连分数的渐近分数分析

elim

记形如 \(\small\dfrac{1}{a_1+\dfrac{1}{a_2+\dfrac{1}{a_3+\cdots}}}\) 的连分数为 \(^1[a_1,a_2,\ldots,a_n,\ldots]\) 其
部分分数 \({\large\frac{p_{n+1}}{q_{n+1}} }= ^1\hspace{-2mm}[a_1,\ldots, a_n,a_{n+1}]=^1\hspace{-2mm}[a_1,\ldots,a_{n-1},a_n+\frac{1}{a_{n+1}}]\)
于是 \(\small\dfrac{p_1}{q_1}=^1\hspace{-1mm}[a_1]=\dfrac{1}{a_1},\;\dfrac{p_2}{q_2}=^1\hspace{-1mm}[a_1,a_2]=\dfrac{1}{a_1\scriptsize +\dfrac{1}{a_2}}=\dfrac{a_2}{a_1a_2+1}\) 假定
\(\small\dfrac{p_n}{q_n}=\dfrac{a_np_{n-1}+p_{n-2}}{a_nq_{n-1}+q_{n-2}},\,\)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 \(\small\begin{pmatrix}p_n&p_{n-1}\\q_n&q_{n-1}\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\displaystyle\prod_{k=1}^n\begin{pmatrix}a_k&1\\1&0\end{pmatrix},\;\;\dfrac{p_n}{q_n}-\dfrac{p_{n-1}}{q_{n-1}}=\dfrac{(-1)^{n-1}}{q_nq_{n-1}}\).
可见 \(\gcd(p_n,q_n)=1\) 即 \(p_n/q_n\)是既约分数.
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.

elim

现在来看 \(_1[b_1,b_2,b_3,\ldots]=\small\dfrac{b_1}{1+\dfrac{b_2}{1+\dfrac{b_3}{1+\cdots}}}\), 即 \(_1[b_1]=\large\frac{b_1}{1}\),
\(_1[b_1,\ldots,b_{n-1},b_n,b_{n+1}]=\,_1[b_1,\ldots,b_{n-1},{\large\frac{b_n}{1+b_{n+1}}}]\) 于是
\(\small\dfrac{p_n}{q_n}=\dfrac{p_{n-1}+b_np_{n-2}}{q_{n-1}+b_nq_{n-2}}\implies\dfrac{p_{n+1}}{q_{n+1}}=\dfrac{p_{n-1}+\frac{b_n}{1+b_{n+1}}p_{n-2}}{q_{n-1}+\frac{b_n}{1+b_{n+1}}q_{n-2}}=\dfrac{p_n+b_{n+1}p_{n-1}}{q_n+b_{n+1}q_{n-1}}\)
可见 \(\small\begin{pmatrix}p_n&p_{n-1}\\q_n&q_{n-1}\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\displaystyle\prod_{k=1}^n\begin{pmatrix}1&1\\b_k&0\end{pmatrix}\) 进而得到
\(\qquad\small\dfrac{p_n}{q_n}-\dfrac{p_{n-1}}{q_{n-1}}=\dfrac{(-1)^{n-1}\prod_1^n b_k}{q_nq_{n-1}}.\) 若 \(\small 0 <|b_k| < \lambda<1,\;\{b_{n+n_0}\} \) 保号,
则 \(\{p_n/q_n\}\) 从两边夹逼无穷连分数\(\,_1[b_1,b_2,b_3,\ldots].\)

