MT【313】特征方程逆用

已知实数$a,b,x,y$满足
\begin{equation}
\left\{ \begin{aligned}
ax+by &= 3 \\
ax^2+by^2&=7\\
ax^3+by^3&=16\\
ax^4+by^4&=42\\
\end{aligned} \right.
\end{equation}
求$ax^5+by^5$的值.

解答:设$a_n=ax^n+by^n$
则$x,y$是二阶齐次线性递推数列$a_{n+2}+pa_{n+1}+qa_n=0$ 的特征方程$t^2+pt+q=0$ 的两个特征根.
代入$a_1=3,a_2=7,a_3=16,a_4=42$得$p=14,q=-38$故$a_5++14a_4-38a_3=0,a_5=20$

 

 

练习:
已知
\begin{equation}
\left\{ \begin{aligned}
x+y+z &= 1 \\
x^2+y^2+z^2&=2\\
x^3+y^3+z^3&=3\\
\end{aligned} \right.
\end{equation}
求$x^5+y^5+z^5$的值

 

 

 

解答:设$a_n=x^n+y^n+z^n$
则$x,y,z$是三阶齐次线性递推数列$a_{n+3}+pa_{n+2}+qa_{n+1}+ra_n=0$ 的特征方程$t^3+pt^2+qt+r=0$ 的三个特征根.
由韦达定理\begin{equation}
\left\{ \begin{aligned}
x+y+z &= -p \\
xy+yz+zx&=q\\
xyz&=-r\\
\end{aligned} \right.
\end{equation}
结合恒等式

$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

易得$p=-1,q=-\dfrac{1}{2},r=-\dfrac{1}{6}$
故$a_{n+3}-a_{n+2}-\dfrac{1}{2}a_{n+1}-\dfrac{1}{6}a_n=0$结合$a_1=1,a_2=2,a_3=3$易得$a_5=6$