Re: [微積] Cos[Cos[Cos...[Cos[x]...]]]]=?
提供另外一個證法：
n
f(x)=cosx , f (x)=cos(cos(cos...cos(x))) 摳n次
n
Prove that for all x€Real number, lim f (x) exists and are the same value
n→∞
<pf>
1.在cosx中，看整個實數相當於看[0,π]
2.對於[π/2,π]這個區間我們有：
-1 =< cosx <= 0
so cos1 =< cos(cosx) <= 1
所以對於[π/2,π]，摳兩次就會到在[0,π/2)內
n
3.只需證明，對於[0,π/2)內所有的x，lim f (x)存在且相等即可
n→∞
1.for all 0.1 > ε > 0 , for all x€[0,π/2-ε],│f'(x)│=│-sinx│<= λ(ε)
where λ(ε) is a constant dependent on ε, and less than 1
so f(x) is a λ(ε)-contraction on [0,π/2-ε]
2.show f：[0,π/2-ε]→[0,π/2-ε]
Since x€[0,π/2-ε]
so 0 < cosx <= 1 , where (0,1] is included by [0,π/2-ε] (ε<0.1)
3.From 1. and 2. , By Contraction mapping principle,
f has an unique fixed pt. p(ε)€[0,π/2-ε] (f(p(ε))=p(ε))
n
and for all x€[0,π/2-ε], lim f (x) = p(ε)
n→∞
(目前這個不動點跟ε有關)
4.Show p(ε) is constant：
if ε=/= ε' , suppose ε<ε', so [0,π/2-ε'] is included by [0,π/2-ε]
Since p(ε') , p(ε) belongs to [0,π/2-ε'] , [0,π/2-ε] respectively
so f(p(ε')) = p(ε') and f(p(ε)) = p(ε)
Since in [0,π/2-ε], the fixed pt. is unique and p(ε'),p(ε)€[0,π/2-ε]
so p(ε')=p(ε)
5.Since ε is arbitrary, we are done.
(因為[0,π/2)是半開區間，任取其中一個點，都能找到一個[0,π/2-ε]包住它)
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12/18 18:44, , 1F
12/18 18:44, 1F
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