和生成函數相似, 遞歸方程也是一種有效的組合計數工具. 下面, 我們對其進行簡要介紹.
[例1] Fibonacci 數列2 2 2 n n n
[解]f n − 1 f_{n-1} f n − 1 n − 1 n-1 n − 1 f n f_{n} f n
{ f 0 = f 1 = 1 f n + 2 = f n + 1 + f n \begin{cases}f_0 = f_1 = 1 \\ f_{n+2} = f_{n+1} + f_{n}\end{cases} { f 0 = f 1 = 1 f n + 2 = f n + 1 + f n
[例2]n n n 1 1 1 { 0 , 1 } \{0,1\} { 0 , 1 }
[解]f n f_n f n n n n { 0 , 1 } \{0,1\} { 0 , 1 } S n S_n S n a n a_n a n
若 a n = 0 a_n = 0 a n = 0 a n a_n a n S n S_n S n n − 1 n-1 n − 1 0 0 0 n n n
若 a n = 1 a_n = 1 a n = 1 0 0 0 1 1 1 n − 2 n-2 n − 2 f n = f n − 1 + f n − 2 . f_n = f_{n-1} + f_{n-2}. f n = f n − 1 + f n − 2 .
[注]爲了避免遞推式和實際情況發生衝突, 我們可人爲定義: f 0 = 1 , f 1 = 2 , f_0 = 1, f_1 =2, f 0 = 1 , f 1 = 2 ,
[例3]R 2 \mathbb{R}^2 R 2 n n n
任何三條直線不過同一點
任何兩條直線不平行
[解]d n d_n d n n n n R 2 \mathbb{R}^2 R 2 d 0 = 1 , d 1 = 2 , d 2 = 4 , d 3 = 7. d_0 = 1, d_1 = 2, d_2 = 4, d_3 = 7. d 0 = 1 , d 1 = 2 , d 2 = 4 , d 3 = 7 . l n l_n l n n − 1 n-1 n − 1 n − 1 n-1 n − 1 l n l_n l n n − 1 n-1 n − 1 { d n = d n − 1 + n d 0 = 1 , n ⩾ 1. \begin{cases}d_n = d_{n-1} + n \\ d_0 = 1\end{cases}, ~~~ n\geqslant 1. { d n = d n − 1 + n d 0 = 1 , n ⩾ 1 .
在上面的例子中, 我們可以看出最終計數問題均被轉化爲求一個遞推方程 f i f_i f i i i i f n − f n − 1 − f n − 2 = g ( x ) . f_n - f_{n-1} - f_{n-2} = g(x). f n − f n − 1 − f n − 2 = g ( x ) .
定義 9.1
設 b 1 , b 2 , ⋯ , b k , a 0 , a 2 , ⋯ , a k − 1 b_1, b_2, \cdots, b_k, ~~~a_0, a_2, \cdots, a_{k-1} b 1 , b 2 , ⋯ , b k , a 0 , a 2 , ⋯ , a k − 1 b k ≠ 0 b_k \neq 0 b k = 0 f ( n ) f(n) f ( n ) { H ( n ) } n = 0 ∞ \{H(n)\}^{\infty}_{n= 0} { H ( n ) } n = 0 ∞ ( 1 ) { H ( 0 ) = a 0 ⋮ H ( k − 1 ) = a k − 1 H ( n ) + b 1 H ( n − 1 ) + ⋯ + b n − k H ( k ) = f ( n ) , n ⩾ k (1)~~\begin{cases}H(0) = a_0 \\~~~~~ \vdots \\ H(k-1) = a_{k-1} \\ H(n) + b_1H(n-1) + \cdots + b_{n-k}H(k) = f(n), ~~ n\geqslant k\end{cases} ( 1 ) ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ H ( 0 ) = a 0 ⋮ H ( k − 1 ) = a k − 1 H ( n ) + b 1 H ( n − 1 ) + ⋯ + b n − k H ( k ) = f ( n ) , n ⩾ k { H ( a ) } n = 0 ∞ \{H(a)\}^{\infty}_{n= 0} { H ( a ) } n = 0 ∞ 常係數線性迴歸序列 , 並稱 ( 1 ) (1) ( 1 ) 常係數線性迴歸方程 . 若 f ( n ) = 0 f(n) = 0 f ( n ) = 0 齊次的 (homogeneous), 否則稱其爲 非齊次的 (nonhomogeneous).
