mm-4711 === Subject: Re: cubic spline question > I guess what I'm asking here is, can I use cubic splines to produce a > formula which can be used to predict points in a trend based on > existing data? æYou can do so with any type of regression line or > curve, so can you predict points with an equation using cubic splines? Extrapolation with splines is a terrible idea. Why is that a terrible idea? That's essentially what I eventually wish to accomplish. Rather than having to store all the data points, I can procedurally approximate them using their cubic spline equation. Why is that a terrible idea? === Subject: Re: cubic spline question > I guess what I'm asking here is, can I use cubic splines to produce a > formula which can be used to predict points in a trend based on > existing data? You can do so with any type of regression line or > curve, so can you predict points with an equation using cubic splines? > Extrapolation with splines is a terrible idea. Why is that a terrible idea? That's essentially what I eventually > wish to accomplish. Rather than having to store all the data points, > I can procedurally approximate them using their cubic spline > equation. Why is that a terrible idea? It is a terrible idea for several reasons. (1) Do you know the data is absolutely accurate? Are you sure that there are not random measurement errors or noise errors at some or all of the data points? Note: if the data *does* have errors, fitting a quadratic or cubic or spline exactly to the data is a very bad idea: you are trying to fit unpredictable random errors by means of exact functions. The problem is that the errors occurred in the past, while you want to predict the future (i.e., the data in regions where you have not yet made measurements). On the other hand, if you are sure the data is error- free (when is it ever?), then go ahead and make an exact fit; but what makes you think that a polynomial (spline or otherwise) is the appropriate form> Why not sines and cosines, or exponentials or logarithms or square roots? (2) As has already been stated, a cubic spline is a piecewise fit, so why would you assume that the last cubic (at the end of the data set) is the one to use? Maybe the cubic spline fit does not represent the true situation at all; it fits a finite set of existing points and thus says nothing much about the true function at other points--- although there are error bounds available that let you get some feeling for the errors. Here is an example: f(t) = 1 + (1/2)*t + (1/4)*sin(t). Say we have N data points of the form [t1,f(t1)], ..., [tN,f(tN)]. Now fit a cubic spline. It would be a disaster to use a cubic in t (spline or otherwise) to extrapolate to large t; that would givee t^3 behaviour, while the actual behaviour is linear in t + a bounded, periodic oscillation. Anyway, data in the real world is almost always error-prone, so fitting exact formulas to it is a mistake. That is why people work wit things like least-squares fits, or least summed absolute error fits, or least maximum error fits, possibly with different importance weights in different data ranges. R.G. Vickson === Subject: extrapolation based on slope posting-account=uyK1mwoAAADPV1QXMRRacGpjEv75SqnN Gecko/2008061015 Firefox/3.0,gzip(gfe),gzip(gfe) Let's say I had various 2D point samples of waveform data. If I wanted a nonlinear slope equation to describe and approximate my points, what would be the best methodology? I'd want the resulting equation to be more succinct than just storing a complete reference of the points. Should I be using polynomial regression, moving average regression, or something altogether different? What I'm looking for here is a balance of brevity and accuracy. === Subject: Re: extrapolation based on slope- > Let's say I had various 2D point samples of waveform data. If I > wanted a nonlinear slope equation to describe and approximate my > points, what would be the best methodology? I'd want the resulting > equation to be more succinct than just storing a complete reference of > the points. Should I be using polynomial regression, moving average regression, or > something altogether different? What I'm looking for here is a balance of brevity and accuracy. A naive and definitely non-expert approach (I am not an expert in waveform data): Try the technique of periodogram analysis, or any other Fourier Analysis - like technique to establish whether you indeed have a periodical or almost-periodical phenomenon at hand. This step yields one or more periodical components with certain confidence intervals if any periodical components are present. Once you have accepted components by their periods you can proceed with conventional curve-fitting techniques, applied of course to the special case of sinusoidal curves. This would amount to combined Fourier analysis and synthesis. Relevant information online: http://en.wikipedia.org/wiki/Periodogram Wikipedia search for periodogram (URL http://en.wikipedia.org/w/index.php?title=Special%3ASearch&search=periodogra m&ns0=1&fulltext=Search ) yields among others a lemma on the pioneer of periodogram analysis: Arthur Schuster. http://en.wikipedia.org/wiki/Arthur_Schuster Good luck: Johan E. Mebius === Subject: Amazon.com with James Harris and Andrew Beckwith posting-account=_VmC_goAAACU-xAuU22JkRqrgFylfYj- 1.1.4322),gzip(gfe),gzip(gfe) Living together in a closed list, James Harris and Andrew Beckwith are self-appointed masters of their domain, who should write a book to be reviewed by their buddies in the closet list. closed list closed list closed list personal space personal space personal space I'm laughing at the superior intellect Hahahahahahahahahahahahahahahahahahahaha! === Subject: Re: Amazon.com with James Harris and Andrew Beckwith > Living together in a closed list, James Harris and Andrew Beckwith > are self-appointed masters of their domain, who should write a book > to be reviewed by their buddies in the closet list. closed list closed list closed list > personal space personal space personal space > I'm laughing at the superior intellect > Hahahahahahahahahahahahahahahahahahahaha! hihi -ap === Subject: Re: Limit > We get int_0^1 (lim n-->oo (x^n + 1)^(1/n) dx) = > int_0^1 x dx = 1/2........ So for 0 < x < 1 you say that lim (x^n + 1)^(1/n) = x? > (Even though (x^n + 1)^(1/n) > 1.) How queer! Victor Meldrew > I don't believe it! For all x > 0, (x+1)/x > 1. But as x --> oo. (x+1)/x --> 1. How Queer! === Subject: Re: Limit > We get int_0^1 (lim n-->oo (x^n + 1)^(1/n) dx) = > int_0^1 x dx = 1/2........ > > So for 0 < x < 1 you say that lim (x^n + 1)^(1/n) = x? > (Even though (x^n + 1)^(1/n) > 1.) > > How queer! > > Victor Meldrew > I don't believe it! For all x > 0, (x+1)/x > 1. But as x --> oo. (x+1)/x --> 1. How Queer! You missed his point: (x^n + 1)^(1/n) cannot approach x for 0 < x < 1, because (x^n + 1)^(1/n) > 1. === Subject: Andrew Beckwith is now Becky posting-account=_VmC_goAAACU-xAuU22JkRqrgFylfYj- 1.1.4322),gzip(gfe),gzip(gfe) Yep, Becky Beckwith. After a much needed operation to remove what looked like a small pimple on Andrew Beckwith's abdomen, but was what James Harris likes to anyone in his darling closed list. The closet list, where your secret perversions and outright lies to yourself and others are maintained in perpetual confidentiality (*guffaw*). The closet list of frauds. Stroking each others' egos, and other The closet list of buffoons, hiding from the outside world so that they can feel free to lie to everyone ... and to lie with each other. Becky Beckwith is their star, their dream, their love, the one who would lie and commit fraud to advance the agenda and puff up the egos of members of the closet list. closed list closed list closed list personal space personal space personal space I'm laughing at the superior intellect Hahahahahahahahahahahahahahahahahahahaha! === Subject: Games Behind that accounts for ties posting-account=THiYzgoAAABR2rxqh7nJGOtLoi4GI-K8 Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) I understand the typical formula for determining 'Games Behind' in sports standing to be: GB = ((W1 - W2) + (L2 - L1)) / 2 Where W1 = Wins by first place team, W2 = Wins by second place team, L2 = Losses by second place team, and L1 = Losses by first place team. That being correct, one, does that account for all teams having played the same number of games, and also how would this formula need to be written to account for ties? Steve === Subject: Re: Games Behind that accounts for ties > I understand the typical formula for determining 'Games Behind' in > sports standing to be: GB = ((W1 - W2) + (L2 - L1)) / 2 Where W1 = Wins by first place team, W2 = Wins by second place team, > L2 = Losses by second place team, and L1 = Losses by first place team. That being correct, one, does that account for all teams having played > the same number of games, and also how would this formula need to be > written to account for ties? > Assuming that a win is worth 1 point in the standings, a tie 1/2 point, a loss 0, and both teams will play the same total number of games, GB is the difference you'd get between the points totals of the two teams at the end of the season if all their remaining games turned out to be ties. Thus a win by team 1 increases GB by 1/2, a loss by team 1 decreases it by 1/2, and a tie has no effect. Yes, the formula does account for teams not having played the same number of games; and yes, it already accounts for ties. However, GB may be misleading in the case of good teams that can be expected to win most of their remaining games. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Games Behind that accounts for ties posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) On Jun 26, 4:53æpm, Robert Israel > I understand the typical formula for determining 'Games Behind' in > sports standing to be: > GB = ((W1 - W2) + (L2 - L1)) / 2 > Where W1 = Wins by first place team, W2 = Wins by second place team, > L2 = Losses by second place team, and L1 = Losses by first place team. > That being correct, one, does that account for all teams having played > the same number of games, and also how would this formula need to be > written to account for ties? > Assuming that a win is worth 1 point in the standings, a tie 1/2 point, a loss > 0, and both teams will play the same total number of games, GB is the > difference you'd get between the points totals of the two teams at the end of > the season if all their remaining games turned out to be ties. æThus > a win by team 1 increases GB by 1/2, a loss by team 1 decreases it by 1/2, > and a tie has no effect. æYes, the formula does account for teams not having > played the same number of games; and yes, it already accounts for ties. However, GB may be misleading in the case of good teams that can be expected > to win most of their remaining games. The Cubs have demonstrated that to be a fallacy. > -- > Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada === Subject: Re: Games Behind that accounts for ties > I understand the typical formula for determining 'Games Behind' in > sports standing to be: > GB = ((W1 - W2) + (L2 - L1)) / 2 > Where W1 = Wins by first place team, W2 = Wins by second place team, > L2 = Losses by second place team, and L1 = Losses by first place team. > That being correct, one, does that account for all teams having played > the same number of games, and also how would this formula need to be > written to account for ties? > Assuming that a win is worth 1 point in the standings, a tie 1/2 point, a loss > 0, and both teams will play the same total number of games, GB is the > difference you'd get between the points totals of the two teams at the end of > the season if all their remaining games turned out to be ties. Thus > a win by team 1 increases GB by 1/2, a loss by team 1 decreases it by 1/2, > and a tie has no effect. Yes, the formula does account for teams not having > played the same number of games; and yes, it already accounts for ties. > However, GB may be misleading in the case of good teams that can be expected > to win most of their remaining games. However, the rule in soccer is to award 3 points for a win and 1 point for a tie. Therefore, GB is not particularly useful in that context. Soccer schedules generally have all teams active on the same day or perhaps over a 2-day period, and therefore it is not necessary to account for different numbers of games having been played. Rainouts are about as common in soccer as they are in the NFL. Hockey, at least in the NHL, also has a special points system that renders the GB statistic fairly meaningless. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: Looking for a Matrix recipe > I would need a general (for any n) recipe to construct > a matrix P with the following properties. 1. P is upper diagonal. > 2. P^2=unity. > 3. P*transpose(P) is a complete n*n Jordan block. I assume you mean not that it _is_ a complete block, but that its Jordan form _has_ one complete block. > Example for n=2: > 1 2I > 0 -1 Your I, I guess, is i = sqrt(-1). I thought it was just Maple that used I for that. > Example for n=3 > 1 0 1 > 0 1 I > 0 0 -1 > Nonexample for n=4 > 1 0 0 a > 0 1 0 b > 0 0 1 c > 0 0 0 -1 > Can't find a,b,c, it's always 3+1 Jordan blocks! I think there are no 4x4 examples. But here's a 5x5: [ 1 0 0 0 a ] [ 0 1 0 0 aI ] [ 0 0 1 b bI ] [ 0 0 0 -1 0 ] [ 0 0 0 0 -1 ] where a and b are nonzero. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Looking for a Matrix recipe > I assume you mean not that it _is_ a complete block, but that its Jordan > form _has_ one complete block. Yup. I'll never learn fluent Mathese :-) > Your I, I guess, is i = sqrt(-1). I thought it was just Maple that used > I for that. Mathematica too :-) > I think there are no 4x4 examples. But here's a 5x5: > [ 1 0 0 0 a ] > [ 0 1 0 0 aI ] > [ 0 0 1 b bI ] > [ 0 0 0 -1 0 ] > [ 0 0 0 0 -1 ] Strange. Is it only possible for prime n (and n=1)? (Yeah, I always jump to conclusions like that :-) It seems that dropping the upper diagonal condition (just included for my idleness) doesn't help either. In fact, even p^2=1 was only an emulation of a possible weaker condition: Call the inverse of p q. The tensors p_ij*q_kl=q_ij*p_kl shall be equal. Obviosly p=q is a solution. Is it the only one, modulo a constant factor in p? -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de Er-a svo gott sem gott kve[CapitalYAcute]a .9al alda sonum, §v.92 a[CapitalYAcute] f.berra veit er fleira drekkur s.92ns til ge[CapitalYAcute]s gumi. === Subject: Re: Farey sequence lattice points > i want to count integral points (a,b,c,d>=0) satisfying ad-bc = 2 So let F(n) be the number of solutions of a d - b c = 2 > with all variables non-negative integers not exceeding n, > and let f(n) be the number of such solutions where one > variable is exactly n. Note that if n > 2 then only one variable can be n so (for n > 2) > f(n) = 4 g(n) where g(n) is the number of solutions with, say, c = n. a d - b n = 2 -> a d = 2 (mod n) If n is odd then the for each a relatively prime to n there's a unique d > so g(n) = phi(n). If n is even then there's still a unique d for each a > relatively prime to n, but there are also 2 solutions for each a such > that gcd(a, n) = 2. I guess that gives g(n) = phi(n) + 2 phi(n/2). > Just as I said in [1], solutions are given by certain neighboring terms > in the Farey sequence Fn and may be interpreted geometrically by using > Pick's Theorem. Namely, for each a/n in Fn, choose its neighbor b/d > with minimal d. Then a/n, 2b/2d is a solution, i.e. a(2d)-(2b)n = > 2 (ad-bn) = +-2. The sign is negative if a/n < b/d in which case you > must flip the sign by using a/n(-)2b/2d = (a-2b)/(n-2d) vs. 2b/2d, > e.g. for 5/8 in F8 = .. 3/5, 5/8, 2/3 .., choose neighbor 2/3 with min > denom, double it, flip sign 5/8(-)4/6 = (5-4)/(8-6) = 1/2; 5*2-1*8 = 2. > If n is even there are additional solutions given by both neighbors > b/d of the fractions a/c in Fn where c = n/2, namely 2a/n, b/d, > where, again, one must flip the sign as above for a larger neighbor. > E.g. for 3/4 in F8 = .. 5/7, 3/4, 4/5 .. we find solutions 6/8, 5/7 > and 6/8, ~4/5, where ~4/5 = 6/8(-)4/5 = (6-4)/(8-5) = 2/3. [...] > Seems to me that all the solutions you get this way have two even terms > (or more). Don't you miss all-odd solutions like 5*7 - 3*11 = 2? No, for 5/11 in F11 = .. 4/9, 5/11, 1/2 .. choose neighbor 1/2 with min denom, double it, flip sign 5/11(-)2/4 = 3/7, same as example 1 above. --Bill Dubuque === Subject: Re: Farey sequence lattice points i want to count integral points (a,b,c,d>=0) satisfying ad-bc = 2 ... > Note that if n > 2 then only one variable can be n so (for n > 2) > f(n) = 4 g(n) where g(n) is the number of solutions with, say, c = n. > a d - b n = 2 -> a d = 2 (mod n) > If n is odd then the for each a relatively prime to n there's a unique d > so g(n) = phi(n). If n is even then there's still a unique d for each a > relatively prime to n, but there are also 2 solutions for each a such > that gcd(a, n) = 2. I guess that gives g(n) = phi(n) + 2 phi(n/2). Just as I said in [1], solutions are given by certain neighboring terms > in the Farey sequence Fn and may be interpreted geometrically by using > Pick's Theorem. Namely, for each a/n in Fn, choose its neighbor b/d > with minimal d. Then a/n, 2b/2d is a solution, i.e. a(2d)-(2b)n = > 2 (ad-bn) = +-2. The sign is negative if a/n < b/d in which case you > must flip the sign by using a/n(-)2b/2d = (a-2b)/(n-2d) vs. 2b/2d, > e.g. for 5/8 in F8 = .. 3/5, 5/8, 2/3 .., choose neighbor 2/3 with min > denom, double it, flip sign 5/8(-)4/6 = (5-4)/(8-6) = 1/2; 5*2-1*8 = 2. If n is even there are additional solutions given by both neighbors > b/d of the fractions a/c in Fn where c = n/2, namely 2a/n, b/d, > where, again, one must flip the sign as above for a larger neighbor. > E.g. for 3/4 in F8 = .. 5/7, 3/4, 4/5 .. we find solutions 6/8, 5/7 > and 6/8, ~4/5, where ~4/5 = 6/8(-)4/5 = (6-4)/(8-5) = 2/3. [...] > Seems to me that all the solutions you get this way have two even > terms (or more). Don't you miss all-odd solutions like 5*7 - 3*11 = 2? No, for 5/11 in F11 = .. 4/9, 5/11, 1/2 .. choose neighbor 1/2 with min > denom, double it, flip sign 5/11(-)2/4 = 3/7, just as I said above. To be clear, this operation on a/c, b/d produces the fraction in Fn that precedes a/c by two indices, since the Farey sequence here is a - 2b a - b a b Fn = ... ------ , ----- , - , - ..., c = n, d < c-d c - 2d c - d c d But I wanted to show that it arises from a _neighbor_ (modulo sign). --Bill Dubuque === Subject: Spectral radius of a nonnegative matrix posting-account=z7MvkwoAAABTr6FuOIl2OaQFQ_irgZzA Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) I was hoping that somebody would direct me to the relevant literature or a proof for the following problems. Given a a nonnegative irreducible real matrix A that is structurally symmetric (i.e., Aij > 0 iff Aji > 0), what is the relationship between its spectral radius (perron root) .a6.84(A), and the spectral radius of A's hermitian (symmetric) part (.a6.84(H(A)))? I know that .a6.84(A) = .a6.84(A'), (where A' is the transpose of A), and it would appear that .a6.84(A) Á.86 .a6.84(H(A)), but am not aware of any theorems about this, or whether this will hold with equality under some conditions. Similarly, I am also interested in the relationship between .a6.84(A) and .a6.84(D(A)), where D(A) = (A.A')^1/2 (square root of Hadamard product of A and A'). Mhasoba === Subject: exotic spheres, Riemannian manifolds Exotic spheres are all sooth differentiable manifolds, but without a canonical metric, I believe; as a understand it, a metric is extra structure added to a differentiable manifold. Ordinary spheres are obviously Riemannian manifolds (they have a Riemannian metric). Can some of these exotic spheres be given a Riemannian metric, thus turning them into Riemannian manifolds, and subsequently embedded (either with or without an explicit parametrization being known or given) into some high-dimensional euclidean space? What about the infinitely many differential structures on R^4 ... Can they be realized as Riemannian manifolds? David Bernier === Subject: Re: exotic spheres, Riemannian manifolds posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) > Exotic spheres are all sooth differentiable manifolds, but without a > canonical > metric, I believe; as a understand it, a metric is extra structure added to > a differentiable manifold. Ordinary spheres are obviously Riemannian manifolds (they have a Riemannian > metric). Can some of these exotic spheres be given a Riemannian metric, > thus turning them into Riemannian manifolds, and subsequently embedded > (either with or without an explicit parametrization being known or > given) into some high-dimensional euclidean space? Every differential manifold can be given (many) Riemannian structures. Therefore all exotic spheres are Riemannian manifolds (in many ways) > What about the infinitely many differential structures on R^4 ... > Can they be realized as Riemannian manifolds? Again, an R^4 endowed with an exotic differential structure has many Riemannian structures. The `exoticness' of exotic differential structures lies rather more deeply than in possible metric structures. -- m === Subject: Re: exotic spheres, Riemannian manifolds, John Milnor > Exotic spheres are all sooth differentiable manifolds, but without a > canonical > metric, I believe; as a understand it, a metric is extra structure added to > a differentiable manifold. > Ordinary spheres are obviously Riemannian manifolds (they have a Riemannian > metric). Can some of these exotic spheres be given a Riemannian metric, > thus turning them into Riemannian manifolds, and subsequently embedded > (either with or without an explicit parametrization being known or > given) into some high-dimensional euclidean space? Every differential manifold can be given (many) > Riemannian structures. Therefore all exotic spheres > are Riemannian manifolds (in many ways) > What about the infinitely many differential structures on R^4 ... > Can they be realized as Riemannian manifolds? Again, an R^4 endowed with an exotic differential > structure has many Riemannian structures. The `exoticness' of exotic differential structures > lies rather more deeply than in possible metric > structures. -- m The American mathematician John Milnor discovered that the ordinary 7-sphere S^7, the boundary of an 8D ball in ordinary Euclidean space, can be enriched to a differentiable manifold in 28 different ways. The underlying structure of S^7 as a metric space is all the same, but there are 28 essentially different ways in which S^7 can be provided with an atlas, i.e. a system of locally 7D Euclidean patches covering the entire S^7 with overlaps and with differentiable coordinate transformations in the overlaps. In 7D things are much more intricate than geographical longitude and latitude for mankind's physical home sphere S^2. Begin exploration on-line for instance at http://en.wikipedia.org/wiki/John_Milnor . Ciao: Johan E. Mebius === Subject: Advantages of directive newsgroups posting-account=ee2apQoAAABJNMlLTFasCJw9Nfo9FmYk Gecko/20020924 AOL/7.0,gzip(gfe),gzip(gfe) HTTP/1.1 cache-ntc-ab08.proxy.aol.com[CFC87448] (Traffic-Server/6.1.5 [uScM]) Recently you have noticed that I have created new newsgroups as part of a company to solve difficult issues related to the newsgroups: http://mathforum.org/kb/thread.jspa?threadID=1763061&tstart=0 There are many advantages to the individuals as well as the public to what I call a directive newsgroup. These are as follows: 1- We will have true identity of the posters clarified; this will help more responsibility on the side of posters (Particularly when a fee system is initiated). 2- Members must sign a contract of regulations regarding codes of conduct in behavioral languages and other issues in their postings. 3- Members are obliged to restrict themselves to the subject matter of their membership, thus in a newsgroup Sci.math, mainly mathematical issues will be discussed. 4-Currently the site is not open to the public, i.e. only I believe many university professors may not like to go public, this system will give them a chance to get involved in a closed environment (this is very essential for the quality of educational system) contract and we need to open up another site for this purpose (like smoking or non smoking sections! Sci.math (public) or Sci.math (non public)). 6-This is an ideal system for teachers to be involved. We will then have separate sites for elementary, middle school math, as well as college math, subject to memberÍs growth (also other subjects). 7- In this case teachers can concentrate on problem solving of their textbooks and all the other issues of this particular newsgroup will focus on issues like elementary school math. 8-The company is global, and many from different countries can join and later on create their own franchises of the company in many languages when the company created its name and logo trademark in the future. 9-At this time the ideas are in their infancy and the sites of Basti Newsgroups are experimental on Google site. 10- Company will hire professionals to direct each section of its sites for the benefit of the memberÍs interests in variety of dimensions. 