作者yhliu (老怪物)
看板Math
标题Re: [分析] Chebyshev inequality and a sequence
时间Fri Jul 8 20:47:09 2011
※ 引述《CFE220 (五行之友)》之铭言：
: How to proof the following Chebyshev inequality without using induction?
: If a_1 ≧ a_2 ≧ ... ≧ a_n and b_1 ≧ b_2 ≧ ... ≧ b_n,
: then
: (Σa_k)(Σ b_k) ≦ n(Σa_k b_k) , the sum is taken from 1 to n.
令 a* = Σa_k/n, b* = Σb_k/n.
则 (Σa_k)(Σ b_k) ≦ n(Σa_k b_k) if and only if
Σ(a_k-a*)(b_k-b*) = Σa_k(b_k-b*) = Σ(a_k-c)(b_k-b*) ≧ 0.
第2个等式对任意 c 都成立.
因 a_k↓ 且 b_k↓.
存在 p 使
b_k-b*≧0 when k≦p, 且 b_k-b*≦0 when k>p.
取 c 介於 a_p 与 a_{p+1} 之间,
则 a_k-c ≧0 when k≦p, 且 a_k-c≦0 when k>p.
故 Σ(a_k-c)(b_k-b*) 之每一项均非负, 其和当然也非负.
: Let a_1, a_2,..., a_p be fixed positive numbers. Consider the sequence
: (a_1)^n + (a_2)^n + ... + (a_p)^n +
: s_n = ────────────────── and x_n = (s_n)^1/n, n in Z .
: p
: Show that the sequence {x_n} is monotonically increasing without using
: induction.
把题目简化一下:
设 A_1,...,A_p 均为正数, A*=ΣA_i/p.
证明: For any r>1, ΣA_i^r/p≧A*^r.
这是 Jensen inequality. 因 f(x)=x^r 为 convex function.
网路或机率论的教本找一下 "Jensen inequality" 即可找到证明.
: I have finished above two questions by induction,
: and hope to proof them by another ways.
: Enjoy it!!!
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◆ From: 125.233.152.175
1F：推 CFE220 :推~ 07/09 07:59
※ 编辑: yhliu 来自: 125.233.153.42 (07/09 20:34)