[例4]( ∗ ) { f n − f n − 1 − f n − 2 = 0 f 0 = f 1 = 1 . (*)~~\begin{cases}f_n - f_{n-1} - f_{n-2} = 0 \\ f_0 = f_1 = 1\end{cases}. ( ∗ ) { f n − f n − 1 − f n − 2 = 0 f 0 = f 1 = 1 .
[解]( ∗ ) (*) ( ∗ ) q n , q ≠ 0 q^{n}, ~~ q\neq 0 q n , q = 0 f n f_n f n q n q^{n} q n q n − q n − 1 − q n − 2 = q n − 2 ( q 2 − q − 1 ) = 0. q^{n} - q^{n-1} - q^{n-2} = q^{n-2}(q^2 - q -1) = 0. q n − q n − 1 − q n − 2 = q n − 2 ( q 2 − q − 1 ) = 0 . q 1 = 1 + 5 2 , q 2 = 1 − 5 2 . q_1 = \frac{1 + \sqrt{5}}{2}, ~~q_2 = \frac{1 - \sqrt{5}}{2}. q 1 = 2 1 + 5 , q 2 = 2 1 − 5 . C 1 ⋅ q 1 n + C 2 ⋅ q 2 n C_1\cdot q_1^{n} + C_2\cdot q_2^{n} C 1 ⋅ q 1 n + C 2 ⋅ q 2 n C 1 = 1 5 ⋅ 1 + 5 2 , C 2 = − 1 5 ⋅ 1 − 5 2 . C_1 = \frac{1}{\sqrt{5}}\cdot \frac{1 + \sqrt{5}}{2}, ~~C_2 = -\frac{1}{\sqrt{5}}\cdot \frac{1 - \sqrt{5}}{2}. C 1 = 5 1 ⋅ 2 1 + 5 , C 2 = − 5 1 ⋅ 2 1 − 5 . f n = 1 5 ⋅ ( 1 + 5 2 ) n + 1 − 1 5 ⋅ ( 1 − 5 2 ) n + 1 , n ⩾ 0. f_n = \frac{1}{\sqrt{5}}\cdot (\frac{1 + \sqrt{5}}{2})^{n+1}-\frac{1}{\sqrt{5}}\cdot (\frac{1 - \sqrt{5}}{2})^{n+1}, ~~ n\geqslant 0. f n = 5 1 ⋅ ( 2 1 + 5 ) n + 1 − 5 1 ⋅ ( 2 1 − 5 ) n + 1 , n ⩾ 0 .
更爲一般地, 設線性常係數遞歸方程H ( n ) + b 1 H ( n − 1 ) + ⋯ + b k H ( n − k ) = 0 ( ∗ ∗ ) H(n) + b_1H(n-1) + \cdots + b_kH(n-k) = 0~~ (**) H ( n ) + b 1 H ( n − 1 ) + ⋯ + b k H ( n − k ) = 0 ( ∗ ∗ ) H ( m ) = q n , q ≠ 0 H(m) = q^{n}, ~~ q\neq 0 H ( m ) = q n , q = 0 q k + b 1 q k − 1 + ⋯ + b k = 0 ( ∗ ∗ ∗ ) q^{k} + b_1q^{k-1} + \cdots + b_{k} = 0 ~~ (***) q k + b 1 q k − 1 + ⋯ + b k = 0 ( ∗ ∗ ∗ ) ( ∗ ∗ ) (**) ( ∗ ∗ ) 特徵多項式 .
