April 29, 2020

Continued Fractions, Part 1: Definitions

Continued fractions seem to be somewhat a minor topic in modern mathematics, and the author thinks it’s one of reasons why he has always been feeling difficulty in finding literature which gives minimal and clean description of them. Therefore, he decided to create a starting point for that purpose.

Important notice: content of this article will be totally taken over by those of Part 5 and following articles.

Introduction

A continued fraction is, conceptually, a fraction of the form a0+b0a1+b1a2+b2, a_0 + \cfrac{b_0}{a_1 + \cfrac{b_1}{a_2 + \cfrac{b_2}{\cdots}}}, where ana_n and bnb_n are taken from a certain algebra depending on the context. We are going to give a precise definition and properties when this algebra is C\C.

Finite Continued Fractions

Let C^\whC denote C{}\C \cup \{\infty\} and C×\C^\times denote C{0}\C \setminus \{0\}.

A finite continued fraction is a fraction represented by the bracket notation [][\,], which is defined by [a0,b0,a1,b1,,an2,bn2,an1]=a0+b0[a1,b1,,an2,bn2,an1],[an1]=an1, [a_0,\underline{b_0},a_1,\underline{b_1},\ldots,a_{n-2},\underline{b_{n-2}},a_{n-1}]=a_0+\frac{b_0}{[a_1,\underline{b_1},\ldots,a_{n-2},\underline{b_{n-2}},a_{n-1}]}, \quad [a_{n-1}]=a_{n-1}, where akC^a_k\in\whC and bkC×b_k\in\C^\times for all k{0,1,,n1}k\in\{0,1,\ldots,n-1\}. For example, [a0,b0,a1,b1,a2,b2,a3]=a0+b0a1+b1a2+b2a3 [a_0,\underline{b_0},a_1,\underline{b_1},a_2,\underline{b_2},a_3]=a_0+\cfrac{b_0}{a_1+\cfrac{b_1}{a_2+\cfrac{b_2}{a_3}}} is a finite continued fraction. Note that we regard [a0]=a0[a_0]=a_0 as a valid finite continued fraction.

Infinite Continued Fractions

Let anC^a_n\in\whC and bnC×b_n\in\C^\times for all nNn\in\N. Note that we include 00 in N\N.

An infinite continued fraction is a sequence κ={cn}nN\kappa=\{c_n\}_{n\in\N} where cnc_n is defined by cn=[a0,b0,a1,b1,,an1],c0=. c_n=[a_0,\underline{b_0},a_1,\underline{b_1},\ldots,a_{n-1}], \quad c_0=\infty. Each term cnc_n is called a convergent. For simplicity, we denote infinite continued fractions like κ=[a0,b0,a1,b1,a2,b2,]=a0+b0a1+b1a2+b2. \kappa=[a_0,\underline{b_0},a_1,\underline{b_1},a_2,\underline{b_2},\ldots]=a_0+\cfrac{b_0}{a_1+\cfrac{b_1}{a_2+\cfrac{b_2}{\cdots}}}.

Remark. One of the reasons why we define c0c_0 to be \infty is that we can always write cn=[a0,b0,a1,b1,,an1,bn1,] c_n=[a_0,\underline{b_0},a_1,\underline{b_1},\ldots,a_{n-1},\underline{b_{n-1}},\infty] and thus c0=[]=c_0=[\infty]=\infty. This will become more natural in a later section where we regard continued fractions as repetitions of Möbius transformations.

Simple Continued Fractions

A continued fraction, finite or infinite, is called simple when all the underlined elements in the bracket notation are 1. And such elements may be omitted from the notation. For example, [a0,a1,a2,a3]=[a0,1,a1,1,a2,1,a3]=a0+1a1+1a2+1a3 [a_0,a_1,a_2,a_3]=[a_0,\underline{1},a_1,\underline{1},a_2,\underline{1},a_3]=a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3}}} and [a0,a1,a2,]=[a0,1,a1,1,a2,1,]=a0+1a1+1a2+1 [a_0,a_1,a_2,\ldots]=[a_0,\underline{1},a_1,\underline{1},a_2,\underline{1},\ldots]=a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{\cdots}}} are simple continued fractions.

