[微積] 連續函數的積分與極限一題

看板Math作者cxcxvv (delta)時間12年前 (2012/01/19 13:30)推噓1(1推 0噓 12→)

留言13則, 4人參與討論串1/5 (看更多)

Suppose f is continuous on [0,1] Define 1 Xn = (n+1)∫x^n f(x) dx 0 Find lim Xn n->infinity 原本以為用積分中值定理就做出來了可是那個c belongs to (0,1)會depend on n 所以其實不能那樣做那這一題要怎麼解呢? -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.24.172.145

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tiwsjia

01/19 14:09, , 1^F

01/19 14:09, 1^F

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tiwsjia

01/19 14:10, , 2^F

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tiwsjia

01/19 14:10, , 3^F

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tiwsjia

01/19 14:10, , 4^F

01/19 14:10, 4^F

謝謝我做看看 p 1.Given 0<p<1, [(n+1)∫x^n f(x) dx ] -> 0 (n->∞) ---------(A) 0 2.Consider x belongs to (p,1) 1 | (n+1)∫x^n f(x) dx - f(1) | --------(B) p choose p s.t. x belongs to (p,1) implies_ 0.5f(1)<f(x)<1.5f(1) 1 1 =>(n+1)∫x^n 0.5f(1) dx < (n+1)∫x^n f(x) dx p p 1 < (n+1)∫x^n 1.5f(1) , write Ln < Mn < Rn p => Ln-f(1) < (B) < Rn-f(1) . . . 1 => lim Ln-f(1) = 0.5f(1) * ∫d(x^n+1) = 0.5f(1) * [1-p^(n+1)] →0 (p->1) n->∞ p 1 lim Rn = 1.5f(1) * ∫d(x^n+1) = 1.5f(1) * [1-p^(n+1)] →0 (p->1) n->∞ p 1 3.Therefore, | (n+1)∫x^n f(x) dx - f(1) | ≦ (A) + (B) 0 1 i.e. lim (n+1)∫x^n f(x) dx = f(1) n->∞ 0 ※ 編輯: cxcxvv 來自: 114.24.172.145 (01/19 14:48)

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recorriendo

01/19 14:43, , 5^F

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recorriendo

01/19 14:43, , 6^F

01/19 14:43, 6^F

※ 編輯: cxcxvv 來自: 114.24.172.145 (01/19 15:02) ※ 編輯: cxcxvv 來自: 114.24.172.145 (01/19 15:05)

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dogy007

01/19 17:28, , 7^F