Cantor Confusion
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From: Andy Smith on 19 Jan 2007 12:29 I don't know about the cos(m! pi x) term, but (bearing in mind I don't have a pilot's licence) I would say that: Lim n->oo {|x|^n} is definitely not 0 for |x|<1, 1 for |x|=1. At a first glance one is tempted to say, well, at any n, let x >= 1-1/(2n). Then for any n, x^n >1/2 (terms of (1-y)^n are monotonic decreasing for y<1 and alternating in sign), so it cannot be true that the lim n->oo of |x|^n = 0 for all |x| <=x<1 But actually that is not very convincing, because 1-1/(2n)->0 as n->oo and it isn't clear how things all wash up. You need to consider the range of x such that x^n<eta for a given eta and n, and see how the limit of that behaves. So, consider some range of x <1 and an eta such that, at a given n, x^n<eta for 0<=x<=xmax<=1. For a given eta and n, xmax = eta^(1/n) (because any x>eta^(1/n) will have x^n>eta). We can consider the difference d between xmax and eta - we wish to make eta->0, but, at a given n, as we reduce eta, we increase xmax. So we need to show that the difference between xmax and eta can -> 1 as n->oo. d = xmax-eta = eta^(1/n) - eta. Or eta =(d+eta)^n For any n, for eta = 0 this has no solution for d other d = 0; For any n, for d = 1, this has no solution for eta. We conclude that the function |x|^n tends to 0 for only a finite range of |x| 0<=|x|<1-delta If you agree, how would you express the argument rigorously? -- Andy Smith
From: Andy Smith on 19 Jan 2007 12:42 No delete previous post (or maybe next one). >eta =(d+eta)^n >For any n, for eta = 0 this has no solution for d other d = 0; >For any n, for d = 1, this has no solution for eta. Need to show that no solution for d<1, which is back where we started. do you think limn->oo |x|^n = 0 for all |x|<1 ? -- Andy Smith
From: David Marcus on 19 Jan 2007 13:06 Michael Press wrote: > In article <MPG.2018a7a73eedebe1989b6d(a)news.rcn.com>, > David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > > Michael Press wrote: > > > Hello. I agree. At least one-hundred-fifty years. > > > Is this a formula? > > > > > > lim_{m -> oo} lim_{n -> oo} [cos (m! * x * pi)]^{2 * n} > > > > These days or in Euler's day? > > Both if you please. I have not seen formula precisely > defined, so I am asking how it is typically thought of, > as in your usage above. I don't think Euler would have considered that to be a formula, but I'm hardly an authority on this. These days: maybe. As you said, "formula" isn't precisely defined. -- David Marcus
From: David Marcus on 19 Jan 2007 13:15 Andy Smith wrote: > No delete previous post (or maybe next one). > > >eta =(d+eta)^n > > >For any n, for eta = 0 this has no solution for d other d = 0; > >For any n, for d = 1, this has no solution for eta. > > Need to show that no solution for d<1, which is back where we started. > > do you think limn->oo |x|^n = 0 for all |x|<1 ? As always in math, the first thing to do is make sure you have precise definitions for everything. In this case, you need to define "lim n-> oo". Let "A" mean "for all" and "E" mean "there exists". The definition of lim n->oo f(n) = L is A e > 0, E N such that if n > N then |f(n) - L| < e. Given this definition, what do you think? Note that it takes people a couple of years until they become comfortable with definitions like this, so you shouldn't be discouraged if you have trouble with it. A good book on this topic is Calculus by Spivak. -- David Marcus
From: David Marcus on 19 Jan 2007 13:21
Andy Smith wrote: > With due rigor, I mean. Obviously when x is rational E some m such that > m! x = integer, cos(m! pi x) = +/- 1, otherwise |cos()|<1 and > |cos()|^(2*n) -> 0. > > But I would expect some hard time especially over whether it is > reasonable to talk about m! as m->oo Why? If m is 10^90, is it hard to talk about (10^90)! ? > and whether lim n->oo {x^n} A |x|<1 > is really 0.00.. and indeed whether that is the same as integer 0. The real number zero is also an integer. Similarly, the real number one is also an integer. The integers are a subset of the real numbers. As for "0.00...", the meaning of such an expression is simply 0 + sum_{n=1}^oo 0 10^{-n}, which is just zero. -- David Marcus |