连分数的渐近分数分析

永远

永远

永远 发表于 2023-5-3 23:48

二楼没有这个连分数转化成级数的分析过程？

elim

【定理】设正整数 \(\small n,\,k\) 满足关系\(\small k=\lfloor\sqrt{n}\rfloor\)有 \(k^2< n < (k+1)^2\)
\(\quad\small\quad\)则 \(\small\sqrt{n}=k+c[{\scriptsize\dfrac{vw}{b},\dfrac{w}{b}, \dfrac{vw}{b},\dfrac{w}{b}, \dfrac{vw}{b},\dfrac{w}{b}, \dfrac{vw}{b},\dfrac{w}{b},\dfrac{vw}{b},\dfrac{w}{b},}\cdots]\)
\(\quad\small\quad\)其中 \(\small c=\gcd(n-k^2,2k), u=\gcd({\frac{n-k^2}{c}},c),\, w={\frac{2k}{c}},\,v={\frac{c}{u}},\)
\(\quad\small\quad b{\scriptsize=\dfrac{n-k^2}{cu}}, x=\sqrt{n},\)
【证】令 \(\scriptsize z :=\dfrac{x-k}{c}=\dfrac{n-k^2}{c(2k+(x-k))}=\dfrac{n-k^2}{c(2k+cz)}=\dfrac{cub}{c(cw+cz)}=\dfrac{b}{vw+vz}\)
\(\qquad{\small\qquad}\scriptsize=\dfrac{b}{vw+\dfrac{vb}{vw+vz}}=\dfrac{b}{vw+\dfrac{b}{w+z}}=\dfrac{b}{vw+\dfrac{b}{v+\dfrac{b}{vw+\dfrac{b}{v+\cdots}}}}.\;\;_\blacksquare\)
注意\(\,{\small n-k^2\le 2k,\;\gcd(b,vw)=1},\;b=\frac{n-k^2}{cu}\le\frac{2k}{cu}=\frac{w}{u}\le w\)
据三楼结果,\(\frac{\sqrt{n}-k}{c}\) 的一般渐近分数由下式给出:\(\big({\small a_n=wv^{\scriptsize\frac{1-(-1)^n}{2}},\; b_n=b}\big)\)
\(\small(^*)\quad\;\;\scriptsize\dfrac{p_n}{q_n}=\dfrac{a_np_{n-1}+bp_{n-2}}{a_nq_{n-1}+bq_{n-2}},\;\;\begin{bmatrix}p_n& p_{n-1}\\q_n&q_{n-1}\end{bmatrix}=\begin{bmatrix}0& 1\\1&0\end{bmatrix}\displaystyle\prod_{k=1}^n\begin{bmatrix}a_k&1\\b&0\end{bmatrix}\)
\({\scriptsize\qquad\;\;\,}\small a_n\ge b,\,q_1=a_1\ge b,\,q_n=a_nq_{n-1}+bq_{n-2}>bq_{n-1}\,(n>1)\)
\({\scriptsize\qquad\;\;\,}\small q_{n+1}q_n\ge q_n(a_{n+1}q_n+bq_{n-1})\ge q_nb(q_n+q_{n-1})\ge b(b+1)q_nq_{n-1}\)
\({\scriptsize\qquad\;\;}\small\ge b^n(b+1)^{n-1} q_1q_0\ge b^n(b+1)^{n-1}\)
\({\small(1)\quad\;\;}\scriptsize\dfrac{p_n}{q_n}-\dfrac{p_{n-1}}{q_{n-1}}=\dfrac{1}{q_nq_{n-1}}\det\begin{bmatrix}p_n&p_{n-1}\\q_n&q_{n-1}\end{bmatrix}=\dfrac{(-1)^{n-1}b^n}{q_nq_{n-1}}(=\delta_n,\;\;|\delta_n|\le\frac{1}{(b+1)^{n-1}})\)
\({\small(2)\quad\;\;}\frac{p_{n+2}}{q_{n+2}}-\frac{p_n}{q_n}{\small=}\frac{{\small(-1)^n}b^n}{q_{n+1}}(\frac{b^2}{q_{n+2}}+\frac{1}{q_n})\;\big(\{\frac{p_{2k}}{q_{2k}}\}\)严格增,\(\{\frac{p_{2k-1}}{q_{2k-1}}\}\)严格降\(\big)\)
\({\small(3)\quad\;\;}\frac{p_{2m}}{q_{2m}}< \frac{p_{2n-1}}{q_{2n-1}},\;{\small\sup}\{\frac{p_{2n}}{q_{2n}}\}=\frac{\sqrt{n}-k}{c}={\small\inf}\{\frac{p_{2n-1}}{q_{2n-1}}\}\)
【例】平方根的无穷连分数公式展开(前\(\small(10)\)个素数)
\(\qquad\begin{array} {|l|l|} \hline
n& k& c& b& w & v & \sqrt{n}=k+c[\frac{vw}{b},\frac{w}{b},\frac{vw}{b},\frac{w}{b},\ldots]]\\ \hline
2  & 1 & 1 & 1 & 2 & 1 & \;\,\sqrt{2}=1+[2,2,2,2,2,2,2\ldots]\\
3  & 1 & 2 & 1 & 1 & 3 & \;\,\sqrt{3}=1+[2,1,2,1,2,1,2,\ldots]\\
5  & 2 & 1 & 1 & 4 & 1 & \;\,\sqrt{5}=1+[4,4,4,4,4,4,4,\ldots]\\
7  & 2 & 1 & 3 & 4 & 1 & \;\,\sqrt{7}=2+[\frac{4}{3},\frac{4}{3},\frac{4}{3},\frac{4}{3},\frac{4}{3},\frac{4}{3}\ldots]\\
11& 3 & 2 & 1 & 3 & 2 & \sqrt{11}=3+[6,3,6,3,6,3,6\ldots]\\
13& 3 & 2 & 1 & 3 & 1 & \sqrt{13}=3+2[3,3,3,3,3,3,\ldots]\\
17& 4 & 1 & 1 & 8 & 1 & \sqrt{17}=4+ [8,8,8,8,8,8,8,\dots]\\
19& 4 & 1 & 3 & 8 & 1 & \sqrt{19}=4+[\frac{8}{3},\frac{8}{3},\frac{8}{3},\frac{8}{3},\frac{8}{3},\frac{8}{3},\ldots]\\
23& 4 & 1 & 7 & 8 & 1 & \sqrt{23}=4+[\frac{8}{7},\frac{8}{7},\frac{8}{7},\frac{8}{7},\frac{8}{7},\frac{8}{7},\ldots]\\
29& 5 & 2 & 1 & 5 & 1 & \sqrt{29}=5+2[5,5,5,5,5,5,5,\ldots]\\
\hline  \end{array}\)

luyuanhong

楼上 elim 的帖子很好！已收藏。