Re: [分析] 幾題高微和線代

[paperbattle (跳針猴王)]

※ 引述《jizhan (天下何思何慮)》之銘言：
: ※ 引述《paperbattle (跳針猴王)》之銘言：
: : 2.Let a_0 = 0 , a_n+1 = 1/(2+a_n) for all n>0
: : 已經知道這數列bounded 但不是monotone
: : 如何證明a_n收斂?
: 有個想法是
: a_2k increasing, a_2k+1 decreasing, a_2k < a_2k+1 for all k>=0
: so
: [a_2k,a_2k+1] contains [a_2(k+1),a_2(k+1)+1] for all k>=0
: by nested intervals theorem
: the intersection of the intervals [a_2k,a_2k+1] contains exactly one point a_
: ∞
: now claim that a_∞ is the limit of the sequence by observing that
: a_2k <= liminf a_n <= limsup a_n <= a_2k+1 for all k>=0
: so the liminf and limsup both belong to all the interval [a_2k,a_2k+1]
: (有辦法直接證明a_n是Cauchy sequence嗎?)

好像可以這樣..

a_n+2

  2+a_n
= ------
  5+2a_n

因為偶數項和奇數項分別是遞增和遞減

所以{a_2n} {a_2n+1}極限存在

然後就可以湊出 s = (2+s) /(5+2s) => 5s + 2s^2 = 2 + s

=> s^2 + 2s - 1 = 0 => s= -1+ sqrt(2)

所以兩個數列的極限值都一樣

{a_n} 收斂到 -1 + sqrt(2)

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◆ From: 122.116.42.100
→ jizhan:在高微課上 應該會需要說明為何最後2行成立
推 toosalt:你必須說他的任意收殮子數列都收到同一點吧,你只try了兩個
推 PhysiAndMath:偶數項跟奇數項都ok不就是全部了嗎@@?
嗯嗯

有個定理是這樣子的

如果{a_n}的兩個subsequence{a_2n+1}和{a_2n}都收斂且極限都是a

{a_n}就會收斂於a
※ 編輯: paperbattle     來自: 122.116.42.100       (01/27 18:10)
推 jizhan:同意 我忘了有這性質 用定義證即可