Generalized Notions of Continued Fractions: Ergodicity and Number Theoretic Applications 103251678X, 9781032516783

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Generalized Notions of Continued Fractions Ancient times witnessed the origins of the theory of continued fractions. Throughout time, mathematical geniuses such as Euclid, Aryabhata, Fibonacci, Bombelli, Wallis, Huygens, or Euler have made significant contributions to the development of this famous theory, and it continues to evolve today, especially as a means of linking different areas of mathematics. This book, whose primary audience is graduate students and senior researchers, is motivated by the fascinating interrelations between ergodic theory and number theory (as established since the 1950s). It examines several generalizations and extensions of classical continued fractions, including generalized Lehner, simple, and Hirzebruch-Jung continued fractions. After deriving invariant ergodic measures for each of the underlying transformations on [0,1] it is shown that any of the famous formulas, going back to Khintchine and Levy, carry over to more general settings. Complementing these results, the entropy of the transformations is calculated and the natural extensions of the dynamical systems to [0,1]2 are analyzed. Features • Suitable for graduate students and senior researchers • Written by international senior experts in number theory • Contains the basic background, including some elementary results, that the reader may need to know beforehand, making it a self-contained volume.

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Generalized Notions of Continued Fractions Ergodicity and Number Theoretic Applications

Juan Fernández Sánchez Jerónimo López-Salazar Codes Juan B. Seoane Sepúlveda Wolfgang Trutschnig

First edition published 2023 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 4 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN CRC Press is an imprint of Taylor & Francis Group, LLC © 2024 Juan Fernández Sánchez, Jerónimo López-Salazar Codes, Juan B. Seoane Sepúlveda, Wolfgang Trutschnig Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright. com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging-in-Publication Data

Names: Fernández Sánchez, Juan (Mathematician), author. | López-Salazar Codes, Jerónimo, author. | Seoane Sepúlveda, Juan B., author. | Trutschnig, Wolfgang, author. Title: Generalized notions of continued fractions : ergodicity and number theoretic applications / Juan Fernández Sánchez, Jerónimo López-Salazar Codes, Juan B. Seoane Sepúlveda, Wolfgang Trutschnig. Description: First edition. | Boca Raton : C&H/CRC Press, 2023. | Series: Chapman and Hall/CRC monographs and research notes in mathematics | Includes bibliographical references. Identifiers: LCCN 2023000909 (print) | LCCN 2023000910 (ebook) | ISBN 9781032516783 (hardback) | ISBN 9781032518183 (paperback) | ISBN 9781003404064 (ebook) Subjects: LCSH: Continued fractions. | Ergodic theory. | Number theory. Classification: LCC QA295 .F46 2023 (print) | LCC QA295 (ebook) | DDC 512.7/2--dc23/eng20230512 LC record available at https://lccn.loc.gov/2023000909 LC ebook record available at https://lccn.loc.gov/2023000910 ISBN: 978-1-032-51678-3 (hbk) ISBN: 978-1-032-51818-3 (pbk) ISBN: 978-1-003-40406-4 (ebk) DOI: 10.1201/9781003404064 Typeset in CMR10 by KnowledgeWorks Global Ltd. Publisher’s note: This book has been prepared from camera-ready copy provided by the authors

Contents

Preface

vii

Authors

xi

1 Generalized Lehner continued fractions

1

2 a-modified Farey series

15

3 Ergodic aspects of the generalized Lehner continued fractions

25

4 The a-simple continued fraction

43

5 The generalized Khintchine constant

57

6 The entropy of the system ([0, 1], B, µa , Ta )

67

7 The natural extension of ([0, 1], B, µa , Ta )

75

8 The dynamical system ([0, 1], B, νa , Qa )

93

9 Generalized Hirzebruch-Jung continued fractions

99

10 The entropy of ([0, 1], B, ϑa , Ha )

113

11 The natural extension of ([0, 1], B, ϑa , Ha )

117

12 A new generalization of the Farey series

131

Bibliography

139

v

Preface

The notion of continued fraction dates back to ancient times. Mathematical geniuses such as Euclid, Aryabhata, Fibonacci, Bombelli, Wallis, Huygens, or Euler contributed to the development of this famous theory which (linking different areas of mathematics) continues to evolve today (see, e.g., [7, 20, 25, 33, 36]). There are numerous ways to generalize continued fractions. One of the aims of this text is to explore and provide an overview of some of these directions. The families of generalized continued fractions considered here depend on a parameter a ∈ (0, +∞) or a ∈ (0, 1] (for the special case of a = 1 one of the constructions boils down to the classical one). Being self-contained with detailed proofs, and having an expository nature, this text can easily serve as a student’s manual for potential senior researchers who, in their junior stage, wish to delve into the field of continued fractions after having some previous solid background on the topic. A large portion of the results presented here already existed in the form of unpublished notes dating back to 2007. Not surprisingly, in the meanwhile plenty of literature, mostly articles, have been appearing. Nevertheless we only learnt about the existence of the paper [12] by Dajani, Kraaikamp and van der Wekken during the process of proof-reading the first draft of this book. For the case of a = 1/N with N ∈ N their article [12] contains several results studied for general a ∈ (0, +∞) or a ∈ (0, 1] in various chapters of this book (including ergodicity, the natural extension as well as the entropy of T1/N and H1/N at different levels of detail). Important related articles concerning approximation coefficients are [11, 29], additional works along similar lines of research questions are, to cite some, [15, 26, 28, 32]. Considering that the number of contributions is continuously growing (and it is hard to keep track of all recent developments and its state of the art) we may have overlooked additional relevant papers. Certainly, credit must be given to the authors of the original works in which the results appeared for the first time. Despite serving as a survey text, this book, however, also seems to contain various novel results (as in, for instance, Chapter 5). Having said this, let us now move on and present the way in which this text is arranged. Without doubt, the interplay between number theory and ergodic theory is potentially one of the most beautiful and fruitful ones throughout mathematics, particularly when it comes to continued fractions. As pointed out by Kac in [23], already in 1951 Ryell-Nardzewski in [37] used the fact that the

vii

viii

Preface   1 1 Gauss map − is measure-preserving and ergodic with respect to the x x 1 1 absolutely continuous probability measure with density on [0, 1] in ln 2 1 + x order to derive the famous Khintchine formula (see [24]) via Birkhoff’s ergodic theorem. Many more remarkable contributions were to follow: the articles by Philipp [34] and Dajani and Kraaikamp [9], contributions by Liardet (see [14, 2]) as well as the books by Einsiedler and Ward (see [13]), by Dajani and Kraainkamp (see [10]) and by Schweiger (see [38]) are just some shining examples. Building upon this interplay, in the current work we revisit continued fractions, consider natural generalizations of Lehner continued fractions, of Farey series, of simple continued fractions and of Hirzebruch-Jung continued fractions, prove that the corresponding dynamical systems are ergodic with respect to some absolutely continuous measures and then show how famous number theoretic formulas going back to Khintchine and Levy carry over to the considered more general settings. More precisely, the rest of this work is organized as follows: Chapter 1 generalizes Lehner continued fractions q0

p0 + p1 +

q1 p2 +

q2 .. .

by replacing the condition (pn , qn ) ∈ {(1, 1), (2, −1)} for each n ∈ N by (pn , qn ) ∈ {(1, a), (2, −1)} for some arbitrary but fixed a ∈ (0, +∞) and by considering the so-called a-Lehner map defined by  1  x if 0 ≤ x ≤ 1+a , 1−ax La (x) =  1−x 1 if 1+a ≤ x ≤ 1. ax Chapter 2 focuses on the closely related concept of so-called a-modified Farey series, the results of which are then used in Chapter 3 to prove the fact that La is measure-preserving and ergodic with respect to the σ-finite measure γa defined by Z 1 dx γa (E) = x E for every Borel set E ⊂ [0, 1]. Using ergodicity, a variation of the famous Khintchine formula (see [24]) is then derived as main result of Chapter 3. Replacing the well-known Gauss map being key in the context of classical continued fractions by the transformation  0 if x = 0, Ta (x) =

 1 1 − 1 − 1 1 − 1 if 0 < x ≤ 1, a x a x

Preface

ix

(where a ∈ (0, +∞) and [z] denotes the integer part of a real number z) yields so-called a-simple continued fractions, i.e., representations of the form 1

x=

a

1 + m1 a + 1 + m2 a +

a 1 + m3 a +

a .. .

with mi ∈ N ∪ {0} for every i ∈ N. Observe that Ta is obtained via iterations of La . In fact, ( n 1 1 La (x) if 1+na < x ≤ 1+(n−1)a for some n ∈ N, Ta (x) = 0 if x = 0, where the exponent n in Lna denotes the usual n times composition. Considering that Ta is measure-preserving and ergodic with respect to the probability measure µa , defined by Z 1 a µa (E) = dλ(x) log(1 + a) E 1 + ax for every Borel set E ⊂ [0, 1], is then shown to allow for a straightforward extension of Khintchine’s and Levy’s famous formulas to a-simple continued fractions. Moreover, as proved in Chapter 5, for almost every x ∈ [0, 1] the constants (mi )i∈N fulfill v u n uY n lim t (1 + mi a) = Ka n→∞

i=1

where Ka denotes the generalized Khintchine constant. Chapter 6 calculates the entropy of the ergodic dynamical system ([0, 1], B, µa , Ta ) for every a ∈ (0, +∞), Chapter 7 considers the natural extension of the latter dynamical system to [0, 1]2 , and Chapter 8 presents a modification of the transformation Ta . In the same fashion as La is defined depending on a parameter a, we can also define another family of transformations, Ca (now with a ∈ (0, 1]), as follows: ( x 1 if 0 ≤ x ≤ 1+a , 1−ax Ca (x) =
1 1 1 if 1+a < x ≤ 1. a 1+a− x Proceeding analogously as in previous chapters, Chapter 9 studies the transformation ( n 1 1 Ca (x) if 1+na < x ≤ 1+(n−1)a for some n ∈ N, Ha (x) = 0 if x = 0. 1  ( 1 1 1 + a − ax + ax − a1 if 0 < x ≤ 1, = 0 if x = 0.

x

Preface

and its interrelation with generalized Hirzebruch-Jung continued fractions. It is shown that Ha is measure-preserving and ergodic with respect to the probability measure ϑa , defined by Z 1 −a ϑa (E) = dλ(x), log(1 − a) E (1 − ax) which is then used to extend the Khintchine and Levy formulas to generalized Hirzebruch-Jung continued fractions. Chapter 10 calculates the entropy of the ergodic dynamical system ([0, 1], B, µa , Ta ) for every a ∈ (0, 1). Chapter 11 considers the natural extension of the latter dynamical system to [0, 1]2 . Finally, Chapter 12 studying a generalization of Farey series concludes the work. The notation used in this book will be standard. Throughout the text, the letter a will always denote a strictly positive real number. Given a set X, a mapping f : X → X and a natural number n, f n will denote the composition of f with itself n times, i.e., f 0 (x) = x, f 1 (x) = f (x) and f n (x) = f (f n−1 (x)) for each x ∈ X. Card(X) stands for the cardinality of the set X. The Borel σ-field on [0, 1] will be denoted by B and the Lebesgue measure on B by λ. The characteristic function of a set B will be written as XB . Other required notation and tools will be introduced on the fly in each of the chapters as needed. Juan Fern´andez S´anchez Jer´onimo L´opez-Salazar Codes Juan Benigno Seoane Sep´ ulveda Wolfgang Trutschnig

Authors

Juan Fern´ andez S´ anchez earned his PhD in mathematics from the University of Almer´ıa (Spain) in 2010. His research interests are in dependence modeling and copulas, dynamical systems, singular functions, and number theory. Jer´ onimo L´ opez-Salazar Codes completed his doctoral work under the supervision of Professors Jos´e Mar´ıa Mart´ınez Ansemil and Socorro Ponte at Universidad Complutense de Madrid (Spain) and obtained his PhD in 2013. He currently works at Universidad Polit´ecnica de Madrid (Spain). His research is mainly devoted to infinite dimensional holomorphy and lineability. Juan B. Seoane Sep´ ulveda earned his first PhD at the Universidad de C´adiz (Spain) jointly with Universit¨ at Karlsruhe (Germany) in 2005. He received his second PhD at Kent State University (Kent, Ohio, USA) in 2006. His main interests include real and complex analysis, operator theory, number theory, mathematical modeling, mathematical biology, the geometry of Banach spaces, the history of mathematics, and lineability. He is the author of over 200 scientific publications, including several books. He is currently a professor at Universidad Complutense de Madrid, where he also holds the position of director of the Master’s in Advanced Mathematics. Wolfgang Trutschnig obtained his PhD at the Vienna University of Technology, Austria, in 2006. He is currently professor for stochastics and director of the IDA Lab at the Paris Lodron University Salzburg (PLUS) and mainly works in dependence modeling and nonparametric statistics with regular excursions to dynamical systems, fractals, and ergodic theory.

xi

Chapter 1 Generalized Lehner continued fractions

In [30] Lehner proved that every real number can be expressed as the sum of an integer and a continued fraction q0

p0 + p1 +

q1 p2 +

q2 .. .

with (pn , qn ) ∈ {(1, 1), (2, −1)} for each n. The aim of this chapter is to generalize this result by considering continued fractions with coefficients (pn , qn ) ∈ {(1, a), (2, −1)} for some fixed positive number a. We first establish several auxiliary lemmas which are then applied in order to prove Theorem 1.8, the main result of this chapter. Considering the mapping La : [0, 1] → [0, 1] (see Figure 1.1), defined by  x 1   1−ax if 0 ≤ x ≤ 1+a (1.1) La (x) =   1−x 1 if ≤ x ≤ 1, ax 1+a each number x in [0, 1] can be expressed in terms of La (x) by

x=

1  −   a

1/a2 1 a +La (x)

 1/a   1 a + La (x)

DOI: 10.1201/9781003404064-1

if 0 ≤ x ≤ if

1 1+a

1 1+a

< x ≤ 1.

1

2

Generalized Notions of Continued Fractions

FIGURE 1.1: The transformation La for a = 2 (black) and a = 5 (magenta). Proceeding analogously and expressing La (x) in terms of L2a (x) obviously yields  i h 1 1/a2  1  − if x, L (x) ∈ 0,  a  1+a 1/a2 2  a  1  a − a +L2a (x)        1 1/a2  1   − if 0 ≤ x ≤ 1+a < La (x) ≤ 1  1/a 1  a  + 1 2  a

x=

a +La (x)

  1/a     1/a2 2   1  a − a +L2a (x)        1/a      1 + 1 1/a2 a +L (x) a

if 0 ≤ La (x) ≤

if x, La (x) ∈



1 1+a

0.

(1.5)

In order to compute Bk2 , we need to distinguish two different cases. On the one hand, if k2 = k1 + 1, then Bk2 = pk2 Bk1 + qk1 Bk1 −1 = Bk1 + aBk1 −1 = 1 + ak1 > 0. On the other hand, if k2 ≥ k1 + 2, then (pk2 −1 , qk2 −1 ) = (2, −1) and we can apply Equation (1.5) to get Bk2 = pk2 Bk2 −1 + qk2 −1 Bk2 −2 = Bk2 −1 − Bk2 −2 = [1 + (k2 − 1 − k1 ) (1 + ak1 )] − [1 + (k2 − 2 − k1 ) (1 + ak1 )] > 0. For i = k2 + 1, . . . , n we may proceed analogously. Lemma 1.4. Suppose that (pn , qn ) ∈ {(2, −1), (1, a)} for all n ≥ 0. If K = {n ∈ N : (pn , qn ) = (1, a)}, then there exist some C > 0 and some n0 ∈ N such that √ n  1 + 1 + 4a Bn > C 2 holds for every n ∈ K such that n ≥ n0 . Proof. Since the result is obvious for finite K, it suffices to assume that this set of indices is infinite. We start with considering three consecutive elements of K, m < n < k. That is, (pm , qm ) = (pn , qn ) = (pk , qk ) = (1, a) and (pi , qi ) = (2, −1) for each i ∈ {m + 1, . . . , n − 1} ∪ {n + 1, . . . , k − 1}. On the one hand, if k = n + 1, then Bk = pk Bk−1 + qk−1 Bk−2 = Bn + aBn−1 . On the other hand, if k > n + 1, we may apply Lemma 1.2 to obtain Bk = pk Bk−1 + qk−1 Bk−2 = Bk−1 − Bk−2 = [Bn + (k − 1 − n) (Bn + aBn−1 )] − [Bn + (k − 2 − n) (Bn + aBn−1 )] = Bn + aBn−1 .

6

Generalized Notions of Continued Fractions

Thus, in both cases, Bk = Bn + aBn−1 follows. Again considering Lemma 1.2, the equality Bn−1 = Bm + (n − 1 − m) (Bm + aBm−1 ) follows and we get Bk = Bn + a [Bm + (n − 1 − m) (Bm + aBm−1 )]
= Bn + aBm + (n − 1 − m) aBm + a2 Bm−1 . Altogether this shows that Bk ≥ Bn + aBm whenever m < n < k are consecutive elements of K. Let n1 < n2 < n3 < · · · be the indices in K. As already proved, the minimum possible value of each Bni is Bni = Bni−1 + aBni−2 . Considering that the solutions of the equation t2 = t + a are √ √ 1 − 1 + 4a 1 + 1 + 4a , t2 = , t1 = 2 2 it follows that the solution of the linear difference equation   yi = yi−1 + ayi−2 y1 = Bn1   y2 = Bn2 is yi = C1 ti1 + C2 ti2 , whereby C1 =

Bn2 − t2 Bn1 , t21 − t1 t2

C2 =

Bn2 − t1 Bn1 . t22 − t1 t2

The constant C1 is positive since t1 > 0 > t2 and Bn > 0 for all n. Hence "  i # t2 i i Bni ≥ yi = C1 t1 + C2 t2 = C1 + C2 ti1 . t1 Since tt12 < 1, there exists i0 such that B ni ≥

C1 i C1 t1 = 2 2



1+

√

1 + 4a 2

i

holds for all i ≥ i0 . Lemma 1.5. Suppose that (pn , qn ) ∈ {(2, −1), (1, a)} for all n ≥ 0 and let An+1 An An and Bn be defined according to Lemma 1.1. Then B ̸ B = holds for n n+1 every n. Moreover, the following implications hold:

Generalized Lehner continued fractions

7

(i) If (pn , qn ) = (1, a) and

An−1 Bn−1




Ak Bk

for all k > n.

(ii) If (pn , qn ) = (1, a) and

An−1 Bn−1

>

An Bn ,

then

An Bn


n.

(iii) If (pn , qn ) = (2, −1) and

An−1 Bn−1




An Bn ,

then

An Bn

>

Ak Bk

for all k > n.

Proof. The determinant formula in Lemma 1.1 implies An An+1 (−1)n+1 q0 q1 · · · qn − = ̸= 0, Bn Bn+1 Bn Bn+1 so

An Bn

̸=

An+1 Bn+1

follows.

In order to prove (i), fix n ∈ N such that (pn , qn ) = (1, a) and hold. Let d denote the cardinality of the set

An−1 Bn−1


0, Bn Bn+1 Bn Bn+1 which proves that An An+1 > Bn Bn+1 An−1 An whenever B n, we define

p∗n+1 = pn +

and

qn pn+1 +

A∗n+1 = p0 + ∗ Bn+1 p1 +

qn+1 ..

. pk +

qk pk+1

q0

.

q1 ..

. pn +

qn ∗ pn+1

8

Generalized Notions of Continued Fractions

An−1 Considering that we already know that B < n−1 follows that A∗ Ak An = > n+1 , ∗ Bn Bn+1 Bk

An Bn

and (pn , qn ) = (1, a), it

which completes the proof of (i). The other statements are obtained analogously. The proof of the following lemma can easily be established by induction. Lemma 1.6. For t ≥ 0 the identity 1

2− 2−

=

1 ..

. 2−

1 + (k + 1)t 1 + kt

1 1+t

holds, whereby the number 2 appears k times in the continued fraction. In particular, if 1 + t = 2, the number 2 appears m = k + 1 times and in this case we have m+1 1 = . 2− 1 m 2− .. 1 . 2− 2 Lemma 1.7. Suppose that (pn , qn ) ∈ {(2, −1), (1, a)} for all n ≥ 0. If An = p0 + Bn p1 +

then there exists some n0 ∈ N with

An Bn

q0

,

q1 ..

. pn−1 +

qn−1 pn

∈ [1, 1 + a] for every n > n0 .

Proof. The proof will be divided in several cases depending on how many pairs (pn , qn ) are equal to (1, a). Case 1: If (pn , qn ) = (2, −1) for every n ≥ 1, then Lemma 1.6 yields An = p0 + Bn p1 +

q0

= p0 +

q1 ..

. pn−1 +

qn−1 pn

q0 n+1 n

= p0 + q0 ·

If (p0 , q0 ) = (1, a), then An n =1+a ∈ [1, 1 + a] Bn n+1 for every n ∈ N. If (p0 , q0 ) = (2, −1), then An n 1 =2− =1+ , Bn n+1 n+1 which belongs to [1, 1 + a] for all n ∈ N fulfilling n ≥

1 a

− 1.

n . n+1

Generalized Lehner continued fractions

9

Case 2: There exists some N ≥ 1 such that (pN , qN ) = (1, a) and (pn , qn ) = (2, −1) for every n ≥ 1 with n ̸= N . If n > N , then Lemma 1.6 implies pN +1 +

qN +1 ..

q . pn−1 + n−1 pn

=2−

1 ..

1 . 2− 2

=

n−N +1 , n−N

so An = p0 + Bn p1 +

q0 ..

. pN −1 +

q0

= p0 +

q1

2−

qN −1 qN pN + n−N +1

.

1 ..

1

. 2− 1+

(n−N )a n−N +1

n−N

Again by Lemma 1.6 we have An = p0 + Bn

q0 (n−N )a n−N +1 )a 1) · (n−N n−N +1

1+N · 1 + (N −

= p0 + q0 ·

(n−N )a n−N +1 (n−N )a n−N +1

1 + (N − 1) · 1+N ·

.

