characteristic equation

Question

Apply the method of the characteristic equation to find an explicit closed formula for the numbers an that satisfy the recurrence equation
an = 2an-1(n-1在a下面) - an-2(n-2在a下面)
for n>= 2, with the initial values a0 = 4 and a1 = 1.

請問一下 為何公式不是 an = αn = A1*(1)n次方 + A2*(1)n次方??

Answer 1

Using characteristics equation:
Let an=αn 代入, 得到 αn = 2αn-1 - αn-2
除以 αn-2 得 α2 = 2α - 1, 或 (α-1)2 = 0
解為 α = 1, 1 (重根)
所以, an = αn = A1*(1)n n*A2*(1)n [註: * 是乘]
現在, 用已知的起始值來決定 A1 及 A2 到底是多少:
Let n = 0: ao = A1 0 = 4
Let n = 1: a1 = A1 A2 = 1
上面兩式可知 A1 = 4, A2 = -3
所以, 答案為 an = 4 - 3n

2007-10-21 18:45:27 補充：
若上面的特性方程式得α = λ1, λ2 且非重根, 則是
an = αn = A1*(λ1)^n 加 A2*(λ2)^n
若是α = λ重根 (如此題), 則是
an = αn = A1*(λ)^n + A2*(λ)^n
(請見
http://en.wikipedia.org/wiki/Recurrence_relation#Solving_generally)

2007-10-21 18:48:03 補充：
對不起, 補充時打錯了, 應是:
若是α = λ重根 (如此題), 則是
an = αn = A1*(λ)^n + n* A2*(λ)^n