11-If you would like many other branches of science and education, please let me know. 12-Teachers may have a separate one contract for their school, enrolling their students in a separate site. These are many advantages of directive newsgroups that current system is lacking. You are welcome to join in the newsgroups. Dr.Mehran Basti CEO, Basti Newsgroups === Subject: Re: Advantages of directive newsgroups --------------------------------------------------------------------- Recently you have noticed that I have created new newsgroups as part > of a company to solve difficult issues related to the newsgroups: There are many advantages to the individuals as well as the public to > what I call a directive newsgroup. These are as follows: 10- Company will hire professionals to direct each section of its > sites for the benefit of the memberÍs interests in variety of > dimensions. > This is self contradictory. What's wrong with your stuff that you don't know how to spell member's? It's spelled with an apostrophy and not memberRs which you spelt with an R or a ctrl_R. Why can't you use the apostrophy key instead like you should do? It is not in my nor the public's interest that you don't or won't. ---- === Subject: Re: Advantages of directive newsgroups > Recently you have noticed that I have created new > newsgroups as part > of a company to solve difficult issues related to > the newsgroups: > > There are many advantages to the individuals as > well as the public to > what I call a directive newsgroup. These are as > follows: > > 10- Company will hire professionals to direct each > section of its > sites for the benefit of the memberÍs interests in > variety of > dimensions. > This is self contradictory. What's wrong with your > stuff that you don't > know how to spell member's? It's spelled with an > apostrophy and not > memberRs which you spelt with an R or a ctrl_R. > Why can't you use the > apostrophy key instead like you should do? It is not > in my nor the > public's interest that you don't or won't. ---- This is not my mistake, but within character encoding and computer languages issues of Math Forum. Once asked them they said I need to write like &rsque before the word to write it correctly. They provided a PDF code guidelines and sometime it is not working or I do not apply, or it was posted in the Google site with word processing. Sorry for inconvenience. M.Basti === Subject: Re: Advantages of directive newsgroups <5174092.1214557386088.JavaMail.jakarta@nitrogen.mathforum.org> This is self contradictory. What's wrong with your stuff that you > don't know how to spell member's? It's spelled with an apostrophy > and not memberRs which you spelt with an R or a ctrl_R. > Why can't you use the > apostrophy key instead like you should do? It is not > in my nor the > public's interest that you don't or won't. This is not my mistake, but within character encoding and computer > languages issues of Math Forum. Once asked them they said I need to write like &rsque before the word to > write it correctly. They provided a PDF code guidelines and sometime it > is not working or I do not apply, or it was posted in the Google site > with word processing. > Is it math forum or your soft ware? To test this, email me a post with the characters single and double quotes, ie ' and . === Subject: Re: Advantages of directive newsgroups Actually I posted my first one on Math forum and did not appear on Google (for many hours). I thought it might not be transferred to Google site. Thus I posted again at Google site. The one at Google site is correct, both ones at Math Forum are wrong printings. The same happened at Sci. math. symbolic site. I do not like to use apostrophes on Math Forum site! By the way welcome to join the newsgroup. M.Basti === Subject: Re: Advantages of directive newsgroups boundary=----=_Part_508_3335108.1214557970759 --------------------------------------------------------------------- For those using Math Forum please see the following PDF codes for computer languages writing. Dr.M.Basti --------------------------------------------------------------------- === Subject: Re: Advantages of directive newsgroups > ... memberÍs ... It's spelled with an apostrophy and not > memberRs which you spelt with an R or a ctrl_R. Why can't you use > the apostrophy key instead like you should do? It is not in my nor the > public's interest that you don't or won't. Hi william, As allready mentioned, it is your newsreader that corrupts the message. I clearly see an apostrophy looking glyph. But your newsreader is unable to read quoted-printable and '=92' is an apostrophy, although an unusual one. PS : the =92 apostrophy is so unusual that I had to change the charset to be able to *send* it. BTW your newsreader packs your message in some unnecessary mime section. -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: Advantages of directive newsgroups <20080627005435.R1283@agora.rdrop.com>mn.da917d86cca986ce.22155@free.fr> Content-ID: <20080627032603.W11436@agora.rdrop.com> --------------------------------------------------------------------- Content-ID: <20080627032603.Y11436@agora.rdrop.com ... member.92s > ... It's spelled with an apostrophe and not > memberRs which you spelt with an R or a ctrl_R. Why can't you use > the apostrophe key instead like you should do? It is not in my nor the > public's interest that you don't or won't. Hi William, > As allready mentioned, it is your newsreader that corrupts the message. > I clearly see an apostrophy looking glyph. > But your newsreader is unable to read quoted-printable and '=92' is > an apostrophy, although an unusual one. > PS : the =92 apostrophy is so unusual that I had to change the > charset to be able to *send* it. > Don't blame me. It's also the lament of others with more modern computers. That ever since ascii was declared 'a not valued added proddick' computers have been plagued by a hoard of formattings and cursed like the Tower of Babel for their pride of attempting to speak the incomprehensible all. > BTW your newsreader packs your message in some unnecessary mime > section. > That's to deal with jerks who like to send 8-bit text. Wish I could turn if off or make to disappear everything but printable characters. === Subject: Re: Advantages of directive newsgroups posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) Recently you have noticed that I have created new newsgroups as part > of a company to solve difficult issues related to the newsgroups: http://mathforum.org/kb/thread.jspa?threadID=1763061&tstart=0 There are many advantages to the individuals as well as the public to > what I call a directive newsgroup. These are as follows: 1- We will have true identity of the posters clarified; this will help > more responsibility on the side of posters (Particularly when a fee > system is initiated). 2- Members must sign a contract of regulations regarding codes of > conduct in behavioral languages and other issues in their postings. 3- Members are obliged to restrict themselves to the subject matter of > their membership, thus in a newsgroup Sci.math, mainly mathematical > issues will be discussed. æ æ æ 4-Currently the site is not open to the public, i.e. only æ æ æ I believe many university professors may not like to go public, > this system will give them a chance to get involved in a closed > environment (this is very essential for the quality of educational > system) contract and we need to open up another site for this purpose (like > smoking or non smoking sections! Sci.math (public) or Sci.math (non > public)). 6-This is an ideal system for teachers to be involved. We will then > have separate sites for elementary, middle school math, as well as > college math, subject to memberÍs growth (also other subjects). 7- In this case teachers can concentrate on problem solving of their > textbooks and all the other issues of this particular newsgroup will > focus on issues like elementary school math. 8-The company is global, and many from different countries can join > and later on create their own franchises of the company in many > languages when the company created its name and logo trademark in the > future. 9-At this time the ideas are in their infancy and the sites of Basti > Newsgroups are experimental on Google site. 10- Company will hire professionals to direct each section of its > sites for the benefit of the memberÍs interests in variety of > dimensions. 11-If you would like many other branches of science and education, > please let me know. 12-Teachers may have a separate one contract for their school, > enrolling their students in a separate site. These are many advantages of directive newsgroups that current system > is lacking. You are welcome to join in the newsgroups. Dr.Mehran Basti CEO, Basti Newsgroups paying contributors to your nonsense newsgroup? The irony! === Subject: Re: Advantages of directive newsgroups posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) Recently you have noticed that I have created new newsgroups as part > of a company to solve difficult issues related to the newsgroups: http://mathforum.org/kb/thread.jspa?threadID=1763061&tstart=0 There are many advantages to the individuals as well as the public to > what I call a directive newsgroup. These are as follows: 1- We will have true identity of the posters clarified; this will help > more responsibility on the side of posters (Particularly when a fee > system is initiated). Because, as we all know, there is a marked correlation between money and responsability! [...rest of essentially white noise removed...] -- m === Subject: Re: Einstein: Jewish Liar, Conman and Plagerist > A theory of Einstein the irrational plagiarist > Christopher Jon Bjerknes [CapitalEth] The Canberra Times > September 19, 2006 The name Einstein evokes images of a good-humoured > genius, who > revolutionised our concepts of space, time, energy, > mass and motion. > Time named Albert Einstein person of the century. > The language > itself has incorporated Einstein into our common > vocabulary as a > synonym for extraordinary brilliance. Many consider > Einstein to have > been the finest mind in recorded human history. That is the popular image, fostered by textbooks, the > media, and hero > worshiping physicists and historians. However, when > one reads the > scientific literature written by Einstein's > contemporaries, a quite > different picture emerges: one of an irrational > plagiarist, who > manipulated credit for their work. > You Jew haters are so cute when you're taking on Einstein. Sort of like a flea crawling up an elephant's leg with rape on its mind. Tom === Subject: Re: Least squares similarity transformation <649bc$485a0937$82a1e228$3323@news2.tudelft.nl> <20080619.020548@whim.org> <20080619.165018@whim.org> <20080620.214215@whim.org> posting-account=xM691AkAAACRhg2rzU0Kd6hZjdls4krD rv:1.8.1.14) Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > Value Decomposition to Polar Decomposition to the page cited earlier: >http://www.whim.org/nebula/math/conformalregress.html>. >Some time ago I ran into a similar problem (in > 2 dimensions) which >had the additional requirement that the transformation be a proper >rotation, with determinant = +1. The best I could suggest if the SVD- >based approach gave a matrix whose determinant was -1 was to minimize >the error by a Jacobi-like sequence of single-plane rotations. What >would you do in this situation? In the least > squares conformal map is computed. I will refer to this as [CR]. If the M computed in [CR] has negative determinant, then we must > reflect the result of P M + R across a hyperplane. The hyperplane > that would least affect the squares minimized in [CR], would be the > hyperplane which has the minimum square distance to the { Q_j }. > The unit normal to this hyperplane is the unit W that minimizes n > --- _ 2 > > < Q - Q , W > [1] > --- j > j=1 Thus we need to find the W so that n > --- _ _ > 0 = > < Q - Q , W > Q - Q , dW > [2] > --- j j > j=1 for all dW under the condition that |W| = 1, that is n > --- > 0 = > < W , dW > [3] > --- > j=1 Using orthogonality, we see that we need to have n > --- _ _ > k W = > < Q - Q , W > ( Q - Q ) [4] > --- j j > j=1 That is, W is a unit eigenvector of n > --- _ T _ > N = > ( Q - Q ) ( Q - Q ) [5] > --- j j > j=1 Note that [1] equals T > W N W [6] and since W is minimizes [1], W must be the eigenvector of N whose > eigenvalue has the smallest absolute value. Thus, the reflection > matrix is T > I - 2 W W [7] Since we can still rescale, define T > M' = r' M ( I - 2 W W ) [8] from which it follows that T 2 T 2 > M' M' = r' M M = ( r' r ) I [9] Similar to [11] in [CR], varying the scale parameter alone yields > a critical point when n > --- _ _ _ > 0 = > < ( P - P ) M' - ( Q - Q ) , ( P - P ) M' > [10] > --- j j j > j=1 Therefore, we can compute r' with n > --- _ 2 > r' > | ( P - P ) M | > --- j > j=1 n > --- _ _ T > = > < Q - Q , ( P - P ) M ( I - 2 W W ) > [11] > --- j j > j=1 Then, using [11] in [CR] and [11] above, we get n > 1 - r' --- _ 2 > ------ > | ( P - P ) M | > 2 --- j > j=1 n > --- _ _ > = > < Q - Q , W > ( P - P ) M , W > [12] > --- j j > j=1 and since the translation must still obey [5] from [CR], we need _ _ > R' = Q - P M' [13] Summary > ------- > Compute M from [CR]. Compute n > --- _ T _ > N = > ( Q - Q ) ( Q - Q ) > --- j j > j=1 Let W be the unit eigenvector of N whose eigenvalue has the smallest > absolute value. Then compute r' from n > 1 - r' --- _ 2 > ------ > | ( P - P ) M | > 2 --- j > j=1 n > --- _ _ > = > < Q - Q , W > ( P - P ) M , W --- j j > j=1 Then compute M' as T > M' = r' M ( I - 2 W W ) Finally, calculate R' using _ _ > R' = Q - P M' That should get us close. Rob Johnson posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Where math is certain it is abstract entity. Where it > applies to the > real world it is not exact. > Mitch Raemsch > Do you understand what Einstein meant by that? æDo > you know where he said it? æDo you even know > the exact wording? æ > As far as the laws of mathematics refer to reality, > they are not certain; and as far as they are certain, > they do not refer to reality. ~ Einstein, Geometry > and Experience, 1921; an essay that expands on an address > he delivered to the Prussian Academy of Sciences that > year. > I am sure I am not the only member of this forum > weary of seeing Einstein quotes tossed about > meaninglessly, without source or attribution. æGeometry > and Experience is a centerpiece of Einstein's philosophy > of space and time. He used it to explain how the > continuum can be derived from the experience of three > dimensional geometry, and concludes, My only aim today > has been to show that the human faculty of visualisation > is by no means bound to capitulate to non-Euclidean > geometry. > That is to say, while the mathematical models of > geometry are certain, the experience of geometry (in > the deep sense of an Einstein, at least) reflects a > certain experience of reality not yet captured in the > model--though Einstein tried to do so until the day > he died. > Tom > I am talking Calculs where you must make infinite calculations to > arive at a real world problem's exact derivative. > Math as abstract entity is exact as in polynomial functions > derivatives. While the math for real world derivatives is only > approximation Tom. > Mitch Raemsch Einstein feared that real world derivities would be excluded! > -- > 'foolsrushin.'- Hide quoted text - - Show quoted text - They are in a sense. They are never completly exact and that is the point. Mitch Raemsch === Subject: OT Humor: Eris, Smoking Banana Peels & New Murphy's Laws Last week while I was thinking about the analytic continuation of the Ackermann function in the complex plane, I fell on my head after stepping on a discarded banana peel which I previously had tried to smoke. The fall resulted in a sudden and dangerous revelation, if you will, which caused the spontaneous birth of dozens of new scientific & non-scientific Murphy's Laws. I suspect the originator of the revelation was Goddess Eris, but I cannot be sure because I was unconscious for several minutes. I don't know if they are correct. I don't even know if they are funny, so some may find them objectionable or even offending. In either case, don't blame me. Blame the banana peel and Goddess Eris. They made me do it. Sorted according to category for your reading convenience and efficient apoplexy. Brace yourselves. The news are bad (as they usually are with Murphy's Laws) ;o) http://ioannis.virtualcomposer2000.com/writing/NewMurphysLaws.html Further laws to be added as providence allows... -- I.N. Galidakis === Subject: Re: Smooth submanifold > On 25 Jun., 20:46, The World Wide Wade On 24 Jun., 21:17, The World Wide Wade If M were a smooth submainfold of R^2, then locally near 0 it would be > the zero set of a smooth function f : U -> R, with U open in R^2, 0 in > U, grad(f)(0) not zero. > > What do such manifolds have to do with zero sets? > > Every smooth 1-dimensional submanifold of R^2 is locally the zero set > of a smooth function as above. Why do you fund this surpsising? > > This sounds > interesting because it connects them with schemes which are a kind of > zero sets of special (polynomial) functions. I have learned how such a > manifold can be treated as a sheaf. But where do 'zero sets' come in > now? > > By the implicit function theorem, M near 0 > would then be the graph of a smooth function g defined in an interval > containing 0. But that can't be. > > My definition of M being a smooth submanifold of R^2 is the following: > For every point x in R^2 exists a chart h:U-->V with x in U and V a > subset of R^2 such that h(Mcap U)=Vcap(R^1x0) and this can be chosen > smooth. As described i have found _one_ atlas with the first property > but to proof that M is not smooth I have to check that for all > atlases. > > S. > > Then g = h^(-1) is a diffeomorphism of V that takes an interval I of 0 > in R x {0} onto a neighborhood of 0 in M. This means we have a smooth > function f : I -> R, with f(0) = 0, such that the map x -> (f(x), > |f(x)|) has nonzero derivative at 0, -><-. How does this be in conflict with the definition? I mean, why should g > homeomorphism from V to U and U is a neighbourhood of 0. Only the > corrdinate-changes of such charts have to be diffeomorphisms! Here, we > only have got a simgle chart h. How can I proceed? > S. If R^2 is endowed with its usual smooth structure, aren't all charts in this structure diffeomorphisms? (Take any such chart and consider the smoothness condition on the transition maps generated by this chart and the identity map on the whole space.) === Subject: Re: Comprehensive Solution Manual for Textbooks posting-account=rLOz6QoAAAAmvEIbrGZd27QhtZqovu5R rv:1.8.1.14) Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) === Subject: Re: Characterization of involutory matrices posting-account=VwzmsAoAAAAZ1SrsELdA_PfudH0-se1O Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) First of all let me thank everyone who has answered. One more question: suppose now that A is also orthonormal AND symmetric. According to a Lemma in Gantmacher we have A=exp(i S) with S symmetric and real. Since A is involutory we can write exp(i S)*exp(i S) = I = exp(0), where 0 is the matrix whose elements are all zeroes. Furthermore, since (i S) and (i S) obviously commute, exp(i 2S) = exp(0). So my question is: is it possible to find a general expression for S? Note that this is equivalent to the scalar equations cos(2 S) = I sin(2 S) = 0 It seems that it will have an infinite number of solutions, like in the scalar case exp(i theta)=1. It also seems that S is independent of A and this looks a bit surprising (at least to me!). On Jun 25, 10:37æpm, Robert Israel > Is there a complete characterization of allinvolutorymatrices > (A*A=I), i.e. the square roots of > the identity? I am especially interested in the real case. The Jordan canonical form of such a matrix is diagonal with diagonal > entries 1 and -1. æSo: A is any matrix of the form S D S^(-1) where D is > diagonal with diagonal entries 1 and -1 and S is invertible. > -- > Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada === Subject: Re: Characterization of involutory matrices > First of all let me thank everyone who has answered. One more > question: suppose now that A is also orthonormal AND symmetric. According to > a Lemma in Gantmacher we have A=3Dexp(i S) with S symmetric and real. Since A is involutory we can write exp(i S)*exp(i S) =3D I =3D exp(0), where 0 is the matrix whose elements are all zeroes. > Furthermore, since (i S) and (i S) obviously commute, exp(i 2S) =3D exp(0). So my question is: is it possible to find a general expression > for S? Note that this is equivalent to the scalar equations cos(2 S) =3D I > sin(2 S) =3D 0 It seems that it will have an infinite number of solutions, > like in the scalar case exp(i theta)=3D1. > It also seems that S is independent of A and this looks a bit > surprising (at least to me!). The eigenvalues of S can be any integer multiples of pi. As I mentioned before, there are two complementary subspaces U and V of R^n, with A u = u for u in U, A v = -v for v in V. In this case (since A is normal) U and V are orthogonal. The eigenvectors of S for even multiples of pi are in U, and the eigenvectors for odd multiples of pi are in V. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: formula for nth power of 3x3 matrix?? (akin to this one for 2x2?) posting-account=x9DlGAoAAAAvZTRvRYG8JjmPJCeyCza7 SV1),gzip(gfe),gzip(gfe) I'd be grateful if someone could tell me the formula for the nth power of a 3x3 matrix, which is akin to this one for the nth power of a 2x2 matrix: A^n = a^n*(A-bI)/(a-b) + b^n*(A-aI)/(b-a) where a, b are the eigenvalues and I is the identity matrix. This is assuming of course that a formula of this type exists for nxn matrices when n>2 :-) Michael === Subject: Re: formula for nth power of 3x3 matrix?? (akin to this one for 2x2?) > I'd be grateful if someone could tell me the formula for the nth power > of a 3x3 matrix, which is akin to this one for the nth power of a 2x2 > matrix: A^n = a^n*(A-bI)/(a-b) + b^n*(A-aI)/(b-a) where a, b are the eigenvalues and I is the identity matrix. This is assuming of course that a formula of this type exists for nxn > matrices when n>2 :-) Assume the m x m matrix A has m distinct eigenvalues lambda_j, j=1..m. Then for any polynomial p, p(A) = sum_{j=1}^m p(lambda_j) g_j(A) where g_j(x) = product_{k <> j} (x - lambda_k)/(lambda_j - lambda_k) is the polynomial of degree m-1 such that g_j(lambda_i) = 1 for i=j, 0 for i <> j. For example, if A is 3x3 with eigenvalues a,b,c, A^n = a^n (A-bI)(A-cI)/((a-b)(a-c)) + b^n (A-aI)(A-cI)/((b-a)(b-c)) + c^n (A-aI)(A-bI)/((c-a)(c-b)) -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: First and Second Fundamental Forms for/of an Implicit Surface <_UN8k.5076$xb2.977@newsfe12.phx> posting-account=a93YEwoAAAClHm9Euy--V39SJpP16NI8 Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > Given a parametric surface P(u,v) = (x(u,v), y(u,v), z(u,v)), > the first fundamental form is > æ {{Dot(Pu,Pu), Dot(Pu,Pv)}, {Dot(Pv,Pu), Dot(Pv,Pv)}} > where Pu and Pv are the partial derivatives with respect to > the parameters u and v. When the surface is the graph of a function, a parameterization > is P(u,v) = (u, v, z(u,v)), in which case Pu = (1, 0, zu) and > Pv = (0, 1, zv). æThe first fundamental form becomes > æ {{1+zu^2, zu*zv}, {zu*zv, 1+zv^2}} Differential Geometry), and everything finally seems to have come together (phew). One last trivial question: I'm doing this in order to compute the principal curvatures (k1, k2) and principal directions (T1, T2) at a any point on the implicit surface. I'm aware that: k1, and k2 are the eigenvalues of the 2 x 2 matrix II/I , where I is the matrix of the first fundamental form, and II the matrix of the second fundamental form; while T1, and T2 are corresponding eigenvectors of (II/I). Obviously, the T1, and T2 are 2 X 1 vectors embedded(?) in a 3D space. What then is the third component of these vectors? Is it 1? === Subject: Re: First and Second Fundamental Forms for/of an Implicit Surface > I'm doing this in order to compute the principal curvatures (k1, k2) > and principal directions (T1, T2) at a any point on the implicit > surface. I'm aware that: k1, and k2 are the eigenvalues of the 2 x 2 > matrix II/I , where I is the matrix of the first fundamental form, and > II the matrix of the second fundamental form; while T1, and T2 are > corresponding eigenvectors of (II/I). Obviously, the T1, and T2 are 2 > X 1 vectors embedded(?) in a 3D space. What then is the third > component of these vectors? Is it 1? You are thinking of the T1 and T2 as 2-tuples in the parameter space. You have to lift them to the surface. See http://www.geometrictools.com/Documentation/PrincipalCurvature.pdf I also have code at my web site to compute the various quantities for implicitly defined surfaces. -- Dave Eberly http://www.geometrictools.com === === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > An irrational interval in the above sense (required to guarantee the > existence of more irrational numbers than rational numbers) is an > interval of only irrational numbers. Definition: An irrational interval (that contains more than one point) > is an interval on the real line that contains a set of irrational > numbers x_1, x_2, ... such that between any two numbers x_1 < x_2 of > the set there is no rational number q with x_1 < q < x_2. Not necessary to say that an irrational interval with more than one > point does not exist on the real line. Nevertheless it must exist, if > there were more irrational numbers than rational numbers. Call this P. You can use exactly the same argument to prove that there are more rationals than irrationals, which we'll call ~P. Since both P and ~P are true (according to you), the argument upon which they are based is flawed. So your proof is meaningless. > There is a simple argument. If here are aleph_0 rationals then we can > take the dense ordering DO and the sequential well-ordering SO. If > aleph_0 is really existing, then we can transform the dense ordering > into the sequential well-ordering. We start with DO and pick rational > numbers according to their index in the well ordering. So we get > [...] > In the last step we obtain the sequential ordering. But in between > there must have been a step which destroys the dense ordering. This is > impossible. Therefre aleph_0 steps do not exist. Therefore aleph_0 > rationals do not exist. There is no actual infijnity. And yet there exist several well-orderings of the rationals, such as the one based on a simple Farey sequence: http://david.tribble.com/text/sequence1.html Perhaps you could explain how this ordering destroys the denseness of the rationals? (Your phrase dense ordering makes no sense.) > The most simple argument against actual infinity is the following: If > we represent the natural numbers in unary, then the list of all aturas > looks as follows: > o > oo > ooo > ... > It is obvious by the symmetry of the triangle that, when going down, > we experience that everything that happens in the first column also > happens in a line. In particular if the first column contains more o's > than any natural number, then also a line must contain more o's than > any natural number. But then this line is no longer representing a > natural number. You are deducing that because the first column is infinite, there must exist a last row that is also infinite. This is clearly a contradiction in terms; a last row cannot exist if the first column is infinite, and likewise the first row cannot be infinite if a last row exists. So your argument is meaningless. 7,000 posts and you're still repeating the same flawed logic you started with at the beginning. Have you learned nothing in all this time? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=yKimjgoAAACk5WwPVD4l9HmbpoR6Hmy4 en),gzip(gfe),gzip(gfe) > An irrational interval in the above sense (required to guarantee the > existence of more irrational numbers than rational numbers) is an > interval of only irrational numbers. > Definition: An irrational interval (that contains more than one point) > is an interval on the real line that contains a set of irrational > numbers x 1, x 2, ... such that between any two numbers x 1 < x 2 of > the set there is no rational number q with x 1 < q < x 2. > Not necessary to say that an irrational interval with more than one > point does not exist on the real line. Nevertheless it must exist, if > there were more irrational numbers than rational numbers. Call this P. æYou can use exactly the same argument to prove > that there are more rationals than irrationals, which we'll call ~P. > Since both P and ~P are true (according to you), the argument > upon which they are based is flawed. æSo your proof is meaningless. Or ZFC is inconsistent. Mission accomplished! === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=U44YcwkAAAAbGXB70Qr7gA3kornmKE4i Gecko/20080325 Ubuntu/7.10 (gutsy) Firefox/2.0.0.13,gzip(gfe),gzip(gfe) It is obvious by the symmetry of the triangle that, when going down, > we experience that everything that happens in the first column also > happens in a line. In particular if the first column contains more o's > than any natural number, then also a line must contain more o's than > any natural number. But then this line is no longer representing a > natural number. You are deducing that because the first column is infinite, there > must exist a last row that is also infinite. This is clearly a > contradiction in terms; ... Indeed. But if you use the i-word, be sure you will never convince a crank of anything. I am fairly sure that WM's notion of completed infinity is pretty much: it's the end of something that never ends. He believes (if you think he's stupid), or claims to believe (if you think he's dishonest) that omega means the number at the end of the naturals. In this case, clearly infinite rows means the last one is number omega, and since a triangle has three sides, this omegath row must indeed be of the same length as its position in the um triangle, viz omega. Therefore, if you reach omega going down you must reach omega to the right. But if the first column *never ends*, then it _is_ hard to understand what his argument is supposed to be. > 7,000 posts and you're still repeating the same flawed logic you > started with at the beginning. Have you learned nothing in all this > time? Plainly not. Brian Chandler http://imaginatorium.org === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > An irrational interval in the above sense (required to guarantee the > existence of more irrational numbers than rational numbers) is an > interval of only irrational numbers. > > Definition: An irrational interval (that contains more than one point) > is an interval on the real line that contains a set of irrational > numbers x_1, x_2, ... such that between any two numbers x_1 < x_2 of > the set there is no rational number q with x_1 < q < x_2. > > Not necessary to say that an irrational interval with more than one > point does not exist on the real line. Nevertheless it must exist, if > there were more irrational numbers than rational numbers. Call this P. You can use exactly the same argument to prove > that there are more rationals than irrationals, which we'll call ~P. > Since both P and ~P are true (according to you), the argument > upon which they are based is flawed. So your proof is meaningless. > There is a simple argument. If here are aleph_0 rationals then we can > take the dense ordering DO and the sequential well-ordering SO. If > aleph_0 is really existing, then we can transform the dense ordering > into the sequential well-ordering. We start with DO and pick rational > numbers according to their index in the well ordering. So we get > [...] > In the last step we obtain the sequential ordering. But in between > there must have been a step which destroys the dense ordering. This is > impossible. Therefre aleph_0 steps do not exist. Therefore aleph_0 > rationals do not exist. There is no actual infijnity. And yet there exist several well-orderings of the rationals, such > as the one based on a simple Farey sequence: > http://david.tribble.com/text/sequence1.html Perhaps you could explain how this ordering destroys the > denseness of the rationals? (Your phrase dense ordering > makes no sense.) > The most simple argument against actual infinity is the following: If > we represent the natural numbers in unary, then the list of all aturas > looks as follows: > o > oo > ooo > ... > It is obvious by the symmetry of the triangle that, when going down, > we experience that everything that happens in the first column also > happens in a line. In particular if the first column contains more o's > than any natural number, then also a line must contain more o's than > any natural number. But then this line is no longer representing a > natural number. You are deducing that because the first column is infinite, there > must exist a last row that is also infinite. This is clearly a > contradiction in terms; a last row cannot exist if the first column > is infinite, and likewise the first row cannot be infinite if a last > row exists. So your argument is meaningless. Shouldn't that be therefore a first COLUMN cannot be infinite if a last row exists? > 7,000 posts and you're still repeating the same flawed logic you > started with at the beginning. Have you learned nothing in all this > time? WM thinks he knows it all and has nothing more to learn. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Nntp-Posting-Host: hera.cwi.nl > > If you consider the complete diagonal of a Cantor list, then there > > must be a last line. So what is the digit and at what position is it > > in the last line? > > Why must there be a last line? > > In order to be sure that no line can be following which may contain > the diagonal number. Eh? According to the definitions there is no last line, and according to the axiom of infinity such a list of lines does exist, so there is no last line. So, what do you mean? That the diagonal is defined using a finite initital segment of the list of lines? > > It seems so but that is only for finite sets. Otherwise: If you > > consider the complete diagonal of a Cantor list, then there must be a > > last line. So what is the digit and at what position is it in the last > > line? > > Why do you keep on thinking that the definition of the Cantor diagonal > is a step by step procedure? In what way does does the definition of > the n-th digit of the diagonal depend on the definitions of the k-th > digits of the diagonal with k < n? > > But when considering a list or the indexes of a number, the n-th digit > is not independent of the n-1st digit. It is. > You cannot know whether a given > large number is a natural (belongs to the sequence of the naturals) > unless you verified it proving that all digits are really digits. Can > you see on the first glance whether this number is a natural? This is completely irrelevant. Cantor states that given a list that list is incomplete. It is in good faith that you indeed do give a list. If you do not give a list the proof will break down. What does that matter? The condition was that you gave a list. If I have a method that when I ask you to give three naturals the result will be a prime, and you give me three naturals and the result is not a prime, that does not mean that my method is wrong, it does mean only that what you gave me is not what I asked for. > And it is a very small number, less than most numbers of Cantor's > list. Yes, so what? So you put something in the list that is not a number, so you fail the requirement for the proof, and so the proof fails in this case. Or do you think that a proof also should imply the verification that what you give is also what is required? That would be beyond nonsense, that would be madness. Consider the following theorem: A natural number divisible by 2 ends with one of the digits 0, 2, 4, 6, 8. Using this consideration that is a false theorem because it would say that > 4627403895720394857329463527416235417263451726345187263451827634512635462316 5 5312563236857333333333333333333333333388888882999901231231929292982937438346 3 4737423428342374623746283746273642736428374628374628374628374628734628734628 3 7428734628374628734623645153415312763618723617315263518273618888888888888888 8 888888882736128363521333l227635162351263517631534153172316352176366666666666 6 6666666666666666644444444444517263517263517623517623517635172635176351762351 7 6351726351273516351763451763512. is divisible by 2. Or do you really think that the theorem is wrong? > > What natural enumerates the last step? > > > > You mean in Cantor's list? I am not sure whether there is a last line. Why do so many people call it Cantor's list? It is not a list Cantor makes, but a list that is given to Cantor. > That is not a step by step procedure. > > Have you found out, without going step by step, whether the number > above is a natural? Yes. :.s/[0-9]//g. But that is irrelevant. > > To be of different definition is necessary for distinct items, but > > obviously it is not sufficient, because there are also different > > definitions for same things. > > Again failing to give a definition for what it means to be distinct. > > But you see that different definition is not suffcient? Yes, also I do not see that it is necessary. Failing a proper definition, I have no idea. Anyhow, your first definition of different definition was clearly incomplete. > That is indeed true with the roots of a 100 degree polynomial for which > the individual roots can not be determined, but that can be proven to > have no multiple zeros. So in what way do those roots not exist? Or > do they exist? (Determining whether a 100 degree polynomial has multiple > roots is indeed easy, although you do not know the actual roots.) > > I agreed already that the roots of your recently given polynomial > exist, at least as ideas, i.e., like most existing real numbers. Did > we have a difference there? Well, what you call ideas, mathematicians call numbers. But apparently you think that the word numbers have some absolute meaning. What is happening is that we apparently (according to you now) have a single definition that defines a host of numbers, namely those 100 roots. And so apparently some definitions may introduce a multitude of numbers. > But as far as I > remember we were discussing about those reals that have no > definition at all. No, we were talking about numbers that come in through a definition that defines a multitude at once without being able to define any one individually. The roots of the polynomial above are an example. in that case (because the polynomial is well-conditioned) we can approximate the roots, but can not designate the roots (each approximation is easily shown to be a non-root). And so the 100 roots are not individually defined, but nevertheless they exist, by a single definition. You are just trying to make numerical mathematics the mainstream, while it heavily depends on what is currently mainstream mathematics. But for some reason you eschew numerical mathematics. Consider that almost all convergence proofs of methods in numerical mathematics depend on mainstream mathematics, and inherently on set theory. -- 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: A consideration concerning the diagonal argument of G. Cantor > On Jun 25, 10:36 pm, Ross A. Finlayson (R is a relation & Axy((x in field(R) & y > in field(R)) -> exactly one of these three disjuncts obtains: in > R or in R or x=y)) > No, that's technical. Right, that surely disqualfies it. MoeBlee No, I agree that's a technical definition of trichotomy in a field, what I was referring to was the technical context of the terms for which definitions are provided. (It's undecideable in ZF whether two arbitrary sets' cardinals are comparable.) Apparently, those notions are quite incomprehensible in general. Now, that's a joke. These dozens of arguments I generally put forward are for the most part quite simple. (Consider, Goedel's theory of theories, where each theorem establishes the theory as one of inconsistent or incomplete: inconsistent or incomplete, where the only possible true fact about the objects not decided is that one of them is consistent and complete. Well-order the reals and no ordinal can, in an extension of nested intervals, have that it is not less than any limit ordinal (that's a funny one). There is no ordinal in ZF large enough to model ZF. ZF's model is not universal, so any supermodel of ZF includes a model of ZF that contains itself. ZF is inconsistent. Classes are defined by their elements. If any function were a bijection between naturals and reals, the natural/unit equivalency function would be the most natural example. Differentials are infinitesimals, as has been well-known for a long time. Etcetera.) Those transfinite recursion schemata are technical, using quite standard definitions, vis-a-vis sets dense in the real numbers, of the continuum of real numbers, the real real numbers. Plainly, there are applications of real analysis, continuum analysis, for which regular set theory is insufficient, besides irrelevant. (Physicists don't use transfinite cardinals, they just care about applications of real numbers, eg in the polydimensional, and they need foundations that are currently nonstandard.) Pure mathematics are applicable. Not my problem. Ross F. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Nntp-Posting-Host: hera.cwi.nl > No, you also need to show that you can attach them to a rational. If you > can not show that it is possible to do what you do (attaching gaps to > rationals) your premissa is invalid. > > Every rational q is a point on the real line. Why do you believe I > could not attach a point q' to this point q? > > Define: q' is a point on the real line next to q if q' has the > following properties > q' > q > q' is not rational. > For all rationals p > q we have p > q'. Such points q' do not exist. The proof is easy. In general, between any pair of points on the real line there are infinitely many rationals and irrationals. > > As my conclusion upto now is that you use your definition of gap > > where a gap is equivalent to an irrational number my question > > becomes again, *how* do you attach a gap to a rational number? > > There is no irrational at the right hand side of a rational number. > > > > You mean we cannot determine it. > > No. I mean that there is no such irrational. Given a specific rational > r and any irrational i such that i > r, there is between r and i > another irrational i1 that is between r and i, and so there is *no* > irrational directly to the right of r > > that we could determine. Yes. But as we cannot determine nearly every > real number (except a subset of the reals of measure zero) I do not > see that this is a problem. The problem is not that we could not determine it but that it does not exist. There are two possibilities: (1) the distance between the two points is 0, but in that case they are equal and so the irrational would be equal to the rational which would be very strange (2) so the distance between the two points is L for some non-negative real number L. There is a rational smaller than L (take 1/ceiling(1/L)), add that rational to the original rational q to get a new rational q_1 between q and the irrational. > > A gap is defined as a position right to a rational number. It can be > > filled by an irrational number, but that irrational is not the gap. > > There is no position immediately to the right of a rational number. And > so your gaps do not exist. > > I defined them above. Defining something does not mean that that something does exist. If I state: define q the rational equal to sqrt(2) that is a proper definition, but q does not exist. Definitions in itself do not bring something in existence. > As with your current definition gaps do not exist I wonder why you keep > on talking about gaps. > > I gave you another example of definable numbers on the same problem: > > By the bijection q <--> q + e (Euler number) we see that there is > one and only one q mapped on the number q + e. Both sets are equal in > this configuration and there is no q that is not in the bijection. Right. > Now remove these two sets from the real line. What remains? For > instance all numbers of the form q + pi. There are no longer > rationals between them (because we have removed them. But re-introducing > step by step (or simultaneously) the pairs of numbers of the form q and > q' + e does not improve the situation, because none of these steps leads to > a reduction of irrational numbers that have no rational between them. Doing it step by step indeed does not do so. Doing them all at once *does* do so. > Every rational q divides an irrational interval into two subintervals, > every irrational q + e joins at least one irrational interval, And this is already wrong. In each step, introducing a q indeed does increase the number of subintervals, but introducing q+e does not necessarily decrease the number of intervals. It will not if it falls in the middle of such an interval. But drawing conclusions from a step by step procedure about a simultaneous procedure is not valid. How many times do you think I already stated that? > Conclusion: Irrational sets, if they exist, cannot be handled. But they do only exist as singletons, and they can be handled pretty well, thank you. > > Yes, perhaps, but the problem remains that that infinitely many steps > > can *not* be done. > > > > After having done them we have an empty DO. > > Perhaps, after having performed the impossible. There is quite a lot > possible if you perform the impossible. > > If you accept, for instance, the finished infinity. But the finished infinity is not something that is obtained in a step by step procedure. It is an axiom: there is an inductive set. Why do you keep stating that it must be a step by step procedure when it is not? > > I need not really perform them. I use Cantor's enumeration of the > > linear ordering of the rationals and define that the n-th step removes > > the n-th rational from the dense ordering. > > Yes, and so you never come to an end. > > The end is given by finishing the infinite linear order. Where you will not get. You can not obtain an infinite set by inserting elements one by one. > > > Therefore also all sets of the form DO{q_1, q_2, q_ 3, ... , > > > q_= n} exist. > > > > Yes, as I said, they do exist. > > > > Then all of them can be enumerated and can be done in zero time, > > defined like Cantor's diagonal. > > Yes and no. Each step depends on the preceding step, that is false with > the diagonal. > > Wrong. None of my steps depends on the preceding step. All are > determined by Cantor's well-ordering of the rationals. For any n I can > assume that n steps have been performed (*every* configuration exists) > and go to the next configuration, just as we go from n to n+1 in > Cantor's list. In the list you need not determine the n-th element to determine the n+1-st element. That is a basic definition of a list, it is a bijection between N and the set of elements of the list, say f, such that when you feed a natural n to f it returns with f(n), i.e. the n-th element of the list. Why should I need to invoke f(n-1) in order to get f(n)? > > What is the last element q? For what n does it have the form > > q_1, q_2, q_3, ..., q_n DO{q_1, q_2, q_3, ..., q_n}? > > Your proof is valid for such set *with* some q_n. > > > > The proof is valid for Cantor's well-ordering of all rational numbers. > > Or Cantor's diagonal proof also is only valid for finite lists. > > The proof above is valid for an *infinite* list. Note that the list: > q_1, q_2, q_3, ..., q_n, DO{q_1, q_2, q_3, ..., q_n} > determines an *infinite* list. > > Yes, q_1, q_2, q_3, ..., q_n, DO{q_1, q_2, q_3, ..., q_n} is one > element of my list of all configurations. This element is infinite > itself as are the entries of Cantor's list. That was not my argument. > But > q_1, q_2, q_3, ..., DO{q_1, q_2, q_3, ...} > is not an element of that list. > > q_1, q_2, q_3, ..., DO{q_1, q_2, q_3, ...} is a configuration, > therefore it belongs to the list of all configurations. It can only be an element of the list if does not come at the end, otherwise the list would not be a list. (Note: lists do not have a last element.) So if it is part of the list it should be at a finite place, and other such sets should follow it... > > Induction does also prove that every set of positive even numbers > > contains numbers that are larger than the cardinal number of that set. > > For positive even natural numbers. But, no, induction does not prove > that. It does prove it only for finite sets, not for arbitrary sets. > > Cantor used it not only for finite segments of his series (5) and (6) > but for the whole series. No, for *elements* of the finite segments and *elements* of the whole series. That is something different. And, but you going beyond that. You have the series: {q_1, q_2, q_3, ..., q_n, DO{q_1, q_2, q_3, ..., q_n}} but you conclude something about q_1, q_2, q_3, ..., DO{q_1, q_2, q_3, ...} which is not an element of the series. > > Nevertheless you were reluctant to accept that for the whole set. Why > > do you now agree to Cantor's same argument? > > Because it is not the same argument. You extend the argument in a way > that makes it invalid. > > It is exactly the conclusion, by induction, from any finite initial > segment of his en passant well-ordering of the phi_nu to the complete > infinite set. No, the induction is not about sets, it is about elements of the sets. > > > Rekursionssatz, Dedekind, 1888, Grundlage der Definition durch > > > Rekursion, zu beweisen mittels vollst=E4ndiger Induktion. > > > > And what do you mean with that? > > > > Recursion requires induction. > > I do not see that from the title. I see only Basics about the definition > through recursion and to proofs through induction. Definitions do not > need proof. > > Rekursion needs proof. It is proven by induction. Therefore every > recursion can be traced back to induction. No, recursion does not need proof in an axiom, so induction is not needed in an axiom. Or do you think that axioms need proof? > > Peano had no inductive set when he defined the inductive set by > > induction. > > What Peano axioms are you using? There are quite a few variants floating > around. > > Originally there were 9 I think. But I need only induction: > > 1 is in N. With n also n + 1 is in n. That is an axiom. As it does not need proof, there is no need for induction. > > And that is definition by recursion, not by induction. > > Using recursion is using induction as using a function f(x) is using x. No, induction is a matter of proof. In an axiom you do not need proof, so you do not need induction. > > Induction is establishing P(1) and then conclusion from P(n) to P(n > > +1). Precisely this is done in the axiom of infinity. > > You are again wrong. > > No. Yes. It is a statement, there is no proof, i.e. no induction, required. > > Why do you think that a method for proving something cannot be used > > for defining something? > > Because you do not need to prove something that you define. > > Induction is a method. This method can be used to prove or to define. No! It is a method of proof. > Do you know the origin of complete induction? Dedekind coined it. > For what purpose do you think did he need it? To prove things I would think. But terminology changes over time. So what was the terminology more than 100 years ago is not necessarily the same as the terminology now. And what is the terminology is in German is not necessarily the same as the terminology in English. But: > Basics about the definition > through recursion and to proofs through induction. So he was clearly not writing about definitions through induction. -- 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: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) But drawing conclusions from a step by step procedure about a simultaneous > procedure is not valid. How many times do you think I already stated that? I see. A set that contains n and with n also contains n+1 is not necessarily infinite. It could also be finite in your opinion. It is not sufficient to show that all steps exist. It is only by chance that the axiom of infinity defines an infinite set. Why do you think that the reals are a set of larger cardinality than the naturals then? Cantor's list shows only one step of infinitely many steps required. That poor diagonal number easily could be put into the bijection of list entries with the naturals, for instance by defining that 0 is a natural number and bijecting the diagonal with 0. No, if it is proved that every step implies the next one in a procedure that never stops, then something is proven for an infinite set. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Every rational q divides an irrational interval into two subintervals, > > every irrational q + e joins at least one irrational interval, And this is already wrong. In each step, introducing a q indeed does > increase the number of subintervals, but introducing q+e does not > necessarily decrease the number of intervals. It will not if it falls > in the middle of such an interval. q + e cannot fall into a rational interval, because such intervals do not exist at any time. Therefore introducing q + e will increase the number of points in an irrational interval. This equilibrium (reducing the number of points of an irrational interval by q and increasing them by q + e guarantees that the irrational intervals, iff they existed, will remain to exist. But drawing conclusions from a step by step procedure about a simultaneous > procedure is not valid. How many times do you think I already stated that? This is not a step by step procedure. You can perform that simultaneously. But the finished infinity is not something that is obtained in a step by > step procedure. It is an axiom: there is an inductive set. Why do you > keep stating that it must be a step by step procedure when it is not? Because the axiom uses induction to define this set. If n exists, then also n+1 exists. > > I need not really perform them. I use Cantor's enumeration of the > > > linear ordering of the rationals and define that the n-th step removes > > > the n-th rational from the dense ordering. > > > Yes, and so you never come to an end. > > The end is given by finishing the infinite linear order. Where you will not get. You can not obtain an infinite set by inserting > elements one by one. unless infinitely many insertions are done. And this can even be done simultaneously. > > Cantor used it not only for finite segments of his series (5) and (6) > > but for the whole series. No, for *elements* of the finite segments and *elements* of the whole series. > That is something different. Darn, can't you read? Cantor said he proved his theorem for the whole sets. And, but you going beyond that. You have the series: > {q_1, q_2, q_3, ..., q_n, DO{q_1, q_2, q_3, ..., q_n}} > but you conclude something about > q_1, q_2, q_3, ..., DO{q_1, q_2, q_3, ...} > which is not an element of the series. It is obtained after having gone through all, assertedly existing, configurations. > > Nevertheless you were reluctant to accept that for the whole set. Why > > > do you now agree to Cantor's same argument? > > > Because it is not the same argument. You extend the argument in a way > > that makes it invalid. > > It is exactly the conclusion, by induction, from any finite initial > > segment of his en passant well-ordering of the phi_nu to the complete > > infinite set. No, the induction is not about sets, it is about elements of the sets. diese beiden Reihen sind abgesehen von der Folge ihrer Glieder identisch dieselben Can you read: The sets (diese beiden Reihen) are identical (except in their sequence). There is a bijection between these two sets. > > > Rekursionssatz, Dedekind, 1888, Grundlage der Definition durch > > > > Rekursion, zu beweisen mittels vollst=E4ndiger Induktion. > > > > > And what do you mean with that? > > > > Recursion requires induction. > > > I do not see that from the title. I see only Basics about the definition > > through recursion and to proofs through induction. Definitions do not > > need proof. > > Rekursion needs proof. It is proven by induction. Therefore every > > recursion can be traced back to induction. No, recursion does not need proof in an axiom, so induction is not needed > in an axiom. Or do you think that axioms need proof? Recursion needs proof. The idea that recursion is meaningful needs proof. > > Using recursion is using induction as using a function f(x) is using x. No, induction is a matter of proof. In an axiom you do not need proof, so > you do not need induction. Mathematics could live without recursion. Recursion is but a special application of induction. > > Do you know the origin of complete induction? Dedekind coined it. > > For what purpose do you think did he need it? To prove things I would think. To prove recursion. