mm-2709 === Subject: Iterative mean I elaborate a lot of numbers. To find the average value I memorize all these values, sum them and then divide for the count. I need another way to do the mean. I'd like something like I can calculate the mean whitout memorize all the numbers in a vector.. Is it possible? === Subject: Re: Iterative mean if you are using a cicle like for i=1 to n...you can do in this way let's suppose that a is the object you want to save the mean in. a=0 for for i=1 to n do ... b=output of your elaboration at iteration i a=a+b/n end do === Subject: Re: Iterative mean The problem in your implementation is that I don't know 'a priori' the size of my vector.. === Subject: Re: Iterative mean allora tieni solo la somma, come vedo ti .8f gi.88 stato suggerito ...(e vedo che .8f pi.9d semplice se scrivo in italiano!!!) === Subject: Re: Iterative mean Given a set of values {x[1], x[2], x[3], ..., x[n]}, define a[i] as the average of all values from x[1] to x[i], so that a[n] is the average of all values in the set. a[i] = (x[1] + x[2] + ... + x[i])/i Let's obtain a[i+1] in terms of a[i] a[i] = (x[1] + x[2] + ... + x[i])/i a[i]*i = x[1] + x[2] + ... + x[i] a[i]*i + x[i+1] = x[1] + x[2] + ... + x[i+1] (a[i]*i + x[i+1])/(i+1) = (x[1] + x[2] + ... + x[i+1])/(i+1) (a[i]*i + x[i+1])/(i+1) = a[i+1] a[i+1] = (a[i]*i + x[i+1])/(i+1) The resulting formula can be described verbally as follows: Given an old average of i values, multiply the old average by the old total count of values which is i, add the new value that will form part of the new average, and divide by the new total count of values which is i+1, to obtain the new average. For example, if already I know that the average of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is 5.5, and I want to calculate the new average after I add 11 to the set, all I do is: Multiply the average 5.5 by 10 (the old total count of values), add the new value 11, and divide by 11 (the new total count of values) to obtain 5.5 * 10 = 55 + 11 = 66 / 11 = 6 which is the new average, that is, the average of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}. I don't even need to know the values in the old set, I only need to know its average and how many values are there. For example, let's say that I know that a set of 20 values average 7. If I add the value 49 to the set, the new average will be 7 * 20 = 140 + 49 = 189 / 21 = 9. Now all I need to remember is that I have a set of 21 values that average 9. Hope that helps. --Tech === Subject: Re: Iterative mean >The resulting formula can be described verbally as follows: >Given an old average of i values, >multiply the old average by the old total count of values which is i, >add the new value that will form part of the new average, >and divide by the new total count of values which is i+1, >to obtain the new average. Or perhaps better, Subtract the old average from the new value, divide by the new count and add to the old average. Mike Guy === Subject: Re: Iterative mean > I elaborate a lot of numbers. To find the average value I memorize all > these values, sum them and then divide for the count. > I need another way to do the mean. I'd like something like I can > calculate the mean whitout memorize all the numbers in a vector.. > Is it possible? Keep track of two numbers: (1) the number of values encountered so far; (2) the sum of all the values encountered so far. If you know in advance how many values are to be encountered, you don't need to keep track of (1). === Subject: Re: Iterative mean U reason man! tnx ;) === Subject: October caption competition Another month's caption to have a go at! You can also find last month's winners, and have a look at the current issue too. J === Subject: Re: Why do Christian schools have religious themed math books? <3q7jptFdiu2bU1@individual.net> P.S. Google Groups only lets you post to 5 groups at a time. Those of >> you who aren't in alt.atheism, free.christians, alt.bible, >> alt.talk.creationism, or sci.math, will not receive this reply. > One thing I hate about web based newsreaders. Yeah, five groups isn't anywhere near enough for the truly annoying. -- Jesse F. Hughes I don't know if you noticed but I had a tremendous drop in confidence concomittant [sic] with a dramatic grip of existential crisis. --- James S. Harris even has better diseases than you === Subject: Re: Two new (???) continued fractions for tan z and atan z Le 03-10-2005, didier.oslo@hotmail.com a .8ecrit : > Could you precise your notations? > I can't grasp the meaning of the > formulae in http://baruchel.thomas.free.fr/ > i.e. what means [a,bar{b,c}] ? OK, it is the notation used for repeating a pattern (usually with an increasing index, here it is n). [a,bar{b,c}] = [a, b_{n=0}, c_{n=0}, b_{n=1}, c_{n=1}, ...] see the notation used for Hurwitz numbers in the book of Perron. Of course you have understood that [a,b,c,d,...] = a + 1/(b + 1/(c + 1/(d + 1/( ... )))) -- Thomas Baruchel write to baruchel at the host called bluebottle dot com .8ecrire .88 baruchel chez l'h.99te nomm.8e bluebottle point com http://baruchel.thomas.free.fr/ === Subject: Re: Two new (???) continued fractions for tan z and atan z <43400ffd$0$23811$626a14ce@news.free.fr> <4340eb70$0$22403$626a14ce@news.free.fr> Try to compare your formulae with the continued fractions given in Abramowitz &Stegun 1965, Handbook of mathematical functions, #4.3.94 and #4.4.43 D. === Subject: Re: Two new (???) continued fractions for tan z and atan z <43400ffd$0$23811$626a14ce@news.free.fr Could you precise your notations? > I can't grasp the meaning of the > formulae in http://baruchel.thomas.free.fr/ > i.e. what means [a,bar{b,c}] ? [a,bar{b,c}] = [a,b,c,b,c,b,c,b,c,...] If b and c are functions of n, then a starting value of n is given (suppose it's 0); then [a,bar{b(n),c(n)}] = [a,b(0),c(0),b(1),c(1),b(2),c(2),...] This is standard notation for continued fractions; for instance, e-1 = [bar{1,1,2n}], starting with n=1. --- Christopher Heckman === Subject: Re: Group generated by two elements >On 2 Oct 2005 08:13:39 -0700, Julien Santini >The 2 proofs presented are similar and they are nice -- elegant even, >but to me, they are both somewhat unsatisfying. What do they teach? >That if you cleverly play around with identities, you can get new >ones? >Don't get me wrong -- it's a fine challenge problem just for fun, but >unless there's a more theoretical way to approach it, I think it's a >poor choice of problem for a preliminary exam. Possibly. But there are a number of interesting ideas from which this problem might have naturally arisen in the context of the course. Balanced presentations (same number of generators and relations) of the trivial (and more generally of finite) groups have for a long time been a topic of interest to group theorists. Probably a simpler o similar example of the same type was doen in the course. Also (and I suspect this is the real source of the problem), the group with just the first of the two relations < a,b | a^-1 b^2 a = b^3 > due to Baumslag and Solitar, was the first example found of a group G with a finite presentation, which is isomorphic to a proper quotient G/N (N != 1) of itself. >If the problem only requires manipulative skill and pattern matching >ability, then it could just as well be done by a machine. That is certainly true for specific examples. The first thing I did on seeing the problem was to check it on the computer, which of course announced within milliseconds that the group was indeed trivial. But it immediately suggests the more general problem, for which values of m and n is the group < a,b | a^-1 b^m a = b^n, b^-1 a^m b = b^n > trivial (or maybe finite). It might be a good idea to use a computer to get some evidence about this by trying small values of m and n, but it would be much more difficult to prove a general result of this kind with a computer. Derek Holt. === Subject: Re: Group generated by two elements > But it >immediately suggests the more general problem, for which values of >m and n is the group >< a,b | a^-1 b^m a = b^n, b^-1 a^m b = b^n trivial (or maybe finite). This can be solved using the methods elsewhere in this thread. They lead to the conclusion that a^(m-n) and b^(m^n) commute. But so do b^(m-n) and a^(m^n). So the group is commutative with order gcd(m, n). Mike Guy === Subject: Re: Group generated by two elements >> But it >>immediately suggests the more general problem, for which values of >>m and n is the group >>< a,b | a^-1 b^m a = b^n, b^-1 a^m b = b^n >trivial (or maybe finite). >This can be solved using the methods elsewhere in this thread. >They lead to the conclusion that a^(m-n) and b^(m^n) commute. >But so do b^(m-n) and a^(m^n). So the group is commutative with >order gcd(m, n). It is not always true that the group is commutative or finite. For example, if m=n, then the relations just say that a^m and b^m lie in the centre of G. So Z(G) is free abelian with generators a^m, b^m, and G/Z(G) is a free product of two cyclic groups of order m. Derek Holt. === Subject: Re: Group generated by two elements >>On 2 Oct 2005 08:13:39 -0700, Julien Santini >>The 2 proofs presented are similar and they are nice -- elegant even, >>but to me, they are both somewhat unsatisfying. What do they teach? >>That if you cleverly play around with identities, you can get new >>ones? >>Don't get me wrong -- it's a fine challenge problem just for fun, but >>unless there's a more theoretical way to approach it, I think it's a >>poor choice of problem for a preliminary exam. >Possibly. But there are a number of interesting ideas from which this >problem might have naturally arisen in the context of the course. >Balanced presentations (same number of generators and relations) of >the trivial (and more generally of finite) groups have for a long time >been a topic of interest to group theorists. Probably a simpler o >similar example of the same type was doen in the course. >Also (and I suspect this is the real source of the problem), the group >with just the first of the two relations >< a,b | a^-1 b^2 a = b^3 due to Baumslag and Solitar, was the first example found of a group G with >a finite presentation, which is isomorphic to a proper quotient G/N (N != 1) >of itself. >>If the problem only requires manipulative skill and pattern matching >>ability, then it could just as well be done by a machine. >That is certainly true for specific examples. The first thing I did on seeing >the problem was to check it on the computer, which of course announced >within milliseconds that the group was indeed trivial. But it >immediately suggests the more general problem, for which values of >m and n is the group >< a,b | a^-1 b^m a = b^n, b^-1 a^m b = b^n > Sorry, I meant < a,b | a^-1 b^m a = b^n, b^-1 a^m b = a^n > of course >trivial (or maybe finite). It might be a good idea to use a computer to >get some evidence about this by trying small values of m and n, but >it would be much more difficult to prove a general result of this kind with >a computer. >Derek Holt. === Subject: Re: help to solve this equation? <7323157.1128207978010.JavaMail.jakarta@nitrogen.mathforum.org Don't you want to know why you're being invited > to let Y(h) = r^h and not something else :-) Exactly - Ignatio claims, with a handwave, that the generality of this solution is obvious. But to someone seeing this for the first time it isn't (or shouldn't be!) obvious at all. === Subject: cool math magic party trick a) magician asks a random woman at a party to pick a number between 1 and 100, to tell nobody, and to memorize it. b) the magician tells the woman to reach deep into his pocket and pull out a piece of paper. c) the magician tells the woman to read what is on the paper. d) the woman says omigod! it's the number I was thinking about! How did the magician do this trick? === Subject: Re: cool math magic party trick >a) magician asks a random >woman at a party to pick a number between 1 and 100, to tell nobody, >and to memorize it. >b) the magician tells the woman to reach deep into his pocket and pull >out a piece of paper. >c) the magician tells the woman to read what is on the paper. >d) the woman says omigod! it's the number I was thinking about! >How did the magician do this trick? The most obvious explanation is that omigod! it's the number I was thinking about! is written on the paper. Derek Holt. === Subject: Re: Discontinuous Derivative > A dramatic example is this: > f(0)=0, > f(x) = x^2 * sin(1/x^2) for all non-zero x > Then f'(0) = 0, Don't you need f(0)=1 for f'(0) to exist? -- Esa Peuha student of mathematics at the University of Helsinki http://www.helsinki.fi/~peuha/ === Subject: Discontinuous Derivative <86p3bnjhsvg.fsf@sirppi.helsinki.fi A dramatic example is this: > f(0)=0, > f(x) = x^2 * sin(1/x^2) for all non-zero x > Then f'(0) = 0, > Don't you need f(0)=1 for f'(0) to exist? You need f(0) = 0 for f to be continuous. If f(0) = 1, then f isn't continuous and f'(0) goes poof. f'(0) = lim(h->0) (f(h) - f(0))/h = lim(h->0) f(h)/h - lim(h->0) f(0)/h = 0 + 0 if f(0) = 0 = 0 +- oo if f(0) = 1 === Subject: Re: Discontinuous Derivative >>A dramatic example is this: >>f(0)=0, >>f(x) = x^2 * sin(1/x^2) for all non-zero x >>Then f'(0) = 0, > Don't you need f(0)=1 for f'(0) to exist? No. taking f(0)=0, (f(h)-f(0))/h = h*sin(1/h^2) since |sin| is bounded by 1, |h*sin(1/h^2)| <= |h|, so the lim as h tends to 0 is itself 0. If you take f(0)=1, then (f(h)-f(0))/h = h*sin(1/h^2) - 1/h which isn't nearly as well behaved as h approaches 0. === Subject: help please Hi! I'm fighting with an integral that is probably easy to solve but i don't know why i can't find the solution! Can anybody help me? int [((x-a)/b)^2 -1][((x-c)/d)^2 -1]exp(-(1/2)*((x-a)/b)^2)exp(-(1/2)*((x-c)/d)^2) Giovanna === Subject: Re: help please > int [((x-a)/b)^2 -1][((x-c)/d)^2 > -1]exp(-(1/2)*((x-a)/b)^2)exp(-(1/2)*((x-c)/d)^2) Don't write it in such a complicated way. With y=(x-a)/b we get: R*int [(y^2 -1][((y-c)/d)^2 -1] exp(-(1/2)*(y^2+((y-c)/d)^2))) Here R and c,d are new constants. Completing the square in the exponent gives: R'*int [(y^2 -1][((y-c)/d)^2 -1] exp(-(1/2)*((y+c')/d')^2))) with R', c', d' new constants. Substituting y -> u= (y+c')/d' gives: R''*int [((u-a'')/b'')^2 -1][((u-c'')/d'')^2 -1]exp(-(1/2)*u^2)) With R'', a'', b'', c'', d'' new constants. Making the substitution u-> w=(1/2)*u^2 gives finally integrals in the form: r'''*int[w^k exp(-w)] with r''', k new constants. These are incomplete Gamma functions. Ciao Karl === Subject: Re: help please <43410d54$0$26235$9b4e6d93@newsread2.arcor-online.net> === Subject: Re: help please <43410d54$0$26235$9b4e6d93@newsread2.arcor-online.net> get in the second exponential exp(-(1/2)*((y-c')/d')^2)? x=bt+a =>exp(-(1/2)*b^2*((y-c')/d')^2)<>const*exp(-(1/2)*((y-c')/d')^2) am i wrong? Giovanna === Subject: Re: Mathematical induction. > suppose you have a proof by induction that some proposition P(n) is > true for all natural numbers. that's good. > now suppose that some proposition P(n) is true for all natural numbers. > is there a proof by induction for that? always? if not, can you give me > an example? and prove it (that there is no proof by induction)? Ah, and I've just posted such a _nice_ problem in a thread that is going to die out if nobody does something about it: Induction with a hard start Han de Bruijn === Subject: Re: Mathematical induction- > For each natural number n (i.e. n=1,2,3,...) let P(n) > be the statement: > x^(n+1) + y^(n+1) = z^(n+1) > has no non-zero integer solutions for x, y and z. > It is known that P(n) is true for each natural number n. > Can you exhibit a proof by mathematical induction? No. Modern proofs like Wiles' are in so-called court room style :-( But you should leave out the n=1 case, because: 3^2 + 4^2 = 5^2 . Han de Bruijn === Subject: Re: preliminary proof that primes of form (n^2)+1 are infinite Re: ... > The first number is prime of 101 but the next number of 10001 has prime > factors of 73,137 and the succeeding number of 1000001 has prime > factors of 101,9901. > Dik, what I want to know is whether every one of these composites has 2 > and only 2 prime factors? The answer is no. The very next one has the factors 101, 3541 and 27961. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: preliminary proof that primes of form (n^2)+1 are infinite Re: > The first number is prime of 101 but the next number of 10001 has prime > > factors of 73,137 and the succeeding number of 1000001 has prime > > factors of 101,9901. > > Dik, what I want to know is whether every one of these composites has 2 > > and only 2 prime factors? > The answer is no. The very next one has the factors 101, 3541 and 27961. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ of form (n^2)+1 are finite or infinite. Primes of form 10^(2k)+1 shows us how a form can be finite of primes because of the large supply of prime factors, ready and able to make all future candidates become composites. What happens to sequence of 101, 10001, 1000001.... probably happens to the sequence 37, 257, 677,.... and likewise the sequence 17, 197, 577, .... and likewise the sequence 101, 401, 1601,.... only they do not happen as fast as 10^(2k)+1. Which is all very good news for my method of Math Induction coupled with Euclid's IP. If primes of form (n^2)+1 is a finite set then it has a proof that it is finite. The Math Induction coupled with Euclid IP would be such a proof, in that it is impossible to arrange or construct case#1 for the initial step of Math Induction. It is impossible to have two of these primes of form (n^2)+1 multiply them together, then multiply by a factor of 2 then add/subtract 1 and get another new prime of form (n^2)+1. It is the Algebraic form of (n^2)+1 that makes it impossible to form the first step of Math Induction. And since we cannot do a Math Induction on primes of this form implies the set is finite. Math Induction does not point to where the last and final prime of (n^2)+1 occurs. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Is the set N of natural numbers well defined? > Limits can be defined in terms of topologies without > need of metrics. > For a sequence of reals, for example, one uses the > cofinite topology on > the set of naturals and the topology generated by > open intervals on the > reals. At this stage, I have a preliminary observation.The definition of the cofinite topology on N presupposes the existence of N;so it is doubtful if one could use that in the limiting process used to assert the existence of N and to define N. All that is available to us is the recursion k={1,2,. . .(k-1)}.If a limit w exists, we would have w= {1,2,3.. w}. -Apoorv === Subject: Re: math disput btwn friends (dimension of vector spaces) >In message , David C. >>What? There is, I thought, one-count-them-one embedding >>of R into C as a subfield. >No, there are infinitely many, unless you assume continuity >or a specific automorphism of C of order 2 (a.k.a. complex >conjugation). You can't identify R as a subfield of C >using only field operations. >>heh-heh. You math guys like to make things sooo complicated. >>Where I come from we don't need no steenking embeddings, >>R _is_ a subfield of C, and hence a complex vector space >>of dimension n _is_ (ok, is, if you want to be picky) a >>real vector space of dimension 2n. >Pertti's ghost is smiling smugly. >>RIP. You know you're really supposed to send Rudolph >>a letter with the offending statement underlined in >>red. (I don't recall what the next step is if he fails >>to acknowledge the error of his ways...) >R X R X R is a subfield of C X C Uh, no it's not. Neither is a field. (At least neither is a field using natural operations...) >but dim(R^3) isn't 2n... ************************ David C. Ullrich === Subject: Re: math disput btwn friends (dimension of vector spaces) <3q2t3pFd25v3U1@individual.net> <0+OXl0DVYGQDFwTU@jboden.demon.co.uk> <3252k11790mu1v6tjf6mk6np0kjsqsvmtc@4ax.com> In message <3252k11790mu1v6tjf6mk6np0kjsqsvmtc@4ax.com>, David C. >>In message , David C. >What? There is, I thought, one-count-them-one embedding >of R into C as a subfield. >>No, there are infinitely many, unless you assume continuity >>or a specific automorphism of C of order 2 (a.k.a. complex >>conjugation). You can't identify R as a subfield of C >>using only field operations. >heh-heh. You math guys like to make things sooo complicated. >Where I come from we don't need no steenking embeddings, >R _is_ a subfield of C, and hence a complex vector space >of dimension n _is_ (ok, is, if you want to be picky) a >real vector space of dimension 2n. >>Pertti's ghost is smiling smugly. >RIP. You know you're really supposed to send Rudolph >a letter with the offending statement underlined in >red. (I don't recall what the next step is if he fails >to acknowledge the error of his ways...) >>R X R X R is a subfield of C X C >Uh, no it's not. Neither is a field. (At least >neither is a field using natural operations...) >>but dim(R^3) isn't 2n... >************************ Sorry - I think I meant to say subspace not a subfield. But I'm not at all sure it's true to say that R^3 is isomorphic to a subspace of C^2... -- Jeremy Boden === Subject: Re: Lipschitz function >>there's a smallest Lipschitz constant k, in the sense that |f(x1) - >>f(x2)| <= k*|x1 - x2| for every x1 and x2 in D and, for every k' < k, >>there exists y1 and y2 in D such that |f(y1) - f(y2)| > k'*|y1 - y2|. >>It's also true that, for every eps>0, there are x1<>x2 in D such that >>||f(x1) - f(x2)/|x1 - x2| - k| < eps. >>I got the following statement: >> What do you mean when you say you got this statement? >>If D is compact, then there are x1 and >>x2<>x1 in D such that |f(x1) - f(x2)| = k*|x1 - x2|. I tried to prove >>this considering the function g(x1,x2) = |f(x1) - f(x2)/|x1 - >>x2|defined for {x1, x2 in D | x1<>x2}, but this set isn't compact. So, >>though sup g = k, the conclusion doesn'follow from this. I'd like some >>help, if the statement is really true. >It's not an unreasonable conjecture. Certainly not - I didn't say it was. In fact I'm not going to tell you the latest date at which I thought it was actually true (never having thought much about the proof.) > It seems like a nice idea; >unfortunately, the fact is false. Perhaps she was given it as a >homework assignment, or came up with it on her own. Or perhaps it was homework, but labelled prove or disprove - that was my guess. Seems like the appropriate reply depends on which of these is the case, hence my request for her to clarify what she meant when she said she got this statement. > (If a homework >assignment, I'm not too impressed by her professor; if she came up with >it on her own, then she's beginning to think like a mathematician, and >I'm impressed.) >A pseudo-disproof: it's true, for every k' < k there is a pair y1 ne >y2 with |f(y1) - f(y2)| > k' |y1-y2|. You might say, therefore, Aha! >Choose sequences u_n and v_n with > |f(u_n) - f(v_n)| >= k_n |u_n - v_n|, u_n ne v_n, >for some sequence k_n which strictly increases to k, and argue on >SUBSEQUENCES of u_n and v_n. But what can happen is that while these >subsequences will converge to elements u, v with > |f(u) - v(v)| >= k |u-v|, >we can no longer guarantee that u ne v. They could converge to the >same thing. >This is a pseudo-disproof because it only points out what could go >wrong. Pointing out a difficulty in a proof doesn't invalidate the >result the proof is trying to prove. But with the hint of what COULD >go wrong, we immediately construct a counterexample: the one I posted >before, namely, f(x) = sin x on the interval [0,1]. By the mean value >theorem we have > |f(x)-f(y)| = |cos(z)| |x-y| <= |x-y| (some z between x and y), >so the Lipschitz constant is <= 1. In fact, cos(z) is almost always < >1 in absolute value, and it's not hard to see that > |sin(x) - sin(y)| < |x - y| (x ne y). >Yet |sin(x) - sin(0)|/|x-0| is as close to 1 as we wish, because >sin(x)/x --> 1 as x --> 0. >--Ron Bruck ************************ David C. Ullrich === Subject: Re: obvious fact about volume >Let A be what Rudin calls a k-cell, that is, a product of k intervals. A >is >a subset of R^k. >Suppose that for each i=1,..,n, A_i is a product of k _open_ intervals, >the >collection >{A_i} is disjoint, and the union of A_i is a subset of A. >Then sum{i=1,...,n}vol(A_i) <= vol(A), where vol is the obvious product of >the lengths of the intervals. >This is almost painfully obvious, but I cannot see a pleasant way to prove >it. >> First consider the special case of what you might call trivial >> decompositions, and then deduce the general case using a >> refinement: >> Take k = 2. Say A = (a,b)x(c,d). Suppose you have numbers >> a = x_0 < ... < x_n = b >> c = y_0 < ... < y_m = c. >> Define A_j,k = (x_{j-1}, x_j) x (y_{k-1}, y_k). >> The A_j,k form what I'm going to call a trivial decomposition >> of A. >> Special Case: If the A_j,k are a trivial decomposition >> of A then vol(A) = sum vol(A_j,k). >> Proof: The distributive law. QED. >> Now in general, say A is a rectangle containing >> disjoint subrectangle A_1, ... A_L. Form a trivial >> decomposition of A by using all the x values that >> arise as a vertical side of any A_j and all the >> y values that arise as a horizontal side of any A_i. >> Then the A_j,k are a trivial decomposition of A. >> And also for each i there is a _subset_ of the >> A_j,k which form a trivial decomposition of >> A_i. Apply the Special Case L+1 times, once >> to the trivial decomposition of A and once >> to the trivial decomposition of each A_i. >Disjointness tells you that different A_i's use up disjoint parts of the sum >from the trivial decomposition. Yes, that's a detail I left to you. And so the result follows... >I was worried that when you removed A_i, there might not be enough left in >the sum >to cover the reamaining A_j's. Not sure what you mean... Oh: You mean until you realized what you just said you were worried that things might overlap, so you'd need one A_j,k for part of two different A_i's? After posting _I_ was worried that you were going to misread my And also for each i there is a _subset_ of the A_j,k, since it wasn't all that well put. Of course I meant a subset of the collection of all the A_j,k; if we had to divide a single A_j,k into subsets the argument wouldn't work. ************************ David C. Ullrich === Subject: Re: obvious fact about volume >>I was worried that when you removed A_i, there might not be enough left in >>the sum >>to cover the reamaining A_j's. > Not sure what you mean... Oh: You mean until you realized > what you just said you were worried that things might > overlap, so you'd need one A_j,k for part of two different > A_i's? > After posting _I_ was worried that you were going to misread > my And also for each i there is a _subset_ of the > A_j,k, since it wasn't all that well put. Of course I meant > a subset of the collection of all the A_j,k; if we had to > divide a single A_j,k into subsets the argument wouldn't > work. I did not misread what you said, I understood the idea. While you are here...do you have any helpful comments to make about Rudin's construction of Lebesgue measure? I'm having trouble dealing with several aspects of it. For instance, vol(E_r) -> Vol(W) (p. 52). It seems intuitively true that if you fill up W with smaller and smaller boxes, the volume of the union will approach the volume of W. But I find volume to be an unwieldy beast to work with, as demonstrated previously. or how about f < W implies that I(f) <= Vol(W), where I is the linear functional constructed in the proof. This is again intuitively clear, but the definition of I is unpleasant. I actually found the Riesz Representation Theorem much easier to follow. > ************************ > David C. Ullrich === Subject: Re: Lebesgue Integral Convergence Let f_n converge to f a.e. Let Int[A](f) denote the integral of f over A. Assume Int[R](f_n) converges to Int[R](f), which is finite. Show that for every measurable set E, we have Int[E](f_n) converging to Int[E](f). Hint: This follows from Fatou's Lemma. You need to apply the lemma twice, once to Int(E)f_n and once to Int(???)f_n. ************************ David C. Ullrich === Subject: Re: Lebesgue Integral Convergence >>Let be a sequence of nonnegative measurable functions defined on R. >>Let f_n converge to f a.e. Let Int[A](f) denote the integral of f over A. >>Assume Int[R](f_n) converges to Int[R](f), which is finite. Show that for >>every measurable set E, we have Int[E](f_n) converging to Int[E](f). > Hint: This follows from Fatou's Lemma. You need to apply the lemma > twice, once to Int(E)f_n and once to Int(???)f_n. I'm only getting that lim inf (Int(E) f_n) = Int(E) f. without fatou for lim sups, I don't see how to make this work. > ************************ > David C. Ullrich === Subject: Re: Lebesgue Integral Convergence >Let be a sequence of nonnegative measurable functions defined on R. >Let f_n converge to f a.e. Let Int[A](f) denote the integral of f over A. >Assume Int[R](f_n) converges to Int[R](f), which is finite. Show that for >every measurable set E, we have Int[E](f_n) converging to Int[E](f). >> Hint: This follows from Fatou's Lemma. You need to apply the lemma >> twice, once to Int(E)f_n and once to Int(???)f_n. > I'm only getting that lim inf (Int(E) f_n) = Int(E) f. > without fatou for lim sups, I don't see how to make this work. >> ************************ >> David C. Ullrich Nevermind. I got it. === Subject: Re: confused about sequence problem. >> So now that your counterexample is broken, try to >> prove the original >> statement. >> quasi >oh, i didnt think of that one. ya, you are right. >now about proving, If n is a positive integer then sup(S) - 1/n < sup(S). Hence... >i cant even use Bolzano-Weierstrass >thm since its not known whether S is bounded . because if >S is bounded and infinite then i can argue that there exists at least one accumulation point ( cluster point ). >and somehow i can relate this point to the limit of some >sequence. >wood ************************ David C. Ullrich === Subject: Re: Serge Lang dead Below is my rejected letter to the New York Times regarding Serge's 9/25 obituary. Sean To the Editor Re: Re: Serge Lang, 78, a Gadfly and Mathematical Theorist, Dies A fact. An opinion. A hole in the ground. To hold the distinction between these three types in whatever arena they might appear, and demand that his students do so as well, was Serge Lang's constant business. He may have hurled chalk by way of punctuation, but that was merely a bonus. I am tempted, in reviewing his obituary, to object to the manner in which serial details align toward the impression that he was a screaming madman, incidentally blessed with a mathematical brilliance that withered when exported to the realms of politics, journalism, or other, softer fields. Rather a large charge, so let me share one opinion that changes nothing: from first meeting him as a Yale undergraduate to the days before he died, I regarded Serge Lang as the warmest and most intellectually generous teacher I have ever known. Those with a more fluent command of mathematics can speak to his specific greatness, but it is hilarious to read that Serge came to be better known for his academic jousts with nonmathematicians on social and political issues than for his work in geometry and the properties of numbers. Better known by whom? Mathematicians? Editors at the New York Times? The lay public clamoring for the hot new thing in algebraic geometry? You write that political scientist Samuel P. Huntington was the focus of Dr. Lang's ire, contributing to the impression that Lang's political engagements were byproducts of eccentricity, zealotry, In his book Political Order in Changing Societies, Huntington classified apartheid-era South Africa as a satisified society. Following Lang's scrupulously documented challenges to this claim, among others, Huntington said the following in a 1987 interview published in the New Republic magazine: The term 'satisfied' has to do with whether or not there are measurable signs that people are satisfied or not with their lot. That lot may be good, fair, or awful; what this particular term is describing is the fact that the people for some reason are not protesting it. When this study...was done in the early 1960's, there had been no major riots, or disturbances [in South Africa]... Times readers may evaluate to what extent Huntington can distinguish, as Lang would say, a fact from a hole in the ground. One of his former students has reminded me that Serge would often flag an important theorem by saying you should absorb it so comprehensively as to remember it three days after you died. Sometimes it was seven. Members of his cc list were, as it happened, receiving his latest mailing several days after his end. Who knows what may be coming next? Suspense. -Sean Westmoreland New York, New York === Subject: Re: Serge Lang dead regarding Serge's 9/25 obituary. > [...] I've heard that highbrow US papers expect contributors, including letter writers, to adopt a staid and measured prose style throughout, verging on soporific, perhaps in a desire to avoid seeming partial or sensationalist. Also, in my (UK) experience, if a letters editor sees a sentence of more than ten words they have to go and lie down for half an hour in a dark room to recover! Some years ago, around the time of a national election, an election (which the incumbent party can do in the UK) in view of oscillating opinion polls is like firing a cannon in one of those old square rigged ships - The leader of the incumbent party may sometimes have to ask for Parliament to be dissolved when their polls are dipping, in the hope that by polling day a couple of weeks later the polls will have risen back to a maximum in their favour! But after the letters editor had got their inky fingers on the text and chopped it up into bite-sized sentences, the letter as published looked like someone had fired a 24 pounder through it! John R Ramsden (jhnrmsdn@yahoo.com.uk) remove m from com before replying === Subject: GRAVITY AND THE PHOTON SUBJECT: GRAVITY IS NOT A FORCE PLANETS ORBIT THE SUN TO CONSERVE TOTAL ENERGY THE FORCE OF GRAVITY IS AN ILLUSION ISBN 0-9644267,LIBRARY OF CONGRESS CAT.CARD NO.94-90554 Gravitational effect is the result of an acceleration of mass. Galileo demonstrated this. Newton assumed that this was caused by a force of gravity between all masses. Was this a correct assumption? Einstein and many other scientists felt that there must be more to gravitation than an attraction at a distance. Action at a distance was considered to be impossible in the absence of a transfer of energy at the speed of light. Hubble then showed that the distant Galaxies were moving away from the earth and that the universe was expanding in all directions. If this is true , What else must be true? 1. The potential energy of the rest of the universe must be decreasing relative to the mass of the earth. It has long been assumed that the first law of thermodynamics, which says that the total energy of the universe is a constant, was a fact of nature. If this is true what then. 2. The kinetic energy of the universe must be increasing at the same rate that the potential energy is decreasing as the universe expands. How is this possible? Masses must be accelerating, because, kinetic energy change is the result of an acceleration. But all orbital masses are accelerating toward the center of the earth or some other mass. Why would this occur otherwise? 