推 KitWoolsey :strick ly@@? 06/15 00:33
※ 編輯: znmkhxrw 來自: 111.243.148.115 (06/15 00:34)
→ znmkhxrw :XD.. 06/15 00:34
推 dreamenjoy :感恩! 06/15 08:48
1 4
Consider f(x) = ─(x + ─)
2 x
x 2
= ─ + ─
2 x
Using derivative , we have for x > 0 ,
f(x) is strictly decreasing , for 0 < x <= 2
f(x) is strictly increasing , for 2 <= x < inf ....(*)
then we see a1 as f(4)
so an = f^n(4) = f(f(.....f(4))) (iterating for n steps)
we find that：
2 <= f(4) = 2.5 <= 4
by (*) , we have：
f(2) = 2 <= f(f(4)) = f^2(4) <= f(4)
so：
2 <= f^2(4) <= f(4)
step by step , since f(2) = 2 , we have：
2 <= f^n(4) <= f^(n-1)(4)
i.e. an <= a(n-1) , so an is decreasing with lower bound 2
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