August 20, 2020

Continued Fractions, Part 5: Redefining Finite Continued Fractions

In this series of articles so far, we have excluded continued fractions where some of so-called partial numerators ($b_k$ in this series) are zero. However, the author has realized that there might be plenty of classical results which include such continued fractions at least as minor cases. In fact, he carelessly described one of such results in Part 3. Therefore, he decided to devise a unified definition of continued fractions with and without zeros as some partial numerators.

Möbius Transformation

Let $\whC$ denote $\C\cup\{\infty\}$ and $\C^\times$ denote $\C\setminus\{0\}$. We include $0$ in $\N$.

A Möbius transformation is a rational function $f:\whC\to\whC$ defined by \[ f(z)=\frac{az+b}{cz+d} \] where $a,b,c,d\in\C$ and \[ ad-bc\ne0. \] In particular, when $c\ne0$, we have \[ f\left(-\frac{d}{c}\right)=\infty \] and \[ f(\infty)=\frac{a}{c}. \] When $c=0$, we have \[ f(\infty)=\infty. \]

Degenerate Möbius Transformation

Let us consider the degenerate version of the Möbius transformation, where \[ ad-bc=0. \] If $c\ne0$ and $cz+d\in\C^\times$, we have \[ \frac{az+b}{cz+d}=\frac{a(cz+d)-(ad-bc)}{c(cz+d)}=\frac{a}{c}. \] Therefore, when $c\ne0$, we define \[ f(z)=\frac{a}{c} \] for all $z\in\whC$. Similarly, when $d\ne0$, we define \[ f(z)=\frac{b}{d} \] for all $z\in\whC$. Of course, these two definitions coincide when $c\ne0$ and $d\ne0$ because $ad-bc=0$.

When $c=d=0$ and $(a\ne0 \vee b\ne0)$, we define \[ f(z)=\infty \] for all $z\in\whC$.

When $a=b=c=d=0$, we consider $f(z)$ cannot be defined.

Finite Continued Fractions

Let $m_{a,b}(z)$ denote a possibly-degenerate Möbius transformation \[ m_{a,b}(z)=a+\frac{b}{z} \] where $(a,b)\in\C^2$. In fact, $m_{a,b}(z)$ is degenerate if and only if $b=0$, where it becomes a constant function \[ m_{a,0}(z)=a. \]

For all $n\in\N$, let $\cC_n$ denote the set of all the length-$n$ sequences of elements of $\C^2$, that is, \[ \cC_n=\{((a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1}))\,|\, (a_k,b_k)\in\C^2\;\text{for all $k\in\{0,1,\ldots,n-1\}$}\}. \] We consider $\cC_0$ has just one element $()$, which is the empty sequence of elements of $\C^2$. Define $\cC$ by \[ \cC=\bigcup_{n=0}^\infty\cC_n. \] We call elements of $\cC$ finite continued fractions.

Define $[\,]:\cC\to\whC$ recursively by \[ \begin{aligned} [((a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1}))]&=m_{a_0,b_0}([((a_1,b_1),\ldots,(a_{n-1},b_{n-1}))]), \\ [()]&=\infty. \end{aligned} \] We call $[\gamma]$ the value of $\gamma\in\cC$. For simplicity, we write \[ \begin{aligned} [((a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1}))]&=[(a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1})], \\ [()]&=[]. \end{aligned} \]

Degenerateness

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

Proposition 5.1. If $\gamma=((a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1}))$ is degenerate, say $b_k=0$, then \[ [\gamma]=[(a_0,b_0),(a_1,b_1),\ldots,(a_k,b_k)]. \]

Proof. When $k=n-1$, it is trivial. Otherwise, we have \[ m_{a_k,b_k}([(a_{k+1},b_{k+1}),\ldots,(a_{n-1},b_{n-1})])=a_k=m_{a_k,b_k}([]) \] and therefore \[ \begin{aligned} [\gamma]&=(m_{a_0,b_0}\circ\cdots\circ m_{a_k,b_k})([(a_{k+1},b_{k+1}),\ldots,(a_{n-1},b_{n-1})]) \\ &=(m_{a_0,b_0}\circ\cdots\circ m_{a_k,b_k})([]) \\ &=[(a_0,b_0),(a_1,b_1),\ldots,(a_k,b_k)]. \end{aligned} \]

Revisions

Kenichi Kondo © 2001-2020