If (p0 , q0 ) = (1, a), then )a 1 + (N − 1) · (n−N An n−N +1 , =1+a· )a Bn 1 + N · (n−N n−N +1

which belongs to [1, 1 + a] for every n ∈ N. Also, if (p0 , q0 ) = (2, −1), then )a 1 + (N − 1) · (n−N An (n − N )a n−N +1 =1+ ≥ 1. =2− (n−N )a Bn n − N + 1 + N (n − N )a 1 + N · n−N +1

In this last case we have a An =1+ < 1 + a, n→∞ Bn 1 + Na lim

An which implies that B ∈ [1, 1 + a] for every n large enough. n Case 3: There exist at least two numbers M > N ≥ 1 such that (pN , qN ) = (pM , qM ) = (1, a). Let N and M be the first indices with those properties, that is, (pn , qn ) = (2, −1) for all n = 1, . . . , N −1 and all n = N +1, . . . , M −1. AN −1 AN AN An According to Lemma 1.5, B >B implies B >B for every n > N and n N N −1 N AM BM


N and

for every n > M . Also, if AM BM

>

An Bn

 min

AN BN


M . Note that AN = p0 + BN p1 +

q0 q1 ..

q0

= p0 + 2−

and that pN +1 +

Hence for

AM BM

. pN −1 +

= p0 + q0 ∈ [1, 1 + a]

1 ..

. 2−

1 1

qN +1 ..

qN −1 pN

q . pM −1 + M −1 pM

=2−

1 ..

1 . 2− 1

= 1.

we get

AM = p0 + BM p1 +

q0

= p0 +

q1

q0

1 2− qN −1 .. 1 . pN −1 + . 2− pN + qN 1+a q0 1 + (N − 1)a = p0 + 1+N a = p0 + q0 · . 1 + Na 1+(N −1)a ..

On the one hand, if (p0 , q0 ) = (2, −1), then a 1 + (N − 1)a AM =1+ ∈ [1, 1 + a]. =2− BM 1 + Na 1 + Na On the other hand, if (p0 , q0 ) = (1, a), then AM 1 + (N − 1)a =1+a· ∈ [1, 1 + a]. BM 1 + Na AM AN and B belong to [1, 1 + a], which shows that We have seen that both B N M An Bn ∈ [1, 1 + a] holds for every n > M . This completes the proof.

Theorem 1.8. Suppose that (pn , qn ) ∈ {(2, −1), (1, a)} for all n ≥ 0. For each n ∈ N, assume that An = p0 + Bn p1 +

An n→∞ Bn

Then the limit lim ∞ (pn )n=0

and (1.3), then

and

q0 q1 ..

. pn−1 +

qn−1 pn

exists and belongs to [1, 1+a]. Moreover, if x ∈ [1, 1+a]

∞ (qn )n=0

are the sequences associated with x by Equation lim

n→∞

An = x. Bn

Generalized Lehner continued fractions

11

Proof. First of all, we define the index sets I and J as follows:   An Ak I= n∈N: < for every k > n , Bn Bk   An Ak J = n∈N: > for every k > n . Bn Bk According to Lemma 1.5, for each n ≥ 1 we have either n ∈ I or n ∈ J, i.e., I ∪ J = N. We distinguish the following three situations: (i) J is finite (or empty) and I is infinite. (ii) I is finite (or empty) and J is infinite. (iii) Both I and J are infinite. First suppose that J is finite or empty, i.e., there is some N ∈ N such that n ∈ I for all n ≥ N . Then AN AN +1 AN +2 < < < ··· BN BN +1 BN +2 and by Lemma 1.5 (pn , qn ) = (2, −1) holds for all n > N . Consequently, n > N implies An = p0 + Bn ..

q0

= p0 +

qN

. pN + pN +1 +

qN +1 ..

q0 ..

2−

q . pn−1 + n−1 pn

According to Lemma 1.6, we have An q0 = p0 + Bn .. qN . pN + 1+n−N n−N An n→∞ Bn

for every n > N , so the limit lim lim

n→∞

exists and we have

An = p0 + Bn p1 +

q0 q1 ..

. pN + qN

.

.

qN

. pN +

1 ..

.2−

1 2

12

Generalized Notions of Continued Fractions

Moreover, using Equation (1.3) and again Lemma 1.6, it follows that x = p0 +

q0 ..

qN

. pN + 2−

= p0 +

1 ..

. 2−

1 Φn+1 (x) a

q0 ..

. pN +

qN 1+(n−N +1)(Φn+1 (x)−1) a 1+(n−N )(Φn+1 (x)−1) a

for every n > N . Considering n → ∞, yields q0

x = p0 + p1 +

q1 ..

= lim

n→∞

An , Bn

. pN + qN

which completes the proof for finite J. If I is finite (or empty), there exists some N ∈ N such that AN AN +1 AN +2 > > > ··· BN BN +1 BN +2 holds. According to Lemma 1.5, (pn , qn ) = (2, −1) for all n > N follows and the proof continues along the same line as the previous case. Finally, assume that both I and J are infinite, i.e., there are infinitely many indexes n with (pn , qn ) = (1, a). If i1 < i2 < i3 < · · · belong to I and j1 < j2 < j3 < · · · to J, then Ai 1 Ai Ai Aj Aj Aj < 2 < 3 < ··· < 3 < 2 < 1 , Bi1 B i2 B i3 Bj 3 Bj 2 Bj1 so both subsequences of convergents are monotone and bounded and we may set Ai Aj α = lim , β = lim i∈I Bi j∈J Bj with α ≤ β. Setting K = {n ∈ N : n ≥ 1 and (pn , qn ) = (1, a)} , we have the following: If n < k are two consecutive elements of K, then (pn , qn ) = (1, a), (pi , qi ) = (2, −1) for each i = n + 1, . . . , k − 1 and (pk , qk ) = (1, a). Hence q0 An = p0 + Bn .. q . pn−1 + n−1 1

Generalized Lehner continued fractions

13

and Ak = p0 + Bk ..

q0

= p0 +

qn−1 a

. pn−1 + 1+

2−

q0 ..

. pn−1 +

1 ..

. 2−

qn−1 a 1+ 1

.

1 1

Ak An That is, we can consider B and B as two consecutive approximants with n k ∗ pn+1 = 1 whenever n and k are two consecutive elements of K. According to Lemma 1.5, n ∈ I implies k ∈ J (and vice versa), so An Ak An Bk − Ak Bn |β − α| ≤ − = Bn Bk Bn Bk Ak An follows. Considering B and B as two consecutive approximants, the detern k minant formula in Lemma 1.1 implies (−1)n+1 q0 q1 · · · qn−1 · a M n+1 ≤ |β − α| ≤ Bn Bk , Bn Bk

where by M = max {1, a}. Using Lemma 1.4 yields M n+1

|β − α| ≤ C2



n+k √ 1+ 1+4a 2

M = 2 C

M

a+

√ 1+ 1+4a 2

M n+1

≤ C2



2n √ 1+ 1+4a 2

!n −→ 0,

n→∞

which proves α = β. Applying Lemma 1.7, it follows that α = lim

n→∞

An ∈ [1, 1 + a]. Bn

It therefore remains to prove x = α. Using Equation (1.3) and Lemma 1.1, for n ∈ K, we get x = p0 +

q0 ..

. pn +

qn Φn+1 (x) a

=

Φn+1 (x)An + qn An−1 Φn+1 (x)An + aAn−1 a a = . n+1 n+1 Φa (x)Bn + qn Bn−1 Φa (x)Bn + aBn−1

Thus, for n ∈ K, it follows that x−

An Φn+1 (x)An + aAn−1 An a (An−1 Bn − An Bn−1 ) a   = n+1 − = Bn Bn Φa (x)Bn + aBn−1 Bn Φn+1 (x)Bn + aBn−1 a a  n+1  · (−1)n q0 q1 · · · qn−1 . = Bn Φa (x)Bn + aBn−1

14

Generalized Notions of Continued Fractions

Therefore, if n < k are two consecutive elements of K, then (pn , qn ) = (1, a) and (pi , qi ) = (2, −1) for each i = n + 1, . . . , k − 1, which implies that x−

Ak a  k+1  · (−1)k q0 q1 · · · qk−1 = Bk Bk Φa (x)Bk + aBk−1 a   · (−1)k q0 q1 · · · qn−1 · a · (−1)k−1−n = Bk Φk+1 (x)B + aB a k k−1 =

a2  k+1  · (−1)n+1 q0 q1 · · · qn−1 . Bk Φa (x)Bk + aBk−1

Ak An In other words, x − B and x − B have opposite signs whenever n and k are n k two consecutive elements of K. Finally, let n1 < n2 < n3 < · · · be the indices of K. Choosing i ∈ N such A that x > Bnni yields i

x


Ani+2 , Bni+2

x
2 and k is odd, then 2k+1 Fa,n =

2k+1 Na,n 2k+1 Da,n

=

2k+2 2k + aNa,n Na,n 2k + aD 2k+2 Da,n a,n

.

Proof. We startnby considering the result for n = 2. The Farey series of second o 1 order is Fa,2 = 0, 1+a , 1 . Hence, 0 Na,2 0 = , 0 Da,2 1

1 Na,2 1 = , 1 Da,2 1+a

2 Na,2 1 = . 2 Da,2 1

For k = 0, we have 2k+1 Na,2 2k+1 Da,2

=

0 2 aNa,2 + Na,2 1 = , 0 2 1+a aDa,2 + Da,2

which coincides with the first equation of the lemma.

20

Generalized Notions of Continued Fractions

Suppose now that the result as fol mholds for an index n ≥ 2 and proceed 2k+1 lows: Considering Fa,n+1 = Fa,n+1 : 0 ≤ m ≤ 2n , the element Fa,n+1 verifies 2k + 1 ≤ 2n , which implies k ≤ 2n−1 − 1. We may thus distinguish four situations: (i) k is even and 0 ≤ k ≤ 2n−2 − 1. (ii) k is odd and 0 ≤ k ≤ 2n−2 − 1. (iii) k is even and 2n−2 ≤ k ≤ 2n−1 − 1. (iv) k is odd and 2n−2 ≤ k ≤ 2n−1 − 1. Suppose that (i) holds. Then Lemma 2.3(i) implies ! 2k 2k
Na,n Na,n 2k 2k Fa,n+1 = fa Fa,n = fa = 2k 2k + D 2k Da,n aNa,n a,n as well as 2k+2 Fa,n+1

=

2k+2 fa Fa,n




2k+2 Na,n

= fa

2k+2 Da,n

! =

2k+2 Na,n 2k+2 2k+2 + Da,n aNa,n

,

and the inductive hypothesis yields 2k+1 Fa,n =

2k+2 2k + Na,n aNa,n 2k + D 2k+2 aDa,n a,n

.

Therefore, 2k+1 Fa,n+1

= =

2k+1 fa Fa,n




= fa

2k+2 2k + Na,n aNa,n

!

2k + D 2k+2 aDa,n a,n

2k+2 2k + Na,n aNa,n 2k + aN 2k+2 + aD 2k + D 2k+2 a2 Na,n a,n a,n a,n

=

2k+2 2k aNa,n+1 + Na,n+1 2k+2 2k aDa,n+1 + Da,n+1

,

which shows that the result also holds for n + 1 whenever k is even and k ≤ 2n−2 − 1. The proof of case (ii) is similar. In situation (iii), writing n

2 −2k−2 Fa,n =

p , q

r 2n −2k Fa,n = , s

and using that 2n−2 ≤ k ≤ 2n−1 − 1, it follows that 2n−1 ≤ 2k < 2k + 2 ≤ 2n holds. Therefore, Lemma 2.3 shows  n  r s 2k 2 −2k Fa,n+1 = ga Fa,n = ga = s ar + s

a-modified Farey series as well as 2k+2 Fa,n+1

= ga



2n −2k−2 Fa,n



21

  p q = ga = . q ap + q

If k is even, then 2n − 2k − 1 = 2k ′ + 1, where k ′ = 2n−1 − k − 1 is odd. Using the already proved case (ii) it follows that ′

2n −2k−1 Fa,n

=

2k′ +1 Fa,n

=

′

2k 2k +2 Na,n + aNa,n 2k′ Da,n

+

2k′ +2 aDa,n

n

=

n

2 −2k−2 2 −2k Na,n + aNa,n 2n −2k−2 Da,n

+

2n −2k aDa,n

=

p + ar , q + as

which yields 2k+1 Fa,n+1

= ga =



2n −2k−1 Fa,n



 = ga

p + ar q + as



2k+2 2k aNa,n+1 + Na,n+1 q + as = . 2k+2 2k ap + a2 r + q + as aDa,n+1 + Da,n+1

We have therefore shown that the result also holds for n + 1 whenever k is even and 2n−2 ≤ k ≤ 2n−1 − 1. The case (iv) can be proved analogously to (iii). The next result extends the classical Farey theorem as stated in [17, Theorem 28]. Theorem 2.5. For n ∈ N, define Ψ : Fa,n → {0, 1} by  0    1
m Ψ Fa,n =  1    0

if if if if

m ≡ 0 (mod m ≡ 1 (mod m ≡ 2 (mod m ≡ 3 (mod

4) 4) 4) 4).

For every m ≤ 2n−1 − 1, set



m m1 m2 m0 Rn,m = Ψ Fa,n + Ψ Fa,n−1 + Ψ Fa,n−2 + · · · + Ψ Fa,1n−1 , where m0 = m and each index mi+1 is chosen depending on the previous one via  mi  if mi is even  2 mi+1 =   mi − 1 if mi is odd. 2 Then the identity m m+1 m+1 m Da,n Na,n − Da,n Na,n = aRn,m

holds.

22

Generalized Notions of Continued Fractions

0 Proof. For n = 1, we have Fa,1 = n−1 2 − 1 is m = 0, it follows that

0 1

1 and Fa,1 =

1 1.

Since the unique m ≤


0 R1,0 = Ψ Fa,1 = 0. Therefore, 0 1 1 0 Da,1 Na,1 − Da,1 Na,1 = 1 = aR1,0 ,

which implies that the result holds for n = 1. Assuming now that it holds for some n − 1, we will prove that it also holds for n and for any m ≤ 2n−1 − 1 as follows: First suppose
that m ≡ 0 (mod 4), i.e., m = 2k for some even number k. m Then Ψ Fa,n = 0, hence Rn,m = Rn−1,m1 = Rn−1,k , and Lemma 2.4 yields m m+1 m+1 m 2k 2k+1 2k+1 2k Da,n Na,n − Da,n Na,n = Da,n Na,n − Da,n Na,n

2k 2k 2k 2k+2 2k 2k+2 = Da,n aNa,n + Na,n − aDa,n + Da,n Na,n 2k 2k+2 2k+2 2k = Da,n Na,n − Da,n Na,n .

Moreover, applying Lemma 2.3, k+1 k+1 m m+1 m+1 m k k Da,n Na,n − Da,n Na,n = Da,n−1 Na,n−1 − Da,n−1 Na,n−1

= aRn−1,k = aRn,m follows, so the result holds for n and m ≡ 0 (mod 4). As next step suppose
that m ≡ 1 (mod 4), i.e., m = 2k + 1 for some even m number k. Then Ψ Fa,n = 1, implying Rn,m = 1 + Rn−1,m1 = 1 + Rn−1,k and Lemma 2.4 yields m m+1 m+1 m 2k+1 2k+2 2k+2 2k+1 Da,n Na,n − Da,n Na,n = Da,n Na,n − Da,n Na,n

2k 2k+2 2k+2 2k+2 2k 2k+2 = aDa,n + Da,n Na,n − Da,n aNa,n + Na,n
2k 2k+2 2k+2 2k = a Da,n Na,n − Da,n Na,n .

Using Lemma 2.3, it follows that k+1 k+1 m m+1 m+1 m k k Da,n Na,n − Da,n Na,n = a Da,n−1 Na,n−1 − Da,n−1 Na,n−1




= a · aRn−1,k = a1+Rn−1,k = aRn,m , so the result also holds for n and m ≡ 1 (mod 4). The proof of the cases m ≡ 2 (mod 4) and m ≡ 3 (mod 4) works analogously (instead of the first equality in Lemma 2.4 the second one must be used).

a-modified Farey series

23

m m Theorem 2.6. If a ∈ N and 1 ≤ m ≤ 2n , then Na,n and Da,n are natural numbers without common factors.

Proof. The result is obvious for a = 1 since Theorem 2.5 states that m m+1 m+1 m Da,n Na,n − Da,n Na,n = 1.

Thus, we may assume that a is a natural number bigger than 1. We again proceed by induction and show that m Da,n ≡ 1 (mod a)

(2.2)

1 0 = 11 , implying holds for all n and m. If n = 1, then Fa,1 = 10 and Fa,1 0 0 Da,1 = Da,1 = 1. If the property in (2.2) holds for some n, then Lemma 2.3 implies that  m m m aNa,n + Da,n ≡ Da,n ≡ 1 (mod a) if 0 ≤ m ≤ 2n−1 m Da,n+1 = aN 2n −m + D2n −m ≡ D2n −m ≡ 1 (mod a) if 2n−1 ≤ m ≤ 2n . a,n a,n a,n

By induction, the property in (2.2) holds for all n and m. m m had a common prime factor p, then and Da,n Finally, if Na,n m m+1 m+1 m Da,n Na,n − Da,n Na,n = aRn,m m ≡ 1 (mod a), it would therefore would imply that p would divide a. Since Da,n m follows that Da,n ≡ 1 (mod p), which is impossible since p was supposed to m . divide Da,n

Chapter 3 Ergodic aspects of the generalized Lehner continued fractions

We start this chapter by recalling basic definitions and theorems needed in the rest of the text for studying ergodicity of some dynamical systems associated with continued fractions. Given a (not necessarily finite) measure space (X, A, µ) and a measurable mapping T : X → X, the
dynamical system (X, A, µ, T ) is said to be measure-preserving if µ T −1 (E) = µ(E) for every E ∈ A. Moreover, the (not necessarily measure-preserving) dynamical system (X, A, µ, T ) is called ergodic if the following property holds: T −1 (E) = E for some E ∈ A implies µ(E) = 0 or µ(X\E) = 0, i.e., if all invariant sets or their complements have measure zero. The following version of the ergodic theorem can be found in [16, pages 18 and 31]. Theorem 3.1. Let (X, A, µ, T ) be a measure-preserving system such that the measure µ is σ-finite. Then for every f ∈ L1 (X, µ), there exists some function f ∗ ∈ L1 (X, µ) such that n−1 1X f (T i (x)) = f ∗ (x) n→∞ n i=0

lim

for almost every x ∈ X. Moreover, the function f ∗ satisfies the following properties: R R (i) If µ(X) < ∞, then f ∗ dµ = f dµ. (ii) If (X, A, µ, T ) is ergodic, then f ∗ is constant almost everywhere. R (iii) If (X, A, µ, T ) is ergodic and µ(X) = 1, then f ∗ (x) = f dµ for almost every x ∈ X. The next theorem is a variation of a result that appears in [4, Section 24, Lemma 2] and states a sufficient condition for a dynamical system to be ergodic. We include its proof for the sake of completeness. Theorem 3.2. Suppose that (X, A, µ) is a measure space, that P and Q are two subsets of A, that T : X → X is a measurable mapping and that c > 0 is a constant fulfilling the following properties: (i) µ ({x}) = 0 for every x ∈ X. DOI: 10.1201/9781003404064-3

25

26

Generalized Notions of Continued Fractions

(ii) For each x ∈ X, the set T −1 ({x}) is at most countable. (iii) Q is an algebra generating the σ-algebra A. (iv) For each E ∈ Q, there exists another set C ⊂ E such that C is a countable disjoint union of elements of P and E\C is at most countable. (v) If A ∈ P, then there exists some index nA µ (A ∩ T −nA (B)) ≥ cµ(A)µ(B) for every B ∈ P.

∈

N such that

Under these conditions the dynamical system (X, A, µ, T ) is ergodic. Proof. We proceed in a similar manner as done in [4, Section 24, Lemma 2]. Suppose S that A ∈ P and B ∈ Q. Then there exists a set C of the form C = i∈I Ci with I being at most countable, Ci ∈ P for every i, Ci ∩ Cj = ∅ if i ̸= j, such that C ⊂ B and B\C is at most countable. By the properties (i) and (ii), the set A ∩ T −nA (B\C) is at most countable and has measure zero. Then


µ A ∩ T −nA (B) = µ A ∩ T −nA (B\C) + µ A ∩ T −nA (C) X
= µ A ∩ T −nA (Ci ) . i∈I

Considering Ci ∈ P for every i and applying the property (v), it follows that
X µ A ∩ T −nA (B) ≥ cµ(A)µ(Ci ) = cµ(A)µ(C) = cµ(A)µ(B). i∈I

In other words, the property (v) holds for every A ∈ P and every B ∈ Q. Given a fixed element A ∈ P, let M denote the collection of all measurable sets B fulfilling µ (A ∩ T −nA (B)) ≥ cµ(A)µ(B). We have already shown that ∞ Q ⊂ M. Moreover, it is straightforward S∞to verify that if (Bn )n=1 is an increas∞ ing sequence of elements of M, then n=1 Bn ∈ M. T∞Furthermore, if (Bn )n=1 is a decreasing sequence of elements of M, then n=1 Bn ∈ M. Altogether this shows that M is a monotone class with Q ⊂ M. Since Q generates the σ-algebra A, the identity M = A follows immediately (see [4, Theorem 3.4]), i.e., the property (v) holds whenever A ∈ P and B is measurable. In order to show that T is ergodic, fix a measurable set E fulfilling T −1 (E) = E. We will prove that either µ(E) = 0 or µ(X\E) = 0. Let N denote the collection of all measurable sets A suchSthat µ (A ∩ E) ≥ cµ(A)µ(E) holds. If A ∈ Q, then there exists some set D = i∈J Di with J being at most countable such that Di ∈ P for every i, Di ∩ Dj = ∅ if i ̸= j, D ⊂ A and A\D is at most countable. We have already seen that the property (v) holds for the sets Di and E. Thus, for each i there exists indices ni ∈ N such that
µ Di ∩ T −ni (E) ≥ cµ(Di )µ(E).