Remark. A widely used notation is [a0;a1,a2,,an1][a_0; a_1, a_2, \ldots, a_{n-1}], which is basically same as [a0,a1,a2,,an1][a_0, a_1, a_2, \ldots, a_{n-1}] in this article. The notation [a0,b0,a1,b1,,an1][a_0,\underline{b_0},a_1,\underline{b_1},\ldots,a_{n-1}] is uncommon.

Non-degenerateness Condition

Let us consider a finite continued fraction [a0,b0,a1,b1,,an2,bn2,an1][a_0,\underline{b_0},a_1,\underline{b_1},\ldots,a_{n-2},\underline{b_{n-2}},a_{n-1}]. If there exists l{0,1,,n1}l\in\{0,1,\ldots,n-1\} such that al=a_l=\infty, then we have [a0,b0,a1,b1,,al1]=[a0,b0,a1,b1,,al1,bl1,al]=[a0,b0,a1,b1,,al1,bl1,al,bl,al+1]==[a0,b0,a1,b1,,al1,bl1,al,bl,al+1,,bn2,an1], \begin{aligned} [a_0,\underline{b_0},a_1,\underline{b_1},\ldots,a_{l-1}]&=[a_0,\underline{b_0},a_1,\underline{b_1},\ldots,a_{l-1},\underline{b_{l-1}},a_l] \\ &=[a_0,\underline{b_0},a_1,\underline{b_1},\ldots,a_{l-1},\underline{b_{l-1}},a_l,\underline{b_l},a_{l+1}] \\ &=\cdots \\ &=[a_0,\underline{b_0},a_1,\underline{b_1},\ldots,a_{l-1},\underline{b_{l-1}},a_l,\underline{b_l},a_{l+1},\ldots,\underline{b_{n-2}},a_{n-1}], \end{aligned} that is, al+1,al+2,,an1a_{l+1},a_{l+2},\ldots,a_{n-1} and bl1,bl,,bn2b_{l-1},b_l,\ldots,b_{n-2} does not affect the value at all. Similarly, for an infinite continued fraction with al=a_l=\infty, we have cl=cl+1=cl+2=c_l=c_{l+1}=c_{l+2}=\cdots and it is almost a finite continued fraction. We call these cases degenerate ones.

Thus, hereafter, we assume the non-degenerateness condition, which is of course defined as an a_n\ne\infty for all nNn\in\N.

Möbius Transformations and Convergents

Let MnM_n be (anbn10)GL2(C)\M{a_n}{b_n}{1}{0}\in\text{GL}_2(\C) and mn:C^C^m_n:\whC\rightarrow\whC the associated Möbius transformation to MnM_n, that is, mn(x)=an+bnx. m_n(x)=a_n+\frac{b_n}{x}. Apparently, cn=(m0m1mn1)(). c_n=(m_0\circ m_1\circ\cdots\circ m_{n-1})(\infty).

Theorem 1.1. We have cn=pn/qnc_n=p_n/q_n where pnp_n and qnq_n are defined by the recurrence relation (1.1a)pn+2=an+1pn+1+bnpn,p1=a0,p0=1,(1.1a)(1.1b)qn+2=an+1qn+1+bnqn,q1=1,q0=0.(1.1b) \begin{aligned} \phantom{\text{(1.1a)}}&\hspace{4em} & p_{n+2}&=a_{n+1}p_{n+1}+b_np_n, & p_1&=a_0, & p_0&=1, & \hspace{4em}&\text{(1.1a)} \\ \phantom{\text{(1.1b)}}& & q_{n+2}&=a_{n+1}q_{n+1}+b_nq_n, & q_1&=1, & q_0&=0. & &\text{(1.1b)} \end{aligned}

