June 26, 2021

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