Re: Cantor Confusion

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
.