不知道怎么老是报错，哎

永远

设\(\small\,(n\in\mathbb{N}^+)\wedge(\sqrt{n}\not\in\mathbb{N}),\;F(n)=\{(m,k)\in\mathbb{Z}\times\mathbb{N}:\,k\mid n-m^2\in(0,n)\}\)
令\(\,{\small\lambda(m,k)=}\lfloor\frac{\sqrt{n}+m}{k}\rfloor,\;m_1=k\cdot\lambda(m,k)-m,\,k_1=\large\frac{n-m_1^2}{k}.\;\)则有
\({\small\sqrt{n}-m_1=}\big(\frac{\sqrt{n}+m}{k}{\small-\lambda(m,k)}\big)k>;0,\;\sqrt{n}+m_1>;\sqrt{n}-m>0\)
\(\therefore\;k\mid n-m^2+2mk\lambda-(k\lambda)^2=n-m_1^2;0,\;(m_1,k_1)\in F(n)\)
\(\therefore\;\frac{k}{\sqrt{n}+m}=\frac{k}{k\lambda+\sqrt{n}+m-k\lambda}=\frac{k}{k\lambda+\sqrt{n}-m_1}=\frac{1}{\lambda+\large\frac{k_1}{\sqrt{n}+m_1}}\)
令\(\,\psi^{\langle k+1\rangle}=\psi(\psi^{\langle k\rangle}),\;(m_j,k_j)=\psi^{\langle j\rangle}(m,k),\;\lambda_j=\lambda(m_{j-1},k_{j-1})\)
定义\(\,[a,b]=a+\large\frac{1}{\lbrack b \rbrack},\,\)则有\(\small\,1\le m,\,L\in\mathbb{N}\,\)使\(\small(m_{u+L},k_{u+L})=(m_u,k_u)\)
\(\,\frac{k}{\sqrt{n}+m}=[0,\lambda_1+\frac{k_1}{\sqrt{n}+m_1}]=[0,\lambda_1,\ldots,\lambda_{u}+\frac{k_u}{\sqrt{n}+m_u}]\)
\(\qquad\cdots=[0,\lambda_1,\ldots,\lambda_{u},\ldots,\lambda_{u+L}+\frac{k_{u+L}}{\sqrt{n}+m_{u+L}}]\)
\(\qquad\cdots=[0,\lambda_1,\ldots,\lambda_{u},\dot{\lambda}_{u+1},\ldots,\dot{\lambda}_{u+L}]\)
\(\qquad\cdots=[0,\lambda_1,\ldots,\lambda_{u},\overline{{\lambda}_{u+1},\ldots,\lambda}_{u+L}]\)
这是因为\(F(n)\)是有限集.
取\(\,m,k\in\mathbb{N}\,\)使\(\small\,m^2< n< (m+1)^2,\,k=n-m^2,\,\)则\(\small\,(m,k)\in F(n).\)
可见\(\,\sqrt{n}=[m;\lambda_1,\ldots,\lambda_u,\overline{\lambda_{u+1},\ldots,\lambda}_{u+L}]\,\)是循环连分数.

永远

确切地说抄e老师的源代码，抄了好多遍还是报错

永远

谁来解释一下

elim

&gt;  换成 >
&lt;   换成 <

永远

elim 发表于 2020-9-1 06:07
&gt;  换成 >
&lt;   换成

请问还有中间定义那行代码怎么改

elim

[a,b,...]=a+\large\frac{1}{[b,...]}

主要是因为论坛软件遇到[b]就想多了…

乘简

\(\,[a,b]=a+\large\frac{1}{[b]},\,\)
\[\,[a,b]=a+\large\frac{1}{[b]},\,\]
\(\,\lbrack a,b\rbrack=a+\large\frac{1}{\lbrack b\rbrack},\,\)

永远

乘简 发表于 2020-9-1 08:38
\(\,[a,b]=a+\large\frac{1}{},\,\)
\[\,[a,b]=a+\large\frac{1}{},\,\]
\(\,\lbrack a,b\rbrack=a+\larg ...

谢谢老师指导

wlc1