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kuing
发表于 2014-4-27 13:28
$\newcommand\led[1]{\left\{\begin{aligned}
#1
\end{aligned}\right.}$
设数列 $\{a_n\}$ 满足 $a_{n+3}=sa_{n+2}+ta_{n+1}+ua_n$ 且 $a_0=2$, $a_1=3$, $a_2=4$, $a_3=6$, $a_4=12$, $a_5=32$，则应有
\[
\led{
6&=4s+3t+2u, \\
12&=6s+4t+3u, \\
32&=12s+6t+4u,
}
\]
解得 $s=5$, $t=-6$, $u=2$，于是
\[a_{n+3}=5a_{n+2}-6a_{n+1}+2a_n,\]
令
\[v^3=5v^2-6v+2,\]
解得
\[v_1=1,v_2=2-\sqrt2,v_3=2+\sqrt2,\]
则由特征方程的理论知 $a_n$ 的通项必能写成
\[a_n=\lambda_1v_1^n+\lambda_2v_2^n+\lambda_3v_3^n,\]
于是应有
\[
\led{
2&=\lambda_1+\lambda_2+\lambda_3, \\
3&=\lambda_1v_1+\lambda_2v_2+\lambda_3v_3, \\
4&=\lambda_1v_1^2+\lambda_2v_2^2+\lambda_3v_3^2,
}
\]
解得
\[\lambda_1=4,\lambda_2=-1-\frac3{2\sqrt2},\lambda_3=-1+\frac3{2\sqrt2},\]
于是我们得到
\[
\led{
\lambda_1+\lambda_2+\lambda_3&=2,\\
\lambda_1v_1+\lambda_2v_2+\lambda_3v_3&=3,\\
\lambda_1v_1^2+\lambda_2v_2^2+\lambda_3v_3^2&=4,\\
\lambda_1v_1^3+\lambda_2v_2^3+\lambda_3v_3^3&=6,\\
\lambda_1v_1^4+\lambda_2v_2^4+\lambda_3v_3^4&=12,\\
\lambda_1v_1^5+\lambda_2v_2^5+\lambda_3v_3^5&=32,
}
\]
与原方程组对比，可知原方程组的解至少有如下六组
\[
\led{
x&=\lambda_1,\\
y&=\lambda_2,\\
z&=\lambda_3,\\
p&=v_1,\\
q&=v_2,\\
r&=v_3,
}
\quad
\led{
x&=\lambda_1,\\
y&=\lambda_3,\\
z&=\lambda_2,\\
p&=v_1,\\
q&=v_3,\\
r&=v_2,
}
\quad
\led{
x&=\lambda_2,\\
y&=\lambda_1,\\
z&=\lambda_3,\\
p&=v_2,\\
q&=v_1,\\
r&=v_3,
}
\quad
\led{
x&=\lambda_2,\\
y&=\lambda_3,\\
z&=\lambda_1,\\
p&=v_2,\\
q&=v_3,\\
r&=v_1,
}
\quad
\led{
x&=\lambda_3,\\
y&=\lambda_1,\\
z&=\lambda_2,\\
p&=v_3,\\
q&=v_1,\\
r&=v_2,
}
\quad
\led{
x&=\lambda_3,\\
y&=\lambda_2,\\
z&=\lambda_1,\\
p&=v_3,\\
q&=v_2,\\
r&=v_1,
}
\]
又因为原方程组是六元六次方程组，既然有此六组解，它们就是全部解。 |
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爪机专用
发表于 2014-4-27 03:10
我敢肯定结果没错