. Maybe what I need to have in the restriction on > the interchange rules is that if you change a zero to a one then you > have to change a one to a zero that is within a given 'nite distance > of the one changed, for example q or q-1 or one for the set with p/q > ones and (p-q)/q zeros. What is the point of all this? It will not provide next s in densely ordered sets, because an essential property of such dense ordering is that there are no next s. You are trying to create something like a 4-sided triangle, i.e., something inherently impossible because it contradicts its own de'nition. === Subject: BlissÍs Theorem Does anyone know of a statement of BlissÍs Theorem on a website? If so, will you please share the link with me? In case there is more than one BlissÍs Theorem, itÍs the theorem that it used to justify the arc length formula for parametric equations being a Riemann integral, when the derivation doesnÍt lead to a Riemann sum. John === Subject: Re: Boolean Algebra - Arithmetic Relationship IÍd like to recommend some reading. For logic gemmetric matrices (for complex matrix, symmetric is > hermitian symmetric). In the case that M is symmetric, try the > singular-value decomposition (or, what amounts to nearly the same thing, > an eigenvalue/eigenvector decomposition). > Hi Johan, be non-symmetircal. In that case, I am actually looking for an optimal approximation M=v1*v1Í+v2*v2Í+... in the minimum MSE sense... Is there any technique for that? -walala === Subject: Re: (matrix analysis)how to convert a matrix to the inner product of two vectors?