\(\Huge^\star\textbf{ 连分数的渐近分数分析}\)

elim

我们来定义一般的连分数 \([r_1,r_2,\ldots,r_n]\). 这里 \(r_k\in\mathbb{Q}-\{0\}\). 首先,
存在唯一的 \(\small a_k\in\mathbb{N},\,b_k\in\mathbb{Z}-\{0\}\,(\gcd(a_k,b_k)=1)\) 使 \(r_k=\large\frac{a_k}{b_k}.\) 其次
定义 \([r_1]:= r_1^{-1},\;[r_1,\ldots,r_k,r_{k+1}]:=[r_1,\ldots,r_{k-1},\frac{a_k+r_{k+1}^{-1}}{b_k}]\).
定义 \([r_1,r_2,r_3,\ldots]:=\displaystyle\lim_{n\to\infty}[r_1,r_2,\ldots,r_n]\) \(({\small a_m+r_{m+1}>0\,(\forall m))}\)
于是 \({\small[r_1,r_2,\ldots,r_n,\ldots]=}\scriptsize\dfrac{b_1}{a_1+\dfrac{b_2}{a_2+\dfrac{b_3}{a_3+\cdots}}}\;\;\big(\gcd(a_m,b_m)=1,\,a_m|b_m|>0\big)\)
称 \(\frac{p_k}{q_k}=[\frac{a_1}{b_1},\ldots,\frac{a_k}{b_k}]\,(\frac{p_0}{q_0}=\frac{0}{1})\) 为所论连分数的第\(k\)个渐近分数.
于是有\(\frac{p_0}{q_0}=\frac{0}{1},\frac{p_1}{q_1}=\frac{b_1}{a_1},\frac{p_2}{q_2}=\frac{a_2b_1}{a_2a_1+b_2}.\) 假定 \(\frac{p_n}{q_n}=\frac{a_np_{n-1}+b_np_{n-2}}{a_nq_{n-1}+b_nq_{n-2}},\)
则 \(\frac{p_{n+1}}{q_{n+1}}=\frac{(a_n+\frac{b_{n+1}}{a_{n+1}})p_{n-1}+b_np_{n-2}}{(a_n+\frac{b_{n+1}}{a_{n+1}})q_{n-1}+b_nq_{n-2}}= \frac{(a_{n+1} a_n+b_{n+1})p_{n-1}+a_{n+1}b_np_{n-2}}{(a_{n+1} a_n+b_{n+1})q_{n-1}+a_{n+1}b_nq_{n-2}}\)
\(\qquad\quad\;=\frac{a_{n+1}(a_np_{n-1}+b_np_{n-2})+b_{n+1}p_{n-1}} {a_{n+1}(a_nq_{n-1}+b_nq_{n-2})+b_{n+1}q_{n-1}}=\frac{a_{n+1}p_n+b_{n+1}p_{n-1}} {a_{n+1}q_n+b_{n+1}q_{n-1}}.\) 所以
\(\small(1)\;\;\begin{bmatrix}p_0\\q_0\end{bmatrix}{\scriptsize:=}\begin{bmatrix}0\\1\end{bmatrix}{\scriptsize,\,}
\begin{bmatrix}p_1\\q_1\end{bmatrix}{\scriptsize:=}\begin{bmatrix}b_1\\a_1\end{bmatrix}{\scriptsize,\,}
\begin{bmatrix}p_n& p_{n-1}\\q_n&q_{n-1}\end{bmatrix}=\begin{bmatrix}0& 1\\1&0\end{bmatrix}\displaystyle\prod_{k=1}^n\begin{bmatrix}a_k&1\\b_k&0\end{bmatrix}\)
\(\small(2)\quad\dfrac{p_n}{q_n}-\dfrac{p_{n-1}}{q_{n-1}}=\dfrac{1}{q_nq_{n-1}}\det\begin{bmatrix}p_n&p_{n-1}\\q_n&q_{n-1}\end{bmatrix}=\dfrac{(-1)^{n-1}\prod_{k=1}^n b_k}{q_nq_{n-1}}\)
\(\small(3)\quad\)若 \(\small a_k,b_k,M\in\mathbb{N}^+,\, b_k\le M\,(\forall k)\), 则 \(\small q_nq_{n-1}\ge(1+b_n)\ge\cdots\)
\(\small\;\qquad\ge\prod_{k=2}^n(1+b_k).\) 于是 \(\small\left|\dfrac{p_n}{q_n}-\dfrac{p_{n-1}}{q_{n-1}}\right|\le b_1\big(\dfrac{M}{1+M}\big)_{\large,}^{n-1}\;\big\{\dfrac{p_n}{q_n}\big\}\)收敛.
【定义】称连分数 \([a_1,a_2,\ldots,a_n,\ldots]\;({\small a_k\in\mathbb{N}^+\;(\forall k)})\) 为简单连分数
\(\small\dfrac{p_n}{q_n}=\dfrac{a_np_{n-1}+p_{n-2}}{a_nq_{n-1}+q_{n-2}},\,\)则 \(\small\dfrac{p_{n+1}}{q_{n+1}}=\dfrac{(a_n+\frac{1}{a_{n+1}})p_{n-1}+p_{n-2}}{(a_n+\frac{1}{a_{n+1}})q_{n-1}+q_{n-2}}=\)
\(\quad\;\;\small=\dfrac{p_n+p_{n-1}/a_{n+1}}{q_n+q_{n-1}/a_{n+1}}=\dfrac{a_{n+1}p_n+p_{n-1}}{a_{n+1}q_n+q_{n-1}}\;\;(p_0=0,\,q_0=1)\)
可见 \(\small\begin{pmatrix}p_n&p_{n-1}\\q_n&q_{n-1}\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\displaystyle\prod_{k=1}^n\begin{pmatrix}a_k&1\\1&0\end{pmatrix},\;\;\dfrac{p_n}{q_n}-\dfrac{p_{n-1}}{q_{n-1}}=\dfrac{(-1)^{n-1}}{q_nq_{n-1}}\).
若 \(\small q_nq_{n-1}\to\infty\)(例如\(a_n \small>\delta>0\)), 则 \(\scriptsize\dfrac{p_{2n}}{q_{2n}}<\dfrac{p_{2n+2}}{q_{2n+2}}<\cdots<\dfrac{p_{2n+1}}{q_{2n+1}}<\dfrac{p_{2n-1}}{q_{2n-1}},\)
\(\small^1[a_1,\ldots,a_n,\ldots]=\displaystyle\lim_{n\to\infty}\frac{p_n}{q_n}=\sum_{n=1}^\infty\big(\frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}}\big)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{q_nq_{n-1}}.\) 收敛.