Proof. Define LnL_n by the recurrence relation Ln+1=LnMn,L0=(1001). L_{n+1}=L_nM_n, \quad L_0=\M{1}{0}{0}{1}. Then m0m1mn1m_0\circ m_1\circ\cdots\circ m_{n-1} is the associated Möbius transformation to LnL_n. Therefore, if we write Ln=(pnrnqnsn)L_n=\M{p_n}{r_n}{q_n}{s_n}, we have cn=pn/qnc_n=p_n/q_n. On the other hand, we have pn+1=anpn+rn,p0=1,qn+1=anqn+sn,q0=0,rn+1=bnpn,r0=0,sn+1=bnqn,s0=1. \begin{aligned} p_{n+1}&=a_np_n+r_n, & p_0&=1, \\ q_{n+1}&=a_nq_n+s_n, & q_0&=0, \\ r_{n+1}&=b_np_n, & r_0&=0, \\ s_{n+1}&=b_nq_n, & s_0&=1. \end{aligned} By eliminating rnr_n and sns_n, we get Ln+1=(pn+1bnpnqn+1bnqn)L_{n+1}=\M{p_{n+1}}{b_np_n}{q_{n+1}}{b_nq_n} and the recurrence relation (1.1a)\text{(1.1a)}, (1.1b)\text{(1.1b)}.

Corollary 1.2. For any nNn\in\N which satisfies cn+1c_{n+1}\ne\infty and cn+2c_{n+2}\ne\infty, we have cn+2cn+1=(1)nbnbn1b0qn+2qn+1. c_{n+2}-c_{n+1}=(-1)^n\frac{b_nb_{n-1}\cdots b_0}{q_{n+2}q_{n+1}}.

Proof. If cn+1c_{n+1}\ne\infty and cn+2c_{n+2}\ne\infty, we have cn+2cn+1=pn+2qn+2pn+1qn+1=pn+2qn+1pn+1qn+2qn+2qn+1=detLn+2bn+1qn+2qn+1 c_{n+2}-c_{n+1}=\frac{p_{n+2}}{q_{n+2}}-\frac{p_{n+1}}{q_{n+1}}=\frac{p_{n+2}q_{n+1}-p_{n+1}q_{n+2}}{q_{n+2}q_{n+1}}=\frac{\det L_{n+2}}{b_{n+1}q_{n+2}q_{n+1}} and detLn+2=detM0detM1detMn+1=(1)nb0b1bn+1. \det L_{n+2}=\det M_0 \det M_1 \cdots \det M_{n+1} = (-1)^n b_0b_1\cdots b_{n+1}.

Remark. Since there is no nNn\in\N for which both cn+1c_{n+1} and cn+2c_{n+2} are \infty, it is tempting to extend the above formula for the cases where either cn+1c_{n+1} or cn+2c_{n+2} is \infty. But the difficulty is we must handle ++\infty, -\infty, and \infty as if they are different elements although they are identical in the view of C^\whC. Therefore, we restrict ourselves to the cases where both are finite.

Appendix

The first few terms of κ={cn}nN\kappa=\{c_n\}_{n\in\N}, written in the form of pn/qnp_n/q_n, are c0=10,c1=a01,c2=a1a0+b0a1,c3=a2a1a0+a2b0+b1a0a2a1+b1,c4=a3a2a1a0+a3a2b0+a3b1a0+b2a1a0+b2b0a3a2a1+a3b1+b2a1,andc5=a4a3a2a1a0+a4a3a2b0+a4a3b1a0+a4b2a1a0+a4b2b0+b3a2a1a0+b3a2b0+b3b1a0a4a3a2a1+a4a3b1+a4b2a1+b3a2a1+b3b1. \begin{aligned} c_0&=\frac{1}{0}, \\ c_1&=\frac{a_0}{1}, \\ c_2&=\frac{a_1a_0+b_0}{a_1}, \\ c_3&=\frac{a_2a_1a_0+a_2b_0+b_1a_0}{a_2a_1+b_1}, \\ c_4&=\frac{a_3a_2a_1a_0+a_3a_2b_0+a_3b_1a_0+b_2a_1a_0+b_2b_0}{a_3a_2a_1+a_3b_1+b_2a_1}, \text{and} \\ c_5&=\frac{a_4a_3a_2a_1a_0+a_4a_3a_2b_0+a_4a_3b_1a_0+a_4b_2a_1a_0+a_4b_2b_0+b_3a_2a_1a_0+b_3a_2b_0+b_3b_1a_0}{a_4a_3a_2a_1+a_4a_3b_1+a_4b_2a_1+b_3a_2a_1+b_3b_1}. \end{aligned}

References

Revisions

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