Re: [分析] 兩題高微
※ 引述《bajifox (嘖)》之銘言：
: 1
: {x_n} be a sequence of non-negative real number
: 1
: satisfying x_(n+1) =< x_n + -----
: n^2
: 則x_n是否一定會收斂?
: 我猜是沒有 因為Cauchy sequence的條件只有一邊
: 但是畫圖想找反例又覺得好像隱隱有遞減= =
: 想請問該怎麼做
: 2
: http://www.lib.ntu.edu.tw/exam/graduate/98/98047.pdf
: (D)
: 原來想說是用反函數定理
: 但是證完每個f'(x)都是invertible完以後卻發現不太對
: 反函數定理都只有在小小的neighborhood
: 就算做到onto(而且我好弱做不到)
: 兩個neighborhood的交集的部分又該怎麼確認他們的f^(-1)是相等的
: 謝謝
By inverse function theorem, f(R^n) is open.
Also {f(x_i)} is cauchy implies {x_i} is cauchy as
|x_i - x_j|C <= |f(x_i)-f(x_j)|.
This implies that f(R^n) is closed.
However, R^n is connected, so f(R^n)=R^n.
On the other hand, clearly, f is injective. Thus, f is globally invertible.
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