→ hsnu310659 : 謝謝原po 04/05 08:21
※ 引述《hsnu310659 (Tang)》之銘言：
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4.
(a)
a_n = a_(n-1) + 2a_(n-2)
a_0 = 1
a_1 = 1
(b)
[ a_n ] = [1 2][ a_(n-1) ]
[a_(n-1)] [1 0][ a_(n-2) ]
[1 2] = P
[1 0]
(c)
Q = [2 1]
[1-1]
Q^(-1) = (-1/3)[-1 -1]
[-1 2]
(d)
R = Q^(-1)PQ
= (-1/3)[-2 -2][2 1]
[ 1 -2][1-1]
= (-1/3)[-6 0]
[ 0 3]
= [2 0]
[0-1]
(e)
應該是a_n吧
[ a_n ] = [1 2]^(n-1)[ a_1 ]
[a_(n-1)] [1 0] [ a_0 ]
= Q [2^(n-1) 0 ] Q^(-1) [1]
[ 0 (-1)^(n-1)] [1]
a_n = (-1/3)[-2^(n+1) + (-1)^(n+1)]
= (2/3)*2^n + (1/3)(-1)^n
: 求解這三題要怎麼做呢～
: 一直寫不出來也跟班上討論過了QQ
: 拜託大家了!!
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