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设无穷数列 ${\displaystyle \{a_n\}}$ 满足 ${\displaystyle x_{n+1}=\frac{x_1^2}{2x_1-(x_1^2+1)x_n}}$, 其中 ${\displaystyle x_1>0}$, 且 ${\displaystyle x_1\ne 1}$.

1. 若 ${\displaystyle x_1=\sqrt{3}}$, 写出 ${\displaystyle x_2,x_4}$, 并求 ${\displaystyle \sum _{i=1}^{2026}\frac{1}{x_i}}$;
2. 证明: ${\displaystyle \{x_n\}}$ 不是单调数列;
3. 已知存在 ${\displaystyle m\in {\mathbb{N}}^{\ast }}$, 使得对任意的 ${\displaystyle k\in {\mathbb{N}}^{\ast }}$, ${\displaystyle x_{m+k}=x_k}$. 证明: ${\displaystyle m}$ 的最小值是奇数, 且 ${\displaystyle x_1,x_2,\cdots ,x_m}$ 的算术平均数小于 ${\displaystyle \frac{1}{2}}$.

1. 解 直接计算得 ${\displaystyle x_2=-\frac{\sqrt{3}}{2}}$, ${\displaystyle x_3=\frac{\sqrt{3}}{4}}$, ${\displaystyle x_4=\sqrt{3}}$. 于是 ${\displaystyle \{x_n\}}$ 是以 ${\displaystyle 3}$ 为周期的周期数列, ${\displaystyle \left\{\frac{1}{x_n}\right\}}$ 自然也是, 从而 $$\sum _{i=1}^{2026}\frac{1}{x_i}=675(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})+\frac{1}{x_1}=\frac{2026\sqrt{3}}{3}.$$

2. 证明 假设 ${\displaystyle \{x_n\}}$ 单调.

   • 如果存在 ${\displaystyle x_k<0}$, 则 ${\displaystyle x_{k+1}=\frac{x_1^2}{2x_1-(x_1^2+1)x_k}>0}$, 于是 ${\displaystyle x_{k+1}>x_k}$, ${\displaystyle x_k<x_1}$, 产生矛盾.
   • 否则, ${\displaystyle \mathrm{\forall }n,x_n>0}$ (因为 ${\displaystyle \{x_n\}}$ 肯定没有零项), 因此要求 ${\displaystyle x_n<\frac{2x_1}{x_1^2+1}}$. 这样, ${\displaystyle \{x_n\}}$ 既单调又有界, 所以收敛, 记极限为 ${\displaystyle t}$. 对递推式取极限: $$t=\frac{x_1^2}{2x_1-(x_1^2+1)t}\Rightarrow (x_1^2+1)t^2-2x_{1}t+x_1^2=0,$$ 这个方程的判别式为 ${\displaystyle \mathrm{\Delta }=4x_1^2-4x_1^2(x_1^2+1)<0}$, 说明极限压根不存在, 产生矛盾!

   综上, ${\displaystyle \{x_n\}}$ 不是单调数列.

3. 证明 设 ${\displaystyle x_1=\tan \theta }$, ${\displaystyle \theta \in (0,\frac{\pi }{2})\backslash \left\{\frac{\pi }{4}\right\}}$. 注意到 $$x_2=\frac{{\tan }^2\theta }{2\tan \theta -({\tan }^2\theta +1)\tan \theta }=\frac{\sin \theta \cos \theta }{2{\cos }^2\theta -1}=\frac{\sin \theta \cos \theta }{\cos 2\theta }.$$ 再结合 ${\displaystyle x_1=\tan \theta =\frac{\sin \theta \cos 0\theta }{\cos 1\theta }}$, 我们大胆猜测 ${\displaystyle x_n=\frac{\sin \theta \cos (n-1)\theta }{\cos n\theta }}$. 因为 $$\begin{array}{rl}& \frac{x_1^2}{2x_1-(x_1^2+1)x_n}=\frac{{\tan }^2\theta }{2\tan \theta -({\tan }^2\theta +1)\frac{\sin \theta \cos (n-1)\theta }{\cos n\theta }}=\frac{\sin \theta }{2\cos \theta -\frac{\cos (n-1)\theta }{\cos n\theta }}\\ =& \frac{\sin \theta \cos n\theta }{2\cos \theta \cos n\theta -\cos [n\theta -\theta ]}=\frac{\sin \theta \cos n\theta }{\cos \theta \cos n\theta -\sin \theta \sin n\theta }=\frac{\sin \theta \cos n\theta }{\cos (n+1)\theta },\end{array}$$ 这样根据数学归纳法我们的猜测成立!
   记 ${\displaystyle m}$ 的最小值为 ${\displaystyle T}$, 则 $$x_{T+1}=x_1\Rightarrow \frac{\sin \theta \cos T\theta }{\cos (T+1)\theta }=\tan \theta \Rightarrow \cos T\theta \cos \theta =\cos (T+1)\theta ,$$ 也即 $$\sin T\theta \sin \theta =0\Rightarrow \sin T\theta =0\Rightarrow T\theta =p\pi ,p\in {\mathbb{N}}^{\ast }.$$ 而 ${\displaystyle T}$ 是最小正周期，也即最小的让 ${\displaystyle T\theta }$ 是 ${\displaystyle \pi }$ 的整数倍的取值, 所以 ${\displaystyle T,p}$ 互质. 如果 ${\displaystyle T}$ 是偶数, 则 ${\displaystyle p}$ 是奇数, 这样 ${\displaystyle \cos \left(\frac{T}{2}\theta \right)=\cos \left(\frac{p}{2}\pi \right)=0}$, 从而 ${\displaystyle x_{\frac{T}{2}}}$ 项的分母为 ${\displaystyle 0}$, 这和"无穷数列"的假设矛盾! 所以 ${\displaystyle T}$ 只能是奇数.
   现在我们计算算术平均值 $$\begin{array}{rl}\frac{1}{m}\sum _{n=1}^m{x}_n& =\frac{1}{m}\sum _{n=1}^m\frac{\sin \theta (\cos n\theta \cos \theta +\sin n\theta \sin \theta )}{\cos n\theta }=\frac{1}{2}\sin \theta +\frac{{\sin }^2\theta }{m}\sum _{n=1}^m\tan n\theta \\ & =\frac{1}{2}\sin 2\theta +\frac{{\sin }^2}{m}\theta \sum _{n=1}^{\frac{m-1}{2}}(\tan n\theta +\tan (m-n)\theta )+\frac{{\sin }^2\theta \tan m\theta }{m}\\ & =\frac{1}{2}\sin 2\theta +\frac{{\sin }^2}{m}\theta \sum _{n=1}^{\frac{m-1}{2}}(\tan n\theta -\tan n\theta )+0\\ & =\frac{1}{2}\sin 2\theta <\frac{1}{2}.\end{array}$$