Re: [中學] 遞迴數列解法一問

看板Math作者WasabiSushi (田馥甄Hebe)時間9年前 (2015/01/30 14:47)推噓2(2推 0噓 5→)

留言7則, 4人參與討論串3/3 (看更多)

※ 引述《dagood (不是姚仁傑喔~~)》之銘言： : A1=1 , An+1=(5An+3)/(An+3) ,求An : A1=1 A2=2 現在令 An = Pn/Qn 所以P1=1,Q1=1,P2=2,Q2=1 : An = Pn/Qn 代入An+1=(5An+3)/(An+3) : 得 Pn+1/Qn+1 = (5Pn+3Qn)/(Pn+3Qn) : 得 Pn+1=5Pn+3Qn : Qn+1=Pn+3Qn : => Qn+2 - 8Qn+1 + 12Qn=0 : 所以設 Qn= a*2^n + b*6*n , n=1,2 代入得 a=5/8 , b=-1/24 : 所以Qn=5/8*2^n - 1/24*6^n 代入Qn+1=Pn+3Qn : 得 Pn=-5/8*2^n - 3/24*6^n : 進而得到 An=(-5/8*2^n - 3/24*6^n)/(-5/8*2^n - 3/24*6^n) : 答案根本不對 : 我要問的是為什麼我這樣的方法算出來會有錯誤呢 : 應該這麼說好了我以前算非線性遞迴都用這各方法做 : 大部分都對,但是遇到了這題卻怎麼做都是錯 : 還是說我這各方法本身就有問題有瑕疵 ? 想請板上朋友解惑 : 想知道這麼做錯誤是發生在哪裡? thx~~ The logical mistake is that P_{n+1}=5P_n+3Q_n and Q_{n+1}=P_n+3Q_n. How do you know that? You cannot obtain the two equations even if P_n and Q_n are coprime because they may differ by + or -. Hence, I do not recommend you to solve this question by setting two variables P and Q. A direct method is that A_{n+1}+1=(6A_n+6)/(A_n+3). Set B_n=1/(A_n+1), and you will obtain B_{n+1}=B_n/3+1/6. Then you can go on to solve it. Notice that an easy proof will show that A_n>0 and hence A_n+1 is never 0, so B_n makes sense. -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 180.154.58.228 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1422600450.A.8B0.html

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