求高阶线性递推数列通项的一般方法

ak , ⋅ ⋅ ⋅, a2 , a1 是不全为零的实常数 , 当 x1 , x2 , ⋅ ⋅ ⋅ , xk 为已知常数时,求数列 {xn } 的通项 xn . 定 理 1 若数列 {α1n },{α2n }, ⋅ ⋅ ⋅,{α kn } 都满 足递推关系① ,则数列 {∑ Ajα jn } 也满足递推
i =1 k
A1 A2 xn = (α1n α 2n ⋅ ⋅ ⋅ akn ) . M A k x1 x 证明 ∵ xk +1 = ( a1 a2 ⋅⋅⋅ ak ) 2 M x k A1 A = (a1 a2 ⋅⋅⋅ ak ) A 2 M A k A1 k k k A2 = (∑ aiα1i ∑ aiα 2i ⋅ ⋅ ⋅ ∑ aiα ki ) M i=1 i=1 i=1 A k ・13・
j =1 j =1 i =1
k
k
k
= ∑[ ak − i+1 ∑ A jα j ( n+ k− i) ] ,
i =1 j =1
k
k
即数列 {∑ Ajα jn } 也满足递推关系①.
j =1
k
定 理 2 设数列 {α1n },{α2n }, ⋅ ⋅ ⋅,{α kn } 都满 足递推关系①,且存在常数 A1 , A2, ⋅ ⋅ ⋅, Ak 使得
[1] 毛金才.高阶线性递推数列通项公式的矩阵求法. 河北理科教学研究.2002.1.
,… , akn = λ
n −1 k
,解方程组②即可求得
A1 , A2, ⋅ ⋅ ⋅, Ak 的值,由定理 1 知 xn = A1α1 n + A2α 2n + ⋅ ⋅ ⋅ + Ak αkn 满足递推关系①,即通项 xn = A1α1 n + A2α 2n + ⋅ ⋅ ⋅ + Ak αkn . 例 1 已知点的序列 An ( xn ,0) , n ∈ N ,其中 x1 = 0, x2 = a (a > 0) , A3 是 线 段 A1 A2 的 中 点 , A4 是 线 段 A2 A3 的 中 点 , … , An 是 线 段 An− 2 An−1 的中点,….(I) 写出 xn 与 xn −1 、 xn − 2 之 间的关系式 (n ≥ 3) ；(II)求 lim xn .
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求高阶线性递推数列 通项的一般方法
山东省安丘市 7571 信箱 邹 明
文 [1] 给 出 了 xn +3 = pxn + 2 + qxn+1 + rxn 型 递推数列通项公式的 Jordan 矩阵求法.本文给 出求一般的高阶线性递推数列通项的初等方 法(为叙述简洁而用矩阵形式),而且更为简捷 和可操作.即已知数列 {xn } 满足 xn+ k = ak xn +k −1 + ⋅ ⋅ ⋅ + a2 xn+! + a1 xn = ∑ ak − i +1xn+ k − i ① , 其 中
n −1 2 n −1 1பைடு நூலகம்
2 lim xn = a . n →∞ 3 例 2 已知 x1 = 0 , x2 = 1 , x3 = 3 , xn+ 3 = 2 xn + 2 + xn +1 − 2 xn ,求 xn . 解 解 λ 3 = 2λ 2 + λ − 2 得 λ1 = 1 , λ2 = − 1 , λ3 = 2 . ∴ α1n = λ1n −1 = 1 , α 2 n = λ2n −1 = (− 1) n −1 , α3 n = λ3n −1 = 2n −1 . A1 + A2 + A3 = 0, 解方程组 A1 − A2 + 2 A3 = 1, 得 A + A + 4A = 3 2 3 1 A1 = −1 , A2 = 0 , A3 = 1 .
