Re: [分析] limsup of cos n
※ 引述《vimrc (vimrc)》之銘言：
: 想請教怎麼求 limsup & liminf of cos n ( n = 1, 2, 3, ... )
: 直覺 limsup 就是 1, liminf 就是 -1
: 可是要怎麼證明呢？
: 謝謝！
出來了!
<Lemma> if
i is irrational number , i > 0
S = {n-mi：n,m are positive integers}
then
0€L(S)
<pf> Suppose not , then there exists e>0 s.t. │n-mi│> e , for all n,m
Take 0<q_n<i , q_n → i , q_n are rational numbers
Write q_n = a_n/b_n , a_n , b_n are positive integers for all n
Obviously, a_n , b_n are not bounded sequence
so there exists q_n_j = a_n_j / b_n_j s.t.
(1) a_n_j , b_n_j goes to inf
(2) q_n_j → i
Hence │n-m*q_n_j│> e , for all n,m and j>=J
= │n-m*(a_n_J / b_n_J)│ > e
take m = b_n_J , n = a_n_J , contradiction
<Theorem> limsup cos(n) = 1 , liminf cos(n) = -1
<pf> Consider S={n-m*(2pi)}
Since 0€L(S) , we have
there exists sequence s_k = n_k - m_k *(2pi) → 0
It's possible to choose such n_k,m_k increasing to inf
so cos(n_k) = cos(n_k - m_k *(2pi)) → cos(0) = 1 by continuity of cosx
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
by periodic property
since n_k is a subsequence of n
so we find a subsequence of cos(n) s.t. cos(n_k) → 1
Finally, since │cos(n)│<= 1, done.
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