Ergodic aspects of the generalized Lehner continued fractions

27

Considering T −1 (E) = E yields the following chain of inequalities: X X
µ (A ∩ E) ≥ µ (D ∩ E) = µ (Di ∩ E) = µ Di ∩ T −ni (E) i∈J

≥

X

i∈J

cµ(Di )µ(E) = cµ(D)µ(E) = cµ(A)µ(E).

i∈J ∞

Hence, A ∈ N and we deduce that Q ⊂ N . Moreover, if S(An )n=1 is an ∞ increasing sequence of elements of N , it is immediate that n=1 An ∈ N . ∞ Also, if (An )n=1 is a decreasing sequence of elements of N , we also have that T ∞ n=1 An ∈ N . This implies that N is a monotone class as well as Q ⊂ N . Again using [4, Theorem 3.4] it follows that N coincides with the whole σalgebra A. In particular, X\E ∈ N , that is, 0 = µ ((X\E) ∩ E) ≥ cµ(X\E)µ(E) holds, so µ(X\E)µ(E) = 0, which completes the proof. The subsequent Lemmas 3.3, 3.4, and 3.5 will be used for constructing a collection P with the properties needed in Theorem 3.2. Lemma 3.3. Given n + 1 pairs (p0 , q0 ), . . . , (pn , qn ) from {(2, −1), (1, a)} define Rp0 ,...,pn , rp0 ,...,pn , Sp0 ,...,pn and sp0 ,...,pn as follows: Rp0 ,...,pn = p0 +

Sp0 ,...,pn = p0 +

q0 p1 +

,

q1 ..

,

q1 ..

Rp0 ,...,pn − 1 , a

. pn + qn q0

p1 +

rp0 ,...,pn =

. pn +

sp0 ,...,pn =

Sp0 ,...,pn − 1 . a

qn 1+a

Then the following assertions hold: (i) Rp0 ,...,pn = Rp0 ,...,pn ,2 . (ii) Sp0 ,...,pn = Rp0 ,...,pn ,1 . (iii) Rp0 ,...,pn−2 ,2,1 = Rp0 ,...,pn−2 ,1,1 . (iv) Sp0 ,...,pn−1 ,2 = Sp0 ,...,pn−1 ,1 . (v) If p0 + · · · + pn is even, then Rp0 ,...,pn < Sp0 ,...,pn . (vi) If p0 + · · · + pn is odd, then Rp0 ,...,pn > Sp0 ,...,pn . (vii) The points rp0 ,...,pn and sp0 ,...,pn are consecutive elements of the Farey series Fa,n+2 .

28

Generalized Notions of Continued Fractions

(viii) If x ∈ Fa,n , then there exist (p0 , q0 ), . . . , (pn , qn ) in {(2, −1), (1, a)} such that x = rp0 ,...,pn . Proof. Properties (i), (ii), (iii), and (iv) are direct consequences of the definitions of Rp0 ,...,pn and Sp0 ,...,pn . In order to prove (v) and (vi), suppose that (pn , qn ) = (2, −1) (the arguments for (pn , qn ) = (1, a) are analogous). According to Lemma 1.1, we have Rp0 ,...,pn−1 ,2 = p0 +

q0 ..

q . pn−1 + n−1 2−1

=

An−1 + qn−1 An−2 Bn−1 + qn−1 Bn−2

and  Sp0 ,...,pn−1 ,2 = p0 +

q0 ..

qn−1 . pn−1 + 1 2− 1+a

=

2−

2−

1 1+a



An−1 + qn−1 An−2

1 1+a



Bn−1 + qn−1 Bn−2

.

As a direct consequence, Rp0 ,...,pn−1 ,2 < Sp0 ,...,pn−1 ,2 if and only if An−1 + qn−1 An−2 < Bn−1 + qn−1 Bn−2

1+2a 1+a An−1 1+2a 1+a Bn−1

+ qn−1 An−2 + qn−1 Bn−2

holds, which is equivalent to 1 + 2a (An−2 Bn−1 − An−1 Bn−2 )qn−1 < (An−2 Bn−1 − An−1 Bn−2 )qn−1 . 1+a Using the determinant formula in Lemma 1.1 yields that Rp0 ,...,pn−1 ,2 < Sp0 ,...,pn−1 ,2 if and only if 1 + 2a · (−1)n−1 q0 · · · qn−1 < (−1)n−1 q0 · · · qn−1 . 1+a Thus, if (pn , qn ) = (2, −1), then Rp0 ,...,pn−1 ,2 < Sp0 ,...,pn−1 ,2 if and only if (−1)n−1 q0 · · · qn−1 < 0. Letting d denote the cardinality of the set {i ∈ {0, . . . , n − 1} : (pi , qi ) = (1, a)} , it follows that p0 + · · · + pn = d + 2(n − d) + pn . On the one hand, for p0 + · · · + pn even and pn = 2 we conclude that d is even as well. Therefore, (−1)n−1 q0 · · · qn−1 = (−1)n−1 · (−1)n−d ad = (−1)d+1 ad < 0,

Ergodic aspects of the generalized Lehner continued fractions

29

implying Rp0 ,...,pn < Sp0 ,...,pn whenever p0 + · · · + pn is even and (pn , qn ) = (2, −1). On the other hand, if p0 + · · · + pn is odd and pn = 2, it follows that d is odd too. Therefore, (−1)n−1 q0 · · · qn−1 = (−1)d+1 ad > 0, which implies Rp0 ,...,pn > Sp0 ,...,pn whenever p0 +· · ·+pn is odd and (pn , qn ) = (2, −1). The proof of the properties (v) and (vi) for (pn , qn ) = (1, a) works analogously. Property (vii) will be proved via induction on n. If n = 0, then r2 = 0 = 1 1 2 0 = Fa,2 and r1 = 1 = Fa,2 , i.e., the result holds for n = 0. Fa,2 , s2 = s1 = 1+a Assuming validity for some natural number n − 1 we now prove the property for n. First suppose that p0 +· · ·+pn is even and that p0 = 2. Then p1 +· · ·+pn is even as well, so rp1 ,...,pn < sp1 ,...,pn . As assumed, there exists m ∈ {0, . . . , 2n } m+1 m such that rp1 ,...,pn = Fa,n+1 and sp1 ,...,pn = Fa,n+1 . Note that Rp0 ,...,pn = 2 −

1 1 1 + 2arp1 ,...,pn =2− = . Rp1 ,...,pn 1 + arp1 ,...,pn 1 + arp1 ,...,pn

Hence rp0 ,...,pn =


rp1 ,...,pn Rp0 ,...,pn − 1 m = = fa (rp1 ,...,pn ) = fa Fa,n+1 a 1 + arp1 ,...,pn

follows, where fa is the function defined in Equation (2.1). Similarly,
m+1 sp0 ,...,pn = fa (sp1 ,...,pn ) = fa Fa,n+1 , m+1 m and using Lemma 2.3, rp0 ,...,pn = Fa,n+2 and sp0 ,...,pn = Fa,n+2 follow. Now consider the case that p0 +· · ·+pn is even and p0 = 1. Then p1 +· · ·+pn is odd, so sp1 ,...,pn < rp1 ,...,pn and there exists some m ∈ {0, . . . , 2n } such that m+1 m sp1 ,...,pn = Fa,n+1 and rp1 ,...,pn = Fa,n+1 . Considering

Rp0 ,...,pn = 1 +

a a =1+ , Rp1 ,...,pn 1 + arp1 ,...,pn

it follows that rp0 ,...,pn =


Rp0 ,...,pn − 1 1 m+1 = = ga (rp1 ,...,pn ) = ga Fa,n+1 , a 1 + arp1 ,...,pn

where ga is the second function defined in Equation (2.1). The identity
m sp0 ,...,pn = ga (sp1 ,...,pn ) = ga Fa,n+1 follows in the same manner. According to Lemma 2.3, we have rp0 ,...,pn = 2n+1 −m−1 2n+1 −m Fa,n+2 and sp0 ,...,pn = Fa,n+2 , which implies rp0 ,...,pn and sp0 ,...,pn are

30

Generalized Notions of Continued Fractions

always consecutive elements of Fa,n+2 whenever p0 + · · · + pn is even. If p0 + · · · + pn is odd the arguments are similar. It remains to prove assertion (viii). If n = 1 and x ∈ Fa,1 , then obviously either x = 0 = r2 or x = 1 = r1 . Suppose now that (viii) holds for an index n. We will show that then it also holds for n + 1. If x ∈ Fa,n+1 , then there exists some y ∈ Fa,n such that either x = fa (y) or x = ga (y). By the induction assumption there exist (p0 , q0 ), . . . , (pn , qn ) with y = rp0 ,...,pn . On the one hand, if x = fa (y), then Rp

,...,p

−1

n 0 rp0 ,...,pn a = x = fa (rp0 ,...,pn ) = 1 + arp0 ,...,pn Rp0 ,...,pn   1 2 − Rp ,...,p −1 R2,p0 ,...,pn − 1 n 0 = = = r2,p0 ,...,pn a a

and on the other hand, if x = ga (y), then 1 1 x = ga (rp0 ,...,pn ) = = 1 + arp0 ,...,pn Rp0 ,...,pn   a 1 + Rp ,...,p − 1 R1,p0 ,...,pn − 1 n 0 = = r1,p0 ,...,pn . = a a This shows that the property (viii) also holds for n + 1 and the proof is complete. Lemma 3.4. Given (p0 , q0 ), . . . , (pn , qn ) in {(2, −1), (1, a)} let rp0 ,...,pn and sp0 ,...,pn be the numbers defined according to Lemma 3.3 and set  [rp0 ,...,pn , sp0 ,...,pn ) if p0 + · · · + pn is even ∆p0 ,...,pn = [s p0 ,...,pn , rp0 ,...,pn ) if p0 + · · · + pn is odd. Then for every x ∈ (0, 1], the interval [0, x) is a countable union of elements in the family n o D = ∆p0 ,...,pn : n ∈ N ∪ {0} and (pi , qi ) ∈ {(2, −1), (1, a)} for all i ≤ n . As a consequence, D generates the Borel σ-algebra in [0, 1). Proof. According to Lemma 3.3, the endpoints of each interval ∆p0 ,...,pn belong to Fa,n+2 , which implies that ∆p0 ,...,pn is contained in [0, 1). For n = 0, we have 0 = r2 < s2 = s1 < r1 = 1 and for n = 1, obviously 0 = r2,2 < s2,2 = s2,1 < r2,1 = r1,1 < s1,1 = s1,2 < r1,2 = 1

Ergodic aspects of the generalized Lehner continued fractions

31

holds. In general, Lemma 3.3 implies that for each n ≥ 0 the points rp0 ,...,pn and sp0 ,...,pn are ordered as follows: 0 = r2,...,2,2,2 < s2,...,2,2,2 = s2,...,2,2,1 < r2,...,2,2,1 = r2,...,2,1,1 < < s2,...,2,1,1 = s2,...,2,1,2 < · · · < r1,2,2,...,2 = 1. Therefore, for each n ≥ 0, we have that o [n ∆p0 ,...,pn : (pi , qi ) ∈ {(2, −1), (1, a)} for all i = 0, . . . , n , [0, 1) = i.e., the intervals form a partition of [0, 1) into consecutive intervals. For 0 < x ≤ 1, the set [ En = {∆p0 ,...,pn : ∆p0 ,...,pn ⊂ [0, x)} is a finite union of elements of D and is contained in [0, x). If x is an endpoint of some interval ∆p0 ,...,pn for some n ∈ N, then obviously [0, x) = En . For the rest of the proof,Swe will assume that x is not an endpoint of ∞ any interval ∆p0 ,...,pn . Then n=1 En is a countable union S∞of elements of D which is contained in [0, x) and we have to show [0, x) = n=1 En . Applying Theorem 1.8 yields An 1 + ax = lim , n→∞ Bn where each convergent is given by An = p0 + Bn p1 +

q0

.

q1 ..

. pn−1 +

qn−1 pn

Consider the set K = {n ∈ N ∪ {0} : (pn , qn ) = (1, a)} . If K was empty, then (pn , qn ) = (2, −1) for every n ≥ 0, hence Lemma 1.6 yields An n+2 = , Bn n+1 so 1 + ax = 1 and x = 0 = r2 . In other words, x would be an endpoint of ∆2 . If K was finite, then we could find some N ∈ N such that (pN , qN ) = (1, a) and (pn , qn ) = (2, −1) for all n > N . In this case, An = p0 + Bn .. . pN −1 +

q0 1+

qN −1 a 2−

1 ..

. 2−

1 2

32

Generalized Notions of Continued Fractions

for every n > N and using Lemma 1.6 An q0 = p0 + Bn .. qN −1 . pN −1 + a 1 + n−N +1 n−N

would follow. The latter implies 1 + ax = lim

n→∞

An q0 = p0 + Bn .. . pN −1 +

= Sp0 ,...,pN −1 , qN −1 1+a

so x = sp0 ,...,pN −1 would be an endpoint of an interval. Therefore, if x is not an endpoint of any interval ∆p0 ,...,pn , the set K must be infinite. As shown in the proof of Theorem 1.8, if n < k are two consecutive elements of K, Ak An then (1 + ax) < B if and only if (1 + ax) > B . Hence, we may choose an n k increasing subsequence An 2 An 3 An 1 < < < ··· Bn1 B n2 Bn3 converging to 1 + ax and (pni , qni ) = (1, a) for all i. Considering An i = p0 + B ni p1 +

q0

= Rp0 ,...,pni −1 ,

q1 ..

. pni −1 + qni −1

the sequence rp0 ,...,pni −1 =

Ani Bni

−1

a

is increasing and converges to x, which implies [0, x) = the proof.

S∞

n=1

En and completes

The next lemma introduces the collection P that fulfills the properties needed in Theorem 3.2. Lemma 3.5. Let D be the class of sets defined in Lemma 3.4 and P be defined as follows n o P = ∆p0 ,...,pn−1 ,1 : n ∈ N, (pi , qi ) ∈ {(2, −1), (1, a)} for every i ≤ n − 1 . Then for every E ∈ D, there exists another set C ⊂ E such that C is a countable disjoint union of elements of P and E\C is at most finite. Proof. Consider an interval ∆p0 ,...,pn ∈ D with pn = 2. We distinguish the two situations where p0 + · · · + pn is even or odd.

Ergodic aspects of the generalized Lehner continued fractions

33

First suppose that p0 + · · · + pn is odd. Using Lemma 3.3, we have ∆p0 ,...,pn = [sp0 ,...,pn , rp0 ,...,pn ) and, considering that p0 +· · ·+pn +1 is even and p0 + · · · + pn + 2 is odd, Lemma 3.3 implies the following chain of inequalities: Sp0 ,...,pn = Rp0 ,...,pn ,1 < Sp0 ,...,pn ,1 = Sp0 ,...,pn ,2 < Rp0 ,...,pn ,2 = Rp0 ,...,pn . As a direct consequence, it follows that ∆p0 ,...,pn = [sp0 ,...,pn , rp0 ,...,pn ) = [rp0 ,...,pn ,1 , sp0 ,...,pn ,1 ) ∪ [sp0 ,...,pn ,2 , rp0 ,...,pn ,2 ) = ∆p0 ,...,pn ,1 ∪ ∆p0 ,...,pn ,2 . Repeating the previous argument yields ∆p0 ,...,pn = ∆p0 ,...,pn ,1 ∪ ∆p0 ,...,pn ,2,1 ∪ ∆p0 ,...,pn ,2,2 = ∆p0 ,...,pn ,1 ∪ ∆p0 ,...,pn ,2,1 ∪ ∆p0 ,...,pn ,2,2,1 ∪ ∆p0 ,...,pn ,2,2,2 . Defining (pi , qi ) = (2, −1) for each i > n implies that for every k ≥ 0 we have ∆p0 ,...,pn =

k [

∆p0 ,...,pn+i ,1 ∪ ∆p0 ,...,pn+k ,2

i=0

and this union is disjoint. Considering Rp0 ,...,pn+k ,2 = p0 +

q0 ..

= p0 +

qn

. pn + 2−

1 ..

q0 ..

= Rp0 ,...,pn ,

. pn + qn

. 2−1

it follows that rp0 ,...,pn+k ,2 = rp0 ,...,pn . Moreover, according to Lemma 1.6, we have q0 q0 Sp0 ,...,pn+k ,2 = p0 + = p0 + , .. .. qn qn . pn + . pn + 1+(k+2)a 1 2− 1+(k+1)a .. 1 . 2− 1+a so Sp0 ,...,pn+k ,2 → Rp0 ,...,pn and sp0 ,...,pn+k ,2 → rp0 ,...,pn as k
→ ∞. This shows that the interval ∆p0 ,...,pn+k ,2 = sp0 ,...,pn+k ,2 , rp0 ,...,pn+k ,2 reduces to the empty set when k → ∞, hence ∆p0 ,...,pn =

∞ [

∆p0 ,...,pn+i ,1

i=0

S∞ and we can set C = i=0 ∆p0 ,...,pn+i ,1 . This completes the proof for the case p0 + · · · + pn being odd.

34

Generalized Notions of Continued Fractions

Assume now that p0 + · · · + pn is even, so ∆p0 ,...,pn = [rp0 ,...,pn , sp0 ,...,pn ). Then Lemma 3.3 implies the following chain of inequalities: rp0 ,...,pn = rp0 ,...,pn ,2 < sp0 ,...,pn ,2 = sp0 ,...,pn ,1 < rp0 ,...,pn ,1 = sp0 ,...,pn and shows ∆p0 ,...,pn = [rp0 ,...,pn ,2 , sp0 ,...,pn ,2 ) ∪ [sp0 ,...,pn ,1 , rp0 ,...,pn ,1 ) = ∆p0 ,...,pn ,2 ∪ ∆p0 ,...,pn ,1 . Repeating the previous argument we obtain ∆p0 ,...,pn = ∆p0 ,...,pn ,2,2 ∪ ∆p0 ,...,pn ,2,1 ∪ ∆p0 ,...,pn ,1 = ∆p0 ,...,pn ,2,2,2 ∪ ∆p0 ,...,pn ,2,2,1 ∪ ∆p0 ,...,pn ,2,1 ∪ ∆p0 ,...,pn ,1 . As done in the previous case, defining (pi , qi ) = (2, −1) for each i > n, it follows that for every k ≥ 0 we have ∆p0 ,...,pn =

k [

∆p0 ,...,pn+i ,1 ∪ ∆p0 ,...,pn+k ,2 .

i=0

Again, we have that rp0 ,...,pn+k ,2 = rp0 ,...,p  n and sp0 ,...,pn+k ,2 →
rp0 ,...,pn as k → ∞, so the interval ∆p0 ,...,pn+k ,2 = rp0 ,...,pn+k ,2 , sp0 ,...,pn+k ,2 reduces to {rp0 ,...,pn } when k → ∞. Therefore, ∆p0 ,...,pn =

∞ [

∆p0 ,...,pn+i ,1 ∪ {rp0 ,...,pn }

i=0

and we may set C =

S∞

∆p0 ,...,pn+i ,1 , which completes the proof. S∞ Lemma 3.6. Suppose that F = n=1 Fa,n and that P is the collection of intervals defined in Lemma 3.5. Let Q denote the collection formed by the empty set and all finite disjoint unions of sets [x, y) and {1} with x, y ∈ F. Then Q is an algebra that generates the Borel σ-field in [0, 1]. Moreover, for every E ∈ Q, there exists another set C ⊂ E such that C is a countable disjoint union of elements of P and E\C is at most finite. i=0

Proof. It is straightforward to check that Q is an algebra in [0, 1]. Moreover, according to Lemma 3.3, the endpoints of an interval ∆p0 ,...,pn belong to the Farey series, which shows ∆p0 ,...,pn ∈ Q. According to Lemma 3.4, the Borel σ-algebra in [0, 1) is generated by the collection of all intervals of the form ∆p0 ,...,pn , so the Borel σ-algebra in [0, 1] is generated by Q as well. Obviously, each element E ∈ Q is of the following types:

Ergodic aspects of the generalized Lehner continued fractions

35

ˆ E = {1}. ˆ E = [x1 , y1 ) ∪ · · · ∪ [xm , ym ), whereby the union is disjoint and xi , yi ∈ F for each i ∈ {1, . . . , m}. ˆ E = [x1 , y1 ) ∪ · · · ∪ [xm , ym ) ∪ {1}.

It is therefore enough to prove that for every interval [x, y) with x, y ∈ F there exists some set C ⊂ [x, y) such that C is a countable disjoint union of elements of P and [x, y)\C is at most finite. Again using Lemma 3.3, x, y ∈ F implies x = rp′0 ,...,p′n and y = rp′′0 ,...,p′′k for some p′0 , . . . , p′n , p′′0 , . . . , p′′k . That is, x and y are endpoints of two intervals of the class D. Considering x = rp′0 ,...,p′n = rp′0 ,...,p′n ,2 = rp′0 ,...,p′n ,2,2 = rp′0 ,...,p′n 2,2,...,2 and y = rp′′0 ,...,p′′k = rp′′0 ,...,p′′k ,2 = rp′′0 ,...,p′′k ,2,2 = rp′′0 ,...,p′′k ,2,2,...,2 , we may assume that n = k. Suppose that x = rp′0 ,...,p′n and y = rp′′0 ,...,p′′n . Then the index n fulfills o 
[n [x, y) = rp′0 ,...,p′n , rp′′0 ,...,p′′n = ∆p0 ,...,pn : ∆p0 ,...,pn ⊂ [x, y) and this union is finite and disjoint. Using Lemma 3.5, for each ∆p0 ,...,pn there exists some set C ⊂ ∆p0 ,...,pn such that C is a countable disjoint union of elements of P and ∆p0 ,...,pn \C is at most finite. This concludes the proof. Lemma 3.7. For x ∈ ∆p0 ,...,pn we have q0

1 + ax = p0 + p1 +

q1 ..

