This post is my note for What is…? seminar on Lüroth expansion.

Definition 1. A Lüroth expansion of a real number $x \in (0, 1]$ is a (possibly) infinite sequence $(a_1, a_2, \cdots, a_n, \cdots)$ with $a_n \in \N$ and $a_n \ge 2$ for all $n \ge 1$ such that $\begin{equation} x = \frac{1}{a_1} + \frac{1}{a_1(a_1-1)a_2} + \cdots + \frac{1}{a_1(a_1-1)\cdots a_{n-1}(a_{n-1}-1)a_n} + \cdots \end{equation}$

By abuse of notation, we will something write the right hand side of Definition 1 as $(a_1, a_2, \cdots, a_n, \cdots)$.

Given a system of representation of real numbers in $(0, 1]$, we need to determine whether these representations are 1 to 1 corresponding the numbers in $(0, 1]$. Aslo we should study the (Lüroth) shift map: $\begin{equation} T: (a_1, a_2, \cdots, a_n, \cdots) \mapsto (a_2, \cdots, a_n, \cdots). \label{20170620-2} \end{equation}$

Let us define $T: (0, 1] \to (0, 1]$. For each $x \in (0, 1]$, define $\begin{equation} T(x) = \left\lfloor\frac{1}{x}\right\rfloor \left(\left\lfloor\frac{1}{x}\right\rfloor+1\right)x - \left\lfloor\frac{1}{x}\right\rfloor, \label{20170620-3} \end{equation}$

where $\left\lfloor\alpha\right\rfloor$ is the largest integer that is no more than $\alpha$.

Lemma 2. $T$ is a well-defined map.

Pf. To check this, we need to show that for each $x \in (0, 1]$, $T(x) \in (0, 1]$. For such an $x$, there exists an integer $n \ge 1$ such that $\frac{1}{n+1} < x \le \frac{1}{n}.$ Then $\left\lfloor\frac{1}{x}\right\rfloor = n$. So, $\begin{equation*} 0 < T(x) = n[(n+1)x-1] = n(n+1)\left(x - \frac{1}{n+1}\right) \le 1. \end{equation*}$ This finishes the proof of the lemma. $\vartriangleleft$

Remark. (1) There is more to say from the proof of the lemma: for each integer $n \ge 1$, $T$ defines a bijective map between $(\frac{1}{n+1}, \frac{1}{n}]$ and $(0, 1]$.

(2) Suppose $x \in (0, 1]$ and $(a_1, \cdots, a_n, \cdots)$ is such that Definition 1 holds. Then $\left\lfloor\frac{1}{x}\right\rfloor = a_1-1$. Hence by definition, we have $\begin{equation*} T(x) = (a_1-1)a_1 \cdot x-(a_1-1) = (a_2, \cdots, a_n, \cdots) \end{equation*}$ This shows that the definition of $T$ in $\eqref{20170620-3}$ make it exactly a shift map in the sense of $\eqref{20170620-2}$.

Theorem 3. There is one and exactly one Lüroth expansion for each $x \in (0, 1]$.

Proof. Let’s prove the uniqueness first. Suppose $(a_1, \cdots, a_n, \cdots)$ is a Lüroth expansion for $x$. Then Definition 1 holds. In the remark above, we see that $x = \frac{1}{a_1}+\frac{1}{a_1(a_1-1)}T(x).$ Therefore, we have $\frac{1}{a_1} < x \le \frac{1}{a_1}+\frac{1}{a_1(a_1-1)} = \frac{1}{a_1-1}.$ Since $a_1$ is an integer, we must have $a_1 = \left\lfloor\frac{1}{x}\right\rfloor+1$. Recursively, we can determine $a_2$ using $T(x)$. In general, we have $\begin{equation} a_n = \left\lfloor\frac{1}{T^{n-1}(x)}\right\rfloor+1 \ge 2. \label{20170620-4} \end{equation}$

This proves the uniqueness.

