求遞迴關係式的解 - 考試
By Puput
at 2013-04-01T13:11
at 2013-04-01T13:11
Table of Contents
T(n)={ 1 , n<=1 ; T(n-1)+1 ,n>1 }
我用 T(n)-T(n-1)=1 -----(1)
T(n-1)-T(n-2)=1 -----(2)
(1)-(2)
= T(n)-2T(n-1)+T(n-2)=0
特徵方程式 r^2-2r+1
解得 r=1
通解為 c1*1^n + c2*1^n
可是這題的 T(n) = n
請問通解跟 T(n) = n 有甚麼關係嗎?
--
我用 T(n)-T(n-1)=1 -----(1)
T(n-1)-T(n-2)=1 -----(2)
(1)-(2)
= T(n)-2T(n-1)+T(n-2)=0
特徵方程式 r^2-2r+1
解得 r=1
通解為 c1*1^n + c2*1^n
可是這題的 T(n) = n
請問通解跟 T(n) = n 有甚麼關係嗎?
--
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