作者xmisery (misery)
看板Grad-ProbAsk
標題Re: [理工] 遞迴求解
時間Tue Mar 24 17:10:52 2009
※ 引述《showforce (秀佛阿~~~~斯)》之銘言：
: Find all soluotions of recurrence relation
: a = 5a - 6a + 2ˆn + 3n
: n n-1 n-2
沒給a(0) a(1)....
假設a(0)=0 a(1)=1
α^2-5α+6=0 α=2,3
ａ(h)n = c0*(2)^n + c1*(3)^n
ａ(p)n = (d0+d1*n)*2^n + d2 + d3*n
代入原式求得d1=-2, d3=3/2, d2=21/4
ａn = ａ(h)n + ａ(p)n = (c0+d0-2n)2^n + c1*3^n + 21/4 + (3/2)*n
令c0+d0=A
ａn = (A-2n)2^n + c1*3^n + 21/4 + (3/2)*n
代入邊界條件
0=a(0)=A+c1+21/4
1=a(1)=2(A-2)+3*c1+27/4
解得A=-14 c1=35/4
ａn = (-14-2n)2^n + (35/4)*3^n + 21/4 + (3/2)*n
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 118.169.210.239
1F：推 s987692:可以問一下那特解釋如何令的嗎? 03/24 18:19
2F：推 Derix:囧...原PO不就說假設的了嗎XD 03/24 19:36
3F：推 s987692:特解的令法,不是初值 03/24 20:59