Re: [理工] [工數]-差分方程式

看板Grad-ProbAsk作者doom8199 (～口卡口卡　修～)時間15年前 (2010/05/02 02:06)推噓6(6推 0噓 3→)

留言9則, 7人參與討論串2/4 (看更多)

※ 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 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.113.141.151

推

iyenn

05/02 13:23, , 1^F

05/02 13:23, 1^F

推

qqac45

05/02 13:53, , 2^F

05/02 13:53, 2^F

推

gn00067504

05/02 18:37, , 3^F

05/02 18:37, 3^F