Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: David Marcus on 19 Jan 2007 23:13 Andy Smith wrote: > In message <JC5BJn.8C1(a)cwi.nl>, Dik T. Winter <Dik.Winter(a)cwi.nl> writes > >In article <ZNAVsEWGCVsFFwdL(a)phoenixsystems.demon.co.uk> Andy Smith > ><Andy(a)phoenixsystems.co.uk> writes: > >... > > > >Why? If m is 10^90, is it hard to talk about (10^90)! ? > > > > > > > Some time back you showed me that sin(pi/x) has a fundamental > > > discontinuity at x=0. > > > >Yes, that is fundamental. > > > > > Doesn't the same thing apply to Lim m->oo {cos(m! pi x)} and if not, can > > > you explain, please? > > > >That function has a whole lot of problematical points. It is not even > >defined for irrational x. So there are a whole lot of discontinuities. > >But they are not fundamental. Defining the function: > > f(x) = lim{m -> oo} cos(m!.pi.x) when x is rational and > > f(x) = 1 when x is irrational > >solves them all and leaves a continuous function. So the discontinuities > >are not fundamental. (Unless someone is able to show that for some x, > >lim{m -> oo} cos(m!.pi.x) is defined.) > > Yes, thanks. But actually the whole point of all of this discussion was > a throwaway remark by David Marcus to the effect that finding a formula > f(x) to generate 1 if x was rational, and 0 if x was irrational would be > difficult (for me), someone generated the smarty formula, and I said > subsequently that I might have been able to generate it, but would not > be able to rigorously justify it. ... Anyway, redefining lim{m -> oo} > cos(m!.pi.x) as a function of rationality of x rather defeats the > purpose of that exercise... You left out the exponent. The situation is different with the 2n exponent. Af first, I didn't notice that you had left it out. -- David Marcus
From: Michael Press on 20 Jan 2007 00:19 In article <GM2CMJZeuXsFFw$f(a)phoenixsystems.demon.co.uk>, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > David Marcus writes > >Andy Smith wrote: > >> > >> Some time back you showed me that sin(pi/x) has a fundamental > >> discontinuity at x=0. > > > >We said two things. Division by zero is not defined. So, the expression > >sin(pi/x) is not defined for x = 0. We also said that the function > > > > f(x) = sin(pi/x), x != 0 > > > >can not be extended to a continuous function that is defined on all of > >R. > > > >> Doesn't the same thing apply to Lim m->oo {cos(m! pi x)} and if not, can > >> you explain, please? > > > >You don't evaluate limits by taking the value you are approaching and > >sticking it in the expression. And, oo isn't a number, so it wouldn't > >even make sense in this case. cos(m! pi x) is defined for all real > >numbers x and all nonnegative integers m. So, it isn't any harder to > >consider the limit as m -> oo as it would be to consider the limit as x > >-> oo. > > > If x is irrational cos(m! pi x) does not have a limit as m->oo, it is > undefined? Remember that the expression currently under discussion is not the original expression, nor a part of it. <rubrum-B45B9E.17135117012007(a)newsclstr02.news.prodigy.com> That expression is lim_{m -> oo} lim_{n -> oo} [cos (m! * x * pi)]^{2 * n} There is a difference. -- Michael Press
From: David Bernier on 20 Jan 2007 04:14 David Marcus wrote: > Andy Smith wrote: >> David Marcus writes >>> Andy Smith wrote: >>>> Some time back you showed me that sin(pi/x) has a fundamental >>>> discontinuity at x=0. >>> We said two things. Division by zero is not defined. So, the expression >>> sin(pi/x) is not defined for x = 0. We also said that the function >>> >>> f(x) = sin(pi/x), x != 0 >>> >>> can not be extended to a continuous function that is defined on all of >>> R. >>> >>>> Doesn't the same thing apply to Lim m->oo {cos(m! pi x)} and if not, can >>>> you explain, please? >>> You don't evaluate limits by taking the value you are approaching and >>> sticking it in the expression. And, oo isn't a number, so it wouldn't >>> even make sense in this case. cos(m! pi x) is defined for all real >>> numbers x and all nonnegative integers m. So, it isn't any harder to >>> consider the limit as m -> oo as it would be to consider the limit as x >>> -> oo. >> If x is irrational cos(m! pi x) does not have a limit as m->oo, it is >> undefined? > > Correct. It does not have a limit. So, if x is irrational, then > > lim_{m->oo) cos(m! pi x) > > is undefined. For x = 2exp(1) = 2e, cos(m! pi*(2e)) is close to 1 for several m>1000. 2e = 2/1 + 2/1 + 2/2! + 2/3! + 2/4! + ... 2/(k!) + ... Say m is large. Then (m!)*2/((m-1)!) = 2m is even, and for the k<= m, (m!)*2/k! is even. But for k = m+1, (m!)*2/(k!) = 2/(m+1), which gets closer to 0 as m increases. I'm not convinced that lim_{m->oo} cos(m! pi * (2e)) is undefined. David Bernier > If that is what you meant a few posts back, then you were correct. Sorry > if I said otherwise. >
