看板 Math 關於我們 聯絡資訊
※ 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. -- 哪一首中文歌的歌詞有 Pasadena? -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 76.94.119.209
keroro321 :推 ! 02/11 09:50
znmkhxrw :請問f是open map的理由是?? 02/11 18:27
znmkhxrw :需要證Df(x)都是nonsingular for all x€R^n ?? 02/11 18:28
Sfly :of course, for using inverse function theorem. 02/11 18:46
herstein :yes.....反函數定理 02/11 20:16
bajifox :謝謝幾位強者的回答^^ 02/11 20:38