# Continued Fractions, Attempt 3, Part 3: Regular Continued Fractions

## 1. Regular Finite Continued Fractions

Let $\gamma$ be a finite continued fraction $((a_0, b_0), (a_1, b_1), \ldots, (a_{n-1}, b_{n-1}))$. We say $\gamma$ is degenerate if and only if there exists $k\in\{0,1,\ldots,n-1\}$ such that $b_k=0$. We say $\gamma$ is regular if and only if it is not degenerate. The empty continued fraction $()$ is regular.

We define the regularization of $\gamma$, denoted by $\overline\gamma$, as follows. When $\gamma$ is regular, $\overline\gamma=\gamma$. When $\gamma$ is degenerate, we have $\exists k_0 \forall l:(b_{k_0}=0)\wedge(0\le l<k_0\rightarrow b_l\ne0)$ and define $\overline\gamma$ by $\overline\gamma=((a_0, b_0), (a_1, b_1), \ldots, (a_{k_0}, 1)).$ Then, for any $\gamma\in\cC$, $\overline\gamma$ is regular and $\lang\overline\gamma\rang=\lang\gamma\rang.$

The implication is that, as far as values are of interest, we need to consider regular continued fractions only.

## 2. Regular Infinite Continued Fractions

Let $\gamma$ be an infinite continued fraction $((a_0, b_0), (a_1, b_1), (a_2, b_2), \ldots)$. We say $\gamma$ is degenerate if and only if there exists $k\in\N$ such that $b_k=0$. We say $\gamma$ is regular if and only if it is not degenerate.

We define the regularization of $\gamma$, denoted by $\overline\gamma$, as follows. When $\gamma$ is regular, $\overline\gamma=\gamma$. When $\gamma$ is degenerate, we have $\exists k_0 \forall l:(b_{k_0}=0)\wedge(0\le l<k_0\rightarrow b_l\ne0)$ and define $\overline\gamma$ by $\overline\gamma=((a_0,b_0),(a_1,b_1),\ldots,(a_{k_0},1)).$ Then, for any $\gamma\in\cC_\infty$,

1. $\overline\gamma\in\cC\cup\cC_\infty$ is regular ,
2. $\overline\gamma$ has value if and only if $\gamma$ has value, and
3. when they have values, $\lang\overline\gamma\rang=\lang\gamma\rang$.

The implication is that, as far as values are of interest, only regular infinite continued fractions are “truly” infinite.

Kenichi Kondo © 2001-2021