关于连分数的余式问题

hejoseph

设 $D>1$，记 $\sqrt D$ 的连分数为 $[a_0,a_1,\cdots,a_n,\cdots]$，以及 $r_j=[a_j,a_{j+1},\cdots]$，那么必定有 $r_j=\frac{\sqrt D+P_j}{Q_j}$，可以确定 $Q_j>0$，那么当 $j\geqslant 1$ 时 $P_j$ 必定是正的吗？

hejoseph

$\sqrt D$ 的第 $k$（$k\geq 1$）个渐近分数是 $p_k/q_k$，则 $P_k=(-1)^{k+1}(p_{k-1}p_{k-2}-Dq_{k-1}q_{k-2})$，$Q_k=(-1)^k(p_{k-1}^2-Dq_{k-1}^2)$，不过没找到 $P_k\leq 0$ 的例子。

hejoseph

设 $D$ 不是完全平方数，$D>1$，$\sqrt{D}$ 的连分数为 $[a_0,a_1,\dots,a_n,\dots]$，令 $r_j=[a_j,a_{j+1},\dots]$，那么必定有 $r_j=\dfrac{\sqrt{D}+P_j}{Q_j}$，可以确定 $Q_j>0$，那么当 $j\geqslant 1$ 时 $P_j$ 一定是正的吗？

$\sqrt{D}$ 的第 $k$（$k\geqslant 1$）个渐近分数是 $\dfrac{p_k}{q_k}$，则 $P_k=(-1)^{k+1}(p_{k-1}p_{k-2}-Dq_{k-1}q_{k-2})$，$Q_k=(-1)^k(p_{k-1}^2-Dq_{k-1}^2)$，不过没找到 $P_k\leqslant 0$ 的例子。

睡神

还是算错了。。。

设$x_k=\dfrac{p_k}{q_k}$为$\sqrt D$的第$k$个渐近分数

易知，当$k$为奇数时，$x_k＞\sqrt D$，且$x_k$单调递减；当$k$为偶数时，$x_k<\sqrt D$，且$x_k$单调递增

令$\varepsilon_k=|x_k-\sqrt D|$，​则$\varepsilon_k$单调递减

​1、当$k$为奇数时，

$x_kx_{k+1}  =(\sqrt D+\varepsilon_k)(\sqrt D-\varepsilon_{k+1})

=D+\sqrt D(\varepsilon_k-\varepsilon_{k+1})-\varepsilon_k\varepsilon_{k+1}

>D+\varepsilon_k-\varepsilon_{k+1}-\varepsilon_k\varepsilon_{k+1}$

(猜测：$\forall k\in N$，都有$\varepsilon_k-\varepsilon_{k+1}-\varepsilon_k\varepsilon_{k+1}>0$)

(另外，可证得$\varepsilon_k+\varepsilon_{k+1}=\dfrac{1}{q_kq_{k+1}}$，不知道有没有用)

​2、当$k$为偶数时，
$x_kx_{k+1}=(\sqrt D-\varepsilon_k)(\sqrt D+\varepsilon_{k+1})=D-\sqrt D(\varepsilon_k-\varepsilon_{k+1})-\varepsilon_k\varepsilon_{k+1}<D$

综上所述，$P_k=(-1)^{k-1}(p_kp_{k+1}-Dq_kq_{k+1})=(-1)^{k-1}q_kq_{k+1}(x_kx_{k+1}-D)＞0$

睡神

记录两个推导出来的结论：

1、$P_k+P_{k+1}=a_k\cdot Q_k$

2、$Q_k\cdot Q_{k+1}+P^2_{k+1}=D$

睡神

好像证出来了，先打个草稿，睡觉~

设$x_k=\dfrac{p_k}{q_k}$为$\sqrt D$的第$k$个渐近分数

易知，当$k$为奇数时，$x_k＞\sqrt D$，且$x_k$单调递减；当$k$为偶数时，$x_k<\sqrt D$，且$x_k$单调递增

令$\varepsilon_k=|x_k-\sqrt D|$，​则$\varepsilon_k$单调递减，且$\varepsilon_k+\varepsilon_{k+1}=\dfrac{1}{q_kq_{k+1}}$

