※ 引述《cxcxvv (delta)》之銘言：
: 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?
寫一下我的做法好了
Since f(x) conti. on [0,1], so exist M > 0, such that |f(x)| < M
given ε>0, choose δ > 0, such that
|f(x)-f(1)| < ε/2, for x in [1-δ, 1]
choose N, such that when n > N, (1-δ)^(n+1) < ε/(4M)
then for n > N,
|Xn -f(1)| = |∫_{0 to 1} (n+1)x^n (f(x)-f(1)) dx|
≦ |∫_{0 to 1-δ} (n+1)x^n (f(x)-f(1)) dx| +
|∫_{1-δ to 1} (n+1)x^n (f(x)-f(1)) dx|
≦ ∫_{0 to 1-δ} (n+1)x^n (|f(x)|+|f(1)|) dx +
∫_{1-δ to 1} (n+1)x^n |f(x)-f(1)| dx
≦ 2M(1-δ)^(n+1) + ε/2
< ε
Thus limXn = f(1)
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 220.132.177.99
※ 編輯: dogy007 來自: 220.132.177.99 (01/20 11:54)
※ 編輯: dogy007 來自: 220.132.177.99 (01/20 13:35)