作者doom8199 (～口卡口卡 修～)
看板Grad-ProbAsk
標題Re: [理工] [工數]-差分方程式
時間Sun May 2 02:06:14 2010
※ 引述《qqac45 (阿治)》之銘言：
: Find the solution of the finite difference equation
: Rn=3(Rn-1)+5(Rn-2)-15(Rn-3)-4(Rn-4)+12(Rn-5)
: with initial conditions
: R0=5,R1=1,R2=3,R3=1 and R4=3
: 想了很久 也不知怎麼解
: 麻煩高手解一下
---
│ +1 -3 -5 +15 +4 -12
+1 │ +1 -2 -7 +8 +12
└────────────────
+1 -2 -7 +8 +12 0
-1 │ -1 +3 +4 -12
└─────────────
+1 -3 -4 +12 0
+2 │ +2 -2 -12
└──────────
+1 -1 -6 0
-2 │ -2 +6
└────────
+1 -3 0
+3 │ +3
└─────
+1 0
令 A[n] = R[n] - 2R[n-1] - 7R[n-2] + 8R[n-3] + 12R[n-4]
B[n] = R[n] - 3R[n-1] - 4R[n-2] + 12R[n-3]
C[n] = R[n] - R[n-1] - 6R[n-2]
D[n] = R[n] - 3R[n-1]
with ┌ A[4] = 48
│ B[3] = 48
│ C[2] = -28
└ D[1] = -14
則原式
→ A[n] = A[n-1]
= A[4]
= 48
→ B[n] - 24 = -(B[n-1] - 24)
= [(-1)^(n-3)]*(B[3] - 24)
= -24*(-1)^n
→ C[n] + m[n] = 2(C[n-1] + m[n-1]) with m[n] = 24 + 8(-1)^n
= 2^(n-2) * (C[2] + m[2])
= 2^n
→ C[n] = -24 - 8*(-1)^n + 2^n
→ D[n] + p[n] = -2(D[n-1] + p[n-1]) with p[n] = 8 - 8(-1)^n - (1/2)2^n
= (-2)^(n-1) * (D[1] + p[1])
= (-1/2)*(-2)^n
→ D[n] = -8 + 8*(-1)^n + (1/2)*2^n + (-1/2)*(-2)^n
→ R[n] + q[n] = 3*(R[n-1] + q[n-1])
with q[n] = -4 - 2(-1)^n + 2^n + (1/5)(-2)^n
= 3^n * (R[0] + q[0])
= (1/5)*3^n
n n n n
→ R[n] = 4 + 2*(-1) - 2 - (1/5)*(-2) + (1/5)*3
----
感覺這題 原po 不是不會算
而是計算量龐大反而不敢親自下筆解
1元5次方程式 和 5元1次聯立方程組 XD
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◆ From: 140.113.141.151
推 iyenn:+1 05/02 13:23
推 qqac45:謝謝大大願意花時間幫解~感恩 05/02 13:53
推 gn00067504:這題好像今年交大光顯有考到是嗎? 05/02 18:37
推 QQkimi:有考到喔...我怎麼沒印象XD 考完就變廢了... 05/02 18:38
推 tzucheng:有看有推 感謝高手解答 我也想了好久都不會 05/02 19:46
→ doom8199:考試考這種題目感覺很沒營養＝＝ , 除非有心算技能加持 05/03 13:31
→ doom8199:不然像我們一般人算這種題目，20分鐘一定跑不掉 OTZ 05/03 13:32
→ honestonly:樓上心算技能Lv:Max exp:100% 05/04 00:02
推 tzucheng:原來周易把差分放在7-5，註明不太會考.... 05/04 00:28