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Page 1: Lisa Lorentzen, Haakon Waadeland Continued Fractions
Page 2: Lisa Lorentzen, Haakon Waadeland Continued Fractions

ATLANTIS STUDIES IN MATHEMATICS FOR ENGINEERINGAND SCIENCE

VOLUME 1

SERIES EDITOR: C.K. CHUI

Page 3: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Atlantis Studies in Mathematics for Engineering and Science

Series Editor:

C.K. Chui, Stanford University, USA

(ISSN: 1875-7642)

Aims and scope of the series

The series ‘Atlantis Studies in Mathematics for Engineering and Science’(AMES) publisheshigh quality monographs in applied mathematics, computational mathematics, and statisticsthat have the potential to make a significant impact on the advancement of engineering andscience on the one hand, and economics and commerce on the other. We welcome submis-sion of book proposals and manuscripts from mathematical scientists worldwide who shareour vision of mathematics as the engine of progress in the disciplines mentioned above.

All books in this series are co-published with World Scientific.

For more information on this series and our other book series, please visit our website at:

www.atlantis-press.com/publications/books

© ATLANTIS PRESS / WORLD SCIENTIFIC

AMSTERDAM – PARIS

Page 4: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Continued FractionsSecond edition

Volume 1: Convergence Theory

Lisa LorentzenHaakon Waadeland

Department of MathematicsNorwegian University of Science and Technology

TrondheimNorway

AMSTERDAM – PARIS

Page 5: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Atlantis Press 29 avenue Laumière 75019 Paris, France

For information on all Atlantis Press publications, visit our website at: www.atlantis-press.com.

CopyrightThis book, or any parts thereof, may not be reproduced for commercial purposes in anyform or by any means, electronic or mechanical, including photocopying, recording orany information storage and retrieval system known or to be invented, without priorpermission from the Publisher.

ISBN: 978-90-78677-07-9 ISSN: 1875-7642

© 2008 ATLANTIS PRESS / WORLD SCIENTIFIC

e-ISBN: 978-94-91216-37-4

Page 6: Lisa Lorentzen, Haakon Waadeland Continued Fractions

v

Preface to the second edition

15 years have passed since the first edition of this book was written. A lot hashappened since then – also in continued fraction theory. New ideas have emergedand some old results have gotten new proofs. It was therefore time to revise ourbook “Continued Fractions with Applications” which appeared in 1992 on Elsevier.

The interest in using continued fractions to approximate special functions has alsogrown since then. Such fractions are easy to program, they have impressive conver-gence properties, and their convergence is often easy to accelerate. They even havegood and reliable truncation error bounds which makes it possible to control theaccuracy of the approximation. The bounds are both of the a posteriori type whichtells the accuracy of a done calculation, and of the a priori type which can be used todetermine the number of terms needed for a wanted accuracy. This important aspectis treated in this first volume of the second edition, along with the basic theory.

In the second volume we focus more on continued fraction expansions of analyticfunctions. There are several beautiful connections between analytic function the-ory and continued fraction expansions. We can for instance mention orthogonalpolynomials, moment theory and Padé approximation.

We have tried to give credit to people who have contributed to the continued fractiontheory up through the ages. But some of the material we believe to be new, at leastwe have found no counterpart in the literature. In particular, we believe that tailsequences play a more fundamental role in this book than what is usual. This wayof looking at continued fractions is very fruitful.

Each chapter is still followed by a number of problems. This time we have markedthe more theoretic ones by ♠. We have also kept the appendix from the first edition.This list of continued fraction expansions of special functions was so well receivedthat we wanted it to stay as part of the book. Finally, we have kept the informalityin the sense that the first chapter consists almost entirely of examples which showwhat continued fractions are good for. The more serious theory starts in Chapter 2.

Lisa Lorentzen carries the main responsibility for the revisions in this second edi-tion. Through the first year of its making, Haakon Waadeland was busy writing ahandbook on continued fractions, together with an international group of people.This left Lisa Lorentzen with quite free hands to choose the contents and the wayof presentation. Still, he has played an important part in the later phases of thework on volume 1. For volume 2 Lisa Lorentzen bears the blame alone.

Trondheim, 14 February 2008

Lisa Lorentzen Haakon Waadeland

Page 7: Lisa Lorentzen, Haakon Waadeland Continued Fractions

vi

Preface

The name Shortly before this book was finished, we sent out a number of copies of Chapter 1,under the name “A Taste of Continued Fractions”. Now, in the process of working ourway through the chapters on a last minute search for errors, unintended omissionsand overlaps, or other unfortunate occurrences, we feel that this title might have beenthe right one even for the whole book. In most of the chapters, in particular in theapplications, a lot of work has been put into the process of cutting, canceling and “non-writing”. In many cases we are just left with a “taste”, or rather a glimpse of the role ofthe continued fractions within the topic of the chapter. We hope that we thereby canopen some doors, but in most cases we are definitely not touring the rooms.