3. Orbital motion could then be the result of the expansion of the universe. The Gravitational illusion could be the result. Based on the first law of thermodynamics The total mass energy of the universe is a constant. (total kinetic (mass) energy plus total potential energy is a constant). m(2 pi L)^2 / t^2 + G (M-m)m / L = A constant. m is any mass say that of the earth. Delta m (2 pi L)^2 / t^2 = - Delta G (M-m)m/L follows mathematically. Delta m 4 pi^2 L^2 /t^2 = Delta - G (M-m)m / L^2 or the modified Newton equation for gravity can be derived,but only when L is the orbital distance. The earth orbit is a result of an energy equilibrium, ( the absence of a change of total energy ) and not the result of a force of gravity between masses. Force of gravity is the resulting illusion assumed by Newton to be a force. If a planet (say earth) moved away from the sun its potential energy would decrease as L increased. Its kinetic energy would decrease because it is no longer accelerating toward the sun in orbital motion. Total energy would have to decrease. A very great change of total energy would have to take place. POTENTIAL ENERGY = G(M-m)m/L KINETIC ENERGY = m(2 pi L)^2/t^2 m(2 pi L)^2/t^2 + G(M-m)m/L = A constant = M G= Gravitational constant; M = total energy of the universe (or effective universe) ; m = mass in question. t = time ; L = radial distance. No mechanism exists for this to occur rapidly. So it could not happen. The magnitudes of kinetic and potential energies of planets and moons travelling in orbital motion are equal and any increase or decrease of orbital distance L results in an equal change in magnitude of both.This is the only value of L where no change of total energy will occur if the value of L changes. At any other distance L, an increase of kinetic energy will be at a different rate than potential energy decreases. Orbital motion conserves total energy. Force of gravity isn't needed to explain orbital motion or any other motion at a distance. GRAVITY MECHANICS AND RESEARCH ON ASTRONOMICAL OCEAN TIDES Copyright 1984 to 2002 Allen C. Goodrich An examination of United States Coast and Geodetic Survey Tidal Data, which was gathered by extensive measurements over long periods of time,was compared with astronomical data showing the phases of the moon at corresponding times for many years. This correlation of the two sets of data revealed a very interesting fact, in a manner that had never before been mentioned in the literature. It is invariably and exactly the lowest tide that exists directly under the full and new moons at deep ocean ports. TABULATED co-op.nos.noaa.gov and space.jpl.nasa.gov DATA: OCEAN TIDES AND PHASES OF THE MOON AT DEEP OCEAN PORT- MYRTLE BEACH LOWEST TIDE (YEARS 1992 AND 1993) 1992 FULL MOON---1992 NEW MOON (at moons highest point in the sky) DATE---TIME(std)-DATE---TIME(std) Mar.18--12:00Mid-Mar.3---12:00Noon Apr.17--12:00Mid-Apr.2---12:00Noon May.17--12:00Mid-May.2---12:00Noon Jun.15--12:00Mid-Jun.29--12:00Noon July.13-12:00Mid-July.29-12:00Noon Aug.12--12:00Mid-Aug.27--12:00Noon Sept.11-12:00Mid-Sept.26-12:00Noon Oct.11--12:00Mid-Oct.26--12:00Noon Nov.10--12:00Mid-Mov.25--12:00noon Dec.10--12:00Mid-Dec.25--12:00noon 1993 FULL MOON---1993 NEW MOON (at moons highest point in the sky) DATE---TIME(sdt)-DATE---TIME(sdt) Jan.8--12:00Mid--Jan.24-12:00Noon Feb.6--12:00Mid--Feb.21-12:00Noon Mar.8--12:00Mid--Mar.23-12:00Noon Apr.6--12:00Mid--Apr.21-12:00Noon May.6--12:00Mid--May.20-12:00Noon Jun.4--12:00Mid--Jun.19-12:00Noon July.3-12:00Mid--Juy.18-12:00Noon Aug.2--12:00Mid--Aug.17-12:00Noon Sep.1--12:00Mid--Sep.16-12:00Noon Sep.30-12:00MId--Oct.15-12:00Noon Oct.30-12:00Mid--Nov.14-12:00Noon Nov.29-12:00Mid--Dec.13-12:00Noon Dec.28-12:00Mid--Jan.12-12:00Noon This was a very interesting discovery because current physics,based on the gravitational theory, discussed in the following U.S.Gov. documents: PREDICT THE OCEAN TIDES http://co-ops.nos.noaa.gov/restles1.html SEE PHASES OF THE MOON FROM EARTH http://space.jpl.nasa.gov/ ,would lead one to believe that,except for many possible reasons, the highest tides tend to be under the full and new moons. The dictionary and encyclopedia as well as physics texts predict this with pictures of the earth and oceans bulging on the side facing the full moon. Of course it never happens as the gravitational theory predicts, and many reasons are given for the discrepancies. CONCLUSION: No discrepancies were found in the occurence of exactly the lowest tide directly under the full and new moons, at deep ocean ports. A lowest tide also occurs on the earth's ocean directly opposite the new and full moons. SIGNIFICANCE: One must admit that this is beyond question one of the most important discoveries of modern physics research. It indicates that a change must be made in the theory of gravitation. One can no longer assume that a force between the moon and the water of the earth's oceans, is causing the ocean tides. The force of gravity must be an illusion caused by some other, more basic, reason. What would this be? If the total energy ( kinetic and potential ) of the universe is assumed to be a constant,from this fundamental equation, many interesting things follow. If the rest of the universe is expanding ( potential energy decreasing) relative to masses, the masses must be shrinking ( increasing in kinetic energy ) (gravitation) relative to the rest of the universe. THE FIRST LAW OF MOTION-(GOODRICH) Copyright 1984 to 2002 ALLEN C. GOODRICH A body (m) continues in a state of rest (equilibrium) or motion in a straight or curved line (equilibrium) as long as no change occurs in its total (kinetic and potential) energy, relative to the rest of the effective universe (M-m), Delta m(2 pi L)^2/t^2 = - Delta K(M-m)m/L equilibrium = no change in the total energy relative to the rest of the effective universe (M-m). ^ = to the power of. Orbital motion complies with this equation. This equation is derived from the fundamental equation of the universe which states that the total energy of the universe is a constant. The sum of kinetic and potential energies is a constant. m(2 pi L)^2/t^2 + K(M-m)m/L = A constant. INERTIA AND MOMENTUM are the properties of a mass that evidence its reluctance to change its total energy, or it is its need to maintain a constant total energy. If it could more easily obtain or lose energy, it would have less inertia or momentum. SEE THE UNIVERSE- A GRAND UNIFIED THEORY OF MASS ENERGY SPECTRUM OF THE BUFFALO ASTRONOMICAL ASSOCIATION INC. NOV.1996 TO FEB.1997 See http://ourworld.cs.com/gravitymechanic2/myhomepage/business.html http://ourworld.cs.com/gravitymechanic2/myhomepage/profile.html TIDES AND GRAVITY MECHANICS http://ourworld.cs.com/gravitymechanic2/myhomepage/resume.html A new theory of gravitation is given, which predicted, stimulated the above research,and is consistent with, the new findings. The universe has been found to be expanding at an accelerating rate as predicted in 1984 by this new theory. === Subject: Solving equations (X^2-V^2)*X - Y*U^2 + 19*U^3*V = 0 16*X*V*U^3 + (X^2-V^2)^2 - 25*U^6 = 0 What are the solutions of the above equations such that X and Y are rational numbers not equal zero U and V are complex number === Subject: Re: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~ ~~~~~~~~ In sci.logic, |-|erc <4340cc8f$0$30203$892e7fe2@authen.white.readfreenews.net>: > NEVER SAY DIEEEEEEEEEEEEEEEeee You just did. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: MCAD 13 upgrade This is just an FYI. I bought MCAD 12 and it was very unstable on my XP machine. Found out there is a free or low-cost upgrade to 13 as below: Owners of the following Mathcad products - who purchased their software after August 15, 2005 - are eligible to receive free or low-cost Mathcad 13 upgrades: .87 Mathcad 12 Single-User New Edition (receive a free software upgrade; you pay only shipping and handling charges; see instructions below) .87 Mathcad 12 Single-User Upgrade Edition Tier 1 (pay a $90 software upgrade cost plus shipping and handling charges; see instructions below) .87 Mathcad 12 Single-User Upgrade Edition Tier 2 (pay a $120 software upgrade cost plus shipping and handling charges; see instructions below) .87 Mathcad 12 Single-User Priced for Education Edition (receive a free software upgrade; you pay only shipping and handling charges; see instructions below) fyi, russ === Subject: Re: THE COMPLETENESS THEOREM > Completeness Theorem : a theorem (godels statement G) exists in every consistent system# > that is only valid in a distributed framework. That doesn't sound like any completeness theorem I've ever run into. What does distributed framework mean in the above? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: THE COMPLETENESS THEOREM : > Completeness Theorem : a theorem (godels statement G) exists in every consistent system# : > that is only valid in a distributed framework. : : That doesn't sound like any completeness theorem I've ever run into. : What does distributed framework mean in the above? : a non-singleton set of formal theorem provers. Each individual theorem prover has a set of theorems it can prove. TP1 proves S1 TP2 proves S2 .. TP1 can't prove the godel_statement_written_in_TP1 (tp1g) TP2 can't prove the godel_statement_written_in_TP2 (tp2g) ... ~(tp1g e S1) ~(tp2g e S2) ... TPx may be able to prove the tpyg, as long as x=/=y. tp1g e S2 (TP2 is a meta language of TP1) Assuming there exists a metalanguage of TP1 S1 C (S1 U S2...) S1 is a proper subset, incomplete! Therefore a non-singleton set of theorem provers (a distributed framework) proves a larger set of formula than a centralised theorem prover. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ You didn't acknowledge this as an equivalent Godel proof. : This sentence has no proof in TNT. : : if its false, it doesn't have no proof in TNT, [1] : it has a proof in TNT [2] : its true [3] : it has no proof in TNT [4] : contradiction [5] : : if its true, it has no proof in TNT [6] : : conclusion: it can't be false, therefore its true. [7] : : TNT is a formal system developped by Hofstadter in his book; TNT : stands for Typographical Number Theory. What you're doing is using a distributed set of individual partial theorem provers to ascertain that 1 G is true, and 2 G, being true, does indeed ascertain that some formula require a distributed set of individual partial theorem provers. Herc === Subject: Re: Q: algorithm for group intersection Content-Length: 2327 Originator: rusin@vesuvius >If I'm posting to the wrong newsgroup, please feel free to point me in >the appropriate direction. >I wonder, what is known about the worse time complexity of algorithms >for permutation group intersection? More precisely, what is the best >(proven) running time of an algorithm solving the following problem: I believe that the best currently known bound for the complexity of intersections of subgroups in permutation groups is n^{c sqrt(n)} = exp{c sqrt(n log n)}. This is proved in L. Babai, W.M. Kantor, and E.M. Luks, Computational complexity and the classification of finite simple groups, Proc. 24th IEEE Symp. on Foundations of Comp. Sci. (9183), 162 - 171. The authors refer to this type of bound as moderately exponential rather than subexponential. The same complexity can be proved for a number of other problems about permutation groups, including centralizers of elements, testing conjugacy of elements, and set stabilizers. I think this also at least approximately the best bound known for the complexity of graph auto- and isomorphism. It is unknown whether the decision versions of any of these problems are polynomial or NP-complete. They are all regarded as plausible candidates for problems that are strictly in between P and NP-complete. >INPUT: >Let G and H be two permutation groups on {1,...,n} given >with their bases and strong generating sets. >OUTPUT: base and strong generating set of the intersection >Gcap H of groups G and H. >I know that the problem is as difficult as the graph isomorphism >problem, so I am curios in non-polynomial algorithms for this problem. >For example, are there known sub-exponential algorithms for this >problem? It is known that graph isomorphism is polynomial-reducible to subgroup intersection (hope I have that the right way round) i.e., if subgroup intersection is polynomial, then so is graph isomorphism. But it is not immediately clear to me that this implies that subgroup intersection is at least as difficult as graph isomorphism. It seems conceivable (although I am no expert in complexity theory) that subgroup intersection could be say exp( n^{1/3} ) and graph isomorphism exp( n^{1/2} ). Derek Holt. === Subject: Re: Continuity of spacetime - Dammit people... Come to think of it, this could account for grainyness of fine structure. Also, it seems that if the space which separates some collection of points is nonexistent, then this is pointing toward continuity. I did'nt think that it would be possible to argue taht spacetime is continuous, but I now believe that it is possible - but you need enough points to have a continuum. If you have a set of points which is uncountably infinite and satisfies all the neccesary requirements from analysis, AND the space which separates those points is nonexistent, then you should have continuous spacetime manifold. This statement could be presented more clearly - just a thought. === Subject: Re: The rate of free fall is g/2 <3q0lskFch263U1@individual.net Consider this: > Given: > v = a*t > s = (a/2)*(t^2). Then > v = 2*s/t > Thus it would seem that the rate of fall is independent of 'a'? > That's a bum steer. Where'd v=at come from? It's the velocity as a function of time for constant acceleration. An acceleration of 32 ft/sec^2 means that the velocity increases by 32 ft/sec every second. That is expressed in v = at with a = 32. - Randy === Subject: Re: The rate of free fall is g/2 <3q0lskFch263U1@individual.net Consider this: > Given: > v = a*t > s = (a/2)*(t^2). Then > v = 2*s/t > Thus it would seem that the rate of fall is independent of 'a'? > That's a bum steer. Where'd v=at come from? > It's the velocity as a function of time for constant acceleration. Velocity is changing; whether the acceleration is constant or not. > An acceleration of 32 ft/sec^2 means that the velocity increases > by 32 ft/sec every second. That is expressed in v = at with > a = 32. > - Randy Average aceleration is expressed as [(vt-vi)/t=2s/t^2]; so [a/2=s/t^2]. Where s is the displacement distance; caused by a net force. Don === Subject: Re: The rate of free fall is g/2 <3q0lskFch263U1@individual.net Consider this: > Given: > v = a*t > s = (a/2)*(t^2). Then > v = 2*s/t > Thus it would seem that the rate of fall is independent of 'a'? That's a bum steer. Where'd v=at come from? > It's the velocity as a function of time for constant acceleration. > Velocity is changing; whether the acceleration is constant or not. Every one of the equations you usually write is valid only for constant acceleration. Most require additional assumptions as well. > An acceleration of 32 ft/sec^2 means that the velocity increases > by 32 ft/sec every second. That is expressed in v = at with > a = 32. > Average aceleration is expressed as [(vt-vi)/t=2s/t^2]; so [a/2=s/t^2]. > Where s is the displacement distance; caused by a net force. Average acceleration can be expressed as (vt-vi)/t. That is true in general. If you use that, you will find a value of 32 ft/sec^2 for gravity. Average accleration can be expressed as 2s/t^2 if the acceleration is constant, but not in general. acceleration is equal to 2s/t^2, and not s/t^2. - Randy === Subject: Re: The rate of free fall is g/2 g/2. >Don > Define rate, and how it differs from acceleration. In classical phyics; where I hang out, and not in the fictional physics of relativity; a good definition of velocity as a rate of motion is found in Newton's First Law. In fact the first law is more about velocity being a force free motion, than it is about inertia - where it leaves us without defining inertia itself. The second law of motion defines changes in velocity as acceleration, and being due to net forces acting on and/or by bodies of matter. Don === Subject: Re: The rate of free fall is g/2 Whereas the acceleration of free fall is g; the rate of free fall is >g/2. Don Define rate, and how it differs from acceleration. In classical phyics; where I hang out, and not in the fictional physics of relativity; a good definition of velocity as a rate of motion is found in Newton's First Law; which in effect states that a body in motion will continue at a constant speed; or at rest; in one direction until a net force causes it to slow down, stop, speed up, or change direction. In any case, including changing direction, a force is required. A change in direction is a change in velocity. > In classical phyics; where I hang out, and not in the fictional physics > of relativity; a good definition of velocity as a rate of motion is > found in Newton's First Law. In fact the first law is more about > velocity being a force free motion, than it is about inertia - where it > leaves us without defining inertia itself. > The second law of motion defines changes in velocity as acceleration, > and being due to net forces acting on and/or by bodies of matter. An acceleration is a rate of change in velocity where an initial velocity (vi) is changed to another velocity (vt) in a period of time (t), and is proportional to the net force (f) causing it: That is [f/(vt-vi)/t=f/a], and is a constant: Written more concisely as 2fs/t^2; which is equal to w/g, and expressed in slugs or kilograms. Forget your dip stick f=dv/dt; that's a crock. > Don === Subject: Re: The rate of free fall is g/2 >>Whereas the acceleration of free fall is g; the rate of free fall is >>g/2. >>Don >> Define rate, and how it differs from acceleration. >In classical phyics; where I hang out, and not in the fictional physics >of relativity; a good definition of velocity as a rate of motion is >found in Newton's First Law; which in effect states that a body in >motion will continue at a constant speed; or at rest; in one direction >until a net force causes it to slow down, stop, speed up, or change >direction. >In any case, including changing direction, a force is required. A >change in direction is a change in velocity. >> In classical phyics; where I hang out, and not in the fictional physics >> of relativity; a good definition of velocity as a rate of motion is >> found in Newton's First Law. In fact the first law is more about >> velocity being a force free motion, than it is about inertia - where it >> leaves us without defining inertia itself. >> The second law of motion defines changes in velocity as acceleration, >> and being due to net forces acting on and/or by bodies of matter. >An acceleration is a rate of change in velocity where an initial Then a rate is an acceleration, and g/2=g? >velocity (vi) is changed to another velocity (vt) in a period of time >(t), and is proportional to the net force (f) causing it: That is >[f/(vt-vi)/t=f/a], and is a constant: Written more concisely as >2fs/t^2; which is equal to w/g, and expressed in slugs or kilograms. Do you really need to introduce forces there to define acceleration? I (vt-vi)/t = a which is true when a(t) is constant, and good enough for sufficiently small t. It's a purely geometrical quantity that could be said of, for instance, a wave velocity where there's no clearly identifiable f pushing a distinct m. Or for the velocity versus time of a dot that some program might cause to move on your computer screen. >Forget your dip stick f=dv/dt; that's a crock. F=dp/dt, not dv/dt. It's the definition of force. -- You're not as dumb as you look. Or sound. Or our best testing indicates. -- Monty Burns to Homer Simpson === Subject: Re: The rate of free fall is g/2 >>Whereas the acceleration of free fall is g; the rate of free fall is >>g/2. >>Don >> Define rate, and how it differs from acceleration. >In classical phyics; where I hang out, and not in the fictional physics >of relativity; a good definition of velocity as a rate of motion is >found in Newton's First Law. In fact the first law is more about >velocity being a force free motion, than it is about inertia - where it >leaves us without defining inertia itself. A body in motion will continue in uniform motion unless acted on by an external force? That's a definition of rate? >The second law of motion defines changes in velocity as acceleration, >and being due to net forces acting on and/or by bodies of matter. Newton's second law is F=dp/dt. Or, as commonly written and assuming mass is constant, F=ma. Acceleration is a=dv/dt, but that's geometrical, it's not part of Newton's laws of motion. -- Things should be made as simple as possible -- but no simpler. -- Albert Einstein === Subject: Springs problem I'm trying to understand a problem in an engineering mathematics textbook. The setup is simple enough - 3 springs, 2 freely sliding masses, and 2 fixed walls, arranged in a straight line: wall, spring 1, mass 1, spring 2, mass 2, spring 3, wall. (Friction effects aren't involved.) Mass 1 has the unit value 1, and mass 2 the value 2. The spring constants (K1, K2, K3) are 1, 2, and 2, respectively. The initial positions of the masses are: mass 1 one unit to the left of its equilibrium position (x_1); mass 2 one unit to the right of its equilibrium position (x_2). The textbook gives the following preliminary equations: M_1*x_1 = F_2 - F_1 = K_2 (x_2 - x_1) - K_1*x_1 M_2*x_2 = -F_3 - F_2 = -K_3*x_2 - K_2 (x_2 - x_1) For the life of me, I can't understand why there's a minus sign in front of F_1. Where am I being thick? Subsidiary question: suppose the middle spring was at its equilibrium extent, while 3 was compressed, and 1 stretched. How would the equations be then? Tim === Subject: Re: Springs problem > I'm trying to understand a problem in an engineering > mathematics textbook. The setup is simple enough - > 3 springs, 2 freely sliding masses, and 2 fixed walls, > arranged in a straight line: wall, spring 1, mass 1, > spring 2, mass 2, spring 3, wall. (Friction effects > aren't involved.) > Mass 1 has the unit value 1, and mass 2 the value 2. > The spring constants (K1, K2, K3) are 1, 2, and 2, > respectively. > The initial positions of the masses are: mass 1 one > unit to the left of its equilibrium position (x_1); mass 2 one > unit to the right of its equilibrium position (x_2). > The textbook gives the following preliminary equations: > M_1*x_1 = F_2 - F_1 = K_2 (x_2 - x_1) - K_1*x_1 > M_2*x_2 = -F_3 - F_2 = -K_3*x_2 - K_2 (x_2 - x_1) > For the life of me, I can't understand why there's a minus > sign in front of F_1. > Where am I being thick? > Subsidiary question: suppose the middle spring was > at its equilibrium extent, while 3 was compressed, and > 1 stretched. How would the equations be then? > Tim F_1 is the force in the spring 1 and is defined as F_1 = K_1*x_1. If x_1 is +ve then F_1 is also +ve and spring 1 is in tension. Mass M_1 is being pull back by the tension force and have an acceleration in the -ve x direction. i.e if x_1 is +ve ==> F_1 is +ve (tension) ==> M_1*x_1 -ve (being pulled towards the -ve x direction) Therefore M_1*x_1 = -F_1 = -K_1*x_1 if only spring 1 is being considered ------------------------------------------------------------ Spring 3, F_3 = K_3*x_2 and M_2*x_2 can be treated similarly. If x_2 is +ve ==> F_3 is +ve (Compression) ==> M2*x_2 is -ve (being pushed in the -ve x direction) Therefore M_2*x_2 = -F_3 = -K_3*X_2 if only spring 3 is being considered ------------------------------------------------------------ Spring 2 is a little bit more complicated. F_2 is defined as F_2 = K_2 (x_2 - x_1) Consider spring 2 on mass M_1 We fix x_2 If x_1 is +ve ==> F = K_2*x_1 is +ve (in compression) ==> M_1 is push back in the -ve x direction ==> M_1*x_1 is -ve So M_1*x_1 = -F = -K_2*x_1 Now fix x_1 If x_2 is +ve ==> F = K_2*x_2 is +ve (in tension) ==> M_1 is pull towards the +ve x direction ==> M_1*x_1 is +ve So M_1*x_1 = -F = K_2*x_2 Combine both effects (consider the effect of spring 2 on M_1) M_1*x_1 = K_2*x_2 - K_2*x_1 = K_2*(x_2 - x_1) = F_2 -------------------------------------------------- Consider spring 2 on mass M_2 Left to you for your exercise. You should be able to show that M_2*x_2 = -K_2*x_2 + K_2*x_1 = -K_2*(x_2 - x_1) = -F_2 === Subject: Re: Springs problem > I'm trying to understand a problem in an engineering > mathematics textbook. The setup is simple enough - > 3 springs, 2 freely sliding masses, and 2 fixed walls, > arranged in a straight line: wall, spring 1, mass 1, > spring 2, mass 2, spring 3, wall. (Friction effects > aren't involved.) > Mass 1 has the unit value 1, and mass 2 the value 2. > The spring constants (K1, K2, K3) are 1, 2, and 2, > respectively. > The initial positions of the masses are: mass 1 one > unit to the left of its equilibrium position (x_1); mass 2 one > unit to the right of its equilibrium position (x_2). > The textbook gives the following preliminary equations: > M_1*x_1 = F_2 - F_1 = K_2 (x_2 - x_1) - K_1*x_1 > M_2*x_2 = -F_3 - F_2 = -K_3*x_2 - K_2 (x_2 - x_1) > For the life of me, I can't understand why there's a minus > sign in front of F_1. > Where am I being thick? > Subsidiary question: suppose the middle spring was > at its equilibrium extent, while 3 was compressed, and > 1 stretched. How would the equations be then? The equations of motion do not depend on the initial conditions. To derive the equations of motion, assume each mass has positive displacement, compute the forces of the springs on the masses, and set the net force on each mass equal to Mx'' for that mass (since F=ma). F1 and F2 (whatever they are) are not needed. The equations in your book are correct. === Subject: Re: Springs problem > I'm trying to understand a problem in an engineering > mathematics textbook. The setup is simple enough - > 3 springs, 2 freely sliding masses, and 2 fixed walls, > arranged in a straight line: wall, spring 1, mass 1, > spring 2, mass 2, spring 3, wall. (Friction effects > aren't involved.) > Mass 1 has the unit value 1, and mass 2 the value 2. > The spring constants (K1, K2, K3) are 1, 2, and 2, > respectively. > The initial positions of the masses are: mass 1 one > unit to the left of its equilibrium position (x_1); mass 2 one > unit to the right of its equilibrium position (x_2). > The textbook gives the following preliminary equations: > M_1*x_1 = F_2 - F_1 = K_2 (x_2 - x_1) - K_1*x_1 > M_2*x_2 = -F_3 - F_2 = -K_3*x_2 - K_2 (x_2 - x_1) > For the life of me, I can't understand why there's a minus > sign in front of F_1. > Where am I being thick? > Subsidiary question: suppose the middle spring was > at its equilibrium extent, while 3 was compressed, and > 1 stretched. How would the equations be then? > The equations of motion do not depend on the initial conditions. To derive > the equations of motion, assume each mass has positive displacement, compute > the forces of the springs on the masses, and set the net force on each mass > equal to Mx'' for that mass (since F=ma). F1 and F2 (whatever they are) are > not needed. The equations in your book are correct. You say, The equations of motion do not depend on the initial conditions. Here's my difficulty. In the textbook there's a diagram of forces. <-----M1-----> <-----M2<----- (Three arrows pointing left, one pointing right.) Above the leftmost arrow is the equation: F_1 = K_1 x_1. Above the rightmost arrow is: F_3 = K_3 x_2. Above the two middle arrows is: F_2 = K_2 (x_2 - x_1) (Before trying to understand this without F_1 and F_2, I'm trying to understand it with them.) Now, why does the rightmost arrow point to the left? Presumably it does so because spring 3 is compressed, so it's pushing towards the left. (And the equation reflects this with a negative sign in -F_3.) But, if so, then, as the first spring is also compressed and is pushing to the right, why does the leftmost arrow point to the left? Tim === Subject: Re: Springs problem > I'm trying to understand a problem in an engineering > mathematics textbook. The setup is simple enough - > 3 springs, 2 freely sliding masses, and 2 fixed walls, > arranged in a straight line: wall, spring 1, mass 1, > spring 2, mass 2, spring 3, wall. (Friction effects > aren't involved.) > Mass 1 has the unit value 1, and mass 2 the value 2. > The spring constants (K1, K2, K3) are 1, 2, and 2, > respectively. > The initial positions of the masses are: mass 1 one > unit to the left of its equilibrium position (x_1); mass 2 one > unit to the right of its equilibrium position (x_2). > The textbook gives the following preliminary equations: > M_1*x_1 = F_2 - F_1 = K_2 (x_2 - x_1) - K_1*x_1 > M_2*x_2 = -F_3 - F_2 = -K_3*x_2 - K_2 (x_2 - x_1) > For the life of me, I can't understand why there's a minus > sign in front of F_1. > Where am I being thick? > Subsidiary question: suppose the middle spring was > at its equilibrium extent, while 3 was compressed, and > 1 stretched. How would the equations be then? >> The equations of motion do not depend on the initial conditions. To >> derive the equations of motion, assume each mass has positive >> displacement, compute the forces of the springs on the masses, and >> set the net force on each mass equal to Mx'' for that mass (since >> F=ma). F1 and F2 (whatever they are) are not needed. The equations >> in your book are correct. > You say, The equations of motion do not depend on the > initial conditions. > Here's my difficulty. > In the textbook there's a diagram of forces. > <-----M1-----> <-----M2<----- > (Three arrows pointing left, one pointing right.) > Above the leftmost arrow is the equation: > F_1 = K_1 x_1. > Above the rightmost arrow is: > F_3 = K_3 x_2. > Above the two middle arrows is: > F_2 = K_2 (x_2 - x_1) > (Before trying to understand this without F_1 and F_2, > I'm trying to understand it with them.) > Now, why does the rightmost arrow point to the left? > Presumably it does so because spring 3 is compressed, > so it's pushing towards the left. (And the equation reflects > this with a negative sign in -F_3.) > But, if so, then, as the first spring is also compressed and > is pushing to the right, why does the leftmost arrow point > to the left? Read my first response much more carefully. Since x1 and x2 are defined positive to the right, move each mass to the right to get the forces, and assume (for convenience) x2>x1. You must ignore specified I.C.; they are irrelevant for the purposes of getting the equations of motion. Thus, Spring 1 is in tension, and its force on m1 is negative (left). Spring 3 is in compression, and its force on m2 is negative (left). All you need to do here is move each mass to the right (positive displacement), and draw a free-body diagram for each mass. === Subject: Re: Springs problem <-----M1-----> <-----M2<----- > (Three arrows pointing left, one pointing right.) Now I'm looking at this in light of your equations: M_1*x_1 = F_2 - F_1 = K_2 (x_2 - x_1) - K_1*x_1 M_2*x_2 = -F_3 - F_2 = -K_3*x_2 - K_2 (x_2 - x_1) I think that I'm right in that this diagram is defining the positive direction for F_1, F_2, F_3, and that the positive direction of x_1 and x_2 is already defined as rightward. So F_1 is positive (and to the left) if x_1 is positive (and to the right). The acceleration on M_1 due to F_1 is negative. That's why F_1 appears with a minus sign. More confusingly, if F_2 is positive it appears as a rightward term in equation 1, but a leftward term in equation 2 (with a minus sign). The convention in both equations is that positive acceleration means to the right. - Randy === Subject: Re: Springs problem mathematics textbook. The setup is simple enough - > 3 springs, 2 freely sliding masses, and 2 fixed walls, > arranged in a straight line: wall, spring 1, mass 1, > spring 2, mass 2, spring 3, wall. (Friction effects > aren't involved.) > Mass 1 has the unit value 1, and mass 2 the value 2. > The spring constants (K1, K2, K3) are 1, 2, and 2, > respectively. > The initial positions of the masses are: mass 1 one > unit to the left of its equilibrium position (x_1); mass 2 one > unit to the right of its equilibrium position (x_2). > The textbook gives the following preliminary equations: > M_1*x_1 = F_2 - F_1 = K_2 (x_2 - x_1) - K_1*x_1 > M_2*x_2 = -F_3 - F_2 = -K_3*x_2 - K_2 (x_2 - x_1) > For the life of me, I can't understand why there's a minus > sign in front of F_1. > Where am I being thick? > Subsidiary question: suppose the middle spring was > at its equilibrium extent, while 3 was compressed, and > 1 stretched. How would the equations be then? > The equations of motion do not depend on the initial conditions. To > derive > the equations of motion, assume each mass has positive displacement, > compute > the forces of the springs on the masses, and set the net force on each > mass > equal to Mx'' for that mass (since F=ma). F1 and F2 (whatever they are) > are > not needed. The equations in your book are correct. > You say, The equations of motion do not depend on the > initial conditions. They shouldn't. > Here's my difficulty. > In the textbook there's a diagram of forces. > <-----M1-----> <-----M2<----- > (Three arrows pointing left, one pointing right.) I'd start out by defining one direction as positive, say to the right. So x1 and x2 are positive displacements to the right relative to their equilibrium positions. > Above the leftmost arrow is the equation: > F_1 = K_1 x_1. The arrow agrees with my convention. If M1 is displaced to the right, spring 1 exerts a force to the left of magnitude K_1*x_1. However, I'd write a minus sign on this force on the convention that positive force = force to the right. > Above the rightmost arrow is: > F_3 = K_3 x_2. Again, this seems to agree with the way I'd define things. If M2 is displaced to the right, then the force from spring 3 is to the left and has magnitude K_3*x_2. But again I'd be inclined to put a minus sign. Either the author is ignoring signs at that point, or has an opposite convention for forces. > Above the two middle arrows is: > F_2 = K_2 (x_2 - x_1) If positive means to the right, then spring 2 is stretched if x_2-x_1 is positive and compressed if x_2-x_1 is negative. If stretched, the force is leftward on M2 and rightward on M1. If x_2 - x_1 < 0, then F_2 is negative and opposite to the direction the arrows are drawn. The directions of all those arrows are consistent with an unspecified motion of x_1 and x_2 to the right such that x_2 > x_1. Perhaps the author is defining positive force to be these directions. At any given point in time, some of the F's might be negative which would mean opposite to these arros. > (Before trying to understand this without F_1 and F_2, > I'm trying to understand it with them.) > Now, why does the rightmost arrow point to the left? > Presumably it does so because spring 3 is compressed, > so it's pushing towards the left. (And the equation reflects > this with a negative sign in -F_3.) That's my read. > But, if so, then, as the first spring is also compressed and > is pushing to the right, why does the leftmost arrow point > to the left? It is compressed based on the initial conditions (x_1 < 0 at t=0), but I think it's showing the forces for positive x_1, i.e. before considering the actual initial conditions. The idea is to define your sign conventions first. - Randy === Subject: Re: Real Number Space >>I agree. Modern math is no different from a game. >Correction: Purely formal mathematics is no different from a game. >> So is modern math purely formal? >Some is, some is applies. It's tough to tell the players in the game without a scorecard. ~v~~ === Subject: Re: Real Number Space >>Some is, some is applies. > It's tough to tell the players in the game without a scorecard. It is tough to tell what is going on when one is a ignorant (and stupid) as you are. Bob Kolker === Subject: Re: Real Number Space >> Like I said, modern math is just a game. >Some modern math is applied. It is formal mathematics which is a game. So which modern math does not rely on formal modern math definitions? ~v~~ === Subject: Re: Real Number Space > So which modern math does not rely on formal modern math definitions? All mathematical systems are based on definitons and postulates. However some systems are specifically formulated to solve problems in the world. A purely formal system is one with no real world interpretation given. Being modern or old is not relevent. Russell and Whitehead's -Principia Mathamntica- an early work in which an attempt is made to derive arithmetic from logic is nearly 100 years old. It is purely formal. It has not real world referents. Modern mathematical analysis of turbulence is as new as yesterday and applied. It is not formal and it is not a game. It is about turbelence. Bob Kolker === Subject: Re: Testable Predictions by HdB > In what way? Your reference only shows a lack of familiarity with error > analysis, I think. Emphasis on Error Analysis all the time ... OK. Suppose I want to carry out an error analysis for my Fibonacci Iterations. I'm not a specialist in these things. But I can find the derivative of x_k as a function of x_0, giving: |dx_k| = |dx_0|/[F(k).x_0 - F(k+1)]^2 Question: is this formula relevant for carrying out an error analysis? A naive conclusion would be that the iterands can have no greater error than the initial value, provided that |F(k).x_0 - F(k+1)| > 1 . Is this a sensible result? And if not, could you please elaborate why. And how. Han de Bruijn === Subject: Re: Testable Predictions by HdB > In what way? Your reference only shows a lack of familiarity with error > analysis, I think. ... > But I can find the derivative of x_k as a function of > x_0, giving: > |dx_k| = |dx_0|/[F(k).x_0 - F(k+1)]^2 > Question: is this formula relevant for carrying out an error analysis? No, it is not relevant. This is the iteration 1/(x-1). In error analysis of such things in general there are two parts relevant: 1. Truncation error 2. Rounding error The first comes in iterative algorithms or somesuch where you represent something infinite by something finite. For instance when you use the Taylor series for some function and truncate it, the error induced by that is the truncation error. But also when you calculate an integral using an n-point formula, the error induced by using that n-point formula is called the truncation error. The rounding error is due only to the use of finite arithmetic. In the above there is no truncation error soe we are only concerned with the rounding error. Whenever x outside the interval [1/2,2] there is no blow-up of the relative error, but it may increase by 0.5 ulp, also the inversion may increase the error by at most 0.5 ulp. So, assuming an exact initial x0, the relative error after n iterations will be at most about n ulp, as long as x keeps out of the interval [1/2,2]. A very stable iteration. Whenever you are in that interval, nothing with sense can be said about the error bound. > A naive conclusion would be that the iterands can have no greater error > than the initial value, provided that |F(k).x_0 - F(k+1)| > 1 . Is this > a sensible result? And if not, could you please elaborate why. And how. As shown above, the relative error will slightly increase on each iteration. And the only condition is: x not in [1/2, 2] for any x. Now I do not feel inclined to do the calculations to get bounds on x0. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Testable Predictions by HdB > > ... > > > Consider the following iterates: x := (x/(1-2.x))^2 . > > > > > > And give us the closed formula(s) for (x) as a function of (k), where k > > > denotes the k'th iterand. Where are the invariant points ? And are they > > > stable or unstable? > > > > Off-hand I would say that the fixed points are 0, 1/4 and 1. Of these > > only 0 is stable. > > > > These are fixed points, indeed. But there are (infinitely) many more, > > in the domain 1/4 < x < oo . In that domain, they come with n-cycles > > of any magnitude (n). Outside this domain there is only one, 0 as you > > said, and that one is stable, indeed. > I would not state that points in a cycle are fixed points. But I am > not surprised that there are cycles, I would have expected them. And, > indeed, the 2-cycles are (3 +- sqrt(5))/2, and those are all. The 3-cycles > are a bit more difficult to display, but they can be written as radicals. > This all due to maple... Reiterating your formula easily shows that > in each re-iteration the new value is the quotient of two polynomials of > some degree of the old value. Equating these shows the cycles. > > Examples: (3/2+sqrt(5)/2) and (3/2-sqrt(5)/2) come with a 2-cycle. > Yup, and those are the only ones. > > I do not know whether there is a closed formula. > > > > The I do not know is relevant four our debate. What about theorems, > > indeed, that we do not know? And yet we have that computer program .. > To be honest, I think there is no closed formula, but that is beside the > point. We were talking about theories providing testable predictions. > You are only showing programs that do test predictions of some unknown > theory. As there is no theory yet, how can you be talking about > theories providing testable predictions? Ah, but that is a philosophical question! Are theories _invented_ or are they _discovered_? Personally, I think the definitions are invented: as the things you want to talk about. And then the theorems ARE just there to be discovered. That's why, in some respects, computer experiments are more powerful than theorems. Because many of the latter are: yet to be discovered. But, as I've said: this is philosophy. > To be honest, I think there is no closed formula You are quite wrong. There IS a closed formula. Well, depends on what's > Any mathematician that has looked shortly on Fibonacci numbers would > know immediately that there is a close connection. So to me it did > not come as a susprise at all. Any mathematician that has looked shortly on trigonometric functions and hyperbolic functions would know immediately that there is a close connection . Sorry, Dik. Just couldn't resist the parroting ... I have revealed / will reveal the secret elswhere in this group: Which is in the thread Induction with a hard start. [ ... snip ... snip ... ] > This may not interested you at all. But, so what? Why? Off topic, or even snipped, doesn't mean uninteresting to me. :-) Han de Bruijn === Subject: Re: Testable Predictions by HdB ... > You are quite wrong. There IS a closed formula. Well, depends on what's > Any mathematician that has looked shortly on Fibonacci numbers would > know immediately that there is a close connection. So to me it did > not come as a susprise at all. > Any mathematician that has looked shortly on trigonometric functions > and hyperbolic functions would know immediately that there is a close > connection . Sorry, Dik. Just couldn't resist the parroting ... Here the connection is much deeper. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Testable Predictions by HdB > [ ... snip ... snip ... ] >> This may not interested you at all. But, so what? > Why? Off topic, or even snipped, doesn't mean uninteresting to me. :-) Hmm, now I see it's _not_ so much off topic. You were surprised when you tested some theory. Until you realized that .. and found a piece of the jigsaw puzzle. This is very much the way I'm working towards results myself. So you have the same strategy as I have. But you don't admit it. Han de Bruijn === Subject: Re: Testable Predictions by HdB > [ ... snip ... snip ... ] > >> This may not interested you at all. But, so what? > > Why? Off topic, or even snipped, doesn't mean uninteresting to me. :-) > Hmm, now I see it's _not_ so much off topic. You were surprised when you > tested some theory. Until you realized that .. and found a piece of > the jigsaw puzzle. This is very much the way I'm working towards results > myself. So you have the same strategy as I have. But you don't admit it. You are wrong. I did not test the theory. The theory was plain, there should exist such a common divisor. What I was trying to do is find an algorithm to *calculate* such a common divisor. And the theory was giving me the proper directions. 7^n for some particular n would non-trivially split in factors in the appropriate quadratic field, and so I looked for such splittings. I found already quite early that it is possible that 7^k might split non-trivially if k is a divisor of n. (7^3 has a non-trivial splitting in Q(-38), but n is 6 in that case.) What I had *not* realised is that when 7^k splits into p and q in Q(D), that if n - k is even there are two non-trivial splittings of 7^n. And one of them is p.7^((n-k)/2) and q.7^((n-k)/2). (The other is based on (p + q.D)^(n/k).) There might even be more if n > 2k. But I found this one with n = 24, k = 12 and D = sqrt(-766). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Testable Predictions by HdB > Ah, Eckard. Did you follow my discussion with him? No. But I want to take a look at it, if you only give me a hint. > I thought we were talking mathematics, not physics. So what is the > relevance? OT: A little bit of Physics would be NO Idleness in Mathematics. That's the mantra, my Leitmotiv. Well, you should know it by now ... > In what way? Your reference only shows a lack of familiarity with error > analysis, I think. Show me how your error analysis will be in favorite of Upwind Schemes instead of central difference schemes. I'm breathlessly awaiting ... > What is the relevance with mathematics and numerical mathematics? > (I may note, again, that this is not my field of interest in numerical > mathematics. You should consult those that are knowledgable here.) Without any sensible result. Go to 'sci.math.num-analysis' and see with your own eyes how knowledgable people are there nowadays: Gimme that free drug which will painlessly solve my problems and the like ... But if this is not your field of interest, then it makes a continuation of the debate between us, on _this_ issue, questionable. I am convinced, namely, that the Upwind question can _not_ be settled without experience from computing. And arguments from physics. And I would have challenged you to argue that I am demonstrably wrong. Han de Bruijn === Subject: Re: Testable Predictions by HdB > Do you know how it is done the standard way? Did you do an error > analysis of your algorithm? The problem with numerical differentiation > is that when you make the stepsize too small the numerical stability > suffers due to rounding errors. This kind of problems do not occur with > numerical integration. The problem with standards is that you have so many to choose from :-( Therefore: No. I don't know what your standard way is and I would like to ask you: please explain what it is. But yes, I have an error analysis of my algorithm. Where the curve f'(x) is smooth (say continuous first and second derivatives) there the error, the difference between the interpolation and the original is given by: rho/2.sigma^2 (signed rho preferred) Where rho = curvature, sigma = spread of the Gaussians. Where the curve is continuous but the first derivative is not, there we have the maximal error: 2.cos(phi/2).sigma/sqrt(2.pi) Where phi is the angle between the tangent on the left and the tangent on the right. Han de Bruijn === Subject: Re: Testable Predictions by HdB > Do you know how it is done the standard way? Did you do an error > analysis of your algorithm? The problem with numerical differentiation > is that when you make the stepsize too small the numerical stability > suffers due to rounding errors. This kind of problems do not occur with > numerical integration. > The problem with standards is that you have so many to choose from :-( > Therefore: No. I don't know what your standard way is and I would like > to ask you: please explain what it is. See for a reasonable description. > But yes, I have an error analysis of my algorithm. Where the curve f'(x) > is smooth (say continuous first and second derivatives) there the error, > the difference between the interpolation and the original is given by: > rho/2.sigma^2 (signed rho preferred) > Where rho = curvature, sigma = spread of the Gaussians. Where the curve > is continuous but the first derivative is not, there we have the maximal > error: > 2.cos(phi/2).sigma/sqrt(2.pi) > Where phi is the angle between the tangent on the left and the tangent > on the right. I would expect to find a stepsize involved. Decreasing the error when the stepsize is decreased until some stage where decreasing the stepsize will *increase* the error. Zie the above reference from page 12 for a discussion. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: maximum of trace Please help me to confirm following inference and comment on my questions: X=C*SÍ, XÍX=SCÍCSÍ, Trace(XÍX)=Trace(SCÍCSÍ)=Trace( C ÍCSÍS) Therefore we should seek CÍC to maximize the trace of (XÍX) [u d v]=svd(SÍS); The trace(XÍX) is maximum when CÍC equals to v*uÍ What property of C(or CÍC) is proportional to the trace(XÍX)? Given certain S, is there any quantitative relationship between C and trace(XÍX) ? Lawrence === Subject: Re: maximum of trace >Please help me to confirm following inference and comment on my questions: >X=C*S.89´.9c, X.89´.9cX=SC.89´.9cCS.89´.9c, Trace(X.89´.9cX)=Trace(SC.89´.9cCS.89´.9c )=Trace(C.89´.9cCS.89´.9cS) Please write your formulas using plain text, no fancy characters (just what's found on a standard keyboard). I can't make heads or tails of what are the constants, what are the variables, what are the constraints, what do you want to maximize? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: maximum of trace Sorry for missing ' in previous email. ? denote ', means upper T. Lawrence === Subject: Re: Empiricism and the Assumption of Truth > > > My, my, now the square root of two is transcendental? >> I just got saying that it is not. The square root of two satisfies the >> polynomial equation x^2 - 2 = 0. The square root of two is algebraic >> therfore not transcendental. However it is irrational. >> Bob Kolker >Once more Lester provides evidence of his inability to >read for comprehension. Why else would he respond to > The square root of two which is an algebraic (non-transcendental) > irrational number has the same property. >With > My, my, now the square root of two is transcendental? >???? >Of course, maybe in Lester speak, 'non' means something >other than what the rest of the world thinks it means. Why else indeed unless Lester is pointing out that metaBob with his modern math gamecube speaks with forked tongue. ~v~~ === Subject: Re: Empiricism and the Assumption of Truth >> >> >> My, my, now the square root of two is transcendental? > I just got saying that it is not. The square root of two satisfies the > polynomial equation x^2 - 2 = 0. The square root of two is algebraic > therfore not transcendental. However it is irrational. > Bob Kolker >>Once more Lester provides evidence of his inability to >>read for comprehension. Why else would he respond to >> The square root of two which is an algebraic (non-transcendental) >> irrational number has the same property. >>With >> My, my, now the square root of two is transcendental? >>???? >>Of course, maybe in Lester speak, 'non' means something >>other than what the rest of the world thinks it means. > Why else indeed unless Lester is pointing out that metaBob with his > modern math gamecube speaks with forked tongue. That does not answer the question of why you would respond to >> The square root of two which is an algebraic (non-transcendental) >> irrational number has the same property. with >> My, my, now the square root of two is transcendental? It is clearly stated that the square root of two is non-transcendental. Your response makes it clear that you thought it was stated that the square root of two is transcendental. There is no 'forked tongue', just a simple statement that the square root of two is non-transcendental, which you read as saying that the square root of two is transcendental. So what part of 'non-transcendental' did you not understand? Stephen === Subject: Re: Empiricism and the Assumption of Truth >> My, my, now the square root of two is transcendental? >I just got saying that it is not. The square root of two satisfies the >polynomial equation x^2 - 2 = 0. The square root of two is algebraic >therfore not transcendental. However it is irrational. I thought you just got through saying the square root of two satisfies the same definition as transcendental in modern math. This is all so confusing. Perhaps you could switch to some kind of formal definition which isn't just a game. ~v~~ === Subject: Re: Empiricism and the Assumption of Truth > I thought you just got through saying the square root of two satisfies > the same definition as transcendental in modern math. This is all so > confusing. Perhaps you could switch to some kind of formal definition > which isn't just a game. Listen carefully. I am using standard mathematical terminology. A trancendental number is a real number the does not satisfy any polynomial in one variable which has rational co-efficients. An algebraic number is a real number that satisfies such a polynomial equaitons. A transcendental real number is a non-algebraic real number. Pi is an example of a trancendental number. Square root of two is an algebraic number. Both are irrational (i.e. not the ratio of integers) and neither have a finite or cyclic fractional expansion in any base. Bob Kolker === Subject: Re: Empiricism and the Assumption of Truth > > I thought you just got through saying the square root of two satisfies > the same definition as transcendental in modern math. This is all so > confusing. Perhaps you could switch to some kind of formal definition > which isn't just a game. > Listen carefully. I am using standard mathematical terminology. A > trancendental number is a real number the does not satisfy any > polynomial in one variable which has rational co-efficients. An > algebraic number is a real number that satisfies such a polynomial > equaitons. A transcendental real number is a non-algebraic real number. > Pi is an example of a trancendental number. Square root of two is an > algebraic number. Both are irrational (i.e. not the ratio of integers) > and neither have a finite or cyclic fractional expansion in any base. > Bob Kolker Although the square root of two does have an ultmately 'cyclic' partial fraction expansion, as do all square roots of non-square naturals. === Subject: Re: Empiricism and the Assumption of Truth > Although the square root of two does have an ultmately 'cyclic' partial Not true. Produce a finite sequence repeated cyclically after a finite non cyclid prefix. Do it for base ten. > fraction expansion, as do all square roots of non-square naturals. False. In any basss the fraction expansion of an irrational number is infinite and non-cyclic. This is a standard theorem in number theory and algebra. Bob Kolker === Subject: A Brief History of Science and Mechanics in the West A Brief History of Science and Mechanics in the West ---------------- Galileo displaces Man from the center of the universe. Arithmetic emerges from the shadows of classical geometry with the symbol for differences first followed by addition. Kepler divines conic section orbits for heavenly bodies but fails to mechanically explain his force of angels which keeps them in orbit. Newton explains the force of angels as a centripetally directed inverse square force he calls gravitation. Newton establishes equal and opposite reaction mechanics but fails to explain angular mechanics in the same terms. Euler treats angular mechanics in a manner analogous to Newton's linear mechanics but fails to treat it on the same basis as Newton's treatment of planetary angular mechanics and as a result angular momentum becomes a vector cross product. Arithmetic is unified on the basis of cardinal numerology, set theory of points, and an implied commonality of commutative operational properties in which differences are nothing but the axiomatic inverse of addition. Planck discovers his famous constant which mediates non vector energy through e=h/t but puzzles science because Planck's constant appears to have units of a vector angular momentum (rr/t) whereas energy is a scalar but the time differential of vector cross product angular momentum is a vector not a scalar. So no one can quite understand what Planck's constant is supposed to represent and all happily agree to bequeath the problem to posterity. Einstein displaces Newtonian inertial space with relative velocity as an inertial frame of reference and posits a relative constant velocity for light in space, c, then posits a dimensional frame of reference contraction to explain the constant relative velocity of light without recognizing all such frames of reference overlap one another in space and different contraction factors cannot apply to different frames. Numerous instantiations of Planck's constant emerge to confound classical mechanics including to the particular chagin of classical angular mechanics inexplicable constant spin characteristics 1/2 h/2pi Mechanics to distinguish it from classical mechanics and a host of bizarre properties adduced to describe various seemingly mechanically unrelated hermit quantum functions. Cosmology firmly replaces Man at the center of the universe with facile interpretations of the isotropic Hubble Red Shift as reflective of longitudinal doppler and universal recessional expansion. And there the history of science and mechanics in the West stops. ~v~~ === Subject: Re: A Brief History of Science and Mechanics in the West > A Brief History of Science and > Mechanics in the West > ---------------- > Galileo displaces Man from the center of the universe. > Arithmetic emerges from the shadows of classical geometry with the > symbol for differences first followed by addition. False. Negative numbers were not readily accepted thus subtraction was considered valid only when the smaller number was subtracted from the large. You are writing Zick history which has little to do with what really happened. > Kepler divines conic section orbits for heavenly bodies but fails to > mechanically explain his force of angels which keeps them in orbit. > Newton explains the force of angels as a centripetally directed > inverse square force he calls gravitation. That was his first cut at explaining motion. He later realized that the acceleration is toward the sun. > Newton establishes equal and opposite reaction mechanics but fails to > explain angular mechanics in the same terms. > Euler treats angular mechanics in a manner analogous to Newton's > linear mechanics but fails to treat it on the same basis as Newton's > treatment of planetary angular mechanics and as a result angular > momentum becomes a vector cross product. Nonsense. > Arithmetic is unified on the basis of cardinal numerology, set theory > of points, and an implied commonality of commutative operational > properties in which differences are nothing but the axiomatic inverse > of addition. > Planck discovers his famous constant which mediates non vector > energy through e=h/t but puzzles science because Planck's constant > appears to have units of a vector angular momentum (rr/t) whereas > energy is a scalar but the time differential of vector cross product > angular momentum is a vector not a scalar. So no one can quite > understand what Planck's constant is supposed to represent and all > happily agree to bequeath the problem to posterity. Both units of action are derived from similar lagrangians. > Einstein displaces Newtonian inertial space with relative velocity as > an inertial frame of reference and posits a relative constant velocity > for light in space, c, then posits a dimensional frame of reference > contraction to explain the constant relative velocity of light without > recognizing all such frames of reference overlap one another in space > and different contraction factors cannot apply to different frames. Einstein posited two things. The laws of physics are the same in all inertial frames and the speed of light is the same constant in all inertial frames. He did not posit a contraction. The so called contraction is in fact a kind of rotation and is no more mysterial than a stick appearing to be of different lengths when view from different angles. > Numerous instantiations of Planck's constant emerge to confound > classical mechanics including to the particular chagin of classical > angular mechanics inexplicable constant spin characteristics 1/2 h/2pi > Mechanics to distinguish it from classical mechanics and a host of > bizarre properties adduced to describe various seemingly mechanically > unrelated hermit quantum functions. That is Hermite, you dunce. And the term Hermite or Hermition refers to a type of self adjoint linear transformation. > Cosmology firmly replaces Man at the center of the universe with > facile interpretations of the isotropic Hubble Red Shift as reflective > of longitudinal doppler and universal recessional expansion. Utter and complete nonsense. Man is not at the center of anything on a cosmic scale. We are small beings on a small planet in a very ordinary galaxy. > And there the history of science and mechanics in the West stops. It goes on and on. Your history is a a travesty and a parody and not a very good one. Bob Kolker === Subject: Re: A Brief History of Science and Mechanics in the West > A Brief History of Science and > Mechanics in the West > ---------------- > Galileo displaces Man from the center of the universe. > Arithmetic emerges from the shadows of classical geometry with the > symbol for differences first followed by addition. > False. Negative numbers were not readily accepted thus subtraction was > considered valid only when the smaller number was subtracted from the > large. > You are writing Zick history which has little to do with what really > happened. > Kepler divines conic section orbits for heavenly bodies but fails to > mechanically explain his force of angels which keeps them in orbit. > Newton explains the force of angels as a centripetally directed > inverse square force he calls gravitation. > That was his first cut at explaining motion. He later realized that the > acceleration is toward the sun. > Newton establishes equal and opposite reaction mechanics but fails to > explain angular mechanics in the same terms. > Euler treats angular mechanics in a manner analogous to Newton's > linear mechanics but fails to treat it on the same basis as Newton's > treatment of planetary angular mechanics and as a result angular > momentum becomes a vector cross product. > Nonsense. > Arithmetic is unified on the basis of cardinal numerology, set theory > of points, and an implied commonality of commutative operational > properties in which differences are nothing but the axiomatic inverse > of addition. > Planck discovers his famous constant which mediates non vector > energy through e=h/t but puzzles science because Planck's constant > appears to have units of a vector angular momentum (rr/t) whereas > energy is a scalar but the time differential of vector cross product > angular momentum is a vector not a scalar. So no one can quite > understand what Planck's constant is supposed to represent and all > happily agree to bequeath the problem to posterity. > Both units of action are derived from similar lagrangians. > Einstein displaces Newtonian inertial space with relative velocity as > an inertial frame of reference and posits a relative constant velocity > for light in space, c, then posits a dimensional frame of reference > contraction to explain the constant relative velocity of light without > recognizing all such frames of reference overlap one another in space > and different contraction factors cannot apply to different frames. > Einstein posited two things. > The laws of physics are the same in all inertial frames and the speed of > light is the same constant in all inertial frames. He did not posit a > contraction. The so called contraction is in fact a kind of rotation and > is no more mysterial than a stick appearing to be of different lengths > when view from different angles. He forgot to tell us what is a law of physics and the fact that inertial systems do not really exist. Furthermore, On one hand you have been claiming all along that SR/GR are models. On the other, you are talking of the epistemology of laws of physics, treating these models as laws of nature. If you think you, or Dr Einstein, can phak with our brains, you are mistaken. You are only phaking with your own. Mike > Numerous instantiations of Planck's constant emerge to confound > classical mechanics including to the particular chagin of classical > angular mechanics inexplicable constant spin characteristics 1/2 h/2pi > Mechanics to distinguish it from classical mechanics and a host of > bizarre properties adduced to describe various seemingly mechanically > unrelated hermit quantum functions. > That is Hermite, you dunce. And the term Hermite or Hermition refers to > a type of self adjoint linear transformation. > Cosmology firmly replaces Man at the center of the universe with > facile interpretations of the isotropic Hubble Red Shift as reflective > of longitudinal doppler and universal recessional expansion. > Utter and complete nonsense. Man is not at the center of anything on a > cosmic scale. We are small beings on a small planet in a very ordinary > galaxy. Funny, really funny. Tell me something, why the moon sky is not as bright as expected? Either they never landed or if they actually did land on it, something is wrong about the science you preach. Has you seen the pictures? > And there the history of science and mechanics in the West stops. > It goes on and on. > Your history is a a travesty and a parody and not a very good one. > Bob Kolker === Subject: Re: A Brief History of Science and Mechanics in the West > He forgot to tell us what is a law of physics and the fact that > inertial systems do not really exist. Furthermore, There are systems that are locally inertial. Ride the Vomit Comet sometimes. Bob Kolker === Subject: Re: A Brief History of Science and Mechanics in the West > A Brief History of Science and > Mechanics in the West > ---------------- > Galileo displaces Man from the center of the universe. [snip Zick crap] God did that, idiot Zick. All Galileo did was watch. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: A Brief History of Science and Mechanics in the West > Mechanics in the West > ---------------- > Galileo displaces Man from the center of the universe. > [snip Zick crap] > God did that, idiot Zick. All Galileo did was watch. Gali also brought it to the attention of all the other men, who couldn't see it for themselves because of the blinders someone had placed over their eyes. Who placed the blinders, I wonder? === Subject: Re: A Brief History of Science and Mechanics in the West > Gali also brought it to the attention of all the other men, who > couldn't see it for themselves because of the blinders someone had > placed over their eyes. Who placed the blinders, I wonder? They did with help from Aristotle and Church tradition. Bob Kolker === Subject: Re: A Brief History of Science and Mechanics in the West >> Mechanics in the West >> ---------------- >> Galileo displaces Man from the center of the universe. >[snip Zick crap] >God did that, idiot Zick. All Galileo did was watch. And saw considerably more than yourself. ~v~~ === Subject: Re: A Brief History of Science and Mechanics in the West > Galileo displaces Man from the center of the universe. What about Copernicus, Kepler, Newton, Darwin, Freud, Shapley and Hubble? === Subject: Re: A Brief History of Science and Mechanics in the West >> Galileo displaces Man from the center of the universe. > What about Copernicus, Kepler, Newton, Darwin, Freud, Shapley > and Hubble? Brevity is the soul of wit. Wit is the soul of the usenet. Empiricism is witless because it takes its assumptions of truth seriously. ~v~~ === Subject: Re: A Brief History of Science and Mechanics in the West > > Galileo displaces Man from the center of the universe. > >> What about Copernicus, Kepler, Newton, Darwin, Freud, Shapley >> and Hubble? > Brevity is the soul of wit. Wit is the soul of the usenet. Empiricism > is witless because it takes its assumptions of truth seriously. Likewise Lester just assumes that his assumptions are true and cannot be proven. He even admits it. So much for Lester's wit. Stephen === Subject: Re: A Brief History of Science and Mechanics in the West > Likewise Lester just assumes that his assumptions are true and > cannot be proven. He even admits it. So much for Lester's wit. You are half right. Bob Kolker === Subject: Re: A Brief History of Science and Mechanics in the West > Brevity is the soul of wit. Wit is the soul of the usenet. Empiricism > is witless because it takes its assumptions of truth seriously. So seriously that it produces a superlative technology which you misuse to spread your pernicious canards. The proof of empiricism is the gadgets. Bob Kolker === Subject: Re: A Brief History of Science and Mechanics in the West > Brevity is the soul of wit. Wit is the soul of the usenet. Empiricism > is witless because it takes its assumptions of truth seriously. > So seriously that it produces a superlative technology which you misuse > to spread your pernicious canards. > The proof of empiricism is the gadgets. Oh, utter crap. All a rationalism need to is provide a justification for sense perception and it can do the exact same thing that empiricism does by simply assuming that sense perception is reliable. === Subject: Re: A Brief History of Science and Mechanics in the West > Oh, utter crap. All a rationalism need to is provide a justification for > sense perception and it can do the exact same thing that empiricism does by > simply assuming that sense perception is reliable. A priorist bull. The only thing we have going for us is our senses and sufficient intelligence to organize their result into a coherent whole. If we lived deaf and blind we could not deduce what reality it. Bob Kolker === Subject: Re: A Brief History of Science and Mechanics in the West >> Galileo displaces Man from the center of the universe. > What about Copernicus, Kepler, Newton, Darwin, Freud, Shapley > and Hubble? Freud? What does he do in this noble company? Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: A Brief History of Science and Mechanics in the West >Galileo displaces Man from the center of the universe. >> What about Copernicus, Kepler, Newton, Darwin, Freud, Shapley >> and Hubble? > Freud? What does he do in this noble company? > Mati Meron | When you argue with a fool, > meron@cars.uchicago.edu | chances are he is doing just the same Nicolaus Copernicus demonstrated that Earth is not, in fact, in the center of the system and the motion of the heavens can be explained without assuming that we are observing from a special position. The implications of Copernicus' work can not be exaggerated. His views challenged the literal interpretation of Scripture, the philosophical and metaphysical foundations of moral theory, and even common sense itself. The result was a massive opposition to his reported ideas. It was the slow, sure acceptance of the heliocentric theory by natural philosophers that ultimately quieted the general clamor, however the name of Copernicus is still a battle cry against the establishment in religion, philosophy and science. with Darwin his exalted place among the creatures of the Earth; with Copernicus man had lost his privileged position in the Universe. The lesson learned by future scientists is that if a theory requires a special origin or viewpoint, then it is not plausible. Almost all cosmological and scientific theories are scrutinized by the Copernican principle. Often interpreted that if an idea requires some special condition, then it is incomplete. Since the time of Copernicus, the Copernican principle has been substantiated even further. Not only has the Earth been dethroned from its central position in the universe, in fact, at the beginning of the 20th century, the astronomer Harlow Shapley demonstrated that our entire solar system is not even at the center of our own Milky Way galaxy. Indeed, it is about two-thirds of the way out from the center. === Subject: Re: A Brief History of Science and Mechanics in the West <0ed0f.387334$_o.364536@attbi_s71 Nicolaus Copernicus demonstrated that Earth is not, in > fact, in the center of the system and the motion of the > heavens can be explained without assuming that we are > observing from a special position. [snip more crappy conjestures] You have been told many times that you copy and paste material from popular science books that neither you understand nor the stupid author. Copernican and Ptolemaic systems are kinematically equivalent. ONLY through a choice of a theory of gravitational interaction it is only possible that one dominates over the other. Newton dynamics were developed in favor of the Copernican system. Special relativity axiom of Relativity casted a doubt on the Newtonian choice. General relativity makes no obvious choice between the two. There are several observational facts that support the hypothesis that the earth is at the center of the universe. -Measurements over a century show a decrease in parallax , from 20% to 200% -If ALL parallaxes decreased with time, then all the sample stars are receding from Earth in all directions -There seems to be a trend for recession to be greatest at the poles and equator -The variation by surface site indicates parallax is relative Earth-star motion; any relative motion with the Sun would have a much larger parallax..... but that is not observed. This supports geocentrism. -The stars are much closer than claimed, to have a variable parallax across observatories. This supports geocentrism. -All the sampled stars are receding from Earth in all directions. This supports geocentrism. And much more. For details see the source of the above material: http://www.catholicintl.com/scienceissues/genesis-galileo4.htm Mike === Subject: Re: A Brief History of Science and Mechanics in the West >>Galileo displaces Man from the center of the universe. > What about Copernicus, Kepler, Newton, Darwin, Freud, Shapley > and Hubble? >> Freud? What does he do in this noble company? >> Mati Meron | When you argue with a fool, >> meron@cars.uchicago.edu | chances are he is doing just the same > Nicolaus Copernicus demonstrated that Earth is not, in > fact, in the center of the system and the motion of the > heavens can be explained without assuming that we are > observing from a special position. > The implications of Copernicus' work can not be > exaggerated. His views challenged the literal > interpretation of Scripture, the philosophical and > metaphysical foundations of moral theory, and even > common sense itself. The result was a massive > opposition to his reported ideas. > It was the slow, sure acceptance of the heliocentric > theory by natural philosophers that ultimately quieted > the general clamor, however the name of Copernicus is > still a battle cry against the establishment in > religion, philosophy and science. > with Darwin his exalted place among the creatures of > the Earth; with Copernicus man had lost his privileged > position in the Universe. > The lesson learned by future scientists is that if a > theory requires a special origin or viewpoint, then it > is not plausible. Almost all cosmological and > scientific theories are scrutinized by the Copernican > principle. Often interpreted that if an idea requires > some special condition, then it is incomplete. > Since the time of Copernicus, the Copernican principle > has been substantiated even further. Not only has the > Earth been dethroned from its central position in the > universe, in fact, at the beginning of the 20th > century, the astronomer Harlow Shapley demonstrated > that our entire solar system is not even at the center > of our own Milky Way galaxy. Indeed, it is about > two-thirds of the way out from the center. Very poetic. But, Freud's standing among preening intellectuals notwhithstanding, there is precious little (if any) that scientists learned from Freud. His work is quite unscientific, broad sweeping generalizations deduced from quite few special cases. Nonfalsifiable, as well. It is not the job of science to dethrone, overturn etc. These things may and do happen through the normal progression of science, but if and when they do happen they're a side effect, not the goal. The job of science is to provide good working models of the obsevable world. Freud doesn't qualify. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: A Brief History of Science and Mechanics in the West > >Galileo displaces Man from the center of the universe. > What about Copernicus, Kepler, Newton, Darwin, Freud, Shapley >> and Hubble? > > Freud? What does he do in this noble company? > > Mati Meron | When you argue with a fool, > meron@cars.uchicago.edu | chances are he is doing just the same >> Nicolaus Copernicus demonstrated that Earth is not, in >> fact, in the center of the system and the motion of the >> heavens can be explained without assuming that we are >> observing from a special position. >> The implications of Copernicus' work can not be >> exaggerated. His views challenged the literal >> interpretation of Scripture, the philosophical and >> metaphysical foundations of moral theory, and even >> common sense itself. The result was a massive >> opposition to his reported ideas. >> It was the slow, sure acceptance of the heliocentric >> theory by natural philosophers that ultimately quieted >> the general clamor, however the name of Copernicus is >> still a battle cry against the establishment in >> religion, philosophy and science. >> with Darwin his exalted place among the creatures of >> the Earth; with Copernicus man had lost his privileged >> position in the Universe. >> The lesson learned by future scientists is that if a >> theory requires a special origin or viewpoint, then it >> is not plausible. Almost all cosmological and >> scientific theories are scrutinized by the Copernican >> principle. Often interpreted that if an idea requires >> some special condition, then it is incomplete. >> Since the time of Copernicus, the Copernican principle >> has been substantiated even further. Not only has the >> Earth been dethroned from its central position in the >> universe, in fact, at the beginning of the 20th >> century, the astronomer Harlow Shapley demonstrated >> that our entire solar system is not even at the center >> of our own Milky Way galaxy. Indeed, it is about >> two-thirds of the way out from the center. >Very poetic. But, Freud's standing among preening intellectuals >notwhithstanding, there is precious little (if any) that scientists >learned from Freud. His work is quite unscientific, broad sweeping >generalizations deduced from quite few special cases. Nonfalsifiable, >as well. Technically there is precious little (if any) that empiricists learned from Freud because his work is quite unempirical. There is often a confusion of terms between science and empiricism. Especially on the part of empiricists. >It is not the job of science to dethrone, overturn etc. These >things may and do happen through the normal progression of science, >but if and when they do happen they're a side effect, not the goal. >The job of science is to provide good working models of the obsevable >world. Freud doesn't qualify. >Mati Meron | When you argue with a fool, >meron@cars.uchicago.edu | chances are he is doing just the same ~v~~ === Subject: Re: A Brief History of Science and Mechanics in the West >Galileo displaces Man from the center of the universe. > What about Copernicus, Kepler, Newton, Darwin, Freud, Shapley >> and Hubble? >Freud? What does he do in this noble company? >Mati Meron | When you argue with a fool, >meron@cars.uchicago.edu | chances are he is doing just the same >> Nicolaus Copernicus demonstrated that Earth is not, in >> fact, in the center of the system and the motion of the >> heavens can be explained without assuming that we are >> observing from a special position. >> The implications of Copernicus' work can not be >> exaggerated. His views challenged the literal >> interpretation of Scripture, the philosophical and >> metaphysical foundations of moral theory, and even >> common sense itself. The result was a massive >> opposition to his reported ideas. >> It was the slow, sure acceptance of the heliocentric >> theory by natural philosophers that ultimately quieted >> the general clamor, however the name of Copernicus is >> still a battle cry against the establishment in >> religion, philosophy and science. >> with Darwin his exalted place among the creatures of >> the Earth; with Copernicus man had lost his privileged >> position in the Universe. >> The lesson learned by future scientists is that if a >> theory requires a special origin or viewpoint, then it >> is not plausible. Almost all cosmological and >> scientific theories are scrutinized by the Copernican >> principle. Often interpreted that if an idea requires >> some special condition, then it is incomplete. >> Since the time of Copernicus, the Copernican principle >> has been substantiated even further. Not only has the >> Earth been dethroned from its central position in the >> universe, in fact, at the beginning of the 20th >> century, the astronomer Harlow Shapley demonstrated >> that our entire solar system is not even at the center >> of our own Milky Way galaxy. Indeed, it is about >> two-thirds of the way out from the center. > Very poetic. But, Freud's standing among preening intellectuals > notwhithstanding, there is precious little (if any) that scientists > learned from Freud. His work is quite unscientific, broad sweeping > generalizations deduced from quite few special cases. Nonfalsifiable, > as well. > It is not the job of science to dethrone, overturn etc. These > things may and do happen through the normal progression of science, > but if and when they do happen they're a side effect, not the goal. > The job of science is to provide good working models of the obsevable > world. Freud doesn't qualify. > Mati Meron | When you argue with a fool, > meron@cars.uchicago.edu | chances are he is doing just the same True, Freud, doesn't qualify, nor should he, but nevertheless, Freud influenced thinking about man's place in nature. === Subject: Re: A Brief History of Science and Mechanics in the West >Galileo displaces Man from the center of the universe. What about Copernicus, Kepler, Newton, Darwin, Freud, Shapley > and Hubble? >Freud? What does he do in this noble company? >>Mati Meron | When you argue with a fool, >>meron@cars.uchicago.edu | chances are he is doing just the same > Nicolaus Copernicus demonstrated that Earth is not, in > fact, in the center of the system and the motion of the > heavens can be explained without assuming that we are > observing from a special position. > The implications of Copernicus' work can not be > exaggerated. His views challenged the literal > interpretation of Scripture, the philosophical and > metaphysical foundations of moral theory, and even > common sense itself. The result was a massive > opposition to his reported ideas. > It was the slow, sure acceptance of the heliocentric > theory by natural philosophers that ultimately quieted > the general clamor, however the name of Copernicus is > still a battle cry against the establishment in > religion, philosophy and science. > with Darwin his exalted place among the creatures of > the Earth; with Copernicus man had lost his privileged > position in the Universe. > The lesson learned by future scientists is that if a > theory requires a special origin or viewpoint, then it > is not plausible. Almost all cosmological and > scientific theories are scrutinized by the Copernican > principle. Often interpreted that if an idea requires > some special condition, then it is incomplete. > Since the time of Copernicus, the Copernican principle > has been substantiated even further. Not only has the > Earth been dethroned from its central position in the > universe, in fact, at the beginning of the 20th > century, the astronomer Harlow Shapley demonstrated > that our entire solar system is not even at the center > of our own Milky Way galaxy. Indeed, it is about > two-thirds of the way out from the center. >> Very poetic. But, Freud's standing among preening intellectuals >> notwhithstanding, there is precious little (if any) that scientists >> learned from Freud. His work is quite unscientific, broad sweeping >> generalizations deduced from quite few special cases. Nonfalsifiable, >> as well. >> It is not the job of science to dethrone, overturn etc. These >> things may and do happen through the normal progression of science, >> but if and when they do happen they're a side effect, not the goal. >> The job of science is to provide good working models of the obsevable >> world. Freud doesn't qualify. >> Mati Meron | When you argue with a fool, >> meron@cars.uchicago.edu | chances are he is doing just the same > True, Freud, doesn't qualify, nor should he, but nevertheless, > Freud influenced thinking about man's place in nature. Yes, I won't argue with this. He has his place on the list of thinkers (but not the list of scientists). Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Modern Math as a Game Modern Math as a Game ~v~~ Bob Kolker just noted that when done formally mathematics cannot be distinguished from a game. And I would agree that modern math is just that. Its axioms are only rules of the game. Its definitions are only permissible moves. It has no geometry and its dimensions are only directions. They only defining winning or losing because none of them is reducible to truth. ~v~~ === Subject: Re: Modern Math as a Game > Modern Math as a Game > ~v~~ > Bob Kolker just noted that when done formally mathematics cannot be > distinguished from a game. And I would agree that modern math is just > that. Not modern math. -Formal- math. There is a lot of applied modern mathematics which is not a game but serious analysis of physical processes. Bob Kolker Its axioms are only rules of the game. Its definitions are only > permissible moves. The inference rules are the moves. > It has no geometry and its dimensions are only > directions. They only defining winning or losing because none of them > is reducible to truth. Nonsense. Formal systems, as such, have no empirical content. Winning means proving a theorem correctly, i.e. from the postulates using just the rules of inference. Bob Kolker === Subject: Re: Modern Math as a Game > Modern Math as a Game > ~v~~ > Bob Kolker just noted that when done formally mathematics cannot be > distinguished from a game. And I would agree that modern math is just > that. Its axioms are only rules of the game. Its definitions are only > permissible moves. It has no geometry and its dimensions are only > directions. They only defining winning or losing because none of them > is reducible to truth. Bob Kolker, just noted? Took him a while to figure out, eh? Here's some stupendous news for both of you geniuses, Z-Dick and Cock-er: If Life itself is a game, (sometimes brutal, othertimes beautiful, but always awe-inspiring), then it follows that *everything* is a game, because everything is part of life. Now, what other profound bits of wisdom do you two love-birds want to share with us? > ~v~~ === Subject: Re: Modern Math as a Game > If Life itself is a game, (sometimes brutal, othertimes beautiful, but > always awe-inspiring), then it follows that *everything* is a game, because > everything is part of life. Within the field of mathematics, the rules are known. The sciences, other field of study, and life itself only allow us to guess at the rules. -- Ron === Subject: Re: Modern Math as a Game On 3 Oct 2005 11:08:52 -0700, Ron Peterson in >> If Life itself is a game, (sometimes brutal, othertimes beautiful, but >> always awe-inspiring), then it follows that *everything* is a game, because >> everything is part of life. >Within the field of mathematics, the rules are known. The sciences, >other field of study, and life itself only allow us to guess at the >rules. Or to regress rules to truth. I think you'll find within the field of math a very great many of the rules are not known at all. The only rules which can be said to be known in math are some of those not inconsistent with basic assumptions of truth in math. And even many of these are speculative. As in life without some reduction to truth I doubt anyone can even know the rules in math until the game is lost. ~v~~ === Subject: Re: Modern Math as a Game >> ~v~~ >> Bob Kolker just noted that when done formally mathematics cannot be >> distinguished from a game. And I would agree that modern math is just >> that. Its axioms are only rules of the game. Its definitions are only >> permissible moves. It has no geometry and its dimensions are only >> directions. They only defining winning or losing because none of them >> is reducible to truth. >Bob Kolker, just noted? Took him a while to figure out, eh? It took nine months of heavy labor to deliver Bob of his pretensions regarding real number lines and game rule definitions in modern math. >Here's some stupendous news for both of you geniuses, Z-Dick and Cock-er: >If Life itself is a game, (sometimes brutal, othertimes beautiful, but >always awe-inspiring), then it follows that *everything* is a game, because >everything is part of life. Well this is certainly genius in action. Did you get your daddy to help you figure out the logic in this empirical assumption of truth? >Now, what other profound bits of wisdom do you two love-birds want to share >with us? You're just tuning in? Maybe you'd be better off just tuning out. ~v~~ === Subject: Re: Modern Math as a Game In message <43413ebd.19872572@netnews.att.net>, Lester Zick > Modern Math as a Game > ~v~~ >Bob Kolker just noted that when done formally mathematics cannot be >distinguished from a game. And I would agree that modern math is just >that. Its axioms are only rules of the game. Its definitions are only >permissible moves. It has no geometry and its dimensions are only >directions. They only defining winning or losing because none of them >is reducible to truth. Why do you distinguish modern from any other kind? -- Richard Herring === Subject: Re: Modern Math as a Game >> Modern Math as a Game >> ~v~~ >>Bob Kolker just noted that when done formally mathematics cannot be >>distinguished from a game. And I would agree that modern math is just >>that. Its axioms are only rules of the game. Its definitions are only >>permissible moves. It has no geometry and its dimensions are only >>directions. They only defining winning or losing because none of them >>is reducible to truth. >Why do you distinguish modern from any other kind? Mostly a matter of gamesmanship, Red. Bob's Nintendo Gamecube theory of reality. Predecessors to modern math were preoccupied with the possibility of truth. Modern math just codified empiricism's assumption of truth in arithmetic form. ~v~~ === Subject: Re: How to decide system response using step response? > Hi all, > Suppose the forced input to the system is f(t), the step response of > the system is a(t) and the output is y(t). > Now we want to find y(t), > I am confused: ... I won't answer your question because I think you can answer it yourself. One way to discern the correct method is applying both to a simple case with known result and observing which method gives that result. When you know which is correct, I think you will easily see why it must be so. Jerry -- Engineering is the art of making what you want from things you can get. .88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N 210N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N21 0N.88N.88N.88N.88N === Subject: Re: what area of math this problem belongs to? >> A4 = min(A1,A2,A3) >> B4_guess = max(B1,B2,B3) >> n4 = ceil(log2(B4_guess+1-A4)) >> B4 = A4+2^n4-1 > Sorry I made a mistake on the problem description. After the 5 >conditions, I should have said this way, >Given the above, I need to > i) Find A4 and B4, so that A4 + 2^n4 - 1 = B4 and [A4,B4] covers >[A1,B1], [A2,B2] BUT NOT [A3,B3]. If I can find A4, n4 and B4, then I >done; if I can not, then > ii) Find [A4,B4] and [A5, B5], so that their union covers [A1,B1], >[A2,B2] BUT NOT [A3,B3]. Actually in this case I know they are [A1,B1] >and [A2,b2]. > The problem is something like, >[A1,B1]=[0,1], [A2,B2]=[4,5] and [A3,B3]=[6,7]. You can see that the >only answer is [A1,B1] and [A2,B2]. For problem like this, a >try-and-fail-and-adjust algorithm is good enough, but the real problem >set is very large, I must have some math workout first. Hmmm, you could change the algorithm to: B4 = max(B1,B2) A4_guess = min(A1,A2) n4 = ceil(log2(B4+1-A4)) A4 = B4+1-2^n4 then check for intersection with [A3,B3] > If you have some idea how to formalize this into some math equation >like what you did in your reply, or point me where I should look for >related solution, I would appreciate very much. It seems your problem involves superposition of intervals... and a little algebra. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Motion > Aren't you the one who said KE is not changed by an external force? > No, I never said such a thing. Can you provide the quote where > you garbled that inference from? I think I can save you some work. I said that in the case of circular motion, the force does not do any work on the object, and it also does not change the KE. The reason the force does not do any work in that situation is that the motion and the force are at right angles. When a force does work, it changes the KE. As Tipler says, the amount of work done is equal to the change in KE. In the case of circular motion, even though there is an external force, the amount of work done = 0, and the change in KE = 0. As I recall, I brought up this example because you were accusing somebody else of saying that force never changes KE, and I was guessing that that person meant circular motion, since you didn't provide an actual quote to support your accusation. - Randy === Subject: Re: Motion > In my universe, if two things are the same by definition, > they're equal. > Yes, but in the real universe, they're not always equal. In the real universe, things that are the same are always equal. > What objects > or quantities or properties are equal to depends on more than just an > equal sign. If you say, These two mountains are equal, the statement > is nonsense. But so is the sentence those two mountains are the same. > For it to make sense, you must qualify it something to > compare the two, height, area, mineral amounts, difficulty in climbing, > etc. If I were to say the heights of those two mountains are them same I would also mean the heights of those two mountains are equal. Ditto for difficulty, mineral content, or surface area. I can't imagine a situation where I would use same but also mean not equal. > That makes no sense. Again, if two things are the same thing, > they're equal. > Yes, in your universe. OK, having rejected logic, you now reject A = A. > By the Tipler quote that says force is the time rate of change > of momentum? Again, how does that prove that F = dp/dt is wrong? > Another one of your bad inferences. > Tipler said that force is the time rate of change of momentum. > That's not an inference, it's in the quote you provided. > Above, you claim the Tipler quote proves me wrong. > In your universe, proving you wrong equals proving F=dp/dt wrong, so > that's why you inferred I said that. What else are we arguing about? What do you think the Tipler quote proves me wrong ABOUT? I see nothing in the Tipler quote to disagree with. > Well, I did, actually. It's true in non-relativistic situations, > but not in general. > But in your universe relativistic situations are reality while in our > real universe, relativistic situations are imaginary. Not only are > your eyes crossed, but your brain is upside down. No comment. Every one of your pronouncements is a gem. > Nor can I see how work done on an object equals change in > KE of the object contradicts force and momentum are not > interchangeable. Aren't you the one who said KE is not changed by an external force? > No, I never said such a thing. Can you provide the quote where > you garbled that inference from? Can you? - Randy === Subject: Re: Motion > No, sorry, Randy. Your eyes must be crossed. Tipler in my quote does > not equate the work done _to_ KE at all. > The work done by an unbalanced force...on...[an object]...equals the > change in the kinetic energy of the...[object] > Subject of sentence: work done on an object. > Object of sentence: change in KE of object > Verb: equals. > You really don't think the passage above says the work done > equals the change in KE? You don't see this sentence equating > work and KE? You don't see work at the start, KE at the > end, and equals in the middle? > I don't see the word on meaning to in any way. The relevant verb is equals. When Tipler says the work EQUALS the change in KE, you don't read this as the work is EQUAL TO the change in KE? Remember (I know your memory is short, but your words are quoted above), you are saying Tipler does not equate work done, to change in KE. In my universe, if I say x = y, I am equating x to y. - Randy === Subject: Re: Find a different basis for ... <3qaiajFdec1pU1@individual.net> <3qbvokFe5u3cU1@individual.net> === Subject: The Ultimate Online Pharmaceutical boundary=----------1A7632A9BFEDB1E by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with SMTP id j93FavL11812 by artesmarciales.com (Qmailv1) with ESMTP id A9D9DECFBB --------------------------------------------------------------------- Get Cipdro, Avanpdia, Protzac and more ONLINE! Vlagqra - $3.3 Lebvitra - $3.3 Cialims - $3.7 Immitrex - $16.4 Flromax - $2.2 Ulqtram - $0.78 Vifoxx - $4.75 Amgbien - $2.2 VaIiunm - $0.97 Sqoma - $3 Xanvax - $1.09 Maeridia - $2.2 If you compare the cost and services of our competitors, you will see that we give you much moreá for less. Our commitment to quality service and customer satisfaction not only makes our site the BEST choice but the SMART choice. * we guarantee 100% top-quality of all products * we offer a money-back guarantee in case the product doesn't work for you or you may be somehow dissatisfied by its performance * we've efficiently streamlined our service, letting you buy from us in a very discreet, non-embarrassing and hassle-free manner The Facts: Our site has lower prices and more services than its competitors for three consecutive years! Our prices are between 24.62 and 37.12% (an average of 30.87%) less than the market average. Visit our website: http://wateverac.com/ === Subject: Cyberchase Crew in online math chat today Today, 2-3 p.m. ET, CYBERCHASE's Matt Wilson (Harry) and Editorial Director Frances Nankin will chat live at Washingtonpost.com to take questions about kids and math and the hit mystery cartoon on PBS KIDS GO! To participate, log on to washingtonpost.com/liveonline. === Subject: a reposted question. With a correction on it. Hey group, I was wonder to you know of any free websites that teaches basic geometry through games or lessons? My school along time ago would not teach disable people this stuff. Well my niece was learning it and have taught me some of the basic of it. Example Like a square shape or the four corners of the square means 4 and when it come to letters, count up the alphabets & you could up with the answer what the letter is then with problem it comes with, example: X+3=24.96.9b24+3=27. I would appreciate it. Tia. Pam-Doggirl3 === Subject: analysis Let f_n be a sequence of lower semicontinuous functions.Show that the function f defined by f(x)= sup f_n (x) is also lower semicontinuous. I know that any extended real-valued function f is called lower semicontinuous at a point y if f(y) does not equal -oo and f(y) is less than or equal to the liminf f(x) as x--->y please help!!! === Subject: analysis Let f be a real valued function with domain E.Prove that f is continuous if and only if for each open set O there is an open set U such that f^-1[O]= ENU i.e. E intersecting U I know since f is continuous that it's inverse image is continuos as well. How would i prove this please? === Subject: real valued function Let F be a closed set of real numbers and f a real valued function which is defined and continuous on F.Show that there is a function g defined and continuous on (-00, 00) such that f(x)= g(x) for each x in F. I know i can start off by taking g to be a linear in each of the intervals of which H is composed. Please help. Reply === Subject: Re: real valued function > Let F be a closed set of real numbers and f a real valued function > which is defined and continuous on F.Show that there is a function g > defined and continuous on (-00, 00) such that f(x)= g(x) for each x in > F. > I know i can start off by taking g to be a linear in each of the > intervals of which H is composed. Before that, you will have to define H. Then remark it is an open set, so is made up as a disjoint union of open intervals. Define g as you say, but incase H has one or two unbounded components you need to say what to do then. Then prove your extension g is continuous. === Subject: Re: real valued function > Let F be a closed set of real numbers and f a real valued function > which is defined and continuous on F.Show that there is a function g > defined and continuous on (-00, 00) such that f(x)= g(x) for each x in > F. > I know i can start off by taking g to be a linear in each of the > intervals of which H is composed. Which intervals are you talking about? Jose Carlos Santos === Subject: Re: distance between two line > Hi Please can someone help me with this problem: > Find the shortest distance between the 2 lines: > L1: (1,0,0) +t(2,3,4) > L2: (2,1,0) +s(1,2,0) To find the shortest distance between the 2 lines in 2 steps. Step 1. Find the unit vector perpendicular to the 2 lines (2,3,4) Cross product (1,2,0) |i j k| |2 3 4| |1 2 0| = (-8,4,1) Unit perpendicular vector = (-8,4,1)/sqrt(8^2+4^2+1^1) = (-8,4,1)/9 Step 2. Dot product between the vector joining the two points on L1 and L2 and the unit perpendicular vector [(2,1,0) - (1,0,0)] dot product (-8,4,1)/9 = |(1,1,0) . (-8,4,1)/9| = |-4/9| = 4/9 === Subject: Re: distance between two line > Hi Please can someone help me with this problem: > Find the shortest distance between the 2 lines: > L1: (1,0,0) +t(2,3,4) > L2: (2,1,0) +s(1,2,0) > B > To find the shortest distance between the 2 lines in 2 steps. > Step 1. Find the unit vector perpendicular to the 2 lines > (2,3,4) Cross product (1,2,0) > |i j k| > |2 3 4| > |1 2 0| > = (-8,4,1) > Unit perpendicular vector = (-8,4,1)/sqrt(8^2+4^2+1^1) > = (-8,4,1)/9 > Step 2. Dot product between the vector joining the two points on L1 and > L2 and the unit perpendicular vector > [(2,1,0) - (1,0,0)] dot product (-8,4,1)/9 > = |(1,1,0) . (-8,4,1)/9| > = |-4/9| > = 4/9 If you are not comfortable with the above method, I have one which is more understandable. Step 1. Find the normal to the two lines and the corresponding plane to the normal Normal = (2,3,4) Cross product (1,2,0) Same as above |i j k| |2 3 4| |1 2 0| = (-8,4,1) An arbitary plane to the normal: -8x + 4x + z = 0 Find the distance from line L1 to the plane. i.e. Distance from the point (1,0,0) to the plane -8x + 4x + z = 0 d1 = (-8*1 + 4*0 + 1*0)/sqrt(8^2+4^2+1^2) = -8/9 keep the sign for now Find the distance from line L2 to the plane. i.e. Distance from the point (2,1,0) to the plane -8x + 4x + z = 0 d2 = (-8*2 + 4*1 + 1*0)/sqrt(8^2+4^2+1^2) = -12/9 keep the sign for now Shortest distance d = |d1 - d2| = |(-8/9) - (-12/9)| = 4/9 To find the shortest distance this way, we do not know what s and t are. ---------------------------------------------------------- Please see below for the method to find s and t. L1: (1,0,0) +t(2,3,4) L2: (2,1,0) +s(1,2,0) Normal vector N to L1 and L2 is N = L1 X L2 = (2,3,4) X (1,2,0) = (-8,4,1) If we look at the line joining the two points at the shortest distance, with proper s and t we have Starting point: (1,0,0) +t(2,3,4) End point : (2,1,0) +s(1,2,0) Vector joining the two points: (-8,4,1) We can write Starting point + Normal Vector = End point ==> (1,0,0) + t(2,3,4) + r(-8,4,1) = (2,1,0) + s(1,2,0) where r is another unknown in the direction of the normal vector 1 + 2t -8r = 2 + s (1) 0 + 3t +4r = 1 + 2s (2) 0 + 4t + r = 0 + 0s (3) 3 equations with 3 unknowns 2 + t - 20r = 3 (4) = 2 x (1) - (2) 2 + 81t = 3 (5) = 20 x(3) + (4) ==> t = 1/81 1 + 8t = 4 + 5s (6) = (1) + 2 x (2) 1 + 8/81 = 4 + 5s Sub t = 1/81 ==> s = -47/81 It looks lengthy but don't worry. In the actual exam, you can omit most of the explanations. === Subject: Re: distance between two line > Hi Please can someone help me with this problem: > Find the shortest distance between the 2 lines: L1: (1,0,0) +t(2,3,4) > L2: (2,1,0) +s(1,2,0) > B > To find the shortest distance between the 2 lines in 2 steps. > Step 1. Find the unit vector perpendicular to the 2 lines > (2,3,4) Cross product (1,2,0) > |i j k| > |2 3 4| > |1 2 0| > = (-8,4,1) > Unit perpendicular vector = (-8,4,1)/sqrt(8^2+4^2+1^1) > = (-8,4,1)/9 > Step 2. Dot product between the vector joining the two points on L1 and > L2 and the unit perpendicular vector > [(2,1,0) - (1,0,0)] dot product (-8,4,1)/9 > = |(1,1,0) . (-8,4,1)/9| > = |-4/9| > = 4/9 > If you are not comfortable with the above method, I have one which is > more understandable. > Step 1. Find the normal to the two lines and the corresponding plane to > the normal > Normal = (2,3,4) Cross product (1,2,0) Same as above > |i j k| > |2 3 4| > |1 2 0| > = (-8,4,1) > An arbitary plane to the normal: > -8x + 4x + z = 0 > Find the distance from line L1 to the plane. > i.e. Distance from the point (1,0,0) to the plane -8x + 4x + z = 0 > d1 = (-8*1 + 4*0 + 1*0)/sqrt(8^2+4^2+1^2) = -8/9 keep the sign for now > Find the distance from line L2 to the plane. > i.e. Distance from the point (2,1,0) to the plane -8x + 4x + z = 0 > d2 = (-8*2 + 4*1 + 1*0)/sqrt(8^2+4^2+1^2) = -12/9 keep the sign for > now > Shortest distance d = |d1 - d2| > = |(-8/9) - (-12/9)| > = 4/9 > To find the shortest distance this way, we do not know what s and t > are. > ---------------------------------------------------------- > Please see below for the method to find s and t. > L1: (1,0,0) +t(2,3,4) > L2: (2,1,0) +s(1,2,0) > Normal vector N to L1 and L2 is > N = L1 X L2 > = (2,3,4) X (1,2,0) > = (-8,4,1) > If we look at the line joining the two points at the shortest distance, > with proper s and t we have > Starting point: (1,0,0) +t(2,3,4) > End point : (2,1,0) +s(1,2,0) > Vector joining the two points: (-8,4,1) > We can write > Starting point + Normal Vector = End point > ==> (1,0,0) + t(2,3,4) + r(-8,4,1) = (2,1,0) + s(1,2,0) > where r is another unknown in the direction of the normal vector > 1 + 2t -8r = 2 + s (1) > 0 + 3t +4r = 1 + 2s (2) > 0 + 4t + r = 0 + 0s (3) > 3 equations with 3 unknowns > 2 + t - 20r = 3 (4) = 2 x (1) - (2) > 2 + 81t = 3 (5) = 20 x(3) + (4) > ==> t = 1/81 > 1 + 8t = 4 + 5s (6) = (1) + 2 x (2) > 1 + 8/81 = 4 + 5s Sub t = 1/81 > ==> s = -47/81 > It looks lengthy but don't worry. In the actual exam, you can omit > most of the explanations. One more point to add. If we use the unit normal then r would be the shortest distance between the two lines. Unit normal = (-8/9,4/9,1/9) ==> (1,0,0) + t(2,3,4) + r(-8/9,4/9,1/9) = (2,1,0) + s(1,2,0) 1 + 2t -8r/9 = 2 + s (1) 0 + 3t +4r/9 = 1 + 2s (2) 0 + 4t + r/9 = 0 + 0s (3) t = 1/81 s = -47/81 0 + 4/81 + r/9 = 0 + 0s (4) sub t = 1/81 into (3) r = -4/9 Take the positive value, so shortest distance = 4/9 === Subject: Re: distance between two line In sci.math, brad sue <151992.1128306585158.JavaMail.jakarta@nitrogen.mathforum.org>: > Hi Please can someone help me with this problem: > Find the shortest distance between the 2 lines: > L1: (1,0,0) +t(2,3,4) > L2: (2,1,0) +s(1,2,0) I'd have to do it the hard way, using calculus. This is of course extreme overkill, but my geometry is a little weak. For starters, one can change the problem so that one of the lines starts at the origin. L1': (0,0,0) +t(2,3,4) L2': (1,1,0) +s(1,2,0) This should make bits of the math slightly easier. Rotations would also work if one can figure out what rigid transformation M can be used to transform the vector (1,2,0) to something that might be, such as (sqrt(5),0,0). Uh...maybe that's not easier... :-) Now write D^2, the square of the distance between two points on the line. D^2(s,t) = (2t-(1+s))^2 + (3t-(1+2s))^2 + (4t)^2 = 29*t^2 + 5*s^2 - 16*s*t + 6*s - 10*t + 2 Under certain conditions, we can minimize D^2(s,t) by minimizing first D^2(s,t) holding t constant, then minimizing the resulting function of t. So first, minimize F_t(s) = D^2(s,t) = 29*t^2 + 5*s^2 - 16*s*t + 6*s - 10*t + 2 Obviously F_t'(s) = 10*s - 16*t - 6 . (This is actually a partial derivative, so I'm abusing the notation here.) Ergo, s = (16*t - 6)/10 = (8*t - 3)/5. We now have to minimize F(t) = F_t(8*t - 3)/5) = 81/5*t^2 - 2/5*t + 1/5. F'(t) = 162/5*t - 2/5; therefore t = 1/81, s=-47/81, and the actual distance is 4/9. This will only give one partial credit since I've not characterized what conditions are required for using this method. I'm not all that up on my analysis, either... :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: distance between two line > <151992.1128306585158.JavaMail.jakarta@nitrogen.mathforum.org>: > Hi Please can someone help me with this problem: > Find the shortest distance between the 2 lines: > L1: (1,0,0) +t(2,3,4) > L2: (2,1,0) +s(1,2,0) If A = (a1,a2,a3) and B = (b1,b2,b3) are vectors then the dot product is A.B = a1*b2 + a2*b2 + a3*b3 Let P(t) = (1,0,0) + t*(2,3,4) = (1+2*t, 3*t, 4*t) and Q(s) = (2,1,0) + s*(1.2.0) = (2+ s, 1 + 2*s, 0) ( P(y) - Q(s) ).( P(y) - Q(s) ) is an expression, quadratic in s and t, rpresenting the square of the distance between P(y) ad Q(s). It can be minimized algebraically by completing the squares for s and t, or with calculus by taking derivatives with respect to s and t. The square root of that minimum is desired shortest distance. === Subject: Re: Monic polynomials of the nth degree! >> 1,2,3,4,5,6,7,8,9,10,11,12,13,.. deg. >>------------------------------------------------- >> 1,1,2,3,5,8,13,21,34,55,89,144,233,377,.. >> 1,2,5,12,29,70... >> 1,3,10,33,109,... >> 1,4,17,72,305,... >> 1,5,26,135,701,... >> I could be wrong but I believe these (column) >> sequences are derived from monic polynomials. >Now you've lost me. Where do you get the 1,2,5,12,29,70? The second term in each of the above sequences including the Fibonacci sequence is the multiplier for the entire sequence. E.G. Given [1,2] for the start of the second sequence -- 2*2+1= 5 2*5+2= 12 2*12+5 =29 2*29+12 =70 etc. Given [1,3] for the start of the third sequence -- 3*3+1=10 3*10+3=33 3*33+10=109 3*109+33=360 etc. given [1,4] next sequence. 