# α-Lu¨ roth展式若干度量性质Some Metric Properties in α-Lu¨ roth Expansions

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For the α-Lüroth expansion, some metric properties, such as “0-1” law, iterated logarithm law of the digits are studied in this paper. As the extension of alternating-Lüroth expansion, the conclusions in this paper include those of alternating-Lüroth case.

1. 引言

${L}_{\alpha }\left(x\right):=\left\{\begin{array}{l}\frac{{t}_{n}-x}{{a}_{n}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in {A}_{n},n\in N\\ 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=0\end{array}$

$x={t}_{{l}_{1}\left(x\right)}+\underset{j=2}{\overset{\infty }{\sum }}{\left(-1\right)}^{j-1}\left(\underset{1\le i\le j}{\prod }{a}_{{l}_{1}\left(x\right)}\right){t}_{{l}_{j}\left(x\right)}$

1) 如果 $\underset{n=1}{\overset{\infty }{\sum }}\frac{\text{1}}{\varphi \left(n\right)}$ 发散，则 $\lambda \left(A\right)=0$,即

2) 如果 $\underset{n=1}{\overset{\infty }{\sum }}\frac{\text{1}}{\varphi \left(n\right)}$ 收敛，则 $\lambda \left\{x\in \left(0,1\right]:{l}_{n}\left(x\right)>\varphi \left(n\right)\text{ }\text{\hspace{0.17em}}i.o.n\right\}=\text{0}$

${L}_{n}\left(x\right)=\mathrm{max}\left\{{l}_{1}\left(x\right),\cdots ,{l}_{n}\left(x\right)\right\}$

$\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\mathrm{log}{l}_{n}\left(x\right)-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}=1$

$\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\mathrm{log}{L}_{n}\left(x\right)-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}=\text{1}$

$\underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\mathrm{log}{l}_{n}\left(x\right)-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}=-\infty$

$\underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\mathrm{log}{L}_{n}\left(x\right)-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}=0$

2. 准备工作

$I\left({l}_{1},{l}_{2},\cdots ,{l}_{k}\right):=\left\{x\in \left[0,1\right):{l}_{i}\left(x\right)={l}_{i}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}1\le i\le k\right\}$

1) 若对于 $\rho >1$，有 $\underset{n\to \infty }{\mathrm{lim}}\frac{{t}_{n}}{{t}_{n+1}}=\rho$ ；则 $\alpha$ 是扩张的；

2) 如果分割 $\alpha$ 的尾部满足幂律，，其中 $\psi :N\to {R}^{+}$ 是一个缓变函数，则 $\alpha$ 是指数 $\theta >0$ 的指数扩张的；

3) 如果n充分大，我们有，则分割 $\alpha$ 最终是递减的。

$\lambda \left(I\left({l}_{1},{l}_{2},\cdots ,{l}_{k}\right)\right)=\underset{i=1}{\overset{k}{\prod }}\lambda \left(I\left({l}_{i}\right)\right)$

${n}^{-\left(1+\theta +\eta \right)}\le {a}_{n}\le {n}^{-\left(1+\theta -\eta \right)}$(2.1)

${n}^{-\left(\theta +\eta \right)}\le {t}_{n}\le {n}^{-\left(\theta -\eta \right)}$(2.2)

$\underset{n\to \infty }{\mathrm{lim}}\frac{\mathrm{log}{a}_{n}}{n}=\underset{n\to \infty }{\mathrm{lim}}\frac{{t}_{n}}{{t}_{n+1}}=-\mathrm{log}\rho$

3. 主要结果的证明

3.1. 定理1.2的证明

${a}_{n}=\left\{\begin{array}{l}{n}_{k}\mathrm{log}{n}_{k}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}k\in N,n={n}_{k}\\ 0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne {n}_{k}\end{array}$

$\lambda \left\{x\in \left(0,1\right]:{l}_{n}\left(x\right)\le {a}_{n}\text{\hspace{0.17em}}i.o.n\right\}=1$

$\frac{\mathrm{log}{l}_{n}-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}\le \frac{\mathrm{log}{a}_{n}-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}$

$\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\mathrm{log}{l}_{n}-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}\le 1$

$\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\mathrm{log}{l}_{n}-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}=1$

$\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{\mathrm{log}{L}_{n}-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}=1$

3.2. 定理1.3的证明

$\underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\mathrm{log}{l}_{n}-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}\le -\infty$

$\theta \in \left(1,+\infty \right)$ 时，证明过程是相同的。然后，我们给出 ${L}_{n}$ 的“0-1”比率。

${b}_{n}=\left\{\begin{array}{l}{n}_{k}{\left(\mathrm{log}{n}_{k}\right)}^{\gamma }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}k\in N,n={n}_{k}\\ 0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne {n}_{k}\end{array}$

${\left(1-\frac{1}{{\left(n{\left(\mathrm{log}n\right)}^{\gamma }\right)}^{\left(\theta -\epsilon \right)}}\right)}^{n}\le \lambda \left({A}_{\gamma }\right)\le {\left(1-\frac{1}{{\left(n{\left(\mathrm{log}n\right)}^{\gamma }\right)}^{\left(\theta +\epsilon \right)}}\right)}^{n}$

