2016-07-12 03:00:32rickj752026d4
證明一個數列極限的問題
標題:
證明一個數列極限的問題
發問:
令g:[a,b]->R 是一個正的且連續函數,令M=sup{g(x)|x 屬於[a,b]}, 證明M=lim(n->inf){S(a到b) (g(x))^n dx}^(1/n) 想了好久,請問一下如何證???
最佳解答:
f is continuous on [a,b], so 存在 t€[a,b] such that f(t)=M>0 又 f is continuous at t , forε >0 , there is δ>0 such that |f(x)-M|<ε whenever x€(t-δ,t+δ) Hence we have M-εlimsup(n->∞){∫_[a,b][f(x)]^ndx}^(1/n)<=M =>limint(n->∞){∫_[a,b][f(x)]^ndx}^(1/n)>=M-ε Since ε was abiriary so M<=liminf(n->∞){∫_[a,b][f(x)]^ndx}^(1/n)<=limsup(n->∞){∫_[a,b][f(x)]^ndx}^(1/n)<=M =>lim(n->∞){∫_[a,b][f(x)]^ndx}^(1/n)=M 2009-04-02 23:54:38 補充： 所有的f(x)都改成g(x) 2009-04-02 23:58:45 補充： 為了要保證極限存在 所以我取limsup 你學過高微 我想你應該看的懂 因為 f為正的連續函數 故 0<=f(x)<=M for all x€[a,b] 0<={∫_[a,b][f(x)]^ndx}^(1/n)<=M*(b-a)^(1/n) =>limsup(n->∞)[∫_[a,b][f(x)]^ndx}^(1/n)<=M 2009-04-03 00:03:34 補充： 第五行的開頭改成: (M-ε)(2δ)^(1/n) 2009-04-03 00:15:52 補充： 其實不用取limsup也可以 直接取lim 就可以了
其他解答:
5D8232B72DC99558
證明一個數列極限的問題
發問:
令g:[a,b]->R 是一個正的且連續函數,令M=sup{g(x)|x 屬於[a,b]}, 證明M=lim(n->inf){S(a到b) (g(x))^n dx}^(1/n) 想了好久,請問一下如何證???
最佳解答:
f is continuous on [a,b], so 存在 t€[a,b] such that f(t)=M>0 又 f is continuous at t , forε >0 , there is δ>0 such that |f(x)-M|<ε whenever x€(t-δ,t+δ) Hence we have M-ε
其他解答:
5D8232B72DC99558