定理 9.1
設常係數迴歸方程 ( ∗ ∗ ) (**) ( ∗ ∗ ) H ( i ) = a i , 0 ⩽ i ⩽ k − 1 H(i) = a_i, ~~~ 0\leqslant i\leqslant k-1 H ( i ) = a i , 0 ⩽ i ⩽ k − 1 ( ∗ ∗ ∗ ) (***) ( ∗ ∗ ∗ ) k k k q 1 , q 2 , ⋯ , q k q_1, q_2, \cdots, q_k q 1 , q 2 , ⋯ , q k c 1 , c 2 , ⋯ , c k c_1,c_2, \cdots, c_k c 1 , c 2 , ⋯ , c k ∑ i = 1 k q i k \sum_{i=1}^{k}q_{i}^{k} ∑ i = 1 k q i k
[證明]q 1 , q 2 , ⋯ , q k q_1,q_2,\cdots, q_k q 1 , q 2 , ⋯ , q k ( ∗ ∗ ∗ ) (***) ( ∗ ∗ ∗ ) k k k c 1 , c 2 , ⋯ , c k c_1,c_2, \cdots, c_k c 1 , c 2 , ⋯ , c k ∑ i = 1 k q i k \sum_{i=1}^{k}q_{i}^{k} ∑ i = 1 k q i k ( ∗ ∗ ) (**) ( ∗ ∗ ) n ∈ { 0 , 1 , ⋯ , k − 1 } n \in \{0,1,\cdots, k-1\} n ∈ { 0 , 1 , ⋯ , k − 1 } ( Q ) { a 0 = c 1 + c 2 + ⋯ + c k a 1 = c 1 q 1 + c 2 q 2 + ⋯ + c k q k ⋯ ⋯ ⋯ ⋯ a k − 1 = c 1 q 1 k − 1 + c 2 q 2 k − 1 + ⋯ + c k q k k − 1 , i . e . [ 1 1 ⋯ 1 q 1 q 2 ⋯ q k ⋮ ⋮ ⋮ q 1 k − 1 q 2 k − 1 ⋯ q k k − 1 ] ⋅ [ c 1 c 2 ⋮ c k ] = [ a 0 a 1 ⋮ a k − 1 ] . (\mathscr{Q})~~\begin{cases}a_0 = c_1 + c_2 + \cdots + c_k \\ a_1 = c_1q_1 + c_2q_2 + \cdots + c_kq_k \\ \cdots ~~ \cdots ~~ \cdots ~~ \cdots \\ a_{k-1} = c_1q_1^{k-1} + c_2q_2^{k-1} + \cdots + c_{k}q_k^{k-1} \end{cases}, ~~~ i.e. \begin{bmatrix} 1& 1& \cdots & 1 \\ q_1& q_2& \cdots & q_k \\ \vdots & \vdots & &\vdots \\ q_1^{k-1} & q_2^{k-1} & \cdots & q_{k}^{k-1}\end{bmatrix} \cdot \begin{bmatrix} c_1 \\c_2\\\vdots\\c_k \end{bmatrix} = \begin{bmatrix}a_0 \\ a_1 \\ \vdots \\ a_{k-1} \end{bmatrix}. ( Q ) ⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ a 0 = c 1 + c 2 + ⋯ + c k a 1 = c 1 q 1 + c 2 q 2 + ⋯ + c k q k ⋯ ⋯ ⋯ ⋯ a k − 1 = c 1 q 1 k − 1 + c 2 q 2 k − 1 + ⋯ + c k q k k − 1 , i . e . ⎣ ⎢ ⎢ ⎢ ⎡ 1 q 1 ⋮ q 1 k − 1 1 q 2 ⋮ q 2 k − 1 ⋯ ⋯ ⋯ 1 q k ⋮ q k k − 1 ⎦ ⎥ ⎥ ⎥ ⎤ ⋅ ⎣ ⎢ ⎢ ⎢ ⎡ c 1 c 2 ⋮ c k ⎦ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎡ a 0 a 1 ⋮ a k − 1 ⎦ ⎥ ⎥ ⎥ ⎤ . ( Q ) (\mathscr{Q}) ( Q ) c 1 , c 2 , ⋯ , c k c_1, c_2, \cdots, c_k c 1 , c 2 , ⋯ , c k ■ \blacksquare ■
設 C \mathscr{C} C R \mathbb{R} R k k k C \mathscr{C} C k + 1 k+1 k + 1
設 C = { b 0 + b 1 k + ⋯ + b k n k ∣ b i ∈ R , 0 ⩽ i ⩽ k , n ∈ N ∪ { 0 } } \mathscr{C} = \{b_0 + b_{1}k + \cdots + b_{k}n^{k} ~|~b_i \in \mathbb{R}, 0\leqslant i\leqslant k, n \in \mathbb{N}\cup \{0\} \} C = { b 0 + b 1 k + ⋯ + b k n k ∣ b i ∈ R , 0 ⩽ i ⩽ k , n ∈ N ∪ { 0 } } C \mathscr{C} C k + 1 k+1 k + 1 1 , n , n 2 , ⋯ , n k 1,n, n^2, \cdots, n^k 1 , n , n 2 , ⋯ , n k
