求助于论坛高手乘简、elim老师，这个论坛底部LaTEX预览输入怎么做到的

永远

求助于论坛高手乘简、elim老师，这个论坛底部LaTEX预览输入怎么做到的

在：header_common里具体怎么编码才能实现这个，我很好奇

求助！！

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.

Ysu2008

要修改网页才能实现.

永远

Ysu2008 发表于 2020-9-1 00:54
要修改网页才能实现.

请看2楼代码如何调整，乱码啦

Ysu2008

永远 发表于 2020-9-1 00:56
请看2楼代码如何调整，乱码啦

单独拿出错的那块代码测试，不行再细分为几个部分测试，直到找到原因。

elim

首先讨论非完全平方数的平方根的一般理论.
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
\({\small\sqrt{n}-m_1=}\big(\frac{\sqrt{n}+m}{k}{\small-\lambda(m,k)}\big)k>0,\;\sqrt{n}+m_1>\sqrt{n}-m>0\)
\(\therefore\;k\mid n-m^2+2mk\lambda-(k\lambda)^2=n-m_1^2>0,\;(m_1,k_1)\in F(n)\)
令\(\,\psi:  (m,k)\overset{\psi}{\mapsto}(m_1,k_1)\;\;\big(\frac{k}{\sqrt{n}+m}=\frac{k}{k\lambda+\sqrt{n}-m_1}=\frac{1}{\lambda+\large\frac{k_1}{\sqrt{n}+m_1}}\big)\)
\((m_j,k_j)=\psi^{\langle j\rangle}(m,k),\;\lambda_j=\lambda(m_{j-1},k_{j-1})\;(\psi^{\langle k+1\rangle}=\psi(\psi^{\langle k\rangle}))\)
定义\(\,[a,b]=a+\large\frac{1}{[\,b]},\,\)则有\(\small\,1\le m,\,L\in\mathbb{N}\,\)使\(\small(m_{u+L},k_{u+L})=(m_u,k_u)\)
\(\,\frac{k}{\sqrt{n}+m}=[0,\lambda_1+\frac{k_1}{\sqrt{n}+m_1}]=[0,\lambda_1,\ldots,\lambda_{u}+\frac{k_u}{\sqrt{n}+m_u}]\)
\(\qquad\cdots=[0,\lambda_1,\ldots,\lambda_{u},\ldots,\lambda_{u+L}+\frac{k_{u+L}}{\sqrt{n}+m_{u+L}}]\)
\(\qquad\quad\;=[0,\lambda_1,\ldots,\lambda_{u},\dot{\lambda}_{u+1},\ldots,\dot{\lambda}_{u+L}]\)
\(\qquad\quad\;=[0,\lambda_1,\ldots,\lambda_{u},\overline{{\lambda}_{u+1},\ldots,\lambda}_{u+L}]\)
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.
命题 若\(\,a_i\in\mathbb{N}^+,\;\alpha=[\overline{a_1,\ldots,a}_L]\;\small(L>1)\,\)
\(\qquad\)则\(\,\alpha\,\)是整系数二次方程的根.
证明：此时\(\,\alpha=[a_1,\ldots,a_L+\alpha].\) 当\(\,L=2\,\)时
\(\,\alpha=a_1+\large\frac{1}{a_2+\alpha}\;\)即\(\,\alpha^2+(a_2-a_1)\alpha-a_1a_2-1=0\)
当\(\,L>2\,\)时\(\,\alpha=[a_1,\ldots,a_L+\alpha]=[a_1,\ldots,a_{L-1}+\frac{1}{a_L+\alpha}]\)
所以据归纳法原理, 命题对一切\(L>1\)成立.
推论1 循环连分数的值具有一般形式\(\,\frac{u\pm\sqrt{|v|}}{w}\;\small(u,v,w\in\mathbb{Z})\)
推论2 若\(\,k,\,n\in\mathbb{N}^+,\, k>2,\,\sqrt[k]{n}\not\in\mathbb{N},\,\)则
\(\qquad\sqrt[k]{n}\,\)的连分数表示是无穷不循环的. 这有点出乎意料.