wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> limit of cosine to the nth at n
        
(Message started by: william wu on Jan 19^th, 2004, 12:13am)

Title: limit of cosine to the nth at n
Post by william wu on Jan 19^th, 2004, 12:13am

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?

Title: Re: limit of cosine to the nth at n
Post by towr on Jan 19^th, 2004, 1:23am

::[hide]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..[/hide]::

Title: Re: limit of cosine to the nth at n
Post by Sir Col on Jan 19^th, 2004, 7:13am

::[hide]
If |a|<1, |a.a^k|<|a^k|, 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 cos^x(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. ???
[/hide]::

Title: Re: limit of cosine to the nth at n
Post by towr on Jan 19^th, 2004, 7:49am

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]) )

Title: Re: limit of cosine to the nth at n
Post by Barukh on Jan 19^th, 2004, 8:13am

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 {a_n} 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 - a_n| < [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).

Title: Re: limit of cosine to the nth at n
Post by Eigenray on Jan 19^th, 2004, 12:50pm

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 |cos^h(h)| > cos^5N(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 N_i[to][infty], h_i[to][infty], such that
|cos^h_i(h_i)| > cos^5N_i(1/N_i).
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.

Title: Re: limit of cosine to the nth at n
Post by Sir Col on Jan 19^th, 2004, 1:31pm

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. :)

Title: Re: limit of cosine to the nth at n
Post by Icarus on Jan 19^th, 2004, 5:50pm

on 01/19/04 at 13:31:52, 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.