elim

现在来看一般的连分数 \(\small\dfrac{b_1}{a_1+\dfrac{b_2}{a_2+\dfrac{b_3}{a_3+\cdots}}},\;\;\dfrac{p_1}{q_1}=[b_1^\lrcorner a_1]=\dfrac{b_1}{a_1},\ldots\),
\(\small\dfrac{p_n}{q_n}=[b_1^\lrcorner a_1,\ldots,b_n^\lrcorner a_n],\;\dfrac{p_{n+1}}{q_{n+1}}=[b_1^\lrcorner a_1,\ldots,b_{n-1}^{\;\lrcorner} a_{n-1},\,b_n^\lrcorner (a_n+{\large\frac{b_{n+1}}{a_{n+1}}})]\)
故有
\((1)\quad\;\;\small\dfrac{p_n}{q_n}=\dfrac{a_np_{n-1}+b_np_{n-2}}{a_nq_{n-1}+b_nq_{n-2}},\;\;\begin{bmatrix}p_n& p_{n-1}\\q_n&q_{n-1}\end{bmatrix}=\begin{bmatrix}0& 1\\1&0\end{bmatrix}\displaystyle\prod_{k=1}^n\begin{bmatrix}a_k&1\\b_k&0\end{bmatrix}\)
\((2)\quad\small\;\;\dfrac{p_n}{q_n}-\dfrac{p_{n-1}}{q_{n-1}}=\dfrac{1}{q_nq_{n-1}}\det\begin{bmatrix}p_n&p_{n-1}\\q_n&q_{n-1}\end{bmatrix}=\dfrac{(-1)^{n-1}\prod_{k=1}^n b_k}{q_nq_{n-1}}\)

一楼二楼的两类连分数都是本贴一般连分数的特例。

elim

例 1 \(x=\small\dfrac{1}{1+x}=\cdots={\scriptsize\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\cdots}}}}=[1,1,1,\ldots]=\dfrac{\sqrt{5}-1}{2}\)
例 2.  令 \(\beta_0 =\pi,\;a_0=\lfloor \beta_0\rfloor,\;\beta_{n+1}=(\beta_n-a_n)^{-1},\,a_{n+1}=\lfloor\beta_{n+1}\rfloor\)
\(\quad\)则 \(\pi\small=a_0+\dfrac{1}{\beta_1}=a_0+\dfrac{1}{a_1+\dfrac{1}{\beta_2}}=\cdots=[3; 7, 15, 1, 292, 1, 1, 1, 2,\ldots]\)
易见对(\(a_k,b_k\in\mathbb{Z}\)), 有理数能且只能表为有限连分数.
不难看出, \((0,\infty)-\mathbb{Q}=\{[a_0;a_1,a_2,\ldots]\mid \{a_n\}\in(\mathbb{N}^+)^{\mathbb{N}^+}\}\)
即简单无穷连分数皆收敛于无理数.
例 3.  令\(\,q_n{\small=2-2^{-n},}\;a_n=\frac{3}{2(2^n-1)},\;b_n\small=1\;(\small\forall n).\)
\(\qquad\)则\(q_n=a_nq_{n-1}+q_{n-2},\;q_nq_{n-1}\to 4\,(n\to\infty)\)
\(\qquad\)可见无穷连分数\(\,^1[\frac{3}{2},\frac{1}{2},\ldots,\frac{3}{2(2^n-1)},\ldots]\) 发散.

elim

\(\small\tanh 1=\dfrac{e^2-1}{e^2+1}=\,^1[1,3,5,\ldots,2n-1,\ldots]=\scriptsize\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{7+\cdots}}}}\)

王守恩

elim 发表于 2023-4-13 11:10
\(\small\tanh 1=\dfrac{e^2-1}{e^2+1}=\,^1[1,3,5,\ldots,2n-1,\ldots]=\scriptsize\dfrac{1}{1+\dfrac{1} ...

这个"0"有点怪怪的，应该去掉。 ContinuedFraction[Coth[1], 41]

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41

luyuanhong

楼上 elim 的帖子很好！已收藏。

王守恩

优美的”数字串“。\(\sqrt[a]{e}\)      a=1,2,3,4,5,6,7,8,9,...

