Continued Fractions, Part 7: Redefining Equivalence Transformation

This is the seventh article in the continued fractions series, following Part 6.

Equivalence Transformation of Finite Continued Fractions

Let $t_n\in\C^\times$ be an arbitrary constant for all $n\in\N$. Define $\tau:\cC\to\cC$ by $\tau(((a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1})))=((t_0a_0,t_1t_0b_0),(t_1a_1,t_2t_1b_1),\ldots,(t_{n-1}a_{n-1},t_nt_{n-1}b_{n-1})).$ We call this the tau transformation of finite continued fractions with coefficients $t_0,t_1,\ldots,t_n$.

Proposition 7.1 For all $\gamma\in\cC$, we have $[\tau(\gamma)]=t_0[\gamma].$

Proof. Let us write $\gamma=((a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1}))$ and prove by induction on $n$. When $n=0$, we have $[\tau(())]=[()]=\infty=t_0\infty=t_0[()].$ When $n>0$, we have \begin{aligned} &\phantom{=}[(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})] \\ &=m_{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})]) \\ &=m_{t_0a_0,t_1t_0b_0}(t_1[(a_1,b_1),\ldots,(a_{n-1},b_{n-1})]). \end{aligned} If $b_0\ne0$, we have \begin{aligned} m_{t_0a_0,t_1t_0b_0}(t_1[(a_1,b_1),\ldots,(a_{n-1},b_{n-1})])&=t_0a_0+\frac{t_1t_0b_0}{t_1[(a_1,b_1),\ldots,(a_{n-1},b_{n-1})]} \\ &=t_0\left(a_0+\frac{b_0}{[(a_1,b_1),\ldots,(a_{n-1},b_{n-1})]}\right) \\ &=t_0[(a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1})]. \end{aligned} If $b_0=0$, we have $m_{t_0a_0,t_1t_0b_0}(t_1[(a_1,b_1),\ldots,(a_{n-1},b_{n-1})])=t_0a_0=t_0[(a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1})].$

$\square$

Because of this proposition, when $t_0=1$, $\tau$ is called the equivalence transformation of finite continued fractions.

Equivalence Transformation of Infinite Continued Fractions

Define $\tau:\cC_\infty\to\cC_\infty$ by $\tau(((a_0,b_0),(a_1,b_1),(a_2,b_2),\ldots))=((t_0a_0,t_1t_0b_0),(t_1a_1,t_2t_1b_1),(t_2a_2,t_3t_2b_2),\ldots).$ We call this the tau transformation of infinite continued fractions with coefficients $t_0,t_1,t_2,\ldots$.

Proposition 7.2 For all $\gamma\in\cC_\infty$, $\tau(\gamma)$ has a value in $\whC$ if and only if $\gamma$ has a value in $\whC$, and when they have, the values have the relation $[\tau(\gamma)]=t_0[\gamma].$

Proof. Define $\kappa=(c_n)_{n\in\N}$ by $c_n=[(a_0,b_0),(a_1,b_1),\ldots,(a_{n-1}, b_{n-1})], \quad c_0=[]$ and define $\kappa'=(c'_n)_{n\in\N}$ by $c'_n=[(t_0a_0,t_1t_0b_0),(t_1a_1,t_2t_1b_1),\ldots,(t_{n-1}a_{n-1}, t_nt_{n-1}b_{n-1})], \quad c'_0=[].$ By Proposition 7.1, we have $c'_n=t_0c_n$ and therefore $[\tau(\gamma)]=\lim_{n\to\infty}c'_n=t_0\lim_{n\to\infty}c_n=t_0[\gamma].$

$\square$

Because of this proposition, when $t_0=1$, $\tau$ is called the equivalence transformation of infinite continued fractions.

Apparently, for all $\gamma\in\cC$, we have $[\tau(\iota(\gamma))]=[\iota(\gamma)]=[\gamma]=[\tau(\gamma)]=[\iota(\tau(\gamma))].$

Simple Continued Fractions

A finite continued fraction $((a_0,b_0),(a_1,b_1),\ldots,(a_{n-1},b_{n-1}))\in\cC$ is called simple if and only if $b_k=1$ for all $k\in\{0,1,\ldots,n-1\}$.

Proposition 7.3. For all non-degenerate $\gamma\in\cC$, there exists an equivalence transformation $\tau:\cC\to\cC$ 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, $t_1=\frac{1}{b_0}, \quad t_2=\frac{b_0}{b_1}, \quad t_3=\frac{b_1}{b_2b_0}, \quad t_4=\frac{b_2b_0}{b_3b_1}.$

$\square$

An infinite continued fraction $((a_0,b_0),(a_1,b_1),(a_2,b_2),\ldots)\in\cC_\infty$ is called simple if and only if $b_n=1$ for all $n\in\N$.

Proposition 7.4. For all non-degenerate $\gamma\in\cC_\infty$, there exists an equivalence transformation $\tau:\cC_\infty\to\cC_\infty$ such that $\tau(\gamma)$ is simple.

Proof. Define $\tau$ by $\tau(((a_0,b_0),(a_1,b_1),(a_2,b_2),\ldots))=((t_0a_0,t_1t_0b_0),(t_1a_1,t_2t_1b_1),(t_2a_2,t_3t_2b_2),\ldots)$ where $t_n\;(n\in\N)$ is define by $t_n=\frac{b_{n-2}b_{n-4}b_{n-6}\cdots}{b_{n-1}b_{n-3}b_{n-5}\cdots}$ like in Proposition 7.3 (in particular, $t_0=1$). Apparently, this $\tau$ satisfies the condition.

$\square$

Another Special Case

Let $\gamma=((a_0,b_0),(a_1,b_1),(a_2,b_2),\ldots)\in\cC_\infty$ where $a_n\in\C^\times$ for all $n\in\N$. Define $\tau:\cC\to\cC$ be the tau transformation with coefficients $\frac{a}{a_0},\frac{a}{a_1},\frac{a}{a_2},\ldots$ where $a\in\C^\times$ is an arbitrary constant. Then, we have $\tau(\gamma)=\left(\left(a,\frac{a^2b_0}{a_1a_0}\right),\left(a,\frac{a^2b_1}{a_2a_1}\right),\left(a,\frac{a^2b_2}{a_3a_2}\right),\ldots\right).$ This form, where all the so-called partial denominators are same, is sometimes useful for discussions about convergence of $\gamma$.