Re: [微積] summable

[Vulpix (Sebastian)]

※ 引述《GSXSP (Gloria)》之銘言：
: Let 0< x_0 < 1 and x_{n+1} = x_n - x_n^2.
: Is x_n summable?

x_{n+1} = x_n - x_n^2 ≦ 1/4 for all n ≧ 0.

and we see that the rest of the sequence is unaffected

        by changing x_0 into  ( 1 - x_0 ).

W.L.O.G., we may assume that x_0 ≦ 1/2.

claim): x_n ≧ x_0/(n+1) for all n.

        the case for n=0 is surely true.

        x_{n+1} = x_n - x_n^2

                ≧ x_0/(n+1) - (x_0)^2/(n+1)^2

                ≧ x_0/(n+1) - x_0/[2*(n+1)^2]

                = x_0 * { (2n+1) / [2*(n+1)^2] }

                ≧ x_0/(n+2)

        so the claim is true, by MIT.

so Σx_n is greater than a harmonic series, and hence diverges.

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推 WES2163818  : 請問那個claim的下界怎麼想到的？
→ Vulpix      : 是算了一些x_n以後，才發現差不多是一個調和數列。
→ GSXSP       : Tks!
推 znmkhxrw    : x_n - x_n^2≧ x_0/(n+1) - (x_0)^2/(n+1)^2 這行
→ znmkhxrw    : 是怎麼來的呀?? 前面正 後面負
→ znmkhxrw    : 是怎麼看出>=的??
→ Vulpix      : 用到x_n<=1/2
→ Vulpix      : 還有f(x)=x-x^2在[0,1/2]上遞增。
推 znmkhxrw    : 喔喔!! 謝謝
※ 編輯: Vulpix (61.230.71.135), 05/29/2017 20:15:20