# Continued Fractions, Part 2: Equivalence Transformation

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

## Introduction

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 $t_n\in\C^\times$ be an arbitrary constant for all $n\in\N$. The map $(a_0,b_0,a_1,b_1,\ldots,a_{n-1})\mapsto(t_0a_0,t_1t_0b_0,t_1a_1,t_2t_1b_1,\ldots,t_{n-1}a_{n-1})$ between tuples is called the equivalence transformation because of the following theorem.

Theorem 2.1. For all $n\in\N$, we have $[t_0a_0,\underline{t_1t_0b_0},t_1a_1,\underline{t_2t_1b_1},\ldots,t_{n-1}a_{n-1}]=t_0[a_0,\underline{b_0},a_1,\underline{b_1},\ldots,a_{n-1}]$

Remark. Usually the equivalence transformation refers to cases where $t_0=1$.

Proof. By induction on $n$. When $n=0$, we have $[\infty]=\infty=t_0\infty=t_0[\infty]$. When $n>0$, we have \begin{aligned} [t_0a_0,\underline{t_1t_0b_0},t_1a_1,\underline{t_2t_1b_1},\ldots,t_{n-1}a_{n-1}]&=t_0a_0+\frac{t_1t_0b_0}{[t_1a_1,\underline{t_2t_1b_1},\ldots,t_{n-1}a_{n-1}]} \\ &=t_0a_0+\frac{t_1t_0b_0}{t_1[a_1,\underline{b_1},\ldots,a_{n-1}]} \\ &=t_0\left(a_0+\frac{b_0}{[a_1,\underline{b_1},\ldots,a_{n-1}]}\right) \\ &=t_0[a_0,\underline{b_0},a_1,\underline{b_1},\ldots,a_{n-1}]. \end{aligned}

For infinite continued fractions, the map $\{a_0,b_0,a_1,b_1,a_2,b_2,\ldots\}\mapsto\{t_0a_0,t_1t_0b_0,t_1a_1,t_2t_1b_1,t_2a_2,t_3t_2b_2,\ldots\}$ between sequences is called the equivalence transformation.

Corollary 2.2. $[t_0a_0,\underline{t_1t_0b_0},t_1a_1,\underline{t_2t_1b_1},\ldots]=t_0[a_0,\underline{b_0},a_1,\underline{b_1},\ldots].$

Proof. Trivial.

## Special Case: Simple Continued Fractions

Any continued fraction can be transformed into a simple continued fraction using the equivalence transformation. By setting $t_0=1$ and solving $t_{n+1}t_nb_n=1$, we obtain $t_n=\frac{b_{n-2}b_{n-4}b_{n-6}\cdots}{b_{n-1}b_{n-3}b_{n-5}\cdots}$ where the both products terminate with $b_0$ or $b_1$ depending on whether $n$ 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 $a_n\ne 0$ for all $n\in\N$, $a_n$ can be transformed to $1$. This is easily achieved by setting $t_n=1/a_n$.

## References

• [1] S. Hitotsumatsu, Introduction to Special Functions, Morikita, 1999.

## Revisions

• July 26, 2020
• Introduce the bracket notation and simplify descriptions.
• May 6, 2020
• Initial publication.

Kenichi Kondo © 2001-2021