​1、当$k$为奇数时，

$x_kx_{k+1}=(\sqrt D+\varepsilon_k)(\sqrt D-\varepsilon_{k+1})=D+\sqrt D(\varepsilon_k-\varepsilon_{k+1})-\varepsilon_k\varepsilon_{k+1}$

而$\varepsilon_k+\varepsilon_{k+1}<x_k-\dfrac{D}{x_k}=\sqrt D+\varepsilon_k-\dfrac{D}{\sqrt D+\varepsilon_k}$

$\Rightarrow  \varepsilon_{k+1}<\sqrt D-\dfrac{D}{\sqrt D+\varepsilon_k}=\dfrac{\sqrt D\varepsilon_k}{\sqrt D+\varepsilon_k}$

$\Rightarrow  \sqrt D\varepsilon_{k+1}+\varepsilon_k\varepsilon_{k+1}<\sqrt D\varepsilon_k$

即$\sqrt D(\varepsilon_k-\varepsilon_{k+1})-\varepsilon_k\varepsilon_{k+1}＞0$成立

所以  $x_kx_{k+1}＞D$
​
2、当$k$为偶数时，
$x_kx_{k+1}=(\sqrt D-\varepsilon_k)(\sqrt D+\varepsilon_{k+1})=D-\sqrt D(\varepsilon_k-\varepsilon_{k+1})-\varepsilon_k\varepsilon_{k+1}<D$

综上所述，$P_k=(-1)^{k-1}(p_kp_{k+1}-Dq_kq_{k+1})=(-1)^{k-1}q_kq_{k+1}(x_kx_{k+1}-D)＞0$

睡神

睡神 发表于 2024-4-2 02:15
好像证出来了，先打个草稿，睡觉~

设$x_k=\dfrac{p_k}{q_k}$为$\sqrt D$的第$k$个渐近分数

$\varepsilon_k+\varepsilon_{k+1}<x_k-\dfrac{D}{x_k}$这一步未得证，化简时应该得到$p_k<q_{k+1}Q_{k+1}$，那会躺在床上，错想成$q_k<q_{k+1}Q_{k+1}$这个了

如何证明$p_k<q_{k+1}Q_{k+1}$？

睡神

好吧，决定重写...

由连分数的基本理论易知：

当$k$为奇数时，$x_k>\sqrt D$，且$x_k$单调递减；

当$k$为偶数时，$x_k<\sqrt D$，且$x_k$单调递增；

令$\varepsilon_k=|x_k-\sqrt D|$，​则$\varepsilon_k$单调递减

以上的理论就不在这证明了

证明：$\because |p_{k+1}q_k-p_kq_{k+1}|=1$

$\therefore |x_{k+1}-x_k|=|\dfrac{p_{k+1}}{q_{k+1}}-\dfrac{p_k}{q_k}|=\dfrac{1}{q_kq_{k+1}}$

$\therefore \varepsilon_k+\varepsilon_{k+1}=|(x_{k+1}-\sqrt D)+(\sqrt D-x_k)|=|x_{k+1}-x_k|=\dfrac{1}{q_kq_{k+1}}$

$\therefore \varepsilon_{k+1}=\dfrac{1}{q_kq_{k+1}}-\varepsilon_k$

$\because \sqrt D=\dfrac{r_{k+1}p_k+p_{k-1}}{r_{k+1}q_k+q_{k-1}}$

$\therefore \varepsilon_k=|\sqrt D-\dfrac{p_k}{q_k}|=\dfrac{1}{q_k(r_{k+1}q_k+q_{k-1})}=\dfrac{1}{q_k[(a_{k+1}+\dfrac{1}{r_{k+2}})q_k+q_{k-1}]}=\dfrac{1}{q_k(q_{k+1}+\dfrac{q_k}{r_{k+2}})}>\dfrac{1}{q_k(q_k+q_{k+1})}$

1、当$k$为奇数时，

$x_kx_{k+1}=(\sqrt D+\varepsilon_k)(\sqrt D-\varepsilon_{k+1})=D+\sqrt D(\varepsilon_k-\varepsilon_{k+1})-\varepsilon_k\varepsilon_{k+1}$