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Nntp-Posting-Host: hera.cwi.nl > > > > Every rational q divides an irrational interval into two subintervals, > > every irrational q + e joins at least one irrational interval, > > And this is already wrong. In each step, introducing a q indeed does > increase the number of subintervals, but introducing q+e does not > necessarily decrease the number of intervals. It will not if it falls > in the middle of such an interval. > > q + e cannot fall into a rational interval, because such intervals do > not exist at any time. Ok, I misread. BTW, apparently you have some proof that that e is irrational. Where is the proof to be found? > But drawing conclusions from a step by step procedure about a simultaneous > procedure is not valid. How many times do you think I already stated that? > > This is not a step by step procedure. You can perform that > simultaneously. Yes, but your conclusions when you perform a step by step procedure are not valid when you perform them all simultaneously. > But the finished infinity is not something that is obtained in a step by > step procedure. It is an axiom: there is an inductive set. Why do you > keep stating that it must be a step by step procedure when it is not? > > Because the axiom uses induction to define this set. If n exists, then > also n+1 exists. No, it uses recursion, not induction. It state there is a specific set where the elements have a certain property. With that statement the set (and also its elements) does exist with the certain property. Induction is a manner of *proof*. > > Yes, and so you never come to an end. > > > > The end is given by finishing the infinite linear order. > > Where you will not get. You can not obtain an infinite set by inserting > elements one by one. > > unless infinitely many insertions are done. And this can even be done > simultaneously. This can *only* be done if at some stage you do infinitely many insertions simultaneously. But your conclusions about the incrementing and decrementing of number of irrational intervals then becomes invalid. > > Cantor used it not only for finite segments of his series (5) and (6) > > but for the whole series. > > No, for *elements* of the finite segments and *elements* of the whole > series. > That is something different. > > Darn, can't you read? Cantor said he proved his theorem for the whole > sets. Meaning for each element of the set. > And, but you going beyond that. You have the series: > {q_1, q_2, q_3, ..., q_n, DO{q_1, q_2, q_3, ..., q_n}} > but you conclude something about > q_1, q_2, q_3, ..., DO{q_1, q_2, q_3, ...} > which is not an element of the series. > > It is obtained after having gone through all, assertedly existing, > configurations. No, because you can not go through all existing configurations one by one. > > Because it is not the same argument. You extend the argument in a way > > that makes it invalid. > > > > It is exactly the conclusion, by induction, from any finite initial > > segment of his en passant well-ordering of the phi_nu to the complete > > infinite set. > > No, the induction is not about sets, it is about elements of the sets. > > diese beiden Reihen sind abgesehen von der Folge ihrer Glieder > identisch dieselben > Can you read: The sets (diese beiden Reihen) are identical (except in > their sequence). There is a bijection between these two sets. The sets are identical. Given two sets A and B, there are in that case two bijections: f: A -> B, for every a in A f(a) is in B. F: {A} -> {B}, f(A) = B. f is not the same as F. > > Rekursion needs proof. It is proven by induction. Therefore every > > recursion can be traced back to induction. > > No, recursion does not need proof in an axiom, so induction is not needed > in an axiom. Or do you think that axioms need proof? > > Recursion needs proof. The idea that recursion is meaningful needs > proof. No. You are clearly not a mathematician. > > Using recursion is using induction as using a function f(x) is using x. > > No, induction is a matter of proof. In an axiom you do not need proof, so > you do not need induction. > > Mathematics could live without recursion. Recursion is but a special > application of induction. Eh? > > Do you know the origin of complete induction? Dedekind coined it. > > For what purpose do you think did he need it? > > To prove things I would think. > > To prove recursion. -- 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: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > > No, you also need to show that you can attach them to a rational. If you > > can not show that it is possible to do what you do (attaching gaps to > > rationals) your premissa is invalid. > > Every rational q is a point on the real line. Why do you believe I > > could not attach a point q' to this point q? > > Define: q' is a point on the real line next to q if q' has the > > following properties > > q' > q > > q' is not rational. > > For all rationals p > q we have p > q'. Such points q' do not exist. The proof is easy. In general, between any > pair of points on the real line there are infinitely many rationals and > irrationals. Not between q and q'. What should exist there? By the last condition, for all rationals p > q we have p > q'. So we can exclude an irrational between q and q'. So there can at most be a rational between q and q'. Assume such a rational. Between two rationals, there exists an irrational. Hence, there cannot be a rational between q and q'. The problem is not that we could not determine it but that it does not exist. > There are two possibilities: > (1) the distance between the two points is 0, but in that case they are equal > and so the irrational would be equal to the rational which would be very > strange > (2) so the distance between the two points is L for some non-negative real > number L. There is a rational smaller than L (take 1/ceiling(1/L)), add > that rational to the original rational q to get a new rational q_1 between > q and the irrational. You forget that your arguing is already true for the rational numbers alone. (1) the distance between the two points is 0, but in that case they are equal and so two different rationals would be equal which would be very strange (2) so the distance between the two points is L for some non-negative real number L. There is a rational smaller than L (take 1/ceiling(1/ L)), add that rational to the original rational q to get a new rational q_1 between q and the irrational. Nevertheless there are points on the real line where no rational is but an irrational. > > A gap is defined as a position right to a rational number. It can be > > > filled by an irrational number, but that irrational is not the gap. > > > There is no position immediately to the right of a rational number. And > > so your gaps do not exist. > > I defined them above. Defining something does not mean that that something does exist. > If I state: > define q the rational equal to sqrt(2) > that is a proper definition, but q does not exist. Definitions in itself > do not bring something in existence. By the way, that's the same with axioms. > As with your current definition gaps do not exist I wonder why you keep > > on talking about gaps. > > I gave you another example of definable numbers on the same problem: > > By the bijection q <--> q + e (Euler number) we see that there is > > one and only one q mapped on the number q + e. Both sets are equal in > > this configuration and there is no q that is not in the bijection. Right. > Now remove these two sets from the real line. What remains? For > > instance all numbers of the form q + pi. There are no longer > > rationals between them (because we have removed them. But re-introducing > > step by step (or simultaneously) the pairs of numbers of the form q and > > q' + e does not improve the situation, because none of these steps leads to > > a reduction of irrational numbers that have no rational between them. Doing it step by step indeed does not do so. Doing them all at once > *does* do so. Why? The argument does not lose anything of its validity when being applied simultaneously. Every rational can divide an irrational interval into two intervals (at most). Every irrational cannot divide a rational interval, hence increases the size of an irrational interval. And if you fail to hit the correct destination of the site of a number in the stress of simultaneous applivcation, then put every number where you like. In no case, the irrational intervals disappear. In no case we can obtain agreement with the real reals on the real line. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Nntp-Posting-Host: hera.cwi.nl ... > Such points q' do not exist. The proof is easy. In general, between any > pair of points on the real line there are infinitely many rationals and > irrationals. > > Not between q and q'. What should exist there? By the last condition, > for all rationals p > q we have p > q'. So we can exclude an > irrational between q and q'. So there can at most be a rational > between q and q'. Assume such a rational. Between two rationals, there > exists an irrational. Hence, there cannot be a rational between q and > q'. Right, and as it is easy to prove, such q' do not exist. > The problem is not that we could not determine it but that it does not > exist. > There are two possibilities: > (1) the distance between the two points is 0, but in that case they are > equal and so the irrational would be equal to the rational which > would be very strange > (2) so the distance between the two points is L for some non-negative real > number L. There is a rational smaller than L (take 1/ceiling(1/L)), > add that rational to the original rational q to get a new rational > q_1 between q and the irrational. > > You forget that your arguing is already true for the rational numbers > alone. I know that, between two rationals there is always another. I would have thought that that was trivial for you, but apparently it is not. > Nevertheless there are points on the real line where no rational is > but an irrational. Right. And, so what? > > > A gap is defined as a position right to a rational number. It can > > > be filled by an irrational number, but that irrational is not the > > > gap. > > > > There is no position immediately to the right of a rational number. > > And so your gaps do not exist. > > > > I defined them above. > > Defining something does not mean that that something does exist. > If I state: > define q the rational equal to sqrt(2) > that is a proper definition, but q does not exist. Definitions in itself > do not bring something in existence. > > By the way, that's the same with axioms. You are wrong. With an additional axiom there are two possibilities: (1) The new axiom makes the system inconsistent or (2) With the new axiom the system is still consistent. When I state: Definition: z is a natural number less than 0. the definition is correct, but there is no z that satisfies the definition. WHen I state: Axiom: z is a natural number less than 0. the system becomes inconsistent, and so that axiom can not be added. (That the system becomes inconsistent is because with this additional axiom we can prove: (1) All natural numbers are greater than 0 and (2) There are natural numbers less than 0. ) > > Now remove these two sets from the real line. What remains? For > > instance all numbers of the form q + pi. There are no longer > > rationals between them (because we have removed them. But re-introducing > > step by step (or simultaneously) the pairs of numbers of the form q and > > q' + e does not improve the situation, because none of these steps > > leads to a reduction of irrational numbers that have no rational > > between them. > > Doing it step by step indeed does not do so. Doing them all at once > *does* do so. > > Why? The argument does not lose anything of its validity when being > applied simultaneously. It does. Because you are again leaping from doing things step by step to doing infinitely many things simultaneously. -- 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: A consideration concerning the diagonal argument of G. Cantor posting-account=U44YcwkAAAAbGXB70Qr7gA3kornmKE4i Gecko/20080325 Ubuntu/7.10 (gutsy) Firefox/2.0.0.13,gzip(gfe),gzip(gfe) If you accept, for instance, the finished infinity. But the finished infinity is not something that is obtained in a step by > step procedure. It is an axiom: there is an inductive set. Why do you > keep stating that it must be a step by step procedure when it is not? Is'nt it because he thinks that finished infinity is at the end of something that has no end? And he's all excited about spotting a contradiction based on that belief... Brian Chandler http://imaginatorium.org === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Nntp-Posting-Host: hera.cwi.nl ... > > But that is something different from what you said earlier. Now, > > apparently, heap number n is distinguished by having n symbols. In > > that case there is an easy bijection: n <-> heap_n for the naturals n. > > > > Yes, every finite set can be distinguished by a finite set of heaps > > with a finite number of symbols. > > Every infinite set cannot be distinguished by a finite set of heaps > > with a finite number of symbols. > > Right. But as there is an infinite set of heaps that does not prove what > you want to prove. > > There is no infinite set of heaps that contains an infinite number of > symbols such that every heap contains only a finite number of symbols. > > Proof: By definition of set and element, the same symbol cannot > exist in any multiplicity larger than 1 in a set. In order to have an > infinite number of symbols within a set of heaps there must be an > infinite number of symbols within one heap. Can you provide a proof of the last line? And in what way is this counter-proof wrong: Consider the set of heaps {H_n} where the heaps are indexed by n. Each heap contains the symbols o_k for k = 1 to n. The union of the H_n contains o_k for every natural number n, so there are infinitely many o_k. > Sorry, I see I was not clear enough in my wording. A set of X heaps > with at least X symbols does not mean that every heap has X symbols. > The set must contain X symbols. To answer your question: We need not > at least three heaps with three symbols, but we need a set of heaps > that contains at least three symbols. If three symbols are contained > in a set of heaps, then they must be in at least one of the heaps. Right. > Three symbols in a set outside of a heap or within different heaps are > not three symbols. Right. As always your statements are correct in the finite case, but not in the infinite case. > Wrong. If there are aleph_0 natural numbers there are also aleph_0 heaps. > Each heap indexed by its natural number. > > That is provably wrong. > Any set of heaps with only a finite number X of symbols can > distinguish only a finite number X of elements. This is a bijection > between sets of X elements and sets of heaps with X symbols. But the set N (of naturals in unary notation) is not part of this bijection, and should not be part of this bijection. The bijection is: heap_n <-> n and as this is valid for every n in N we have a bijection. And I wonder why N itself also should be in the bijection (which you are assuming). > > (There > > are only heaps with a number of symbols that is not larger than every > > finite number). > > Where in my statement above do I allude to a heap that has a number of > symbols that is larger than every finite number? Remeber the axiom of > infinity (which you state you obey). An inductive set does exist. > When there is a last element in a set it is by definition not an inductive > set, because that last element does not have a successor. > > There is not only the axiom, but there is the theorem that this set > has a cardinal number that is larger than any natural number. I start > from the axiom and this theorem. I accept an infinite set with > cardinal number aleph_0 > n for every n in N. I come to the > unavoidable conclusion that this set, if formed by heaps, contains at > least one heap with aleph_0 elements. You *wrongly* come to that conclusion. There is an easy bijection between heaps and natural numbers: heap_n <-> n, where n is a natural number and heap_n is the heap containing n symbols. > So either the theorem is false or the axiom or the interpretation of > there exists. What is wrong is your conclusion. > > And remember also: Any number of symbols in a set of heaps > > is in one heap by definition of heap and number of symbols. > > I do not think so. You have to prove that using a proper definition of > heap. > > See above. By proper definition of set, there cannot be more symbols > in a set than are in at least one heap of the set. And this is wrong. But it is exactly the same wrong conclusion you were making with your use of FISONs. > Heaps prevent the > reduction of many equal symbols like oooooooooooooo to one symbol o in > the set. The set > {o, oo, ooooooo} contains three heaps with 7 symbols. The set > {o,o,o,o,o,o,o,o,o,o} = {o} contains one symbol. And again a *finite* example (where your statement is true) to prove an infinite example. Finite examples do not prove what is happening in the infinite. -- 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: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > ... > > > But that is something different from what you said earlier. Now, > > > apparently, heap number n is distinguished by having n symbols. In > > > that case there is an easy bijection: n <-> heap_n for the naturals n. > > > > Yes, every finite set can be distinguished by a finite set of heaps > > > with a finite number of symbols. > > > Every infinite set cannot be distinguished by a finite set of heaps > > > with a finite number of symbols. > > > Right. But as there is an infinite set of heaps that does not prove what > > you want to prove. > > There is no infinite set of heaps that contains an infinite number of > > symbols such that every heap contains only a finite number of symbols. > > Proof: By definition of set and element, the same symbol cannot > > exist in any multiplicity larger than 1 in a set. In order to have an > > infinite number of symbols within a set of heaps there must be an > > infinite number of symbols within one heap. Can you provide a proof of the last line? That's the definition of heap and set. A heap prevents the reduction that is automatically executed by a set. > And in what way is this > counter-proof wrong: > Consider the set of heaps {H_n} where the heaps are indexed by n. > Each heap contains the symbols o_k for k = 1 to n. Yes. > The union of the H_n contains o_k for every natural number n, so > there are infinitely many o_k. Yes, there are as many o's as there are natural numbers. But if there is a number aleph_0 of natural numbers, then there must be aleph_0 o's in that set. And by definition of set and heap these aleph_0 o's must be within one heap. > Sorry, I see I was not clear enough in my wording. A set of X heaps > > with at least X symbols does not mean that every heap has X symbols. > > The set must contain X symbols. To answer your question: We need not > > at least three heaps with three symbols, but we need a set of heaps > > that contains at least three symbols. If three symbols are contained > > in a set of heaps, then they must be in at least one of the heaps. Right. > Three symbols in a set outside of a heap or within different heaps are > > not three symbols. Right. As always your statements are correct in the finite case, but not > in the infinite case. If I give correct bijections for all finite cases, then it is impossible, that one part of the bijection becomes wrong in the infinite case while the other part remains true. But that is what you claim. > Wrong. If there are aleph_0 natural numbers there are also aleph_0 heaps. > > Each heap indexed by its natural number. > > That is provably wrong. > > Any set of heaps with only a finite number X of symbols can > > distinguish only a finite number X of elements. This is a bijection > > between sets of X elements and sets of heaps with X symbols. But the set N (of naturals in unary notation) is not part of this bijection, > and should not be part of this bijection. The bijection is: > heap_n <-> n > and as this is valid for every n in N we have a bijection. And I wonder > why N itself also should be in the bijection (which you are assuming). My bijection is between initial segments of natural numbers and heaps. The union of all initial segments is the complete set of natural numbers. That's why the union cannot exist. > > (There > > > are only heaps with a number of symbols that is not larger than every > > > finite number). > > > Where in my statement above do I allude to a heap that has a number of > > symbols that is larger than every finite number? Remeber the axiom of > > infinity (which you state you obey). An inductive set does exist. > > When there is a last element in a set it is by definition not an inductive > > set, because that last element does not have a successor. > > There is not only the axiom, but there is the theorem that this set > > has a cardinal number that is larger than any natural number. I start > > from the axiom and this theorem. I accept an infinite set with > > cardinal number aleph_0 > n for every n in N. I come to the > > unavoidable conclusion that this set, if formed by heaps, contains at > > least one heap with aleph_0 elements. You *wrongly* come to that conclusion. There is an easy bijection between > heaps and natural numbers: heap_n <-> n, where n is a natural number and > heap_n is the heap containing n symbols. If natural numbers have a unary representation, then they *are* heaps. Then heap_n = n (identity enforces bijectability). > So either the theorem is false or the axiom or the interpretation of > > there exists. What is wrong is your conclusion. Which step? > > And remember also: Any number of symbols in a set of heaps > > > is in one heap by definition of heap and number of symbols. > > > I do not think so. You have to prove that using a proper definition of > > heap. > > See above. By proper definition of set, there cannot be more symbols > > in a set than are in at least one heap of the set. And this is wrong. Right or wrong my set theory? You think the definition of sets and heaps must be wrong (can a definition be wrong?) because otherwise set theory must be wrong? Yes, that's right. > But it is exactly the same wrong conclusion you were > making with your use of FISONs. Exactly the same, but not wrong. > Heaps prevent the > > reduction of many equal symbols like oooooooooooooo to one symbol o in > > the set. The set > > {o, oo, ooooooo} contains three heaps with 7 symbols. The set > > {o,o,o,o,o,o,o,o,o,o} = {o} contains one symbol. And again a *finite* example (where your statement is true) to prove an > infinite example. Finite examples do not prove what is happening in the > infinite. Why then do you think that Cantor's diagonal proof is valid in the infinite? Admittedly every finite initial segment of the diagonal differs from the corresponding finite initial segment of the list. But finite examples do not prove what is happening in the infinite, i.e., in the complete list. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Nntp-Posting-Host: hera.cwi.nl ... > > There is no infinite set of heaps that contains an infinite number of > > symbols such that every heap contains only a finite number of symbols. > > > > Proof: By definition of set and element, the same symbol cannot > > exist in any multiplicity larger than 1 in a set. In order to have an > > infinite number of symbols within a set of heaps there must be an > > infinite number of symbols within one heap. > > Can you provide a proof of the last line? > > That's the definition of heap and set. A heap prevents the reduction > that is automatically executed by a set. No, according to the definition the same symbol cannot exist in any multiplicity larger than 1 in a set. It does not state that in order to have an infinite number of symbols within a set of heaps there must be an infinite number of symbols within one heap. So can you a proof of that? > And in what way is this > counter-proof wrong: > Consider the set of heaps {H_n} where the heaps are indexed by n. > Each heap contains the symbols o_k for k = 1 to n. > > Yes. > > The union of the H_n contains o_k for every natural number n, so > there are infinitely many o_k. > > Yes, there are as many o's as there are natural numbers. But if there > is a number aleph_0 of natural numbers, then there must be aleph_0 o's > in that set. And by definition of set and heap these aleph_0 o's must > be within one heap. In that case tell me where in the definition that is stated. Again, there are aleph_0 o's in the set of heaps {H_n}. As the H_n are all different from each other, the set {H_n} contains aleph_0 H's. The collection of o's in all those H_n's together is also aleph_0, but there is no H_n that contains them all. I may note that union{n in N} H_n is *not* a H_n, just as union{n in N} {1, ..., n} is *not* a finite initial segment and as union{n in N} [1/n, 1] is *not* a closed interval on the real line. > > Sorry, I see I was not clear enough in my wording. A set of X heaps > > with at least X symbols does not mean that every heap has X symbols. > > The set must contain X symbols. To answer your question: We need not > > at least three heaps with three symbols, but we need a set of heaps > > that contains at least three symbols. If three symbols are contained > > in a set of heaps, then they must be in at least one of the heaps. > > Right. > > > Three symbols in a set outside of a heap or within different heaps are > > not three symbols. > > Right. As always your statements are correct in the finite case, but not > in the infinite case. > > If I give correct bijections for all finite cases, then it is > impossible, that one part of the bijection becomes wrong in the > infinite case while the other part remains true. But that is what you > claim. Tell me what the bijection *is*. I state that line_n <-> H_n is a bijection between all natural numbers and all H_n's. So, what is missing in that bijection? > But the set N (of naturals in unary notation) is not part of this > bijection, and should not be part of this bijection. The bijection is: > heap_n <-> n > and as this is valid for every n in N we have a bijection. And I wonder > why N itself also should be in the bijection (which you are assuming). > > My bijection is between initial segments of natural numbers and heaps. > The union of all initial segments is the complete set of natural > numbers. That's why the union cannot exist. Strange. I thought you wished a bijection between the finite initial segments. Apparently you also want to include their union, and so you want a bijection between {N, 1, ..., } and the heaps themselves. Why? In what sense is N a line in your triangle in unary notation? And why can I not for that bijection use union{n in N} heap_n (which is not a finite heap)? > You *wrongly* come to that conclusion. There is an easy bijection between > heaps and natural numbers: heap_n <-> n, where n is a natural number and > heap_n is the heap containing n symbols. > > If natural numbers have a unary representation, then they *are* heaps. > Then heap_n = n (identity enforces bijectability). So, now you state that, indeed, there is a bijection between heaps and natural numbers. > > See above. By proper definition of set, there cannot be more symbols > > in a set than are in at least one heap of the set. > > And this is wrong. > > Right or wrong my set theory? You think the definition of sets and > heaps must be wrong (can a definition be wrong?) because otherwise set > theory must be wrong? Yes, that's right. Where in the definition it is stated that there cannot be more symbols in a set than there are in at least one heap of the set? > But it is exactly the same wrong conclusion you were > making with your use of FISONs. > > Exactly the same, but not wrong. Exactly the same and wrong. union{n in N} FISON(n) = N. If it is not N you should be able to give a natural number k that is in the union and not in N or a natural number k that is in N but not in any FISON(n). > > Heaps prevent the > > reduction of many equal symbols like oooooooooooooo to one symbol o in > > the set. The set > > {o, oo, ooooooo} contains three heaps with 7 symbols. The set > > {o,o,o,o,o,o,o,o,o,o} = {o} contains one symbol. > > And again a *finite* example (where your statement is true) to prove an > infinite example. Finite examples do not prove what is happening in the > infinite. > > Why then do you think that Cantor's diagonal proof is valid in the > infinite? Admittedly every finite initial segment of the diagonal > differs from the corresponding finite initial segment of the list. But > finite examples do not prove what is happening in the infinite, i.e., > in the complete list. Do you not see the difference that in Cantor's proof something is proved for each finite n and nothing more? You wish to prove something for some infinite n based on what is happening for each finite n. I.e. you know something about sets of n heaps for each finite n. But from that you want to conclude something about a set of infinitely many heaps. -- 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: A consideration concerning the diagonal argument of G. Cantor posting-account=-eQqtQoAAACZVM-kNEsOn3k7GSvoJoS4 CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > [WM] easily ranks as among the most > intellectually dishonest persons I've ever discussed mathematics with. I'm glad to see that I am not the only one harbouring this sentiment. