下面圖檔的做法哪裡有誤呢?我認為是增根。還有圖檔的作法，利用了小技巧。因為根本不知道是否收斂?
10、 給定二階方陣，
A = \left[ {\begin{array}{*{20}{c}}2&7\\1&{ - 4}\end{array}} \right]，若
A\left[ {\begin{array}{*{20}{c}}
{{x_{n - 1}}}\\{{y_{n - 1}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{x_n}}\\{{y_n}}\end{array}} \right]，
n \in N，
n \ge 2，求
\mathop {\lim }\limits_{n \to \infty } \frac{{{x_n}}}{{{y_n}}} =
解 (1)
A = \left[ {\begin{array}{*{20}{c}}2&7\\1&{ - 4}\end{array}} \right],
|A - \lambda I| = \left| {\begin{array}{*{20}{c}}{2 - \lambda }&7\\1&{ - 4 - \lambda }\end{array}} \right| = {\lambda ^2} + 2\lambda - 15 = \left( {\lambda + 5} \right)\left( {\lambda - 3} \right) = 0 \Rightarrow \lambda = - 5\;,\;\lambda = 3
(2)
\lambda = - 5
\left[ {\begin{array}{*{20}{c}}
7&7\\
1&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0\\
0
\end{array}} \right] \Rightarrow \;\;{v_1} = \left( {1, - 1} \right)
(3)
\lambda = 3
\left[ {\begin{array}{*{20}{c}}
{ - 1}&7\\
1&{ - 7}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0\\
0
\end{array}} \right] \Rightarrow \;\;{v_2} = \left( {7,1} \right)
(4)
AP=PD ~~~
A = PD{P^{ - 1}}
(5)
\begin{array}{l}
P = \left[ {\begin{array}{*{20}{c}}
1&7\\
{ - 1}&1
\end{array}} \right]\;\;,\;\;{P^{ - 1}} = \frac{1}{8}\left[ {\begin{array}{*{20}{c}}
1&{ - 7}\\
1&1
\end{array}} \right]\;\;\;,\;\;\\
A = \frac{1}{8}\left[ {\begin{array}{*{20}{c}}
1&7\\
{ - 1}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{ - 5}&0\\
0&3
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&{ - 7}\\
1&1
\end{array}} \right]\\
\left[ {\begin{array}{*{20}{c}}
{{x_n}}\\
{{y_n}}
\end{array}} \right] = {A^{n - 1}}\left[ {\begin{array}{*{20}{c}}
{{x_1}}\\
{{y_1}}
\end{array}} \right]\\
\;\;\;\;\;\;\; = {\left( {\frac{1}{8}} \right)^{n - 1}}\left[ {\begin{array}{*{20}{c}}
1&7\\
{ - 1}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{{\left( { - 5} \right)}^{n - 1}}}&0\\
0&{{3^{n - 1}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&{ - 7}\\
1&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{x_1}}\\
{{y_1}}
\end{array}} \right]\\
\;\;\;\;\;\;\; = {\left( {\frac{1}{8}} \right)^{n - 1}}\left[ {\begin{array}{*{20}{c}}
{{{\left( { - 5} \right)}^{n - 1}}\left( {{x_1} - 7{y_1}} \right) + 7{{\left( 3 \right)}^{n - 1}}\left( {{x_1} + {y_1}} \right)}\\
{ - {{\left( { - 5} \right)}^{n - 1}}\left( {{x_1} - 7{y_1}} \right) + {{\left( 3 \right)}^{n - 1}}\left( {{x_1} + {y_1}} \right)}
\end{array}} \right]\\
\\
\\
\frac{{{x_n}}}{{{y_n}}} = \frac{{\frac{{{{\left( { - 5} \right)}^{n - 1}}}}{{{{\left( { - 5} \right)}^{n - 1}}}}\left( {{x_1} - 7{y_1}} \right) + 7\frac{{{{\left( 3 \right)}^{n - 1}}}}{{{{\left( { - 5} \right)}^{n - 1}}}}\left( {{x_1} + {y_1}} \right)}}{{ - \frac{{{{\left( { - 5} \right)}^{n - 1}}}}{{{{\left( { - 5} \right)}^{n - 1}}}}\left( {{x_1} - 7{y_1}} \right) + \frac{{{{\left( 3 \right)}^{n - 1}}}}{{{{\left( { - 5} \right)}^{n - 1}}}}\left( {{x_1} + {y_1}} \right)}}\\
\;\;\;\; = \frac{{\left( {{x_1} - 7{y_1}} \right) + 7\frac{{{{\left( 3 \right)}^{n - 1}}}}{{{{\left( { - 5} \right)}^{n - 1}}}}\left( {{x_1} + {y_1}} \right)}}{{ - 1\left( {{x_1} - 7{y_1}} \right) + \frac{{{{\left( 3 \right)}^{n - 1}}}}{{{{\left( { - 5} \right)}^{n - 1}}}}\left( {{x_1} + {y_1}} \right)}}
\end{array}
{x_1},{y_1}為定值，
{\left( {\frac{{ - 3}}{5}} \right)^{n - 1}}逼近於 0，當n逼近無限大
所以 當 n逼近於無限大
\frac{{{x_n}}}{{{y_n}}}逼近於
-1
\displaystyle \Bigg[\; \matrix{2 & 7 \cr 1 & -4} \Bigg]\; \Bigg[\; \matrix{x_{n-1} \cr y_{n-1}} \Bigg]\;=\Bigg[\; \matrix{x_n \cr y_n} \Bigg]\;
\displaystyle \Rightarrow \cases{x_n=2x_{n-1}+7y_{n-1} \cr y_n=x_{n-1}-4y_{n-1}}
\displaystyle \frac{x_n}{y_n}=\frac{2x_{n-1}+7y_{n-1}}{x_{n-1}-4y_{n-1}} (同除以
y_{n-1} )
\displaystyle =\frac{2 (\frac{x_{n-1}}{y_{n-1}})+7}{(\frac{x_{n-1}}{y_{n-1}})-4} (取極限)
令
\displaystyle \lim_{n \to \infty}\frac{x_n}{y_n}=\lim_{n \to \infty}\frac{x_{n-1}}{y_{n-1}}=t
\displaystyle t=\frac{2t+7}{t-4} \Rightarrow t=7,-1
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本帖最後由 bugmens 於 2014-5-22 05:04 AM 編輯 ]