$=D+(\dfrac{p_k}{q_k}-\varepsilon_k)(2\varepsilon_k-\dfrac{1}{q_kq_{k+1}})-\varepsilon_k(\dfrac{1}{q_kq_{k+1}}-\varepsilon_k)$

$=D-\varepsilon^2_k+\dfrac{2p_k}{q_k}\varepsilon_k-\dfrac{p_k}{q^2_kq_{k+1}}>D-\dfrac{1}{q^2_k(q_k+q_{k+1})^2}+\dfrac{2p_k}{q_k}\dfrac{1}{q_k(q_k+q_{k+1})}-\dfrac{p_k}{q^2_kq_{k+1}}$

$=D+\dfrac{-q_{k+1}+2p_kq_{k+1}(q_k+q_{k+1})-p_k(q_k+q_{k+1})^2}{q^2_kq_{k+1}(q_k+q_{k+1})^2}$

$=D+\dfrac{-q_{k+1}+p_kq^2_{k+1}-p_kq^2_k}{q^2_kq_{k+1}(q_k+q_{k+1})^2}\ge D+\dfrac{-q_{k+1}+p_k(q_k+1)q_{k+1}-p_kq^2_k}{q^2_kq_{k+1}(q_k+q_{k+1})^2}$

$=D+\dfrac{q_{k+1}(p_k-1)+p_kq_k(q_{k+1}-q_k)}{q^2_kq_{k+1}(q_k+q_{k+1})^2}\ge D$

2、当$k$为偶数时，

$x_kx_{k+1}=(\sqrt D-\varepsilon_k)(\sqrt D+\varepsilon_{k+1})=D-\sqrt D(\varepsilon_k-\varepsilon_{k+1})-\varepsilon_k\varepsilon_{k+1}<D$

综上所述，$P_k=(-1)^{k+1}(p_{k-1}p_{k-2}-Dq_{k-1}q_{k-2})=(-1)^{k+1}q_{k-1}q_{k-2}(x_{k-1}x_{k-2}-D)>0$

睡神

睡神 发表于 2024-3-22 22:12
还是算错了。。。

设$x_k=\dfrac{p_k}{q_k}$为$\sqrt D$的第$k$个渐近分数

也可以接2楼的，证明$\varepsilon_k-\varepsilon_{k+1}-\varepsilon_k\varepsilon_{k+1}>0$

证明如下：

$\varepsilon_k+\varepsilon_{k+1}=\dfrac{1}{q_kq_{k+1}},\varepsilon_k>\dfrac{1}{q_k(q_k+q_{k+1})}$的证明如上楼，略

$\varepsilon_k-\varepsilon_{k+1}-\varepsilon_k\varepsilon_{k+1}=2\varepsilon_k-\dfrac{1}{q_kq_{k+1}}-\varepsilon_k(\dfrac{1}{q_kq_{k+1}}-\varepsilon_k)$

$=\varepsilon^2_k+\dfrac{2q_kq_{k+1}-1}{q_kq_{k+1}}\varepsilon_k-\dfrac{1}{q_kq_{k+1}}>\dfrac{1}{q^2_k(q_k+q_{k+1})^2}+\dfrac{2q_kq_{k+1}-1}{q^2_kq_{k+1}(q_k+q_{k+1})}-\dfrac{1}{q_kq_{k+1}}$

$=\dfrac{q_{k+1}+(2q_kq_{k+1}-1)(q_k+q_{k+1})-q_k(q_k+q_{k+1})^2}{q^2_kq_{k+1}(q_k+q_{k+1})^2}$

$=\dfrac{q_kq^2_{k+1}-q^3_k-q_k}{q^2_kq_{k+1}(q_k+q_{k+1})^2}\ge \dfrac{q_kq_{k+1}(q_k+1)-q^3_k-q_k}{q^2_kq_{k+1}(q_k+q_{k+1})^2}$

$=\dfrac{q^2_k(q_{k+1}-q_k)+q_k(q_{k+1}-1)}{q^2_kq_{k+1}(q_k+q_{k+1})^2}\ge 0$

hejoseph

谢谢，看了没什么问题了。这类数论的问题对我来说还是不太熟悉。

睡神

hejoseph 发表于 2024-4-8 14:38
谢谢，看了没什么问题了。这类数论的问题对我来说还是不太熟悉。

连分数难，数论更难，我也不懂