※ 引述《suzzdicon (舒玆迪控)》之銘言：
: a_(n+1) = 1/(4+a_n)
: a_0 = 1/4
: 證明:a_n收斂
1
Consider f(x) = ──── , x€[0,1]
4 + x
then we find：
(1) f：[0,1] → [0,1]
-1 1
(2) f'(x) = ────── , and │f'(x)│<= ── < 1 for all x€[0,1]
(4 + x)^2 16
Hence f(x) is a λ-contraction
so from (1),(2) , we can apply contraction mapping principle
1
it says there exists only one fixed point {a} , s.t. a = ──── (取正的！)
4 + a
and for all x€[0,1] , lim f^n(x) = a ----(*)
n→∞
1
Now consider x = a_1 = ──
4
1
so f^n(a_1) = f^(n-1)f(a_1) = f^(n-1)(─────) = f^(n-1)(a_2)
4 + a_1
= .... = a_(n+1)
Since (*) tells us lim f^n(a_1) = a => lim a_(n+1) = a
n→∞ n→∞
其實用contraction的話 不只證收斂 值也順便找出來
1
其實就是在證明解 x = ─── 的正確性而已
4 + x
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