n −1 ∴ xn = A1α1 n + A −1 . 2 α2 n + A 3α 3 n = 2 (ii)若方程③有重根,设有 r (1 < r < k ) 重根
,
λ1 = λ2 = ⋅ ⋅ ⋅ = λr ,则取 α1 n = λ1n −1 , α 2n = ( n − 1) ⋅λ1n −1 , α3 n = ( n − 2) λ1n −1 , ⋅ ⋅ ⋅, arn = (n − r + 1) λ1n −1 , 据定理 1 知 {α 2n },{α 3n}, ⋅ ⋅ ⋅,{α rn } 也满足①,对 于多组重根的情形也做类似处理 ,对于非重 根的 αin 的取法则同情形(i ).依据定理 2 得通 项 xn = A1α1 n + A2α 2n + ⋅ ⋅ ⋅ + Ak αkn . 例 3 已知 x1 = 1 , x2 = 2 , x3 = 5 , x4 = 6 , xn+ 4 = 2 xn +3 − 2 xn +1 + xn ,求 xn . 解 解特征方程 λ 4 = 2λ 3 − 2λ + 1 得 λ1 = λ2 = λ3 = 1, λ4 = −1 . ∴ α1n = λ1n −1 = 1 , α 2 n = ( n − 1)λ1n −1 = n − 1 , α3 n = ( n − 2)λ1n −1 = n − 2 , α 4 n = λ4n −1 = (− 1) n −1 . A1 − A3 + A4 = 1, A + A2 − A4 = 2, 解方程组 1 得 A1 + 2 A2 + A3 + A4 = 5, A1 + 3 A2 + 2 A3 − A4 = 6 A1 = 0 , A2 = 5 / 2 , A3 = −1/2, A4 = 1 / 2 . ∴ xn = A1α1 n + A 2α2 n + A 3α3 n + A 4α 4 n 1 3 = 2n + ( −1) n−1 − . 2 2 参考文献
[1] 陈龙安.创造性思维教与学.中国轻工业出版社, 1999.6. [2] 陈丽琴.如何在课堂教学中培养学生的创新意识. 中学数学研究,2002.6. [3] 张大华.培养学生创新意识和能力的再认识.中学 数学研究,2002.8. [4] 沈一凡.数学教学中创新素质的培养.中学数学教 学,2002.6. [5] 吴以浩.计算机辅助数学教学的误区及反思.中学 数学教学参考,2001.6. [6] 宋殿良等.多媒体辅助数学教学的实践与思考.中 学数学杂志(高中),2000.4.
j =1 k
关系①.其中 A1 , A2, ⋅ ⋅ ⋅, Ak 为任意常数. 证明 ∵ α j ( n+ k ) = ∑ ak − i+1α j ( n+ k− i) ( j = 1,2, ⋅⋅⋅, k ) ,
i =1 k
∴ ∑ Ajα j ( n+k ) = ∑∑ Aj ak −i +1α j ( n+ k −i )
n →∞
(2000 年春季高考题) x +x 解 (I)当 n ≥ 3 时, xn = n−1 n− 2 ; 2 1 1 1 (II)解 λ 2 = λ + 得 λ1 = 1 、 λ2 = − . 2 2 2 1 n −1 ∴ α1n = λ1n −1 = 1 、 α 2n = λ2 = (− )n−1 , 2 A1 + A2 = 0, 2 2 解 得 A1 = a, A2 = − a, 1 3 3 A − A =a 1 2 2 2 1 ∴ xn = A1α1 n + A2α2 n = a[1 − ( − ) n−1 ], 3 2
活地应用各种有效的数学分析方法 .(3) 迁移 能力强 ,能举一反三. 多媒体教学中应注重知 识间比较 ,通过比较展现数学知识间的联系 与区别,及对其应用加以总结、 归纳,注意把知 识模块化 ,通过多媒体将知识模块化演示. 如 1 1 由 x ∈ R + , 则 x + ≥ 2 , 给出 x2 + 2 + 2 的 x x +1 最值 ,让学生观察 ,寻找问题的解决方法,通过 这样的教学设计 ,有利于培养学生的发散思 维能力 ,同时也可使学生联想能力得以增强 , 产生新方法,有了创新. 参考文献
x1 x2 = M x k A1 α11 α 21 A α α 22 A 2 ②,其中 A = 12 M ⋅⋅⋅ ⋅⋅⋅ α A 1 k α2 k k ⋅ ⋅ ⋅ αk1 ⋅ ⋅ ⋅ αk2 , ⋅⋅⋅ ⋅⋅⋅ ⋅ ⋅ ⋅ α kk
= A1 ∑ aiα1i + A2 ∑ aiα 2i + ⋅ ⋅ ⋅ + Ak ∑ aiα ki
i =1 i =1 i =1
k
k
k
= A1α1( k +1) + A2α2( k +1) + ⋅ ⋅ ⋅ + Ak α k ( k +1) , ⋅⋅⋅, 依次递推得 xn = A1α1 n + A2α 2n + ⋅ ⋅ ⋅ + Ak αkn . 下面寻求数列 {α1n },{α2n }, ⋅ ⋅ ⋅,{α kn } ,并确 定 A1 , A2, ⋅ ⋅ ⋅, Ak ,从而依据定理 2 求出通项. 将 {α1n },{α 2 n}, ⋅ ⋅ ⋅, {α kn } 取为等比数列 1, λ , λ 2 , ⋅ ⋅ ⋅, λ n−1 , ⋅ ⋅ ⋅, (λ ≠ 0) 代入①得: λ n + k −1 = ak λ n + k −2 + ⋅ ⋅ ⋅ + a2 λ n + a1 λ n −1 即 λ k = ak λ k −1 + ⋅ ⋅ ⋅ + a2 λ + a1 ( λ ∈ C ) , ③ ③称为递推数列①的特征方程 ,其根称 为特征方程③的特征根. (i)若方程③有 k 个不等的根 λ1 , λ2 , ⋅ ⋅ ⋅, λk , 则得数列 {α1n },{α2n }, ⋅ ⋅ ⋅,{α kn } 为 : a1 n = λ a2 n = λ