. pn +

.

qn 1 + aLn+1 (x) a

Proof. We again proveh the result by induction on n. For n = 0, if (p0 , q0 ) =  1 x (2, −1) and x ∈ ∆2 = 0, 1+a , we have La (x) = 1−ax , which implies 1 q0 = p0 + . 1 + aLa (x) 1 + aLa (x) h  1 If n = 0, (p0 , q0 ) = (1, a) and x ∈ ∆1 = 1+a , 1 , then La (x) = 1 + ax = 2 −

1 + ax = 1 +

a q0 = p0 + . 1 + aLa (x) 1 + aLa (x)

1−x ax ,

so

36

Generalized Notions of Continued Fractions

Suppose now that the statement holds for a natural number n − 1. If x ∈ ∆p0 ,...,pn , then x ∈ ∆p0 ,...,pn−1 and, by the induction hypothesis, it follows that q0 1 + ax = p0 + . (3.1) q1 p1 + .. qn−1 . pn−1 + 1 + aLna (x) Moreover, 1 + ax is contained in the interval formed by the following two points: q0 q0 Rp0 ,...,pn = p0 + , Sp0 ,...,pn = p0 + . .. .. qn−1 qn−1 . pn−1 + . pn−1 + qn pn + qn pn + 1+a qn . Therefore, 1 + aLna (x) belongs to the interval formed by pn + qn and pn + 1+a On the one hand, if (pn , qn ) = (2, −1), then 1 + aLna (x) is a point between 1 1 and 2 − 1+a , so 1 0 ≤ Lna (x) ≤ , 1+a

which implies that Ln+1 (x) = a 1 + aLna (x) = 2 −

Ln a (x) 1−aLn a (x)

as well as

1 qn = pn + . n+1 1 + aLa (x) 1 + aLn+1 (x) a

Using Equation (3.1), yields 1 + ax = p0 +

q0 ..

qn−1 . pn−1 + 1 + aLna (x)

= p0 +

q0 ..

. pn−1 +

qn−1 qn pn + 1 + aLn+1 (x) a

.

On the other hand, if (pn , qn ) = (1, a), then 1 + aLna (x) is a point between qn a pn + 1+a = 1 + 1+a and pn + qn = 1 + a, so 1 ≤ Lna (x) ≤ 1, 1+a which implies that Ln+1 (x) = a 1 + aLna (x) = 1 +

1−Ln a (x) aLn a (x)

and

a qn = pn + . 1 + aLn+1 (x) 1 + aLn+1 (x) a a

Using Equation (3.1) shows 1 + ax = p0 +

q0 ..

. pn−1 +

qn−1 1 + aLna (x)

so the statement also holds for n.

= p0 +

q0 ..

. pn−1 +

qn−1 qn pn + 1 + aLn+1 (x) a

,

Ergodic aspects of the generalized Lehner continued fractions

37

The proof of the subsequent Theorem 3.8 builds upon the results contained in [4, Section 24] and [10, Theorem 3.5.1]. Theorem 3.8. Let La be the function defined according to Equation (1.1). If B denotes the σ-algebra of Borel sets of [0, 1] and λ the Lebesgue measure on B, then the dynamical system ([0, 1], B, λ, La ) is ergodic. Proof. We prove the result by applying Theorem 3.2 with P being the collection of sets given in Lemma 3.5 and Q the algebra considered in Lemma 3.6. Fix an interval ∆p0 ,...,pn ∈ P, that is, (pn , qn ) = (1, a), and let An and Bn be the coefficients associated to (p0 , q0 ), . . . , (pn , qn ) according to the formulas from Lemma 1.1. Consider the function f : [0, 1] → R, defined by f (t) = p0 +

q0 ..

. pn−1 +

qn−1 qn pn + 1 + at

=

(1 + at)An + qn An−1 . (1 + at)Bn + qn Bn−1

Checking that f is well-defined can easily be done inductively by showing that Bn + qn Bn−1 > 0 holds. Obviously, B1 + q1 B0 = p1 + q1 ≥ 1. Assuming that Bi + qi Bi−1 > 0 for all i < n, yields Bn + qn Bn−1 = (pn Bn−1 + qn−1 Bn−2 ) + qn Bn−1 = (pn + qn ) Bn−1 + qn−1 Bn−2 ≥ Bn−1 + qn−1 Bn−2 > 0. As a consequence, (1 + at)Bn + qn Bn−1 ≥ Bn + qn Bn−1 > 0 holds for every t ∈ [0, 1]. The sign of f ′ is constant, so either f is increasing or f is decreasing on the full interval [0, 1]. We prove the result for the case that f is increasing, since the other case can be shown analogously. First note that f (0) = Rp0 ,...,pn and f (1) = Sp0 ,...,pn , so   f (0) − 1 f (1) − 1 , . ∆p0 ,...,pn = a a Furthermore, the length of ∆p0 ,...,pn is given by |f (1) − f (0)| 1 (1 + a)An + qn An−1 An + qn An−1 λ (∆p0 ,...,pn ) = = · − a a (1 + a)Bn + qn Bn−1 Bn + qn Bn−1 qn (An Bn−1 − An−1 Bn )

. = (1 + a)Bn + qn Bn−1 Bn + qn Bn−1

38

Generalized Notions of Continued Fractions

Using the determinant formula from Lemma 1.1, the latter simplifies to λ (∆p0 ,...,pn ) =

|q0 · · · qn |

. (1 + a)Bn + qn Bn−1 · Bn + qn Bn−1

Suppose that [u, v) is another element in P and let us determine the −(n+1) measure of the intersection ∆p0 ,...pn ∩ La ([u, v)). Using Lemma 3.7, x ∈ ∆p0 ,...pn implies q0

1 + ax = p0 + p1 +

q1 ..

. pn +


= f Ln+1 (x) . a

qn 1 + aLn+1 (x) a

−(n+1)

(x) < v. Therefore, ([u, v)), it follows that u ≤ Ln+1 Moreover, for x ∈ La a −(n+1) ([u, v)) if and only if 1 + ax = f (t) for some t ∈ [u, v). x ∈ ∆p0 ,...,pn ∩ La −(n+1) ([u, v)) is the interval formed by f (u)−1 That is, the set ∆p0 ,...,pn ∩ La a and f (v)−1 . Since f is increasing, we have a   f (u) − 1 f (v) − 1 ∆p0 ,...,pn ∩ L−(n+1) , . ([u, v)) = a a a The measure of the latter set is given by f (v) − f (u) λ ∆p0 ,...,pn ∩ = a 1 (1 + av)An + qn An−1 (1 + au)An + qn An−1 = · − a (1 + av)Bn + qn Bn−1 (1 + au)Bn + qn Bn−1 qn (An Bn−1 − An−1 Bn )(v − u)

. = (1 + av)Bn + qn Bn−1 (1 + au)Bn + qn Bn−1 

L−(n+1) ([u, v)) a



The determinant formula in Lemma 1.1 implies that   λ ∆p0 ,...,pn ∩ L−(n+1) ([u, v)) = a =

|q0 · · · qn |(v − u)

(1 + av)Bn + qn Bn−1 · (1 + au)Bn + qn Bn−1

|q0 · · · qn |(v − u)

(1 + a)Bn + qn Bn−1 · (1 + a)Bn + qn Bn−1 Bn + qn Bn−1 = λ (∆p0 ,...,pn ) · (v − u) · . (1 + a)Bn + qn Bn−1 ≥

Ergodic aspects of the generalized Lehner continued fractions

39

Recalling that (pn , qn ) = (1, a) since ∆p0 ,...,pn ∈ P, we finally get Bn + aBn−1 (1 + a)Bn + aBn−1 Bn + aBn−1 ≥ λ (∆p0 ,...,pn ) · (v − u) · (1 + a)(Bn + Bn−1 ) min{1, a} ≥ λ (∆p0 ,...,pn ) · λ ([u, v)) · . 1+a

  λ ∆p0 ,...,pn ∩ L−(n+1) ([u, v]) ≥ λ (∆p0 ,...,pn ) · (v − u) · a

According to Theorem 3.2, the dynamical system ([0, 1], B, λ, La ) is ergodic. The previous theorem implies the following stronger version of topological transitivity of continuous mappings as studied, e.g., in [42]. Thereby the set OLa (x) = x, La (x), L2a (x), . . . will be referred to as La -orbit of x. Theorem 3.9. For λ-almost every x ∈ [0, 1], the La -orbit OLa (x) of x is dense in [0, 1]. Proof. Let I1 , I2 , . . . denote an enumeration of all non-empty open subintervals of [0, 1] with rational endpoints. For every interval Ij , ergodicity of La implies the existence of a set Λj ∈ B with λ(Λj ) = 1 such that for every x ∈ Λj , we have n−1
1X lim XIj Lia (x) = λ(Ij ) > 0 n→∞ n i=0 T∞ (XIj denotes the indicator of the set Ij ). Setting Λ = j=1 Λj ∈ B yields λ(Λ) = 1 and the orbit of each x ∈ Λ is obviously dense in [0, 1] since x ‘visits’ each set Ij via La infinitely often. Theorem 3.10. Let γa be the σ-finite measure on [0, 1] defined by Z 1 γa (E) = dx E x for every E ∈ B. Then La is γa -preserving and the dynamical system ([0, 1], B, γa , La ) is ergodic. Proof. Since B is generated by the collection of all closed subintervals of [0, 1], it suffices to prove that La preserves the measure of [r, s] whenever 0 ≤ r < 1 , then s ≤ 1 (see [4, Section 24, Lemma 1]). On the one hand, if 0 ≤ x ≤ 1+a 1 1 − ax > 0. Hence, if 0 ≤ x ≤ 1+a and r ≤ La (x) ≤ s, then r≤ which implies that

x ≤ s, 1 − ax

r s ≤x≤ . 1 + ra 1 + sa

40

Generalized Notions of Continued Fractions

On the other hand, if

1 1+a

< x ≤ 1 and r ≤ La (x) ≤ s, then 1−x ≤ s, ax

r≤ which implies that

1 1 ≤x≤ . 1 + sa 1 + ra Therefore, L−1 a ([r, s])

   s 1 1 r , ∪ , = 1 + ra 1 + sa 1 + sa 1 + ra 

and the γa -measure of the latter set calculates to    
s 1 1 r , + γ , γa L−1 ([r, s]) =γ a a a 1 + ra 1 + sa 1 + sa 1 + ra     s r = log − log 1 + sa 1 + ra     1 1 + log − log 1 + ra 1 + sa = log(s) − log(r) = γa ([r, s]). This shows that La is γa -preserving. It has already been proved in Theorem 3.8 that ([0, 1], B, λ, La ) is ergodic. Since obviously γa (E) = 0 if and only if λ(E) = 0, it follows that ([0, 1], B, γa , La ) is also ergodic. ∞

Theorem 3.11. For every x ∈ [0, 1], let (bi )i=1 be the sequence assigned to x via  2 1   a if 0 ≤ Lia (x) ≤ 1+a bi =   1 if 1 < Li (x) ≤ 1. a a 1+a Then there exists a subset E in [0, 1] with λ(E) = 1 such that for every x ∈ E and α ̸= 0, we have !1/α n 1X α 2 lim b = n→∞ n i=1 i a as well as lim

n→∞

p n

b1 b2 · · · bn =

2 . a

Proof. Define the function f by

f (x) =

  0

if 0 ≤ x ≤

 1

if

1 1+a

1 1+a

< x ≤ 1.

Ergodic aspects of the generalized Lehner continued fractions

41

Then according to Theorem 3.1, there exists some f ∗ ∈ L1 ([0, 1], γa ) such that n−1
1X lim f Lia (x) = f ∗ (x) n→∞ n i=0 holds for almost every x ∈ [0, 1]. Since the system ([0, 1], B, γa , La ) is ergodic, Theorem 3.1 implies that f ∗ is constant almost everywhere. Considering γa ([0, 1]) = +∞ and that f ∗ is integrable, it follows that f ∗ must be zero γa -almost everywhere. This yields the existence of a Borel set E in [0, 1] fulfilling γa ([0, 1]\E) = 0 such that for every x ∈ E n n−1

1X 1X f Lia (x) = lim f Lia (x) = 0. n→∞ n n→∞ n i=1 i=0

lim

The fact that γa ([0, 1]\E) = 0 implies that λ ([0, 1]\E) = 0, so λ(E) = 1. Fix x ∈ E and set   1 Ix = i ∈ {1, . . . , n} : bi = a as well as

 Jx =

i ∈ {1, . . . , n} : bi =

2 a

 .



Note that f Lia (x) = 0 if bi = 2/a, while f Lia (x) = 1 if bi = 1/a. It therefore follows that n X


f Lia (x) = Card (Ix ) .

i=1

For x ∈ E, we have lim

n→∞

Card (Ix ) = 0. n

Therefore, n

1X α lim bi = lim n→∞ n n→∞ i=1

1 i∈Ix aα

P

n

P

i∈Jx

+ lim


2 α a

n

n→∞ α

1 Card (Ix ) 2 n − Card (Ix ) · lim + α · lim α n→∞ n→∞ a n a n 2α = α. a =

This implies n

lim

n→∞

for every α ̸= 0.

1X α b n i=1 i

!1/α =

2 a

42

Generalized Notions of Continued Fractions If α = 1, then lim

n→∞

b1 + · · · + bn 2 = n a

and if α = −1, then 1 b1

lim

+ ··· +

1 bn

!−1

n

n→∞

=

2 a

follows. It is well-known that the geometric mean is bounded between the harmonic mean and the arithmetic mean, so lim

n→∞

which completes the proof.

p n

b1 b2 · · · bn =

2 , a

Chapter 4 The a-simple continued fraction

In this chapter we again consider a real number a > 0 and study expansions of a point x ∈ (0, 1) in the form of 1

x=

a

1 + m1 a +

a

1 + m2 a +

1 + m3 a +

a .. .

with mi ∈ N∪{0} for every i ∈ N. We will call expressions of this type a-simple continued fractions and work with the function Ta : [0, 1] → [0, 1], defined by  0 if x = 0 Ta (x) = (4.1)

 1 1 − 1 − 1 1 − 1 if 0 < x ≤ 1, a x a x where [t] denotes the integer part of a number t (see Figure 4.1 for two examples). Theorem 4.1. Suppose that x ∈ [0, 1], that n ∈ N and that Tai (x) ̸= 0 for every i = 0, . . . , n − 1. Furthermore, for each i ∈ {1, . . . , n} set    1 1 mi = −1 a Tai−1 (x) and let An and Bn be defined recursively as follows: (i) A0 = 0, A1 = 1, An = (1 + mn a)An−1 + aAn−2 . (ii) B0 = 1, B1 = 1 + m1 a, Bn = (1 + mn a)Bn−1 + aBn−2 . Then we have 1

x=

=

a

1 + m1 a + 1 + m2 a +

a ..

An + aTan (x)An−1 . Bn + aTan (x)Bn−1

. 1 + mn a + aTan (x) An . n→∞ Bn

and if Tan (x) ̸= 0 for every n ∈ N, then x = lim DOI: 10.1201/9781003404064-4

43

44

Generalized Notions of Continued Fractions

FIGURE 4.1: The transformation Ta for a = 2 (left panel) and for a = (right panel).

1 2

Proof. First of all note that as a direct consequence of the determinant formula given in Lemma 1.1, we have An Bn−1 − An−1 Bn = (−a)n−1 ,

(4.2)

an identity which will be used frequently throughout this chapter. The definition of mi implies   1 1 Tai (x) = − 1 − mi , a Tai−1 (x) hence Tai−1 (x) =

1 1 + mi a + aTai (x)

holds for each i = 1, . . . , n. Altogether this shows x=

1 = 1 + m1 a + aTa (x) 1 + m1 a +

1 a 1 + m2 a + aTa2 (x)

1

=

.

a

1 + m1 a + 1 + m2 a +

a ..

. 1 + mn a + aTan (x)

If Tan (x) = 0, then Lemma 1.1 yields 1

x= 1 + m1 a +

a ..

. 1 + mn a

=

An + aTan (x)An−1 An = . Bn Bn + aTan (x)Bn−1

The a-simple continued fraction

45

If Tan (x) ̸= 0, again by Lemma 1.1, it follows that 1

x= 1 + m1 a +

=

=

a ..

1 Tan (x) An 1 T n (x) Bn a

. 1 + mn a +

+ aAn−1 + aBn−1

a 1/Tan (x)

An + aTan (x)An−1 . Bn + aTan (x)Bn−1

Suppose now that Tan (x) ̸= 0 for every n. Then the coefficients mn , An and Bn are well-defined for every n ∈ N. For each n ∈ N we have An An−2 (1 + mn a)An−1 + aAn−2 An−2 − = − Bn Bn−2 (1 + mn a)Bn−1 + aBn−2 Bn−2 (1 + mn a)[An−1 Bn−2 − An−2 Bn−1 ] = Bn Bn−2 and Equation (4.2) yields An An−2 (−a)n−2 (1 + mn a) − = . Bn Bn−2 Bn Bn−2 Thus, if n is even, then An An−2 − > 0, Bn Bn−2 while if n is odd, then An An−2 − < 0. Bn Bn−2 Using Equation (4.2), for n even we have An An−1 (−a)n−1 − = < 0. Bn Bn−1 Bn Bn−1 Therefore, A2 A4 A6 A5 A3 A1 < < < ··· < < < , B2 B4 B6 B5 B3 B1 which implies the existence of α = lim

n→∞

A2n , B2n

β = lim

n→∞

A2n−1 B2n−1

and shows that α ≤ β. Considering Bn−1 > Bn−2 , we additionally have An An−1 An Bn−1 − An−1 Bn |β − α| ≤ − = Bn Bn−1 Bn Bn−1 n−2 (−a)n−1 (1 + mn a) ≤ a · (−a) = , Bn Bn−1 Bn Bn−2

(4.3)

46

Generalized Notions of Continued Fractions

so Equation (4.3) yields An An−2 |β − α| ≤ a − −→ a |α − α| = 0. Bn Bn−2 n→∞ n even We have shown α = β and it remains to prove that x = α = β. Considering x=

An + aTan (x)An−1 Bn + aTan (x)Bn−1

yields x−

An An + aTan (x)An−1 An aTan (x) · [An−1 Bn − An Bn−1 ] = − = Bn Bn + aTan (x)Bn−1 Bn Bn2 + aTan (x)Bn Bn−1 n n−1 n aT (x) · (−(−a) ) Ta (x)(−a)n = 2a = Bn + aTan (x)Bn Bn−1 Bn2 + aTan (x)Bn Bn−1

and it follows that x > consequence,

An Bn

α = lim

if n is even and x
f (1) = . Bn Bn + taBn−1 This shows that the interval Im1 ,...,mn is well-defined in each situation. Using Lemma 1.1, we have 1 1 + m1 a +

=

a ..

An Bn

. 1 + mn a

as well as 1 1 + m1 a +

=

=

a ..

. 1 + mn−1 a +

(1 + (mn + 1)a)An−1 + aAn−2 (1 + (mn + 1)a)Bn−1 + aBn−2

a 1+(mn +1)a

(1 + mn a)An−1 + aAn−2 + aAn−1 An + aAn−1 = . (1 + mn a)Bn−1 + aBn−2 + aBn−1 Bn + aBn−1

The a-simple continued fraction

49

Therefore, if n is odd, then Im1 ,...,mn is the interval     1 + m1 a +

1

,

a

..

 1

1 + m1 a +

. 1+mn−1 a+ 1+(man +1)a

a

..

. 1+mn−1 a+ 1+ma n a

and if n is even, then Im1 ,...,mn is the interval     1 + m1 a +

1

,

a

..

  

 1

1 + m1 a +

. 1+mn−1 a+ 1+ma n a

 . 

a

..

. 1+mn−1 a+ 1+(man +1)a

These two expressions for Im1 ,...,mn will be used in the rest of the proof. As next step we tackle the following identity: Im1 ,...,mn = ∆2, . . . , 2,1,2, . . . , 2,1,...,2, . . . , 2,1 . | {z } m1

| {z }

(4.4)

| {z }

m2

mn

If n = 1, then the number 2 appears m1 times and, using Lemma 3.3, it follows that ∆2,...,2,1 = [s2,...,2,1 , r2,...,2,1 ). Lemma 1.6 implies S2,...,2,1 = 2 −

1 ..

. 2−

as well as

1 a 1 + 1+a 1

R2,...,2,1 = 2 −

..

. 2−

=

= 1 1+a

1 + (m1 + 1) · 1 + m1 ·

a 1+a

a 1+a

1 + (m1 + 1)a . 1 + m1 a

This yields s2,...,2,1 = and

S2,...,2,1 − 1 1 = a 1 + (m1 + 1)a

R2,...,2,1 − 1 1 = , a 1 + m1 a   1 1 , = Im1 = 1 + (m1 + 1)a 1 + m1 a

r2,...,2,1 = so ∆2,...,2,1

follows. Suppose now that (4.4) holds for some index n − 1 and let us prove that the same is true for n. In order to simplify the notation, we write R = R2, . . . , 2,1,2, . . . , 2,1,...,2, . . . , 2,1 , | {z } m1

| {z } m2

| {z } mn

r=

R−1 , a

50

Generalized Notions of Continued Fractions S = S2, . . . , 2,1,2, . . . , 2,1,...,2, . . . , 2,1 , | {z }

| {z }

m1

m2

| {z }

m2

R′ − 1 , a

s′ =

S′ − 1 . a

mn

S ′ = S2, . . . , 2,1,2, . . . , 2,1,...,2, . . . , 2,1 , | {z }

m2

r′ =

| {z }

m3

| {z }

S−1 , a

mn

R′ = R2, . . . , 2,1,2, . . . , 2,1,...,2, . . . , 2,1 , | {z }

s=

| {z }

| {z }

m3

mn

Suppose that n is even. Then r < s and s′ < r′ by Lemma 3.3 and Lemma 1.6 implies R=2−

1 ..