4*4+1=17 4*17+4=72 etc. And so on ---->oo There is no direct closed form for this as far as solving by column the monic polynomial that would give you the column sequence that would start with a Fibonacci number. Using this table, as I recall, I have solved up too monic degree 7 but no higher to find each term in the degree 7 column sequence. I am sure you will have a complex way to calculate each term in these column sequences after you see how they are constructed by row. That's why I said in my OP that some of these monic polynomials are much more difficult than others such as the closed form method that creates each term of a certain nth degree monic poly. with ease. Dan === Subject: Re: Monic polynomials of the nth degree! <33159101.1128355841580.JavaMail.jakarta@nitrogen.mathforum.org> 1,2,3,4,5,6,7,8,9,10,11,12,13,.. deg. >>------------------------------------------------- >> 1,1,2,3,5,8,13,21,34,55,89,144,233,377,.. >> 1,2,5,12,29,70... >> 1,3,10,33,109,... >> 1,4,17,72,305,... >> 1,5,26,135,701,... >> I could be wrong but I believe these (column) >> sequences are derived from monic polynomials. >Now you've lost me. Where do you get the 1,2,5,12,29,70? > The second term in each of the above sequences > including the Fibonacci sequence is the multiplier > for the entire sequence. > E.G. > Given [1,2] for the start of the second sequence -- > 2*2+1= 5 > 2*5+2= 12 > 2*12+5 =29 > 2*29+12 =70 > etc. > Given [1,3] for the start of the third sequence -- > 3*3+1=10 > 3*10+3=33 > 3*33+10=109 > 3*109+33=360 > etc. > given [1,4] next sequence. > 4*4+1=17 > 4*17+4=72 > etc. > And so on ---->oo OK, why didn't you say so in the first place? So the k'th member of the sequence that starts 1, r is x_{r,k} where x_{r,1} = 1, x_{r,2} = r and x_{r,k+1} = r x_{r,k} + x_{r,k-1}. The solution of this recurrence is x_{r,k} = (r^2+4)^{-1/2} (((r+sqrt(r^2+4))/2)^k - ((r-sqrt(r^2+4))/2)^k) = 2^{1-k} (r^2+4)^(-1/2) sum_{j=1..k, j odd} (k choose j) r^(k-j) (r^2+4)^{j/2} = 2^(1-k) sum_{i=0}^floor((k-1)/2} (k choose (2i+1)) r^(k-2i-1) (r^2+4)^i Yes, for each column k you have a monic polynomial of degree k-1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Monic polynomials of the nth degree! >>Where x= 1,2,3,4..x for each monic polynomial >>below. >??? What do you mean? >>x^3 + 1x^1 - 10x^0 +1 = a monic 3rd degree. >>x^4 + 2x^2 - 10x^1 +1 = a monic 4th degree. >>x^5 + 3x^3 - 10x^2 +1 = a monic 5th degree. >>x^6 + 4x^4 - 10x^3 +1 etc. >>x^7 + 5x^5 - 10x^4 +1 >>x^n + (n-2)x^(n-2) - 10x^(n-3)+1 = a monic nth >>degree. >OK, so these are monic polynomials of each degree. So >what? >Where n--->oo. >>In other words a closed form method. >What method? For doing what? >>The question I have is, why is this so simple? >Why is what simple? >>Normally anything above a degree 5 is hard to >>solve. >You didn't solve them. >>E.G. >>A 13th degree monic polynomial and its associated >>sequence. >>x^13 + 11x^11 - 10x^10 + 1 = a monic 13th degree. >>1,20479,2952449,102760447,1660156249,16446799871, >>115814852089,633507676159,2852189640017, >>10999999999999,37401766274641,114547123814399, >>321210286008169,835367443570687, >>2035575878906249,4686118557581311,.. >These seem to be the values of >x^13 + 11*x^11 - 10*x^10 - 1 at x=0,1,2,... >>Where its 13 delta row is a repeated 13!. >>This is true for all these sequences where >>they represent the factorial of the degree >>in the last delta row >0. >You're taking the difference sequences? Yes, for any >polynomial a_n x^n + ... + a_0 of degree n, after n >differences you'll get all n! a_n. This is easy to >prove. >>A much more complex monic polynomial that has no >>known equation for degrees higher than 8 are >>taken from a Fibonacci matrix where the first >>row is the Fib. sequence. >That's not a polynomial. Hint: F_{n+1} - F_n = F_{n-1} >by definition of the Fibonacci sequence. I know its not but the associated sequences along with the first term(a Fibonacci number) is. Extracting each sequence column wise! E.G 1,2,3,4,5,6,7,8,9,10,11,12,13,.. deg. ------------------------------------------------- 1,1,2,3,5,8,13,21,34,55,89,144,233,377,.. 1,2,5,12,29,70... 1,3,10,33,109,... 1,4,17,72,305,... 1,5,26,135,701,... I could be wrong but I believe these (column) sequences are derived from monic polynomials. Dan === Subject: Re: Monic polynomials of the nth degree! <18382166.1128341281062.JavaMail.jakarta@nitrogen.mathforum.org>A much more complex monic polynomial that has no >>known equation for degrees higher than 8 are >>taken from a Fibonacci matrix where the first >>row is the Fib. sequence. >That's not a polynomial. Hint: F_{n+1} - F_n = F_{n-1} >by definition of the Fibonacci sequence. > I know its not but the associated sequences along > with the first term(a Fibonacci number) is. > Extracting each sequence column wise! > E.G > 1,2,3,4,5,6,7,8,9,10,11,12,13,.. deg. > ------------------------------------------------- > 1,1,2,3,5,8,13,21,34,55,89,144,233,377,.. > 1,2,5,12,29,70... > 1,3,10,33,109,... > 1,4,17,72,305,... > 1,5,26,135,701,... > I could be wrong but I believe these (column) > sequences are derived from monic polynomials. Now you've lost me. Where do you get the 1,2,5,12,29,70? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: How do we measure the speed of the light/radio arriving from distant galaxies <01U_e.40679$sx2.4368@fed1read02> <433C106A.188F@nf.sympatico.ca> Relur a htiw? === Subject: Re: What manifolds allow an everywhere parallel vector field? In , on 10/02/2005 at 08:25 PM, lrudolph@panix.com (Lee Rudolph) said: >I don't get the i.e. there; but then again, I don't >really know what you mean by everywhere parallel. My reading is that he's not familiar with the concept of a connection. Can someone supply him with the URL of a good introduction to parallel transport? -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: What manifolds allow an everywhere parallel vector field? > In , on 10/02/2005 > at 08:25 PM, lrudolph@panix.com (Lee Rudolph) said: >I don't get the i.e. there; but then again, I don't >really know what you mean by everywhere parallel. > My reading is that he's not familiar with the concept of a connection. > Can someone supply him with the URL of a good introduction to parallel > transport? _Elementary Topics in Differential Geometry_ by Thorpe http://www.amazon.com/exec/obidos/tg/detail/-/0387903577/qid=1128376459/s r=8-1/ref=pd_bbs_1/102-8056077-5548929?v=glance&s=books&n=507846 is a good introduction for someone who has had a course in advanced calculus. One good point of the book is that he doesn't get hung up on the definitions of differential manifolds - only inverse images of regular values (ala Lagrange multipliers) are considered. (One bad point of the book is that he doesn't go over general cases of differential manifolds.) Holonomy groups are introduced as an exercise in the section on parallel transport - an exercise worth doing. -- Jeffrey Rolland === Subject: Re: What manifolds allow an everywhere parallel vector field? 10/02/2005 at 05:03 PM, john_ramsden@sagitta-ps.com said: >The only 2D orientable manifold that allows an everywhere parallel >vector field (i.e., no vanishing points) has genus 1, i.e. smoothly >deformable into a one-hole torus. ITYM The only compact 2D orientable manifold; the plane R^2 allows such a field, as does the cylinder RxS^1. Also, your parenthetical note is incorrect; parallelism is defined in terms of a connection. 10/03/2005 >I meant, and perhaps should have done (given my >shaky knowledge of differential geometry terms), >by a picture of a surface covered with small >arrows, it is well known that on a sphere the >arrows cannot be combed to be parallel >everywhere (the hairy ball theorem); The theorem says something stronger than that. >That is what I meant by everywhere parallel >(intrinsicly to the manifold); but if this is >the wrong term I'd be glad to learn the correct >way of expressing the notion. Google for connection or for parallel transport. Note that there exist inequivalent connections on the same manifold. >Presumably this is the result you give below, >and everywhere parallel corresponds to >nowhere zero, No. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: What manifolds allow an everywhere parallel vector field? everywhere parallel vector field (i.e., no > vanishing points) > I don't get the i.e. there; but then again, I don't > really know what you mean by everywhere parallel. > Do you have some sort of connection, Riemannian metric, > or whatever? I would have *thought* that by no > vanishing points you would have meant no points > at which the vector field is 0, but that seems > (to me) to have nothing much to do with any notion > of parallel. I meant, and perhaps should have done (given my shaky knowledge of differential geometry terms), by a picture of a surface covered with small arrows, it is well known that on a sphere the arrows cannot be combed to be parallel everywhere (the hairy ball theorem); but on a torus they can. That is what I meant by everywhere parallel (intrinsicly to the manifold); but if this is the wrong term I'd be glad to learn the correct way of expressing the notion. I have gleaned that for orientable surfaces _only_ the torus allows an everywhere parallel combing. A sphere (as noted above) and surfaces of genus >= 2 do not. Presumably this is the result you give below, and everywhere parallel corresponds to nowhere zero, possibly after some suitable mapping of the vector field. But I am also interested in the corresponding question for non-orientable surfaces and for higher-dimensional compact manifolds, namely in which do there exist everywhere parallel (in my possibly incorrect but hopefully now clear terminology) stackings of submanifolds? I suppose it suffices to state the result only for submanifolds of codimension 1. > has genus 1, i.e. smoothly deformable into a > one-hole torus. > Is a similar result for higher dimensional > manifolds known? > If what you really want is not no points at which the > vector field is 0, then ignore the rest of this post. > A compact manifold M with empty boundary has a nowhere-vanishing > vectorfield if and only if the Euler characteristic of every > component of M is 0. A non-compact manifold M with empty boundary > has a nowhere-vanishing vectorfield if and only if each of the > compact components of M has Euler characteristic 0. A manifold > with boundary has a nowhere-vanishing vectorfield if and only > if its interior (i.e., the complement of its boundary) has one. > And how about nonorientable manifolds? > The theorems just stated don't require orientability hypotheses. > All of them follow easily (the non-compact and non-empty-boundary > cases by the cheap trick of shoving the difficulties off to > infinity or across the boundary) from the case of a compact, > connected manifold with empty boundary. To find a proof of > that case, you might do a Google (or Wikipedia) search for > Poincare-Hopf. (The direction non-zero vectorfield implies > zero Euler characteristic is the easier of the two.) John R Ramsden (jhnrmsdn@yahoo.com.uk) remove m from com to reply === Subject: Re: What manifolds allow an everywhere parallel vector field? >>I don't get the i.e. there; but then again, I don't >>really know what you mean by everywhere parallel. > I meant, and perhaps should have done (given my > shaky knowledge of differential geometry terms), > by a picture of a surface covered with small > arrows, it is well known that on a sphere the > arrows cannot be combed to be parallel > everywhere (the hairy ball theorem); but > on a torus they can. > That is what I meant by everywhere parallel OK, so all you're asking for is a nowhere vanishing vector field on the manifold. This maybe happens more often than you might want. For example, *all* three dimensional manifolds allow this. In fact, three manifolds all admit a parallelization, which means that there are three nowhere vanishing vectors fields which are everywhere linearly independent. (This is much harder to do.) In general, Lee gave the result: >>A compact manifold M with empty boundary has a nowhere-vanishing >>vectorfield if and only if the Euler characteristic of every >>component of M is 0. As an example of that, every compact odd-dimensional manifold has a nowhere vanishing vector field. === Subject: Classifieds Site http://BayPortal.net === Subject: Re: Birthday odds? beginner question | > As I understand it, the odds of calculating the same birthday in a | > group of varying size make an assumption. They start calculations | > assuming that in each succeding calculation in the formula, P = | > 1-(364/365 X 363/354 X 362/365 ...) that for each succeeding | > iteration(?) there were no matches in the previous trials, and that the | > progression is simply as stated. | > Is this correct? | > Is there a more precise calculation? | > Is my question valid? | | If the number of terms is supposed to be the number of people then it | should be 1-(365/365 X 364/365 X 363/365 X 362/365 ...). For example, | if there are two people in a group then the probability of their | sharing a birthday is 1-(365/365 X 364/365) = 1/365. I assume that | 354 for 365 is just a typo. I'm confused. If there is one person in the group, the probability that I share a birthday with myself is 1-365/365? Androcles, birthdayless. | | The assumptions are that every year has 365 days (i.e. leap years are | ignored), and that the people's birthdays are uniformly distributed | throughout the year (i.e. everybody has exactly 1/365 chance of having | a birthday on any given day). The formula given is exact under these | assumptions; there is no more precise way to do it. | | I am not entirely sure what you mean by no matches in the previous | trials. If you mean something like why doesn't the formula take into | account cases where some of the people have matching birthdays? then | the answer is that it isn't meant to. We specifically want to exclude | those cases. The expression 365/365 X 364/365 X 363/365 X 362/365 ... | is the probability that all those people will all have DIFFERENT | birthdays. Then it's subtracted from one to give the probability that | at least two people will share a birthday. | === Subject: Re: Birthday odds? beginner question >| > As I understand it, the odds of calculating the same birthday in a >| > group of varying size make an assumption. They start calculations >| > assuming that in each succeding calculation in the formula, P = >| > 1-(364/365 X 363/354 X 362/365 ...) that for each succeeding >| > iteration(?) there were no matches in the previous trials, and that > the >| > progression is simply as stated. >| | > Is this correct? >| | > Is there a more precise calculation? >| | > Is my question valid? >| If the number of terms is supposed to be the number of people then it >| should be 1-(365/365 X 364/365 X 363/365 X 362/365 ...). For example, >| if there are two people in a group then the probability of their >| sharing a birthday is 1-(365/365 X 364/365) = 1/365. I assume that >| 354 for 365 is just a typo. > I'm confused. If there is one person in the group, the probability > that I share a birthday with myself is 1-365/365? If there is one person in the group, the probability that there are two of you with the same birthday is 1-365/365. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Birthday odds? beginner question | | >| > As I understand it, the odds of calculating the same birthday in a | >| > group of varying size make an assumption. They start calculations | >| > assuming that in each succeding calculation in the formula, P = | >| > 1-(364/365 X 363/354 X 362/365 ...) that for each succeeding | >| > iteration(?) there were no matches in the previous trials, and that | > the | >| > progression is simply as stated. | >| > | >| > Is this correct? | >| > | >| > Is there a more precise calculation? | >| > | >| > Is my question valid? | >| | >| If the number of terms is supposed to be the number of people then it | >| should be 1-(365/365 X 364/365 X 363/365 X 362/365 ...). For example, | >| if there are two people in a group then the probability of their | >| sharing a birthday is 1-(365/365 X 364/365) = 1/365. I assume that | >| 354 for 365 is just a typo. | | > I'm confused. If there is one person in the group, the probability | > that I share a birthday with myself is 1-365/365? | | | If there is one person in the group, the probability that there are two | of you with the same birthday is 1-365/365. Ah... I see. So the probability that no people in the group share the same birthday is 1 - (), is it? Androcles. | | -- | Dave Seaman | Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. | === Subject: Re: Birthday odds? beginner question | > As I understand it, the odds of calculating the same birthday in > | >| > group of varying size make an assumption. They start calculations > | >| > assuming that in each succeding calculation in the formula, P = > | >| > 1-(364/365 X 363/354 X 362/365 ...) that for each succeeding > | >| > iteration(?) there were no matches in the previous trials, and > that > | > the > | >| > progression is simply as stated. > | >| | >| > Is this correct? > | >| | >| > Is there a more precise calculation? > | >| | >| > Is my question valid? > | >| > | >| If the number of terms is supposed to be the number of people then > it > | >| should be 1-(365/365 X 364/365 X 363/365 X 362/365 ...). For > example, > | >| if there are two people in a group then the probability of their > | >| sharing a birthday is 1-(365/365 X 364/365) = 1/365. I assume that > | >| 354 for 365 is just a typo. > | > I'm confused. If there is one person in the group, the probability > | > that I share a birthday with myself is 1-365/365? > | If there is one person in the group, the probability that there are > two > | of you with the same birthday is 1-365/365. > Ah... I see. > So the probability that no people in the group share the same birthday > is > 1 - (), is it? The number of terms in the product is the number of people in the group, not the number who share birthdays. Possibly you are asking what happens if there are no people in the group, when the formula gives a probability of 1 - () that two or more will share a birthday. This may appear wrong, but actually () is equal to one, not zero, so 1 - () = 1 - 1 = 0, which is correct. This is the old product of zero terms equals one chestnut, exemplified by x^0 = 1 and 0! = 1. It might be easier to see if we let n represent the number of people in the group, and write the probability of two or more people sharing birthdays as 1 - 365!/(365^n*(365-n)!). Setting n = 0 we get 1 - 365!/(365^0*365!) = 0, as expected. === Subject: Re: Birthday odds? beginner question | > | | > | >| > As I understand it, the odds of calculating the same birthday in | > a | > | >| > group of varying size make an assumption. They start calculations | > | >| > assuming that in each succeding calculation in the formula, P = | > | >| > 1-(364/365 X 363/354 X 362/365 ...) that for each succeeding | > | >| > iteration(?) there were no matches in the previous trials, and | > that | > | > the | > | >| > progression is simply as stated. | > | >| > | > | >| > Is this correct? | > | >| > | > | >| > Is there a more precise calculation? | > | >| > | > | >| > Is my question valid? | > | >| | > | >| If the number of terms is supposed to be the number of people then | > it | > | >| should be 1-(365/365 X 364/365 X 363/365 X 362/365 ...). For | > example, | > | >| if there are two people in a group then the probability of their | > | >| sharing a birthday is 1-(365/365 X 364/365) = 1/365. I assume that | > | >| 354 for 365 is just a typo. | > | | > | > I'm confused. If there is one person in the group, the probability | > | > that I share a birthday with myself is 1-365/365? | > | | > | | > | If there is one person in the group, the probability that there are | > two | > | of you with the same birthday is 1-365/365. | > Ah... I see. | > So the probability that no people in the group share the same birthday | > is | > 1 - (), is it? | | The number of terms in the product is the number of people in the | group, not the number who share birthdays. Possibly you are asking what | happens if there are no people in the group, when the formula gives a | probability of 1 - () that two or more will share a birthday. This | may appear wrong, but actually () is equal to one, not zero, so 1 - | () = 1 - 1 = 0, which is correct. It is? I'll inform the Christian Church of your discovery, they've lacked a year zero for ... 2000+ years. | | This is the old product of zero terms equals one chestnut, Not a walnut or brazilnut, then, so you won't be needing the nutcrackers: (). 1- . Better tell Tchaikowski. :-) Androcles. | exemplified by x^0 = 1 and 0! = 1. It might be easier to see if we let | n represent the number of people in the group, and write the | probability of two or more people sharing birthdays as 1 - | 365!/(365^n*(365-n)!). Setting n = 0 we get 1 - 365!/(365^0*365!) = 0, | as expected. | === Subject: Female equation Female equation at http://www.washtimes.com/metro/20051002-093823-4290r.htm === Subject: Female equation Female equation at http://www.washtimes.com/metro/20051002-093823-4290r.htm === Subject: Re: Female equation > Female equation > at http://www.washtimes.com/metro/20051002-093823-4290r.htm A computer scientist, who probably is a geek, shouldn't decry being categorized as belonging to a group that has been labeled as consisting of geeks. -- Ron === Subject: Super harmonic functions? When I took complex analysis, I learned that a harmonic function is one that satisfies Poisson's equation: nabla^2 f = 0 Wikipedia says that a subharmonic function is one that satisfies: nabla^2 f >= 0 Is there a similarly *simple* definition of a super harmonic function? Is there a geometric significance to super harmonic functions that I can use to visualize their general properites? === Subject: Re: Uniform sampling of a sphere Nils' Code was very helpful. === Subject: One-point union How can I show that the one-point union of S^2 and S^1, which I will denote S^2 */ S^1, is homotopically equivalent to the union of S^2 and A, which I will denote S^2 U A, where A is the line segment joining the north and south poles of S^2? I have tried to visualize this and don't know if I have the right picture I need to construct explicit maps f,g where f : S^2 */ S^1 ---> S^2 U A and g : S^2 U A ---> S^2 */ S^1, such that f o g is homotopic to the identity and g o f is homotopic to the identity. Visually, if I take S^2 union A, I can imagine grabbing A at the south pole, moving it around towards the north pole and joining the ends of the line to form a circle such that the intersection of the circle and the sphere is one point (at the north pole). So I have identified the south pole with the north pole in this manner, and it seems like this was all continuous. I was thinking that since I wasn't getting anywhere with constructing explicit functions f and g, I might try to think about stereographic projection, but I am still lost. Kevin === Subject: Re: One-point union > How can I show that the one-point union of S^2 and S^1, which I will denote > S^2 */ S^1, is homotopically equivalent to the union of S^2 and A, which I > will denote S^2 U A, where A is the line segment joining the north and south > poles of S^2? > I have tried to visualize this and don't know if I have the right picture I > need to construct explicit maps f,g where f : S^2 */ S^1 ---> S^2 U A and g > : S^2 U A ---> S^2 */ S^1, such that f o g is homotopic to the identity and > g o f is homotopic to the identity. > Visually, if I take S^2 union A, I can imagine grabbing A at the south pole, > moving it around towards the north pole and joining the ends of the line to > form a circle such that the intersection of the circle and the sphere is one > point (at the north pole). So I have identified the south pole with the > north pole in this manner, and it seems like this was all continuous. > I was thinking that since I wasn't getting anywhere with constructing > explicit functions f and g, I might try to think about stereographic > projection, but I am still lost. > Kevin Can you show that the figure-eight space is homotopy equivalent to the theta space? Munkres' approach in Topology, 2nd edition, p. 362, might provide one idea for you: find a third space, of which both of yours are deformation retracts. Another idea: instead of looking at S^2 U A, choose a point x on S^2 and consider S^2 U B_x, where B_x is a curve (in the interior of the sphere, say) connecting that point to the south pole. Now let x move toward the south pole. When x reaches the south pole, S^2 U B_x will be the one-point union of S^2 and S^1. Now find a map from S^2 U A to S^2 U B_x for each x (and I would do it so that it's surjective and so the map restricted to S^2 is the identity). In particular, when x is the south pole, this gives you a map from S^2 U A to S^2 v S^1. Try to find a map from S^2 v S^1 back to S^2 U A in a similar way, and use the motion of x from the north pole to the south to define at least one of your homotopies. -- J. H. Palmieri Associate Professor of Mathematics University of Washington Box 354350, Seattle, WA 98195-4350 palmieri@math.washington.edu http://www.math.washington.edu/~palmieri/ === Subject: Missing equality operator Are there any interesting number theories derived from the basic arithmetic axioms while omitting the equality (reflexive) operator? I.e., can we create an arithmetic system with the usual '+', 'x', an '<' operators and the usual '0' and '1' identities, but omit the '=' equality operator? I was thinking about this in the context of Douglas Adam's Bistromathics (http://hhgproject.org/entries/bistromathics.html), wherein a recipriversexclusion number is a number that is equal to anything except itself. Naturally, this led to thinking about excluding the '=' equality operator from a hypothetical number theory, so that (conveniently?) numbers could not ever be equal to themselves. === Subject: Re: Set Theoretical approach to compression <3no9coF2hao3U1@individual.net> <3okmijF6dmc8U1@individual.net> logical error. please supply data relating to the required necessity of someone being unequal, and hence inline for the short straw. (c)2005 Unpaid Intelligence Incorporated Number Theory and Entropy Connection PrimePi[x]=Bound[1/log[y]*Sum[(PartEnt[y]^n)*Diff[-PartEnt[y],y,n],n,0,Infin ity],y,2,x] where.... PartEnt[y]=y*log[y] === Subject: Re: Set Theoretical approach to compression <3no9coF2hao3U1@individual.net> <3okmijF6dmc8U1@individual.net> light knows where it will end up when it starts, and arrives when it is traveling at the speed of light time stops therefore the light leaves and arrives at the same time. === Subject: Re: Divisors = b1 *b2 *b3 ..*(b1 +b2 +b3+...) >Let a(m,r) be the number of ways that: >divisors of m = (product{k=1 to r} b(k)) * (sum{k=1 to r} b(k)), >where {b(k)} consists only of r positive integers. >Different orderings of the same integers are counted separately. By the previous sentence I mean, if there are n(k) k's in a particular {b(k)}, then the number of reorderings of this sequence is r!/(n(1)!n(2)!n(3)!...), where sum{k>=1} n(k) = r. >So, for example, for m= 6, >1 = 1*(1), >2 = 1*1*(1+1), >3 = 1*1*1*(1+1+1), >6 = 1*1*1*1*1*1*(1+1+1+1+1+1) = > 1*2*(1+2) = 2*1*(2+1). >So we have a(6,2) = 3, for 1*1*(1+1), 1*2*(1+2), and 2*1*(2+1). >In other words, if c(m,r) = the number of ways that: >m = (product{k=1 to r} b(k)) * (sum{k=1 to r} b(k)), >then a(m,r) = sum{k|m} c(k,r). >As a math puzzle I ask: >What, then is the average, for a given r, of the a(m,r)'s? >Rigorously, what is >limit{n->oo} (1/n) sum{k=1 to n} a(k,r) ? >I will give my answer in a few days if no one else gets it sooner. >Leroy Quet === Subject: Divisors = b1 *b2 *b3 ..*(b1 +b2 +b3+...) Let a(m,r) be the number of ways that: divisors of m = (product{k=1 to r} b(k)) * (sum{k=1 to r} b(k)), where {b(k)} consists only of r positive integers. Different orderings of the same integers are counted separately. So, for example, for m= 6, 1 = 1*(1), 2 = 1*1*(1+1), 3 = 1*1*1*(1+1+1), 6 = 1*1*1*1*1*1*(1+1+1+1+1+1) = 1*2*(1+2) = 2*1*(2+1). So we have a(6,2) = 3, for 1*1*(1+1), 1*2*(1+2), and 2*1*(2+1). In other words, if c(m,r) = the number of ways that: m = (product{k=1 to r} b(k)) * (sum{k=1 to r} b(k)), then a(m,r) = sum{k|m} c(k,r). As a math puzzle I ask: What, then is the average, for a given r, of the a(m,r)'s? Rigorously, what is limit{n->oo} (1/n) sum{k=1 to n} a(k,r) ? I will give my answer in a few days if no one else gets it sooner. Leroy Quet === Subject: Re: Help with tensors (re-post) > I don't think so. Using only numbers, how can you tell me >> which way is north? >Tom asked the same question. The problem is that there isn't enough >context to parse the question. North is north, of course. > Yes, that's a name for a direction. It's *not* a number. >Which way is north from where? Which way are you facing? > That's the point: I can't answer that question using numbers > alone. >These same problems exist with any quantity, not just directions. > That's why it's wrong (usually) to say that physical quantities are > numbers. That's a good point Daryl, sorry if I misspoke. Physical quantities like vectors, or their components, are more than just numbers - they have units and other associated baggage. === Subject: Re: Help with tensors (re-post) <3qaqfuFe6ku0U1@individual.net> <3qb21uFdsvgiU1@individual.netAnd that's a vector, not a covector. > For several years, I made a point when teaching linear algebra > to introduce (linear) scales, which are to covectors what > oriented line segments are to vectors. > Misner, Thorne and Wheeler use something like that > in their pictorial representation of vectors > and one-forms in the big black book: a vector > of a given length passes through hyperplanes > a given distance apart, ringing a bell each time > it pierces a plane. The number of rings of the bell > is the pairing. Alas, I think more readers > found it tiresome than helpful :-/ > Agreed, for all that's good in MTW, I defer to > Weinberg, (Roberts aside), because analogies can > be confusing for the sophicated reader. He said tiresome, not confusing. There's nothing whatsoever confusing about this analogy. Possibly MTW communicated it in a way that you found confusing; their prose style is a bit scattershot, not everyone's cup of tea. But the analogy itself is nevertheless quite nice, and the pictures in MTW are worth thousands of words. You should give it another try. (If the problem is you don't like bells, then think of clicking some hand counter or such.) > Better use the truth, and admit not everybody will > grasp GR. So I stopped in the theory for the > dinks, if you want soap go elsewhere, I think that's > Weinbergs PoV. What is untrue about the analogy? I would be interested in your pointing out anything in Weinberg that is inconsistent with it. I think you won't find anything. === Subject: Re: Help with tensors (re-post) <3qaqfuFe6ku0U1@individual.net> <3qb21uFdsvgiU1@individual.net> Mr. Russell, this is an ettiquette reply. >And that's a vector, not a covector. > For several years, I made a point when teaching linear algebra > to introduce (linear) scales, which are to covectors what > oriented line segments are to vectors. Misner, Thorne and Wheeler use something like that > in their pictorial representation of vectors > and one-forms in the big black book: a vector > of a given length passes through hyperplanes > a given distance apart, ringing a bell each time > it pierces a plane. The number of rings of the bell > is the pairing. Alas, I think more readers > found it tiresome than helpful :-/ > Agreed, for all that's good in MTW, I defer to > Weinberg, (Roberts aside), because analogies can > be confusing for the sophicated reader. > He said tiresome, not confusing. > There's nothing whatsoever confusing about this > analogy. Possibly MTW communicated it in a way that > you found confusing; their prose style is a bit > scattershot, not everyone's cup of tea. But the > analogy itself is nevertheless quite nice, and the > pictures in MTW are worth thousands of words. You > should give it another try. > (If the problem is you don't like bells, then think > of clicking some hand counter or such.) If one chooses a *strict* text for a reference, I doubt anyone would accept in a heated GR argument the bing-bong analogy. If MTW wants to appeal to an audience using comics, that's fine, but I think that disrespects the work, that is too fluffy, Weinberg doesn't do it. Perhaps MTW is attempting to cover a wider readership, ok fine, but Weinberg is more refined, and I'll say classy. > Better use the truth, and admit not everybody will > grasp GR. So I stopped in the theory for the > dinks, if you want soap go elsewhere, I think that's > Weinbergs PoV. > What is untrue about the analogy? I would be > interested in your pointing out anything in Weinberg > that is inconsistent with it. I think you won't > find anything. Yes I can! Sound does not transfer in a vacuum, GR analogies are for potato heads. I repeat, if your going serious GR, tough it out and don't expect pablum, crutches or training wheels. GR is a fine legal document exposing the universe, and should remain as naked as possible. Ken S. Tucker === Subject: Re: Help with tensors (re-post) Ken S. Tucker says... >If one chooses a *strict* text for a reference, >I doubt anyone would accept in a heated GR argument >the bing-bong analogy. If MTW wants to appeal to an >audience using comics, that's fine, but I think that >disrespects the work, that is too fluffy, Weinberg >doesn't do it. Perhaps MTW is attempting to cover >a wider readership, ok fine, but Weinberg is more >refined, and I'll say classy. Well, MTW covers *very* advanced material. It isn't actually possible for anyone other than serious students to get much out of MTW. That makes the folksy quality of MTW a little weird. I think that the point is to puncture pretentions, rather than get a wider audience. It's actually like Conway's mathematics books, such as On Numbers and Games or Winning Ways for Your Mathematical Plays. The language sounds very whimsical and folksy, but the mathematics is completely rigorous and not for wimps. >> What is untrue about the analogy? I would be >> interested in your pointing out anything in Weinberg >> that is inconsistent with it. I think you won't >> find anything. >Yes I can! Sound does not transfer in a vacuum, How does MTW lead anyone to think that sound travels through the vacuum? >GR analogies are for potato heads. Or for anyone who is trying to understand it. >I repeat, if your going serious GR, tough it >out and don't expect pablum, crutches or training >wheels. Some would say that the use of components *is* a crutch, that the conflation of vectors and 1-forms is a simplification to make things easier for some people who can't handle differential geometry. >GR is a fine legal document exposing the universe, >and should remain as naked as possible. That's exactly what the discussion about tensors is about: what is the most basic way to understand tensors. How do you make tensors as naked as possible? Weinberg takes one approach, where the bottom line is components and the manipulation of components. MTW takes the other approach, that the bottom line is the geometry. One man's heart of the matter is another man's fluff. -- Daryl McCullough Ithaca, NY === Subject: Re: Help with tensors (re-post) <3qaqfuFe6ku0U1@individual.net> <3qb21uFdsvgiU1@individual.net> (Sorry for my earlier blank reply) > Mr. Russell, this is an ettiquette reply. >And that's a vector, not a covector. > For several years, I made a point when teaching linear algebra > to introduce (linear) scales, which are to covectors what > oriented line segments are to vectors. Misner, Thorne and Wheeler use something like that > in their pictorial representation of vectors > and one-forms in the big black book: a vector > of a given length passes through hyperplanes > a given distance apart, ringing a bell each time > it pierces a plane. The number of rings of the bell > is the pairing. Alas, I think more readers > found it tiresome than helpful :-/ Agreed, for all that's good in MTW, I defer to > Weinberg, (Roberts aside), because analogies can > be confusing for the sophicated reader. > He said tiresome, not confusing. > There's nothing whatsoever confusing about this > analogy. Possibly MTW communicated it in a way that > you found confusing; their prose style is a bit > scattershot, not everyone's cup of tea. But the > analogy itself is nevertheless quite nice, and the > pictures in MTW are worth thousands of words. You > should give it another try. > (If the problem is you don't like bells, then think > of clicking some hand counter or such.) > If one chooses a *strict* text for a reference, > I doubt anyone would accept in a heated GR argument > the bing-bong analogy. Like I said, if the bongs bother you, take them out. There is still a lesson in this chapter (what is it, 2?) that you desperately need to learn. If MTW wants to appeal to an > audience using comics, that's fine, but I think that > disrespects the work, that is too fluffy, Weinberg > doesn't do it. Perhaps MTW is attempting to cover > a wider readership, ok fine, but Weinberg is more > refined, and I'll say classy. I'll take it that you're one of the people Low talked about, who felt they were being condescended to. As I said, MTW's not for everyone. However, you are certainly wrong about their intended readership -- i.e. if you actually *read* MTW, you find out pretty fast that it is not some sort of GR for Dummies. A great deal of sophistication is expected from the reader. > Better use the truth, and admit not everybody will > grasp GR. So I stopped in the theory for the > dinks, if you want soap go elsewhere, I think that's > Weinbergs PoV. > What is untrue about the analogy? I would be > interested in your pointing out anything in Weinberg > that is inconsistent with it. I think you won't > find anything. > Yes I can! Sound does not transfer in a vacuum, > GR analogies are for potato heads. And vectors don't make bells ring or pierce anything. Do you have any idea how silly you are being? I repeat, if > your going serious GR, tough it out and don't expect > pablum, crutches or training wheels. GR is a fine > legal document exposing the universe, and should > remain as naked as possible. Training wheels can be a great help to those who need them -- a need that the person in question might not even recognize. The proof is in seeing where your bicycle gets you. You do a lot of talking about how fast you are riding, but the scenery around you doesn't seem to change much... === Subject: Re: Help with tensors (re-post) >> Perhaps MTW is attempting to cover >>a wider readership, ok fine, but Weinberg is more >>refined, and I'll say classy. > I'll take it that you're one of the people