逆序的連分數

hbghlyj

例如$\cfrac17=\cfrac{1}{6+\cfrac11}= [ 0; 6, 1 ],\cfrac27=\cfrac{1}{3+\cfrac1{1+\cfrac{1}{1}}} = [ 0; 3, 1, 1 ]$
\[
\begin{array}{lll}
\cfrac{1}{7} & = [ 0; 7 ] & = [ 0; 6, 1 ] \\
\cfrac{2}{7} & = [ 0; 3, 2 ] & = [ 0; 3, 1, 1 ] \\
\cfrac{3}{7} & = [ 0; 2, 3 ] & = [ 0; 2, 2, 1 ] \\
\cfrac{4}{7} & = [ 0; 1, 1, 3 ] & = [ 0; 1, 1, 2, 1 ] \\
\cfrac{5}{7} & = [ 0; 1, 2, 2 ] & = [ 0; 1, 2, 1, 1 ] \\
\cfrac{6}{7} & = [ 0; 1, 6 ] & = [ 0; 1, 5, 1 ] \\
\end{array}
\]
發現6,1的逆序是1,6
3,1,1的逆序是1,1,3
2,2,1的逆序是1,2,2
1,1,2,1的逆序是1,2,1,1
有何規律？

hbghlyj

\[
\begin{array}{lll}
\frac{1}{8} & = [ 0; 8 ] & = [ 0; 7, 1 ] \\
\frac{2}{8} & = [ 0; 4 ] & = [ 0; 3, 1 ] \\
\frac{3}{8} & = [ 0; 2, 1, 2 ] & = [ 0; 2, 1, 1, 1 ] \\
\frac{4}{8} & = [ 0; 2 ] & = [ 0; 1, 1 ] \\
\frac{5}{8} & = [ 0; 1, 1, 1, 2 ] & = [ 0; 1, 1, 1, 1, 1 ] \\
\frac{6}{8} & = [ 0; 1, 3 ] & = [ 0; 1, 2, 1 ] \\
\frac{7}{8} & = [ 0; 1, 7 ] & = [ 0; 1, 6, 1 ] \\
\end{array}
\]
發現7,1的逆序是1,7
3,1的逆序是1,3
2,1,1,1的逆序是1,1,1,2
有何規律？

hbghlyj

\[
\begin{array}{lll}
\frac{1}{9} & = [ 0; 9 ] & = [ 0; 8, 1 ] \\
\frac{2}{9} & = [ 0; 4, 1, 1 ] & = [ 0; 4,2 ] \\
\frac{3}{9} & = [ 0; 3 ] & = [ 0; 2, 1 ] \\
\frac{4}{9} & = [ 0; 2,4] & = [ 0; 2, 3, 1 ] \\
\frac{5}{9} & = [ 0; 1, 1, 4] & = [ 0; 1, 1,3,1 ] \\
\frac{6}{9} & = [ 0;  1, 2 ] & = [ 0; 1, 1, 1] \\
\frac{7}{9} & = [ 0; 1, 3, 2 ] & = [ 0; 1, 3,1, 1 ] \\
\frac{8}{9} & = [ 0; 1, 8 ] & = [ 0; 1, 7, 1 ] \\
\end{array}
\]
發現8,1的逆序是1,8
4,1,1的逆序是1,1,4
1,2的逆序是2,1
1,3,2的逆序是2,3,1
有何規律？

hbghlyj

例如 $\color{magenta}{[1;1,1,2]} =\cfrac85 =\cfrac AB$
和 $[1;1,1] =\cfrac32 =\cfrac CD$，
$\implies\color{magenta}{[2;1,1,1]}= \cfrac83 = \cfrac AC$ 是顺序相反的

hbghlyj

hbghlyj 发表于 2024-2-12 09:45
例如 $\color{magenta}{[a_0;a_1,\dots,a_n]} =\cfrac AB$
和 $[a_0;a_1,\dots,a_{n-1}]=\cfrac CD$，
$\implies[a_n;a_{n-1},\dots,a_0]=\cfrac AC$

如何證明呢

睡神

先解答第一个问题，即使不一定对…第二个问题感觉方向差不多，就懒得动手了…数学大菜鸟一个，错了勿怪哈

记$A=\{\displaystyle\frac{1}{q},\frac{2}{q},\cdots ,\frac{p}{q},\cdots ,\frac{q-1}{q}\}$

下证：若$ \displaystyle\frac{p}{q}=[0;a_1,a_2,\cdots ,a_n]\in A$，则$ [0;a_n,a_{n-1},\cdots ,a_1]\in A$

证明：记$ \displaystyle\frac{p}{q}$的渐进分数为$ \displaystyle\frac{p_k}{q_k}$，且$(p_k,q_k)=1$，则$q=q_n$

而$q_k=a_kq_{k-1}+q_{k-2},q_{-1}=0,q_0=1$为分母的递推关系

$ \Longrightarrow \displaystyle\frac{q_k}{q_{k-1}}=a_k+\displaystyle\frac{q_{k-2}}{q_{k-1}}=a_k+\displaystyle \frac{1}{\frac{q_{k-1}}{q_{k-2}}}$

$ \Longrightarrow \displaystyle\frac{q_{k-1}}{q_k}=\displaystyle \frac{1}{a_k+\displaystyle \frac{1}{\frac{q_{k-1}}{q_{k-2}}}}=[0;a_k,a_{k-1},\cdots ,a_1]$

当$k=n$时，$ \displaystyle\frac{q_{n-1}}{q_n}=[0;a_n,a_{n-1},\cdots ,a_1]=\displaystyle\frac{q_{n-1}}{q}\in A $

尝试码一下代码，原图片就删啦

睡神

hbghlyj 发表于 2024-2-12 19:22
如何證明呢

第三行应该是：$\color{magenta}{[a_n;a_{n-1},\dots,a_0]} =\cfrac AC$