具体数学第二版第三章习题

Posted 2020-12-29 jianglangcaijin

46 （1）证明：

首先有$2n(n+1)=left lfloor 2n(n+1)+frac{1}{2} ight floor=left lfloor 2(n^{2}+n+frac{1}{4}) ight floor=left lfloor 2(n+frac{1}{2})^{2} ight floor$

其次，令$n+ heta =(sqrt{2}^{l}+sqrt{2}^{l-1})m=(1+frac{sqrt{2}}{2})sqrt{2}^{l}m$,$n^{‘}+ heta^{‘} =(sqrt{2}^{l+1}+sqrt{2}^{l})m=(1+sqrt{2})sqrt{2}^{l}m$

如果$l$为偶数，那么$n+ heta =(1+frac{sqrt{2}}{2})k,n^{‘}+ heta^{‘} =(1+sqrt{2})k$.所以$ heta =left { frac{sqrt{2}}{2}k ight }, heta^{‘} =left { sqrt{2}k ight }$,所以$ heta$和$ heta^{‘}$的关系是要么$ heta^{‘}=2 heta$($left lfloor sqrt{2}k ight floor$为偶数)，要么$ heta^{‘}=2 heta-1$($left lfloor sqrt{2}k ight floor$为奇数)；

如果$l$为奇数，那么$n+ heta =(1+sqrt{2})k,n^{‘}+ heta^{‘} =(2+sqrt{2})k$,这时候$ heta = heta^{‘}$

最后,假设要证明的式子成立，那么$n^{‘}=left lfloor sqrt{2n(n+1)} ight floor$

$=left lfloor sqrt{left lfloor 2(n+frac{1}{2})^{2} ight floor} ight floor$

$=left lfloor sqrt{2}(n+frac{1}{2}) ight floor$ (这一步参见公式$3.9$)

$=left lfloor sqrt{2}left ( (1+frac{sqrt{2}}{2})sqrt{2}^{l}m - heta +frac{1}{2} ight ) ight floor$

$=left lfloor n^{‘}+ heta^{‘}+sqrt{2}(frac{1}{2}- heta) ight floor$

所以只要证明$0leq heta^{‘}+sqrt{2}(frac{1}{2}- heta)<1$

首先当$ heta= heta^{‘}$时成立，

其次，如果$ heta^{‘}=2 heta-d$时($d=0$或者$d=1$)，那么$0leq heta^{‘}+sqrt{2}(frac{1}{2}- heta)<1$

$Leftrightarrow 0leq heta^{‘}+sqrt{2}(frac{1}{2}-frac{ heta^{‘} +d}{2})<1$

$Leftrightarrow 0leq heta^{‘}(2-sqrt{2})+sqrt{2}(1-d)<2$

最后这个式子明显成立

（2）由于$Spec(1+frac{sqrt{2}}{2}),Spec(1+sqrt{2})$是一个划分，所以对于任何一个$a$一定存在唯一的$(l,m)$使得$a=(sqrt{2}^{l}+sqrt{2}^{l-1})m$,这时候$L_{n}=left lfloorleft (  sqrt{2}^{l+n} -sqrt{2}^{l+n-1} ight )m ight floor$