Low talked > about, who felt they were being condescended to. And, I should add, only in terms of some of the presentation. The content is pretty damned solid. > i.e. if you actually *read* MTW, you find out pretty > fast that it is not some sort of GR for Dummies. > A great deal of sophistication is expected from the > reader. Too true. Weinberg is more refined is a weird thing to say. === Subject: Re: Help with tensors (re-post) <3qaqfuFe6ku0U1@individual.net> <3qb21uFdsvgiU1@individual.net (Sorry for my earlier blank reply) > Mr. Russell, this is an ettiquette reply. >And that's a vector, not a covector. > For several years, I made a point when teaching linear algebra > to introduce (linear) scales, which are to covectors what > oriented line segments are to vectors. Misner, Thorne and Wheeler use something like that > in their pictorial representation of vectors > and one-forms in the big black book: a vector > of a given length passes through hyperplanes > a given distance apart, ringing a bell each time > it pierces a plane. The number of rings of the bell > is the pairing. Alas, I think more readers > found it tiresome than helpful :-/ Agreed, for all that's good in MTW, I defer to > Weinberg, (Roberts aside), because analogies can > be confusing for the sophicated reader. He said tiresome, not confusing. There's nothing whatsoever confusing about this > analogy. Possibly MTW communicated it in a way that > you found confusing; their prose style is a bit > scattershot, not everyone's cup of tea. But the > analogy itself is nevertheless quite nice, and the > pictures in MTW are worth thousands of words. You > should give it another try. (If the problem is you don't like bells, then think > of clicking some hand counter or such.) > If one chooses a *strict* text for a reference, > I doubt anyone would accept in a heated GR argument > the bing-bong analogy. > Like I said, if the bongs bother you, take them > out. There is still a lesson in this chapter (what > is it, 2?) that you desperately need to learn. > If MTW wants to appeal to an > audience using comics, that's fine, but I think that > disrespects the work, that is too fluffy, Weinberg > doesn't do it. Perhaps MTW is attempting to cover > a wider readership, ok fine, but Weinberg is more > refined, and I'll say classy. > I'll take it that you're one of the people Low talked > about, who felt they were being condescended to. > As I said, MTW's not for everyone. However, you are > certainly wrong about their intended readership -- > i.e. if you actually *read* MTW, you find out pretty > fast that it is not some sort of GR for Dummies. > A great deal of sophistication is expected from the > reader. > Better use the truth, and admit not everybody will > grasp GR. So I stopped in the theory for the > dinks, if you want soap go elsewhere, I think that's > Weinbergs PoV. What is untrue about the analogy? I would be > interested in your pointing out anything in Weinberg > that is inconsistent with it. I think you won't > find anything. > Yes I can! Sound does not transfer in a vacuum, > GR analogies are for potato heads. > And vectors don't make bells ring or pierce anything. > Do you have any idea how silly you are being? > I repeat, if > your going serious GR, tough it out and don't expect > pablum, crutches or training wheels. GR is a fine > legal document exposing the universe, and should > remain as naked as possible. > Training wheels can be a great help to those who > need them -- a need that the person in question might > not even recognize. The proof is in seeing where > your bicycle gets you. You do a lot of talking about > how fast you are riding, but the scenery around you > doesn't seem to change much... Sure, BTW, did you see in MTW a quote attributed to Einstein circa 1970, ok that's the only typo I found. But that proves I read it carefully, and it's ok with me. I was quite enlightened by their use of bold algeba in place of tensors as Roberts advocates, but for me it's a bit advanced as I posted yesterday on Generalizing contravariance and covariance. GR is not for sissy's, and Weinberg councils a professional appraoch that I respect. Even better (IMO) but dated is P.G. Bergmann's. Once my new office is operational I will buy all the newest books, because I'm certainly obsolete. You take care and enjoy my friend Ken S. Tucker === Subject: Re: Help with tensors (re-post) <3qaqfuFe6ku0U1@individual.net> <3qb21uFdsvgiU1@individual.net Mr. Russell, this is an ettiquette reply. >And that's a vector, not a covector. > For several years, I made a point when teaching linear algebra > to introduce (linear) scales, which are to covectors what > oriented line segments are to vectors. Misner, Thorne and Wheeler use something like that > in their pictorial representation of vectors > and one-forms in the big black book: a vector > of a given length passes through hyperplanes > a given distance apart, ringing a bell each time > it pierces a plane. The number of rings of the bell > is the pairing. Alas, I think more readers > found it tiresome than helpful :-/ Agreed, for all that's good in MTW, I defer to > Weinberg, (Roberts aside), because analogies can > be confusing for the sophicated reader. > He said tiresome, not confusing. > There's nothing whatsoever confusing about this > analogy. Possibly MTW communicated it in a way that > you found confusing; their prose style is a bit > scattershot, not everyone's cup of tea. But the > analogy itself is nevertheless quite nice, and the > pictures in MTW are worth thousands of words. You > should give it another try. > (If the problem is you don't like bells, then think > of clicking some hand counter or such.) > If one chooses a *strict* text for a reference, > I doubt anyone would accept in a heated GR argument > the bing-bong analogy. If MTW wants to appeal to an > audience using comics, that's fine, but I think that > disrespects the work, that is too fluffy, Weinberg > doesn't do it. Perhaps MTW is attempting to cover > a wider readership, ok fine, but Weinberg is more > refined, and I'll say classy. > Better use the truth, and admit not everybody will > grasp GR. So I stopped in the theory for the > dinks, if you want soap go elsewhere, I think that's > Weinbergs PoV. > What is untrue about the analogy? I would be > interested in your pointing out anything in Weinberg > that is inconsistent with it. I think you won't > find anything. > Yes I can! Sound does not transfer in a vacuum, > GR analogies are for potato heads. I repeat, if > your going serious GR, tough it out and don't expect > pablum, crutches or training wheels. GR is a fine > legal document exposing the universe, and should > remain as naked as possible. > Ken S. Tucker === Subject: Re: Help with tensors (re-post) <3qaqfuFe6ku0U1@individual.net> <3qb21uFdsvgiU1@individual.net> It exposes the Universe for extreme cases only. Cybernetics rules for normal systems, such as the developing Global Brain, which is the transformative evolutionary development. === Subject: Re: Help with tensors (re-post) >Misner, Thorne and Wheeler use something like that >in their pictorial representation of vectors >and one-forms in the big black book: a vector >of a given length passes through hyperplanes >a given distance apart, ringing a bell each time >it pierces a plane. The number of rings of the bell >is the pairing. Alas, I think more readers >found it tiresome than helpful :-/ >>Agreed, for all that's good in MTW, I defer to >>Weinberg, (Roberts aside), because analogies can >>be confusing for the sophicated reader. > He said tiresome, not confusing. That's right: it wasn't the pairing of vectors and hyperplanes that was the issue--it was feeling condescended to by the 'bongs of the bell' description. But MTW is a sufficently big book that whoever you are, you'll probably find something to love, and something else to hate. === Subject: Re: Help with tensors (re-post) Robert Low says... >That's right: it wasn't the pairing >of vectors and hyperplanes that was >the issue--it was feeling condescended >to by the 'bongs of the bell' description. >But MTW is a sufficently big book that whoever >you are, you'll probably find something to love, >and something else to hate. What I found really weird about MTW was the combination of the light-hearted style with style and some of the diagrams, you would almost guess that it was a mass-market science for dummies book, with friendly illustrations and non-threatening, folksy language like bongs of a bell. However, the actual *content* is extremely advanced, covering differential geometry, spinors, cosmology, etc. -- Daryl McCullough Ithaca, NY === Subject: Re: Help with tensors (re-post) > Robert Low says.. > What I found really weird about MTW was the > combination of the light-hearted style with > the heavy-duty mathematics. I think that that dissonance is at the root of a lot of the trouble. It's a bloody hard book, full of highly technical stuff---but the way it's written, you'd think it was a superficial popular book. At least, until you try to read it properly. For me, it's the book in that old gag: I'm going to start reading MTW again soon. I started it last year, too. === Subject: Re: Help with tensors (re-post) >> Robert Low says.. >> What I found really weird about MTW was the >> combination of the light-hearted style with >> the heavy-duty mathematics. >I think that that dissonance is at the root >of a lot of the trouble. It's a bloody hard >book, full of highly technical stuff---but >the way it's written, you'd think it >was a superficial popular book. At least, >until you try to read it properly. The reading style just didn't put me in the mind-set to work hard at it. I had better luck with Bergmann-- more old-fashioned, more stodgy, straight to the point. >For me, it's the book in that old gag: >I'm going to start reading MTW again soon. >I started it last year, too. I've always told myself that I'd get back to MTW some day. But there are other titles that I could stuff into that gag. -- Let us learn to dream, gentlemen, then perhaps we shall find the truth... But let us beware of publishing our dreams before they have been put to the proof by the waking understanding. -- Friedrich August Kekul.8e === Subject: Re: Help with tensors (re-post) >>And it doesn't help that grad f is called a >>vector in just about every course where the >>concept is introduced... > The fact that rectilinear and orthogonal coordinate bases are used > throughout lots of introductory courses enables one to sweep under the > rug lots of stuff. Sweep under the carpet or even say things that are just downright misleading. A vector doesn't have covariant components and contravariant components, though it's possible to find texts that say it does. Given a metric, a vector has an associated covector: given a basis, one has components for the vector, an associated basis for the dual space, and components for the associated covector. But all the distinctions get forgotten in Euclidean space. It's a shame that doing the familiar special case first can lead to so much 'unlearning' being required. (Especially when the unlearning doesn't happen.) === Subject: Re: Help with tensors (re-post) >And it doesn't help that grad f is called a >vector in just about every course where the >concept is introduced... >> The fact that rectilinear and orthogonal coordinate bases are used >> throughout lots of introductory courses enables one to sweep under the >> rug lots of stuff. >Sweep under the carpet or even say things that are just >downright misleading. A vector doesn't have covariant >components and contravariant components, though it's possible >to find texts that say it does. Given a metric, a vector >has an associated covector: given a basis, one has components >for the vector, an associated basis for the dual space, and >components for the associated covector. But all the distinctions >get forgotten in Euclidean space. Exactly. >It's a shame that doing the familiar special case first >can lead to so much 'unlearning' being required. (Especially >when the unlearning doesn't happen.) Yes, here is the problem. The special case is taught first, since it is simple. Now, pedagogically, it makes sense to start with a simple case first, then build on it to get to more complex ones. But this is so only if the simple case leads in the direction of the more complex ones. In this case, and in the way the simple case is taught, it leads astray. Instead of being able to proceed from it towards the more complex ones, one has to do (as you say) lots of unlearning first. Now, it doesn't have to be this way. Lots of the needed stuff can be introduced in Euclidean space, by considering non-orthogonal axes, and within a linear algebra course by considering (left and right) eigenvector spaces of non-symmetric matrices. But this is rarely done. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Help with tensors (re-post) >>And it doesn't help that grad f is called a >>vector in just about every course where the >>concept is introduced... > The fact that rectilinear and orthogonal coordinate bases are used > throughout lots of introductory courses enables one to sweep under the > rug lots of stuff. >>Sweep under the carpet or even say things that are just >>downright misleading. A vector doesn't have covariant >>components and contravariant components, though it's possible >>to find texts that say it does. Given a metric, a vector >>has an associated covector: given a basis, one has components >>for the vector, an associated basis for the dual space, and >>components for the associated covector. But all the distinctions >>get forgotten in Euclidean space. >Exactly. >>It's a shame that doing the familiar special case first >>can lead to so much 'unlearning' being required. (Especially >>when the unlearning doesn't happen.) >Yes, here is the problem. The special case is taught first, since it >is simple. Now, pedagogically, it makes sense to start with a simple >case first, then build on it to get to more complex ones. But this is >so only if the simple case leads in the direction of the more complex >ones. In this case, and in the way the simple case is taught, it >leads astray. Instead of being able to proceed from it towards the >more complex ones, one has to do (as you say) lots of unlearning >first. >Now, it doesn't have to be this way. Lots of the needed stuff can be >introduced in Euclidean space, by considering non-orthogonal axes, and >within a linear algebra course by considering (left and right) eigenvector >spaces of non-symmetric matrices. But this is rarely done. And you can do a lot of physics without ever considering such things. The distinctions appear, to a certain extent, in crystallography, anything relativistic that uses a metric instead of ict, and any time non-Cartesian (not just non-orthogonal) coordinate systems come up. But you can fake that out by following a few mathematical recipies by rote and still do a lot of physics without going into the covector details. About the only time it's really important is when you're doing general relativity. And despite the discussions here and the shelf space taken up in the small science sections of chain bookstores, general relativity is a very small part of physics. -- A good plan executed right now is far better than a perfect plan executed next week. -Gen. George S. Patton === Subject: Re: Help with tensors (re-post) >And it doesn't help that grad f is called a >vector in just about every course where the >concept is introduced... >> The fact that rectilinear and orthogonal coordinate bases are used >> throughout lots of introductory courses enables one to sweep under the >> rug lots of stuff. >Sweep under the carpet or even say things that are just >downright misleading. A vector doesn't have covariant >components and contravariant components, though it's possible >to find texts that say it does. Given a metric, a vector >has an associated covector: given a basis, one has components >for the vector, an associated basis for the dual space, and >components for the associated covector. But all the distinctions >get forgotten in Euclidean space. >>Exactly. >It's a shame that doing the familiar special case first >can lead to so much 'unlearning' being required. (Especially >when the unlearning doesn't happen.) >>Yes, here is the problem. The special case is taught first, since it >>is simple. Now, pedagogically, it makes sense to start with a simple >>case first, then build on it to get to more complex ones. But this is >>so only if the simple case leads in the direction of the more complex >>ones. In this case, and in the way the simple case is taught, it >>leads astray. Instead of being able to proceed from it towards the >>more complex ones, one has to do (as you say) lots of unlearning >>first. >>Now, it doesn't have to be this way. Lots of the needed stuff can be >>introduced in Euclidean space, by considering non-orthogonal axes, and >>within a linear algebra course by considering (left and right) eigenvector >>spaces of non-symmetric matrices. But this is rarely done. >And you can do a lot of physics without ever considering such things. The >distinctions appear, to a certain extent, in crystallography, anything >relativistic that uses a metric instead of ict, and any time non-Cartesian >(not just non-orthogonal) coordinate systems come up. But you can fake >that out by following a few mathematical recipies by rote and still do a >lot of physics without going into the covector details. About the only >time it's really important is when you're doing general relativity. And >despite the discussions here and the shelf space taken up in the small >science sections of chain bookstores, general relativity is a very small >part of physics. Yes, this is very true, especially the part regarding GR being a very small part of physics (and of significant interest only to a very small subset of physicists). And I would guess that the majority of physicists can go through their whole professional career without ever having to make distinctions between vectors and covectors. Still, I personally found (over time) these distinctions to be useful and aiding understanding (and, in a way, more natural). So, I agree, if one doesn't do GR, one can live without them. But it really doesn't cost that much more effort to learn them. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Help with tensors (re-post) >>And it doesn't help that grad f is called a >>vector in just about every course where the >>concept is introduced... > The fact that rectilinear and orthogonal coordinate bases are used > throughout lots of introductory courses enables one to sweep under the > rug lots of stuff. >>Sweep under the carpet or even say things that are just >>downright misleading. A vector doesn't have covariant >>components and contravariant components, though it's possible >>to find texts that say it does. Given a metric, a vector >>has an associated covector: given a basis, one has components >>for the vector, an associated basis for the dual space, and >>components for the associated covector. But all the distinctions >>get forgotten in Euclidean space. >Exactly. >>It's a shame that doing the familiar special case first >>can lead to so much 'unlearning' being required. (Especially >>when the unlearning doesn't happen.) >Yes, here is the problem. The special case is taught first, since it >is simple. Now, pedagogically, it makes sense to start with a simple >case first, then build on it to get to more complex ones. But this is >so only if the simple case leads in the direction of the more complex >ones. In this case, and in the way the simple case is taught, it >leads astray. Instead of being able to proceed from it towards the >more complex ones, one has to do (as you say) lots of unlearning >first. >Now, it doesn't have to be this way. Lots of the needed stuff can be >introduced in Euclidean space, by considering non-orthogonal axes, and >within a linear algebra course by considering (left and right) eigenvector >spaces of non-symmetric matrices. But this is rarely done. >>And you can do a lot of physics without ever considering such things. The >>distinctions appear, to a certain extent, in crystallography, anything >>relativistic that uses a metric instead of ict, and any time non-Cartesian >>(not just non-orthogonal) coordinate systems come up. But you can fake >>that out by following a few mathematical recipies by rote and still do a >>lot of physics without going into the covector details. About the only >>time it's really important is when you're doing general relativity. And >>despite the discussions here and the shelf space taken up in the small >>science sections of chain bookstores, general relativity is a very small >>part of physics. >Yes, this is very true, especially the part regarding GR being a very >small part of physics (and of significant interest only to a very >small subset of physicists). And I would guess that the majority of >physicists can go through their whole professional career without ever >having to make distinctions between vectors and covectors. >Still, I personally found (over time) these distinctions to be useful >and aiding understanding (and, in a way, more natural). So, I agree, >if one doesn't do GR, one can live without them. But it really >doesn't cost that much more effort to learn them. Let's see if I can find that quote in my .siglist... The hardest conviction to get into the mind of the beginner is that the education he is receiving in college is not a medical course but a life course for which the work of a few years under teachers is but a preparation. -- Sir William Osler Physics programs have already been criticized as trying to turn out miniature theorists when most physicists are experimentalists. That requires a whole different set of skills which, in my experience, coursework doesn't do much to prepare you for. And so there are people who've been working productively and publishing papers for a decade that can't really answer a lot of the stuff that comes up in this newsgroup. They never need to know it. Physics programs have also been criticized for the time it takes to graduate. You have to cut some stuff out. One person wants differential geometry in there, another wants fluid mechanics, others think more coursework on applied and experimental methods should be included, I'd argue that a more comprehensive introduction to group theory should be included *before* it must be applied in the homework for other courses, Reany wants philosophy, Shead thinks they should include a two-year sequence on F=[m]a... can't have it all. But then it was kind of like an epiphany when I realized the k-space vectors in crystallography were the covectors of general relativity. It took me four or five forays into general relativity, on my own time since I'd never taken coursework on it, before it started to click into place for me. And I'd hit field theory a number of times since classes finished up and have a different understanding of it now than I had then. I don't know what's really important enough that it should be included in the coursework at the cost of other things that must be taken out. But you can't just finish up classes and declare your education complete. You have to keep at it. -- What's another word for thesaurus? -- Steven Wright Let me look in my synonymicon. -- Thaddeus Stout === Subject: Re: Help with tensors (re-post) > And you can do a lot of physics without ever considering such things. The > distinctions appear, to a certain extent, in crystallography, anything > relativistic that uses a metric instead of ict, and any time non-Cartesian > (not just non-orthogonal) coordinate systems come up. But you can fake > that out by following a few mathematical recipies by rote and still do a > lot of physics without going into the covector details. Sure you can. Horses for courses. Of course, it's always nice to know what you're actually doing, but as long as you're aware of the limitations of a simplified approach, there's nothing wrong in using one where it works. For that matter, you'll look long and hard in differential geometry texts for the words 'covariant' or 'contravariant', except in the phrase 'covariant derivative'. Of course, that's not because the distinction is lost on geometers: rather, they are so little concerned with the transformation properties of components it just isn't something they bother mentioning out loud much. > general relativity is a very small > part of physics. Well, OK. But it's pretty fundamental, and messes about with people's preconceptions of how space and time work---which is, of course, why it attacts loonies like a candle flame attracts moths. === Subject: Re: Help with tensors (re-post) <3qc8dhFe4dekU1@individual.net> <3qdmogFeemadU1@individual.net And you can do a lot of physics without ever considering such things. The > distinctions appear, to a certain extent, in crystallography, anything > relativistic that uses a metric instead of ict, and any time non-Cartesian > (not just non-orthogonal) coordinate systems come up. But you can fake > that out by following a few mathematical recipies by rote and still do a > lot of physics without going into the covector details. > Sure you can. Horses for courses. Of course, it's always nice > to know what you're actually doing, but as long as you're aware > of the limitations of a simplified approach, there's nothing > wrong in using one where it works. > For that matter, you'll look long and hard in differential > geometry texts for the words 'covariant' or 'contravariant', > except in the phrase 'covariant derivative'. Of course, > that's not because the distinction is lost on geometers: > rather, they are so little concerned with the transformation > properties of components it just isn't something they > bother mentioning out loud much. I think Mr. Low is exceeding his authority. Mathematician's are very interested in a superb developement of this problem. > general relativity is a very small > part of physics. > Well, OK. But it's pretty fundamental, > and messes about with people's preconceptions > of how space and time work---which is, of > course, why it attacts loonies like a > candle flame attracts moths. So Mr. Low posts ......... Ken S. Tucker === Subject: Re: Help with tensors (re-post) >> And you can do a lot of physics without ever considering such things. The >> distinctions appear, to a certain extent, in crystallography, anything >> relativistic that uses a metric instead of ict, and any time non-Cartesian >> (not just non-orthogonal) coordinate systems come up. But you can fake >> that out by following a few mathematical recipies by rote and still do a >> lot of physics without going into the covector details. >Sure you can. Horses for courses. Of course, it's always nice >to know what you're actually doing, but as long as you're aware >of the limitations of a simplified approach, there's nothing >wrong in using one where it works. >For that matter, you'll look long and hard in differential >geometry texts for the words 'covariant' or 'contravariant', >except in the phrase 'covariant derivative'. Of course, >that's not because the distinction is lost on geometers: >rather, they are so little concerned with the transformation >properties of components it just isn't something they >bother mentioning out loud much. The other side that you always have to consider is: how much time would it take to teach this to the students, and what instruction would they have to give up in order to learn it? Something like that you would't just dismiss by declaring they don't need to know it. It's one of those thousand and one things that they should know. But it might be omitted because it's hard enough as it is to push them through the undergrad program in five years, because the grad students take long enough as it is before getting to and finishing their research. Maybe it can and should be introduced in a short time, just to make them aware of it. Like in a mathematical methods class, or in the E&M class when they're learning about vectors in non-Cartesian coordinate systems. But that might cause more confusion than understanding, especially if it's not reinforced throughout the rest of the cirriculum. >> general relativity is a very small >> part of physics. >Well, OK. But it's pretty fundamental, >and messes about with people's preconceptions >of how space and time work---which is, of >course, why it attacts loonies like a >candle flame attracts moths. No doubt. It's just a little funny that the parts of physics that loom largest in the public's mind are the least in the practicing community. -- When the fool walks through the street, in his lack of understanding he calls everything foolish. -- Ecclesiastes 10:3, New American Bible === Subject: Re: Help with tensors (re-post) > The other side that you always have to consider is: how much time would it > take to teach this to the students, and what instruction would they have > to give up in order to learn it? I guess I'd regard it as specialist material: certainly ought to be part of the late undergrad curriculum for anybody doing theoretical physics. But I don't see it as being something to spend a huge amount of time on. Vector spaces and their duals and a bit about tensor products is a couple of lectures in a 'linear algebra for physics' type of course. (Surely it's more useful than solving linear constant coefficient ODEs with standard right hand sides by variation of parameters...) > No doubt. It's just a little funny that the parts of physics that loom > largest in the public's mind are the least in the practicing community. both of great public interest, and neither of much significance in workaday science. === Subject: Re: Help with tensors (re-post) <3qdmogFeemadU1@individual.net> <3qdpgpFeca5mU1@individual.net The other side that you always have to consider is: how much time would it > take to teach this to the students, and what instruction would they have > to give up in order to learn it? > I guess I'd regard it as specialist material: certainly > ought to be part of the late undergrad curriculum for > anybody doing theoretical physics. But I don't see it > as being something to spend a huge amount of time on. > Vector spaces and their duals and a bit about tensor > products is a couple of lectures in a 'linear algebra > for physics' type of course. (Surely it's more useful > than solving linear constant coefficient ODEs with standard > right hand sides by variation of parameters...) > No doubt. It's just a little funny that the parts of physics that loom > largest in the public's mind are the least in the practicing community. > both of great public interest, and neither of much > significance in workaday science. Yes, it's a very specialized thing working as a theoretician, it would be a thing a student would need to commit too. I have a calling that way. Ken === Subject: Re: Help with tensors (re-post) >> The other side that you always have to consider is: how much time would it >> take to teach this to the students, and what instruction would they have >> to give up in order to learn it? >> I guess I'd regard it as specialist material: certainly >> ought to be part of the late undergrad curriculum for >> anybody doing theoretical physics. But I don't see it >> as being something to spend a huge amount of time on. >> Vector spaces and their duals and a bit about tensor >> products is a couple of lectures in a 'linear algebra >> for physics' type of course. (Surely it's more useful >> than solving linear constant coefficient ODEs with standard >> right hand sides by variation of parameters...) >> No doubt. It's just a little funny that the parts of physics that loom >> largest in the public's mind are the least in the practicing community. >> both of great public interest, and neither of much >> significance in workaday science. >Yes, it's a very specialized thing working as a >theoretician, it would be a thing a student would >need to commit too. I have a calling that way. and study materials instead. -- What are the possibilities of small but movable machines? They may or may not be useful, but they surely would be fun to make. -- Richard P. Feynman, 1959 === Subject: Re: Help with tensors (re-post) >> The other side that you always have to consider is: how much time would it >> take to teach this to the students, and what instruction would they have >> to give up in order to learn it? >I guess I'd regard it as specialist material: certainly >ought to be part of the late undergrad curriculum for >anybody doing theoretical physics. But I don't see it >as being something to spend a huge amount of time on. >Vector spaces and their duals and a bit about tensor >products is a couple of lectures in a 'linear algebra >for physics' type of course. (Surely it's more useful >than solving linear constant coefficient ODEs with standard >right hand sides by variation of parameters...) Well, you might include it in E&M instruction with the other vector math, or in the Mathematical Methods course. I'm just not sure how long it would take. I banged my head into GR quite a number of times before I finally started to get it. There were certain conceptual issues regarding differential geometry and doing physics with it that I just didn't work through quickly, but then it might have come a lot sooner if I'd read Taylor and Wheeler's book a lot sooner. I'd be happy to include that stuff in the graduate material if the opportunity cost was low enough. But then, there were a lot of issues with QFT, and even QM, that didn't click in for me until long after classes had finished. I always felt like I was treading water just well enough to slow my sinking until the end of the semester. Maybe I just wasn't very bright, or maybe school moves a bit too quickly to develop a solid understanding. >> No doubt. It's just a little funny that the parts of physics that loom >> largest in the public's mind are the least in the practicing community. >both of great public interest, and neither of much >significance in workaday science. Quantum mechanics is the bread and butter of modern physicists! Generally as it applies to problems in condensed matter, optics, and other also gets more shelf-space at Waldenbooks than its workaday importance would warrant. -- Will we be suturing the anus? === Subject: Re: Induction with a hard start > In case nobody comes up with a suitable hard start, I'll publish > it tomorrrow. Don't think it will be necessary though: this _is_ > 'sci.math', huh ? Enjoy ! A correct solution has be presented by M.J.T. Guy, but I want it to be cast in the original setting. Remember that x >= 0 in the sequel. 1 1 x(n) = ------------------ ==> x(0) = -------------- 2 + 2.cos(2^n.phi) 2 + 2.cos(phi) Thus we can solve (phi) from: cos(phi) = 1/(2.x(0)) - 1 . Meaning that -1 <= 1/(2.x(0)) - 1 <= +1 ==> x(0) >= 1/4 . 1 1 x(n) = ------------------- ==> x(0) = --------------- 2 + 2.cosh(2^n.psi) 2 + 2.cosh(psi) Thus we can solve (psi) from: cosh(psi) = 1/(2.x(0)) - 1 Meaning that 1 <= 1/(2.x(0)) - 1 ==> 0 < x(0) <= 1/4 . At last, we have a solution x(0) = 0 , which always remains the same. Thus the closed formula for the iterations x := (x/(1-2.x))^2 has a splitpersonality. It is the triginometric formula for x(0) >= 1/4 and it is the hyperbolic formula for 0 < x(0) <= 1/4 . For x(0) = 0 it is just 0 . Oh yeah, and the iterands remain where they are. There is NO escape from either 0 < x <= 1/4 or x >= 1/4 (or x = 0). Han de Bruijn === Subject: Re: Induction with a hard start ... > 1 1 > x(n) = ------------------ ==> x(0) = -------------- > 2 + 2.cos(2^n.phi) 2 + 2.cos(phi) ... > 1 1 > x(n) = ------------------- ==> x(0) = --------------- > 2 + 2.cosh(2^n.psi) 2 + 2.cosh(psi) ... > Thus the closed formula for the iterations x := (x/(1-2.x))^2 has > a splitpersonality. It is the triginometric formula for x(0) >= 1/4 > and it is the hyperbolic formula for 0 < x(0) <= 1/4 . For x(0) = 0 > it is just 0 . Oh yeah, and the iterands remain where they are. There > is NO escape from either 0 < x <= 1/4 or x >= 1/4 (or x = 0). Indeed. Do remember that cos(i.x) = cosh(x). So actually they are the same formulas, except that the argument is real in one case and imaginary in the other case. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Induction with a hard start >> Could someone please provide an example of a proof by induction in which >> proving the statement for n =3D 1 is clearly harder than going from n to >> n + 1? >Consider the following interations: x := (x/(1-2.x))^2 . >Start with x = x(0) , then: x(1) := (x(0)/(1-2.x(0)))^2 > x(2) := (x(1)/(1-2.x(1)))^2 > x(3) := (x(2)/(1-2.x(2)))^2 > and so on and so forth .... >Since _any_ negative value x(0) will be turned into a positive value >x(1) and remain so for x(2), x(3) ... , we can restrict attention to >positive (or zero) values only: the interval x >=3D 0 (where x in R). >Question: given a starting value x(0), is there a closed formula for >the iterands x(n) ? Substitute y = 1/(2x) - 1. Then the iteration becomes y := 2y^2 -1, and we only need to consider y(0) >= -1. If y(0) = cos(phi) for some phi, y(n) = cos(2^n phi). If y(0) = cosh(psi) for some psi, y(n) = cosh(2^n psi). y(0) can take both forms if and only if y(0) = 1, with phi = psi = 0. So the only difficulty in setting up the initial condition is deciding whether y(0) <= 1 or y(0) >=1. Not what I'd call clearly hard. Mike Guy === Subject: Re: Induction with a hard start Could someone please provide an example of a proof by induction in which >> proving the statement for n =3D 1 is clearly harder than going from n to >> n + 1? >Consider the following interations: x := (x/(1-2.x))^2 . >Start with x = x(0) , then: x(1) := (x(0)/(1-2.x(0)))^2 > x(2) := (x(1)/(1-2.x(1)))^2 > x(3) := (x(2)/(1-2.x(2)))^2 > and so on and so forth .... >Since _any_ negative value x(0) will be turned into a positive value >x(1) and remain so for x(2), x(3) ... , we can restrict attention to >positive (or zero) values only: the interval x >=3D 0 (where x in R). >Question: given a starting value x(0), is there a closed formula for >the iterands x(n) ? > Substitute y = 1/(2x) - 1. Then the iteration becomes y := 2y^2 -1, > and we only need to consider y(0) >= -1. > If y(0) = cos(phi) for some phi, y(n) = cos(2^n phi). > If y(0) = cosh(psi) for some psi, y(n) = cosh(2^n psi). > y(0) can take both forms if and only if y(0) = 1, with phi = psi = 0. > So the only difficulty in setting up the initial condition is > deciding whether y(0) <= 1 or y(0) >=1. Not what I'd call clearly hard. Uhm, not if you're such a handsome Guy, I admit :-) Han de Bruijn === Subject: Re: pointed cone exercise longer. - Martin === Subject: Re: Complex analysis question,, > Let f be entire. Evaluate > integral_[0,2pi] f [ z_0 + re^(it) ] e^(kit) dt > for k is a positive integer. > I really would like to turn this integral into > something of the form > integral_gamma f(z) dz > where gamma is a closed curve. Then the integral > would be zero. > But I can't seem to do that. Alternatively, I have > that the function > f [ z_0 + re^(it) ] e^(kit) > is f (z_0 + r) at both time t = 0 and time t = 2pi. > Is this enough for me to say that the integral can be > expressed as the > integral of an entire function over a closed curve > since at the beginning of > time and the end of time it is the same value? Then > the integral is 0? > James Gamma(t) = z_0 + re^(it) ((Gamma(t) - z_0)/r)^(k-1) = e^((k-1)it) is a polynomial function of Gamma(t). Does that help? === Subject: Re: Complex analysis question,, >>Let f be entire. Evaluate >>integral_[0,2pi] f [ z_0 + re^(it) ] e^(kit) dt >>for k is a positive integer. >>I really would like to turn this integral into something of the form >>integral_gamma f(z) dz >>where gamma is a closed curve. Then the integral would be zero. > Hint: the closed curve is z = z_0 + r e^{it} for t = 0 to 2 pi. > Now, what is dz? Hi Robert, I have tried this already. The problem is that in the integral there is e^(kit) rather than e^(kit). If I assume k = 1 the problem is easy. If I let gamma be the closed curve you tell me, I don't see how to reduce the integral to integral f(gamma(t)) gamma ' (t) dt. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re: Complex analysis question,, Let f be entire. Evaluate >>integral_[0,2pi] f [ z_0 + re^(it) ] e^(kit) dt >>for k is a positive integer. >>I really would like to turn this integral into something of the form >>integral_gamma f(z) dz >>where gamma is a closed curve. Then the integral would be zero. > Hint: the closed curve is z = z_0 + r e^{it} for t = 0 to 2 pi. > Now, what is dz? > Hi Robert, > I have tried this already. The problem is that in the integral there is > e^(kit) rather than e^(kit). If I assume k = 1 the problem is easy. If I > let gamma be the closed curve you tell me, I don't see how to reduce the > integral to integral f(gamma(t)) gamma ' (t) dt. e^(kit) = e^((k-1)it) e^(it) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Complex analysis question,, >> > Hint: the closed curve is z = z_0 + r e^{it} for t = 0 to 2 pi. >> Now, what is dz? >> Hi Robert, >> I have tried this already. The problem is that in the integral there is >> e^(kit) rather than e^(kit). If I assume k = 1 the problem is easy. If >> I >> let gamma be the closed curve you tell me, I don't see how to reduce the >> integral to integral f(gamma(t)) gamma ' (t) dt. > e^(kit) = e^((k-1)it) e^(it) I'm really sorry but I don't see how that helps. Do I use induction on k? If Gamma(t) = z_0 + re^(it), then int_[0,2pi] f [ z_0 + re^(it) ] e^(kit) dt = 1/(ir) * int_Gamma f(Gamma(t)) * Gamma'(t) * e^((k-1)it) dt = 1/(ir) * int_Gamma f(z) e^((k-1)it) dz If k = 1, the integral is 0. Induct on k? I don't see the induction. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re: Complex analysis question,, > Hint: the closed curve is z = z_0 + r e^{it} for t = 0 to 2 pi. >> Now, what is dz? >> Hi Robert, >> I have tried this already. The problem is that in the integral there is >> e^(kit) rather than e^(kit). If I assume k = 1 the problem is easy. If >> I >> let gamma be the closed curve you tell me, I don't see how to reduce the >> integral to integral f(gamma(t)) gamma ' (t) dt. > e^(kit) = e^((k-1)it) e^(it) > I'm really sorry but I don't see how that helps. Do I use induction on k? No. Just express e^((k-1)it) in terms of z = z_0 + r e^{it}, so you are integrating g(z) dz for a certain function g. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: modern math is failure -- back to the drawing board Randy, Therer is a world of difference between learning the theories in college and the practical application in a real business situation. Even group projects done in college do little to prepare you for the real business world. Managers are hired on the bases on their degree and only during the first year of working for a company can you tell if they aare any good at their job CS112, mmn22 > you contradict yourself. how can the hiring managers be poorly > informed if they inform themselves > by looking at where these job candidate went to school? > Because that in itself is very little information. > - Randy === Subject: Calculator Glitch re fractional powers My Casio fx-991MS calculator has a problem with raising negative bases to fractional powers. I know these are only real for fractional powers that have odd denominators when reduced to lowest terms, but... If I enter ( - 8 ) ^ ( 1 / 3 ) = I get the correct answer for the cube root of -8, -2 But if I enter the same key sequence, but raise -8 to the 2/3 power, it says Math Error. Why would it be able to calculate the one exponential, but not the other? If I enter the exponent as 3/9, it works, and 6/9 doesn't. (-243) raised to the 0.2 power works, but raised to the 0.4 power doesn't. Any ideas? === Subject: Re: Calculator Glitch re fractional powers >My Casio fx-991MS calculator has a problem with raising negative bases >to fractional powers. >I know these are only real for fractional powers that have odd >denominators when reduced to lowest terms, but... >If I enter ( - 8 ) ^ ( 1 / 3 ) = >I get the correct answer for the cube root of -8, -2 >But if I enter the same key sequence, but raise -8 to the 2/3 power, it >says Math Error. >Why would it be able to calculate the one exponential, but not the >other? >If I enter the exponent as 3/9, it works, and 6/9 doesn't. >(-243) raised to the 0.2 power works, but raised to the 0.4 power >doesn't. >Any ideas? It might run an error checker on the result before presenting it. (-8)^(1/3) = -2 (-2)^(3/1) = -8 good, show -2 (-8)^(2/3) = 4 (4)^(3/2) = 8 bad, show Math Error This hypothesis predicts that .6 will work as an exponent but .8 will not. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Calculator Glitch re fractional powers (-243)^n Works for n=0.2, gives -3 Gives MathError for 0.4, 0.6, 0.8 Works for n = 1.0 Gives MathError for n = 1.2, 1.4, 1.6, 1.8 Works for n 2.0 [CapitalEth].87.99.8fnewssvr33.news.prodigy.com!newsdbm03.n ews.prodigy.com!newsdst01.news.prodigy.com!newsmst01b.news.prodigy.com!prodi g y.com!newscon06.news.prodigy.com!prodigy.net!news-feed01.roc.ny.frontiernet. n et!nntp.frontiernet.net!cox.net!news-xfer.cox.net!news.lightlink.com!rahul.n e t!wasp.rahul.net!rahul.net!support1.mathforum.org!mail.mvmvj.com!info!mailpo s t!not-for-mail ®.98.95.92info@mvmvj.com ë.8cÖ.97.8d.98.95[O Tilde][Eth].97sci.math î.9b.89.90.8c.8b.99.bc ÛåÛñÛ[CapitalEGra ve]Û§ÁæÁ[UDoubl eDot]Á¡¬å$B!0!|!K$G$9(B ì.98.8d.87.94.8e.9c[ AAcute].99.8e.95.94The Math Forum í.8e.94.8c.97.bc ±.a6 î.8c.94.8a.8c.98mail@support1.m athforum.org ê.8c.97.97.87.8d.8c[Hyph en].83Ä<20051003135523.10158.qmail@mail.mvmvj .com> ëëï[CapitalEth][Hyphen][CapitalEth ].95.97.99.8e.94.8d[Hyphen][CapitalEG rave].95.97.99.bc .97.9b[Eth][Eth].95.98.99±[Hyph en].b2¨.92.87.99.8f.be[IDouble Dot].98.9b.92¨.95.98[ CCedilla] ¯[Hyphen]Ì.95.92[Eth][IGrav e].87.8e.94.99.97[Hyphen]ï[IDoubleDo t].bc .94.8cÖ.97Ë.97.9b[Eth][Et h].95.98.99±¨.92[A Acute].99.8f.be.95.98.9b.92[Register edTrademark].95.98 ¯[Hyphen]ñ.8c.8b.8c.8e .9a.8c.8afrom mail.mvmvj.com ([58.180.215.135]) .89.9d .97.9b[Eth][Eth].95.98.99±[Regi steredTrademark].92.87.99.8f.be.95.98 .9b.92¨.95.98.8d ¬ü¨±±¨ .a6øü¨±±[Registere dTrademark].a6øï.8f.8c ê.87.99.8f ®.95.98.9b.92 Ûñ.8c.9a.8e.97.8e[ IDoubleDot].94.bc ±¨ primary) with SMTP id j93KR1L07320 16:27:05 -0400 ¯[Hyphen]ñ.8c.8b.8c.8e .9a.8c.8a(qmail 10159 invoked by uid 509); 3 Oct 2005 22:55:23 +0900 ¯[Hyphen]ï.95post-sci-math@mathforum. org ¯.98.8c.benewsmst01b.news.prodigy.com sci.math:839165 ÛåÛ¢Ûæ[Curre ncy]ÀÛèíÛ æ.85È.9aÛ[ CapitalIHat]Ä.beÛëÛ» ê.a6ÛÛê.9c[ CapitalIHat].98ÛÂÈÄ[ Currency]ÌÛ®Û Û[Curre ncy]fiÛ.b9Á£¬[Capita l AHat] ÛåÛ.83Û.a6[Curren cy][Micro]Û.93ÛfiÛ.b9[Cu r rency]ÇÁ©¬å ÛåÁï[CapitalADoubleDo t].be¬å.83Ä ±¡±.b2±.b2Û[ CapitalAHa t]ÁÁÛñÛ éÛ§ÁÍ[CapitalUH at] [Micro]Û.97¢ê¬å ÛåÁ¯Û[Cap italIDoubleDot]ÛüÛ.87Û[C apitalThorn] ÛáÛ®¢.9a[Parag raph ].87ÛÛÛëÛ[C apital CCedilla]´î´Ì´ø´.90 ÛáÛfiÛá[C ur rency]ÀÁ»¬å Ûåë.83Û[LeftGui llemet]ÛÌÛÀ[Capital UHat][EAcu te]ìÌÛáÛ[Ca pitalThorn]ÛÈÛ.97Û[Le ftG uillemet]Á©æ.85[Ri ghtGuillemet].9aåïÛ[CapitalATi lde]Û®Ûfi[Capita lUHat].b9´[C apitalIAcute]Á.9dÁô2 50å Û墬.bc£.bd[Micro][CapitalARi ng]òêè®[UDoubleD ot]Û.98.b2.87Û¨[Curren cy]fiÛ.b9Ûéê .9cë.98.bc.95.bd.9fÛ [Micro]Û.93ÛfiÛ.b9[DownE x clamation]£¬å Û墬.81.81[Capital IDoubleDot]ÌÛì´.87[Do wnExclamation].bc´.91çÖÀ ¨ü.8c.b2Ä.82 275¬å$BHV9fG'>ZI,?(B)$B!#(B .8f.99.99[Eth].bcøøÖ[CapitalODoubleD ot]Ö[Regist eredTrademark].90.9b.92[Eth].89.b2[RegisteredTrademark ].94.8c.99øÀ.91.95.98 .8c.91.95.98 $B$R$H$_MM$N4uK>$h$j$$J$?$NEPO?A4$FL5NA$G$9!#(B Ûåê.9cë.982 72.95.bd.9f¬å ¬ñ.8c.be.9b.97.87.93 ç.8a.98.8c.97.97 ©.8e.98.87.94.87.8e[CapitalAGr ave].90.9b.92[Eth].89.b2¨.94[ ARing].99 === Subject: Game Of b1 *b2 *b3 ..*(b1 +b2 +b3+...) Here is a game I recently came up with, inspired by the question of how many ways are there to write any particular positive integer m as m = (product{k=1 to r} b(k)) * (sum{k=1 to r} b(k)), where {b(k)} consists only of r positive integers. 2 players. There are a couple variations on the same idea. Simplest variation first: Each player secretly comes up with an r, a selection of r positive integers {b(k)}, and a positive integer m where m = (product{k=1 to r} b(k)) * (sum{k=1 to r} b(k)). m must not have been used earlier in the game (if playing plural rounds), and m must not be greater than a pre-agreed-on integer (such as, say, 300). Next, each player reveals their m and r to the other player. The winner of the game/round is the first player to come up with (and write down) any set of b(k)'s which get m = (product{k=1 to r} b(k)) * (sum{k=1 to r} b(k)), where r and m are those revealed by the player's opponent. The b's found by a player need not be the same the player's opponent used, but they still must get the m offered by the opponent. In the more complex version, there are a pre-agreed-on number of rounds. Players reveal an r and m, as before, in each round. Only the first player to find and write down a workable set of b's gets any points on a particular round. A player's points are added up for his/her final score. The players may find a set of r b's that get any DIVISOR d of m, instead of getting m itself. A player, only if she/he wins a round (by writing down the integers first), gets d points added to her/his score. Sample round: Player 1 comes up with {4,3,1,1} for 4*3*1*1*(4+3+1+1) = 108. Player 2 comes up with {6,2,2} for 6*2*2*(6+2+2) = 240. So, player 1 reveals that player 2 must work with an m of 108 and an r of 4. And player 2 reveals that player 1 must work with an m of 240 and an r of 3. Now, the players are playing the variation where they may try for divisors of m. So player 1 gets 1*1*4*(1+1+4) = 24, a divisor of 240. Player 2 gets 1*1*3*4*(1+1+3+4) = 108. (And player 2 happened to get 108 with the same integers used by player 1 to get 108.) so player 1 gets 24 points added to his score, player 2 gets 0 points on this round. Leroy Quet === Subject: Re: GF(16) and irreducible polynomials <27067619.1128216474736.JavaMail.jakarta@nitrogen.mathforum.org HI, > For question 1)First ,an irreducible polynomial of degree 4 in GF(2) must not have a root in GF(2) and must not be > a product of 2 irreducible polynomials of degree 2. > So ,say polynomial p,has not root 0 and 1. > Then p can not be a product q * r ,q and r ireducibles of degree 2.Which polynomials are irreducible of degree 2 in GF(2)? I found only one,namely X^2+X+1. Its square is x^4+x^2+1. So we have to choose one polynomial,without roots 0 or 1, and different than x^4+x^2+1. > For example X^4+x+1 or x^4+x^3+1 .Actually these are the only ones. > Now factorize GF(2)[X]/(X^4+X+1). > which is a field over Gf(2) of dimension 4. > with basis 1,x,x^2,x^3. > So its elements are a+bx+cx^2+dx^3. > where the coefficients are in GF(2). > Now you add them and multiply them as you want ,using that characteristic of the field is 2 and x^4+x+1 = 0 > 2)regarding question 2 ,gf(4)is the splitting field of > x^4-x=x(x-1)(x^2+x+1) > So it is composed by 0,1,a,a+1 where a^2+a+1=0 > Now you should find a polynomial which doesn't have any of these elements as solution > For example q=x^2+ax+1. > q(0)=1,q(1)=1,q(a)=1,q(a+1)=a. > Bogdan Would GF[16] = GF[4]/(x^2+1+1). This is the one I'm having the most difficulty with. I have a good understanding of GF(2)[X]/(X^4+X+1). I just don't 'get' how things are working as well with GF[4]. Jeff === Subject: Re: infinity <4339894A.EC6@nf.sympatico.ca> <3ptiaeFc63p2U1@individual.net> <433C5135.54D5@nf.sympatico.ca> universe and the tachyon universe. The Big Bang, measured to be so >many years ago, is actually infinitely many years ago, > No it is not. > and as we learn >more about it we'll discover ever longer spans. > Wrong again. >As entropy increases, >it goes towards a state where entropy's infinite > 3rd strike. You're out. >and the tachyon >universe is a point. > Is that where the Borg live? >Then it becomes the Big Bang of the tachyon >universe, the tachyon universe becomes the regular universe, and round >and round and round she goes. > OK. > Then I wonder, what was there before the Big Bang? If all matter and > energy of the universe comprised some protoplasmal point at the Big > Bang, and the universal stopwatch started ticking, well, I am not very > familiar with the current cosmological models of the nature of space > and time at the Big Bang. One of the things that you're presuming is that time is an axis that extends to minus infinity. That is, you're presuming that every event has a before. Modern cosmology doesn't presume that. > What if the Big Bang of our universe took place in basically some > region of a still larger cosmiverse, and there are other universes, in > terms the entire region and all contents as from a Big Bang? Basically > that's consideration of larger structures than bordered by the > wavefront of light from the Big Bang formation event of our universe. > Why do some galaxies appear to be going in directions that do not seem > to match their apparent age and formation? Where did they come from? > Maybe at the Big Bang, the tachyon universe was infinite, in extent and > entropy, and as the universe goes flat the tachyon universe goes to a > point? > A tachyon is basically saying, there's a mathematical object that is > basically the opposite of a photon, but it goes faster than light so it > can't be in the universe, so it's in the tachyon universe. They might > be purely mathematical, but to them so are the photons. It's kind of > like the projection of an instanton to all dimensions. > I wonder about basically whether time scales vary in cosmological terms > pointing to the Big Bang. > There've only been predicted and experimental masses of the quarks for > some thirty years, or perhaps longer I don't know, and over that time > they've decreased considerably. > Ross === Subject: Re: infinity <4339894A.EC6@nf.sympatico.ca> <3ptiaeFc63p2U1@individual.net> <433C5135.54D5@nf.sympatico.ca> extends to minus infinity. That is, you're presuming that every event > has a before. Modern cosmology doesn't presume that. Yeah, OK, I agree. It's kind of like: say the entire universe of everything physical began with the Big Bang. Then, it seems consensus currently has there not being a Big Crunch, instead towards heat death of the universe, or cold death. Then, somehow, that becomes another Big Bang, because, when everything is all together then that's the same in some sense as there being nothing anywhere, because of short circuits in logic, and the Liar paradox. That's like saying in the context of infinity that infinity, ... is zero, or infinity, is negative one. It's similar in consideration to the rationalization of temperature, in terms of negative temperature, degrees Kelvin. There is no such thing as less than absolute zero, right? That would seem to be the intuitive notion, what with that all temperatures are measured in terms of positive degrees Kelvin, compared to absolute zero, the cessation of correctly. Where that might seem to be the case, there is the notion among physicists I hear that negative temperatures are actually hotter than any finite positive temperature. It's like the notion that there is no first point on a circle. Select one, the origin is anywhere. It's like the projectively extended reals, with a point at infinity that is negative infinity. Why are there three space dimensions? What did Feynman mean by that? So, just like modern cosmological models might not have moments before the Big Bang, there aren't temperatures below absolute zero or naturals less than zero. Yet, in some senses, it's possible to rationalize why there are. Good day, Ross === Subject: Re: infinity <1128013231.36c8ad3d37d1e02e74f9a3fe23c5906c@teranews> If we can have N, N^2, N^3 etc as 1:000...000, 1:000...000:000...000 and > 1:000...000:000...000:000...000, then we should be able to express what > fraction of these is contained in 1 unit: 0.000...001, 0.000...000:000...001 > and 0.000...000:000...000:000...001. So, there is a theoretical benefit to > it. Tony tells us that his infinite/infinitesimal numbers are such that N = 0:999...999.0 and 1/N = 0:0.000...001 Since N has a most significant digit (presumably the ceil(log(N))th digit), and since 1/N has a least significant digit, it would be reasonable to assume that these well-known constants also have a least significant digit: pi = 0:3.1415926535897932384626433832795...? e = 0:2.7182818284590452353602874713527...? sqrt(2) = 0:1.4142135623730950488016887242097...? phi = 0:0.6180339887498948482045868343656...? Perhaps Tony can shed some light on those least sigificant digits I've marked as '?' above. === Subject: Re: infinity <4339894A.EC6@nf.sympatico.ca> <3ptiaeFc63p2U1@individual.net> all wrong rubbish > [hanson] > hmmm... rubbish?.... not deity like?.....hhmmm... OK, then, > listen hogie: Instead of making such a short quasi intellectual > discourse into own rubbish just demonstrate to us a convincing > counter proof. Go up to the fifth floor, open a window and do > step out. When you arrive at and encounter the close proximity > of the electromagnetic cushion of the calcium-silico aluminate atom > network, kindly do report back with your exiting findings, of how > you convinced/managed/overpowered the deity-like properties > of gravity to act otherwise. > Good luck on your voyage down! http://egroups.com/message/free_energy/18907 === Subject: Re: When negative numbers are real >> Try to tell someone owing money on a mortage that negative numbers >> don't exist. >Wall St. c1928 >The sidewalk littered with the corpses of those who had thought that >way previously. THEY discovered, to their demise, that in nature, you >cannot take 3 coins away, when only two existed. >I have a long standing offer to bare my arse on this ng, if anyone can >provide an example of a less than nothing (zero) entity which occurs >in nature (read reality). >In nature, there exist 1 or 0 ONLY! (-1) refers to less than a >greater, and without exception, when a calculation leads to a net less I have -3 coins in my pocket because, unwilling to be straight-jacketed by mindless conformity, I am comfortable with associating each coin with the left side of the number line rather than the right side. The only reason to bare your arse on this newsgroup is because you think positive numbers have some kind of physical reality that negative numbers don't. Neither have a physical reality. They are associated with stuff by human decisions. You can count with positive integers once you decide that something needs to be counted and what criteria define a unit, but they don't have to be positive or real. That's up to you. -- Suppose you were an idiot... And suppose you were a member of Congress... But I repeat myself. - Mark Twain === Subject: Short definition of category? short that it is misleading. I'm not sure why he tells us that, since he goes on to define category in the usual way. But, naturally, I wondered what that short and misleading definition was (in 1963). [1] The Theories of Functors and Models in The Theory of Models, Proceedings... eds Addison, Henkin, Tarski. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Short definition of category? |so short that it is misleading. I'm not sure why he tells us that, |since he goes on to define category in the usual way. But, |naturally, I wondered what that short and misleading definition was |(in 1963). | |[1] The Theories of Functors and Models in The Theory of Models, |Proceedings... eds Addison, Henkin, Tarski. just as a wild guess, he might be talking about defining a category as a special kind of semigroup containing a special element called undefined, having certain properties. i think that he dicusses this somewhere in his book abelian categories. i don't see any interesting way offhand here to remove the subjectiveness in the concept of shortness of a definition, but personally it doesn't strike me as especially shorter than any of dozens of other possible definitions of category. -- [e-mail address jdolan@math.ucr.edu] === Subject: can this be a proper Laplace transform? sin(s)/(s^2+4), can it be a proper Laplace transform? why? ----------------- === Subject: Re: can this be a proper Laplace transform? > sin(s)/(s^2+4), > can it be a proper Laplace transform? > why? > ----------------- Is it for a test? === Subject: Re: can this be a proper Laplace transform? === Subject: Warp, Wormhole & Time Travel Lecture 5 Star Gate Warp Drive, Wormholes & Time Travel (a technical book under construction) Lectures in 21st Century Physics Quotes below in Lecture 5 are by David Thouless unless otherwise specified. Super Cosmos is now at http://amazon.com These rough notes below to appear (refined) in book Star Gate: Metric Engineering Warp Drive for Wormhole Space & Time Travel. Lecture 5 Topological Concepts in Physics (ref: Topological Quantum Numbers in Nonrelativistic Physics by David Thouless) The More is different (P.W. Anderson) Emergence of Order. This includes both gravity and consciousness in different physical systems that have key formal similarities. Roger Penrose has also suggested a deep connection between gravity and consciousness. Roger's ideas based on quantum gravity collapse of tiny differences in the geometrodynamic field are different from what I am proposing. There is no collapse in Bohmian quantum theory nor in the emergent macro-quantum theory of local order parameters in both the off-shell physical vacuum (for gravity) and in the on-shell brain of higher organisms. Indeed the Landau-Ginzburg phase rigidity of the local non-unitary order parameter is an effective defense against the environmental decoherence degradation that Max Tegmark used against the Penrose-Hameroff model of micro-tubule consciousness in open non-equilibrium living systems. The latter must have signal nonlocality in violation of unitary nonlocal micro-quantum theory. Penrose does not use the concept of a giant macro-quantum order parameter. The space in which an order parameter resides is ... the quotient set of the ... symmetry group of the disordered phase and of the symmetry group of the ordered phase. This quotient means ... the set of all different residual symmetry groups. For example the quotient set SO(3)/SO(2), where SO(n) is the group of proper rotations in n dimensions, is the set of all different possible symmetry axes ... and is equivalent to the sphere S^2 which we have used to describe the directions of magnetization of the isotropic ferromagnet. The topological properties ... are related both to the topology of the space in which the material that possesses an ordered phase is confined, whether it is simply-connected or has holes in it, for example, and to the nature of the order parameter. Ex. 5 The Hedgehog of the NASA Pioneer Anomaly but keeping constant magnitude (Thouless below) a_g = - cH(t) ~ 1 nanometer per second per second pointing back to Sun on scale of 10 to 10^2 AU H(t) = a(t)^-1da(t)/dt in FRW metric (unexpected small scale residue of large-scale cosmology) There is a directional order parameter inside the physical vacuum as well. For an isotropic magnetic material contained in the space between an inner and outer spherical surface the magnetization might point always outwards from the inner surface, taking up the direction (but keeping constant magnitude) of the electric field which would be produced by an electric charge inside the inner surface ... No continuous deformation of the magnetization, keeping the magnitude constant (decoherence-proof generalized phase rigidity, e.g. space-time stiffness c^4/8piG ~ 10^19Gev per 10^-33 cm G-string tension) can turn this into a state of uniform magnetization. This is a topological configuration of the magnetization known as a hedgehog. p. 7 Hedgehog defect in the order parameter is not possible in a planar 2D thin film. Bose-Einstein condensation ... responsible for superfluidity ... unordered state its eigenvalues are all of most of order unity. This is necessary for micro-quantum unitarity with signal locality in what Antony Valentini calls sub-quantal heat death. The eigenvalues of the density matrix are the Born ensemble probabilities that only obtain in sub-quantal heat death. As soon as one of the eigenvalues exceed unity the orthodox quantum probability Ansatz breaks down! coherent superposition decomposed in simultaneous eigenstates |j> |psi> = Sum over j |j> of a set of commuting observables for some total experimental arrangement is rho = |psi> These are the eigenvalues of the density matrix rho. Note that the Trace of a matrix is the sum of the eigenvalues of the matrix. Each projection operator filters out one of the eigenvalues. The determinant is the product of the eigenvalues. Each coefficient in the secular polynomial of the matrix is a sum of distinct products of the eigenvalues in increasing order from Trace to Determinant. Obviously, when one of the eigenvalues exceeds unity the sub-quantal heat death approximation breaks down completely and signal nonlocality is no longer forbidden. This is the Achilles Heel of Lenny Susskind's argument for the recovery of quantum information from evaporating black holes even if there was world enough and time to wait that long - there ain't. In the superfluid state there is a macroscopic eigenvalue no of order N, and the corresponding eigenvector |PSIo> can be regarded as the condensate wave function. The value of no/N for superfluid helium at low temperatures seems to be of the order of 10% and shrinks continuously to zero as the temperature approaches the superfluid transition temperature. This order parameter should not be confused with the the superfluid density of the two-fluid model, which is equal to the total fluid density in the zero temperature limit. In the case of the universe this suggests that the vacuum condensate density is ~ 4%, i.e. no/N ~ 0.04 because both the dark energy and the dark matter constitute 96% of the stuff of the world (negative and positive zero point pressures respectively) outside of the condensate coherent vacuum wave. Slides from Rocky Kolb http://www-conf.slac.stanford.edu/ssi/2005/program.htm Think vacuum condensate when you see superfluid below. The superfluid is a giant coherent macro-quantum wave in ordinary 3D space condensate is a mathematically similar giant local wave made out of virtual off-shell bosons (e.g. virtual electron-positron pairs) inside the vacuum. Multiple-Connectivity If the superfluid is confined to a region which is not simply connected, such as the interior of a torus, there may be metastable states in which the phase of the condensate wave function changes in a non-trivial way around a path that cannot be shrunk continuously to zero. It is this sort of consideration that ... led Onsager to the idea of quantized circulation [vorticity quanta] in superfluid 4He. This method of defining the order parameter was generalized by Yang and given the name off-diagonal-long-range-order (ODLRO). In the case of fermions equilibrium state of a superconductor the eigenvector corresponding to to the macroscopic eigenvalue is an S-state spin singlet function of the relative coordinates, and is independent of the center of mass coordinate except near the boundaries: this represents the Bose condensation of a ^1S0 electron pair. For 3He the pair is in a triplet P-state, with the spin and orbital angular momentum coupled together in different ways in the A and B phases ... Again for the singlet S superfluid fermion system there can be states in non simply connected geometries in which the phase of the center of mass dependence of the order parameter changes by a multiple of 2pi around the system. ... In the quantum Hall effect .. the concept of an order parameter is not so clear ... Nevertheless, the ideas of topological quantum numbers run through discussions of the subject. Not only is the Hall conductance e^2/h quanta of physical dimension speed itself a topological quantum number, but the charge carriers themselves ... Is the aether an ODLRO 4D Diff(4) covariant supersolid? Remember local Diff(4) is simply the rigid global 4D translation group T4 locally gauged. The compensating gauge potential here is B = (hG/c^3)^1/2'd'(Goldstone Phase of Vacuum Higgs Field) Here comes Gennady Shipov's torsion field theory extension of Einstein's 1915 GR. There are two important order parameters in a solid, which are the position of the actual unit cell with respect to an ideal unit cell. p. 10 This is B above, i.e. the curvature field 1-form part of the Einstein-Cartan tetrad field. and the orientation of the unit cell. This is S the compensating torsion field potential 1-form from locally gauging the 4D rotation Lorentz group O(1,3) that generalizes O(3) for the 3D solid. T = dS + W/S + S/(1 + B + S) = torsion field 2-form (dislocations) W = spin connection 1-form determined by B from dB + W/(1 + B) = 0 R = dW + W/W is disclination geodesic deviation tidal curvature 2-form. Homotopy Classes Simplest space is contractible, i.e. continuously shrinkable to a point. The space between the two spheres in the hedgehog is simply connected for closed 1-D loops that can all be shrunk to a point, but is not so for all closed surfaces. The 2D torus has closed paths that cannot be shrunk to a point and that wind around the torus. Assume a continuous order parameter associated with each point in space. Around each closed path there is a continuous change of the order parameter. Homotopy classes classify such continuous changes - mapping of the loops onto the order parameter space - according to whether they can be continuously deformed into each other or not. ... The winding numbers w are integers, and are formally defined by w = (1/2pi)('&'2|'d'Theta) = (1/2pi)(2'|d'd'Theta) Theta = Im ln(Order Parameter) for a complex scalar order parameter or a planar vector field. These must be the same for all loops that can be continuously deformed into each other, and are additive when loops are strung together. The winding numbers round paths that go once around the system are ... all we need to know. This process defines the homotopy group PI1 for the order parameter p. 13 Note that the winding number is a property of the order parameter space G/H as one closed 2pi path is taken around some obstruction in physical 3D space that contains a zero of the order parameter. Order parameters are LOCAL in physical space. This is why emergent More is different macro-physics is local even though the micro-quantum substratum is nonlocal with a complex web of Einstein-Podolsky-Rosen entanglements in higher dimensional configuration space. Indeed, this is the real solution of the Schrodinger Cat paradox. For the U(1) order parameter, PI1 is the group of integers. The A phase of superfluid 3He & liquid crystals have finite groups for PI1. One can also define the 2D homotopy group PI2 in terms of the behavior of the order parameter on a simple closed surface. This is trivial for the U(1) order parameter (complex scalar field). The quantum Hall effect is more complicated, use the mapping of the surface of a two-torus onto a complex projective space ... The topological invariant that describes such a mapping defines its Chern class, and this turns out to be the Hall conductance in units of e^2/h. The order parameter can have discontinuities in physical space. Curvature and torsion are examples of such discontinuities in the World Crystal Super-Solid. If the singularity cannot be removed by smoothing the order parameter, then ... it is topologically stable. The topological stability of these defects depends on homotopy class. p.14 In the case of the U(1) order parameter, the topological stability of a string vortex requires a non-trivial winding number = number of circulation-vorticity quanta. The winding number ... gives the number of complete turns the phase ... makes on a circuit around a closed path in physical space that encircles the defect. The ac Josephson effect involves matching rotations of the relative phases of superconductor wave functions with the temporal oscillations of an ac circuit. p. 15 Low-power metric engineering the local curvature field B = (hG/c^3)^1/2'd'(Phase of Vacuum) to steer the timelike geodesic glide path of the ship from the ship with weightless warp drive (WWD) requires a non-vanishing torsion field potential S and the above Josephson effect in which the stiff macro-quantum carrier phase of the vacuum is modulated by a high Tc signal phase (probably using an anyon condensate in a thin 2D film on the fuselage of the saucer). The signal phase, in turn, is controlled by the EM potential A ~ J using the Bohm-Aharonov effect in tiny nano-solenoids in the 2D layer. The stability of the Bohm hidden variable extended electron (& quark) micro-geon with strong short-range gravity induced directly by the dark zero point energy density core needs a point defect with non-trivial PI2 whose quantized flux quanta is the electric U(1) charge itself, +-e/3 & +-2e/3 for quarks & -e for electrons. Lecture 4 Fractional Quantum Hall Effect for 2D Anyons Note that the Hodge* as used in most elementary applications apparently only works for static curved metrics and needs to be modified for rotating Kerr metrics with gravimagnetism H = (g01,g02,g03) as in frame-dragging in standard 1915 GR with zero torsion. What about (anti)self-dual instantons in Euclidean metric? (vacuum tunneling) Given a 2-form, e.g. curvature, EM field, torsion, then relative to some metric define a * The 2-form is self-dual if *F = +F it is anti self-dual if *F = - F *^2 = +1 for Euclidean signature ++++ *^2 = - 1 for Lorentzian signature - +++ where with causal light cones F = F+ + F- *F+- = +-iF+- Self-interacting nonlinear Yang-Mills eqs in ++++ have (anti)self-dual instanton solutions. Can we generalize this to the N roots of unity? Closed and exact forms: If dB = 0, B is a closed form. If B = dA B is an exact form. All exact forms are closed, but not all closed forms are exact. If we have a p-form A B = 'd'A ('&'(p+1)|A) = (p+1|'d'A) = (p+1|B) =/= 0 Where '&'(p+1) is a non-bounding p-cycle in a multiply-connected p+1 manifold. For example, if p = 2, then &'3 is a closed 2D surface AKA