定理 9.2
設 q q q ( ∗ ∗ ∗ ) (***) ( ∗ ∗ ∗ ) k k k c 0 , c 1 , ⋯ , c k − 1 c_0,c_1,\cdots, c_{k-1} c 0 , c 1 , ⋯ , c k − 1 c 0 q 0 n + c 1 n q 0 n + ⋯ + c k − 1 n k − 1 q 0 n c_0q_0^{n} + c_1nq_0^{n} + \cdots + c_{k-1}n^{k-1}q_0^{n} c 0 q 0 n + c 1 n q 0 n + ⋯ + c k − 1 n k − 1 q 0 n ( ∗ ∗ ) (**) ( ∗ ∗ )
[證明]P ( q ) = q n + b 1 q n − 1 + ⋯ + b k q n − k P(q) = q^n + b_1q^{n-1} + \cdots + b_kq^{n-k} P ( q ) = q n + b 1 q n − 1 + ⋯ + b k q n − k q 0 ≠ 0 q_0 \neq 0 q 0 = 0 P ( q ) = 0 P(q) = 0 P ( q ) = 0 k k k q 0 q_0 q 0 P ( q ) = 0 P(q) = 0 P ( q ) = 0 i i i 0 ⩽ i ⩽ k − 1 0\leqslant i\leqslant k-1 0 ⩽ i ⩽ k − 1 P q ( i ) = n ! ( n − i ) ! q n − i + b 1 ( n − 1 ) ! ( n − i − 1 ) ! q n − i + 1 + ⋯ + b k ( n − k ) ! ( n − k − i ) ! q n − k − i P_{q}^{(i)} = \frac{n!}{(n-i)!}q^{n-i} + b_1\frac{(n-1)!}{(n-i-1)!}q^{n-i+1} + \cdots + b_k\frac{(n-k)!}{(n-k-i)!}q^{n-k-i} P q ( i ) = ( n − i ) ! n ! q n − i + b 1 ( n − i − 1 ) ! ( n − 1 ) ! q n − i + 1 + ⋯ + b k ( n − k − i ) ! ( n − k ) ! q n − k − i i . e i.e i . e 0 ⩽ i ⩽ k + 1 , n ( n − 1 ) ⋯ ( n − i + 1 ) q 0 n 0\leqslant i\leqslant k+1, ~~~n(n-1)\cdots(n-i+1)q_{0}^{n} 0 ⩽ i ⩽ k + 1 , n ( n − 1 ) ⋯ ( n − i + 1 ) q 0 n ( ∗ ∗ ) (**) ( ∗ ∗ ) ( ∗ ∗ ) (**) ( ∗ ∗ ) d 1 , d 2 , ⋯ , d k d_1,d_2,\cdots, d_k d 1 , d 2 , ⋯ , d k d 1 q 0 n + d 2 n q 0 n + ⋯ + d k n ( n − 1 ) ⋯ ( n − k + 1 ) q 0 n d_1q_0^n + d_2nq_0^n + \cdots + d_kn(n-1)\cdots(n-k+1)q_0^n d 1 q 0 n + d 2 n q 0 n + ⋯ + d k n ( n − 1 ) ⋯ ( n − k + 1 ) q 0 n ( ∗ ∗ ) (**) ( ∗ ∗ ) n n n { r 0 + r 1 n + ⋯ + r k − 1 n k − 1 ∣ r 0 , r 1 , ⋯ , r k − 1 ∈ R } \{r_0 + r_1n + \cdots + r_{k-1}n^{k-1} ~~|~~ r_0, r_1, \cdots, r_{k-1} \in \mathbb{R}\} { r 0 + r 1 n + ⋯ + r k − 1 n k − 1 ∣ r 0 , r 1 , ⋯ , r k − 1 ∈ R } k k k { 1 , n , ⋯ , n k − 1 } \{1,n,\cdots, n^{k-1}\} { 1 , n , ⋯ , n k − 1 } { 1 , n , ⋯ , n ( n − 1 ) ⋯ ( n − k + 1 ) ) } \{1,n,\cdots, n(n-1)\cdots (n-k+1))\} { 1 , n , ⋯ , n ( n − 1 ) ⋯ ( n − k + 1 ) ) }
定理 9.3
