## 常係數線性遞迴關係式 (下)

$a_n=d_1a_{n-1}+d_2a_{n-2}+\cdots+d_ka_{n-k}+F(n),~~n\ge k$

$a_n=d_1a_{n-1}+d_2a_{n-2}+\cdots+d_ka_{n-k}$

$\displaystyle a^{(p)}_n=d_1a^{(p)}_{n-1}+d_2a^{(p)}_{n-2}+\cdots+d_ka^{(p)}_{n-k}+F(n)$

$\displaystyle b_n=d_1b_{n-1}+d_2b_{n-2}+\cdots+d_kb_{n-k}+F(n)$

$\displaystyle b_n-a^{(p)}_n=d_1(b_{n-1}-a^{(p)}_{n-1})+d_2(b_{n-2}-a^{(p)}_{n-2})+\cdots+d_k(b_{n-k}-a^{(p)}_{n-k})$

$a_n=2a_{n-1}+3n$

$a_n^{(p)}=\alpha_1 n+\alpha_0$

\begin{aligned} \alpha_1 n+\alpha_0&=2(\alpha_1(n-1)+\alpha_0)+3n\\ &=(2\alpha_1+3)n+2(\alpha_0-\alpha_1), \end{aligned}

$\displaystyle a_n=-3n-6+7\cdot 2^n,~~n\ge 0$

$a_n=a_{n-1}+6a_{n-2}+5^n$

$a_n^{(p)}=\alpha_1 5^n$

$\alpha_1 5^n=\alpha_15^{n-1}+6\alpha_15^{n-2}+5^n$

$25\alpha_1=5\alpha_1+6\alpha_1+25$，解得 $\alpha_1=\frac{25}{14}$，所以 $a_n^{(p)}=\frac{25}{14}\cdot 5^n$。寫出關聯齊次遞迴關係式 $a_n=a_{n-1}+6a_{n-2}$，由特徵方程 $t^2=t+6$ 解出二根，$3$$-2$，因此齊次解為 $a_n^{(h)}=c_13^n+c_2(-2)^n$。合併以上結果，通項可表示為

$\displaystyle a_n=\frac{25}{14}\cdot 5^n+c_1 3^n+c_2(-2)^n$

$\displaystyle \sum_{i=0}^{r+1}(-1)^i\binom{r+1}{i}f(i)=\sum_{i=0}^{r+1}(-1)^i\binom{r+1}{i}q(n-i)=0$

$\displaystyle \sum_{i=0}^{r+1}\binom{r+1}{i}(-s)^iq(n-i)s^{n-i}=\sum_{i=0}^{r+1}\binom{r+1}{i}(-s)^iF(n-i)=0$

$\displaystyle \sum_{i=0}^{r+1}\binom{r+1}{i}(-s)^it^{r+1-i}=(t-s)^{r+1}=0$

$\displaystyle (t^k-d_1t^{k-1}-d_2t^{k-2}-\cdots-d_k)(t-s)^{r+1}=0$

$a_n=5a_{n-1}+F(n)$

\begin{aligned} a_n-5a_{n-1}&=F(n)\\ -2s(a_{n-1}-5a_{n-2})&=-2sF(n-1)\\ s^2(a_{n-2}-5a_{n-3})&=s^2F(n-2), \end{aligned}

$\displaystyle F(n)-2sF(n-1)+s^2F(n-2)=0$

$\displaystyle a_n-5a_{n-1}-2s(a_{n-1}-5a_{n-2})+s^2(a_{n-2}-5a_{n-3})=0$

$a_n=\alpha_05^n+(\alpha_1+\alpha_2n)s^n$

$s=5$

$a_n=(\alpha_0+\alpha_1n+\alpha_2n^2)s^n$

$\displaystyle (\alpha_1+\alpha_2n)3^n=5(\alpha_1+\alpha_2(n-1))3^{n-1}+(2n+1)3^n$

\displaystyle\begin{aligned} 3\alpha_1&=5\alpha_1-5\alpha_2+3\\ 3\alpha_2&=5\alpha_2+6, \end{aligned}

$\displaystyle a_n=-(9+3n)3^n+\alpha_05^n$

$a_n=d_1a_{n-1}+d_2a_{n-2}+\cdots+d_ka_{n-k}+F(n),~~n\ge k$

$\displaystyle a_n^{(p)}=n^mp(n)s^n$

$t^k=d_1t^{k-1}+d_2t^{k-2}+\cdots+d_k$

[1] Wolfram MathWorld：Backward Difference

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