第14届XMO双线性递推数列

设数列 $\{a_n\}$, $\{b_n\}$满足 $a_0=b_0=1$,且对于任意自然数 $n$,均有:
$$\begin{aligned}
a_{n+1}&=\frac{6}{5}a_n-\frac{3}{5}b_n
-a_n(a_n^2+b_n^2),\\
b_{n+1}&=\frac{3}{5}a_n+\frac{6}{5}b_n
-b_n(a_n^2+b_n^2).
\end{aligned}$$
求数列 $\{a_n+b_n\}$ 的通项公式.

解.对递推式平方得:
$$\begin{aligned}
a_{n+1}^{2}&=\frac{9}{25}(2a_{n}-b_{n})^{2}
-\frac{6}{5}a_{n}(2a_{n}-b_{n})(a_{n}^{2}
+b_{n}^{2})+a_{n}^{2}(a_{n}^{2}+b_{n}^{2})^{2},\\
b_{n+1}^{2}&=\frac{9}{25}(a_{n}+2b_{n})^{2}
-\frac{6}{5}b_{n}(a_{n}+2b_{n})(a_{n}^{2}
+b_{n}^{2})+b_{n}^{2}(a_{n}^{2}+b_{n}^{2})^{2},
\end{aligned}$$
两式相加得:
$$a_{n+1}^{2}+b_{n+1}^{2}=\frac{9}{5}(a_{n}^{2}
+b_{n}^{2})-\frac{12}{5}(a_{n}^{2}+b_{n}^{2})^{2}
+(a_{n}^{2}+b_{n}^{2})^{3}.$$

定义数列 $c_{n}=a_{n}^{2}+b_{n}^{2}$, 则 $c_{0}=2$, 可以通过上式归纳出 $c_{n}=2$, 于是 $a_{n}^{2}+b_{n}^{2}=2$.

此时递推式化简为:
$$
a_{n+1}=-\frac{4}{5}a_{n}-\frac{3}{5}b_{n},\quad b_{n+1}=\frac{3}{5}a_{n}-\frac{4}{5}b_{n}.$$

## 解法一 (三角法).

记 $\theta = \arccos\left(-\frac{4}{5}\right)$,
则 $\cos \theta = -\frac{4}{5}$, $\sin \theta = \frac{3}{5}$, 再假设 $a_n = \sqrt{2} \cos \alpha_n$, $b_n = \sqrt{2} \sin \alpha_n$,于是 $\alpha_0 = \frac{\pi}{4}$,且
$$\begin{aligned}
a_{n+1} &= \sqrt{2}(\cos \theta \cos \alpha_n - \sin \theta \sin \alpha_n) = \sqrt{2} \cos(\alpha_n + \theta),\\
b_{n+1} &= \sqrt{2}(\sin \theta \cos \alpha_n + \cos \theta \sin \alpha_n) = \sqrt{2} \sin(\alpha_n + \theta).
\end{aligned}$$
即 $\alpha_{n+1} = \alpha_n + \theta$,于是 $\alpha_n = \alpha_0 + n \theta = n \theta + \frac{\pi}{4}$,那么
$$\begin{aligned}
a_n &= \sqrt{2} \cos\left(n \theta + \frac{\pi}{4}\right) = \cos n \theta - \sin n \theta,\\
b_n &= \sqrt{2} \sin\left(n \theta + \frac{\pi}{4}\right) = \cos n \theta + \sin n \theta.
\end{aligned}$$
于是 $a_n + b_n = 2 \cos n \theta = 2 \cos\left(n \arccos\left(-\frac{4}{5}\right)\right)$.

## 解法二 (线性递推).

由
$$
a_{n+1}=-\frac{4}{5}a_{n}-\frac{3}{5}b_{n},\quad b_{n+1}=\frac{3}{5}a_{n}-\frac{4}{5}b_{n}$$
可得
$$\begin{aligned}
a_{n+2}&=-\frac{4}{5}a_{n+1}-\frac{3}{5}b_{n+1}=-\frac{4}{5}a_{n+1}-\frac{3}{5}\left( \frac{3}{5}a_n-\frac{4}{5}b_n \right)\\
&=-\frac{4}{5}a_{n+1}-\frac{9}{25}a_n+\frac{12}{25}b_n=-\frac{4}{5}a_{n+1}-\frac{9}{25}a_n+\frac{12}{25}\times \frac{5}{3}\left( -a_{n+1}-\frac{4}{5}a_n \right)\\
&=-\frac{4}{5}a_{n+1}-\frac{9}{25}a_n
-\frac{4}{5}a_{n+1}-\frac{16}{25}a_n
=-\frac{8}{5}a_{n+1}-a_n,
\end{aligned}$$
其特征方程为$x^2=-\frac{8}{5}x-1$,解得$x=-\frac{4}{5}\pm \frac{3}{5}\mathrm{i}$.

于是
$$
a_n=A\left( -\frac{4}{5}+\frac{3}{5}\text{i} \right) ^n+B\left( -\frac{4}{5}-\frac{3}{5}\text{i} \right) ^n,
$$
再由$a_0=1$, $a_1=-\frac{7}{5}$可得
$A+B=1$, $A\left( -\frac{4}{5}+\frac{3}{5}\text{i} \right) +B\left( -\frac{4}{5}-\frac{3}{5}\text{i} \right) =-\frac{7}{5}$,解得$A=\frac{1+i}{2},B=\frac{1-i}{2}$.