假設 ( ∗ ∗ ∗ ) (***) ( ∗ ∗ ∗ ) q 1 , q 2 , ⋯ , q r q_1,q_2,\cdots, q_r q 1 , q 2 , ⋯ , q r l 1 , l 2 , ⋯ , l r l_1, l_2, \cdots, l_r l 1 , l 2 , ⋯ , l r { q 1 n , n q 1 n , ⋯ , n l 1 − 1 q 1 n q 2 n , n q 2 n , ⋯ , n l 2 − 1 q 2 n ⋮ ⋮ ⋮ q n n , n q n n , ⋯ , n l r − 1 q n n . \begin{cases}q_1^{n}, & nq_1^{n}, & \cdots, n^{l_{1}-1}q_{1}^{n} \\ q_2^{n}, & nq_2^{n}, & \cdots, n^{l_{2}-1}q_{2}^{n} \\ ~\vdots & ~~~\vdots & ~~~~~\vdots \\ q_n^{n}, & nq_n^{n}, & \cdots, n^{l_{r}-1}q_{n}^{n}\end{cases}. ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ q 1 n , q 2 n , ⋮ q n n , n q 1 n , n q 2 n , ⋮ n q n n , ⋯ , n l 1 − 1 q 1 n ⋯ , n l 2 − 1 q 2 n ⋮ ⋯ , n l r − 1 q n n . ∑ i = 1 k ( ∑ j = 1 l C i j n j − 1 ) q n \sum_{i=1}^{k}(\sum_{j=1}^{l}C_{ij}n^{j-1})q^n ∑ i = 1 k ( ∑ j = 1 l C i j n j − 1 ) q n ( ∗ ∗ ) (**) ( ∗ ∗ ) 通解 .
[例5]{ t n } n = 0 ∞ \{t_n\}_{n = 0}^{\infty} { t n } n = 0 ∞ n n n { 0 , 1 , 2 } \{0,1,2\} { 0 , 1 , 2 } 20 20 2 0 t n t_n t n
[解]S = a 0 a 1 ⋯ a n S = a_0a_1\cdots a_n S = a 0 a 1 ⋯ a n 20 20 2 0 n n n { 0 , 1 , 2 } \{0,1,2\} { 0 , 1 , 2 }
若 a n = 1 a_n = 1 a n = 1 2 2 2 a 0 a 1 ⋯ a n − 1 a_0a_1\cdots a_{n-1} a 0 a 1 ⋯ a n − 1 n − 1 n-1 n − 1 n − 1 n-1 n − 1 2 , 1 2,1 2 , 1 n n n n n n 1 1 1 2 2 2 2 ⋅ t n + 1 2\cdot t_{n+1} 2 ⋅ t n + 1
若 a n = 0 a_n = 0 a n = 0 a n − 1 ≠ 2 a_{n-1} \neq 2 a n − 1 = 2 n − 1 n-1 n − 1 2 2 2 t n − 1 − t n − 2 t_{n-1} - t_{n-2} t n − 1 − t n − 2
於是可得: t n = 3 t n − 1 − t n − 2 t_n = 3t_{n-1} - t_{n-2} t n = 3 t n − 1 − t n − 2 i . e i.e i . e t n − 3 t n − 1 − t n − 2 = 0 , t ⩾ 3 t_n - 3t_{n-1} - t_{n-2} = 0, ~~ t\geqslant 3 t n − 3 t n − 1 − t n − 2 = 0 , t ⩾ 3 q 1 = 3 + 5 5 , q 2 = 3 − 5 5 q_1 = \frac{3+\sqrt{5}}{5}, ~ ~ q_2 = \frac{3-\sqrt{5}}{5} q 1 = 5 3 + 5 , q 2 = 5 3 − 5 1 1 1
故有一般形式:C 1 ⋅ 3 + 5 5 + C 2 ⋅ 3 − 5 5 , n ⩾ 1. C_1\cdot \frac{3+\sqrt{5}}{5} + C_2\cdot \frac{3-\sqrt{5}}{5}, ~~ n\geqslant 1. C 1 ⋅ 5 3 + 5 + C 2 ⋅ 5 3 − 5 , n ⩾ 1 . C 0 = 1 C_0 = 1 C 0 = 1 C 1 , C 2 C_1, C_2 C 1 , C 2
定理 9.4
設序列 { H n } n = 0 ∞ \{H_n\}^{\infty}_{n=0} { H n } n = 0 ∞ ( ∗ ) H ( n ) = b 1 H ( n − 1 ) − ⋯ − b k H ( n − k ) + f ( n ) , n ⩾ k (*) ~~~~~H(n) = b_1H(n-1) - \cdots -b_kH(n-k) + f(n), ~~ n\geqslant k ( ∗ ) H ( n ) = b 1 H ( n − 1 ) − ⋯ − b k H ( n − k ) + f ( n ) , n ⩾ k H ( 0 ) , H ( 1 ) , ⋯ , H ( n ) H(0), H(1), \cdots, H(n) H ( 0 ) , H ( 1 ) , ⋯ , H ( n ) ( ∗ ) (*) ( ∗ ) H ( n ) ‾ \overline{H(n)} H ( n ) H ∗ ( n ) H^*(n) H ∗ ( n ) H ( n ) ‾ + H ∗ ( n ) \overline{H(n)} + H^*(n) H ( n ) + H ∗ ( n ) ( ∗ ) (*) ( ∗ )