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=oTDIagkAAACTxHurtPutBWvNQS8ZCNO9 Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > [WM] easily ranks as among the most > intellectually dishonest persons I've ever discussed mathematics with. I'm glad to see that I am not the only one harbouring this sentiment. I don't know; on sci.math/sci.logic, Lester Zick is at least a contender. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > > But we can make it very simple: > > o > > oo > > ooo > > ... > > What ever (a number) is contained in he first column, the same > > (number) is contained in a line. I see only the symbols o. What do you mean here? Do you mean the > number of symbols? If so, why is that the case? It is still simple > assertion. For every partial triangle o o oo o oo ooo ... we have height / width = 1. This yields a sequence 1,1,1,... with limit 1. If the infinite can be determined by the finite, i.e., if it is not real nonsense in Wolkenkuckucksheim, then the limit of that sequence cannot be other than 1. > The triangle is (in suitable coordinates) simply a representation of > > the function f(x) = x. There is no proof required that f(x) is the > > same as x. Makes no sense. f(x) is the length or width of line number x. > > No, not in principle. In principle no prime with more than 10^100 bits > > information contents is available. Your principle is not the mathematical principle. But contrary to modern mathematics, it is true. > > > The question is whether there is a fixed number of lines. This > > > > fixed number is not a natural, but it is defined by a bijection > > > > with all naturals. > > > > > Eh? In that bijection that fixed number is *not* bijected to a > > > natural. > > > What makes you think so? > > > > It is larger than every natural. > > > Yes, and it is not in the bijection. What makes you think that it *is* > > in the bijection? > > The *number* of line is in bijection with the lines. Can you give the bijection you are meaning, so that it is clear that the > total number of lines is in the bijection? The number of the first n lines is the width of the n-th line. > > That is Cantor's understanding of a function. One set can be a > > > function of the other. > > > Can you explain it, because I do not understand it. The only way I can > > understand it is that given two sets A and B, there is a function f with > > domain B and range A. Why do you copy this text again and again: You asked what was meant by a set being an eindeutige Funktion > And please correct the error in that quote the next time, otherwise people Sorry, I corrected it in my files but I copied it from an older posting. Here is the corrected version: Zun.8achst aber will ich meine Betrachtungen wieder auf die linearen Punktmengen beschr.8anken und einen Beweis f.9fr den Satz entwickeln, da¤ alle linearen perfekten Mengen gleiche M.8achtigkeit haben oder, was dasselbe hei¤t, da¤ je zwei solche Mengen in eine Beziehung zueinander gesetzt werden k.9annen, welcher gem.8a¤ gewisserma¤en die eine als eine [umkehrbar] eindeutige Funktion der andern betrachtet werden kann. The (corrected) version of the quote translated: > Next however I will point my observations again to the linear sets of > points and give a proof for the theorem that all linear perfect sets > have the same cardinality or, what is the same, that two such sets > are in a relation with each other that in a certain sense one can be > looked at that one is a [invertible] function of the other. Note the in a certain sense. Cantor uses this gewisserma¤en quite frequently, for instance: Abstrahiert man sowohl von der Beschaffenheit der Elemente, wie auch von der Ordnung ihres Gegebenseins, so erh.8alt man die Kardinalzahl oder M.8achtigkeit der Menge, einen Allgemeinbegriff, in welchem die Elemente, als sogenannte Einsen, gewisserma¤en organisch ineinander derartig zu einem einheitlichen Ganzen verwachsen sind, da¤ keine vor der anderen ein bevorzugtes Rangverh.8altnis hat. Bei endlichen Mengen fallen die beiden Momente M.8achtigkeit und Anzahl gewisserma¤en zusammen, die Anzahl oder Ordnungszahl von M ist also s.8amtlichen wohlgeordneten Mengen desselben Typus gemeinsam, gewisserma¤en dasjenige, was ihnen allen immanent ist. warum bei endlichen Mengen Ordnungszahl und Kardinalzahl gewisserma¤en zusammenfallen, Allerdings kann omega gewisserma¤en als die Grenze angesehen werden, welcher die ver.8anderliche endliche Zahl n zustrebt and many more findings of gewisserma¤en (= so to speak) in Cantor's collected works. > He had another understanding of function than set theorists. Enlgish speaking mathematicians perhaps. The concept of function has > slightly changed in the course of the years. In the above text he seems to identify the range of a function with the function. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Nntp-Posting-Host: hera.cwi.nl > > What ever (a number) is contained in he first column, the same > > (number) is contained in a line. > > I see only the symbols o. What do you mean here? Do you mean the > number of symbols? If so, why is that the case? It is still simple > assertion. > > For every partial triangle > o > > o > oo > > o > oo > ooo > ... > > we have height / width = 1. This yields a sequence 1,1,1,... with > limit 1. Eh, I never did say that the number of lines was not equal to the number of columns, so why do you state this? > If the infinite can be determined by the finite, i.e., if it is not > real nonsense in Wolkenkuckucksheim, then the limit of that sequence > cannot be other than 1. I never stated otherwise. But assuming there is a line containing infinitely many symbols is the same as assuming that there is a last line. And that is the same as assuming the negation of the axiom of infinity. > > The triangle is (in suitable coordinates) simply a representation of > > the function f(x) = x. There is no proof required that f(x) is the > > same as x. > > Makes no sense. > > f(x) is the length or width of line number x. And so the assumption that there is a line containing aleph-0 elements is the same as the assumtion that there is a aleph-0'th line, which is the same as the negation of the axiom of infinity. > > No, not in principle. In principle no prime with more than 10^100 bits > > information contents is available. > > Your principle is not the mathematical principle. > > But contrary to modern mathematics, it is true. That is irrelevant, we are talking mathematics, not what is possible in the physical world. In the physical world the length of the diagonal of a square with each side 1 meter does not exist, apparently. Your vision is extremely close to that of Kronecker. But he nevertheless contributed a lot to algebraic number theory (stating even that algebraic numbers were numbers). > > Yes, and it is not in the bijection. What makes you think that it > > *is* in the bijection? > > > > The *number* of line is in bijection with the lines. > > Can you give the bijection you are meaning, so that it is clear that the > total number of lines is in the bijection? > > The number of the first n lines is the width of the n-th line. So you refuse to give the bijection? (I assume that 'n' above is a natural number, and so the number of the infinitely many lines is not in that bijection.) > Why do you copy this text again and again: > > You asked what was meant by a set being an eindeutige Funktion Yes, and I still ask. Cantor does not state it. > And please correct the error in that quote the next time, otherwise > > Sorry, I corrected it in my files but I copied it from an older > posting. Here is the corrected version: already corrected the omission in my translation. So why post it again? Is that for the education of those readers of this newsgroup that are not able to read German? > The (corrected) version of the quote translated: > Next however I will point my observations again to the linear sets of > points and give a proof for the theorem that all linear perfect sets > have the same cardinality or, what is the same, that two such sets > are in a relation with each other that in a certain sense one can be > looked at that one is a [invertible] function of the other. > > Note the in a certain sense. > > Cantor uses this gewisserma¤en quite frequently, for instance: Yes, and so? It's meaning is in a certain sense. Nothing else. > Enlgish speaking mathematicians perhaps. The concept of function has > slightly changed in the course of the years. > > In the above text he seems to identify the range of a function with > the function. No, what he does is in a certain sense identify the range of a function with the function. And in strict mathematics, what he is meaning (I think) is that when we have two sets A and B and a function f: A -> B, we can look at B as in a sense as a function of A. This does *not* mean that he thinks that A is in the domain of the function. -- 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: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > æ> æ> What ever (a number) is contained in he first column, the same > æ> æ> (number) is contained in a line. > æ> æ> I see only the symbols o. æWhat do you mean here? æDo you mean the > æ> number of symbols? æIf so, why is that the case? æIt is still simple > æ> assertion. > æ æ> For every partial triangle > æ> o > æ æ> o > æ> oo > æ æ> o > æ> oo > æ> ooo > æ> ... > æ æ> we have height / width = 1. This yields a sequence 1,1,1,... with > æ> limit 1. Eh, I never did say that the number of lines was not equal to the number > of columns, so why do you state this? You said that there are more columns than symbols in a line. That is wrong. The sequence number of lines / symbols in one line is 1 and has limit 1. æ> If the infinite can be determined by the finite, i.e., if it is not > æ> real nonsense in Wolkenkuckucksheim, then the limit of that sequence > æ> cannot be other than 1. I never stated otherwise. æBut assuming there is a line containing infinitely > many symbols is the same as assuming that there is a last line. æAnd that is > the same as assuming the negation of the axiom of infinity. No, it is the same as proving the inconsistency of this axiom with mathematics. The axiom could also state that there are infinitely many symbols in one heap. What would be changed? æ> æ> The triangle is (in suitable coordinates) simply a representation of > æ> æ> the function f(x) = x. There is no proof required that f(x) is the > æ> æ> same as x. > æ> æ> Makes no sense. > æ æ> f(x) is the length or width of line number x. And so the assumption that there is a line containing aleph-0 elements is > the same as the assumtion that there is a aleph-0'th line, which is the > same as the negation of the axiom of infinity. The conclusion from the infinite number of lines on the infinite line is not an assumption. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Nntp-Posting-Host: hera.cwi.nl ... > Eh, I never did say that the number of lines was not equal to the number > of columns, so why do you state this? > > You said that there are more columns than symbols in a line. That is > wrong. > The sequence number of lines / symbols in one line is 1 and has limit > 1. Yes, the limit is 1, and so? As there is no line with number aleph_0, there is also no single line with aleph_0 symbols. That the limit is 1 does not bring an aleph_0'th line with aleph_0 symbols in existence. > > If the infinite can be determined by the finite, i.e., if it is not > > real nonsense in Wolkenkuckucksheim, then the limit of that sequence > > cannot be other than 1. > > I never stated otherwise. But assuming there is a line containing > infinitely many symbols is the same as assuming that there is a last > line. And that is the same as assuming the negation of the axiom of > infinity. > > No, it is the same as proving the inconsistency of this axiom with > mathematics. Assuming is not proving. You *assume* that there is a line containing infinitely many symbols. You do not prove it. > The axiom could also state that there are infinitely many symbols in > one heap. What would be changed? If you mean with the use of your triangle? In that case the axiom would be inconsistent in itself. > > > The triangle is (in suitable coordinates) simply a representation > > > of the function f(x) = x. There is no proof required that f(x) > > > is the same as x. > > > > Makes no sense. > > > > f(x) is the length or width of line number x. > > And so the assumption that there is a line containing aleph-0 elements is > the same as the assumtion that there is a aleph-0'th line, which is the > same as the negation of the axiom of infinity. > > The conclusion from the infinite number of lines on the infinite line > is not an assumption. In that case *prove* for once that if there are infinitely many lines there is also an infinitieth line! Until now you give this only as assertions of prove it from unproven assumptions that are equivalent. -- 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: A consideration concerning the diagonal argument of G. Cantor posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU 5.1),gzip(gfe),gzip(gfe) > æ> But we can make it very simple: > æ> o > æ> oo > æ> ooo > æ> ... As this thread passes its 7000th post (on sci.math, not on sci.logic), I ask the same question I ask every thousand posts, namely why is this thread so long? Some posters didn't like the answer I gave last time, so let me attempt to give a better answer. The reason that this thread is long is because WM doesn't merely claim that ZFC is counterintuitive, but because he believes that it is inconsistent . In other words, he is directly attacking ZFC. Some posters have accused me of being too charitable towards WM and the other so-called cranks, because I attempt to shape their thoughts into a rigorous theory when no such theory is even possible. The reason for my charity is that I really want to see a discussion between the proponents of ZFC and those of another rigorous theory, to compare and contrasts the theorems provable when different axioms are assumed. But WM's theory doesn't qualify as a rigorous theory. The problem is that no one really wants to post a rigorous alternative to ZFC, or discuss such a theory. Posters only want either to attack or to defend ZFC. Suppose I were to come up with a rigorous theory which, for example, incorporates both heaps and unary notation. We notice how decimal notation can be a function from Z (the integers) to the set {0,1,2,3,4,5,6,7,8,9} (i.e., the von Neumann natural 10), so unary notation be thought of as a function whose domain is a natural number and whose image is a singleton. Of course, this is not WM's theory, and I would be rightly called charitable for attempting to ascribe such a theory to someone like WM. Indeed, we would be writing unary notation in terms of functions and therefore sets, whereas WM wants us to use heaps without mentioning sets at all -- since he rejects not only ZFC, but set theory in general. separate from this one, with unary notation and the primitive heap as the cornerstones of this theory. Suppose I claim this theory as my own theory, with no mention of WM at all in that thread, so that no one can accuse me of being charitable to WM when discussing this theory. The question is, would that thread last 7000 posts, with a heated debate between the proponents of ZFC and me, as the proponent/inventor of this new theory? The answer is, of course not. Indeed, I'd be fortunate if that thread would last seven posts, much less 7000! Why not? For one thing, I'm posting via Google Groups, so that half the members of this board would automatically killfile the thread anyway. But suppose, rather than myself, someone with a full-priced newsreader proposed the new theory. I bet that the thread still wouldn't last anywhere near 7000 posts. The real reason is that by proposing a new theory, the thread would no longer be directly attacking ZFC. Posters may villify WM all they want for not being rigorous enough and call him a crank and other ad hominem terms, but the truth is that if WM really were rigorous, he'd be ignored. The only way to be noticed is to attack ZFC. This is what WM and many others have discovered. If one doesn't like ZFC, making an alternative theory is futile. If one wants to see a real discussion, one that spans significantly many posts, then one must attack ZFC and claim that they have a proof of its inconsistency, no matter how flimsy. In a way, this makes since, because no matter how beautiful a new theory may be, mathematicians will never replace ZFC unless the current theory is proved inconsistent. The fact that the longest threads by far on sci.math have been started by attackers of ZFC prove this to be the case. And so this should answers my question. The reason that this thread is long is that WM is directly attacking ZFC. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > The reason that this thread is long is because WM doesn't > merely claim that ZFC is counterintuitive, but because I can prove that it is inconsistent. For instance, if we remove all rationals q and all irrational numbers q + pi from the real line, then there should uncountably many real number remain. That is an interval containing only irrational numbers. Returning a rational q will divide an irrational interval into two irrational intervals. Returning the corresponding irrational number of the form q + pi will increase the number of irrationals in one of the irrational intervals existing on the real line (because there is no interval of only rational umbers that could be subdivided by q + pi). After finishing this process we have unavoidably irrational intervals remaining. As we can prove that there are no irrational intervals on the real line, the idea of complete infinite sets like all q or all q + pi or all reals has been shown to be false. The reason for my charity is that I really want to see a > discussion between the proponents of ZFC and those of > another rigorous theory, to compare and contrasts the > theorems provable when different axioms are assumed. But > WM's theory doesn't qualify as a rigorous theory. ZFC is not rigorous. Its proponents claim that it was rigorous but it is simply wrong. > whereas WM wants us > to use heaps without mentioning sets at all -- since he rejects > not only ZFC, but set theory in general. No. Finite set theory is very useful. Further I do not reject sets as I enumerate sets by heaps, even infinitely many sets. separate from this one, with unary notation and the primitive > heap as the cornerstones of this theory. Suppose I claim this > theory as my _own_ theory, with no mention of WM at all in > that thread, so that no one can accuse me of being charitable > to WM when discussing this theory. A piece of advice: You must not lose heart if a fool calls you foolish. Why not? For one thing, I'm posting via Google Groups, so that > half the members of this board would automatically killfile the > thread anyway. I am also posting via Google Groups. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=U44YcwkAAAAbGXB70Qr7gA3kornmKE4i Gecko/20080325 Ubuntu/7.10 (gutsy) Firefox/2.0.0.13,gzip(gfe),gzip(gfe) (I'm adding numbering at the end of relevant sections...) > The reason that this thread is long is because WM doesn't > merely claim that ZFC is counterintuitive, but because I can prove that it is inconsistent. For instance, if we remove all > rationals q and all irrational numbers q + pi from the real line, then > there should uncountably many real number remain. (1) That is an interval > containing only irrational numbers. (2) Returning a rational q will divide > an irrational interval into two irrational intervals. Returning the > corresponding irrational number of the form q + pi will increase the > number of irrationals in one of the irrational intervals existing on > the real line (because there is no interval of only rational umbers > that could be subdivided by q + pi). (3) After finishing this process we have unavoidably irrational intervals > remaining. (4) As we can prove that there are no irrational intervals on > the real line, (5) the idea of complete infinite sets like all q or all q > + pi or all reals has been shown to be false. (6) Am I reading this right (insofar as any of this could be called right)?? It seems that (5) corresponds to the true statement that there is no set of reals containing two or more irrationals (r1 and r2) s.t. there is no rational q with r1 < q < r2. Such a set appears to be what WM means by an irrational interval. Meanwhile, (1) is true, if it means that the set difference of the reals and (rationals U rationals+pi) is an uncountable set. (Though I have no idea of the significance of the pi+q subset.) (2) is obviously true of any set difference between a set of reals and a set including (all) the rationals. The section (3) about returning completely escapes me - perhaps someone can translate into German and back again? So the contradiction is achieved by stating that a set including only irrationals is an interval - although we earlier agreed that between any two irrationals is a rational, now there aren't, because we removed them. What a stunning piece of logic! May as well... Proof that group theory is inconsistent. Let K be the Klein 4-group, {1, i, j, k}. Now remove j, leaving {1, i, k}. But ik = j which is no longer there, so this set of group elements is not closed under multiplication, which is a contradiction. QED As to the metadiscussion, this appears not to be a sign of stupidity, nor or intellectual dishonesty, but simply of barking madness. Brian Chandler http://imaginatorium.org === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU 5.1),gzip(gfe),gzip(gfe) [...] Sorry, a few typos there: > In a way, this makes [sense], because no matter how beautiful > a new theory may be, mathematicians will never replace ZFC > unless the current theory is proved inconsistent. The fact > that the longest threads by far on sci.math have been started > by attackers of ZFC prove[s] this to be the case. > And so this should [answer] my question. The reason that this > thread is long is that WM is directly attacking ZFC. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Sorry, I corrected it in my files but I copied it from an older > posting. Here is the corrected version: Zun.8achst aber will ich meine Betrachtungen wieder auf die linearen > Punktmengen beschr.8anken und einen Beweis f.9fr den Satz entwickeln, da¤ > alle linearen perfekten Mengen gleiche M.8achtigkeit haben oder, was > dasselbe hei¤t, da¤ je zwei solche Mengen in eine Beziehung zueinander > gesetzt werden k.9annen, welcher gem.8a¤ gewisserma¤en die eine als eine > [umkehrbar] eindeutige Funktion der andern betrachtet werden kann. > > The (corrected) version of the quote translated: > Next however I will point my observations again to the linear sets of > points and give a proof for the theorem that all linear perfect sets > have the same cardinality or, what is the same, that two such sets > are in a relation with each other that in a certain sense one can be > looked at that one is a [invertible] function of the other. > > Note the in a certain sense. > Enlgish speaking mathematicians perhaps. The concept of function has > slightly changed in the course of the years. In the above text he seems to identify the range of a function with > the function. Then by modern standards, he is wrong. What he undoubtedly means, properly expressed in modern terms, is that there is an invertable function from any one such linear perfect set to any other. In any event, WM cannot simultaneously claim Cantor as an authority for what is correct and as an exemplar of what is wrong in set theory without showing that WM is both a knave and a fool. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > ... > > But that is something different from what you said earlier. Now, > > apparently, heap number n is distinguished by having n symbols. In > > that case there is an easy bijection: n <-> heap_n for the naturals n. > > > Yes, every finite set can be distinguished by a finite set of heaps > > with a finite number of symbols. > > Every infinite set cannot be distinguished by a finite set of heaps > > with a finite number of symbols. > > Right. But as there is an infinite set of heaps that does not prove what > you want to prove. There is no infinite set of heaps that contains an infinite number of > symbols such that every heap contains only a finite number of symbols. There is an infinite set of heaps all different and each marked by a different number of repetitions of single symbol, but no heap in that set of heaps representing the number of heaps in that set. Wm's argument for finiteness amounts to no more than It is finite because I say that it has to be finite. