Re: [微積] 兩題定積分的極限

看板Math作者znmkhxrw (QQ)時間13年前 (2012/08/15 17:36)推噓3(3推 0噓 6→)

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※ 引述《ChocoRs (巧克先生)》之銘言： : 2. : 2x 2c : 　　　　 2t / e 　\　　　　　　　　　　　　　　　 e : lim　∫ |－－－| dx 我用積分均值定理=> lim　－－ * t = 0 ,c在(t,2t) : t→0 t　|　x | t→0　ｃ : \ / : ~ : 可是答案是ln 2 ，不知道問題出在哪 : 以上,麻煩大家咧~ 1.for t > 0 n (2x)^k f_n(x) = Σ ──── converges to e^(2x) uniformly on [t,2t] k=0 k! so f_n(x)/x conv. to [e^(2x)]/x uniformly on [t,2t] 2t 2t by theorem, ∫ [e^(2x)]/x dx = lim ∫ f_n(x)/x dx ---(●) t n→∞ t 2t 2t Now calculating lim ∫ [e^(2x)]/x dx = lim lim ∫ f_n(x)/x dx t→0 t t→0 n→∞ t if lim , lim can exchange, done. t→0 n→∞ (Theorem：for g_n：(0,1] → R if (1) for each n, lim g_n(t) exists t→0 (2) for each t, lim g_n(t) exists uniformly n→∞ then lim lim g_n(x) exists n→∞ t→0 and equals to lim lim g_n(x) ) t→0 n→∞ 2t by the theorem above , we need to show that g_n(t) = ∫ f_n(x)/x dx conv. uni. t 2t Observe that g_n(t) conv. to g(t) = ∫ [e^(2x)]/x dx by(●) t and f_n(x) conv. to e^(2x) uni. on [0,2] so take sufficient N, │f_n(x) - e^(2x)│< ε , for all x€[0,2] Now estimating │g_n(t) - g(t)│ , t€(0,1] 2t =│∫ f_n(x)/x - [e^(2x)]/x dx│ t 2t │f_n(x) - e^(2x)│ ≦∫ ────────── dx t x ≦ ε* ln2 , when n≧N so g_n(t) conv. to g(t) uniformly. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 1.171.12.77

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znmkhxrw

08/15 17:37, , 1^F

08/15 17:37, 1^F

※ 編輯: znmkhxrw 來自: 1.171.12.77 (08/15 17:38)

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ntust661

08/15 17:42, , 2^F

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suhorng

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