[Assignment ]
2. Solve the recurrence equation r n + 2 = r n + 1 + 2 r n r_{n+2} = r_{n+1} + 2r_{n} r n + 2 = r n + 1 + 2 r n r 0 = 1 r_{0} = 1 r 0 = 1 r 2 = 3 r_2 = 3 r 2 = 3 r 2 r_2 r 2 r 1 r_1 r 1
[Solution]q n − q n − 1 − 2 q n − 2 = 0 q^n - q^{n-1} - 2q^{n-2} = 0 q n − q n − 1 − 2 q n − 2 = 0 q 1 = 2 , q 2 = − 1. q_1 = 2, q_2 = -1. q 1 = 2 , q 2 = − 1 . q n = C 1 ⋅ ( 2 ) n + C 2 ⋅ ( − 1 ) n . q_n = C_1\cdot (2)^n + C_2\cdot(-1)^n. q n = C 1 ⋅ ( 2 ) n + C 2 ⋅ ( − 1 ) n . { C 1 + C 2 = 1 4 C 1 + C 2 = 3 ⇒ { C 1 = 1 3 C 2 = 2 3 . \begin{cases}C_1 + C_2 = 1 \\ 4C_1 + C_2 = 3 \end{cases} \Rightarrow \begin{cases}C_1 = \frac{1}{3} \\ C_2 = \frac{2}{3}\end{cases}. { C 1 + C 2 = 1 4 C 1 + C 2 = 3 ⇒ { C 1 = 3 1 C 2 = 3 2 . r n = 2 n + 2 ⋅ ( − 1 ) n 3 . r_n = \frac{2^n + 2\cdot (-1)^n}{3}. r n = 3 2 n + 2 ⋅ ( − 1 ) n .
3. Find the general solution of the recurrence equation g n + 2 = 3 g n + 1 − 2 g n g_{n+2} = 3g_{n+1} − 2g_n g n + 2 = 3 g n + 1 − 2 g n
[Solution]q n + 2 − 3 q n − 1 − 2 q n = 0 q^{n+2} - 3q^{n-1} - 2q^{n} = 0 q n + 2 − 3 q n − 1 − 2 q n = 0 q 1 = 2 , q 2 = 1. q_1 = 2, q_2 = 1. q 1 = 2 , q 2 = 1 . g n = C 1 ⋅ ( 2 ) n + C 2 . g_n = C_1\cdot (2)^n + C_2. g n = C 1 ⋅ ( 2 ) n + C 2 .
4. Solve the recurrence equation h n + 3 = 6 h n + 2 − 11 h n + 1 + 6 h n h_{n+3} = 6h_{n+2} - 11h_{n+1} + 6h_n h n + 3 = 6 h n + 2 − 1 1 h n + 1 + 6 h n h 0 = 3 , h 1 = 2 h_0 = 3, h_1 = 2 h 0 = 3 , h 1 = 2 h 2 = 4 h_2 = 4 h 2 = 4
[Solution]q n + 3 − 6 q n + 2 + 11 q n + 1 − 6 q n = 0 q^{n+3} - 6q^{n+2} + 11q^{n+1} - 6q^{n} = 0 q n + 3 − 6 q n + 2 + 1 1 q n + 1 − 6 q n = 0 q 1 = 1 , q 2 = 2 , q 3 = 3. q_1 = 1, q_2 = 2, q_3 = 3. q 1 = 1 , q 2 = 2 , q 3 = 3 . q n = C 1 + C 2 ⋅ 2 n + C 3 ⋅ 3 n . q_n = C_1 + C_2\cdot 2^n + C_3\cdot 3^{n}. q n = C 1 + C 2 ⋅ 2 n + C 3 ⋅ 3 n . { C 1 + C 2 + C 3 = 3 C 1 + 2 C 2 + 3 C 3 = 2 C 1 + 4 C 2 + 9 C 3 = 4 ⇒ { C 1 = 6 C 2 = − 5 C 3 = 2 . \begin{cases}C_1 + C_2 +C_3 = 3 \\ C_1 + 2C_2 + 3C_3 = 2 \\ C_1 + 4C_2 + 9C_3 = 4 \end{cases} \Rightarrow \begin{cases}C_1 = 6 \\ C_2 = -5 \\ C_3 = 2\end{cases}. ⎩ ⎪ ⎨ ⎪ ⎧ C 1 + C 2 + C 3 = 3 C 1 + 2 C 2 + 3 C 3 = 2 C 1 + 4 C 2 + 9 C 3 = 4 ⇒ ⎩ ⎪ ⎨ ⎪ ⎧ C 1 = 6 C 2 = − 5 C 3 = 2 . q n = 6 − 5 ⋅ 2 n + 2 ⋅ 3 n . q_n = 6 - 5\cdot 2^n + 2\cdot 3^n. q n = 6 − 5 ⋅ 2 n + 2 ⋅ 3 n .
