limit points points of {Cos(n)}
Dionisio Fleitas
I have a question, if you can help me I really appreciate it.
I want to prove that the set of all adherent points of the
set { cos(n) ; n=0,1,2,...} is the interval [-1,1].
Actually, I'd like to show that for any x in [-1,1],
there exists a subsequence of {cos(n)} that
converges to x.
Dion Fleitas
dlf...@omega.uta.edu
Jim P. Ferry
In article <34F9E644...@omega.uta.edu>, Dionisio Fleitas <dlf...@omega.uta.edu> writes:
|> I have a question, if you can help me I really appreciate it.
|> I want to prove that the set of all adherent points of the
|> set { cos(n) ; n=0,1,2,...} is the interval [-1,1].
|> Actually, I'd like to show that for any x in [-1,1],
|> there exists a subsequence of {cos(n)} that
|> converges to x.
Consider this: for any real number y you can choose an integer k to make
y + 2 pi k arbitrarily close to an integer.
-Jim Ferry
Robert Israel
In article <34F9E644...@omega.uta.edu>,
Dionisio Fleitas <dlf...@omega.uta.edu> wrote:
>I have a question, if you can help me I really appreciate it.
>I want to prove that the set of all adherent points of the
>set { cos(n) ; n=0,1,2,...} is the interval [-1,1].
>Actually, I'd like to show that for any x in [-1,1],
>there exists a subsequence of {cos(n)} that
>converges to x.
This is easy, given that Pi is irrational. Show that the positive
integers mod 2 Pi are dense in [0, 2 Pi) (hint: given epsilon, take
N > 2 Pi/epsilon and note that there are two points of the set
{0,1,...,N} mod 2 Pi with distance < epsilon; use multiples of
the difference of these to show that all of [0, 2 Pi) is within
epsilon of the positive integers mod 2 Pi). Then map onto [-1,1]
using cos.
Robert Israel isr...@math.ubc.ca
Department of Mathematics (604) 822-3629
University of British Columbia fax 822-6074
Vancouver, BC, Canada V6T 1Z2
Robin Chapman
In article <34F9E644...@omega.uta.edu>,
Dionisio Fleitas <dlf...@omega.uta.edu> wrote:
>
> I have a question, if you can help me I really appreciate it.
> I want to prove that the set of all adherent points of the
> set { cos(n) ; n=0,1,2,...} is the interval [-1,1].
> Actually, I'd like to show that for any x in [-1,1],
> there exists a subsequence of {cos(n)} that
> converges to x.
Set x = cos(y) where say 0 <= y <= pi. For real t define [t] as
t - 2m pi where the integer m is chosen so that 0 <= [t] < 2pi. Then
cos(t) = cos([t]). If suffices to find an increasing sequence n_1, n_2, ...
of positive integers with [n_i] --> y as i --> infinity. To construct this
it suffices to show that each subinterval I of [0,pi] with positive
length contains [n] for infinitely many integers n. If this interval
is [a - e, a + e] we need to find infinitely many pairs (n, m) of integers
with n > 0 and |n - 2m pi - a | <= e. This follows since pi is irrational.
By, say, Dirichlet's theorem on approximation by rationals, there are positive
integers r and s with |r - 2s pi| < e. As pi is irrational r - 2s pi isn't
zero. If r - 2s pi > 0 then |k(r - 2s pi) - a| < e for some positive
integer k. If r - 2s pi < 0 then 0 < 1 + u(r - 2s pi) < e for some positive
integer u, and we can replace r + 2s pi by this to complete the proof.
Robin Chapman "256 256 256.
Department of Mathematics O hel, ol rite; 256; whot's
University of Exeter, EX4 4QE, UK 12 tyms 256? Bugird if I no.
r...@maths.exeter.ac.uk 2 dificult 2 work out."
http://www.maths.ex.ac.uk/~rjc/rjc.html Iain M. Banks - Feersum Endjinn
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Victor S. Miller
>>>>> "Dion" == Dionisio Fleitas <dlf...@omega.uta.edu> writes:
Dion> I have a question, if you can help me I really appreciate it. I
Dion> want to prove that the set of all adherent points of the set {
Dion> cos(n) ; n=0,1,2,...} is the interval [-1,1]. Actually, I'd
Dion> like to show that for any x in [-1,1], there exists a
Dion> subsequence of {cos(n)} that converges to x.
This follows easily from the Weyl's theorem using the fact that pi is
irrational, and cos is a uniformly continuous function. Weyl's
theorem says that if alpha is irrational then the set of fractional
parts of n alpha is uniformly distributed in the interval [0,1] --
i.e. if [a,b] is any subinterval of [0,1], M > 0 is an integer, and
c(M) = the number of 1 <= n <= M such that the fractional part of n
alpha is in [a,b], then lim c(M)/M = b-a (the length of the interval).
--
Victor S. Miller | " ... Meanwhile, those of us who can compute can hardly
vic...@ccr-p.ida.org | be expected to keep writing papers saying 'I can do the
CCR, Princeton, NJ | following useless calculation in 2 seconds', and indeed
08540 USA | what editor would publish them?" -- Oliver Atkin
Philippe Gaucher
In article (Dans l'article) <34F9E644...@omega.uta.edu>, Dionisio
Fleitas <dlf...@omega.uta.edu> wrote (馗rivait)�:
> I have a question, if you can help me I really appreciate it.
> I want to prove that the set of all adherent points of the
> set { cos(n) ; n=0,1,2,...} is the interval [-1,1].
> Actually, I'd like to show that for any x in [-1,1],
> there exists a subsequence of {cos(n)} that
> converges to x.
G={(cos(n),sin(n))}={exp(in)}=Z is a subgroup of S^1=R/2*PI*Z
with the inclusion map exp(in) -> n mod 2*PI. Since R/Z is
compact, a discrete subgroup of R/Z is finite. So G is dense
in S^1 => {cos(n)} is dense in [-1,1].
pg.
Gareth McCaughan
Dionisio Fleitas wrote:
> I have a question, if you can help me I really appreciate it.
> I want to prove that the set of all adherent points of the
> set { cos(n) ; n=0,1,2,...} is the interval [-1,1].
> Actually, I'd like to show that for any x in [-1,1],
> there exists a subsequence of {cos(n)} that
> converges to x.
This looks worryingly like homework, so just a hint:
pi is irrational.
--
Gareth McCaughan Dept. of Pure Mathematics & Mathematical Statistics,
gj...@dpmms.cam.ac.uk Cambridge University, England.
David Farmer
A slightly harder challenge:
Determine whether on not |cos(n)|^n is dense in [0,1].
Dave
[
Moderator's note: Using the fact that pi, like any irrational, can be
approximated by rationals n/m with | pi - n/m| < 1/m^2, it's not hard to
show that 1 is a limit point of |cos(n)|^n. For |cos(n)|^n to be dense,
it seems to me that you'll want somewhat worse approximations: for any
0 < a < b <1, you want infinitely many (m,n) such that
a/m^(3/2) <= | pi - n/m | <= b/m^(3/2)
Are there any results in this direction?
]