每日一题[1729]摆动数列

称一个复数数列 $\{z_n\}$ 为“有趣的”,若 $|z_1|=1$,且对任意正整数 $n$,均有 $4z_{n+1}^2+2z_nz_{n+1}+z_n^2=0$.求最大的常数 $C$,使得对一切有趣的数列 $\{z_n\}$ 及任意正整数 $m$,具有 $|z_1+z_2+\cdots+z_m|\geqslant C$.

答案    $\dfrac{\sqrt 3}3$.

解析    根据题意,有\[z_{n+1}=\dfrac{-1\pm\sqrt 3{\rm i}}{4}z_n\iff z_{n+1}=\left(\pm\dfrac{2\pi}3:\dfrac 12\right)\cdot z_n.\] 记 $q_1=\left(\dfrac{2\pi}3:\dfrac 12\right)$,$q_2=\left(-\dfrac{2\pi}3:\dfrac 12\right)$.取 $z_1=1$,且\[z_{n+1}=\begin{cases} q_1\cdot z_n,&2\nmid n,\\ q_2\cdot z_n,&2\mid n,\end{cases}\]则\[\begin{split}\lim_{m\to+\infty}(z_1+z_2+\cdots+z_m)&=(0:1)+\left(\dfrac{2\pi}3:\dfrac 12+\dfrac 18+\cdots\right)+\left(-\dfrac{2\pi}3:\dfrac 14+\dfrac 1{16}+\cdots\right)\\ &=(0:1)+\left(\dfrac{2\pi}3:\dfrac 23\right)+\left(-\dfrac{2\pi}3:\dfrac 13\right)\\ &=\dfrac 12+\dfrac{\sqrt 3}6{\rm i},\end{split}\]因此\[\lim_{m\to+\infty}|z_1+z_2+\cdots+z_m|=\dfrac{\sqrt 3}3.\] 接下来证明 $C=\dfrac{\sqrt 3}3$ 符合题意. 考虑\[|z_n+z_{n+1}|=|z_n|\cdot\left|1+\dfrac{z_{n+1}}{z_n}\right|=\dfrac1{2^{n-1}}\cdot \left|1+\left(\pm\dfrac{2\pi}3:\dfrac 12\right)\right|=\dfrac{\sqrt 3}{2^n},\]于是当 $m=1$ 时,有 $|z_1|=1>\dfrac{\sqrt 3}3$; 当 $m$ 为偶数时,有\[\begin{split} |z_1+z_2+\cdots+z_m|&\geqslant |z_1+z_2|-|z_3+z_4|-\cdots-|z_{m-1}+z_m|\\ &>\dfrac{\sqrt 3}2-\left(\dfrac{\sqrt 3}8+\dfrac{\sqrt 3}{32}+\cdots\right)\\ &=\dfrac{\sqrt 3}3,\end{split}\]当 $m$ 为不小于 $3$ 的奇数时,有\[\begin{split} |z_1+z_2+\cdots+z_m|&\geqslant |z_1+z_2|-|z_3+z_4|-\cdots-|z_{m-2}+z_{m-1}|-|z_m|\\ &>\dfrac{\sqrt 3}2-\left(\dfrac{\sqrt 3}8+\dfrac{\sqrt 3}{32}+\cdots+\dfrac{\sqrt 3}{2^{m-2}}\right)-\dfrac{1}{2^m}\\ &>\dfrac{\sqrt 3}2-\left(\dfrac{\sqrt 3}8+\dfrac{\sqrt 3}{32}+\cdots+\dfrac{\sqrt 3}{2^{m-2}}+\dfrac{\sqrt 3}{2^m}\right)\\ &>\dfrac{\sqrt 3}3,\end{split}\] 综上所述,符合题意的最大常数为 $\dfrac{\sqrt 3}3$.

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