5. Find the general solutions to each recurrence equation with characteristic polynomial as below:
( q − 4 ) 3 ( q + 1 ) ( q − 7 ) 4 ( q − 1 ) = 0 (q-4)^3(q+1)(q-7)^4(q-1) = 0 ( q − 4 ) 3 ( q + 1 ) ( q − 7 ) 4 ( q − 1 ) = 0 ( q − 5 ) 2 ( q + 5 ) 3 ( q − 1 ) 4 ( q + 1 ) ( q − 4 ) 3 = 0 (q-5)^2(q+5)^3(q-1)^4(q+1)(q-4)^3 = 0 ( q − 5 ) 2 ( q + 5 ) 3 ( q − 1 ) 4 ( q + 1 ) ( q − 4 ) 3 = 0
[Solution]
The roots of the characteristic function is 4 , − 1 , 7 , 1 4, -1, 7, 1 4 , − 1 , 7 , 1 3 , 1 , 4 , 1 3, 1, 4, 1 3 , 1 , 4 , 1 q n = C 1 ⋅ ( 4 n + n ⋅ 4 n + n 2 ⋅ 4 n ) + C 2 ⋅ ( − 1 ) n + C 3 ⋅ ( 7 n + n ⋅ 7 n + n 2 ⋅ 7 n + n 3 ⋅ 7 n ) + C 4 . q_n = C_1\cdot (4^n + n\cdot 4^n + n^2\cdot 4^n) + C_2\cdot (-1)^n + C_3\cdot (7^n + n\cdot 7^n + n^2\cdot 7^n + n^3\cdot 7^n ) + C_4. q n = C 1 ⋅ ( 4 n + n ⋅ 4 n + n 2 ⋅ 4 n ) + C 2 ⋅ ( − 1 ) n + C 3 ⋅ ( 7 n + n ⋅ 7 n + n 2 ⋅ 7 n + n 3 ⋅ 7 n ) + C 4 .
The roots of the characteristic function is 5 , − 5 , 1 , − 1 , 4 5, -5, 1, -1, 4 5 , − 5 , 1 , − 1 , 4 2 , 3 , 4 , 1 , 3 2, 3, 4, 1, 3 2 , 3 , 4 , 1 , 3 q n = C 1 ⋅ ( 5 n + n ⋅ 5 n ) + C 2 ⋅ { ( − 5 ) n ⋅ ∑ i = 0 2 n i } + C 3 ⋅ ( ∑ i = 0 3 n i ) + C 4 + C 5 ⋅ { ( 4 ) n ⋅ ∑ i = 0 2 n i } . q_n = C_1\cdot (5^n + n\cdot 5^n) + C_2\cdot \{(-5)^n\cdot \sum_{i = 0}^{2}n^{i}\} + C_3\cdot (\sum_{i = 0}^{3}n^{i}) + C_4 + C_5\cdot \{(4)^n\cdot \sum_{i = 0}^{2}n^{i}\}. q n = C 1 ⋅ ( 5 n + n ⋅ 5 n ) + C 2 ⋅ { ( − 5 ) n ⋅ i = 0 ∑ 2 n i } + C 3 ⋅ ( i = 0 ∑ 3 n i ) + C 4 + C 5 ⋅ { ( 4 ) n ⋅ i = 0 ∑ 2 n i } .