From: Andy Smith on 20 Jan 2007 06:08 David Marcus writes >> >> I can see that I am irritating you. > >No, you aren't. > >> It is not deliberate. It is a >> consequence of a misunderstanding on my part of something fundamental. > >I'm sure! > >> You say, is there such a y? Well, if we consider the mapping >> w = Lim n->oo x^n, 0<=x<=1 , we can consider that as the limit of a >> succession of mappings. > >This is simply a language problem. The meaning of what is written is not >what you think. It is possible to write what you are thinking: see >below. > >An analogy would be if I tell you what the word "cat" means and you ask >why can't a cat be the animal that barks. It could be, but it isn't. We >have another name for that animal. > >> So, for example we can think of some >> representative points located in x Perhaps I should have said in {x} 0<=x<=1 then. > >You can't say "in x". x is a number. We don't know exactly which, but it >is still a number. So, just as you can't say "in 7", you can't say "in >x". "x" is a name for a number, the number x. Similarly, "7" is a name >for the number 7. > >> e.g x - 0,1/4,1/2,1 on [0,1]. After >> x^2 we get >> 0,1/16,1/4,1 and after x^n these points map to 0,1/(4^n),(1/2^n), 1. >> But we still have a continuum of points in [0,1] after n iterations, and >> each of these map back to an original location in x. And that has to be >> true after n->oo > >There is no "after n -> oo". > >> - the line is infinitely elastic. Otherwise you could >> define a space between two reals in x? >> >> There is then an inverse mapping from the points in w in [0,1] back to >> their original locations in x? And these would all describe the y that >> you request? > >Name a number y with |y| < 1 that will work. You can't. > >The following is from Chapter 24 of the book "Calculus" by Spivak. > >"Let {f_n} be a sequence of functions defined on the set A. Let f be a >function which is also defined on A. Then f is called the uniform limit >of {f_n} on A if for every e > 0 there is some [natural number] N such >that for all x in A, > > if n > N, then |f(x) - f_n(x)| < e. > >"We also say that {f_n} converges uniformly to f on A, or that f_n >approaches f uniformly on A. > >"As a contrast to this definition, if we only know that > > f(x) = lim_{n->oo} f_n(x) for each x in A, > >then we say that {f_n} converges pointwise to f on A. Clearly, uniform >convergence implies pointwise convergence (but not conversely!)." > >Another way to write that {f_n} converges uniformly to f on A is > > lim_{n->oo} sup_{x in A} |f_n(x) - f(x)| = 0. > >Let f_n(x) = x^n. Let f(x) = 0. Then {f_n} converges pointwise to f on >[0,1), but not uniformly. > Thanks for taking the time. Spivak should be here soon ... My mental problem with the limit of x^n was that viewed as mapping of points I couldn't see how you could split [0,1] into [0,1) and 1 by any sort of limit process - I had a mental image of x^n as a y-axis and x as the x-axis. At n = 1 it is a line, then progressively curving close to the x-axis as n increases, but always continuous. Viewed from the y-axis, there always seemed to be an inverse mapping. But I can see now that this is far too simplistic, and I accept the proof. Ta. -- Andy Smith
From: Ralf Bader on 20 Jan 2007 13:26
Carsten Schultz wrote: > Virgil schrieb: >> Each path in the complete infinite binary tree is essentially a >> function from N to {0,1}, i.e., an infinite binary sequence of left(0) >> or right(1) branchings. > > There is a level at which WM knows this. > >> But Cantor's diagonal proof, as originally stated, proved that the set >> of all such sequences not to be countable. > > And to WM this is obviously wrong. Now he is trying to find a way to > let others see this too. Of course, his arguments are never valid. But > he is looking for a way to state the ever same wrong arguments that > would convince others. There must be one, because to him it seems all > so clear! Momentarily his tool of choice seems to be this binary tree. But it is not really difficult to see what Mückenheim "sees". Take sequences of digits. Mückenheim just can not imagine that such a sequence is irregular forever. For him, those sequences always get constant (or periodic, which in principle is the same) from some finite index onwards. In Mückenheim's mind, if such a notion comes up, then, no matter how it is dressed up (e.g., as a path in that tree), it *is* constant/periodic. As a little child, for some time my mind was in a state where I could not imagine the look of a surface of water without there being a opposite coastline visible on the horizon. This disappeared when I saw the sea for the first time, or maybe a picture of it. I don't know whether there is anything one could show Mückenheim so that his mental barrier is removed. It is kind of interesting that such mental barriers exist - psychologically, not mathematically. >> So if WM's tree has only countably many paths, it must be leaving most >> paths out. >> >> An several people have presented analyses showing where WM's omissions >> occur. > > Of course. > > The real sad thing is that he teaches mathematics courses. Therefore > while making a fool out of himself here, he also disgraces his school. Mathematicians of stature generally seem not to care about this. "It is useless to talk to those people" (like Mückenheim) and, unspokenly, useless to talk about them, is something I heard. And I think this stance is not quite inept, if I compare it with the kind of large-scale stupefication brought about by the German "Rechtschreibreform" (see http://www.sprachforschung.org/ about this). Some time ago I found a book in a library, written by a certain Max Woitschach in the 80s of the last century. This author who also held a kind of professorship at a "Fachhochschule" also couldn't resist to entertain the public with his collected pseudomathematical misconceptions. So Mückenheim has predecessors - crackpottery rarely seems to be original. Ralf |