conergent subsequences of sin n?
Fickdichhh
I am currently in my second semester of Real Analysis and I was posed a problem
just for fun that I've been working on but with limited success:
Prove that there exists a convergent subsequence {xnk} of
{sin n} (where n is a natural number) such that {xnk} converges for every real
number in [-1, 1].
My first thought was to show that the function sin n with the domain the
natural numbers is equivanlent to the function sin n with the domain the
natural numbers modulo 2*pi. This was easy to show.
I ran into trouble however, when I tried to show that the range of values of {N
modulo 2*pi} contains an infinite quantity of unique values. If this is true I
can show by some manipulation of the Bolzano-Weierstrauss theorem that there is
a Cauchy subsequence of {sin n} that converges for every number in [-1, 1].
But this entire proof hinges on my assumptions regarding the properties of the
set of Natural numbers modulo 2 pi.
Can anyone point me in the right direction or show me what errors I have made?
I would greatly appreciate it! :-)
fickdichhh
If you reply, please send me an email as well.
Fickdichhh
I realized there was in error when I wrote my last post:
>>I ran into trouble however, when I tried to show that the >>range of values
of {N
>>modulo 2*pi} contains an infinite quantity of unique >>values.
This should read "contains an infinite quantity of unique IRRATIONAL values.
Sorry :-)
fickdichhh
Haran Pilpel
(e-mailed and posted)
Fickdichhh wrote:
>
> I am currently in my second semester of Real Analysis and I was posed a problem
> just for fun that I've been working on but with limited success:
>
> Prove that there exists a convergent subsequence {xnk} of
> {sin n} (where n is a natural number) such that {xnk} converges for every real
> number in [-1, 1].
I think what you mean is:
Show that for every x in [-1, 1], there exists a convergent subsequence
{sin(n(k))} (where k goes from 1 to inf) of {sin(n)} (where n goes from
1 to inf) which converges to x.
> I ran into trouble however, when I tried to show that the range of values of {N
> modulo 2*pi} contains an infinite quantity of unique values. If this is true I
> can show by some manipulation of the Bolzano-Weierstrauss theorem that there is
> a Cauchy subsequence of {sin n} that converges for every number in [-1, 1].
It does, indeed, contain an infinite number of unique values. To prove
this, note the following:
1. sin(x) = sin(y) means that (modulo 2pi) x=y or x=pi-y.
2. pi is irrational, and therefore (prove!) there do not exist natural
numbers n and m such that n = m*pi.
If you still have difficulties with this section, send me an e-mail.
[Note that actually _proving_ pi to be irrational is not that easy...but
I assume you don't need to do that.]
I don't really see, though, how the Bolzano-Weierstrass theorem could be
used to conclude the proof from this point. It would show the
_existence_ of a (partial) limit, but unless I misunderstood the
problem, it wouldn't show the limit to be x.
Actually, I would solve this somewhat differently (I will use e instead
of epsilon in the following):
What you need to show is that, for some arbitrary x in [-1, 1], and an
arbitrary e>0, there is an n such that sin(n) is in the e-neigbhourhood
of x. [This is trivially equivalent to the definition of partial limit.]
Since sin(x) is continuous in x, it is sufficient to show that, for
arbitrary e, there is some n such that, modulo 2pi, |n-arcsin(x)| < e.
What remains to be shown is that, for every x in [-1, 1], there is an n
such that |n-x| < e (mod 2pi). I won't prove this, but here's a hint:
try dividing the line segment [0,2pi] into sections of size e, and use
the pigeonhole principle.
Feel free to e-mail me if you still have trouble with this.
Quite a fun little problem, BTW!
> If you reply, please send me an email as well.
I did, but AOL seems to have been bouncing my messages recently...
Hope this helps,
Haran
--
e^(Pi*i) + 1 = 0 (Euler)
William A. McWorter Jr.
Two integers, n and m, are congruent modulo 2pi if and only if m-n is an INTEGRAL
multiple of pi, which, of course, it cannot be; no integral multiple of pi is an
integer.
William
Fickdichhh wrote:
> I am currently in my second semester of Real Analysis and I was posed a problem
> just for fun that I've been working on but with limited success:
>
> Prove that there exists a convergent subsequence {xnk} of
> {sin n} (where n is a natural number) such that {xnk} converges for every real
> number in [-1, 1].
>
> My first thought was to show that the function sin n with the domain the
> natural numbers is equivanlent to the function sin n with the domain the
> natural numbers modulo 2*pi. This was easy to show.
>
> I ran into trouble however, when I tried to show that the range of values of {N
> modulo 2*pi} contains an infinite quantity of unique values. If this is true I
> can show by some manipulation of the Bolzano-Weierstrauss theorem that there is
> a Cauchy subsequence of {sin n} that converges for every number in [-1, 1].
>
> But this entire proof hinges on my assumptions regarding the properties of the
> set of Natural numbers modulo 2 pi.
> Can anyone point me in the right direction or show me what errors I have made?
>
> I would greatly appreciate it! :-)
> fickdichhh
>
slim potter
Apparently some people don't even understand the question. I've often
toyed with using a pi based number system, circle radius pi area
pi^3=1^3 modulo pi, circle diameter pi circumference pi^2= 1^2 mod pi.
I never took it much further than idle musings.
dbe...@frii.com
The problem statement is a little confusing. Perhaps you want a
subsequence of {sin nx} that converges for every real number x in
[0.1]. I suggest that you clear up the statement and post it here
and in sci.math.
David Besser