推 ericakk :感恩你!!祝福你這學期all pass喔!! ^_^ 05/15 18:48
※ 編輯: yueayase (36.236.229.213), 03/20/2015 21:24:11
※ 引述《ericakk (ericakk)》之銘言：
: 已知平面上的點排列：P_1(x_1,y_1)、P_2(x_2,y_2)、...、P_n(x_n,y_n)
: 其中各點座標定義如下：
: x_1=1 , y_1=0, x_n+1= 3/5 x_n + 2/5 y_n , y_n+1= 2/5 x_n + 3/5 y_n ,
: 試求：（1）x_n + αy_n =β(x_n-1 + αy_n-1)的正實數α、β？
: （2）lim p_n = (a,b)，則(a,b)？
: n→∞
(1) x(n) + αy(n) = (3/5 + 2α/5)x(n-1) + (2/5 + 3α/5)y(n-1)
=> 3/5 + 2α/5 = β --- (1) , 2/5 + 3α/5 = αβ --- (2)
(1)+(2) => 1 + α = β(1 + α) => (β - 1)(1 + α) = 0 --- (3)
根據(1), (3)
=> α = -1 , β = 1/5 or α = 1, β = 1
因為α、β都是正實數, 所以 α = 1, β = 1
(2) 根據(1), x(n) + y(n) = x(n-1) + y(n-1) = ... = x(1) + y(0) = 1
=> y(n) = 1 - x(n)
又 x(n) = 3/5 x(n-1) + 2/5 y(n-1)
=> x(n) = 1/5 x(n-1) + 2/5
找出 x(n) - γ = 1/5(x(n-1) - γ) 的 γ
=> x(n) = 1/5 x(n-1) + 4γ/5
=> γ = 1/2
x(n) - 1/2 x(n-1) - 1/2 x(2) - 1/2 n-1
=> -------------- * ------------- * ... * ------------ = (1/5)
x(n-1) - 1/2 x(n-2) - 1/2 x(1) - 1/2
n-1
=> x(n) = 1/2 + 1/2 * (1/5)
n-1
y(n) = 1 - x(n) = 1/2 - 1/2 *(1/5)
所以 lim p_n = lim (x(n), y(n)) = (1/2, 1/2)
n→∞ n→∞
其實用α = -1 , β = 1/5 也可以解, 寫出一些關鍵:
n-1 n-1
x(n) - y(n) = (1/5) (x(1)-x(0)) = (1/5)
n-1
=> x(n) = y(n) + (1/5) , y(n) = 2/5 x(n-1) + 3/5 y(n-1)
n-2
=> y(n) = 2/5 * (1/5) + y(n-1)
n-2 k
=> y(n) = 2/5 Σ (1/5)
k=0
n-2 k n
x(n) = 2/5 Σ (1/5) + (1/5)
k=0
∞ k n
lim x(n) = 2/5 Σ (1/5) + lim (1/5) = 2/5 * [1/(1-1/5)] = 1/2
n->∞ k=0 n->∞
∞ k
lim y(n) = 2/5 Σ (1/5) = 1/2
n->∞ k=0
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