For the existence of such Lüroth expansion for $x$, we set $a_n$ by $\eqref{20170620-4}$. We need to verify Definition 1. But by the definition of $T$ as in $\eqref{20170620-3}$, $x = \frac{1}{\left\lfloor\frac{1}{x}\right\rfloor+1} + \frac{1}{\left(\left\lfloor\frac{1}{x}\right\rfloor+1\right)\left\lfloor\frac{1}{x}\right\rfloor}T(x) = \frac{1}{a_1}+\frac{1}{a_1(a_1-1)}T(x).$ Recursively, we get $\begin{equation*} x = \frac{1}{a_1} + \frac{1}{a_1(a_1-1)a_2} + \cdots + \frac{1}{a_1(a_1-1)\cdots a_{n-1}(a_{n-1}-1)a_n}T^n(x) \end{equation*}$ Using the facts that $0 < T^n(x) \le 1$ and $a_n \ge 2$ for all n, $\begin{equation*} \lim_{n \to \infty} \left|x - \left(\frac{1}{a_1} + \frac{1}{a_1(a_1-1)a_2} + \cdots + \frac{1}{a_1(a_1-1)\cdots a_{n-1}}\right)\right| = 0. \end{equation*}$ This verifies Definition 1. $\square$

Next, we are going into an important property of the Lüroth shift $T$: $T$ is ergodic with respect to the usual Haar measure $\mu$ on $(0, 1]$. Instead of proving it, we are going to explore some interesting properties from this fact together with Birkhoff’s Ergodic Theorem.

Theorem 4 (Birkhoff’s Ergodic Theorem). Let $f: (0, 1] \to \R$ be a measurable function. Then for almost every $x \in (0, 1]$, $\begin{equation} \lim_{n \to \infty} \frac{1}{n}\sum_{i=0}^{n-1} f \circ T^i(x) = \int f \, d\mu. \label{20170620-5} \end{equation}$

Example 1. Let $A$ be some measurable set in $(0, 1]$, and define $\begin{equation*} f(x) = \left\{\begin{array}{ll} 1, & x \in A \\ 0, & x \not\in A \end{array}\right.. \end{equation*}$ Then Theorem 4 is amount to say that “time-averages” is the same as “space-averages” almost everywhere for the Lüroth shift $T$. Indeed, this is a general property for all ergodic maps.

Example 2. Let $k$ be an integer such that $k \ge 2$. Define $\begin{equation*} f(x) = \left\{\begin{array}{ll} 1, & a_1(x) = k \\ 0, & \text{otherwise} \end{array}\right., \end{equation*}$ where $a_1(x)$ is the first term of the Lüroth expansion of $x$. The left hand side of $\eqref{20170620-5}$ calculate the average appearance of $k$, and the right hand side is equal to $\frac{1}{k(k-1)}$ since $f(x)=1$ if and only if $x \in (\frac{1}{k-1}, \frac{1}{k}]$. Therefore, the average appearance of $k$ is $\frac{1}{k(k-1)}$, for $k \ge 2$.

Example 3. Define $f(x) = \ln\left(\left\lfloor\frac{1}{x}\right\rfloor+1\right).$ Suppose $x = (a_1, \cdots, a_n, \cdots)$ is a nice point in $(0, 1]$. By $\eqref{20170620-4}$, the left hand side of $\eqref{20170620-5}$ becomes $\lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^{n} \ln a_i.$ For the right hand side, $\begin{equation*} \begin{split} \int_0^1 f(x) \, dx &= \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}} \ln(n+1) \, dx \\ &= \sum_{n=1}^\infty \frac{\ln(n+1)}{n(n+1)} \approx 1.22638 \end{split} \end{equation*}$ Therefore, $\lim_{n \to \infty} \sqrt[n]{\prod_{i=1}^n a_i} \approx e^{1.22638}.$