{{1, 00, 1, 1, 02, 1, 1, 04, 1, 1, 06, 1, 1, 08, 1, 1, 010, 1, 1, 012, 1, 1, 014, 1, 1, 016, 1, 1, 018}},
{{1, 01, 1, 1, 05, 1, 1, 09, 1, 1, 13, 1, 1, 17, 1, 1, 021, 1, 1, 025, 1, 1, 029, 1, 1, 033, 1, 1, 037}},
{{1, 02, 1, 1, 08, 1, 1, 14, 1, 1, 20, 1, 1, 26, 1, 1, 032, 1, 1, 038, 1, 1, 044, 1, 1, 050, 1, 1, 056}},
{{1, 03, 1, 1, 11, 1, 1, 19, 1, 1, 27, 1, 1, 35, 1, 1, 043, 1, 1, 051, 1, 1, 059, 1, 1, 067, 1, 1, 075}},
{{1, 04, 1, 1, 14, 1, 1, 24, 1, 1, 34, 1, 1, 44, 1, 1, 054, 1, 1, 064, 1, 1, 074, 1, 1, 084, 1, 1, 094}},
{{1, 05, 1, 1, 17, 1, 1, 29, 1, 1, 41, 1, 1, 53, 1, 1, 065, 1, 1, 077, 1, 1, 089, 1, 1, 101, 1, 1, 113}},
{{1, 06, 1, 1, 20, 1, 1, 34, 1, 1, 48, 1, 1, 62, 1, 1, 076, 1, 1, 090, 1, 1, 104, 1, 1, 118, 1, 1, 132}},
{{1, 07, 1, 1, 23, 1, 1, 39, 1, 1, 55, 1, 1, 71, 1, 1, 087, 1, 1, 103, 1, 1, 119, 1, 1, 135, 1, 1, 151}},
{{1, 08, 1, 1, 26, 1, 1, 44, 1, 1, 62, 1, 1, 80, 1, 1, 098, 1, 1, 116, 1, 1, 134, 1, 1, 152, 1, 1, 170}},
{{1, 09, 1, 1, 29, 1, 1, 49, 1, 1, 69, 1, 1, 89, 1, 1, 109, 1, 1, 129, 1, 1, 149, 1, 1, 169, 1, 1, 189}},
{{1, 10, 1, 1, 32, 1, 1, 54, 1, 1, 76, 1, 1, 98, 1, 1, 120, 1, 1, 142, 1, 1, 164, 1, 1, 186, 1, 1, 208}}}

Table[ContinuedFraction[Power[E, (a)^-1], n], {a, 2, 11}, {n, 29, 29}]

王守恩

优美的”数字串“。

\(1,\displaystyle\sum_{k=0}^{\infty}\frac{1+a}{k!}\ \ \ \ a=0,1,2,3,4,5,6,7,8,9,...\)
     \(e,\ 2e,\ 3e,\ 4e,\ 5e,\ 6e,\ 7e,\ 8e,\ 9e,\ 10e, ...\)

\(2,\displaystyle\sum_{k=0}^{\infty}\frac{k+a}{k!}\ \ \ \ a=0,1,2,3,4,5,6,7,8,9,...\)
     \(e,\ 2e,\ 3e,\ 4e,\ 5e,\ 6e,\ 7e,\ 8e,\ 9e,\ 10e ...\)

\(3,\displaystyle\sum_{k=1}^{\infty}\frac{k+a}{k!}\ \ \ \ a=0,1,2,3,4,5,6,7,8,9,...\)
     \(e,2e-1,3e-2,4e-3,5e-4,6e-5,...\)

\(4,\displaystyle\sum_{k=1}^{\infty}\frac{k-a}{k!}\ \ \ \ a=0,1,2,3,4,5,6,7,8,9,...\)
     \(1,\ 2-e,\ 3-2e,4-3e,5-4e,6-5e,...\)

\(5,\displaystyle\sum_{k=0}^{\infty}\frac{k-a}{k!}\ \ \ \ a=0,1,2,3,4,5,6,7,8,9,...\)
     \(e,0,-e,-2e,-3e,-4e,-5e,-6e,-7e, ...\)
.............

永远

主贴烂尾了？？？可惜了