May 6, 2020

Continued Fractions, Part 2: Equivalence Transformation

This is the second article in the continued fractions series, following Part 1.


It is intuitive to think \[ a_0+\cfrac{b_0}{a_1+\cfrac{b_1}{a_2+\cfrac{b_2}{\cdots}}}=a_0+\cfrac{t_1b_0}{t_1a_1+\cfrac{t_2t_1b_1}{t_2a_2+\cfrac{t_3t_2b_2}{\cdots}}} \] where $t_1, t_2, t_3, \ldots$ are arbitrary constants in $\C^\times$. This is formalized as follows.

Equivalence Transformation

Let $t_k\in\C^\times$ be an arbitrary constant for any $k\in\N$. Define $a_k'$, $b_k'$, and $m_k'$ by \[ a_k'=t_ka_k, \quad b_k'=t_{k+1}t_kb_k, \quad m_k'(x)=a_k'+\frac{b_k'}{x}. \] And define $c_k'$ by \[ c_k'=(m_0'\circ m_1'\circ\cdots\circ m_{k-1}')(\infty), \quad c_0' = \infty. \]

This transformation $\{c_k\}\mapsto\{c_k'\}$ between continued fractions is called equivalence transformation (usually $t_0$ is set to be $1$), because of the following theorem.

Theorem. $c_k'=t_0c_k$ for any $k\in\N$.

Proof. Define $c_{k,l}$ and $c_{k,l}'$ where $(k,l)\in\N^2$ satisfies $0\le l\le k$ by \[ \begin{aligned} c_{k,l}&=(m_l\circ m_{l+1}\circ\cdots\circ m_{k-1})(\infty), & c_{k,k}&=\infty, \\ c_{k,l}'&=(m_l'\circ m_{l+1}'\circ\cdots\circ m_{k-1}')(\infty), & c_{k,k}'&=\infty. \\ \end{aligned} \] We now want to prove $c_{k,l}'=t_kc_{k,l}$ by induction on $l$. When $l=k$, this is trivially true. When $l<k$, assuming $c_{k,l+1}'=t_kc_{k,l+1}$, we have \[ c_{k,l}'=a_l'+\frac{b_l'}{c_{k,l+1}'}=t_la_l+\frac{t_{l+1}t_lb_l}{t_{l+1}c_{k,l+1}}=t_l\left(a_l+\frac{b_l}{c_{k,l+1}}\right)=t_lc_{k,l} \] even when $c_{k,l+1}$ is $0$ or $\infty$. Therefore, $c_{k,l}'=t_kc_{k,l}$ holds and especially $c_k'=c_{k,0}'=t_0c_{k,0}=t_0c_k$.

Simple Continued Fraction

A continued fraction is called simple if $b_k=1$ for all $k\in\N$.

Any continued fraction can be transformed into a simple continued fraction using equivalence transformation. By setting $t_0=1$ and solving $b_k'=1 \iff t_{k+1}t_kb_k=1$, we have \[ t_k=\frac{b_{k-2}b_{k-4}b_{k-6}\cdots}{b_{k-1}b_{k-3}b_{k-5}\cdots} \] where the both products terminate with $b_0$ or $b_1$ depending on whether $k$ is even or odd. For example, \[ t_1=\frac{1}{b_0}, \quad t_2=\frac{b_0}{b_1}, \quad t_3=\frac{b_1}{b_2b_0}, \quad t_4=\frac{b_2b_0}{b_3b_1}. \]

Another Special Case

When $a_k\ne 0$ for all $k\in\N$, $a_k'$ can be made to be $1$. This is easily achieved by setting $t_k=1/a_k$.


Kenichi Kondo © 2001-2020