# Continued Fractions, Attempt 3, Part 5: Special Types of Continued Fractions

## 1. Simple Continued Fractions

A finite continued fraction $((a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1}))$ is called simple if and only if $b_k=1$ for any $k\in\{0,1,\ldots,n-1\}$. Similarly, an infinite continued fraction $((a_0,b_0),(a_1,b_1),(a_2,b_2),\ldots)$ is called simple if and only if $b_k=1$ for any $k\in\N$.

Theorem 5.1. For any regular $\gamma\in\cC$, there exists an equivalence transformation $\tau$ such that $\tau(\gamma)$ is simple.

Proof. Let us write $\gamma=((a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1}))$ and $\tau(\gamma)=((t_0a_0,t_1t_0b_0),(t_1a_1,t_2t_1b_1),\ldots,(t_{n-1}a_{n-1},t_nt_{n-1}b_{n-1})), \quad t_0=1.$ By solving the equation $t_{k+1}t_kb_k=1 \quad (k\in\{0,1,\ldots,n-1\}),$ we obtain $t_k=\frac{b_{k-2}b_{k-4}b_{k-6}\cdots}{b_{k-1}b_{k-3}b_{k-5}\cdots}$ where the numerator is a product of $\lfloor k/2\rfloor$ numbers and the denominator is a product of $\lfloor(k+1)/2\rfloor$ numbers. For example, $t_1=\frac{1}{b_0}, \quad t_2=\frac{b_0}{b_1}, \quad t_3=\frac{b_1}{b_2b_0}, \quad t_4=\frac{b_2b_0}{b_3b_1}.$

Theorem 5.2. For any regular $\gamma\in\cC_\infty$, there exists an equivalence transformation $\tau$ such that $\tau(\gamma)$ is simple.

Proof. Similar to Theorem 5.1.

## 2. Simple-denominator Continued Fractions

We call a finite continued fraction $((a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1}))$ simple-denominator if and only if $a_k=1$ for any $k\in\{0,1,\ldots,n-1\}$. Similarly, we call an infinite continued fraction $((a_0,b_0),(a_1,b_1),(a_2,b_2),\ldots)$ simple-denominator if and only if $a_k=1$ for any $k\in\N$. Strictly speaking, we should call them simple-partial-denominator, but we choose brevity here. Also, we may call simple continued fractions simple-numerator.

Let $\gamma=((a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1}))\in\cC$ where $a_k\in\C^\times$ for any $k\in\{0,1,\ldots,n-1\}$. Let $a_l\in\C^\times$ be an arbitrary constant for any $l\in\{n,n+1,n+2,\ldots\}$. Define $\tau:\cC\to\cC$ to be a tau transformation with $t_k=a_k^{-1}$ for any $k\in\N$. Then, we have $\tau(\gamma)=\left(\left(1,\frac{b_0}{a_1a_0}\right),\left(1,\frac{b_1}{a_2a_1}\right),\ldots,\left(1,\frac{b_{n-1}}{a_na_{n-1}}\right)\right),$ which is simple-denominator.

Let $\gamma=((a_0,b_0),(a_1,b_1),(a_2,b_2),\ldots)\in\cC_\infty$ where $a_k\in\C^\times$ for any $k\in\N$. Define $\tau:\cC_\infty\to\cC_\infty$ to be a tau transformation with $t_k=a_k^{-1}$ for any $k\in\N$. Then, we have $\tau(\gamma)=\left(\left(1,\frac{b_0}{a_1a_0}\right),\left(1,\frac{b_1}{a_2a_1}\right),\left(1,\frac{b_2}{a_3a_2}\right),\ldots\right),$ which is simple-denominator.