论文标题
部分可观测时空混沌系统的无模型预测
Increasing rate of weighted product of partial quotients in continued fractions
论文作者
论文摘要
令$ [a_1（x），a_2（x），\ cdots，a_n（x），\ cdots] $为$ x \ in [0,1）$的持续分数扩展。在本文中，我们研究了加权产品$ a^{t_0} _n（x）a^{t_1} _ {n+1}（x）（x）\ cdots a^{t_m} _ {权重。更确切地说，让$φ：\ Mathbb {n} \ to \ Mathbb {r} _+$为$φ（n）/n \ to \ infty $ as $ n \ to \ infty $。对于任何$（t_0，\ cdots，t_m）\ in \ mathbb {r}^{m+1} _+$ with $ t_i \ geq 0 $和至少一个$ t_i \ neq0 \（0 \ leq i \ leq m） $$ \ usewissline {e}（\ {t_i \} _ {i = 0}^m，φ）= \ left \ {x \ in [0,1）：\ liminf \ liminf \ limits_ {n \ to \ to \ to \ forty} \ dfrac {\ dfrac {\ log \ left（a^{t_0} _n（x）a^{t_1} _ {n+1}（x）（x）\ cdots a^{t_m} _ {n+m}（n+m}（x）\ right）}} {φ（n）} = 1 \ right \} $ \ right \} $。在$（t_0，\ cdots，t_m）\ in \ mathbb {r}^{m+1} _+$的情况下，$ 0 <t_0 \ t_0 \ leq t_1 \ leq t_1 \ leq \ cdots \ cdots \ leq t_m $，我们还获得了集合的hausdorff dimension \ set f in set \ begin begin quith \ edline {e}（\ {t_i \} _ {i = 0}^m，φ）= \ left \ {x \ in [0,1）：\ limsup \ lims_ {n \ to \ in \ to \ in \ to \ infty} \ dfrac {\ dfrac {\ log \ left（a^{t_0} _n（x）a^{t_1} _ {n+1}（x）（x）\ cdots a^{t_m} _ {n+m}
Let $[a_1(x),a_2(x),\cdots,a_n(x),\cdots]$ be the continued fraction expansion of $x\in[0,1)$. In this paper, we study the increasing rate of the weighted product $a^{t_0}_n(x)a^{t_1}_{n+1}(x)\cdots a^{t_m}_{n+m}(x)$ ,where $t_i\in \mathbb{R}_+\ (0\leq i \leq m)$ are weights. More precisely, let $φ:\mathbb{N}\to\mathbb{R}_+$ be a function with $φ(n)/n\to \infty$ as $n\to \infty$. For any $(t_0,\cdots,t_m)\in \mathbb{R}^{m+1}_+$ with $t_i\geq 0$ and at least one $t_i\neq0 \ (0\leq i\leq m)$, the Hausdorff dimension of the set $$\underline{E}(\{t_i\}_{i=0}^m,φ)=\left\{x\in[0,1):\liminf\limits_{n\to \infty}\dfrac{\log \left(a^{t_0}_n(x)a^{t_1}_{n+1}(x)\cdots a^{t_m}_{n+m}(x)\right)}{φ(n)}=1\right\}$$ is obtained. Under the condition that $(t_0,\cdots,t_m)\in \mathbb{R}^{m+1}_+$ with $0<t_0\leq t_1\leq \cdots \leq t_m$, we also obtain the Hausdorff dimension of the set \begin{equation*} \overline{E}(\{t_i\}_{i=0}^m,φ)=\left\{x\in[0,1):\limsup\limits_{n\to \infty}\dfrac{\log \left(a^{t_0}_n(x)a^{t_1}_{n+1}(x)\cdots a^{t_m}_{n+m}(x)\right)}{φ(n)}=1\right\}.\end{equation*}