1 . 2− 1 + Ra′

Hence r=

=

1 + (m1 + 1) · 1 + m1 · Ra′

a R′

a . R ′ + m1 a

=1+

R−1 1 1 = ′ = a R + m1 a 1 + m1 a + ar′

follows. Deriving s=

1 1 + m1 a + as′

can be done analogously. Since we assumed that (4.4) holds for n − 1, which is odd, it follows that [s′ , r′ ) = ∆2, . . . , 2,1,...,2, . . . , 2,1 = Im2 ,...,mn | {z } m2

| {z } mn

  = 

 1 1 + m2 a +

,

a

..

1 1 + m2 a +

. 1+mn−1 a+ 1+(man +1)a

  

a

..

. 1+mn−1 a+ 1+ma n a

so  ∆2, . . . , 2,1,...,2, . . . , 2,1 = [r, s) = | {z } m1

| {z }

1 1 , 1 + m1 a + ar′ 1 + m1 a + as′



mn

  = 

 1 1 + m1 a +

a

..

. 1+mn a

,

1 1 + m1 a +

a

..

  

. 1+(mn +1)a

= Im1 ,...,mn . We have shown that Equation (4.4) holds for n too. Since the argument can be easily adapted to the case of n being odd, the proof of the Lemma is complete.

The a-simple continued fraction

51

As it was done in the proof of Theorem 3.8, for establishing the following result we follows the ideas going back to [4, Section 24] and [10, Theorem 3.5.1]. Theorem 4.4. Letting B denote the σ-algebra of Borel sets in [0, 1] and λ the Lebesgue measure on B, the dynamical system ([0, 1], B, λ, Ta ) is ergodic. Proof. We prove the result via applying Theorem 3.2 with P being the collection of sets studied in Lemma 4.3. Fix m ∈ N as well as non-negative integers m1 , m2 , . . . , mn and let An and Bn be defined according to the formulas from Theorem 4.1. We will assume that n is even, since the arguments for n odd are essentially the same. Then the interval Im1 ,...,mn is given by   An An + aAn−1 , Im1 ,...,mn = Bn Bn + aBn−1 and the function f (t) =

An + taAn−1 Bn + taBn−1

is increasing. The measure of Im1 ,...,mn calculates to An + aAn−1 An a · |An−1 Bn − An Bn−1 | = , λ (Im1 ,...,mn ) = − Bn + aBn−1 Bn Bn (Bn + aBn−1 ) and using Equation (4.2), we have λ (Im1 ,...,mn ) =

an . Bn (Bn + aBn−1 )

(4.5)

Suppose that [u, v) is another element of P. If x ∈ Im1 ,...,mn , then x=

An + aTan (x)An−1 = f (Tan (x)) . Bn + aTan (x)Bn−1

Moreover, if x ∈ Ta−n ([u, v)), then u ≤ Tan (x) < v, so x ∈ Im1 ,...,mn ∩ Ta−n ([u, v)) if and only if x = f (t) for some t ∈ [u, v). Considering that n is even, the function f is increasing and we have Im1 ,...,mn ∩ Ta−n ([u, v)) = [f (u), f (v)). Computing the measure of the latter set yields
λ Im1 ,...,mn ∩ Ta−n ([u, v)) = |f (v) − f (u)| An + vaAn−1 An + uaAn−1 = − Bn + vaBn−1 Bn + uaBn−1 a (An−1 Bn − An Bn−1 ) (v − u) . = (Bn + vaBn−1 ) (Bn + uaBn−1 )

52

Generalized Notions of Continued Fractions

Using Equation (4.2), it follows that
λ Im1 ,...,mn ∩ Ta−n ([u, v)) =

an (v − u) . (Bn + vaBn−1 ) (Bn + uaBn−1 )

Since all terms in the last expression are positive and 0 ≤ u < v ≤ 1 holds, we conclude that
λ Im1 ,...,mn ∩ Ta−n ([u, v)) ≥

an (v − u) . (Bn + aBn−1 ) (Bn + aBn−1 )

Using Equation (4.5) and Bn > Bn−1 , we finally obtain Bn Bn + aBn−1 1 ≥ λ (Im1 ,...,mn ) · λ([u, v)) · . 1+a


λ Im1 ,...,mn ∩ Ta−n ([u, v)) ≥ λ (Im1 ,...,mn ) · λ([u, v)) ·

Applying Theorem 3.2, it follows that the dynamical system ([0, 1], B, λ, Ta ) is ergodic. The next result presents an generalization of the Gauss measure for which the mapping Ta is measure-preserving. Theorem 4.5. Let µa be the probability measure on B defined by Z 1 a µa (E) = dx. log(1 + a) E 1 + ax Then the function Ta is measure-preserving on ([0, 1], B, µa ). Moreover, the dynamical system ([0, 1], B, µa , Ta ) is ergodic. Proof. In order to prove that Ta is measure-preserving, it suffices to show that
µa Ta−1 (I) = µa (I) for each interval I in [0, 1] (see [4, Section 24, Lemma 1]). For 0 ≤ r < s ≤ 1, we obviously have µa ([r, s]) = as well as Ta−1 ([r, s]) =

log(1 + as) − log(1 + ar) log(1 + a)

∞  [ k=0

 1 1 , . 1 + ak + as 1 + ak + ar

The a-simple continued fraction Having this, it follows that  ∞ log 1 + X
µa Ta−1 ([r, s]) =

a 1+ak+ar



 − log 1 +

53

a 1+ak+as



log(1 + a)

k=0 ∞

=

=

Xh 1 log(1 + a(k + 1) + ar) − log(1 + ak + ar) log(1 + a) k=0 i − log(1 + a(k + 1) + as) + log(1 + ak + as) − log(1 + ar) + log(1 + as) = µa ([r, s]), log(1 + a)

which shows that Ta is measure-preserving on ([0, 1], B, µa ). We already know from Theorem 4.4 that ([0, 1], B, λ, Ta ) is ergodic, hence, considering that µa (E) = 0 if and only if λ(E) = 0, it follows that ([0, 1], B, µa , Ta ) is ergodic too. The following two theorems extend analogous results going back to Khintchine in [24]. Theorem 4.6. There exists a Borel subset E of [0, 1] fulfilling λ(E) = 1 such that for every x ∈ E we have n

1X mi = +∞, n→∞ n i=1 lim

∞

whereby (mi )i=1 is the sequence corresponding to x according to Theorem 4.1. Proof. For every N ∈ N, define the function fN : [0, 1] → [0, 1] by   0 if 0 ≤ x ≤ 1+(N1+1)a       1 N if 1+(N1+1)a < x ≤ 1+N  a  fN (x) = ...    1 1  < x ≤ 1+a 1 if 1+2a       1 < x ≤ 1. 0 if 1+a According to Theorem 3.1, there exists some Borel subset EN of [0, 1] fulfilling µa (EN ) = 1 as well as n
1X fN Tai−1 (x) = n→∞ n i=1

Z

lim

1

fN dµa 0

for every x ∈ EN . Since F is countable we may assume without loss of generality that EN ∩ F = ∅. Suppose that x ∈ EN . Then x ∈ / F, so Tai (x) = ̸ 0 for

54

Generalized Notions of Continued Fractions

every i ∈ N, each coefficient mi is well-defined, and 1 1 < Tai−1 (x) ≤ . 1 + (mi + 1)a 1 + mi a
Note that for
mi ≤ N we have fN Tai−1 (x) = mi , while for mi > N we have fN Tai−1 (x) = 0. This shows Z 1 n n
1X 1X i−1 mi ≥ lim fN Ta (x) = fN dµa lim n→∞ n n→∞ n 0 i=1 i=1 for every x ∈ EN and the latter integral can be bounded from below as follows:   1 Z 1 N N Z X X 1+ak 1 1 , k dµa = k · µa fN dµa = 1 1 + a(k + 1) 1 + ak 0 k=1 k=1 1+a(k+1)     a a N log 1 + 1+ak − log 1 + 1+a(k+1) X k· = log(1 + a) k=1   N X k (1 + a(k + 1))2 = · log log(1 + a) (1 + ak) · (1 + a(k + 2)) k=1   N X k a2 = · log 1 + . log(1 + a) (1 + ak) · (1 + a(k + 2)) k=1

If 0 ≤ t ≤ 1, then log(1 + t) ≥ 2t , so Z

1

fN dµa ≥ 0

N X k=1

k a2 · log(1 + a) 2(1 + ak)(1 + a(k + 2)) N

≥

X a2 k . 2 log(1 + a) (1 + a(k + 2))2 k=1

T∞

Setting E = N =1 EN yields µa (E) = 1 as well as λ(E) = 1. Moreover, for every x ∈ E and every N ∈ N, we have Z 1 n N X a2 k 1X mi ≥ fN dµa ≥ . lim n→∞ n 2 log(1 + a) (1 + a(k + 2))2 0 i=1 k=1

Since the series

∞ P k=1

k (1+a(k+2))2

n P 1 mi n n→∞ i=1

is divergent, we conclude that lim

=

+∞ holds for every x ∈ E and the proof is complete. Theorem 4.7. There exist a constant C > 0 and a Borel set E in [0, 1] fulfilling λ(E) = 1 and for every x ∈ E, we have n

1X 1 = C, n→∞ n 1 + mi a i=1 lim

The a-simple continued fraction

55

∞

where (mi )i=1 is the sequence corresponding to x according to Theorem 4.1. Proof. Consider the function f : [0, 1] → [0, 1], defined by  1 1 1  1+ak if 1+a(k+1) < x ≤ 1+ak for some k ∈ N ∪ {0} f (x) = 0 if x = 0. According to Theorem 3.1, there exists some Borel set E in [0, 1]\F fulfilling µa (E) = 1 such that Z 1 n
1X lim f Tai−1 (x) = f dµa n→∞ n 0 i=1 holds for every x ∈ E. Then obviously λ(E) = 1 and for x ∈ E we have that every mi is well-defined and fulfills 1 1 < Tai−1 (x) ≤ , 1 + (mi + 1)a 1 + mi a
so f Tai−1 (x) =

1 1+mi a .

Therefore,

n n
1X 1 1X f Tai−1 (x) = = lim n→∞ n 1 + mi a n→∞ n i=1 i=1

Z

lim

1

f dµa 0

holds for every x ∈ E and the integral on the right-hand side is finite because f is bounded and µa is a probability measure. The subsequent result extends a theorem going back to Levy [31]. Theorem 4.8. For each k ∈ N ∪ {0}, there exists a Borel set Ek in [0, 1] fulfilling λ(Ek ) = 1 such that     a a log 1 + 1+ak − log 1 + 1+a(k+1) Card {i : 1 ≤ i ≤ n, mi = k} lim = n→∞ n log(1 + a) ∞

holds for every x ∈ Ek , where (mi )i=1 is the sequence corresponding to x according to Theorem 4.1. Proof. The result is a consequence of the ergodic theorem. In fact, consider the function f , defined by  1 1 1 if 1+a(k+1) < x ≤ 1+ak f (x) = 0 in other case. Then there exists some Borel subset Ek of [0, 1]\F fulfilling µa (Ek ) = 1 such that Z 1 n
1X lim f Tai−1 (x) = f dµa n→∞ n 0 i=1

56

Generalized Notions of Continued Fractions

holds for every x ∈ Ek . Then λ(Ek ) = 1 follows and given x ∈ Ek we have

f Tai−1 (x) = 1 if mi = k, while f Tai−1 (x) = 0 if mi = ̸ k. This implies n X


f Tai−1 (x) = Card {i : 1 ≤ i ≤ n and mi = k} ,

i=1

as well as Card {i : 1 ≤ i ≤ n, mi = k} lim = n→∞ n

Z 0

1

f dµa 

 1 1 , = µa 1 + a(k + 1) 1 + ak     a a − log 1 + 1+a(k+1) log 1 + 1+ak = , log(1 + a) which completes the proof.

Chapter 5 The generalized Khintchine constant

This chapter focuses on generalizing the famous Khintchine’s constant. Theorem 5.1. There exists a Borel set E in [0, 1] fulfilling λ(E) = 1 such that v u n  log(1+ak) ∞  2 log(1+a) Y uY (1 + a(k + 1)) n lim t (1 + mi a) = n→∞ (1 + ak) · (1 + a(k + 2)) i=1 k=1

∞

holds for every x ∈ E, where (mi )i=1 is the sequence corresponding to x according to Theorem 4.1. Proof. Consider the function f : [0, 1] → [0, 1], defined by  1 1 log(1 + ak) if 1+a(k+1) < x ≤ 1+ak for some k ∈ N ∪ {0} f (x) = 0 if x = 0. The function f is integrable with respect to the measure µa : Z 1 Z 1 Z 1 1 a a f dµa = f (x) · dx ≤ f (x)dx log(1 + a) 0 1 + ax log(1 + a) 0 0 ∞ X a2 log(1 + ak) = < +∞ log(1 + a) (1 + ak)(1 + a(k + 1)) k=1

(this last series is finite by the integral test). Then, according to Theorem 3.1, there exists a Borel set E in [0, 1]\F fulfilling µa (E) = 1 such that for x ∈ E Z 1 n n
1X 1X i−1 lim log(1 + mi a) = lim f Ta (x) = f dµa n→∞ n n→∞ n 0 i=1 i=1   ∞ X 1 1 = log(1 + ak) · µa , 1 + a(k + 1) 1 + ak k=1   ∞ X (1 + a(k + 1))2 log(1 + ak) = · log log(1 + a) (1 + ak)(1 + a(k + 2)) k=1

DOI: 10.1201/9781003404064-5

57

58

Generalized Notions of Continued Fractions

holds. As a direct consequence, λ(E) = 1 and v u n  log(1+ak) ∞  log(1+a) Y uY (1 + a(k + 1))2 n t lim (1 + mi a) = n→∞ (1 + ak)(1 + a(k + 2)) i=1 k=1

for every x ∈ E follows. The constant Ka =

∞  Y k=1

(1 + a(k + 1))2 (1 + ak)(1 + a(k + 2))

 log(1+ak) log(1+a) (5.1)

will be referred to as generalized Khintchine constant. Our main objective for this chapter is to establish some identities involving Ka which generalize the ones that Shanks and Wrench derived for K1 in [39]. We start with two auxiliary lemmas which will be used in the sequel. Lemma 5.2. If −1 < x < 1, then 2

log 1 − x




∞ X −x2n = . n n=1

Moreover, for these x, the identity ∞ X −dn 2n x log(1 − x) log(1 + x) = n n=1

holds, whereby the coefficient dn is given by dn =

2n−1 X i=1

(−1)i+1 > 0. i

Proof. The first statement is a direct consequence of the well-known fact that log(1 − x) =

∞ X −xm m m=1

holds for every −1 < x < 1. In order to derive the second one, first note that ! ! ∞ ∞ X X −xm (−1)k+1 xk log(1 − x) log(1 + x) = · m k m=1 k=1

=

∞ X

X

n=2 m+k=n m≥1, k≥1

∞ n−1 (−1) n X X (−1)k n x = x . mk (n − k)k n=2 k

k=1

The generalized Khintchine constant

59

As next step we show that the coefficient of xn is zero if n is odd. Suppose therefore, that n = 2N + 1 for some integer N ≥ 0. Then we obviously have n−1 X k=1

N 2N X X (−1)k (−1)k (−1)k = + (n − k)k (2N + 1 − k)k (2N + 1 − k)k k=1

=

N X k=1

=

N X k=1

k=N +1 N

k

X (−1)2N +1−i (−1) + (2N + 1 − k)k i=1 i(2N + 1 − i) N

X (−1)k (−1)i − = 0, (2N + 1 − k)k i=1 i(2N + 1 − i)

so the Taylor series of log(1 − x) log(1 + x) at 0 only contains powers with even exponent, which yields log(1 − x) log(1 + x) =

∞ 2n−1 X X n=1 k=1

∞ X (−1)k −dn 2n x2n = x , (2n − k)k n n=1

whereby each coefficient dn is given by dn =

2n−1 X k=1

(−1)k+1 n . (2n − k)k

As a direct consequence, we have X X n n dn = − (2n − k)k (2n − k)k k odd

=

=

n  X

n X r=1

=

n−1

X n n + (2n − 2p + 1)(2p − 1) q=1 (2n − 2q)2q p=1

p=1

=

k even

n X

1 1 + 4n − 4p + 2 4p − 2 n X

 −

n−1 X

n−1 X q=1

1 1 + 4n − 4q 4q



n−1

X 1 1 1 1 + − − 4r − 2 p=1 4p − 2 s=1 4s q=1 4q

n X

n−1 2n−1 X 1 X (−1)i+1 1 − = . 2p − 1 q=1 2q i p=1 i=1

This completes the proof. In the following results we will work with the Hurwitz zeta function, which is defined as ∞ X 1 ζ(s, a) = , (k + a)s k=0

60

Generalized Notions of Continued Fractions

whereby a is a strictly positive real and s is a complex number with Re(s) > 1. For main properties of the ζ function, we refer the reader to [1, Chapter 12]. Lemma 5.3. Suppose that a and b are positive real numbers, that x ∈ [0, 1] and that Γ denotes the gamma function. Then the following identities hold: (i) log(1 + a) =

∞ P n=1

1 n



ζ 2n, a1 − a2n and the series on the right-hand

side converges.  2   ∞ Q x = (ii) 1 − n+b n=1

Γ2 (1+b) Γ(1+b+x)Γ(1+b−x) .

Proof. First of all notice that for n ∈ N, we have 2 !      n n X X a a a =− log 1 + · 1− − log 1 − 1 + ak 1 + ak 1 + ak k=1

k=1

n X   2 log(1 + ak) − log(1 + a(k + 1)) − log(1 + a(k − 1)) = k=1

= log(1 + a) + log(1 + an) − log(1 + a(n + 1))   1 + an = log(1 + a) + log . 1 + a(n + 1) Considering n → ∞, this shows log(1 + a) = −

∞ X

 log 1 −

k=1

a 1 + ak

2 ! .

Furthermore, Lemma 5.2 implies log(1 + a) =

 2n ∞ X ∞ X 1 a . n 1 + ak n=1

k=1

Considering that every summand is positive, we may interchange order and obtain 2n X ∞ ∞  ∞ ∞ X a 1X 1 1X = log(1 + a) =
2n 1 n 1 + ak n n=1 n=1 k=1 k=1 a + k     ∞ X 1 1 2n = ζ 2n, −a , n a n=1 which proves (i). In order to show (ii), we use the identity Γ(z) =

∞ e−γz Y  z −1 z/n 1+ e , z n=1 n

The generalized Khintchine constant

61

where γ denotes the Euler constant (see [1, Chapter 12]), to obtain −1 ∞  1+b e−γ(1+b) Y n + 1 + b Γ(1 + b) = e n , 1 + b n=1 n  ∞  Y 1 n + 1 + b + x − 1+b+x n , = (1 + b + x)eγ(1+b+x) e Γ(1 + b + x) n n=1  ∞  Y 1 n + 1 + b − x − 1+b−x γ(1+b−x) n . = (1 + b − x)e e Γ(1 + b − x) n n=1 Altogether this yields ∞

Γ2 (1 + b) (1 + b)2 − x2 Y (n + 1 + b)2 − x2 = Γ(1 + b + x)Γ(1 + b − x) (1 + b)2 n=1 (n + 1 + b)2 Y   ∞  x2 x2 1− = 1− (1 + b)2 n=1 (n + 1 + b)2 !  2 ∞ Y x = 1− n+b n=1 and the proof is complete. Theorem 5.4. For every a > 0, the generalized Khintchine constant satisfies     ∞ X −1 a a log(Ka ) = log 1 − · log 1 + . log(1 + a) 1 + ak 1 + ak k=1

Proof. Considering Ka =

∞  Y k=1

(1 + a(k + 1))2 (1 + ak)(1 + a(k + 2))

yields log(Ka ) =

∞ X log(1 + ak) k=1

where

log(1 + a)

 log(1+ak) log(1+a)

log(tk ),

(5.2)

(1 + a(k + 1))2 (1 + ak)(1 + a(k + 2)) for each k ≥ 1. It is straightforward to check that tk > 1 for every k ∈ N, so log(tk ) > 0. Thus, every summand in the following series is positive and we may again interchange the order of summation. For each pair (n, N ) of natural numbers with n < N , we have   N X 1 + a(n + 1) 1 + a(N + 1) log(tk ) = log (tn · · · tN ) = log · , 1 + an 1 + a(N + 2) tk =

k=n

62

Generalized Notions of Continued Fractions

implying

∞ X

 log(tk ) = log

k=n

1 + a(n + 1) 1 + an

 .

Substituting this identity yields     ∞ X 1 + an 1 + a(n + 1) log · log = 1 + a(n − 1) 1 + an n=1   X ∞ ∞ X 1 + an = log · log(tk ) 1 + a(n − 1) n=1 k=n   ∞ X k X 1 + an = log · log(tk ) 1 + a(n − 1) k=1 n=1 ! ∞ k ∞ X Y X 1 + an = log · log(tk ) = log(1 + ak) · log(tk ). 1 + a(n − 1) n=1 k=1

k=1

According to Equation (5.2), the last series is equal to log(1 + a) log(Ka ), i.e.,     ∞ X 1 + an 1 + a(n + 1) log · log = log(1 + a) log(Ka ). 1 + a(n − 1) 1 + an n=1 Altogether this shows ∞ X

   1 + a(n − 1) 1 + a(n + 1) · log 1 + an 1 + an n=1     ∞ X a a =− log 1 − · log 1 + , 1 + an 1 + an n=1

log(1 + a) log(Ka ) = −



log

which completes the proof. Theorem 5.5. Setting dn =

2n−1 P i=1

(−1)i+1 i

for each n ∈ N, the following iden-

tity holds:     ∞ X dn 1 2n log(1 + a) log(Ka ) = ζ 2n, −a . n a n=1 Proof. According to Theorem 5.4, we have     ∞ X a a log(1 + a) log(Ka ) = − log 1 − · log 1 + , 1 + ak 1 + ak k=1

so Lemma 5.2 implies  2n ∞ X ∞ X dn a log(1 + a) log(Ka ) = . n 1 + ak n=1 k=1

The generalized Khintchine constant

63

Interchanging the summation order therefore yields  2n X ∞ X ∞ ∞ ∞ X dn a dn X 1 log(1 + a) log(Ka ) = =
1 2n n 1 + ak n n=1 k=1 n=1 k=1 k + a     ∞ X dn 1 = ζ 2n, − a2n , n a n=1 which completes the proof. Theorem 5.6. For every a > 0, the generalized Khintchine constant satisfies the following identity:

log(1 + a) log(Ka ) = log(1 + a) −

k ζ 2n, 1 −a2n P ( a) ∞ log(1 + a) − n X n=1

2k(2k + 1)

k=1

.

