Given $\beta\in(1,2)$ and $x\in[0,\frac{ 1}{ \beta-1}]$, a sequence $(\epsilon_{ i})_{ i=1}^{ \infty}\in\{ 0,1\}^{ \mathbb{ N}}$ is called a $\beta$-expansion for $x$ if $$x=\sum_{ i=1}^{ \infty}\frac{ \epsilon_{ i}}{ \beta^{ i}}.$$ In a recent article the author studied the quality of approximation provided by the finite sums $\sum_{ i=1}^{ n}\epsilon_{ i}\beta^{ -i}$ \cite{ Bak}. In particular, given $\beta\in(1,2)$ and $\Psi:\mathbb{ N}\to\mathbb{ R}_{ \geq 0},$ we associate the set $$W_{ \beta}(\Psi):=\bigcap_{ m=1}^{ \infty}\bigcup_{ n=m}^{ \infty}\bigcup_{ (\epsilon_{ i})_{ i=1}^{ n}\in\{ 0,1\}^{ n}}\Big[\sum_{ i=1}^{ n}\frac{ \epsilon_{ i}}{ \beta^{ i}},\sum_{ i=1}^{ n}\frac{ \epsilon_{ i}}{ \beta^{ i}}+\Psi(n)\Big].$$ Alternatively, $W_{ \beta}(\Psi)$ is the set of $x\in \mathbb{ R}$ such that for infinitely many $n\in\mathbb{ N},$ there exists a sequence $(\epsilon_{ i})_{ i=1}^{ n}$ satisfying the inequalities $$0\leq x-\sum_{ i=1}^{ n}\frac{ \epsilon_{ i}}{ \beta^{ i}}\leq \Psi(n).$$ If $\sum_{ n=1}^{ \infty}2^{ n}\Psi(n)<\infty$ then $W_{ \beta}(\Psi)$ has zero Lebesgue measure. We call a $\beta\in(1,2)$ approximation regular, if $\sum_{ n=1}^{ \infty}2^{ n}\Psi(n)=\infty$ implies $W_{ \beta}(\Psi)$ is of full Lebesgue measure within $[0,\frac{ 1}{ \beta-1}]$. The author conjectured in \cite{ Bak} that almost every $\beta\in(1,2)$ is approximation regular. In this paper we make a significant step towards proving this conjecture. The main result of this paper is the following statement: given a sequence of positive real numbers $(\omega_{ n})_{ n=1}^{ \infty},$ which satisfy $\lim_{ n\to\infty} \omega_{ n}=\infty$, then for Lebesgue almost every $\beta\in(1.497\ldots,2)$ the set $W_{ \beta}(\omega_{ n}\cdot 2^{ -n})$ is of full Lebesgue measure within $[0,\frac{ 1}{ \beta-1}]$. Here the sequence $(\omega_{ n})_{ n=1}^{ \infty}$ should be interpreted as a sequence tending to infinity at a very slow rate.
Categories: math.NT math.DS