Re: [理工] 101台科離散
※ 引述《white8824 (hypocrisy*)》之銘言：
: http://www-o.ntust.edu.tw/~lib/pdf/Master/101/m1010802.pdf
: 題目第三題 文法 求Kleene Closure A* 的遞迴
: 有人有解出來這題嗎@@?
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a(n) = a(n-1) +a(n-2) + a(n-3) + 2*a(n-4)
with [ a(1), a(2), a(3), a(4)] = [ 1, 2, 4, 9]
=> a(n) + a(n-1) + a(n-2) + a(n-3) = 2*[a(n-1) + a(n-2) + a(n-3) + a(n-4)]
= 2^(n-4) *[9 + 4 + 2 + 1]
= 2^n
=> a(n) + a(n-2) + (-2/3)2^n = -[a(n-1) + a(n-3) + (-2/3)*2^(n-1)]
= (-1)^(n-3) *[4 + 1 + (-2/3)*2^3]
= (1/3)*(-1)^n
n n n-2 n-2
=> a(n) + (-8/15)2 + (-1/6)*(-1) = -[a(n-2) + (-8/15)2 + (-1/6)(-1) ]
n/2
┌ (3/10)(-1) if n:even
= │
│ (n-1)/2
└ (1/10)(-1) if n:odd
= (3/10)cos(nπ/2) + (1/10)sin(nπ/2)
n n
即 a(n) = (8/15)2 + (1/6)(-1) + (3/10)cos(nπ/2) + (1/10)sin(nπ/2)
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※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.113.211.139
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剛剛試算一下解四元一次方程式，好像比想樣中簡單 = =ll
n n
令 a(n) = c1*2 + c2*(-1) + c3*cos(nπ/2) + c4*sin(nπ/2)
所以 ┌ 2*c1 - c2 + c4 = 1 ____ (1)
│ 4*c1 + c2 - c3 = 2 ____ (2)
│ 8*c1 - c2 - c4 = 4 ____ (3)
└ 16*c1 + c2 + c3 = 9 ____ (4)
┌ (1)+(3) => ┌ 10*c1 - 2*c2 = 5 => ┌ c1 = (8/15)
└ (2)+(4) └ 20*c1 + 2*c2 = 11 └ c2 = (1/6)
再由 (1)(2) 式可得 ┌ c3 = 4*c1 + c2 - 2 = (32/15) + (1/6) - 2 = 3/10
└ c4 = -2*c1 + c2 - 1 = (-16/15) + (1/6) + 1 = 1/10
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看來我上面的做法有點繞遠路 QQ
※ 編輯: doom8199 來自: 140.113.211.139 (02/25 01:32)
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