Proof. Theorem 5.5 implies ∞ X

2n−1 X

!


ζ 2n, a1 − a2n log(1 + a) log(Ka ) = · n n=1 i=1 !
 ∞ n−1 X X 1 ζ 2n, a1 − a2n 1 = 1− − · 2k 2k + 1 n n=1 k=1

∞ ∞ n−1 X ζ 2n, a1 − a2n X X ζ 2n, a1 − a2n 1 = − · . n 2k(2k + 1) n n=1 n=1 (−1)i+1 i

k=1

Therefore, using Lemma 5.3 and interchanging the sums it follows that
∞ ∞ X X ζ 2n, a1 − a2n 1 log(1 + a) log(Ka ) = log(1 + a) − · . 2k(2k + 1) n k=1 n=k+1

Again applying Lemma 5.3 yields the desired expression via log(1 + a) log(Ka ) = ∞ X

"∞ #

k X ζ 2n, 1 − a2n X ζ 2n, a1 − a2n 1 a = log(1 + a) − · − 2k(2k + 1) n=1 n n n=1 k=1 " #
∞ k X X ζ 2n, a1 − a2n 1 = log(1 + a) − · log(1 + a) − . 2k(2k + 1) n n=1 k=1

Theorem 5.7. For every a > 0, the generalized Khintchine constant satisfies

!   Z 1 Γ 1 + a1 + x Γ 1 + a1 − x Ka dx
log(1 + a) log = log . 1 2 2 x(1 + x) Γ 1+ a 0

64

Generalized Notions of Continued Fractions

Proof. First notice that according to Theorem 5.5, we have !
∞ 2n−1 X X (−1)i+1 ζ 2n, a1 − a2n log(1 + a) log(Ka ) = · . i n n=1 i=1 Therefore, defining the function f by f (x) =

2n−1 X i=1

(−1)i+1 i x, i

yields 2n−1 X i=1

Z 1 Z 1 2n−1 X (−1)i+1 ′ = f (1) − f (0) = f (x)dx = (−x)i−1 dx i 0 0 i=1 Z 1 2n−1 Z 1 x 1 + x2n−1 dx = log(2) + dx = 1+x 0 1+x 0

as well as ∞  X


 ζ 2n, a1 − a2n x2n−1 log(1 + a) log(Ka ) = log(2) + dx · n 0 1+x n=1

Z ∞ ∞ X ζ 2n, 1 − a2n X 1 ζ 2n, 1 − a2n x2n−1 a a = log(2) + · dx. n n 1+x n=1 n=1 0 Z

1

Using Lemma 5.3, we have log(1 + a) log(Ka ) = log(2) log(1 + a) +

∞ Z X n=1

0

1


ζ 2n, a1 − a2n x2n−1 · dx, n 1+x

hence  log(1 + a) log

Ka 2

 =

∞ Z X n=1

0

1


ζ 2n, a1 − a2n x2n−1 · dx n 1+x

follows. Moreover, for every n ∈ N and every x ∈ [0, 1], we have
ζ 2n, a1 − a2n x2n−1 · ≥ 0. n 1+x Using monotone convergence and interchanging sum and the integral (see [4, Theorem 16.6]), we conclude that
  Z 1X ∞ ζ 2n, a1 − a2n x2n−1 Ka log(1 + a) log = · dx 2 n 1+x 0 n=1 Z 1 ∞ X ∞ X 1 x2n · = dx
1 2n 0 x(1 + x) n=1 k=1 k + n a  2n Z 1 ∞ X ∞ X 1 1 x = · · dx. k + a1 0 x(1 + x) k=1 n=1 n

The generalized Khintchine constant

65

Applying Lemma 5.2, we have  log(1 + a) log

Ka 2



Z

1

= 0

Z

1

= 0

"  2 # ∞ X −1 x · log 1 − dx x(1 + x) k + a1 k=1 "∞  2 !# Y −1 x · log 1− dx, x(1 + x) k + a1 k=1

which using Lemma 5.3 implies  log(1 + a) log

Ka 2



Z

1

= 0

Z = 0

This completes the proof.

1

" Γ2 1 + −1
· log 1 x(1 + x) Γ 1+ a +x Γ "
Γ 1 + a1 + x Γ 1 · log x(1 + x) Γ2 1 +

1 a

#




1+

1 a

1+
1

1 a

a


dx −x
# −x dx.

Chapter 6 The entropy of the system ([0, 1], B, µa, Ta)

Recall that given a finite partition α = {A1 , . . . , An } of a probability space (X, A, µ), the entropy H(α) of α is given by H(α) = −

n X

µ(Ai ) · log(µ(Ai ))

i=1

(with the convention 0 · log(0) = 0) and can be interpreted as quantification of the uncertainly of an experiment with outcomes {A1 , . . . , An }. Letting α ∨ β denote the join of the partitions α = {A1 , . . . , An } and β = {B1 , . . . , Bm }, defined by α ∨ β = {Ai ∩ Bj : 1 ≤ i ≤ n, 1 ≤ j ≤ m}  −1 as well as T (α) = T −1 (Ai ) : 1 ≤ i ≤ n , then following [42] the entropy h(T, α) of a µ-preserving transformation T with respect to the partition α is defined as ! N_ −1 1 −i h(T, α) = lim H T (α) . N →∞ N i=0 Finally, the Kolmogorov–Sinai metric or measure-theoretic entropy of the dynamical system (X, A, µ, T ) is defined as the worst-case scenario over all finite partitions, i.e., h(T ) = sup h(T, α). (6.1) α

For more details and additional background on the entropy of dynamical systems, we refer to [10, Chapter 6] and [42]. Theorem 6.1. The entropy of the dynamical system ([0, 1], B, µa , Ta ) is given by 2Li2 (−a) h(Ta ) = − log(a) − , log(1 + a) where Li2 denotes the dilogarithm function, defined by Z z log(1 − t) Li2 (z) = − dt. t 0

DOI: 10.1201/9781003404064-6

67

68

Generalized Notions of Continued Fractions n o 1 Proof. For x ∈ / {0, 1} ∪ 1+na : n ∈ N , the derivative of Ta at x is Ta′ (x) = −1 ax2 . Therefore, applying Rohlin formula (see [35, Theorem 12.10]), the entropy of ([0, 1], B, µa , Ta ) is given by

Z h(Ta ) =

1

log |Ta′ (x)| dµa (x)

0

and it follows that h(Ta ) =

1 log(1 + a)

Z

− log(a) log(1 + a)

Z

1

 log

0

1 ax2

 ·

a dx 1 + ax

1

2 a dx − 1 + ax log(1 + a) 0 Z 1 2 a log(x) = − log(a) − dx. log(1 + a) 0 1 + ax

=

Z

1

0

a log(x) dx 1 + ax

Expressing the last integral in terms of Li2 yields Z 1 Z 1 Z −a a log(x) log(1 + ax) log(1 − t) dx = − dx = − dt = Li2 (−a), 1 + ax x t 0 0 0 which altogether shows that h(Ta ) = − log(a) −

2 Li2 (−a). log(1 + a)

Lemma 6.2. If {En : n ∈ N} is a sequence of Borel sets in [0, 1] such that λ(En ) > 0 for every n ∈ N and lim λ(En ) = 0, then we have n→∞

lim

n→∞

log (µa (En )) = 1. log (λ(En ))

Proof. On the one hand, obviously Z Z a a aλ(En ) µa (En ) = dx ≤ dx = log(1 + a) En (1 + ax) log(1 + a) En log(1 + a) holds and, on the other hand, we have Z a aλ(En ) µa (En ) ≥ dx = . (1 + a) log(1 + a) En (1 + a) log(1 + a) This shows

aλ(En ) aλ(En ) ≤ µa (En ) ≤ (1 + a) log(1 + a) log(1 + a)

The entropy of the system ([0, 1], B, µa , Ta )

69

as well as   a log + log (λ(En )) ≤ log (µa (En )) (1 + a) log(1 + a)   a ≤ log + log (λ(En )) . log(1 + a) Dividing by log (λ(En )) we therefore obtain i i h h a a log (1+a) log(1+a) log log(1+a) log (µa (En )) +1≤ ≤ + 1, log (λ(En )) log (λ(En )) log (λ(En )) which, considering λ(En ) → 0 and log (λ(En )) → −∞, implies that lim

n→∞

log (µa (En )) = 1. log (λ(En ))

The next three theorems extend the corresponding identities going back to Levy (see [10, Proposition 3.5.5]). Theorem 6.3. Let F be the collection of all points belonging to the a-modified Farey series and for each x ∈ [0, 1]\F let In (x) denote the collection of all points y ∈ [0, 1]\F fulfilling       1 1 1 1 −1 = −1 a Tai (y) a Tai (x) for every i = 0, . . . , n − 1. Then log (λ(In (x))) 2Li2 (−a) = log(a) + n→∞ n log(1 + a) lim

holds or almost every x ∈ [0, 1]. Proof. We will apply the well-known Shannon-Mcmillan-Breiman Theorem (see, e.g., [3, Section 13] or [35, Corollary 12.11.1]), which implies that the entropy of ([0, 1], B, µa , Ta ) is given by h(Ta ) = lim

n→∞

− log (µa (In (x))) n

for almost every x. The result now follows by applying Lemma 6.2 and Theorem 6.1 via lim

n→∞

− log (λ(In (x))) − log (µa (In (x))) = lim = h(Ta ) n→∞ n n 2Li2 (−a) . = − log(a) − log(1 + a)

70

Generalized Notions of Continued Fractions

Theorem 6.4. There exists a Borel set E in [0, 1] with λ(E) = 1 fulfilling the An following property: if x ∈ E and B is the nth convergent of the expansion of n x as an a-simple continued fraction, then the following identity holds: An log x − B 2Li2 (−a) n = log(a) + . lim n→∞ n log(1 + a) Proof. According to Theorem 6.3, there exists some Borel subset E of [0, 1]\F with λ(E) = 1 fulfilling log (λ(In (x))) 2Li2 (−a) = log(a) + n→∞ n log(1 + a) lim

for every x ∈ E. Let x ∈ E and let n ∈ N be an even number. Theorem 4.1 implies An + aTan (x)An−1 x= Bn + aTan (x)Bn−1 and in the proof of Lemma 4.3, it has already been shown that for even n the function An + taAn−1 f (t) = Bn + taBn−1 is increasing. As a direct consequence we have An An + aAn−1 = f (0) < f (Tan (x)) < f (1) = , Bn Bn + aBn−1 so

An An + aAn−1 0, Bn+1 + aBn Bn (Bn+1 + aBn )Bn (Bn+1 + aBn )Bn

The entropy of the system ([0, 1], B, µa , Ta )

71

which altogether shows that An An+1 + aAn < < x. Bn Bn+1 + aBn It follows that An An an An+1 + aAn ≥ = − x − Bn Bn Bn+1 + aBn Bn (Bn+1 + aBn ) an ≥ Bn+1 (Bn+1 + aBn ) and we have proved that An an ≤ − x ≤ . Bn+1 (Bn+1 + aBn ) Bn Bn (Bn + aBn−1 ) an

Showing the very same inequalities for odd n works analogously. Taking into account that, according to Equation (4.5), λ(In (x)) =

an Bn (Bn + aBn−1 )

holds, it follows that λ(In+1 (x)) an = , a Bn+1 (Bn+1 + aBn ) which, in turn, yields λ(In+1 (x)) An ≤ − x ≤ λ(In (x)) a Bn as well as log (λ(In+1 (x))) − log(a) ≤ n

An log B − x n n

≤

log (λ(In (x))) . n

Considering that x ∈ E, it finally follows that An log B − x log (λ(In (x))) 2Li2 (−a) n lim = lim = log(a) + , n→∞ n→∞ n n log(1 + a) which completes the proof. Theorem 6.5. There exists some Borel subset E of [0, 1] with λ(E) = 1 An fulfilling the following property: if x ∈ E and B is the nth convergent of the n expansion of x as an a-simple continued fraction, then lim

n→∞

log(Bn ) −Li2 (−a) = . n log(1 + a)

72

Generalized Notions of Continued Fractions

Proof. According to Theorem 6.3, there exists some Borel subset E of [0, 1]\F with λ(E) = 1 such that lim

n→∞

log (λ(In (x))) 2Li2 (−a) = log(a) + n log(1 + a)

holds for every x ∈ E. For the rest of the proof, suppose that x ∈ E. Using Equation (4.5), we have λ(In (x)) = as well as

an , Bn (Bn + aBn−1 )

an+1 . Bn+1 (Bn+1 + aBn ) yields the following two chains of inequalities:

λ(In+1 (x)) = Considering Bn < Bn+1

λ(In+1 (x)) ≤

an+1 ≤ a · λ(In (x)), (1 + a)Bn2  n+1  a log (1+a)B 2 log(a) + λ(In (x)) n ≤ . n n

log (λ(In+1 (x))) ≤ n Since x ∈ E, it follows that  n+1  a log (1+a)B 2 λ(In (x)) 2Li2 (−a) n = lim = log(a) + , lim n→∞ n→∞ n n log(1 + a) as well as lim

n→∞

(n + 1) log(a) − log(1 + a) − 2 log(Bn ) 2Li2 (−a) = log(a) + , n log(1 + a)

which finally yields lim

n→∞

log(Bn ) −Li2 (−a) = . n log(1 + a)

The last theorem of this chapter generalizes Loch’s Theorem (see [10, Section 6.2.2]). Theorem 6.6. Let x = 0.x1 x2 x3 · · · be the decimal representation of a number x ∈ [0, 1). For every m ∈ N, consider Pm (x) := {y ∈ [0, 1) : y1 = x1 , . . . , ym = xm } and let nm denote the maximum of all n ∈ N fulfilling Pm (x) ⊂ In (x). Then lim

m→∞

− log(10) nm = 2 (−a) m log(a) + 2Li log(1+a)

holds for almost every x ∈ [0, 1).

The entropy of the system ([0, 1], B, µa , Ta )

73

Proof. Theorem 6.3 shows that lim

n→∞

log (λ(In (x))) 2Li2 (−a) = log(a) + n log(1 + a)

holds for almost every x ∈ [0, 1). Moreover, it is clear that Pm (x) = [z, z + 10−m ), where z = 0.x1 · · · xm . As a consequence, λ(Pm (x)) = 10−m , so log (10−m ) log (λ(Pm (x))) = lim = − log(10). m→∞ m→∞ m m lim

Having this and using the Dajani and Fieldsteel theorem (see [8, Theorem 4]), the result now follows via lim log(λ(Pmm (x))) nm − log(10) m→∞ lim = = . log(λ(I (x))) n 2 (−a) m→∞ m lim log(a) + 2Li n log(1+a) n→∞

Chapter 7 The natural extension of ([0, 1], B, µa, Ta)

Let us suppose that (X, A, µ, T ) and (Y, C, ν, S) are two measure-preserving dynamical systems with µ and ν being probability measures. Then the system (X, A, µ, T ) is said to be a natural extension of (Y, C, ν, S) if T is invertible (outside a set of measure zero) and there exists a surjective measurable mapping Ψ : X → Y with the following properties: (i) Ψ ◦ T = S ◦ Ψ.
(ii) µ Ψ−1 (E) = ν(E) for every E ∈ C (i.e., the push-forward of µ via Ψ coincides with ν).  (iii) A is the σ-algebra generated by T n ◦ Ψ−1 (E) : E ∈ C, n ∈ N ∪ {0} . It is known that every system (Y, C, ν, S) admits a natural extension, which is unique up to isomorphism. Moreover, the natural extension (X, A, µ, T ) is ergodic if and only if (Y, C, ν, S) is ergodic (see [6, page 28]). The next lemmas will be needed to obtain the natural extension of the dynamical system ([0, 1], B, µa , Ta ). They build upon the transformation T a : [0, 1] × [0, 1] → [0, 1] × [0, 1], defined as   1 T a (x, y) = Ta (x), , 1 + mx a + ay whereby mx denotes the first coefficient corresponding to x via Theorem 4.1, i.e., 
  1 1 −1 if x ̸= 0 a x mx = 0 if x = 0. Lemma 7.1. Consider k1 , . . . , kn ∈ N ∪ {0} and let Ik1 ,...,kn = [r, s) denote the interval defined in Lemma 4.3. If 0 ≤ u < v ≤ 1, then
−1 T a (u, v) × (r, s) =     1 1 1 1 1 1 = , × − − k1 , − − k1 1 + ak1 + av 1 + ak1 + au as a ar a holds. DOI: 10.1201/9781003404064-7

75

76

Generalized Notions of Continued Fractions
−1 Proof. Suppose that (x, y) ∈ T a (u, v)×(r, s) and that n is even. According to Lemma 4.3, the endpoints of the interval Ik1 ,...,kn are given by 1

r= 1 + k1 a + Considering

..

1

s= 1 + k1 a +

. 1 + kn a

a ..

.

. 1 + (kn + 1)a

1 1+mx a+ay

∈ (r, s) implies     1 1 1 1 − 1 < mx + y < −1 , a s a r

so k1 +

,

a

1 ..

< mx + y < k1 +

. 1 + (kn + 1)a

(7.1)

1 ..

≤ k1 + 1

. 1 + kn a

follows. As a consequence, we have k1 < mx + y < k1 + 1, and, taking into account that k1 and mx are integers and y ∈ [0, 1], we conclude mx = k1 and the inequalities in (7.1) yield   1 1 1 1 y∈ − − k1 , − − k1 . as a ar a Finally, since Ta (x) ∈ (u, v) and mx = k1 , we have   1 1 − 1 − k1 < v u < Ta (x) = a x as well as

 1 1 x∈ , , 1 + ak1 + av 1 + ak1 + au  
1 1 so (x, y) ∈ 1+ak11 +av , 1+ak11 +au × as − a1 − k1 , ar − a1 − k1 .  
1 1 Suppose now that (x, y) ∈ 1+ak11 +av , 1+ak11 +au × as − a1 −k1 , ar − a1 −k1 . Then obviously   1 1 k1 + u < − 1 < k1 + v ≤ k1 + 1, a x

so



   1 1 mx = −1 = k1 a x

as well as Ta (x) =

1 a



 1 − 1 − k1 ∈ (u, v) x

1 1 follows. Finally, using the fact that y ∈ as − a1 − k1 , ar − that 1 1 = ∈ (r, s). 1 + mx a + ay 1 + ak1 + ay

1 a


− k1 , it follows

The natural extension of ([0, 1], B, µa , Ta )

77

This shows that T a (x, y) ∈ (u, v) × (r, s) and the proof is complete for n even. The case of n can be handled analogously and the only change concerns the definition of r and s. ◦

In what follows let I denote the interior of an interval I. Lemma 7.2. For all non-negative integers m1 , m2 , . . . , mn and k1 , k2 , . . . , kℓ , the following identity holds:   ◦ ◦ n ◦ T a I kn ,kn−1 ,...,k1 ,m1 ,m2 ,...,mℓ × (0, 1) = I m1 ,m2 ,...,mℓ × I k1 ,k2 ,...,kn . Proof. We only prove the result for n+ℓ being even, since the case of odd n+ℓ can be shown analogously. According to Lemma 4.3, Ikn ,...,k1 ,m1 ,...,mℓ = [r, s) holds, whereby 1

r= 1 + kn a +

a ..

a

. 1 + k1 a + 1 + m1 a +

and

a ..

. 1 + mℓ a

1

s= 1 + kn a +

.

a ..

a

. 1 + k1 a + 1 + m1 a +

a ..

. 1 + (mℓ + 1)a

◦

Suppose that x ∈ I kn ,...,k1 ,m1 ,...,mℓ and 0 < y < 1. Then it follows that       1 1 1 1 1 1 −1 < −1 < −1 , a s a x a r so 1

kn + 1 + kn−1 a +

..

a

. 1 + k1 a + 1 + m1 a +


0 > t2 and we have   n  t2 n n Bn,n ≥ bn,n = C1 t1 + C2 t2 = C1 + C2 tn1 . t1 Considering tt21 < 1, there exists some n0 ∈ N such that Bn,n

C1 n C1 ≥ t = 2 1 2



1+

√

n

1 + 4a 2

(7.2)

for all n ≥ n0 . Without loss of generality we may assume that n0 is sufficiently large so that an ε · C12
(7.3) < n 1 4 2 +a holds for all n ≥ n0 . Given y ∈ [0, 1] and n ≥ n0 define z by 1

z= 1 + mn a +

.

a ..

. 1 + m2 a +

a 1 + m1 a + ay

Applying Lemma 7.4 yields

n

n

T a (x, y) − T a (x, 0) = ∥(Tan (x), z) − (Tan (x), φn (x))∥ = |z − φn (x)| . Note that both z and φn (x) are contained in the closure of the interval Imn ,mn−1 ,...,m1 defined in Lemma 4.3, the endpoints of which are An,n = Bn,n 1 + mn a +

1

= φn (x)

a ..

. 1 + m2 a +

a 1 + m1 a

and An,n + aAn,n−1 = Bn,n + aBn,n−1 1 + mn a +

1 a ..

. 1 + m2 a +

a 1 + (m1 + 1)a

.

