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+a1+a2+⋯b2b1b0,
where an and bn 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.
Finite Continued Fractions
Let C denote C∪{∞} and C× denote C∖{0}.
A finite continued fraction is a fraction represented by the bracket notation [], which is defined by
[a0,b0,a1,b1,…,an−2,bn−2,an−1]=a0+[a1,b1,…,an−2,bn−2,an−1]b0,[an−1]=an−1,
where ak∈C and bk∈C× for all k∈{0,1,…,n−1}. For example,
[a0,b0,a1,b1,a2,b2,a3]=a0+a1+a2+a3b2b1b0
is a finite continued fraction. Note that we regard [a0]=a0 as a valid finite continued fraction.
Infinite Continued Fractions
Let an∈C and bn∈C× for all n∈N. Note that we include 0 in N.
An infinite continued fraction is a sequence κ={cn}n∈N where cn is defined by
cn=[a0,b0,a1,b1,…,an−1],c0=∞.
Each term cn is called a convergent. For simplicity, we denote infinite continued fractions like
κ=[a0,b0,a1,b1,a2,b2,…]=a0+a1+a2+⋯b2b1b0.
Remark. One of the reasons why we define c0 to be ∞ is that we can always write
cn=[a0,b0,a1,b1,…,an−1,bn−1,∞]
and thus c0=[∞]=∞. 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+a1+a2+a3111
and
[a0,a1,a2,…]=[a0,1,a1,1,a2,1,…]=a0+a1+a2+⋯111
are simple continued fractions.
Remark. A widely used notation is [a0;a1,a2,…,an−1], which is basically same as [a0,a1,a2,…,an−1] in this article. The notation [a0,b0,a1,b1,…,an−1] is uncommon.
Non-degenerateness Condition
Let us consider a finite continued fraction [a0,b0,a1,b1,…,an−2,bn−2,an−1]. If there exists l∈{0,1,…,n−1} such that al=∞, then we have
[a0,b0,a1,b1,…,al−1]=[a0,b0,a1,b1,…,al−1,bl−1,al]=[a0,b0,a1,b1,…,al−1,bl−1,al,bl,al+1]=⋯=[a0,b0,a1,b1,…,al−1,bl−1,al,bl,al+1,…,bn−2,an−1],
that is, al+1,al+2,…,an−1 and bl−1,bl,…,bn−2 does not affect the value at all. Similarly, for an infinite continued fraction with al=∞, we have cl=cl+1=cl+2=⋯ 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=∞
for all n∈N.
Möbius Transformations and Convergents
Let Mn be (an1bn0)∈GL2(C) and mn:C→C the associated Möbius transformation to Mn, that is,
mn(x)=an+xbn.
Apparently,
cn=(m0∘m1∘⋯∘mn−1)(∞).
Theorem 1.1. We have cn=pn/qn where pn and qn are defined by the recurrence relation
(1.1a)(1.1b)pn+2qn+2=an+1pn+1+bnpn,=an+1qn+1+bnqn,p1q1=a0,=1,p0q0=1,=0.(1.1a)(1.1b)
Proof. Define Ln by the recurrence relation
Ln+1=LnMn,L0=(1001).
Then m0∘m1∘⋯∘mn−1 is the associated Möbius transformation to Ln. Therefore, if we write Ln=(pnqnrnsn), we have cn=pn/qn. On the other hand, we have
pn+1qn+1rn+1sn+1=anpn+rn,=anqn+sn,=bnpn,=bnqn,p0q0r0s0=1,=0,=0,=1.
By eliminating rn and sn, we get Ln+1=(pn+1qn+1bnpnbnqn) and the recurrence relation (1.1a), (1.1b).
Corollary 1.2. For any n∈N which satisfies cn+1=∞ and cn+2=∞, we have
cn+2−cn+1=(−1)nqn+2qn+1bnbn−1⋯b0.
Proof. If cn+1=∞ and cn+2=∞, we have
cn+2−cn+1=qn+2pn+2−qn+1pn+1=qn+2qn+1pn+2qn+1−pn+1qn+2=bn+1qn+2qn+1detLn+2
and
detLn+2=detM0detM1⋯detMn+1=(−1)nb0b1⋯bn+1.
Remark. Since there is no n∈N for which both cn+1 and cn+2 are ∞, it is tempting to extend the above formula for the cases where either cn+1 or cn+2 is ∞. But the difficulty is we must handle +∞, −∞, and ∞ as if they are different elements although they are identical in the view of C. Therefore, we restrict ourselves to the cases where both are finite.
Appendix
The first few terms of κ={cn}n∈N, written in the form of pn/qn, are
c0c1c2c3c4c5=01,=1a0,=a1a1a0+b0,=a2a1+b1a2a1a0+a2b0+b1a0,=a3a2a1+a3b1+b2a1a3a2a1a0+a3a2b0+a3b1a0+b2a1a0+b2b0,and=a4a3a2a1+a4a3b1+a4b2a1+b3a2a1+b3b1a4a3a2a1a0+a4a3a2b0+a4a3b1a0+a4b2a1a0+a4b2b0+b3a2a1a0+b3a2b0+b3b1a0.
References
[1] S. Hitotsumatsu, Introduction to Special Functions, Morikita, 1999.
Revisions
August 20, 2020
Stop revising; see Part 5 and following articles for renewed descriptions.
July 26, 2020
Simplify the introduction order of notions.
July 16, 2020
Describe finite continued fractions, introduce the bracket notation [], and various minor improvements.