于是
$$
a_n=\frac{1+\text{i}}{2}\left( -\frac{4}{5}+\frac{3}{5}\text{i} \right) ^n+\frac{1-\text{i}}{2}\left( -\frac{4}{5}-\frac{3}{5}\text{i} \right) ^n.
$$
故
$$\begin{aligned}
b_n+a_n &=\left( -\frac{5}{3}a_{n+1}-\frac{4}{3}a_n \right) +a_n=-\frac{5}{3}a_{n+1}-\frac{1}{3}a_n\\
&=-\frac{5}{3}\left[ \frac{1+\text{i}}{2}\left( -\frac{4}{5}+\frac{3}{5}\text{i} \right) ^{n+1}+\frac{1-\text{i}}{2}\left( -\frac{4}{5}-\frac{3}{5}\text{i} \right) ^{n+1} \right]\\
&\quad -\frac{1}{3}\left[ \frac{1+\text{i}}{2}\left( -\frac{4}{5}+\frac{3}{5}\text{i} \right) ^n+\frac{1-\text{i}}{2}\left( -\frac{4}{5}-\frac{3}{5}\text{i} \right) ^n \right]\\
&=\left( -\frac{4}{5}+\frac{3}{5}\text{i} \right) ^n+\left( -\frac{4}{5}-\frac{3}{5}\text{i} \right) ^n\\
&=\mathrm{e}^{\mathrm{i}n\theta}
+\mathrm{e}^{-\mathrm{i}n\theta}=2\cos \left( n\theta \right),
\end{aligned}$$
其中$\tan\theta=-\frac{3}{4}$.

## 解法三(复数法).

令$z_n=a_n+b_n\mathrm{i}$,于是
$$\begin{aligned}
z_{n+1}&=a_{n+1}+b_{n+1}\mathrm{i}=\left( -\frac{4}{5}a_n-\frac{3}{5}b_n \right) +\left( \frac{3}{5}a_n-\frac{4}{5}b_n \right) \mathrm{i}\\
&=\left( -\frac{4}{5}+\frac{3}{5}\mathrm{i} \right) a_n-\left( \frac{3}{5}+\frac{4}{5}\mathrm{i} \right) b_n=\left( -\frac{4}{5}+\frac{3}{5}\mathrm{i} \right) \left( a_n+b_n\mathrm{i} \right)\\
&=\left( -\frac{4}{5}+\frac{3}{5}\mathrm{i} \right) z_n,
\end{aligned}$$
于是
$$\begin{aligned}
z_n&=\left( -\frac{4}{5}+\frac{3}{5}\mathrm{i} \right) z_{n-1}=\cdots =\left( -\frac{4}{5}+\frac{3}{5}\mathrm{i} \right) ^nz_0=\left( -\frac{4}{5}+\frac{3}{5}\mathrm{i} \right) ^n\left( 1+\mathrm{i} \right)\\
&=\sqrt{2}\mathrm{e}^{i\left( n\theta +\frac{\pi}{4} \right)}=\sqrt{2}\left[ \cos \left( n\theta +\frac{\pi}{4} \right) +\mathrm{i}\sin \left( n\theta +\frac{\pi}{4} \right) \right],
\end{aligned}$$
于是
$$
a_n+b_n=\sqrt{2}\left[ \cos \left( n\theta +\frac{\pi}{4} \right) +\sin \left( n\theta +\frac{\pi}{4} \right) \right] =2\cos n\theta,
$$
其中$\tan\theta=-\frac{3}{4}$.

## 解法四 (矩阵法).

由
$$
a_{n+1}=-\frac{4}{5}a_{n}-\frac{3}{5}b_{n},\quad b_{n+1}=\frac{3}{5}a_{n}-\frac{4}{5}b_{n}$$
可得
$$
\left( \begin{array}{c}
a_{n+1}\\
b_{n+1}\\
\end{array} \right) =\left( \begin{matrix}
-\frac{4}{5}& -\frac{3}{5}\\
\frac{3}{5}& -\frac{4}{5}\\
\end{matrix} \right) \left( \begin{array}{c}
a_n\\
b_n\\
\end{array} \right) =\cdots =\left( \begin{matrix}
-\frac{4}{5}& -\frac{3}{5}\\
\frac{3}{5}& -\frac{4}{5}\\
\end{matrix} \right) ^{n+1}\left( \begin{array}{c}
a_0\\
b_0\\
\end{array} \right).
$$

记$\cos\theta=-\frac{4}{5}$, $\sin\theta=\frac{3}{5}$,则
$$
\left( \begin{matrix}
-\frac{4}{5}& -\frac{3}{5}\\
\frac{3}{5}& -\frac{4}{5}\\
\end{matrix} \right) ^n=\left( \begin{matrix}
\cos \theta& -\sin \theta\\
\sin \theta& \cos \theta\\
\end{matrix} \right) ^n=\left( \begin{matrix}
\cos n\theta& -\sin n\theta\\
\sin n\theta& \cos n\theta\\
\end{matrix} \right),
$$
因此
$$
\left( \begin{array}{c}
a_n\\
b_n\\
\end{array} \right) =\left( \begin{matrix}
-\frac{4}{5}& -\frac{3}{5}\\
\frac{3}{5}& -\frac{4}{5}\\
\end{matrix} \right) ^n\left( \begin{array}{c}
a_0\\
b_0\\
\end{array} \right) =\left( \begin{matrix}
\cos n\theta& -\sin n\theta\\
\sin n\theta& \cos n\theta\\
\end{matrix} \right) \left( \begin{array}{c}
1\\
1\\
\end{array} \right),
$$
故$a_n=\cos n\theta -\sin n\theta ,b_n=\cos n\theta +\sin n\theta$,因此$a_n+b_n=2\cos n\theta$,其中$\tan\theta=-\frac{3}{4}$.