wormhole mouth (portal) that is not a complete boundary of the interior 3D space. There is at least one other '&'3 closed surface some where-when in our universe or in a parallel universe next door in hyperspace. There can be many such '&'3 in a NONLOCAL multi-pronged network of Star Gate wormhole tunnels held open by negative pressure positive zero point dark energy density distributions. The general theorem is (&(p+1)|A) = Sum over all ('&'(p+1)|A) = Sum over all (p+1|'d'A) = (p+1|dA) In the special case that dA = 0 then the final term on RHS = 0. Of course there is no reason why dA = 0 is necessary. You can have dA =/= 0. Restricting ourselves to a LOCAL wormhole mouth where ('&'(p+1)|A) = (p+1|'d'A) = (p+1|B) =/= 0 Let p = 1 ('&'2|A) = (2|'d'A) Let A = 'd'(Theta) Theta = 0-form phase of a LOCAL macro-quantum order parameter valued in the complex plane. Single-valuedness of the local order parameter around the closed loop ('&'2| that is not the complete boundary of the inner area (2| then gives us quantized Flux without flux. Consider winding around the closed non-bounding loop ('&'2| once. Imagine that the order parameter Goldstone phase changes by 2pi/n (n an integer) in order parameter space G/H for 1 winding of 2pi around the non-bounding closed loop in physical 3D space. In that case, the order parameter changes by e^i2pi/n in G/H vacuum manifold space for each 2pi circuit around the Theta Goldstone phase singularity (e.g. vortex core string) in 3D space. In the simplest case n = 1, Then for N windings in physical 3D space we have the nonlocal Bohm-Aharonov effect ('&'(p+1)|A) = (p+1|'d'A) = (p+1|B) = 2piN of flux quanta through the closed loop in the normal core in a stationary state. We can actually pump non-integer flux through, but the state will not be stationary, it will shake off the excess flux in the effect. Suppose n = 2, then for N windings ('&'(p+1)|A) = (p+1|'d'A) = (p+1|B) = piN This is a macro-quantum spinor condensate. For n = 3 ('&'(p+1)|A) = (p+1|'d'A) = (p+1|B) = 2piN/3 suggestive of the fractional Quantum Hall Effect for high Tc 2D films of anyon condensate with para-statistics of the normal fluctuations? Lecture 3 Gravity energy is nonlocal and Yilmaz et-al are wrong. The Bianchi identities generalize d^2 = 0 to D^2 = 0 when there are gauge field connections A, W etc. present. For example: D = d + A/ for SU(2) & SU(3) internal symmetry non-Abelian Yang-Mills gauge force models of the parity-violating weak hypercharge and parity-conserving strong gluon forces respectively in the standard model. It is the weak hypercharge group SU(2) that is spontaneously broken to produce the small inertia of the leptons and quarks and to produce emergent gravity via my original formula B = (hG/c^3)^1/2d(Goldstone Phase) Here Goldstone Phase is the mean SU(2) hypercharge phase from using the Trace operation (details later on). The large mass of the hadrons comes from the kinetic energy of the confined quarks in dark energy bags. In 1915 General Relativity, locally gauging T4 -> Diff(4): D = d + W/ Where the spin connection W obeys T = De = d(1 + B) + W/(1 + B) = 0 Therefore dB + W/(1 + B) = 0 B is the curvature tetrad field that corresponds to disclination topological defects in the Vacuum ODLRO Manifold G/H and in Hagen Kleinert's World Crystal Lattice. Gennady Shipov's torsion field is when the Lorentz group is locally gauged in addition to T4. This adds 6 new scalar fields that act like extra space dimensions, i.e. 10D space-time. Just as locally gauging T4 brings in the compensating 1-form curvature disclination connection B, provided SU(2) hypercharge spontaneously breaks in the inflation, so does locally gauging the Lorentz group O(1,3) bring in the compensating 1-form torsion dislocation connection S where the non-vanishing torsion 2-form T is now T = dS + W/S + S/(1 + B + S) Note the torsion-curvature coupling terms W/S and S/B. The 1915 GR curvature 2-form is R = DW = dW + W/W The Einstein-Hilbert dark energy vacuum action is the 4-form R/(1 + B)/(1 + B)+ /zpf(1 + B)/(1 + B)/(1 + B)/(1 + B) The 1915 Bianchi identity is, for /zpf = 0 DR = 0 i.e. (d + W/)R = dR + W/R = 0 The Einstein field equation is D*R(Geometry) = *J(Matter) D^2R = D*J(Matter) = 0 is the local conservation stress-energy current density. Note D*J(Matter) = d*J(matter) + W/*J(Matter) W/*J(matter) = d*j(vacuum-matter) *j(vacuum-matter) is a 3-form that is the pseudo-tensor of the (matter-vacuum coupling) Nonlocality of total gravity energy is multiply-connected Flux without flux, i.e. (3|D*R) =/= 0 When *J(matter) = 0 everywhere-when i.e. *R = *DW D^2 = 0 but DD =/= 0 i.e. Flux without flux from multiple connectivity of 3-Manifold slices of space-time. If no S, then /zpf is really a constant, i.e. Cosmological Constant made small by vacuum ODLRO. When S =/= 0, then /zpf is a local scalar field, indeed the one we need for metric engineering warp and wormhole with the Josephson homodyne detection method. This is in complete analogy to Maxwell's U(1) equations dF = 0 d*F = *J d^2*F = d*J = 0 And the Yang-Mills SU(2) & SU(3) equations DF = 0 D = d + A/ D*F = *J D^2F = D*J = 0 The complete theory with torsion and U(1)xSU(2)xSU(3) is clearly the same template where now D' = d + W/ + S/ + A/ + C/ A is U(1)xSU(2)xSU(3) or whatever internal G symmetry group works. W & S are from locally gauging the 10-parameter space-time symmetry Poincare group. If we go to the 15 parameter conformal group we will get an additional conformal or Penrose twistor connection. If we go to Kaluza-Klein extra-dimensions and Grassmann fermion dimensions of supersymmetry (not observed) then everything generalizes in the same template. Now however R' = D'(W + S + C + A ...) D'R' = 0 D'*R' = *J' As well as the Yang-Mills equations F' = D'A DF' = 0 D*F' = *j etc. Similarly for S, B & C? Lecture 2 (Revised Draft #2) Ex 4 Mechanical Model of a Phase Singularity Simplest case. Imagine a plane. Pick an origin O. Use polar coordinates, (r,theta) for arbitrary moving point P. Pick a point 0' with fixed coordinates (a, chi). Draw a circle of radius b < a centered at O' with coordinates (b,phi) Let point P move around this circle whose center O' is displaced from origin O. Obviously when a =/= 0 the total theta angle integral of the 1-form dtheta swept out in one complete circuit round the circle is ZERO. Basically theta oscillates. Note that the angle theta depends on the angles chi and phi. Half of the movement is clockwise and then counter-clockwise for dtheta on successive semicircles as P winds around the circumference of the displaced circle. This is most easily seen intuitively all at once when O' is vertical compared to O (on y-axis ordinate). Note what happens when you move the circle to different locations on the plane. Draw tangents from O to the circle in different locations. In contrast, when a = 0, or alternatively, b > a the total angle integral of dtheta is 2pi. Homework Problem Use trigonometry to make an algebraic proof. ------------------- For 3D flat metric, the Hodge * is with the right-hand rule convention *dx/dy = dz *dy/dz = dx *dz/dx = dy Left-hand rule is *dy/dx = dz *dz/dy = dx *dx/dz = dy Parity transformation interchanges left and right hand rules in 3D. (x,y,z) -> (-x,-y,-z) SU(2)hypercharge breaks parity symmetry and it also may be the origin of inertia and gravity. p-forms |p) = (p!)^-1Fuv ... dx^u/dx^v ... p-factors, p =< n = dim of manifold. Important formula |p)/|q) = (-1)^(pq)|q)/|p) The exterior product / of forms is a parallelepiped in the co-tangent n-dim space of constant phase wave fronts in contrast to the tangent For R^3 A = Axdx + Aydy + Azdz F = dA = ( Az,y - Ay,z)dy/dz + (Az,x - Ax,z)dz/dx + (Ax,y - Ay,x)dx/dy 2-form independent of metric *F = *dA = ( Az,y - Ay,z)dx + (Az,x - Ax,z)dy + (Ax,y - Ay,x)dz * dual 1-form in 3D manifold with a metric specified. Note, if A = df F = dA = d^2f = 0 Therefore ( Az,y - Ay,z) = 0 etc , is ordinary partial derivative i.e. mixed second order partial derivatives of the 0-form f commute in that case. However, in the case of a phase-singularity, there is some kind of region in the manifold where the mixed partials of the 0-form Goldstone phase of the local macro-quantum coherent vacuum order parameter Higgs field in our primary application to physics of this formalism do not commute. This is a topological defect in the vacuum manifold G/H, where I write A = 'd'f d'd' =/= d^2 = 0 due to multiply-connected manifolds F = dA =/= 0 e.g. non-integrable anholonomic multi-valued gauge transformation of Hagen Kleinert AKA Flux without flux see also the related idea of the nonlocal Bohm-Aharonov effect using Feynman amplitude Wilson loop operators. In 3 space d|0) is gradient of a function, i.e. scalar field d|1) is curl of a vector field d|2) is divergence of a vector field B = Bxydx/dy + Byzdy/dz + Bzxdz/dx dB = (Bxy,z + Byz,x + Bzx,y)dx/dy/dz Static 4D Metrics without gravimagnetism (non-rotating spacetimes) & without gravity waves (c = 1 convention) here (curved metric) = g = -dt^2 + 3^g The toy model wormhole is of this form. We need positive dark zero point energy density with negative pressure to keep the wormhole open. There is no event horizon in this wormhole. It's not a black hole! A metric allows the symmetric inner product { , }. Classical energy density of the EM field in the absence of sources is (1/2)[{E,E} + {B,B}] The Lagrangian density is (1/2)[{E,E} - {B,B} E = (Ftx, Fty, Ftz) electric field B = (Fyz, Fzx, Fxy) magnetic field F = B + E/dt F & B are 2-forms E is a 1-form We need a classical EM stress-energy density tensor T to compute T ~ &(Dynamical Action)/&(metric) & is functional derivative of classical Lagrangian field theory (not w = (pressure/energy density) Note, the above is classical without any quantum zero point fluctuations. w = +1/3 for classical far-field radiation with only 2 transverse polarizations. For example, the cosmic black body radiation has w = +1/3 http://www-conf.slac.stanford.edu/ssi/2005/lec_notes/Kolb1/kolb1new_Page_05_ jpg.htm It's wrong to use w = +1/3 for vacuum zero point energy that bends spacetime absolutely. This is an error in SED used by HRP. The Casimir force is not important for metric engineering Weightless Warp Drive and Wormhole Time Travel to The Past (using old wormholes made at the beginning of our local universe and even connecting to the parallel universes of Super Cosmos). Equivalence principle + local Lorentz invariance imply w = -1 for all kinds of zero point energy (isotropically distributed). That is the Zero Point Energy Stress-Energy Current Density Tensor tuv(ZPF) diagonal is for Energy Density dE/dV & Pressure P w(ZPF) = P/(dE/dV) = -1 from EEP & LI of GR (dE/dV,-dE/dV,-dE/dV,dE/dV), Therefore, since w = -1 Trace is -2(dE/dV) = (1 + 3w)(dE/dV) Compare to EM radiation where w = P/(dE/dV) = + 1/3 ( dE/dV, +(1/3)dE/dV, +(1/3)dE/dV, +(1/3)dE/dV), i.e. Trace = +2dE/dV = (1 + 3w)(dE/dV) Note that dark energy is P < 0 & dE/dV > 0. Dark Matter is P > 0 & dE/dV < 0 Assuming here above, of course, compact dark matter sources like the Galactic Halos when P > 0 and measurements by external observers. If we stick in Casimir plates to break translational symmetry or somehow break the rotational symmetry (rotating superconducting disks that phase lock to the vacuum Goldstone phase?) There is analogy here to homodyne detection of quantum information with continuous variables where the local oscillator in a beam splitter is like the vacuum ODLRO field. (Rev Mod Phys, p. 513, April 2005) Squeezed vacuum states in quantum optics is when one quadrature of the zero point virtual photons has less vacuum noise than does the other quadrature which has excess noise to compensate. http://www.cco.caltech.edu/~qoptics/squeeze.html Then ( dE/dV, +a(dE/dV), + b(dE/dV), + c(dE/dV)), the trace is now (1 + a + b + c)dE/dV 1 + 3w = 1 + a + b + c w = (a + b + c)/3 This is quintessence when w < -1/3 and it can perhaps be done with the Shipov torsion field. Phantom energy with the Big Rip tearing the fabric of the Universe apart is when w < -1. Lecture 1 on Cartan Forms I am using John Baez's Ch 4 of Gauge Fields, Knots and Gravity for the standard ideas. All the local physical observables in classical gauge force field theories are examples of Eli Cartan's differential forms, e.g., Au, Fuv, ju. The integrals of forms over manifolds are premetrical until we define a Hodge * operation taking a p-form to a N-p form for N-dim manifolds. The p-forms are very much like Bishop BerkeleyÍs ñghosts of departed quantities.î They ñare neither finite, nor .83 infinitely small, nor yet nothing.î The 1-forms are dual to tangent vector fields on the manifold. A vector picture of quantum theory. The 1-form (AKA ñcotangent vectorî) is like a stack of wave fronts (AKA ñlittle hyperplanesî) of small extent as in Fig. 1 p. 45 (Baez) also in MTWÍs ñGravitation.î ñThe bigger df is, the more tightly packed the hyperplanes are.î Given a Cartesian coordinate basis of tangent vector fields {,u} and a dual basis of 1-forms {dx^u}, then duality here is dx^u,v = 1v^u = Kronecker delta NxN identity matrix. Given a 1-form df and any vector field v, the directional real number df(v) ñcounts how many little hyperplanes in the stack df the vector v crosses.î Linearity is built in as a postulate. The Cartan forms are invariants of local coordinate LNIF transformations Diff(N). Diff(N) is what you get when you locally gauge the global ND translation group. In 1915 GR the Cartan forms are also invariant under the local LIF Lorentz transformations O(1,3). In general this would be O(N) pre-signature. That is TNxO(N). / is the exterior product. Obviously we have a kind of quasi-algebra equivalent to dissecting an N-1 simplex or ñbraneî also giving partially ordered (non-Boolean?) lattices with the 0-form on bottom and the N-form on the top. N = 1 (dx), i.e. 1 N = 2 (dx, dy, dx/dy = -dy/dx), i.e. 3 N = 3 (dx,dx,dz, dx/dy, dx/dz, dy/dz, dx/dy/dz), i.e 7 N = 4 (dx,dy,dz,dt, dx/dy, dx/dz, dy/dx, dt/dx, dt/dy,dt/dz, dx/dy/dz, dt/dx/dy, dt/dx/dz, dt/dy/dz, dt/dx/dy/dz), i.e. 15 If we include the 0-form we have 2, 4, 8, 16, i.e. 2^N elements in the quasi-algebra that suggests the Clifford Algebras. There are obviously N!/p!(N-p)! p-forms in N space. This is also like an information space of N c-Bit Shannon Boolean strings. Obviously there will be some kind of matrix representation. For example N = 2 should correspond to the 3 Paul 2x2 spin matrices with the unit matrix. Therefore, there is a connection to U(1)xSU(2) here. N = 3 should have something to do with the 8 SU(3) matrices, and N = 4 obviously connects with the Dirac algebra and possibly U(4) especially when we complexify each real number space-time dimension and even go beyond that to quaternions & octonians. Classical gauge force theories include Maxwell's U(1) electromagnetic theory, Yang-Mills theories of the SU(2) weak and SU(3) strong forces of the leptons and quarks in the standard model and Einstein's theory of gravity (General Relativity, 1915 AKA GR) provided you do not work at the symmetric metric tensor level guv(x), but work at the square root 1-form tetrad e level. Note that Einstein's local equivalence principle is simply (curved metric ) = e(flat metric)e where e is the Einstein-Cartan 1-form tetrad field. You can write e = 1 + B B = curvature tetrad field Since the forms are local frame invariant this decomposition is objective. Global Special Relativity 1905 AKA SR is when B = 0 everywhere-when. Note that the (curved) metric has linear in B elastic terms and nonlinear quadratic in B plastic terms (H. Kleinert), i.e., (curved metric) = (flat metric) + 1(flat metric)B + B(flat metric)1 + B(flat metric)B The B^2 terms show that the gravity field is self-interacting like the SU(2) & SU(3) gauge fields, but unlike the U(1) Maxwell EM field. The Cartan exterior derivative operator d on forms generalizes the gradient, curl and divergence. Together with its dual boundary operator & on co-forms, there is a generalization of Stokes & Gauss's theorems to N-dimensional manifold integrations with multiple-connectivity (e.g. wormholes). The p-dim form |p) is the thing integrated. The dual co-form (p| is the manifold on which the integral is done. I use a variation on the Dirac bra-ket notation. The basic integration theorem, is like the adjoint operation in quantum theory, i.e. (&(p+1)|p) = (p+1|d|p) The two identities d^2 = 0 &^2 = 0 are analogous to the antisymmetric Pauli exclusion principle in quantum field theory where a^2 = 0 a*^2 = 0 a* creates a fermion, a destroys a fermion. However, we use the notations ïdÍ and ï&î partially introduced by John Ex. 1: ïdthetaÍ = (xdy [CapitalEth] ydx)/(x^2 + y^2) for the polar angle ñthetaî where dr = (xdy + ydx)/(x^2 + y^2) x = rcos(theta) y = rsin(theta) The 1-form ïdthetaÍ above is closed, but not exact. In effect this means dÍdÍ =/= 0 ONLY when the integral is over a non-bounding co-form (AKA non-bounding cycle). Therefore, for this particular example, there is phase (theta) ambiguity at the origin r = 0. When the closed loop integral (&Í2|ÍdÍthetaÍ) = (2Í|dÍdthetaÍ) =/= 0 encircles the origin r = 0 it does not vanish. Note that if the closed loop integral of ïdthetaÍ does not encircle the branch point r = 0, it will vanish. In this sense, ïdthetaÍ is closed, but not exact and &2Í is not a true boundary because of the ñholeî at r = 0. Note that the co-form (2Í| is the area enclosed by the loop &Í2 minus the ñholeî at r = 0. If we extend this to cylindrical coordinates, then we have a vortex core string provided we have a local U(1) complex order parameter PSI(r,theta,z) such that PSI(0,theta,z) = 0 PSI(r,theta,z) = PSI(r, theta + 2pi, z) for equilibrium ñstationary statesî when the closed system relaxes expelling excess flux in the Meissner effect. In that case, (&Í2|ÍdÍthetaÍ) = (2Í|dÍdthetaÍ) = 2piN N = +-1, +-2 .83. N = winding number around the string vortex core on the z-axis. To review, the rigorous theorem is (&(p+1)|p) = (p+1|dp) Where (&(p+1)| is a true boundary, which means (&^2(p+1)| = 0 When |p) is exact, that means |p) = |d(p-1)) and |dp) = |d^2(p-1)) = 0 However, when the topology of the co-form manifold is multiply connected we can have closed p-manifolds, AKA ñnon-bounding p-cyclesî, (&Í(p+1)| that are not true boundaries together with non-exact p-forms |dÍ(p-1)) such that (&Í(p+1)|dÍ(p-1) = (p+1Í|ddÍ(p-1)) =/= 0 The non-bounding p-cycles are p-dim wormhole mouths or ñStar Gate Portalsî that are ñThrough The Looking Glassî Darkly as it were, down the Rabbit Hole in Hyperspace. Ex. 2: Consider the 3D space-like metric of a static spherically symmetric non-rotating uncharged wormhole Star Gate is (3-metric) = dr^2 + f(r)^2(dtheta^2 + sin^2thetadphi^2) Where f(r) is the wormhole shape function. Each wormhole mouth looks like a closed spherical surface of radius R where R = f(r*) df(r*)/dr = 0 d^2f(r*)/dr^2 > 0 This closed S2 surface is not a complete boundary (&3| enclosing a 3-space because it has a twin wormhole mouth somewhere-when else perhaps in a parallel universe next door in hyperspace. Therefore, all wormhole mouths for actual time travel to distant places in negligible proper time for the traveler are really (&Í3| not (&3|. Furthermore, the curved tetrad field B = (hG/c^3)^1/2ÍdÍ(Goldstone Phase) is not exact, i.e. in a 1-D loop around the wormhole mouth S2 surface with the multiply-connected quasi StokeÍs theorem (1Í|B) = (&Í3|dB) =/= 0 This is the curved tetrad flux through the closed loop that ñcutsî the spherical wormhole mouth. Ex. 3: Given in cylindrical coordinates the vortex string along the z-axis ïdÍtheta = (xdy [CapitalEth] ydx)/(x^2 + y^2) For any closed loop (1Í| = (&Í2| around the z-axis (1Í|ÍdÍtheta) = (2Í|dÍdÍtheta) =/= 0 i.e. Nonlocal Bohm-Aharonov ñFlux without fluxî Given the above wormhole 3-metric define a Hodge * operation, with the non-exact 2-form 3*ÍdÍtheta = (xdy/dz + ydz/dx + zdx/dy)/f(r)^3 Where now we have a multiply-connected quasi-divergence Gauss theorem (3Í|d3*ÍdÍtheta) = (&Í3|3*ÍdÍtheta)=/= 0 when (&Í3| is a wormhole portal. There is now a radial 3*ÍdÍtheta flux through the wormhole closed surface in addition to the ïdÍtheta circulation around a closed loop that cuts the wormhole closed surface that is not a complete boundary. === Subject: prisoner's dilemma game I once spent some time trying to devise a simple two-player gambling game based on the prisoner's dilemma. The basic idea is as follows. Each player at each turn independently chooses to be greedy or ungreedy (what's the opposite of greedy?). This choice could be implemented in any number of ways; the mechanism is not important. Following the principles of the prisoner's dilemma, the outcome at each turn should be as follows. A chooses... B chooses... Outcome Ungreedy Ungreedy Indifferent for both A and B Ungreedy Greedy B benefits at A's expense Greedy Ungreedy A benefits at B's expense Greedy Greedy Bad for both A and B Any mechanism of payments between player A and player B is allowed, as well as any payments into any pots to be distributed later somehow. I couldn't find any way to make this into a viable game. The hardest part seems to be figuring out what should happen in the bad-bad situation. If it could be made to work I think it would be great fun (for a while anyway). Any ideas anyone? === Subject: Fault-tolerant telephone trees I am membership secretary of a metal detecting club. Our outings are invariably and inevitably arranged at short notice (say three or four days), and all the members need to be alerted within that period of time. I have inherited a simple in-series system whereby the organizer alerts four people, each of whom then calls one new person each, and so on. This system does not work very well, because of people being out when phoned, or people who just aren't very reliable at passing on the message. I'm trying to design a better way of doing it, still based on telephone (less than half the members have email). It seems to me that by stipulating that each person should get and make two calls rather than one, it should be possible to design a solution which is much more efficient (greatly reduces the maximum number of calls anyone can be from the start) and fault-tolerant (because a missed or an unmade call does not cut off information flow). It occurs to me that there will undoubtedly be a body of theory to call upon (indeed, for the modest numbers of nodes I have to deal with, there may well be an optimum solution already documented). Could Andy === Subject: Re: Some questions on logic <4340ce55$0$30203$892e7fe2@authen.white.readfreenews.net> Actually 1=0 === Subject: Re: Some questions on logic <4340ce55$0$30203$892e7fe2@authen.white.readfreenews.net> Off the meds again? === Subject: Re: Standard Deviation of PISA > All of which proves that the 285 PISA score for nigger girls in Peru, > the 333 TIMSS score for niggers in South Africa, the 427 IAEP score for > niggers in Mozambique, the 17 ACT math score for niggers in Washington, > DC, (as well as their 320 SAT math score, and their 231 NAEP math All of which proves you do not understand the basic principles of test design and validation. James Powell === Subject: Re: Standard Deviation of PISA >It's even better than that. >By our own English definition of race, you cannot distinguish it from >Tribe. So each of the 12 Tribes of Israel are actually the 12 races >of Israel [since they each have a separate patriarch]. >BUT--each HOUSE is a different tribe as well. And the Edomites who're >descendants of Esau, the Moabites who're descendants of Moab, and the >Ammonites who're descendants of Ammon, are all separate and distinct >races. >AND--there were different Hebrew words for the entity who made the >behemah in Genesis 1:25, the entity who created the first man Adam in >Genesis 1:27, and the entity who formed the last man Adam in Genesis >2:7. >And after all that, as if that's not enough, it was a DIFFERENT entity >who impregnated Rebekah with Jacob, who HATED Esau because Esau was not >his progeny, and LOVED Jacob because Jacob was His firstborn: >And thou shalt say unto Pharaoh, Thus saith the LORD, Israel is my son, >even my firstborn: Exodus 4:22 >http://jacobisrael.us/jacob.htm >There are more than 40 races described in the Holy Bible--and God loved > only ONE of them! >Why? Because only ONE were descendants of HIS firstborn [and it aint' >jews]. >John Knight No surprise. Darwin and his buddies thought there may be as many as 63 races of humans. They, at that time were trying to decide whether or not the various races ought to be classified as different species. Man has been studied more carefully than any other animal, and yet there is the greatest possible diversity amongst capable judges whether he should be classed as a single species or race, or as two (Virey), as three (Jacquinot), as four (Kant), five (Blumenbach), six (Buffon), seven (Hunter), eight (Agassiz), eleven (Pickering), fifteen (Bory de St-Vincent), sixteen (Desmoulins), twenty-two (Morton), sixty (Crawfurd), or as sixty-three, according to Burke. This diversity of judgment does not prove that the races ought not to be ranked as species, but it shews that they graduate into each other, and that it is hardly possible to discover clear distinctive characters between them. Charles Darwin (Descent of Man Chapter VII - On the Races of Man) Now what gets me is that these humanists who hate religion and proclaim science as their god, (as did Darwin since he turned from Jesus) are the very first to point to Christians and holler you racist you! How absurd. How ridiculous. How self-destrutive these liberals are in evolving into traitors of their own (white) race. Here's Darwin pointing out his status as an X-christian (as if it's some sort of badge of honor) in the begriming paragraph of his very own autobiography: During these two years I was led to think much about religion. Whilst on board the Beagle I was quite orthodox, and I remember being heartily laughed at by several of the officers (though themselves orthodox) for quoting the Bible as an unanswerable authority on some point of morality. I suppose it was the noveltry of the argument that amused them. But I had gradually come, by this time, to see that the Old Testament from its manifestly false history of the world, with the Tower of Babel, the rainbow at sign, etc., etc., and from its attributing to God the feelings of a revengeful tyrant, was no more to be trusted than the sacred books of the Hindoos, or the beliefs of any barbarian. - Charles Darwin (The Autobiography of Charles Darwin) pleasure talking to you, Jd === Subject: Re: Standard Deviation of PISA <0001HW.BF5E426C0027D38CF0284550@news.giganews.com> <5uj3k15jn9mtl75f6toh15770dcsfa1vda@4ax.com> Yes, it is ironic that Darwin disputes the very Darwinists who quote him in their desperate attempts to claim that all men are descendants It's mathematically and scientifically IMPOSSIBLE for niggers and Whites to have a common ancestor, the Holy Bible claims that they do NOT have a common ancestor--and Darwin himself admits in that one passage that there are numerous races (AND that he didn't have a clear enough understanding of the scientific method to deduce a theory in the first place). On an average every species must have same number killed year with year by hawks, by cold, & c.--even one species of hawk decreasing in number must affect instantaneously all the rest. The final cause of all this wedging must be to sort out proper structure.... One may say there is a force like a hundred thousand wedges trying to force every kind of adapted structure in the gaps in the oeconomy of nature, or rather forming gaps by thrusting out weaker ones. What a miserable supposition. What a limited understanding of probabilities and statistics this man possessed. This is not science. This is not based on solid math. This is a statement of a supposition which is irresponsibly presented as a theory, which even Darwin himself questioned, and ultimately rejected. This is his entire proof that natural selection exists. Why would anyone accept his notion that an entire species can transform itself into another species simply by thrusting out weaker ones? He concludes that the only way that the South American rhea can so closely resemble an African ostrich is that they both evolved from another species. And from this, Darwinists and evolutionists make the giant leap that man evolved from monkeys!!? http://jacobisrael.us/evolution.htm Fortunately, even with half a decade of teaching evolution [and refusing to teach the Word of God], 91% of Americans STILL reject this moron's theory. John Knight === Subject: Re: Standard Deviation of PISA >#3 is particularly interesting seeing that we have Bob here pretending >to be a Christian who says Jesus (and all humans) decended from apes. >>Depending on how one defines the term race, it might very well be >>possible to show that several races of Christians are currently >>occupying U.S. soil. >>Later, >>Jd >They may be right that THEY are descendants of apes. The nigger DNA is >closer to apes than to Israelites. The physical charcteristics of apes >are more similar to apes than they are to Israelites. The IQ of Koko >the Gorilla of 85 is higher than the IQ of most sub-Saharan niggers by >more than the IQ of many Israelites is higher than that of Koko. But >of course this is an insult to apes, as they: >1) Know who their daddy is. >2) Don't murder each other at genocidal rates. >3) Don't cost us honkies $800 BILLION per year in social transfer >payments [read: WELFARE] >http://jacobisrael.us/blackcosts.htm >John Knight [chuckle!] But on the serious side, it's a shame that many white folks have become traitors to their own race ain't it? Jd === Subject: Re: Standard Deviation of PISA > > Why are Korean and Gaelic the same language? More than 3,000 years > ago, Hebrew was spoken all the way from Ireland to Phoenicia to Judaea > to Korea, and the odds are good that several of the hundred plus > Indian languages are Hebrew. >Indian as in feather or dot? Because if it's >feather, then Lakota obviously isn't. > Indian as in India. > Even American Indians know this isn't India. > To be here as long as they were, American Indians have to be REAL > stupid to never develop a written language. Even the Indians of India, > whose average earnings are $100 per month, were smart enough to develop > several written languages (or at least smart enough to inherit written > languages from the original developers and maintain it). Which part of Mayan seems to be giving you trouble? -- cary === Subject: Re: Standard Deviation of PISA >> > < standard written Chinese, so Mandarin speakers typically can read >> < such communication without much difficulty. >>All literate people write in standard written Chinese! >My earlier post noted that official Chinese sources only claim 85% >literacy. No doubt, as with American weaknesses in functional >literacy there are many counted as literate that are not in fact >functionally literate. On the other hand, getting data on such a >phenomenon without the assistance of the government would be >difficult. > Even if all Whites in this country were functionally literate, there > are NO niggers, and NO latrinos (who together make up 25% of the > population) who're literate, much less functionally literate. Yeah, yeah, yeah...tell it to John McWhorter and Richard Rodriguez. -- cary === Subject: Re: Standard Deviation of PISA >Why are you so disconnected from the real world all the time? With few >exceptions, those who take the SAT know why they're planning to go to >college: About 3/4 of all kids change majors at least once before graduating (citing notes from a seminar I took back in college) and around 1/2 of all kids don't graduate from college. Thus, what a kid thinks they want to do at age 17 when taking the SAT has relatively little to do with what they will be doing 5 years later. lojbab -- lojbab lojbab@lojban.org Bob LeChevalier, Founder, The Logical Language Group (Opinions are my own; I do not speak for the organization.) Artificial language Loglan/Lojban: http://www.lojban.org === Subject: Re: The Black Hole in a Spatially Compact Spacetime > If matter falls into a black hole at the suface any light emitted there > would be infinitely redshifted. This would be true if any was emitted from there. None is. > So General Relativitiy predicts Energyless light. At a time infinitely far in the future. Who cares. > Dim holes are the extreme of gravity - light redshifted only by a > finite amount. it doesn't get there unless you wait forever. Bob -- Things should be described as simply as possible, but no simpler. A. Einstein === Subject: Re: The Black Hole in a Spatially Compact Spacetime > If matter falls into a black hole at the suface any light emitted there > would be infinitely redshifted. > So General Relativitiy predicts Energyless light. This is the obvious > reason for no black holes. This infinity is its failure. > Dim holes are the extreme of gravity - light redshifted only by a > finite amount. Do you honestly think repeating yourself makes things true? === Subject: Re: The Black Hole in a Spatially Compact Spacetime Yes, gravitation is the energy I'm talking about. Where does it come from? John === Subject: Re: The Black Hole in a Spatially Compact Spacetime In sci.math, Sam Wormley : >>Take the spacetime cylinder for example. All >>observers agree on the simultaneity of events. > No >>Clock rates everywhere, in this instance, are >>all physically tied together. > No >> You obviously don't understand SxR. >> http://www.everythingimportant.org/viewtopic.php?t=605 >> http://www.everythingimportant.org/relativity/simultaneity.htm > Shubert says ...my global theorem reveals, there is an absolute > past, present and future. > You are in correct conflict with Special Relativity. SR enjoys > experimental verification. Shubert does not. Errm....that phrasing might be in slight need of clarification? :-) (I could say something about incorrect conflicts such as the War in Iraq but that's probably fodder for a different newsgroup...) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: The Black Hole in a Spatially Compact Spacetime > In sci.math, Sam Wormley > : >Take the spacetime cylinder for example. All >observers agree on the simultaneity of events. >> No >Clock rates everywhere, in this instance, are >all physically tied together. >> No >You obviously don't understand SxR. >http://www.everythingimportant.org/viewtopic.php?t=605 >http://www.everythingimportant.org/relativity/simultaneity.htm >> Shubert says ...my global theorem reveals, there is an absolute >> past, present and future. >> You are in correct conflict with Special Relativity. SR enjoys >> experimental verification. Shubert does not. > Errm....that phrasing might be in slight need of clarification? :-) > (I could say something about incorrect conflicts such as the War > in Iraq but that's probably fodder for a different newsgroup...) My point was that Henry and Albert see events in a different order when one is moving with respect to the other... there is no absolute ordering of events... order can be different for different observers. === Subject: Re: riemann vs lebesgue integrals > Actually, *every* Riemann-integrable function is also > Lebesgue-integrable and the two integrals are equal. This does not > apply to improper integrals though. What is an example of an improper Riemann integral that is not equal to the corresponding Lebesgue integral? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: riemann vs lebesgue integrals >I was wondering what the formal definition of the lebesgue integral was and how it differs from the riemann integral. In general, when doing an integral (as asked in classes) we are using the riemann inegral, correct? Which formal definition? I prefer not using the term definition, but rather characterization, as they are equivalent. The approach best showing the difference is to start with a vector space S, closed with respect to absolute value, of simple functions, which are usually taken to be the step functions, although continuous functions will give the same result. On this set, it is assumed that an integral I is defined, which is a positive linear operator. Then the extension I* defined by v = I*(f) if for all e > 0 there exists a function g in S, and non-negative h_j in S such that |I*(g) - v| < e, |f-g| < sum h_j, sum I(h_j| < e. If the sum if required to be finite, or equivalently one, it is the Riemann extension. If countable is allowed, it is the Lebesgue extension. This does not deal with improper or limit integrals. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: riemann vs lebesgue integrals >> In general, when doing an integral (as asked in classes) we are > using the riemann inegral, correct? > Depends on the class (and the function). But for continuous functions, > for example, the two values agree. But, yes, the Fundamental Theorem > of Calculus is about Riemann integrals. There is a stronger version of the FTOC that applies specifically to Lebesgue integrals. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.