$\begin{array}{l}\lambda \left\{\underset{m\ge 1}{\cap }\underset{n\ge m}{\cup }x\in \left[0,1\right):{L}_{n}\left(x\right)\le n{\left(\mathrm{log}n\right)}^{\gamma }\right\}\\ \ge \underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}\lambda \left\{x\in \left[0,1\right):{L}_{n}\left(x\right)\le n{\left(\mathrm{log}n\right)}^{\gamma }\right\}\\ \ge \underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}{\left(1-\frac{1}{{\left(n{\left(\mathrm{log}n\right)}^{\gamma }\right)}^{\left(\theta -\epsilon \right)}}\right)}^{n}\end{array}$

$\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}{\left(1-\frac{1}{{\left(n{\left(\mathrm{log}n\right)}^{\gamma }\right)}^{\left(\theta -\epsilon \right)}}\right)}^{n}\ge \underset{n\to \infty }{\mathrm{lim}}{\left(1-\frac{1}{{\left({n}^{\left[\frac{\text{2}}{\theta }\right]}{\left(\left[\frac{\text{2}}{\theta }\right]\mathrm{log}n\right)}^{\gamma }\right)}^{\left(\theta -\epsilon \right)}}\right)}^{n}\ge \underset{n\to \infty }{\mathrm{lim}}{\left(1-\frac{1}{{\left({n}^{\frac{\text{2}}{\theta }}{\left(\frac{\text{2}}{\theta }\mathrm{log}n\right)}^{\gamma }\right)}^{\left(\theta -\epsilon \right)}}\right)}^{n-p\left(n\right)}=1$

$\underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\mathrm{log}{L}_{n}-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}\le \gamma$

$\underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\mathrm{log}{L}_{n}-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}\le 0$

$\stackrel{¯}{{A}_{\frac{\tau }{\theta }}}=\left\{x\in \left[0,1\right):{L}_{n}\left(x\right)\le {n}^{{}_{\frac{1}{\theta }}}{\left(\mathrm{log}n\right)}^{-\frac{\tau }{\theta }},{L}_{n+1}\left(x\right)\le {\left(n+1\right)}^{{}^{{}_{\frac{1}{\theta }}}}{\left(\mathrm{log}\left(n+1\right)\right)}^{-\frac{\tau }{\theta }}\right\}$

$\begin{array}{c}\lambda \left(\stackrel{¯}{{A}_{\frac{\tau }{\theta }}}\right)=\lambda \left\{x\in \left[0,1\right):{L}_{n}\left(x\right)\le {n}^{{}_{\frac{1}{\theta }}}{\left(\mathrm{log}n\right)}^{-\frac{\tau }{\theta }},{L}_{n+1}\left(x\right)\le {\left(n+1\right)}^{{}^{{}_{\frac{1}{\theta }}}}{\left(\mathrm{log}\left(n+1\right)\right)}^{-\frac{\tau }{\theta }}\right\}\\ \le {\left(1-\frac{1}{{\left({n}^{\frac{\theta +\epsilon }{\theta }}{\left(\mathrm{log}n\right)}^{\gamma }\right)}^{-\frac{\tau \left(\theta +\epsilon \right)}{\theta }}}\right)}^{n}\frac{1}{{\left(n+1\right)}^{\frac{\theta -\epsilon }{\theta }}{\left(\mathrm{log}\left(n+1\right)\right)}^{-\frac{\tau \left(\theta +\epsilon \right)}{\theta }}}\\ \le {\text{e}}^{-\frac{1}{{\left(\mathrm{log}n\right)}^{-\tau }}}\cdot \frac{{\left(\mathrm{log}n\right)}^{\tau }}{n}\\ \le {\text{e}}^{-\frac{1}{{\left(\mathrm{log}n\right)}^{\tau }}}\cdot {n}^{\tau -1}\\ \le {\text{e}}^{-\frac{1}{{n}^{\tau }}}\cdot {n}^{\tau -1}\end{array}$

${\text{e}}^{-\frac{1}{{n}^{\tau }}}\cdot {n}^{\tau -1}=\underset{k=1}{\overset{\infty }{\sum }}{\left(-1\right)}^{k}\frac{1}{k!{n}^{k\tau }}\cdot {n}^{\tau -1}=\underset{k=1}{\overset{\infty }{\sum }}{\left(-1\right)}^{k}\frac{1}{k!{n}^{\left(k-1\right)\tau +1}}$

$\underset{n=1}{\overset{\infty }{\sum }}\lambda \left(\stackrel{¯}{{A}_{\frac{\tau }{\theta }}}\right)=\underset{n=1}{\overset{\infty }{\sum }}\frac{1}{{k}_{0}!{n}^{\left({k}_{0}-1\right)\tau }}\le +\infty$

$\lambda \left\{{L}_{n}\left(x\right)\le {n}^{\frac{1}{\theta }}{\left(\mathrm{log}n\right)}^{-\frac{\tau }{\theta }}\text{\hspace{0.17em}}\text{ }i.o.n\right\}=0$

$\underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\mathrm{log}{L}_{n}-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}\ge -\frac{\tau }{\theta }$

$\underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{\mathrm{log}{L}_{n}-\mathrm{log}n}{\mathrm{log}\mathrm{log}n}\ge -\text{0}$

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