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > A number of lines cannot be distinguished by a set of heaps, unless at > > least one heap contains as many symbols as the set contains lines. > > More symbols than are in at least one heap cannot exist in a set of > > heaps. > > Again simple assertion. You just claimed to have understood that every symbol belongs to a > line. But we can make it very simple: > o > oo > ooo > ... What ever (a number) is contained in he first column, the same > (number) is contained in a line. The triangle is (in suitable coordinates) simply a representation of > the function f(x) = x. There is no proof required that f(x) is the > same as x. There is, however, proof required that there is a finite upper bound on the set of heaps for which that f(x) = x holds. And WM keeps claiming that there is such an upper bound, so he owes us proof. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > If every gap between two rational numbers is filled by one irrational > only, then obviously there is an alternating sequence rational, > irrational, rational, ... and the cardinalities are equal. Therefore > there must be irrational intervals > > This might make sense if I knew what an irrational interval is. I > thought > from earlier posts you meant by this expression all the irrationals > in an interval, > An irrational interval in the above sense (required to guarantee the > existence of more irrational numbers than rational numbers) is an > interval of only irrational numbers. Definition: An irrational interval (that contains more than one point) > is an interval on the real line that contains a set of irrational > numbers x_1, x_2, ... such that between any two numbers x_1 < x_2 of > the set there is no rational number q with x_1 < q < x_2. That definition cannot be instantiated. The set of objects it defines is empty. There cannot be two real different points (their rationality is irrelevant) having no rational real points between them. Not necessary to say that an irrational interval with more than one > point does not exist on the real line. Nevertheless it must exist, if > there were more irrational numbers than rational numbers. Another false claim. When WM asserts such wild claims let him prove them. WM has no notion of what density means. > but now I have the feeling you mean an interval that > contains only irrationals, which of course doesn't make sense. Of course not. That proves that there are not more irrational than > rationals. How does it manage do anything so false? > Actually you know that, and think that this implies there > can't be more than a countable number of irrationals in any > interval, altho why that latter follows from the former is > beyond my ken, so to speak. Deplorable. WM is! > > I'm confused. We know that every gap between two rational numbers > is filled by at least a countable infinity of irrationals, right? Every gap between two rational numbers that we can name is filled with > infinitely many numbers, yes. > And every gap between two irrational numbers is filled by at > (most) a countable infinity of rationals? > > Therefore it doesn't make sense to talk about the numbers alternating > rational, irrational, rational, irrational, etc. For the numbers > to alternate would mean there was a rational followed 'immediately' > by an irrational, and we know that's impossible. (but you knew this, > see above). You talk about numbers that we can identify. Of course it is > impossible to identify two adjacent numbers in a dense ordering. That impossibility occurs because between any two members of a dense ordering there are infinitely many others. So such adjacency is a non-starter in dense sets. WM's diet seems to be exclusively made up of such non-starters. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > 1) The list is spanned by the lines. > 2) There is no chance to have a column outside of a line. > 3) Every line starts at the first column. > > How about: > Every line is finite. > Each line is longer than the one above it. > There is no last line. By the symmetry of the triangle we obtain an even simpler truth: Whatever (a number) is contained in the first column, the same > (number) must be contained in at least one line. The only number, or anything else, in any column is o, which is contained in every line. > > There is no chance to circumvent this fact. > If you want to define it by a bijection: There must be at least one > line in the set > > o > oo > ooo > ... > > that contains enough o's such that every natural number can be mapped > on a different o. If that were true, WM should be able to prove it. But since WM has not even tried to prove it, there is no evidence that it need be true and good reason to thing it false. Among other things, unless there is a last line which cannot be made longer by appending another o, there need be no line which is not exceeded by another. And there cannot be a last line. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > Die Peano-Axiome sind eine andere Formulierung der Definition der > > nat.9frlichen Zahlen. > > All together, not only the last one. > > No, induction is a way to prove something. > > > That is not in contradiction with the fact that it is also in use to > > define something. > > Oh. The standard terminology is that it is a way to prove something, are > you using non-standard terminology? The standard terminology is: Die Peano-Axiome sind eine andere > Formulierung der Definition der nat.9frlichen Zahlen They define the > natural numbers, not a method of proof. In ZF, the axiom of infinity , along with other axioms produces the necessary minimal inductive set without invoking any Peano axiom, which axioms then are theorems about naturals produces without their use. > Stop. I do not need different steps. I consider Cantor's well- > ordering. I define once and for all that in that sequence the > rationals have to be picked from the dense ordering simultaneously - > all or any subset that I want. What is the difference to Cantor's > diagonal proof? That one can reorder the dense set of rationals into a list ordered set of rationals no one denies. But only if one does it all at one fell swoop rather that one element at a time. > Then why do you claim it? > > > There is no jumping > > required if all steps exist. > > If you get there at all with steps there must be a last step, otherwise > you would never get there. So, what is the last step in your sequece > of transpositions? If you consider the complete diagonal of a Cantor list, then there > must be a last line. WM claims that infinite sequences must have last elements? The whole point of a sequence being infinite is that it does not have any last element. > So what is the digit and at what position is it > in the last line? What last line would that be? > > Never is wrong. After an infinity of steps, defined for instance by a > > sequence, you come to an end. > > No, when you come to an end there must be a last step you have done. It seems so but that is only for finite sets. Otherwise: If you > consider the complete diagonal of a Cantor list, then there must be a > last line. Then WM is again claiming that a sequence defined to have no last element must have a last element. WM's attitude towards definitions apparently is that they may be ignored if one doesn't like their consequences. > So what is the digit and at what position is it in the last line? Any last digit in a sequence with no last digit is effectively determined by the digits that follow it. > > What natural enumerates the last step? You mean in Cantor's list? I am not sure whether there is a last line. Then why go on insisting that there is one? > And so you are not understood. Do you wish to be not understood? I see that not everybody is able to understand everything. And WM leads the list of those who don't. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > But they need not be > identified because it is sufficient to know that they exist. That also justifies all those irrationals that WM hates so much. So by WM's own arguments, we do not need to identify all of the members of any set of numbers in order to have them all exist. > > > > It is enough to know that in a linear order the > > > number > > > of gaps is the same as the number rationals and the latter is > > > countable. > > > > But you prove that with aa'bb'cc'... which is nonsense as you cannot > > attach gaps to rational numbers. > > > I did it. > > As my conclusion upto now is that you use your definition of gap where a > gap is equivalent to an irrational number my question becomes again, > *how* do you attach a gap to a rational number? There is no irrational > at the right hand side of a rational number. You mean you we cannot determine it. We can determine of any irrational that it is not at the right side of any rational. Given any rational, q, and any greater irrational, r, then density gives q < (q+r)/2 < r, where (q+r)/2 is also irrational. Thus for every irrational, r, r is not at the right side of q. Now those with any competence at mathematics at all would have seen that immediately that for themselves, so WM's lack of such competence puts him outside the mathematical pale. A gap is defined as a position right to a rational number. It can be > filled by an irrational number, but that irrational is not the gap. But between any rational number and any irrational number there are infinitely many other rationals and infinitely many other irrationals, so any position to the right of a rational cannot be filled by either a rational or an irrational unless we have another gap between each rational and its gap. So that every rational in WM's GAPology is the lower bound of an endless chain of gaps preceding gaps. > No. I use the same definition in all cases as I use the same > definition of the rationals. That definition is impossible to instantiate without self-contradictions all over the place. > If gaps change meaning or number than the > rational do the same. That is just the fundament of my argument. Wm's argument requires either infinitely many gaps for each rational or infinitely many numbers, both rational and irrational, between a rational and its gap, at least in the standard real number order. Or both. === Subject: Re: JSH: Probabilistic quadratic residue solving posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > An off-shoot of my surrogate factoring research is a probabilistic > method to solve quadratic residues, as given > k^2 = q mod p > when p is an odd prime, and q is a quadratic residue modulo p, you > find k. > The technique requires introduction of a few additional variables > starting with T, where > T = 2q + np > where n is an odd natural number so you have T = 2q mod p, but T - 2q > is also forced to have p as a factor. > Next you find z, where with integer factors f 1 and f 2 where f 1*f 2 > = T: > z = (f 1 + f 2)/2 > and now finally you try for an answer for k, with > k = 3^{-1}(2z) mod p. > The method is probabilistic because if I've got the analysis right you > have a 50% probability of getting the right k, for each z that you > try. æChecking is done by looking at k^2 mod p, to see if you get q. > Example: Let q=2, p=17 so T = 2(2) mod 17 = 4 mod 17. > Here T=21 does not work, but T = 55 = 5(11), so z = 8 and the answer > then from > 3k = 2(8) mod 17, is k = 11 mod 17. > Is there any use for such a technique? > James Harris /* notes on JSH's probabilistic square root mod p */ Given prime p and quadratic residue q modulo p, > we seek k s.t. k^2 = q mod p. JSH claims that the following procedure has 50% > chance of success: Choose odd multiplier n to get T = 2q + np. Factor T: f 1 * f 2 = 2q + np [Here I simplify JSH's formulation, in that he > refers to the sum of factors f 1 + f 2 as 2z.] Define z = f 1 + f 2 and k = z/3 mod p. What has this to do with a square root of q? > Well, for k^2 = q mod p, it would require: k^2 = (f 1^2 + 2 f 1*f 2 + f 2^2)/9 mod p æ æ = (f 1^2 + 4 q + f 2^2)/9 mod p 5q = f 1^2 + f 2^2 mod p A priori I can see no reason this should be > likely. /* Example: æFind k^2 = 3 mod 73 */ Since 3 is a quadratic residue mod 73 if and only if 73 is a quadratic > residue mod 3 (by quadratic reciprocity), and 73 = 1 mod 3, then 3 is > a quadratic residue mod 73. æAlso, 5q = 15. Now taking: n = 1: T = 6 + 73 = 79 prime æ(1 + 79^2) mod 73 = 37 n = 3: T = 6 + 219 = 225 = 3*75 = 5*45 = 9*25 = 15 * 15 æ(1 + 225^2) mod 73 = 37 > æ(9 + 75^2) mod 73 = 13 > æ(25 + 45^2) mod 73 = 6 > æ(81 + 25^2) mod 73 = 49 > æ 450 mod 73 = 12 n = 5: T = 6 + 365 = 371 = 7 * 53 æ(1 + 371^2) mod 73 = 37 > æ(49 + 53^2) mod 73 = 11 n = 7: T = 6 + 511 = 517 = 11 * 47 æ(1 + 517^2) mod 73 = 37 > æ(11^2 + 47^2) mod 73 = 67 n = 9: T = 6 + 657 = 663 = 3*13*17 æ(1 + 663^2) mod 73 = 37 > æ(9 + 221^2) mod 73 = 13 > æ(13^2 + 51^2) mod 73 = 69 > æ(17^2 + 39^2) mod 73 = 58 n = 11: T = 6 + 803 = 809 prime æ(1 + 809^2) mod 73 = 37 None of the smallish multipliers n = 1 to 11 > gives a factorization T = f 1 * f 2 that > produces k such that k^2 = 3 mod 73. > k = 21, so maybe k has to be coprime to q? But if so, why? Did you try that at random or find it? Searching for a case that wouldn't work? JSH === Subject: Re: Calculus doesn't apply to the unchanging curve > On Jun 21, 2:23 pm, The World Wide Wade The circle is an unchanging curve. It has no derivative. > Mitch Raemsch > It's becoming clear you don't know what a derivative is. > There is no changing slope for a circle. > Mitch Raemsch There are changing slopes when changing tangent lines to a circle, and > while such lines have slopes, circles don't. Now you've encouraged the crank. Tut tut. === Subject: Re: Calculus doesn't apply to the unchanging curve posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/20070622 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) One moment you are talking about a curve, the other about a point. Which is it? === Subject: Re: Calculus doesn't apply to the unchanging curve posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/20070622 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) I don't give a crap. > A zero dimensional object has no slope or rate of change of any kind > therefore no derivative. > But a CURVE is NOT zero-dimensional. Tell me why you think it is. === Subject: Re: Calculus doesn't apply to the unchanging curve posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/20070622 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) >On Jun 25, 12:47 pm, The World Wide Wade So your definition of curvature is change of curvature? >Yes calculus is about the change. > C'mon BURT, answer my question: Have you ever had a calculus course? > Possible answers are yes or no. > --Lynn Figure it out for yourself Lynn. Calculus cannot handle the unchanging > curve. Calculus has been called the science of change. It is that > simple. > Yes it can. Slope of hemicircle with respect to the x-axis: f(x) = sqrt(r^2 - x^2) f'(x) = slope(x) = -x/sqrt(r^2 - x^2). Debunk. The definition of derivative, or slope as you call it, does not depend on curvature. In fact curvature is defined USING the derivative. === Subject: Re: Calculus doesn't apply to the unchanging curve <316468.1214252937402.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/20070622 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > mitch: > something of zero dimension cannot have slope. > Which is kind of like saying that a dog is not a cat. > What are you trying to say, and why should anyone be > interested? > Tom A circle or any round geometry is an unchanging curve and therefore > can't be treated by calculus. > ??? What, you mean that calculus only applies when the curve is morphing over time? Regardless, how does this have to do with zero dimension? A circle is 1-dimensional or 2-dimensional depending on how you look at it (as from the view of someone on it vs the view of it as a plane figure). Nobody would call it 0-dimensional, though. === Subject: Re: Calculus doesn't apply to the unchanging curve <12192596.1214052186086.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/20070622 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > There are changing slopes when changing tangent lines to a circle, and > while such lines have slopes, circles don't.- Hide quoted text - > - Show quoted text - >Mitch Raemsch > Not! The slope of a curve at a point a is *defined* as f'(a), > presuming it exists. Circles and other curves have slopes except at > points where the tangent line is vertical. > --Lynn > In my day, curves could have derivatives at some points, but they were > slopes of the appropriate tangent lines , not slopes of the curves > themselves. > When did that change?- Hide quoted text - > - Show quoted text - Slope of the tangent line must be from two points. All slopes require > two points. The closer together the more accurate. Unfortunatly you > cannot get two points infinitely close without an infinitude of > calculations. Therefore the answer you get is always an approximation. > Who says you need an infinite number of calculations to get the points infinitely close together? You don't need billions of calculations to get them 1 billionth of a unit apart, do you? > There can be no slope for a single 0 dimensionaal point because there > is no change in height to mathematically measure. > === Subject: Re: Calculus doesn't apply to the unchanging curve <12192596.1214052186086.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/20070622 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > On Jun 21, 2:23 pm, The World Wide Wade The circle is an unchanging curve. It has no derivative. > Mitch Raemsch > It's becoming clear you don't know what a derivative is. There is no changing slope for a circle. > Yes there is. Slope is measured relative to a line. If I draw a line through the center of a circle, for example, then one can measure the slope of the circle at various points relative to the line. Are you telling me a dome (half-circle with respect to the ground) does not have a variable slope? === Subject: Re: Calculus doesn't apply to the unchanging curve posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/20070622 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > On Jun 18, 12:49 pm, The World Wide Wade A circle; a sphere; an hypersphere are all round unchanging curvature. > Calculus only deals with a changing curve. > Mitch Raemsch > Wrong. Next ... Changing curvature is the domain of Calculus. > Changing NUMBERS are the domain of Calculus. CURVING curves (i.e. anything not a straight line -- circles included) are the domain of Calculus. A circle is round, not straight. === Subject: Re: Calculus doesn't apply to the unchanging curve posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > On Jun 18, 12:49 pm, The World Wide Wade A circle; a sphere; an hypersphere are all round unchanging curvature. > Calculus only deals with a changing curve. > Mitch Raemsch > Wrong. Next ... > Changing curvature is the domain of Calculus. Changing NUMBERS are the domain of Calculus. CURVING curves (i.e. anything not a straight line -- circles > included) are the domain of Calculus. A circle is round, not > straight.- Hide quoted text - - Show quoted text - You are stupid for saying that calculus is not about change in curves. I don't think you know what calculus is from that comment. Mitch Raemsch === Subject: Re: Calculus doesn't apply to the unchanging curve posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/20070622 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > On Jun 18, 12:49 pm, The World Wide Wade A circle; a sphere; an hypersphere are all round unchanging curvature. > Calculus only deals with a changing curve. > Mitch Raemsch > Wrong. Next ... > Changing curvature is the domain of Calculus. Changing NUMBERS are the domain of Calculus. CURVING curves (i.e. anything not a straight line -- circles > included) are the domain of Calculus. A circle is round, not > straight. And straight lines *too*, but curves are where it's really more useful. === Subject: Re: Calculus doesn't apply to the unchanging curve posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) >On Jun 25, 12:47æpm, The World Wide Wade So your definition of curvature is change of curvature? >Yes calculus is about the change. > C'mon BURT, answer my question: Have you ever had a calculus course? > Possible answers are yes or no. > --Lynn >Figure it out for yourself Lynn. Calculus cannot handle the unchanging >curve. Calculus has been called the science of change. It is that >simple. Still won't answer, eh? OK, here's what I figure: Possibility 1. You have had calculus and obviously it went right over > you head. So in this case you are pontificating about a subject you > don't understand. Possibility 2. You haven't had a calculus course. So in this case also > you are pontificating about a subject you don't understand. Conclusion: You don't know what you are talking about. >Go look at your sheepskins Lynn. My sheepskins say B.S., M.S, and Ph.D. all in math. What do yours say? --Lynn- Hide quoted text - - Show quoted text - Put those sheepskins on the wall so you can remind yourself lynn. === Subject: Re: Calculus doesn't apply to the unchanging curve >On Jun 25, 12:47æpm, The World Wide Wade So your definition of curvature is change of curvature? >Yes calculus is about the change. > C'mon BURT, answer my question: Have you ever had a calculus course? > Possible answers are yes or no. > --Lynn >Figure it out for yourself Lynn. Calculus cannot handle the unchanging >curve. Calculus has been called the science of change. It is that >simple. > Still won't answer, eh? OK, here's what I figure: > Possibility 1. You have had calculus and obviously it went right over > you head. So in this case you are pontificating about a subject you > don't understand. > Possibility 2. You haven't had a calculus course. So in this case also > you are pontificating about a subject you don't understand. > Conclusion: You don't know what you are talking about. >Go look at your sheepskins Lynn. > My sheepskins say B.S., M.S, and Ph.D. all in math. What do yours say? > --Lynn >Put those sheepskins on the wall so you can remind yourself lynn. OK BURT. I may be a bit slow but it is becoming more and more apparent from your replies that you must be about 13 years old and are practicing the art of trolling. You are doing pretty well at it too, given that this thread currently has 49 entries. Keep up the good work. Bye bye. --Lynn I hope to die peacefully in my sleep, like Grandpa did. Not screaming in terror like his passengers. === Subject: Re: Calculus doesn't apply to the unchanging curve posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/20070622 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) >On Jun 25, 12:47 pm, The World Wide Wade So your definition of curvature is change of curvature? >Yes calculus is about the change. > C'mon BURT, answer my question: Have you ever had a calculus course? > Possible answers are yes or no. > --Lynn >Figure it out for yourself Lynn. Calculus cannot handle the unchanging >curve. Calculus has been called the science of change. It is that >simple. > Still won't answer, eh? OK, here's what I figure: > Possibility 1. You have had calculus and obviously it went right over > you head. So in this case you are pontificating about a subject you > don't understand. > Possibility 2. You haven't had a calculus course. So in this case also > you are pontificating about a subject you don't understand. > Conclusion: You don't know what you are talking about. >Go look at your sheepskins Lynn. > My sheepskins say B.S., M.S, and Ph.D. all in math. What do yours say? > --Lynn- Hide quoted text - > - Show quoted text - Put those sheepskins on the wall so you can remind yourself lynn. Take this egofighting crap somewhere else. === Subject: Re: Calculus doesn't apply to the unchanging curve > So your definition of curvature is change of curvature?- Hide quoted text > - > > - Show quoted text - Yes calculus is about the change. Using the definition of curvature you have given, how do you compute the curvature of the parabola y = x^2 at the point (1,1)? === Subject: Number with fermat prime 2^(2^n) + 1 Hello teacher~ Let F_n = 2^(2^n) + 1. (n = 0, 1, 2, 3, ...) Show that (F_m, F_n) = 1 for m =/= n. ------------------------------------------------------- Lemma) (F_0)(F_1)(F_2)...(F_(n-1)) = (F_n) - 2 , (n >= 1) pf) F_0 = 3, F_1 = 5 so, (F_0) = (F_1) - 2. Suppose that (F_0)(F_1)(F_2)...(F_(n-1)) = (F_n) - 2 is true. (F_0)(F_1)(F_2)...(F_(n-1))(F_n) = {(F_n) - 2}(F_n) = {2^(2^n) - 1}{2^(2^n) + 1} = 2^(2^(n+1)) - 1 = (F_(n+1)) - 2 -------------------------------------------------------- Let n > m , d = (F_m, F_n). Since (F_0)(F_1)(F_2)...(F_m)...(F_(n-1)) = (F_n) - 2 , so, d | 2. so, d = 1 or 2. Since F_n is odd, d = 1. === Subject: I TELL YOU THE 8-TH WONDERES OF THE WORLD . YOU JUST CLICK AND JOIN HERE. YOU SEE WHAT WILL HAPPAN IN YOUR LIFE. WWW.ONLINESECRETS1000.BLOGSPOT.COM posting-account=6zQagQoAAABni3rZnT9qJkW_QauhBHbj I TELL YOU THE 8-TH WONDERES OF THE WORLD . YOU JUST CLICK AND JOIN HERE. YOU SEE WHAT WILL HAPPAN IN YOUR LIFE. WWW.ONLINESECRETS1000.BLOGSPOT.COM === Subject: Rubik's Cube group with maximum order 1980? posting-account=G-HjYAoAAAAXYcAmzajOfGn3wg7eU7rl Gecko/20080404 Iceweasel/2.0.0.14 (Debian-2.0.0.14-2),gzip(gfe),gzip(gfe) It's easy to know that by applying a specific sequence of moves (composite of L,R,F,B,U,D and their inverse) repeatedly, the rubik's cube will return to its original configuration, for example, if we apply RUR'U' to the start configuration six times, the cube will return to the start configuration. So I was confused because some people around me said that all possible sequences of moves will return in less than or equal to 1980 times. I guess that he means that all elements of the cube group have their orders not more than 1980, is that right? Do we have any method to prove this? === Subject: Re: Rubik's Cube group with maximum order 1980? > It's easy to know that by applying a specific sequence of moves > (composite of L,R,F,B,U,D and their inverse) repeatedly, the rubik's > cube will return to its original configuration, for example, if we > apply RUR'U' to the start configuration six times, the cube will > return to the start configuration. > So I was confused because some people around me said that all > possible sequences of moves will return in less than or equal to 1980 > times. I guess that he means that all elements of the cube group > have their orders not more than 1980, is that right? Do we have any > method to prove this? According to Christoph Bandelow, Inside Rubik's Cube and Beyond, page 51, the maximum order is 1260. But no proof is given. David Singmaster, Notes on Rubik's Magic Cube, agrees with the 1260 number, and discusses it in some detail, especially pages 50-51. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Rubik's Cube group with maximum order 1980? posting-account=-PngCgkAAAD2yUjosqWv1Nf1lkqWP4lp rv:1.8.1.14) Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) On 27 Jun, 07:49, Gerry Myerson It's easy to know that by applying a specific sequence of moves > (composite of L,R,F,B,U,D and their inverse) repeatedly, the rubik's > cube will return to its original configuration, for example, if we > apply RUR'U' to the start configuration six times, the cube will > return to the start configuration. > So I was confused because some people around me said that all > possible sequences of moves will return in less than or equal to 1980 > times. I guess that he means that all elements of the cube group > have their orders not more than 1980, is that right? Do we have any > method to prove this? According to Christoph Bandelow, Inside Rubik's Cube and Beyond, > page 51, the maximum order is 1260. But no proof is given. David Singmaster, Notes on Rubik's Magic Cube, agrees with > the 1260 number, and discusses it in some detail, especially > pages 50-51. -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email) It is straightforward to compute the conjugacy classes using GAP or Magma. The highest order of an element is indeed 1260. The LCM of the orders is 55400, so every element raised to that power gives the identity. Derek Holt. === Subject: Re: Rubik's Cube group with maximum order 1980? posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008061712 Fedora/3.0-1.fc9 Firefox/3.0,gzip(gfe),gzip(gfe) > On 27 Jun, 07:49, Gerry Myerson It's easy to know that by applying a specific sequence of moves > (composite of L,R,F,B,U,D and their inverse) repeatedly, the rubik's > cube will return to its original configuration, for example, if we > apply RUR'U' to the start configuration six times, the cube will > return to the start configuration. > æ æ So I was confused because some people around me said that all > possible sequences of moves will return in less than or equal to 1980 > times. I guess that he means that all elements of the cube group > have æ their orders not more than 1980, is that right? Do we have any > method to prove this? > According to Christoph Bandelow, Inside Rubik's Cube and Beyond, > page 51, the maximum order is 1260. But no proof is given. > David Singmaster, Notes on Rubik's Magic Cube, agrees with > the 1260 number, and discusses it in some detail, especially > pages 50-51. > -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email) It is straightforward to compute the conjugacy classes using GAP or > Magma. The highest order of an element is indeed 1260. The LCM of the orders is 55400, so every element raised to that power > gives the identity. Derek Holt. Do you have the `canonical' permutation representation of the group at hand? I'd save me some typing ;-) ... -- m === Subject: Re: Rubik's Cube group with maximum order 1980? posting-account=-PngCgkAAAD2yUjosqWv1Nf1lkqWP4lp Gecko/20080410 SUSE/2.0.0.14-3.1 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) On 27 Jun, 13:32, Mariano Su.87rez-Alvarez > On 27 Jun, 07:49, Gerry Myerson It's easy to know that by applying a specific sequence of moves > (composite of L,R,F,B,U,D and their inverse) repeatedly, the rubik's > cube will return to its original configuration, for example, if we > apply RUR'U' to the start configuration six times, the cube will > return to the start configuration. > So I was confused because some people around me said that all > possible sequences of moves will return in less than or equal to 1980 > times. I guess that he means that all elements of the cube group > have their orders not more than 1980, is that right? Do we have any > method to prove this? > According to Christoph Bandelow, Inside Rubik's Cube and Beyond, > page 51, the maximum order is 1260. But no proof is given. > David Singmaster, Notes on Rubik's Magic Cube, agrees with > the 1260 number, and discusses it in some detail, especially > pages 50-51. > -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email) > It is straightforward to compute the conjugacy classes using GAP or > Magma. The highest order of an element is indeed 1260. > The LCM of the orders is 55400, so every element raised to that power > gives the identity. > Derek Holt. Do you have the `canonical' permutation representation > of the group at hand? I'd save me some typing ;-) ... -- m Here are the generators on 48 points, copied from http://www.gap-system.org/Doc/Examples/rubik.html ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19), ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35), (17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11), (25,27,32,30)(26,29,31,28)( 3,38,43,19)( 5,36,45,21)( 8,33,48,24), (33,35,40,38)(34,37,39,36)( 3, 9,46,32)( 2,12,47,29)( 1,14,48,27), (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)(16,24,32,40) Derek Holt. === Subject: Re: Rubik's Cube group with maximum order 1980? posting-account=G-HjYAoAAAAXYcAmzajOfGn3wg7eU7rl Gecko/20080404 Iceweasel/2.0.0.14 (Debian-2.0.0.14-2),gzip(gfe),gzip(gfe) On Jun 27, 2:49 pm, Gerry Myerson It's easy to know that by applying a specific sequence of moves > (composite of L,R,F,B,U,D and their inverse) repeatedly, the rubik's > cube will return to its original configuration, for example, if we > apply RUR'U' to the start configuration six times, the cube will > return to the start configuration. > So I was confused because some people around me said that all > possible sequences of moves will return in less than or equal to 1980 > times. I guess that he means that all elements of the cube group > have their orders not more than 1980, is that right? Do we have any > method to prove this? According to Christoph Bandelow, Inside Rubik's Cube and Beyond, > page 51, the maximum order is 1260. But no proof is given. David Singmaster, Notes on Rubik's Magic Cube, agrees with > the 1260 number, and discusses it in some detail, especially > pages 50-51. -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email) Well, I really appreciate your reply, and I have to admit that I did not see the two references with great significance you mentioned above. I searched extensively for the two books, but with no result. So if possible, could you tell me where I can find copies of them, I will be very grateful. === Subject: Re: Rubik's Cube group with maximum order 1980? posting-account=H1y7YgoAAADzGQwbcYaL9UvwttgsjOjp AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.1 Safari/525.18,gzip(gfe),gzip(gfe) > On Jun 27, 2:49 pm, Gerry Myerson According to Christoph Bandelow, Inside Rubik's Cube and Beyond, > page 51, the maximum order is 1260. But no proof is given. > David Singmaster, Notes on Rubik's Magic Cube, agrees with > the 1260 number, and discusses it in some detail, especially > pages 50-51. Well, I really appreciate your reply, and I have to admit that I did > not see the two > references with great significance you mentioned above. I searched > extensively for the > two books, but with no result. So if possible, could you tell me where > I can find copies > of them, I will be very grateful. http://www.abebooks.com/servlet/SearchResults?kn=rubik+singmaster&x=58&y=10 -- GM === Subject: Number with expression. Hello teacher~ When a = b+c = d+e+f , Express to form of a with a(ad + bc + df) + be(c + f) + (d + e)(f^2) + ce(c + f) + f^3. -------------------------------------------------------- is this possible ? a(ad + bc + df) + ae(c + f) + a(f^2) = a(ad + bc + df + ce + ef + f^2) = a(ad + bc + ce + af) = ?? === Subject: Re: Number with expression. posting-account=ogMREwkAAAC5xUr8sg7heGtsvzzF18LA Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > Hello teacher~ When æa = b+c = d+e+f , Express to form of a with > a(ad + bc + df) + be(c + f) + (d + e)(f^2) + ce(c + f) + f^3. -------------------------------------------------------- > is this possible ? a(ad + bc + df) + ae(c + f) + a(f^2) > = a(ad + bc + df + ce + ef + f^2) > = a(ad + bc + ce + af) > = ?? I do not see a way to simplify this further. I suggest you check the original problem (if you got it from a text), as it seems that if the original expressions was a(ad + be + df) +... (note the change in the second term in the parentheses), then this would simplify all the way to a^3. If the problem is correct as originally stated, I would suggest maybe adding and subtracting abe to your final statement, to arrive at a^3 + ab(c + e), which may be a more useful form. === Subject: Re: Number with expression. > Hello teacher~ > When a = b+c = d+e+f , > Express to form of a with > a(ad + bc + df) + be(c + f) + (d + e)(f^2) + ce(c + f) + f^3. > -------------------------------------------------------- > is this possible ? > a(ad + bc + df) + ae(c + f) + a(f^2) > = a(ad + bc + df + ce + ef + f^2) > = a(ad + bc + ce + af) > = ?? I do not see a way to simplify this further. I suggest you check the > original problem (if you got it from a text), as it seems that if the > original expressions was a(ad + be + df) +... (note the change in the > second term in the parentheses), then this would simplify all the way > to a^3. > If the problem is correct as originally stated, I would suggest maybe > adding and subtracting abe to your final statement, to arrive at a^3 + > ab(c + e), which may be a more useful form. a^3 + ab(c - e)... Maybe, it looks like wrong problem. === Subject: IIT JEE Previous Years Solved Papers By Different Authors posting-account=fdpItwoAAACflDymrrJ4xDjHeMFbRGop Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) 2007 : Paper1Eng.pdf http://rapidshare.com/files/25108467/Paper1Eng.pdf Paper2Eng.pdf http://rapidshare.com/files/25108946/Paper2Eng.pdf 2008 : PaperU11.pdf http://rapidshare.com/files/107322602/PaperU11.pdf Paper-2_final.pdf http://rapidshare.com/files/107322962/Paper-2_final.pdf === Subject: Limit point, open, closed posting-account=hbAh5QoAAACK7L2mtLy1aRgxLvWNQZe1 Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) Let E be a set. Prove that if E' = {x : x is a limit point of E}, E' is closed. === Subject: Re: Limit point, open, closed > Let E be a set. Prove that if E' = {x : x is a limit point of E}, E' > is closed. Nice game: you have to find the proof and ask me to do it for you. So, in return, I will ask you to prove that the complement of E' is open. Good trick, right? After that I have my proof and in consequence you've got your's. A fair game. Rainer === Subject: Re: Limit point, open, closed > Let E be a set. Prove that if E' = {x : x is a limit point of E}, E' > is closed. Maybe, T_1 is necessary. I already proved before. === Subject: Re: MinMax Technology <6df4e$48634746$82a1e228$19253@news1.tudelft.nl> <20080626030216.R16222@agora.rdrop.com> <71f87$4863ac8c$82a1e228$22526@news1.tudelft.nl > Let X be a finite set of M reals x_i : X = {x_1,x_2, .. ,x_i, .. ,x_M} > Let Y be a finite set of N reals y_j : Y = {y_1,y_2, .. ,y_j, .. ,y_N} In order to determine whether X is a subset of Y, we could compare the > elements. Then X is a subset of Y, X <= Y , if and only if for each of > the x_i one of them is equal to y_j : > What if I first ordered each set from smallest to largest? How would you do that, with e.g. elements in the Euclidean plane ? > You were discussing linear orders, not I. I was suggesting however, that you extend your ponderings unto multi-dimensional logics. For what you ask, you'll want to linear order R^2 which can be done, not with the product order, but the lexicographical order. > The rest of your comment certainly hold water, and I'll come back to it > You're welcome. You've hit upon one of my favorite interests, namely order theory. It's exceeding more complex than what you'd expect from the simplicity of (partial) orders. I've been revising my notes upon ordering a partition of an ordered set and it's applications to lattices, ordered groups and lattice ordered groups. Maybe I'll get it together enough to present to the group. Again, I'll be wanting to know why it too is not a counter example to William's Metatheorm: Whatever math I dream up is already old hat. === Subject: Re: i am looking for a couple ALMOST impossible math problems-Alan B Fabian posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > i am a relatively new teacher and i want something to really > push my advanced math class. i have a couple amazingly in-touch > students that are ready for some advanced material far beyond > the rest. i wanted to see if anyone had some stuff for me > to give to them. I appreciate anything you have to offer. > this is an advanced high school class. Of the several thousand posts I've made to sci.math and a few other internet-archived groups in the past 9 years, several hundred of these discuss topics that would likely be of interest to some of your students. I've included a few of these posts below, but they only represent a very quick search of my posts for a handful of topics. Incidentally, if it's of any interest, for several years in the 1990s I taught at one of the top few math/science academies in the U.S. and some of the topics/ideas in my posts have actually been used in my classes (often as extra credit assignments or supplementary work for near-olympiad level to olympiad level students). Inspired mathematical induction examples http://mathforum.org/kb/thread.jspa?messageID=5965460 http://mathforum.org/kb/message.jspa?messageID=5966861 http://mathforum.org/kb/message.jspa?messageID=5968462 Rationalizing 1 / [ sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56) ] Which is larger, e^pi or pi^e? http://mathforum.org/kb/message.jspa?messageID=6016525 BIG NUMBERS #1, #2, #3 A paradox in differentiating x^2 = x+x+...+x (x-times) http://mathforum.org/kb/message.jspa?messageID=676045 http://mathforum.org/kb/message.jspa?messageID=5219431 Derivative of rational powers of x using limit definition http://mathforum.org/kb/message.jspa?messageID=6017409 A lot of examples, remarks, and references for L'Hopital's rule http://tinyurl.com/2comud Limits of the form infinity - infinity http://mathforum.org/kb/message.jspa?messageID=5928321 The graph of x^2 + (x^2 + 1)(y^2 - 10 - x)^2 = 100 http://mathforum.org/kb/message.jspa?messageID=5846292 Implicit differentiation and evolute of a curve http://mathforum.org/kb/message.jspa?messageID=6128959 Estimates of partial sums of harmonic series http://mathforum.org/kb/message.jspa?messageID=4725124 Examples of finding ODEs for families of curves Number of vertices, edges, 2-faces, 3-faces, etc. of an n-cube Various associativity relations (examples, proofs, references) Dave L. Renfro === Subject: Re: i am looking for a couple ALMOST impossible math problems-Alan B Fabian posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) [omitted] would serve for strong students in a trigonometry class: The approximation sin(x) = x - (1/6)x^3 http://mathforum.org/kb/message.jspa?messageID=6189492 Dave L. Renfro === Subject: Re: i am looking for a couple ALMOST impossible math problems-Alan B Fabian <31448564.1214504950498.JavaMail.jakarta@nitrogen.mathforum.org>, > i am a relatively new teacher and i want something to really push my advanced > math class. i have a couple amazingly in-touch students that are ready for > some advanced material far beyond the rest. i wanted to see if anyone had > some stuff for me to give to them. > I appreciate anything you have to offer. > this is an advanced high school class. Here are some books you might find useful: Hungarian Problem Book I and Hungarian Problem Book II. Don't worry, they're in English, translated by Elvira Rapaport. Originally published by Singer, I'm pretty sure they've been republished by the MAA. The William Lowell Putnam Mathematical Competition. There are three books in this series. The 2nd was something of a cut'n'paste job, the other two are more substantial volumes. The Inquisitive Problem Solver, by Vaderlind, Guy, and Larson. Two books by Peter Winkler: Mathematical Puzzles and Mathematical Mind-Benders. Halmos, Problems for Mathematicians Young and Old. Kornhauser, Velleman, and Wagon, Which Way did the Bicycle Go? Steven Barr has written some nice books of puzzles, as has Ross Honsberger. Then there's always Dudeney, and Sam Loyd. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: i am looking for a couple ALMOST impossible math problems-Alan B Fabian >i am a relatively new teacher and i want something to really push > my advanced math class. i have a couple amazingly in-touch > students that are ready for some advanced material far beyond > the rest. i wanted to see if anyone had some stuff for me to give > to them. I appreciate anything you have to offer. this is an > advanced high school class. Google will work for you. Or you can start here: http://mathforum.org Or here: http://www.artofproblemsolving.com/ === Subject: Re: i am looking for a couple ALMOST impossible math problems-Alan B Fabian >i am a relatively new teacher and i want something to really push > my advanced math class. i have a couple amazingly in-touch > students that are ready for some advanced material far beyond > the rest. i wanted to see if anyone had some stuff for me to give > to them. I appreciate anything you have to offer. this is an > advanced high school class. Google will work for you. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Topology with boundary f^-1(B) > Hello teacher~ Let f : X -> Y. cl[f^{-1}(B)] subset f^{-1}[cl(B)] for all B subset Y. >== b[f^{-1}(B)] subset f^{-1}[b(B)] for all B subset Y. May as well get it all done at once. Notes on Continuity f continuous when for all open U, f^-1(U) open iff for all base sets U, f^-1(U) open iff for all subbase sets U, f^-1(U) open iff for all closed K, f^-1(K) closed That's the basic stuff on continuity. In addition is the useful f continuous iff for all A, f(cl A) subset cl f(A) iff for all B, cl f^-1(B) subset f^-1(cl B) iff for all B, bd f^-1(B) subset f^-1(bd B) iff for all B, f^-1(int B) subset int f^-1(B) The way I went about proving this was to show each implied the next and that the last implied the first, that f is continuous. As a stray though, one notices that the third and fifth are immediately equivalent by DeMorgan's laws of complementation, extended to a duality for topological spaces.d === Subject: Re: Topology with boundary f^-1(B) > Let f : X -> Y. > cl[f^{-1}(B)] subset f^{-1}[cl(B)] for all B subset Y. >==> b[f^{-1}(B)] subset f^{-1}[b(B)] for all B subset Y. It means that > f is continuous >== f^{-1} is open map. No. If f is not a bijection, then f^-1 is not a map from Y to X. OTOH, when f is a bijection, the conclusion is immediate. >== f[int(B)] subset int[f(B)] >== f[cl(B)] subset cl[f(B)] >== cl[f^{-1}(B)] subset f^{-1}[cl(B)] >== b[f^{-1}(B)] subset f^{-1}[b(B)] > === Subject: Re: Topology with boundary f^-1(B) Let f : X -> Y. cl[f^{-1}(B)] subset f^{-1}[cl(B)] for all B subset Y. >== b[f^{-1}(B)] subset f^{-1}[b(B)] for all B subset Y. > It means that > f is continuous >==> f^{-1} is open map. No. If f is not a bijection, then f^-1 is not a map from Y to X. > OTOH, when f is a bijection, the conclusion is immediate. Yes, you're right. Maybe, I need more proof. f : X -> Y is continuous <==> cl[f^{-1}(B)] subset f^{-1}[cl(B)] for B subset Y. pf) (==>) Since f is continuous, f^{-1}[cl(B)] is closed. Since f^{-1}(B) subset f^{-1}[cl(B)], and cl[f^{-1}(B)] is the intersection of all closed sets containing f^{-1}(B), so, cl[f^{-1}(B)] subset f^{-1}[cl(B)]. (<==) Let B be a closed set of Y. I must show that f^{-1}(B) is closed in X. cl[f^{-1}(B)] subset f^{-1}[cl(B)] = f^{-1}(B) so, cl[f^{-1}(B)] = f^{-1}(B). so, f^{-1}(B) is closed in X. -------------------------------------------------------------- f : X -> Y is continuous <==> f^{-1}(int B) subset int[f^{-1}(B)] for B subset Y. pf) (==>) Since f is continuous, f^{-1}(int B) is open in X. Since f^{-1}(int B) subset f^{-1}(B), and int[f^{-1}(B)] is the union of all open sets contained in f^{-1}(B). so, f^{-1}(int B) subset int[f^{-1}(B)]. (<==) Let B be a open set of Y. I must show that f^{-1}(B) is open in X. f^{-1}(B) = f^{-1}(int B) subset int[f^{-1}(B)]. so, f^{-1}(B) = int[f^{-1}(B)] so, f^{-1}(B) is open in X. --------------------------------------------------------------- It means that f : X -> Y is continuous <==> f[cl(A)] subset cl[f(A)] for A subset X <==> cl[f^{-1}(B)] subset f^{-1}[cl(B)] for B subset Y. <==> f^{-1}(int B) subset int[f^{-1}(B)] <==> b[f^{-1}(B)] subset f^{-1}[b(B)] For reference, Let f : X -> Y. f is open map <==> f(int A) subset int[f(A)] for A subset X. f is closed map <==> cl[f(A)] subset f[cl(A)] for A subset X. === Subject: Re: How many periodic functions are there? posting-account=SZpKQgoAAAAQ5dLxToYzarELzrJFwm0a Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > Not quite, but close. >And where can I read about this? > The magic words you are looking for are Fourier series and Fourier > analysis. >So, how close did I get? >I've looked up Fourier series on wikipedia (I know it's not 100 >percent reliable, >just for start). It is saying In mathematics, a Fourier series >decomposes a periodic >function into a sum of simple oscillating functions, namely sines and >cosines. >Well, a Fourier series decomposes a periodic function into a sum of >sines and cosines. >Question: Is this statement true for *every* periodic function? >I guess that's my initial question. > There are certain conditions on the function; not just any function > will do (you can have functions that are discontinuous at every point > and periodic; generally, those will not work), but the conditions > are generally considered mild, in that most 'nice' periodic > functions will work. Also, the sum is in fact an infinite sum; that > means that the function is ->approximated<- by a sum of sines and > cosines, and adding more terms yields closer approximations. I believe > the convergence need not be uniform (so you cannot in general ensure > that a partial sum will be epsilon-close to the original function at > all points, but for each point you can ensure that a certain partial > sum will be epsilon close at that point). Actually the Fourier series for a continuous function need > not even converge pointwise (to the original function or to > anything else)! It's uniformly summable to the original > function by any of various summability methods. And if f > is just a little better than continuous, for example if there > exist c < infinity and a > 0 such that |f(x) - f(y)| <= c |x - y|^a then the Fourier series of f converges uniformly to f. So in fact you have uniform convergence for the continuous > functions that are likely to come up unless you're looking > for trouble, but strictly speaking what you said isn't so. -- > David C. Ullrich Another one of my naive (and silly) questions: What is the motivation of writing a, say continuous, function as Fourier series? What is the benefit? === Subject: Re: How many periodic functions are there? posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > Another one of my naive (and silly) questions: > What is the motivation of writing a, say continuous, > function as Fourier series? What is the benefit? Expansions in powers of x had been widely used in the 1600s and 1700s (binomial series for fractional exponents, trig. functions, etc.). However, Fourier (between 1805 to 1822) found himself dealing with multiple-angle sines and/or cosine expansions of functions for a method he developed to solve the heat equation (an equation he derived from physical and mathematical principles). For Fourier's actual work, see the English translation of http://www.archive.org/details/analyticaltheory00fourrich [53 MB .pdf file] In particular, glance over Sections 169-235 on pp. 135-269. Dave L. Renfro === Subject: Re: let's clear that up > One Universe equals two YinYnag ???? I think you have misunderstood the concept. yin yang is a relationship not two things. This relationship can not be put into words so easily, if at all, so it is represented by it's polar opposites. Sort of like having a snazzy graphic of +1 and -1 representing the infinite points between them through zero. === Subject: Re: let's clear that up > ssbh.jpg) Galaxies: 1st Cold and Dark 2nd Warm and Bright or 1st Warm > and Dark and 2nd Cold and Bright depending on what was the angle > between the pro-anti heat axis with the balance-chaos brightness axis. > in this so called Parallel Multi-Universe with TATO VASE DOOM27 in the > center there are countless numbers of Universes. Another excluded middle. Even the Eastern thing is bi-polar. === Subject: Name this enumeration type What's the mathematical name of the enumeration type below? I think this is neither combination nor permutation, so then what is this one called? Generated by a call to a function f(v=5,k=3): (the first column is just the index starting at 0 and it isn't part of the generated rows). 0: 0 0 0 1: 0 0 1 2: 0 0 2 3: 0 0 3 4: 0 0 4 5: 0 1 1 6: 0 1 2 7: 0 1 3 8: 0 1 4 9: 0 2 2 10: 0 2 3 11: 0 2 4 12: 0 3 3 13: 0 3 4 14: 0 4 4 15: 1 1 1 16: 1 1 2 17: 1 1 3 18: 1 1 4 19: 1 2 2 20: 1 2 3 21: 1 2 4 22: 1 3 3 23: 1 3 4 24: 1 4 4 25: 2 2 2 26: 2 2 3 27: 2 2 4 28: 2 3 3 29: 2 3 4 30: 2 4 4 31: 3 3 3 32: 3 3 4 33: 3 4 4 34: 4 4 4 I'm looking for two computational methods: 1.) given the index create the corrosponding row 2.) given the row compute the corrosponding index TIA === Subject: Re: Name this enumeration type Another term might by partition ... Here, into 3 parts of size at most 4. http://mathworld.wolfram.com/Partition.html === Subject: Re: Name this enumeration type posting-account=ogMREwkAAAC5xUr8sg7heGtsvzzF18LA Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > What's the mathematical name of the enumeration type below? > I think this is neither combination nor permutation, > so then what is this one called? Generated by a call to a function f(v=5,k=3): > (the first column is just the index starting at 0 > and it isn't part of the generated rows). ? ?0: ? ? ?0 ?0 ?0 > ? ?1: ? ? ?0 ?0 ?1 > ? ?2: ? ? ?0 ?0 ?2 > ? ?3: ? ? ?0 ?0 ?3 > ? ?4: ? ? ?0 ?0 ?4 > ? ?5: ? ? ?0 ?1 ?1 > ? ?6: ? ? ?0 ?1 ?2 > ? ?7: ? ? ?0 ?1 ?3 > ? ?8: ? ? ?0 ?1 ?4 > ? ?9: ? ? ?0 ?2 ?2 > ? 10: ? ? ?0 ?2 ?3 > ? 11: ? ? ?0 ?2 ?4 > ? 12: ? ? ?0 ?3 ?3 > ? 13: ? ? ?0 ?3 ?4 > ? 14: ? ? ?0 ?4 ?4 > ? 15: ? ? ?1 ?1 ?1 > ? 16: ? ? ?1 ?1 ?2 > ? 17: ? ? ?1 ?1 ?3 > ? 18: ? ? ?1 ?1 ?4 > ? 19: ? ? ?1 ?2 ?2 > ? 20: ? ? ?1 ?2 ?3 > ? 21: ? ? ?1 ?2 ?4 > ? 22: ? ? ?1 ?3 ?3 > ? 23: ? ? ?1 ?3 ?4 > ? 24: ? ? ?1 ?4 ?4 > ? 25: ? ? ?2 ?2 ?2 > ? 26: ? ? ?2 ?2 ?3 > ? 27: ? ? ?2 ?2 ?4 > ? 28: ? ? ?2 ?3 ?3 > ? 29: ? ? ?2 ?3 ?4 > ? 30: ? ? ?2 ?4 ?4 > ? 31: ? ? ?3 ?3 ?3 > ? 32: ? ? ?3 ?3 ?4 > ? 33: ? ? ?3 ?4 ?4 > ? 34: ? ? ?4 ?4 ?4 I'm looking for two computational methods: > 1.) given the index create the corrosponding row > 2.) given the row compute the corrosponding index TIA This appears to be a listing of what is sometimes called the multisets of size three, from a set of 5 elements. A multiset is a set where a repeated element is counted every time it appears. For illustrative purposes: the SET {3,4,4} is equal to the SET {3,4}, both have cardinality 2 However, the MULTISET {3,4,4} is not equal to the MULTISET {3,4}, the latter being a proper sub-multiset of the former. The number of multisets of size k from a set of size v (using the parameter names from your example) is equivalent to the number of ways to place k indistinguishable balls into v distinguishable urns. Off-hand I don't know of an indexing function for a general (v,k)- which is what you are looking for- but if you try to write a simple pseudocode program to list them, that might give you an idea. === Subject: Re: Primes of the form n!+1 > For what n is n!+1 a prime? How many primes of the form n!+1 are there? > Perhaps would be of > interest. > Although there is no proof, we expect that there are infinitely many > because sum(1/log(n!