特征根求教

nttz

教材中给出讨论就是代入，发现满足原来的递归方程就完事了，怎么来的呢？

tommywong

例如$a_n=c_1 q_1^n+c_2 q_2^n$
首先要證明特解的線性組合都是解
然後要證明$a_n=q_1^n,~a_n=q_2^n$都是方程的特解

tommywong

其實我覺得由特解得出通解不怎麼好，一來不知道怎麼得出特解，二來不知道是否已經找到了所有特解。

我試一下重新推一次看看：

$a_{n+1}=qa_n+b_n$
$\displaystyle \boxed{a_n=q^{n-n_0} a_{n_0}+\sum_{r=0}^{n-n_0-1}q^{n-n_0-1-r} b_{n_0+r}}$

(i) $a_{n+2}-(q_1+q_2)a_{n+1}+q_1q_2a_n=0$

$(a_{n+2}-q_1a_{n+1})-q_2(a_{n+1}-q_1a_n)=0$
$(a_{n+2}-q_1a_{n+1})=q_2(a_{n+1}-q_1a_n)$
$a_{n+1}-q_1a_n=(a_{n_0+1}-q_1a_{n_0})q_2^{n-n_0}$
$a_{n+1}=q_1a_n+(a_{n_0+1}-q_1a_{n_0})q_2^{n-n_0}$
$\displaystyle a_n=q_1^{n-n_0} a_{n_0}+\sum_{r=0}^{n-n_0-1}q_1^{n-n_0-1-r}(a_{n_0+1}-q_1a_{n_0})q_2^r$
$\displaystyle a_n=q_1^{n-n_0} a_{n_0}+(a_{n_0+1}-q_1a_{n_0})\frac{q_1^{n-n_0}-q_2^{n-n_0}}{q_1-q_2}$
$\displaystyle \boxed{a_n=\frac{a_{n_0+1}-a_{n_0} q_2}{q_1-q_2}q_1^{n-n_0}+\frac{a_{n_0} q_1-a_{n_0+1}}{q_1-q_2}q_2^{n-n_0}}$

(ii) $a_{n+2}-2qa_{n+1}+q^2 a_n=0$

$(a_{n+2}-qa_{n+1})-q(a_{n+1}-qa_n)=0$
$(a_{n+2}-qa_{n+1})=q(a_{n+1}-qa_n)$
$a_{n+1}-qa_n=(a_{n_0+1}-qa_{n_0})q^{n-n_0}$
$a_{n+1}=qa_n+(a_{n_0+1}-qa_{n_0})q^{n-n_0}$
$\displaystyle a_n=q^{n-n_0} a_{n_0}+\sum_{r=0}^{n-n_0-1}q^{n-n_0-1-r}(a_{n_0+1}-qa_{n_0})q^r$
$a_n=q^{n-n_0} a_{n_0}+q^{n-n_0-1}(a_{n_0+1}-qa_{n_0})(n-n_0)$
$\boxed{a_n=q^{n-n_0}[a_{n_0}+(a_{n_0+1}q^{-1}-a_{n_0})(n-n_0)]}$

(iii) $\displaystyle a_n=\frac{a_{n_0+1}-a_{n_0} q_2}{q_1-q_2}q_1^{n-n_0}+\frac{a_{n_0} q_1-a_{n_0+1}}{q_1-q_2}q_2^{n-n_0},~q_1=re^{i\theta},~q_2=re^{-i\theta}$

$\displaystyle a_n=\frac{a_{n_0+1}-a_{n_0} (re^{-i\theta})}{re^{i\theta}-re^{-i\theta}}(re^{i\theta})^{n-n_0}+\frac{a_{n_0} (re^{i\theta})-a_{n_0+1}}{re^{i\theta}-re^{-i\theta}}(re^{-i\theta})^{n-n_0}$

$\displaystyle a_n=r^{n-n_0}\left(\frac{a_{n_0+1}-a_{n_0} re^{-i\theta}}{re^{i\theta}-re^{-i\theta}}e^{i(n-n_0)\theta}+\frac{a_{n_0} re^{i\theta}-a_{n_0+1}}{re^{i\theta}-re^{-i\theta}}e^{-i(n-n_0)\theta}\right)$

$\displaystyle \boxed{a_n=r^{n-n_0}\left(a_{n_0}\cos(n-n_0)\theta
+\frac{a_{n_0+1}-a_{n_0}r\cos\theta}{r\sin\theta}\sin(n-n_0)\theta\right)}$