Author |
Topic: limit of cosine to the nth at n (Read 994 times) |
|
william wu
wu::riddles Administrator
Gender: 
Posts: 1291
|
 |
limit of cosine to the nth at n
« on: Jan 19th, 2004, 12:13am » |
 Quote  Modify
|
Does the following limit exist?
lim n[to][infty] cosn(n), where n is a natural number
If so, what's the limit? If not, why not?
|
| « Last Edit: Jan 19th, 2004, 12:15am by william wu » |
 IP Logged |
[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: limit of cosine to the nth at n
« Reply #1 on: Jan 19th, 2004, 1:23am » |
Quote Modify
|
::If the limit exists it would probably be 0, because |cos(n)| < 1 for all n. It can however get arbitrarily close to 1, so I'm not sure. I'd guess the first, but since that would seem obvious perhaps it's the second..::
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Sir Col
Uberpuzzler
impudens simia et macrologus profundus fabulae
Gender:
Posts: 1825
|
|
Re: limit of cosine to the nth at n
« Reply #2 on: Jan 19th, 2004, 7:13am » |
Quote Modify
|
::
If |a|<1, |a.ak|<|ak|, therefore an[to]0 as n[to][infty].
As towr pointed out, if n is a natural number, then it cannot be a multiple of pi, so |cos(n)|<1. Hence cosn(n)[to]0 as n[to][infty].
I suppose an interesting question would be to consider the limit of cosx(x), as x[to][infty].
I suspect that no limit exists, simply because, cos(x)=[pm]1, for x=k[pi]. It even becomes difficult (for me) to imagine the function being continuous for very large values of x. 
::
|
|
IP Logged |
mathschallenge.net / projecteuler.net
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: limit of cosine to the nth at n
« Reply #3 on: Jan 19th, 2004, 7:49am » |
Quote Modify
|
I think you are forgetting one thing, or at the very least omit to proof it isn't the case.
|cos(n)| could go faster to 1 than the power drives it down to zero.
For instance like (1-2^-n)^n; (1-2^-n) is allways smaller than 1, but goes to 1 in the limit faster than the power can drive it back down.
In the case of cos(n) you can get increasingly better approximations of multiples of 2[pi] (f.i. just take round(2m[pi]) )
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: limit of cosine to the nth at n
« Reply #4 on: Jan 19th, 2004, 8:13am » |
Quote Modify
|
Sir Col, I think the original question is not so simple… It’s very important to recall what the definition of the limit of the sequence is, so let me give it here: The sequence {an} has the limit A as n [to][infty] if for every positive number [epsilon] there exists an integer N (depending on [epsilon]) such that |a - an| < [epsilon] for all n [ge] N.
The words “for all” are essential here. Now, if I give you an [epsilon], say, 2-10, could you find such an N? towr has indicated that this may be not the case since cos(n) may be arbitrarily close to 1. We know that cos(n) = cos(n mod 2[pi]), and the argument of the latter – as was shown on another thread – is uniformly distributed in (0, 1).
|
|
IP Logged |
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
|
Re: limit of cosine to the nth at n
« Reply #5 on: Jan 19th, 2004, 12:50pm » |
Quote Modify
|
By Dirichlet's Approximation Theorem, for any positive integer N, there are integers h,k, 0<k<=N such that |k[pi]-h| < 1/N.
Thus 1 > |cos(h)| > cos(1/N), and also, 0 < h < k[pi] + 1/N < 5N.
Thus |cosh(h)| > cos5N(1/N) ([to]1).
The next claim is that h may be chosen arbitrarily large:
To see this, fix M > 0. Then [epsilon] := min { |k[pi]-h| : 0 < h < M, 0 < k < (h+1)/[pi] } > 0, so by picking N > 1/[epsilon], the corresponding h must be larger than M.
Thus there is a sequence Ni[to][infty], hi[to][infty], such that
|cosh_i(hi)| > cos5N_i(1/Ni).
Taking limits shows that limsup |cosn(n)| = 1.
(In particular, the sequence cosn(n) gets arbitrarily close to 1 infinitely often.)
A similar argument, however, show that liminf |cosn(n)| = 0, thus lim cosn(n) does not exist.
|
|
IP Logged |
|
|
|
Sir Col
Uberpuzzler
impudens simia et macrologus profundus fabulae
Gender:
Posts: 1825
|
|
Re: limit of cosine to the nth at n
« Reply #6 on: Jan 19th, 2004, 1:31pm » |
Quote Modify
|
Thanks for correcting me, guys; I see what you're both driving at. Just when we thought we'd found n0 such that |cosn(n)|<[epsilon], for all n[ge]n0, some later value of n is above it.
I can see that this problem required tools that are outside of my limited mathematical repertoire. 
|
| « Last Edit: Jan 19th, 2004, 1:34pm by Sir Col » |
IP Logged |
mathschallenge.net / projecteuler.net
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: limit of cosine to the nth at n
« Reply #7 on: Jan 19th, 2004, 5:50pm » |
Quote Modify
|
on Jan 19th, 2004, 1:31pm, Sir Col wrote:| I can see that this problem required tools that are outside of my limited mathematical repertoire. |
|
Probably not, though you did make that one oversight. It is a fairly common mistake in this sort of situation.
But the Dirichlet Approximation Theorem, as hi-falutin as the name sounds, is nothing more than the observation that the rational numbers are dense in the Reals (i.e., they are arbitrarily close to every real number). The more exacting version that EigenRay develops is just noting that restricting the rational to denominators greater than an arbitrary fixed N still leaves a dense set. Both of these are things I believe you are already familiar with, as you have discussed similar things in the past.
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print