86

Generalized Notions of Continued Fractions

Using Equation (4.5), the measure of Imn ,mn−1 ,...,m1 fulfills
λ Imn ,mn−1 ,...,m1 =

an an ≤ 2, Bn,n (Bn,n + aBn,n−1 ) (Bn,n )

hence applying Equations (7.2), (7.3), and (7.4), we obtain


n

n

T a (x, y) − T a (x, 0) = |z − φn (x)| ≤ λ Imn ,mn−1 ,...,m1 ≤ 4a

≤ C12



n

2n √ 1+ 1+4a 2

(7.4)

an 2

(Bn,n )

n


0 define the rectangle S = [r, s] × [u − ε, v + ε] ⊂ [0, 1] × [0, 1]. (if u = 0, then set S = [r, s] × [0, v + ε]; if v = 1, then S = [r, s] × [u − ε, 1]; if u = 0 and v = 1, then S = [r, s] × [0, 1]). As a consequence of the ergodic theorem 3.1, there exists some Borel set Ωε in [0, 1] × [0, 1] with µa (Ωε ) = 1 such that n−1  Z 1  i 1X XS d¯ µa = µa (S) (7.5) lim XS T a (x, y) = n→∞ n 0 i=0 for every (x, y) ∈ Ωε .

The natural extension of ([0, 1], B, µa , Ta )

87

Suppose that (x, y) ∈ Ωε . Then Lemma 7.7 implies the existence of some n0 ∈ N such that

n

n

T a (x, y) − T a (x, 0) < ε n

n

holds for every n ≥ n0 . Thus, if n > n0 and T a (x, 0) ∈ R, then T a (x, y) ∈ S and n−1  n−1   i  i X X XS T a (x, y) . XR T a (x, 0) ≤ i=n0

i=n0

Considering n0 −1  i  1 X XR T a (x, 0) = 0, n→∞ n i=0

lim

it follows that n−1 n−1  i   i  1X 1X XR T a (x, 0) ≤ lim XS T a (x, y) . n→∞ n n→∞ n i=0 i=0

lim

(7.6)

Using the equations (7.5) and (7.6), if (x, y) ∈ Ωε , then we have Z v+ε Z s n−1  i  1X 1 a XR T a (x, 0) ≤ µa (S) = dxdy n→∞ n log(1 + a) (1 + axy)2 u−ε r i=0   log (1+as(v+ε))·(1+ar(u−ε)) (1+ar(v+ε))·(1+as(u−ε)) = . log(1 + a) lim

A similar argument shows that for S ′ = [r, s] × [u + ε, v − ε], there exists a Borel set Ω′ε in [0, 1] × [0, 1] with µa (Ω′ε ) = 1 such that for (x, y) ∈ Ω′ε Z v−ε Z s n−1   i 1X 1 a ′ lim XR T a (x, 0) ≥ µa (S ) = dxdy n→∞ n log(1 + a) (1 + axy)2 u+ε r i=0   (1+as(v−ε))·(1+ar(u+ε)) log (1+ar(v−ε))·(1+as(u+ε)) = log(1 + a) ∞

holds. Choose a sequence (εn )n=1 of positive numbers such that lim εn = 0. n→∞ Then the Borel set ∞ \
ΩR = Ωεn ∩ Ω′εn n=1

has measure 1 and if (x, y) ∈ ΩR , then   (1+asv)(1+aru) n−1  i  log (1+arv)(1+asu) 1X lim XR T a (x, 0) = = µa (R). n→∞ n log(1 + a) i=0

88

Generalized Notions of Continued Fractions ∞

∞

∞

∞

Let (rn )n=1 , (sn )n=1 , (un )n=1 and (vn )n=1 be dense sequences in [0, 1] containing the points 0 and 1 and set R = {[rm , sn ] × [ui , vj ] : m, n, i, j ∈ N, rm < sn , ui < vj } . Then D=

\

ΩR ,

R∈R

as countable intersection of Borel sets, is a Borel set too and has measure 1. Given a (not necessarily closed) rectangle M in [0, 1] × [0, 1] with sides parallel to the axes and given any δ > 0, we can find R1 , R2 ∈ R such that R1 ⊂ M ⊂ R2 and µa (M ) − δ < µa (R1 ) ≤ µa (M ) ≤ µa (R2 ) < µa (M ) + δ. If (x, y) ∈ D, then n−1 n−1    i  i 1X 1X XM T a (x, 0) ≥ lim XR1 T a (x, 0) = µa (R1 ) n→∞ n n→∞ n i=0 i=0

lim

> µa (M ) − δ follows. Similarly, if (x, y) ∈ D, then n−1 n−1  i   i  1X 1X XM T a (x, 0) ≤ lim XR2 T a (x, 0) = µa (R2 ) n→∞ n n→∞ n i=0 i=0

lim

< µa (M ) + δ holds. Since these inequalities are fulfilled for every δ > 0, we finally get that n−1  i  1X XM T a (x, 0) = µa (M ) n→∞ n i=0

lim

for every (x, y) ∈ D. As next step, let us define E = π(D), where π(x, y) = x. The set E is Lebesgue measurable since it is the continuous image of a Borel set (see [5, Corollary 1.10.9]). Moreover, using µa (D) = 1, it follows that the Lebesgue i−1 measure of D is 1 as well, so λ(E) = 1. In addition, if x ∈ E and zi = T a (x, 0) for each i ∈ N, then the equalities n n−1  i  1X 1X XM (zi ) = lim XM T a (x, 0) = µa (M ) n→∞ n n→∞ n i=1 i=0

lim

hold for every rectangle M in [0, 1] × [0, 1].

(7.7)

The natural extension of ([0, 1], B, µa , Ta )

89

∞

In order to prove that the sequence (zi )i=1 is µa -uniformly distributed, choose any continuous function f on [0, 1]×[0, 1]. By the definition of Riemann p P integral, for every ε > 0 there exist two step functions g = bk XMk and h =

q P

k=1

ck XNk such that p, q ∈ N, bk , ck ∈ R for every k, M1 , . . . , Mp and

k=1

N1 , . . . , Nq are rectangles that form two partitions of [0, 1] × [0, 1], g ≤ f ≤ h and Z 1Z 1 ε log(1 + a) (h(u, v) − g(u, v))dudv < . a 0 0 As a direct consequence, it follows that ZZ Z 1Z 1 a dudv f d¯ µa = f (u, v) · log(1 + a)(1 + auv)2 [0,1]×[0,1] 0 0 Z 1Z 1 a mℓ and mi = ki if i < ℓ. ˆ ki = mi for every i ∈ {1, . . . , N } and N < n.

We start with the first case and assume that kℓ > mℓ and mi = ki if i < ℓ.

Generalized Hirzebruch-Jung continued fractions

103

Then the set kN [

C1 =



! Jk1 ,...,kN −1 ,i

[

∪

i=1



kN −1 −1





kℓ+1 −1

Jk1 ,...,kN −2 ,i  ∪ · · · ∪ 

i=1

[

Jk1 ,...,kℓ ,i 

i=1





  = x,  1 + k1 a −

1 a

..

. 1+kℓ−1 a− 1+(ka−1)a

   

ℓ

Skℓ −1 is a finite disjoint union of elements of P. Setting C2 = i=m Jm1 ,...,mℓ−1 ,i ℓ +1 and then considering that ki = mi if i < ℓ, it follows that     C1 ∪ C2 = x,  1 + k1 a −

1

   

a

..

a . 1+kℓ−1 a− 1+m

ℓa





  = x,  1 + m1 a −

1

  . 

a

..

a . 1+mℓ−1 a− 1+m

ℓa

If zℓ is the right endpoint of this last interval and Eℓ+1 = then    C1 ∪ C2 ∪ {zℓ } ∪ Eℓ+1 = x,  1 + m1 a −

S∞

i=1+mℓ+1



1

  . 

a

..

Jm1 ,...,mℓ ,i ,

. 1+mℓ a− 1+ma

ℓ+1 a

If S∞zℓ+1 is the right endpoint of this last interval and Eℓ+2 i=1+mℓ+2 Jm1 ,...,mℓ+1 ,i , then    C1 ∪ C2 ∪ {zℓ , zℓ+1 } ∪ Eℓ+1 ∪ Eℓ+2 = x,  1 + m1 a −

=

 1 a

..

. 1+mℓ+1 a− 1+ma

  . 

ℓ+2 a

Proceeding in the same manner shows the existence of points zℓ , . . . , zn−1 ∈ W and sets Eℓ+1 , . . . , En such that each Ei is a countable disjoint union of

104

Generalized Notions of Continued Fractions

elements of P and C1 ∪ C2 ∪ {zℓ , . . . , zn−1 } ∪ Eℓ+1 ∪ · · · ∪ En =    = x,

1

  = [x, y). 

a

1 + m1 a −

..

. 1+mn−1 a− 1+ma n a

Moreover, the previous union is disjoint. Finally, consider the case N < n and ki = mi for every i ∈ {1, . . . , N }. Then x=

1 1 + m1 a −

,

a

..

. 1+mN −1 a− 1+ma

y=

1 1 + m1 a −

.

a

..

Na

. 1+mn−1 a− 1+ma n a

For each j ∈ {N, . . . , n − 1}, setting xj =

1 1 + m1 a −

∈ W,

a

..

a . 1+mj−1 a− 1+m

ja

it follows that  [x, y) = {xN , . . . , xn−1 } ∪ 



∞ [

Jm1 ,...,mN ,i  ∪

i=1+mN +1

 ∪

∞ [

 Jm1 ,...,mN +1 ,i  ∪ · · · ∪

i=1+mN +2

∞ [

! Jm1 ,...,mn−1 ,i

,

i=1+mn

which completes the proof. Lemma 9.3. Let 0 < a ≤ 1. Let W and P be the sets defined in Lemma 9.2 and Q the collection of the empty set and all finite disjoint unions of sets [x, y) and {1} with x, y ∈ W . Then Q is an algebra that generates the Borel σalgebra in [0, 1]. Moreover, for every element E ∈ Q, there exists some C ⊂ E such that C is a countable disjoint union of elements of P and E\C is at most countable. Proof. It is straightforward that Q is an algebra in [0, 1]. Moreover,  to verify ∞ An of convergents of x given in Theorem 9.1 if x ∈ (0, 1], the sequence Bn n=1 An , n→∞ Bn

is increasing and fulfills x = lim

so

 ∞  [ An [0, x) = 0, Bn n=1

Generalized Hirzebruch-Jung continued fractions

105

is a countable union of elements of Q and Q generates the Borel σ-algebra in [0, 1]. Notice that each element E ∈ Q is of one of the following types: ˆ E = {1}. ˆ E = [x1 , y1 )∪· · ·∪[xm , ym ), whereby the union is disjoint and xi , yi ∈ W for each i ∈ {1, . . . , m}. ˆ E = [x1 , y1 ) ∪ · · · ∪ [xm , ym ) ∪ {1}.

According to Lemma 9.2, for every interval [x, y) with x, y ∈ W , there exist D ⊂ W and C such that C is a countable disjoint union of elements of P and [x, y) = D ∪ C. Therefore, for every E ∈ Q, there exists a set C ⊂ E such that C is a countable disjoint union of elements of P and E\C is contained in the countable set W . Theorem 9.4. For every a ∈ (0, 1), the dynamical system ([0, 1], B, λ, Ha ) is ergodic. Proof. We follows the line of argumentation as in the proof of Theorem 4.4 and proceed as follows. Let P and Q be the collections of sets defined in Lemmas 9.2 and 9.3 and f be the function considered in the proof of Lemma 9.2, i.e., An − taAn−1 f (t) = . Bn − taBn−1 We already know that f is increasing and that Jm1 ,...,mn = [f (0), f (1)) holds. Using Equation (9.3), the measure of a set of the type Jm1 ,...,mn is given by λ (Jm1 ,...,mn ) =

An − aAn−1 An an − = . Bn − aBn−1 Bn (Bn − aBn−1 )Bn

(9.5)

Suppose that [u, v) is another element of the class P. By Theorem 9.1, if x ∈ Jm1 ,...,mn , then x=

An − aHan (x)An−1 = f (Han (x)) Bn − aHan (x)Bn−1

follows. Moreover, x ∈ Ha−n ([u, v)) implies u ≤ Han (x) < v, which shows that x ∈ Jm1 ,...,mn ∩ Ha−n ([u, v)) if and only if x = f (t) for some t ∈ [u, v), i.e., Jm1 ,...,mn ∩ Ha−n ([u, v)) = [f (u), f (v)).

106

Generalized Notions of Continued Fractions

By Equation (9.3), computing the measure of Jm1 ,...,mn ∩ Ha−n ([u, v)) yields
λ Jm1 ,...,mn ∩ Ha−n ([u, v)) = f (v) − f (u) An − vaAn−1 An − uaAn−1 = − Bn − vaBn−1 Bn − uaBn−1 an (v − u) = (Bn − vaBn−1 ) (Bn − uaBn−1 ) an (v − u) , ≥ Bn2 so using Equation (9.5) we obtain
Bn − aBn−1 λ Jm1 ,...,mn ∩ Ha−n ([u, v)) ≥ λ (Jm1 ,...,mn ) · λ([u, v)) · . Bn If 0 < a < 1 then Bn − aBn Bn − aBn−1 > = 1 − a > 0, Bn Bn implying
λ Jm1 ,...,mn ∩ Ha−n ([u, v)) ≥ λ (Jm1 ,...,mn ) · λ([u, v)) · (1 − a). According to Theorem 3.2, the system ([0, 1], B, λ, Ha ) is ergodic. Theorem 9.5. Suppose that a ∈ (0, 1) and let ϑa denote the measure on the Borel subsets of [0, 1], defined by Z 1 −a ϑa (E) = dx. log(1 − a) E 1 − ax Then the transformation Ha is measure-preserving on ([0, 1], B, ϑa ) and the dynamical system ([0, 1], B, ϑa , Ha ) is ergodic. Proof. The ϑa -measure of an interval [r, s] ⊂ [0, 1] is given by ϑa ([r, s]) =

log(1 − as) − log(1 − ar) . log(1 − a)

Moreover, we have Ha−1 ([r, s])

=

∞  [ k=1

 1 1 , , 1 + ka − ra 1 + ka − sa

Generalized Hirzebruch-Jung continued fractions

107

which implies
ϑa Ha−1 ([r, s]) =

 ∞ log 1 − X

a 1+ka−sa



 − log 1 −

a 1+ka−ra



log(1 − a)   1 (1 + (k − 1)a − sa) · (1 + ka − ra) = log log(1 − a) (1 + ka − sa) · (1 + (k − 1)a − ra) k=1   1 1 − sa = · log . log(1 − a) 1 − ra
Therefore, ϑa Ha−1 ([r, s]) = ϑa ([r, s]) for each interval [r, s] in [0, 1] and Ha preserves ϑa (see [4, Section 24, Lemma 1]). Since ([0, 1], B, λ, Ha ) is ergodic and ϑa (E) = 0 if and only if λ(E) = 0, the ergodicity of ([0, 1], B, ϑa , Ha ) follows. k=1

∞ X

The subsequent results extend the famous Khintchine and Levy formulas to the case of generalized Hirzebruch-Jung continued fractions. The proof will be similar to those established in Chapter 4. Theorem 9.6. If 0 < a < 1, then there exists a Borel set E in [0, 1] with λ(E) = 1 such that n 1X mi = +∞ lim n→∞ n i=1 ∞

for every x ∈ E, where (mi )i=1 is the sequence associated to x via Theorem 9.1. Proof. For each N ∈ N define the function fN on [0, 1] by  1 0 if 0 ≤ x ≤ 1+N a fN (x) = 1 k if k ∈ {1, . . . , N } and 1 < x ≤ 1+ka 1+(k−1)a . h i 1 1 Recall that if x ∈ (0, 1], then mi = 1 + aH i−1 − a , so (x) a

1 1 < Hai−1 (x) ≤ . 1 + mi a 1 + (mi − 1)a
As a direct consequence,
mi ≤ N implies fN Hai−1 (x) = mi , while for mi > N we have fN Hai−1 (x) = 0. Therefore, it follows that n

n


1X 1X mi ≥ lim fN Hai−1 (x) . lim n→∞ n n→∞ n i=1 i=1 According to the ergodic theorem, there exists some Borel set EN ⊂ (0, 1] with ϑa (EN ) = ϑa ([0, 1]) = 1 such that Z 1 n
1X lim fN Hai−1 (x) = fN dϑa n→∞ n 0 i=1

108

Generalized Notions of Continued Fractions T∞ holds for every x ∈ EN . Setting E = N =1 EN yields ϑa (E) = 1 as well as λ(E) = 1. Moreover, for every x ∈ E and every N ∈ N, we have Z 1 n 1X lim mi ≥ fN dϑa . n→∞ n 0 i=1 As next step we derive a lower bound for the latter integral and first consider   1 Z 1 N Z N X X 1+(k−1)a 1 1 fN dϑa = , k dµa = k · ϑa 1 1 + ka 1 + (k − 1)a 0 k=1 1+ka k=1     a a N − log 1 − 1+ka log 1 − 1+(k−1)a X = k· log(1 − a) k=1   N X k a2 = · log 1 − . log(1 − a) (1 + (k − 1)a)2 k=1

Note that for 0 ≤ t < 1 we have log(1 − t) ≤ 1

Z

fN dϑa ≥ 0

N X k=1

≥

−t 2 ,

so

−a2 k · log(1 − a) 2(1 + (k − 1)a)2

N X −a2 k · . 2 log(1 − a) (1 + (k − 1)a)2 k=1

Altogether this yields the following inequalities for every x ∈ E and every N ∈ N: Z 1 n N X −a2 k 1X mi ≥ fN dϑa ≥ · . lim n→∞ n 2 log(1 − a) (1 + (k − 1)a)2 0 i=1 k=1

Since the series

∞ P k=1

k (1+(k−1)a)2

is divergent the result follows.

Theorem 9.7. If 0 < a < 1, then there exist a constant C > 0 and a Borel set E in [0, 1] with λ(E) = 1 such that n

1X 1 =C n→∞ n 1 + mi a i=1 lim

∞

holds for every x ∈ E, where (mi )i=1 is the sequence corresponding to x via Theorem 9.1. Proof. Define the function f by  1 1  1+ka if 1+ka Bn , Equations (9.5) and (9.3) yield 1−a an+1 1−a · λ(Jn+1 (x)) = · a a (Bn+1 − aBn )Bn+1 (1 − a)an an An+1 An ≤ = = − (Bn − aBn )Bn+1 Bn Bn+1 Bn+1 Bn An ≤x− . Bn Thus, we obtain 1−a An · λ(Jn+1 (x)) ≤ x − ≤ λ(Jn (x)) a Bn and then log

1−a a




+ log (λ(Jn+1 (x))) ≤ n

 log x −

An Bn



n

≤

log (λ(Jn (x))) . n

Considering that x ∈ E, it finally follows that   An log x − B log (λ(Jn (x))) 2Li2 (a) n lim = lim = log(a) + n→∞ n→∞ n n log(1 − a) and the proof is complete. Theorem 10.5. If 0 < a < 1, then there exists some Borel set E ⊂ [0, 1] with λ(E) = 1 such that for every x ∈ E, the identity lim

n→∞

log(Bn ) −Li2 (a) = n log(1 − a)

An holds, where B is the nth convergent of the expansion of x as a generalized n Hirzebruch-Jung continued fraction according to Theorem 9.1.

Proof. According to Theorem 10.3, there exists a Borel set E in [0, 1] with λ(E) = 1 and lim

n→∞

log (λ(Jn (x))) 2Li2 (a) = log(a) + n log(1 − a)

for every x ∈ E. Let x ∈ E be arbitrary but fixed. As shown in Equation (9.5), we have an λ(Jn (x)) = , (Bn − aBn−1 )Bn

116

Generalized Notions of Continued Fractions

and the identity λ(Jn+1 (x)) =

an+1 (Bn+1 − aBn )Bn+1

follows in the same manner. Considering Bn+1 > Bn > Bn−1 , it therefore follows that λ(Jn+1 (x)) ≤

an+1 a an+1 ≤ = · λ(Jn (x)), 2 (1 − a)Bn (1 − a)(Bn − aBn−1 )Bn 1−a

which shows log

log (λ(Jn+1 (x))) ≤ n



an+1 2 (1−a)Bn

n



log ≤



a 1−a



+ λ(Jn (x)) n

.

Since x ∈ E we have log lim



an+1 2 (1−a)Bn

n

n→∞

 = log(a) +

2Li2 (a) , log(1 − a)

so lim

n→∞

(n + 1) log(a) − log(1 − a) − 2 log(Bn ) 2Li2 (a) = log(a) + , n log(1 − a)

which implies the result. We conclude this chapter with a Loch type theorem, whose proof is analogous to the one of Theorem 6.6. Theorem 10.6. Suppose that 0 < a < 1. Let x = 0.x1 x2 x3 · · · be the decimal representation of a number x ∈ [0, 1). Given m ∈ N, set Pm (x) := {y ∈ [0, 1) : y1 = x1 , . . . , ym = xm } , and let nm be defined as the maximum of all n ∈ N with Pm (x) ⊂ Jn (x). Then lim

m→∞

nm − log(10) = 2Li2 (a) m log(a) + log(1−a)

holds for almost every x ∈ [0, 1).

Chapter 11 The natural extension of ([0, 1], B, ϑa, Ha)

Define H a : [0, 1] × [0, 1] → [0, 1] × [0, 1] by   1 , H a (x, y) = Ha (x), 1 + mx a − ay where mx is the first coefficient associated to x given in Theorem 9.1, i.e., ( 1  1 + ax − a1 if x ̸= 0 mx = 0 if x = 0. Lemma 11.1. Let k1 , . . . , kn ∈ N and Jk1 ,...,kn = [r, s) be the interval defined according to Lemma 9.2. If 0 ≤ u < v ≤ 1, then we have
−1 H a (u, v) × (r, s) =     1 1 1 1 1 1 , = × k1 + − , k1 + − . 1 + ak1 − au 1 + ak1 − av a ar a as
−1 Proof. Consider (x, y) ∈ H a (u, v) × (r, s) . According to Lemma 9.2, the endpoints of the interval Jk1 ,...,kn are r=

Using

1 1 + k1 a −

1 1+mx a−ay

,

a

..

s=

1 1 + k1 a −

. 1+kn a

.

a

..

. 1+(kn −1)a

∈ (r, s), this shows that 1 a



1 −1 r

 > mx − y >

1 a



 1 −1 , s

(11.1)

i.e., k1 −

1 ..

> mx − y > k1 −

. 1 + kn a

DOI: 10.1201/9781003404064-11

1 ..