+1)) diverges. > Why does that matter? > Prime Number Theorem. The probability that a randomly chosen integer > near X, for large X, is prime is 1/log(X). >Has it ever ocured that sum{1/log(f(x))} diverges and that are only >finitely many number x such that f(x) is a prime? >Probabalistically, if f(n) is random and not constrained in some way >that affects the probability of f(n) being prime, then the divergence >of > oo > --- 1 > > --------- > --- log(f(n)) > n=1 >and the converse of the Borel-Cantelli Lemma > infinitely many n. >This does not mean that it is true, just that it has a 100% chance of >being true. See > This last statement is nonsense. All one can say is that the current intuitive probability arguments, if they are correct in this case, would make the probability at 100%. -- 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: Primes of the form n!+1 > For what n is n!+1 a prime? How many primes of the form n!+1 are the= >re? > Perhaps would be= > of > interest. > Although there is no proof, we expect that there are infinitely many > because sum(1/log(n!+1)) diverges. > Why does that matter? >Prime Number Theorem. The probability that a randomly chosen integer >near X, for large X, is prime is 1/log(X). In fact, the intuitive estimate of the probability is much greater. Since the number is prime to 2, this gives and extra factor of 2, prime to 3, of 3/2, and so on for any finite number of small primes. If one can push this to the limit, my calculations say the probability of primality should be on the order of 1. Note: This is a probabilistic argument, and may or may not indicate what is happening. -- 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: Primes of the form n!+1 87tzfgxf27.fsf@nonospaz.fatphil.org> posting-account=a6woBRAAAADpNFZJBA7ZBx35zXaKmaP4 Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) On Jun 26, 1:48æpm, Phil Carmody For what n is n!+1 a prime? How many primes of the form n!+1 are there? > Perhaps would be of > interest. > Although there is no proof, we expect that there are infinitely many > because sum(1/log(n!+1)) diverges. > Why does that matter? > Prime Number Theorem. æThe probability that a randomly chosen integer > near X, for large X, æis prime is 1/log(X). > Has it ever ocured that sum{1/log(f(x))} diverges and that are only > finitely many number x such that f(x) is a prime? There are no numbers such that f(x) is prime, as f(x) is > undefined. Phil > -- > -- Microsoft voice recognition live demonstration I think he's asking if it is possible to define f(x) such that sum(1/log(f(x)) diverges (summming over the positive integers x), yet only finitely many integers x exists s.t. f(x) is prime. Of course the answer is yes, e.g. f(x) = 2x. --c === Subject: Re: Primes of the form n!+1 87tzfgxf27.fsf@nonospaz.fatphil.org> posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/20070622 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) On Jun 26, 12:02 pm, Robert Israel > For what n is n!+1 a prime? How many primes of the form n!+1 are > there? > Perhaps would > be of > interest. > Although there is no proof, we expect that there are infinitely many > because sum(1/log(n!+1)) diverges. > Why does that matter? > Prime Number Theorem. The probability that a randomly chosen integer > near X, for large X, is prime is 1/log(X). > Has it ever ocured that sum{1/log(f(x))} diverges and that are only > finitely many number x such that f(x) is a prime? > There are no numbers such that f(x) is prime, as f(x) is > undefined. My guess is that he was thinking of f(x) = x!. ^^^^^^^^^ Just a nitpick: f(x) = x! + 1, you mean, right? Since x! is _never_ prime except for x = 2. :) === Subject: Re: Primes of the form n!+1 > For what n is n!+1 a prime? How many primes of the form n!+1 are > there? > Perhaps would > be of > interest. > Although there is no proof, we expect that there are infinitely many > because sum(1/log(n!+1)) diverges. > Why does that matter? > Prime Number Theorem. The probability that a randomly chosen integer > near X, for large X, is prime is 1/log(X). > Has it ever ocured that sum{1/log(f(x))} diverges and that are only > finitely many number x such that f(x) is a prime? > There are no numbers such that f(x) is prime, as f(x) is > undefined. My guess is that he was thinking of f(x) = x!. Heck, if we're talking about heuristics, why not involve guessing too! Even I can do that kind of maths. > The answer is that the fact sum(1/log(n!+1)) diverges is only one part of > the reason we expect that there are infinitely many, the other part being > that we don't know any reason to expect n!+1 to be divisible by anything in > particular. Yup. However, if the saysero thinks that Bob Silverman would come up with the same argument about the number of primes of the form n!, then it's clear he knows both nothing about Bob Silverman, and nothing about how a reasonable person would construct a heuristic. Knowledge about unusual divisibility in the n!+1 case, however, isn't zero. We know it's not divisible by vast swathes of primes - every prime less than n. So Bob's heuristic is quantitatively pessimistic, by something like a loglog term in the numerator. However, it was of course a qualitative heuristic, not a quantitative one, so quantitative tweaks are irrelevant. Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: Primes of the form n!+1 87tzfgxf27.fsf@nonospaz.fatphil.org> posting-account=lHNboAoAAACyasQ0uqX7OeM_tLuWGoQp CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) On Jun 26, 2:02æpm, Robert Israel > For what n is n!+1 a prime? How many primes of the form n!+1 are > there? > Perhaps would > be of > interest. > Although there is no proof, we expect that there are infinitely many > because sum(1/log(n!+1)) diverges. > Why does that matter? > Prime Number Theorem. æThe probability that a randomly chosen integer > near X, for large X, æis prime is 1/log(X). > Has it ever ocured that sum{1/log(f(x))} diverges and that are only > finitely many number x such that f(x) is a prime? > There are no numbers such that f(x) is prime, as f(x) is > undefined. My guess is that he was thinking of f(x) = x!. The answer is that the fact sum(1/log(n!+1)) diverges is only one part of > the reason we expect that there are infinitely many, the other part being > that we don't know any reason to expect n!+1 to be divisible by anything in > particular. > -- > Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Hide quoted text - Almost. We do know that any divisor is greater than n...... However, as n-->oo the set of numbers that can't be a divisor obtains density 0 among all integers less than n! (I'm not telling you anything you don't know. This is for the more general audience) === Subject: Re: Primes of the form n!+1 > On Jun 26, 2:02æpm, Robert Israel > For what n is n!+1 a prime? How many primes of the form n!+1 are > there? > Perhaps would > be of > interest. > Although there is no proof, we expect that there are infinitely many > because sum(1/log(n!+1)) diverges. > Why does that matter? > Prime Number Theorem. æThe probability that a randomly chosen integer > near X, for large X, æis prime is 1/log(X). > Has it ever ocured that sum{1/log(f(x))} diverges and that are only > finitely many number x such that f(x) is a prime? > There are no numbers such that f(x) is prime, as f(x) is > undefined. > My guess is that he was thinking of f(x) = x!. > The answer is that the fact sum(1/log(n!+1)) diverges is only one part of > the reason we expect that there are infinitely many, the other part being > that we don't know any reason to expect n!+1 to be divisible by anything in > particular. Almost. We do know that any divisor is greater than n...... > However, as n-->oo the set of numbers that can't be a divisor > obtains density 0 among all integers less than n! Out-by-one in the negatives! (Or out-by-one in the density.) > (I'm not telling you anything you don't know. This is for the more > general audience) The general audience who are on the same page, or at least the same book, as you, Yes. A fair proportion, alas, haven't even entered the right wing of the library. Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: Primes of the form n!+1 87tzfgxf27.fsf@nonospaz.fatphil.org> <87ej6jykc7.fsf@nonospaz.fatphil.org> posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) On Jun 26, 4:09æpm, Phil Carmody On Jun 26, 2:02æpm, Robert Israel > For what n is n!+1 a prime? How many primes of the form n!+1 are > there? > Perhaps would > be of > interest. > Although there is no proof, we expect that there are infinitely many > because sum(1/log(n!+1)) diverges. > Why does that matter? > Prime Number Theorem. æThe probability that a randomly chosen integer > near X, for large X, æis prime is 1/log(X). > Has it ever ocured that sum{1/log(f(x))} diverges and that are only > finitely many number x such that f(x) is a prime? > There are no numbers such that f(x) is prime, as f(x) is > undefined. > My guess is that he was thinking of f(x) = x!. > The answer is that the fact sum(1/log(n!+1)) diverges is only one part of > the reason we expect that there are infinitely many, the other part being > that we don't know any reason to expect n!+1 to be divisible by anything in > particular. > Almost. æWe do know that any divisor is greater than n...... > However, as n-->oo the set of numbers that can't be a divisor > obtains density 0 among all integers less than n! Out-by-one in the negatives! (Or out-by-one in the density.) > (I'm not telling you anything you don't know. æThis is for the more > general audience) The general audience who are on the same page, or at least > the same book, as you, Yes. A fair proportion, alas, haven't > even entered the right wing of the library. Some of us aren't even in the right city. Phil > -- > -- Microsoft voice recognition live demonstration- Hide quoted text - - Show quoted text - === Subject: Re: Einstein s Lamentation That Special Relativity Was Not A Fundamental Theory > Right, but it's not an attempt to learn. What he's doing is > setting up an impossible standard for Greg, as if Greg is > wrong unless he can convince Ken. >As is the standard for many cranks here, as though physical >understanding were an entitlement rather than the result of hard work. > Can you imagine how demeaning it is to be called a crank by a self > righteous hypocrite like you? Obviously you're badly in need of some > elementary Canadian sensitivity training and anger management. Well, I'm not in the least angry, so I'm not sure what anger >management would offer in the present case. Is a pig's ass pork? Of course you're angry. You're hostile, bitter, and enraged that people don't take your opinions as the fact you proclaim but rather prefer to think for themselves. > As for sensitivity >training, you may have a point. I invite you to illustrate the proper >Canadian way of educating someone like Seto that I have characterized >as a crank. I look forward to following your model. Seto? Why would I care about your opinion of Seto? Why would I care about your opinions of anything. Am I supposed to do your thinking for you now? You're used to browbeating students into submission. You're just a spiteful, misguided, academic misanthrope. Canadians don't expect to succeed in life.They just learn to sing God Save the Queen and make the best of it. Europeans on the other hand worship futility. > Here's looking at me, Kid. YWH Thou shalt kill file all others before me. === Subject: Re: Einstein s Lamentation That Special Relativity Was Not A Fundamental Theory > What happpened to your Quint Essential name? > Nothing, Dutchess. > Any reason why you hide in anonimity? > No anonymity, Different personas, multiple personalities. Yes, anonymity, different personas, multiple personalities, >same coward. Hearken unto me, you stupid cow. You won't even define the terms you use yet I'm the coward? YWH Thou shalt kill file all others before me. === Subject: relationship between geometry of fractals and geometry of general relativity theory????????? posting-account=yMk5SQoAAADCmqm4d5kvJsQWXGMWgn4i I need a little guidance with regard to the mathematics of both fractals and relativity theory. It has been a couple decades since I opened a book on either, that that was out of curiosity as my training is in ecology, and I earned a living developing software. Hence a need for something of a refresher. Pointers to open source code. or pseudocode that I can readily transform into fortran or C++. that would be useful would be GREATLY appreciated (because I have no budget for this). Anyway, I am working on a project involving analysis of market data, and of course you're all aware of Mandelbrot's work on the misbehaviour of the markets. There are couple things I want to examine in the near term. 1) How to generate a (multi)fractal brownian motion? Generating a random walk is trivially easy. In C++, I'd use the mersenne twister generator, from the boost library, to produce the sequence of random values, and then proceed to produce the random walk (or use them in whatever model that needs random values). How would I modify this to produce a fractal random sequence or a multifractal random sequence? 2) As you undoubtedly know, there are a number of algorithms available for estimating the Hurst exponent. Is there a way to simulate a time series for which the true value for the Hurst exponent is known, and which can therefore be used to test the algorithms that are available to estimate it? I recall reading the estimation of the Hurst exponent is problematic because the algorithms available give different results. Is that true? If so, is there any chance that this is due to them actually estimating different, but possibly related, properties? 3) One of the things I found most interesting, when decades ago I was studying physics (as breadth courses out of curiousity), was relativity theory and the notion of the equivalence of frames of reference. On looking at finance data, it seems that idea, stripped of the physics, would be especially useful as prices of commodities and stocks and ETFs, &c. are defined in terms of currencies, and currencies themselves float relative to each other and how different national economies are performing relative to each other. Trying, therefore, to find meaning in the prices is a nightmare. Has there been any work done trying to relate the math of fractals to the math of relativity theory? If so, can you point me to web resources (or email me electronic copies of relevant papers: to my gmail account: r.ted.byers@gmail.com)? If not, can you provide some guidance on how best to proceed? It seems to me that there must be some relation between the mathematics you use when there is an equivalence between reference systems with neither absolute space nor absolute time, and the mathematics of fractals. I am just not sure how to go about finding it, and putting it into a context that I can use to actually analyse market data. Guidance on this would be greatly appreciated, as would contact information for anyone who is working on anything like this. Ted === Subject: Re: algebraic functions Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I think the following would helpful: 1. Can someone point me to an algorithm which decides whether a radical expression (nested roots and polynomials) is 0? 2. Where can I read about the compositional structure of polynomials: 2.1 For two polynomials p,q under which conditions are there polynomials r,s such that r o p = s o q or p o r = q o s? 2.2 Is there some kind of unique prime decomposition for polynomials regarding the composition operation? === Subject: MACAS 2 - Conference proceedings Content-Length: 279 Originator: bergv@math.uiuc.edu (Maarten Bergvelt) MACAS 2 - Conference proceedings are now available: Proceedings of the 2nd International Symposium of Mathematics and its Connections to the Arts and Sciences (MACAS 2), Centre for Science and Mathematics Education, University of Southern Denmark, Odense, Denmark. === Subject: Re: Limits and continuity >Can someone help me with this : Lim(n->inf) nSin(2(pi)(e)n!) I believe that limit does not exist. One of the guys tried to prove that the limit is 2(pi) by expanding e. >One of the guys tried to prove that the limit is 2(pi) by expanding e. Rohan, he was right. Let (e)n! = I + f, where I is the integer part and f is the fraction. Then sin(2(pi)(e)n!) = sin(2(pi)f). f = 1/(n+1) + 1/(n+1)(n+2) + ... So sin(2(pi)(e)n!) may be approximated by sin(2(pi)/(n+1)) which tends to 2(pi)/(n+1) in the limit. >My only doubt in this is that how do you prove that 1/(n+1) + 1/(n+1)(n+2) .... where this sequence itself contains infinite terms and has each term tending to 0, how is it that its limit is 0, because thats what the whole thing depends upon. If someone can do that more rigorously, it wud be great. >1/(n+1) + 1/(n+1)(n+2) .... < 1/n + 1/(n^2) + ... = (1/n)(1/(1-1/n)) = 1/(n-1) to which you are replying. Why not just 1 --- n+1 1 1 1 1 1 1 1 1 1 1 < --- + --- --- + --- --- --- + --- --- --- --- + ... n+1 n+1 n+2 n+1 n+2 n+3 n+1 n+2 n+3 n+4 1 1 1 1 1 1 1 1 1 1 < --- + --- --- + --- --- --- + --- --- --- --- + ... n+1 n+1 n+1 n+1 n+1 n+1 n+1 n+1 n+1 n+1 1 = - n The Sandwich Theorem says that 2pi 2pi lim n sin( --- ) <= lim n sin(2pi e n!) <= lim n sin( --- ) n->oo n+1 n->oo n->oo n Using lim_{x->0} sin(x)/x = 1, we get 2pi n n+1 2pi lim n sin( --- ) = 2pi lim --- --- sin( --- ) = 2 pi n->oo n+1 n->oo n+1 2pi n+1 and 2pi n 2pi lim n sin( --- ) = 2pi lim --- sin( --- ) = 2 pi n->oo n n->oo 2pi n Therefore lim n sin(2pi e n!) = 2 pi n->oo Rob Johnson take out the trash before replying === Subject: solution manual pdf /htmp/word posting-account=sUmyJwoAAADNSyRfupeIkvPE3_taOX2R CLR 1.1.4322),gzip(gfe),gzip(gfe) i want solutiona manual for integrated electronics ,by millman,halkias also electronic devices and circuits by millman,halkias .can any one send it for me.please to my mail :san2k7@gmail.com === Subject: Re: Programming skills required by mathematicians? <4b56a$486b4114$82a1e228$22581@news1.tudelft.nl> posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) >As for gotos, I find them easy to use, >and spaghetti code is best for spaghetti algorithms. > Sure, gotos are easy to write, but what's usually important > for code is that it be easy to read and easy to maintain, > and gotos are anathema to that. I think nowadays nobody, except Herman, will argue in favour of gotos. FWIW, I would. In fact, my idea is that the two poles, LLL and HLL, should finally split, where LLL is the only true programming possible, while HLL is to become the high-end side of user interfaces. In any case, there may be lots of reasons to favour gotos, all invariably coming from those who have been programming for a life, at least in the sense of getting back to the roots of our practices and problems, and cleaning up to an otherwise unreachable next phase. > On the other hand, I find that the Object Oriented Programming paradigm > has given rise to bad programming practices, such as putting more and > more software layers on top of each other Yep, that is an easy example. What has gone wrong there? -LV > , so many that even a GigaHerz > processor can hardly plough through them, in execution time. Han de Bruijn === Subject: Re: Programming skills required by mathematicians? posting-account=7VnrmgkAAADfdiWnpCJtxxF3tpMHF0Y0 1.0.3705; .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) >com>, æ æ æ æ æ æ æ æ æ æ æ æ ..................... > Computers have been used in number theory, algebraic geometry, > computational problems which are more abstract than concrete, > as well as more practical problems. =A0My technical reports, > available from the Purdue Statistics Department, are a combination > of an odd type of numerical analysis, and involve using Boolean > operations on the bit patter of floats, something which is not > in any HLL to my knowledge. > Efficient random number procedures can be mainly done in HLL, > but not with full efficiency. =A0And I doubt that anyone can > get a full accuracy efficient procedure to generate > f(x) =3D x - ln(1+x) to full machine precision without using > the hardware structure. =A0Also, on most machines, multiple > precision (I consider what is called double to be single) > arithmetic is a mess because of the forced normalization of > floats and the hidden bit. =A0The factor may be higher than 10. > This is partly due to the HLLs, which have discounted fixed > point arithmetic. >If you were an applied mathematician, I would recommend Fortran and >Algol only because there are many legacy programs written in these >languages. =A0In particular, the Collected Algorithms from ACM Vol I to >III are mostly in Algol and Fortran. > At that time, those were the only moderately well known HLLs. > Machine independence is fine, but one must take into account > the machine to get efficiency. >With modern compilers, one can easily mix a HLL with assembly >language. æFor example, Visual C++ object code can be disassembled >into assembly language, modified, and then recompiled back to object >code. This is not good with the usual assemblers. æOften, > the compiled code is optimized in such a way that > it is hard to follow. æOptimization is needed later, > not at compile time. Few current assembler languages are concise enough > for easy coding. æTyping 50 characters when 5 would > suffice leads to errors. æSomething like the Cray > assembler language with simplifications and overloading > is needed. >For mental exercise and development, I recommend learning Forth which >uses Reverse Polish Notation (RPN), i.e. operands precede operators, >as in reality. æForth and RPN foster efficient programming and >Forth can be modified to do anything because it is easily extensible. >Some modified versions directly use the PC FPU stack very efficiently. RPN is moderately easy to write, and is my choice for > a calculator. æI am less fond of the stack idea of > Forth for a computer. æI do know how to juggle registers, > and providing the appropriate information for a machine > which offers many registers is quite different from doing > good programming on a stack machine. -- > This address is for information only. æI do not claim that these views > are those of the Statistics Department or of Purdue University. > Herman Rubin, Department of Statistics, Purdue University > hru...@stat.purdue.edu æ æ æ æ Phone: (765)494-6054 æ FAX: (765)494-0558- Hide quoted text - - Show quoted text - prevent fragmentation of heap memory, the heap must be treated like a stack, i.e., last in - first out (LIFO). About 30 years ago, I modified Forth for matrix operations. The actual matrices were allocated in heap memory, but the operand references to those matrices were placed on the Forth stack which, along with RPN, forced LIFO allocation of matrices in the heap. Now, I find this experience useful when writing image processing programs in C++, because heap fragmentation is often a problem when processing large images. === Subject: Re: Programming skills required by mathematicians? <48626818$0$13948$afc38c87@news.optusnet.com.au> <48626DC4.1050101@electrooptical.net> <486374CD.AAE31BA4@tesco.net> <2JqdnfnOOu2SmvnVnZ2dnUVZ_trinZ2d@supernews.com> posting-account=W7RQ6gkAAACLDC5JWapQU1UV8ot8KkAS 1.0.3705; Dealio Toolbar 3.1.1),gzip(gfe),gzip(gfe) On Jun 26, 5:16æpm, Phil Hobbs [...] æFor the theoreticians I know, it would be how to hammer in a > nail without bending it. ;) > The answer to that problem is to use a hammer to knock in screws: screws > being thicker than nails are less likely to bend. Spoken like a theoretician. æ;) So, very true. Since it took real engineers to invent explosive bolts. The purpose being to prove to idiot theoreticians that the only difference between a screw and a nail is the s. It has nothing to do with science idiots and any amount of carpenters. > Phil Hobbs === Subject: Re: Programming skills required by mathematicians? <48626818$0$13948$afc38c87@news.optusnet.com.au> <48626DC4.1050101@electrooptical.net> <486374CD.AAE31BA4@tesco.net> <2JqdnfnOOu2SmvnVnZ2dnUVZ_trinZ2d@supernews.com> posting-account=7VnrmgkAAADfdiWnpCJtxxF3tpMHF0Y0 1.0.3705; .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) On Jul 3, 5:30æam, zzbun...@netscape.net [...] æFor the theoreticians I know, it would be how to hammer in a > nail without bending it. ;) > The answer to that problem is to use a hammer to knock in screws: screws > being thicker than nails are less likely to bend. > Spoken like a theoretician. æ;) æ æSo, very true. Since it took real engineers to invent explosive > bolts. > æ æThe purpose being to prove to idiot theoreticians that the only > difference > æ æbetween a screw and a nail is the s. It has nothing to do with > æ æscience idiots and any amount of carpenters. > Phil Hobbs- Hide quoted text - - Show quoted text - What you may not appreciate is that most papers published in electrical engineering journals (and for all I know all engineering journals) represent PURE research in that they are of no foreseeable practical use. It is like a carpenter authoring an in depth mathematical analysis of using a rock to hammer a screw. It gets published only because it has never been done before and there are no obvious technical errors. === Subject: Re: Programming skills required by mathematicians? > RPN is moderately easy to write, and is my choice for > a calculator. I am less fond of the stack idea of > Forth for a computer. I do know how to juggle registers, > and providing the appropriate information for a machine > which offers many registers is quite different from doing > good programming on a stack machine. FORTH has been my choice once, only for the reason that it can run on very, very small machines: http://hdebruijn.soo.dto.tudelft.nl/www/muziek/midicode.htm But for the rest, FORTH is a truly _dreadful_ programming language, as all stack oriented languages; PostScript is another terrible example. The reason is that data stacks are always, yes: always, in wrong order. Han de Bruijn === Subject: Re: Programming skills required by mathematicians? As for gotos, I find them easy to use, >and spaghetti code is best for spaghetti algorithms. >Sure, gotos are easy to write, but what's usually important >for code is that it be easy to read and easy to maintain, >and gotos are anathema to that. Hah! They are much easier to read than following indentations. > Several levels of indentations especially. Yes ! Sometimes even running off 80 columns IBM punch cards ! Especially with those TAB (8 characters) wide indentations. Many, if not all, Java programmers suffer from this indentation syndrome. It seens to be a part of their little culture. Such deep indentations can be prevented easily, though, with proper modularization. And 2 characters wide indentations offer much better readability - they are pretty standard in my culture, which is Delphi Pascal. Don't think that gotos are a better way out .. > Assembler language and the machine use gotos. The user > cannot do things efficiently without telling the machine > what to do. One instruction I find missing on the modern machines I > have seen is a conditional skip of a few instructions; I > have never seen more than one. This would enable no loss > of pipelining. The compiler could count how many instructions > in a large number of cases from code. How about: if then else end; Then the instructions are skipped if OK is true, that _is_ conditionally, not ? Another (loop) construct in (Delphi) Pascal: while true do begin OK := ; if OK then Continue; end; So what's your problem ? Loss of pipelining ? Han de Bruijn