July 18, 2021

Continued Fractions, Attempt 3, Part 7: An Example of Infinite Continued Fractions

1. An Example of Infinite Continued Fractions

Let $a, b$ be constants in $\C$ and $\gamma$ an infinite continued fraction $((a,b),(a,b),(a,b),\ldots)$.

Theorem 7.1. Define $S_1,S_2\subset\C^2$ by \[ \begin{aligned} S_1&=\left\{(x,y)\in\C^2\;\Bigg|\;x\ne0 \text{ and } \frac{y}{x^2}\in\C\setminus\left\{h\in\R\;\bigg|\;h<-\frac{1}{4}\right\}\right\}, \\ S_2&=\left\{(x,y)\in\C^2\;\big|\;y=0\right\}. \end{aligned} \] Then, $\gamma$ has value if $(a,b)\in S_1\cup S_2$ and the value is \[ \lang\gamma\rang=\begin{cases} a\left(\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{b}{a^2}}\right) & \text{if } (a,b)\in S_1, \\ a & \text{if } (a,b)\in S_2, \end{cases} \] where the argument of the square root lies in $(-\pi/2, \pi/2]$. Note that the two values coincide when $(a,b)\in S_1\cap S_2$.

Lemma 7.2. Let $\alpha, \beta$ be the two roots of $z^2-az-b$. Define $S_3, S_4\subset \C^2$ by \[ \begin{aligned} S_3&=\left\{(x,y)\in\C^2\;\Bigg|\;x\ne0 \text{ and } \frac{y}{x^2}\in\C\setminus\left\{h\in\R\;\bigg|\;h\le-\frac{1}{4}\right\}\right\}, \\ S_4&=\left\{(x,y)\in\C^2\;\big|\;x\ne0 \text{ and } \frac{y}{x^2}=-\frac{1}{4}\right\}. \end{aligned} \] Then, we have $S_1=S_3\sqcup S_4$. When $(a,b)\in S_3$, $\alpha\ne\beta$. When $(a,b)\in S_4$, $\alpha=\beta=a/2$.

Proof of Lemma 7.2. Trivial.

Proof of Theorem 7.1. When $(a,b)\in S_2$, $\gamma$ is degenerate and $\lang\gamma\rang=a$. Hereafter, we assume $(a,b)\in S_1\setminus S_2$, which means $\gamma$ is regular.

Let $c_n$ the $n$-th convergent of $\gamma$. We have $c_n=p_n/q_n$, where $p_n$ and $q_n$ are defined by the recurrence relation \[ \begin{aligned} p_{n+2}&=ap_{n+1}+bp_n, & p_1&=a, & p_0&=1, \\ q_{n+2}&=aq_{n+1}+bq_n, & q_1&=1, & q_0&=0. \end{aligned} \] When $(a,b)\in S_3\setminus S_2$, we have $\alpha \ne \beta$ and \[ \begin{aligned} p_{n+2}-\alpha p_{n+1}&=\beta(p_{n+1}-\alpha p_n)=\beta^{n+1}(p_1-\alpha p_0), \\ p_{n+2}-\beta p_{n+1} &=\alpha(p_{n+1}-\beta p_n)=\alpha^{n+1}(p_1-\beta p_0). \end{aligned} \] Eliminating $p_{n+2}$, we get \[ \begin{aligned} p_{n+1}&=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}p_1-\frac{\alpha^{n+1}\beta-\alpha\beta^{n+1}}{\alpha-\beta}p_0 \\ &=\frac{p_1-\beta p_0}{\alpha-\beta}\alpha^{n+1}-\frac{p_1-\alpha p_0}{\alpha-\beta}\beta^{n+1} \\ &=\frac{\alpha^{n+2}-\beta^{n+2}}{\alpha-\beta}. \end{aligned} \] Similarly, we have \[ q_{n+1}=\frac{q_1-\beta q_0}{\alpha-\beta}\alpha^{n+1}-\frac{q_1-\alpha q_0}{\alpha-\beta}\beta^{n+1}=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}. \] Therefore, by interpreting $0^0=1$, we obtain \[ c_n=\frac{p_n}{q_n}=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha^n-\beta^n} \] for any $n\in\N$. We can explicitly write $\alpha$ and $\beta$, without loss of generality, as \[ \alpha=a\left(\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{b}{a^2}}\right), \quad \beta=a\left(\frac{1}{2}-\sqrt{\frac{1}{4}+\frac{b}{a^2}}\right). \] Noticing that $\Re\sqrt{1/4+b/a^2} > 0$, we have $|\alpha| > |\beta|$ and thus \[ \lim_{n\to\infty}c_n=\lim_{n\to\infty}\frac{1-(\beta/\alpha)^{n+1}}{1-(\beta/\alpha)^n}\alpha=\alpha. \]

When $(a,b)\in S_4\setminus S_2=S_4$, we have $\alpha=\beta=a/2$ and \[ p_{n+2}-\alpha p_{n+1}=\alpha^{n+1}(p_1-\alpha p_0) \Longleftrightarrow \frac{p_{n+2}}{\alpha^{n+2}}-\frac{p_{n+1}}{\alpha^{n+1}}=\frac{p_1}{\alpha}-p_0=1, \] and thus $p_n=(n+1)\alpha^n$. Similarly, we have \[ \frac{q_{n+2}}{\alpha^{n+2}}-\frac{q_{n+1}}{\alpha^{n+1}}=\frac{q_1}{\alpha}-q_0=\frac{2}{a} \] and thus $q_n=2n\alpha^n/a$. Finally, we obtain \[ \lim_{n\to\infty}c_n=\lim_{n\to\infty}\frac{a(n+1)}{2n}=\frac{a}{2}=a\left(\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{b}{a^2}}\right). \]

Corollary 7.3. $\gamma$ has value if and only if $(a,b)\in S_1\cap S_2$.

Proof. Assume $b\ne0$. When $a=0$, we have \[ c_n=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha^n-\beta^n}=\frac{1-(-1)^{n+1}}{1-(-1)^n}\sqrt{b}. \] Therefore, $(c_n)_{n\in\N}=(\infty,0,\infty,0,\ldots)$ does not converge in $\whC$ and $\gamma$ does not have value.

When $a\ne0$ and $b/a^2\in\{h\in\R\;|\;h<-1/4\}$, we have $\alpha\ne\beta$ and $\beta=\overline{\alpha}$. We can write $\alpha$ in polar form as $\alpha=r(\cos\theta+i\sin\theta)$, and assume $0<\theta<\pi/2$ without loss of generality. Then, we have \[ c_n=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha^n-\beta^n}=\frac{r\sin (n+1)\theta}{\sin n\theta}. \] If $\sin n\theta\ne0$, $c_n=r(\cos \theta+\sin\theta\cot n\theta)$. If $\sin n\theta=0$, $c_n=\infty$ and also $r(\cos \theta+\sin\theta\cot n\theta)=\infty$ because $\sin \theta\ne0$ and $\cot n\theta=\infty$. Therefore, for any $n\in\N$, we have \[ c_n=r(\cos \theta+\sin\theta\cot n\theta). \] This implies that $(c_n)_{n\in\N}$ does not converge in $\whC$ and $\gamma$ does not have value.

Kenichi Kondo © 2001-2021