≥ k1 − 1.

. 1 + (kn − 1)a

117

118

Generalized Notions of Continued Fractions

As a consequence, k1 > mx −y > k1 −1 holds and since k1 and mx are integers and y ∈ [0, 1], it follows that mx = k1 . Hence using the inequalities in (11.1) yields   1 1 1 1 y ∈ k1 + − , k1 + − . a ar a as Finally, since Ha (x) ∈ (u, v) and mx = k1 , we have   1 1 1 1 1 1 Ha (x) = 1 + − + − = k1 + − ∈ (u, v), a ax ax a a ax implying 

 1 1 , . 1 + ak1 − au 1 + ak1 − av  
1 1 , k1 + a1 − as . This shows (x, y) ∈ 1+ak11 −au , 1+ak11 −av × k1 + a1 − ar  
1 1 Suppose now that (x, y) ∈ 1+ak11 −au , 1+ak11 −av × k1 + a1 − ar , k1 + a1 − as . Then 1 1 k1 − 1 ≤ k1 − v < − < k1 − u ≤ k1 ax a as well as   1 1 − = k1 mx = 1 + ax a x∈

follows and we obtain   1 1 1 1 1 1 − + − = k1 + − ∈ (u, v). a ax ax a a ax
1 1 Moreover, the fact that y ∈ k1 + a1 − ar , k1 + a1 − as implies Ha (x) = 1 +

1 1 = ∈ (r, s), 1 + mx a − ay 1 + ak1 − ay which proves that H a (x, y) ∈ (u, v) × (r, s). Lemma 11.2. If m1 , m2 , . . . , mn and k1 , k2 , . . . , kℓ are natural numbers, then the following identity holds:   ◦ ◦ ◦ n H a J kn ,kn−1 ,...,k1 ,m1 ,m2 ,...,mℓ × (0, 1) = J m1 ,m2 ,...,mℓ × J k1 ,k2 ,...,kn . Proof. According to Lemma 9.2, Jkn ,...,k1 ,m1 ,...,mℓ is the interval [r, s) whose endpoints are 1

r= 1 + kn a −

a ..

. 1 + k1 a −

a 1 + m1 a −

a ..

. 1 + mℓ a

The natural extension of ([0, 1], B, ϑa , Ha ) and

119

1

s= 1 + kn a −

.

a ..

. 1 + k1 a −

a a

1 + m1 a −

..

. 1 + (mℓ − 1)a

◦

Suppose that x ∈ J kn ,...,k1 ,m1 ,...,mℓ = (r, s) and that 0 < y < 1. Then we have       1 1 1 1 1 1 −1 < −1 < −1 , a s a x a r implying kn − 1 < kn −


0, there exists some n0 ∈ N such that

n

n

H a (x, y) − H a (x, 0) < ε for every y ∈ [0, 1] and every n ≥ n0 .

126

Generalized Notions of Continued Fractions h  i 1 Proof. For each n ∈ N, consider mn = 1 + a1 H n−1 − 1 and for each (x) a

i ∈ {0, . . . , n}, let An,i and Bn,i be defined as follows: ˆ An,0 = 0, An,1 = 1, An,i = (1 + mn−i+1 a) An,i−1 − aAn,i−2 if 2 ≤ i ≤ n. ˆ Bn,0 = 1, Bn,1 = 1 + mn a, Bn,i = (1 + mn−i+1 a) Bn,i−1 − aBn,i−2 if 2 ≤ i ≤ n.

The following facts are straightforward to prove by induction on n: ˆ Bn,i > Bn,i−1 ≥ 1. ˆ Bn,i − aBn,i−1 ≥ 1.

Indeed, Bn,1 > Bn,0 = 1 and if we assume that Bn,i−1 > Bn,i−2 , then Bn,i = (1 + mn−i+1 a) Bn,i−1 − aBn,i−2 > (1 + mn−i+1 a) Bn,i−1 − aBn,i−1 ≥ Bn,i−1 . Moreover, we have Bn,1 − aBn,0 = 1 + mn a − aBn,0 ≥ 1. If we assume that Bn,i−1 − aBn,i−2 ≥ 1, then Bn,i − aBn,i−1 = [(1 + mn−i+1 a) Bn,i−1 − aBn,i−2 ] − aBn,i−1 = Bn,i−1 − aBn,i−2 + (mn−i+1 − 1) aBn,i−1 ≥ Bn,i−1 − aBn,i−2 ≥ 1 follows. Given ε > 0, let n0 ∈ N be such that an < ε for every n ≥ n0 . Furthermore, given any y ∈ [0, 1] and any n ≥ n0 , define z as follows: 1

z= 1 + mn a −

.

a ..

. 1 + m2 a −

a 1 + m1 a − ay

Both z and ψn (x) belong to the closure of the interval Jmn ,mn−1 ,...,m1 , whose endpoints are An,n = Bn,n 1 + mn a −

1

= ψn (x)

a ..

. 1 + m2 a −

a 1 + m1 a

and An,n − aAn,n−1 = Bn,n − aBn,n−1 1 + mn a −

1 a ..

. 1 + m2 a −

a 1 + (m1 − 1)a

.

The natural extension of ([0, 1], B, ϑa , Ha )

127

Using Lemma 11.4 and Equation (9.5), we obtain

n

n

H a (x, y) − H a (x, 0) = ∥(Han (x), z) − (Han (x), ψn (x))∥ = |z − ψn (x)|
≤ λ Jmn ,mn−1 ,...,m1 =

an , Bn,n (Bn,n − aBn,n−1 )

which, considering that Bn,n ≥ 1 and Bn,n − aBn,n−1 ≥ 1, yields the desired result:

n

n

H a (x, y) − H a (x, 0) ≤ an < ε.

Applying Lemma 11.6 instead of Lemma 7.7, it is straightforward to mimic the proof of Theorem 7.8 in order to obtain the following version of Jager’s theorem for the mapping Ha associated to the generalized Hirzebruch-Jung continued fraction: Theorem 11.7. If 0 < a < 1, then there exists a Lebesgue measurable set E ⊂ [0, 1] with λ(E) = 1 such that n−1  i  1X XB H a (x, 0) = ϑa (B) lim n→∞ n i=0

for every x ∈ E and every Borel set B ⊂ [0, 1] × [0, 1] fulfilling ϑa (∂B) = 0. Theorem 11.8. Suppose that 0 < a < 1 and define the function g by   (1−a)−ta if 0 ≤ t ≤ 1 − a log(1−a) g(t) = 1 + t−1−log(t) if 1 − a < t ≤ 1. log(1−a) Then there exists a Lebesgue measurable set E ⊂ [0, 1] with λ(E) = 1 such that Card {i : 1 ≤ i ≤ n and Θi (x) ≤ t} lim = g(t) n→∞ n for every x ∈ E and every t ∈ [0, 1]. Proof. We proceed analogous to the proof of Theorem 7.10. Let E be the subset given by Theorem 11.7 and suppose that x ∈ E. Considering the function f (u, v) = (1−a)u 1−auv and the set At = {(u, v) ∈ [0, 1] × [0, 1] : f (u, v) ≤ t} , Lemma 11.4 shows Θi (x) =

 i 
(1 − a)Hai (x) = f Hai (x), ψi (x) = f H a (x, 0) . i 1 − aHa (x)ψi (x)

128

Generalized Notions of Continued Fractions i

Thus, Θi (x) ≤ t if and only if H a (x, 0) ∈ At . We therefore obtain n  i  Card {i : 1 ≤ i ≤ n, Θi (x) ≤ t} 1X lim = lim XAt H a (x, 0) , n→∞ n→∞ n n i=1

hence, applying Theorem 11.7, lim

n→∞

Card {i : 1 ≤ i ≤ n, Θi (x) ≤ t} = ϑa (At ) n

follows. Thus, we have to prove that ϑa (At ) = g(t). The equality is obvious if t = 0, since A0 = {0} × [0, 1] and ϑa (A0 ) = 0 = g(0). We therefore assume that 0 < t ≤ 1. The set At is limited by a hyperbola, since the equation (1 − a)u =t 1 − auv is equivalent to   1−a 1 u v+ = . ta a The following graphs depicts the set At when 0 < t ≤ 1 − a:

After computing several elementary integrals, we obtain the following result if 0 < t ≤ 1 − a: t Z 1 Z 1−a+tav 1 −a ϑa (At ) = dudv log(1 − a) 0 0 (1 − auv)2 −ta = = g(t). (1 − a) log(1 − a) Next assume that 1 − a < t ≤ 1 holds, in which case the set At is of the following form:

The natural extension of ([0, 1], B, ϑa , Ha )

In this case, the measure of At is given by ϑa (At ) =

1 log(1 − a)

Z

0

1 + log(1 − a) =1+

t−1+a ta

Z

Z

1

0 1 t−1+a ta

Z

−a dudv+ (1 − auv)2 t 1−a+tav

0

t − 1 − log(t) = g(t), log(1 − a)

which completes the proof.

−a dudv (1 − auv)2

129

Chapter 12 A new generalization of the Farey series

In this final chapter, we present a generalization of the Farey series different from the one studied in Chapter 2. Recall that the function fa is defined by fa (x) =

x 1 + ax

and let ha : [0, 1] → [0, 1] denote the function given by ha (x) =

1 . 1 + a − ax

If n = 1, we define Ga,1 = {0, 1}. If n ≥ 2, we set Ga,n = fa (Ga,n−1 ) ∪ ha (Ga,n−1 ) . Note that both fa and ha are increasing and fa (1) = ha (0), so fa (x) < ha (y) if (x, y) ̸= (1, 0). As a consequence, it can be proved that the cardinality of n−1 for the Ga,n is 1 + 2n−1 for each n ∈ N. We write Gm a,n with m = 0, . . . , 2 elements of Ga,n and suppose that n−1

0 = G0a,n < G1a,n < G2a,n < · · · < G2a,n = 1. Each element Gm a,n can be written as a quotient of polynomials in Z[a]: Gm a,n =

m Na,n . m Da,n

Lemma 12.1. For every n ∈ N, the following assertions hold:
m (i) If 0 ≤ m ≤ 2n−1 , then Gm a,n+1 = fa Ga,n .   m−2n−1 (ii) If 2n−1 ≤ m ≤ 2n , then Gm . a,n+1 = ha Ga,n m (iii) If 0 ≤ m ≤ 2n−1 , then G2m a,n+1 = Ga,n .

Proof. The elements of Ga,n fulfill n−1

0 = G0a,n < G1a,n < G2a,n < · · · < G2a,n = 1. DOI: 10.1201/9781003404064-12

131

132

Generalized Notions of Continued Fractions

Since fa and ha are increasing we obtain  n−1 

fa G0a,n < fa G1a,n < · · · < fa G2a,n = fa (1) = ha (0) =  n−1 


= ha G0a,n < ha G1a,n < ha G2a,n < · · · < ha G2a,n , implying
G0a,n+1 = fa G0a,n ,
n−1 G2a,n+1+1 = ha G1a,n ,


G1a,n+1 = fa G1a,n ,

...


n−1 G2a,n+1+2 = ha G2a,n ,

 n−1  2n−1 Ga,n+1 = fa G2a,n ,

...

 n−1  n G2a,n+1 = ha G2a,n .

These equalities imply (i) and (ii). Property (iii) can easily be proved via induction. The property obviously holds for n = 1, since we have G0a,2 = 0 = G0a,1 ,

G2a,2 = 1 = G1a,1 .

Assuming that the result is valid for some index n, we can prove that it also n−2 holds for n + 1 as follows. , then (i) implies
On the one hand, if 0 ≤ m ≤ 2 2m 2m that Ga,n+1 = fa Ga,n and by the induction hypothesis we have

2m m m G2m a,n+1 = fa Ga,n = fa Ga,n−1 = Ga,n . On the other hand, if 2n−2 ≤ m ≤ 2n−1 , then (ii) implies that G2m a,n+1 =   2m−2n−1 ha Ga,n and the induction hypothesis yields     n−2 2m−2n−1 G2m = ha Gm−2 = Gm a,n+1 = ha Ga,n a,n . a,n−1 This shows that the result also holds for n + 1 and completes the proof of (iii). Lemma 12.2. If n ≥ 2 and 0 ≤ m ≤ 2n−2 − 1, then G2m+1 = a,n

2m+1 Na,n 2m+1 Da,n

=

2m 2m+2 aNa,n + Na,n 2m + D 2m+2 aDa,n a,n

.

o n 1 , 1 . The unique m fulfilling m ≤ 2n−2 −1 Proof. If n = 2, then Ga,2 = 0, 1+a is m = 0, so 2m + 1 = 1. Then G0a,2 = as well as

0 Na,2 0 = 0 , 1 Da,2

G2a,2 =

2 Na,2 1 = 2 1 Da,2

2 0 + Na,2 aNa,2 1 = = G1a,2 0 2 aDa,2 + Da,2 a+1

follows, and the result is valid for n = 2.

A new generalization of the Farey series

133

Assume now that the result holds for some index n ≥ 2 and proceed as follows. First note that p/q ∈ [0, 1] implies     p p p q fa = , ha = . q ap + q q q + a(q − p) Suppose that m ≤ 2n−2 − 1. Then 2m + 2 ≤ 2n−1 and Lemma 12.1 yields ! 2m 2m 2m
Na,n+1 Na,n Na,n 2m 2m = G = f G = f = a a a,n+1 a,n 2m 2m 2m + D 2m Da,n+1 Da,n aNa,n a,n and 2m+2 Na,n+1 2m+2 Da,n+1

=

G2m+2 a,n+1

=

fa G2m+2 a,n




= fa

2m+2 Na,n

!

2m+2 Da,n

=

2m+2 Na,n 2m+2 2m+2 . aNa,n + Da,n

Using the induction hypothesis, we therefore obtain G2m+1 = a,n

2m+2 2m + Na,n aNa,n 2m + D 2m+2 aDa,n a,n

and G2m+1 a,n+1 = fa Ga,n


2m+1

= =

2m+2 2m + Na,n aNa,n

= fa

!

2m + D 2m+2 aDa,n a,n

2m 2m+2 aNa,n + Na,n 2m + aN 2m+2 + aD 2m + D 2m+2 a2 Na,n a,n a,n a,n 2m+2 2m aNa,n+1 + Na,n+1 2m+2 2m aDa,n+1 + Da,n+1

,

which shows that the result also holds for n + 1 whenever m ≤ 2n−2 − 1. Finally, suppose that 2n−2 ≤ m ≤ 2n−1 − 1. Then 2m ≥ 2n−1 follows. Simplifying notation, we will write n−1

G2m−2 a,n

=

p , q

n−1

2m+2−2 Ga,n

r = . s

Then Lemma 12.1 implies the following equalities:     p q 2m 2m−2n−1 Ga,n+1 = ha Ga,n = ha = q q + a(q − p) and

  r s 2m+2−2n−1 G2m+2 = h G = h = . a a a,n a,n+1 s s + a(s − r)

Using the induction hypothesis, we obtain G2m+1−2 a,n

n−1

=

ap + r , aq + s

134

Generalized Notions of Continued Fractions

so G2m+1 a,n+1

= ha =



2m+1−2n−1 Ga,n



 = ha

ap + r aq + s



aq + s aq + s = . aq + s + a(aq + s − ap − r) a(q + a(q − p)) + s + a(s − r)

This shows that the statement also holds for n + 1 whenever 2n−2 ≤ m ≤ 2n−1 − 1. Next result provides yet another generalization of the classical Farey theorem. P Theorem 12.3. Let m = mi · 2i be the binary expansion of an integer number m with 0 ≤ m ≤ 2n−1 − 1 and set R = Card{i : mi = 1}. Then the following identity holds: m+1 m m m+1 Na,n Da,n − Na,n Da,n = aR .

Proof. For n = 1, we have G0a,1 =

0 Na,1 0 = , 0 Da,1 1

G1a,1 =

1 Na,1 1 = . 1 Da,1 1

If m = 0, then R = 0, so 1 0 0 1 Na,1 Da,n − Na,n Da,n = 1 = aR .

That is, the statement is valid for n = 1. Assume now that the statement holds for a natural number n − 1. Then according to Lemma 12.1, 0 ≤ m ≤ 2n−2 − 1 implies ! m m
Na,n−1 Na,n−1 m m Ga,n = fa Ga,n−1 = fa = m m m Da,n−1 aNa,n−1 + Da,n−1 and Gm+1 a,n

=

fa Gm+1 a,n−1




= fa

m+1 Na,n−1 m+1 Da,n−1

! =

m+1 Na,n−1 m+1 m+1 aNa,n−1 + Da,n−1

.

We therefore obtain m+1 m m m+1 Na,n Da,n − Na,n Da,n =



m+1 m+1 m+1 m m m aNa,n−1 + Da,n−1 = Na,n−1 aNa,n−1 + Da,n−1 − Na,n−1 m+1 m+1 m m = Na,n−1 Da,n−1 − Na,n−1 Da,n−1 .

Using the induction hypothesis, the last expression is equal to aR , i.e., the result also holds for n whenever m ≤ 2n−2 − 1.

A new generalization of the Farey series

135

Finally, suppose that the result is valid for n − 1 and that 2n−2 ≤ m ≤ − 1. Letting

n−1

2

m=

n−2 X

i

mi · 2 ,

n−2

m−2

=

i=0

n−2 X

ki · 2i

i=0

denote the binary expansions of m and m − 2n−2 , respectively, then mn−2 = 1 because m ≥ 2n−2 and m − 2n−2 = m0 + 2m1 + · · · + 2n−3 mn−3 , so kn−2 = 0 and ki = mi for every i ≤ n − 3. As a consequence, if R = Card {i : mi = 1} and R′ = Card {i : ki = 1}, it follows that R′ = R − 1. Writing n−2 r p m+1−2n−2 Ga,n−1 = , Gm−2 = , a,n−1 q s then Lemma 12.1 yields     q p m−2n−2 Gm = h G = h = a a a,n a,n−1 q q + a(q − p) and

  r s m+1−2n−2 . Gm+1 = = ha a,n = ha Ga,n−1 s s + a(s − r) ′

The induction hypothesis shows rq − ps = aR , so m+1 m m m+1 Na,n Da,n − Na,n Da,n = s(q + a(q − p)) − q(s + a(s − r)) ′

= a(rq − ps) = a · aR = aR . That is, the result is also valid for n whenever 2n−2 ≤ m ≤ 2n−1 − 1. S∞ Theorem 12.4. Suppose that 0 < a ≤ 1. If x ∈ (0, 1] and x ∈ / n=1 Ga,n , then x admits a unique infinite expansion of the form 1

x= 1 + m1 a −

(12.1)

a 1 + m2 a −

a 1 + m3 a −

a ..

.

S∞ with mi ≥ 1 for every i ∈ N. If x ̸= 0, x ̸= 1 and x ∈ n=1 Ga,n , then x admits two expansions of the form (12.1), one of them is finite and the other one is infinite. Proof. According to Theorem 9.1, every x ∈ (0, 1] admits an expansion of the form (12.1). Suppose that 1

x= 1 + m1 a −

a a 1 + m2 a − .. .

1

= 1 + k1 a −

a 1 + k2 a −

a .. .

136

Generalized Notions of Continued Fractions

are two infinite expansions of x with mi ≥ 1 and ki ≥ 1 for every i ∈ N. Then we have 1 1 1 1 − = m1 − a = k1 − a , ax a 1 + m2 a − 1 + k2 a − .. .. . . implying 

 1 1 − = m1 − 1 = k1 − 1. ax a

Therefore, m1 = k1 and by induction we obtain that mn = kn for every n ∈ N. As next step we prove that if a point x ∈ (0, 1] has a finite expansion of the type 1 x= a 1 + m1 a − .. a . 1 + mN −1 a − 1 + mN a S∞ with mi ≥ 1, then x ∈ n=1 Ga,n . If N = 1, then 1 = fa x= 1 + m1 a



1 1 + (m1 − 1)a



∞ [

= fam1 (1) ∈

Ga,n .

n=1

Assuming that the result holds for N − 1, we have that 1

y= 1 + m2 a −

which implies x=

∈

a ..

. 1 + mN −1 a −

a 1 + mN a

∞ [

Ga,n ,

n=1

∞ [ 1 = ha (y) ∈ Ga,n 1 + m1 a − ay n=1

and the result also holds for N . Next we will prove that every x ∈ Ga,n \{0, 1} admits a finite expansion of the form (12.1). The statement is true for n = 2, since in that case we 1 have x = 1+a . Therefore, assume that n ≥ 2 and that every y ∈ Ga,n \{0, 1} admits a finite expansion. If x ∈ Ga,n+1 , then there exists some y ∈ Ga,n such that either x = fa (y) or x = ha (y). By the induction hypothesis, there exist natural numbers N, m1 , . . . , mN such that 1

y= 1 + m1 a −

.

a ..

. 1 + mN −1 a −

a 1 + mN a

A new generalization of the Farey series On the one hand, if x = fa (y) and q = 1/y, then   1 1 1 x = fa = = q a+q 1 + (m1 + 1)a − .. . 1+m

137

,

a

N −1 a

−

a 1 + mN a

and on the other hand, if x = ha (y), then x=

1 = 1 + a − ay 1+a−

1 a 1 + m1 a −

a ..

. 1 + mN −1 a −

a 1 + mN a

follows. Therefore, x admits a finite expansion and the statement is valid for n + 1. Finally, considering Han (1) = 1 for every n ∈ N, the Hirzebruch-Jung expansion of 1 given by Theorem 9.1 is 1

1=

.

a

1+a−

a

1+a−

a .. .

1+a− Therefore, a finite expansion of a point x ∈

S∞

n=1

Ga,n \{0, 1}

1

x= 1 + m1 a −

a ..

. 1 + mN −1 a −

a 1 + mN a

can also be written as an infinite one: 1

x= 1 + m1 a −

a ..

. 1 + mN −1 a −

..

. 1 + mN −1 a −

= 1 + m1 a −

a 1 + (mN + 1)a − a 1 a a 1 + (mN + 1)a −

a 1+a−

a 1+a−

This completes the proof.

a .. .

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