The chapters Each chapter starts with some introductory information, “About this chapter”. Thepurpose is not to tell about the contents in detail. That has been done elsewhere.What we want is to tell about the intention of the chapter, and thereby also to adjustthe expectations to the right (moderate) level. Each chapter ends with a referencelist, reflecting essentially literature used in preparing that particular chapter. As aresult, books and papers will in many cases be referred to more than once in the book.On the other hand, those who look for a complete, updated bibliography on the fieldwill look in vain. To present such a bibliography has not been one of the purposesof the book.

The authors The two authors are different in style and approach. We have not made an effort tohide this, but to a certain extent the creative process of tearing up each other’s draftsand telling him/her to glue it together in a better way (with additions and omissions)may have had a certain disguising effect on the differences. This struggling type ofcooperation leaves us with a joint responsibility for the whole book. The way we thendistribute blame and credit between us is an internal matter.

The treasure chest Anybody who has lived with and loved continued fractions for a long time will alsohave lived with and loved the monographs by Perron, Wall and Jones/Thron. Actuallythe love for continued fractions most likely has been initiated by one or more of thesebooks. This is at least the case for the authors of the present book, and more so: thesethree books have played an essential role in our lives. The present book is in no way anattempt to replace or compete with these books. To the contrary, we hope to urge thereader to go on to these sources for further information.

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Preface vii

For whom? We are aiming at two kinds of readers: On the one hand people in or near mathemat-ics, who are curious about continued fractions; on the other hand senior-graduate levelstudents who would like an introduction (and a little more) to the analytic theoryof continued fractions. Some basic knowledge about functions of a complex variable, alittle linear algebra, elementary differential equations and occasionally a little dash ofmeasure theory is what is needed of mathematical background. Hopefully the studentswill appreciate the problems included and the examples. They may even appreciatethat some examples precede a properly established theory. (Others may dislike it.)

Words of gratitude We both owe a lot to Wolf Thron, for what we have learned from him, for inspirationand help, and for personal friendship. He has read most of this book, and his remarks,perhaps most of all his objections, have been of great help for us. Our gratitude alsoextends to Bill Jones, his closest coworker, to Arne Magnus, whose recent deathstruck us with sadness, and to all other members of the Colorado continued fractioncommunity. Here in Trondheim Olav Njåstad has been a key person in the field, andwe have on several occasions had a rewarding cooperation with him.

Many people, who had received our Chapter I, responded by sending friendly andencouraging letters, often with valuable suggestions. We thank them all for theirinterest and kind help.

The main person in the process of changing the hand-written drafts to a camera-ready copy was Leiv Arild Andenes Jacobsen. His able mastering of LaTeX, incombination with hard work, often at times when most people were in bed, hasleft us with a great debt of gratitude. We also want to thank Arild Skjølsvold andIrene Jacobsen for their part of the typing job. We finally thank Ruth Waadeland,who made all the drawings, except the LaTeX-made ones in Chapter XI.

The Department of Mathematics and Statistics, AVH, The University of Trondheimgenerously covered most of the typing expenses. The rest was covered by ElsevierScience Publishers. We are most grateful to Claude Brezinski and Luc Wuytack forurging us to write this book, and to Elsevier Science Publishers for publishing it.

Trondheim, December 1991,

Lisa Lorentzen Haakon Waadeland

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ix

Contents

Preface to the second edition vPreface vi

1 Introductory examples 11.1 Basic concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.1 Prelude to a definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.3 Computation of approximants . . . . . . . . . . . . . . . . . . . . . . 101.1.4 Approximating the value of K(an/bn) . . . . . . . . . . . . . . . . . 11

1.2 Regular continued fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.2.2 Best rational approximation . . . . . . . . . . . . . . . . . . . . . . . . 171.2.3 Solving linear diophantine equations . . . . . . . . . . . . . . . . . . 21 1.2.4 Grandfather clocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.2.5 Musical scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.3 Rational approximation to functions . . . . . . . . . . . . . . . . . . . . . . . 251.3.1 Expansions of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.3.2 Hypergeometric functions . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.4 Correspondence between power series and continued fractions . . . . . 30 1.4.1 From power series to continued fractions . . . . . . . . . . . . . . 30 1.4.2 From continued fractions to power series . . . . . . . . . . . . . . 33 1.4.3 One fraction, two series; analytic continuation . . . . . . . . . . . 33 1.4.4 Padé approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1.5 More examples of applications . . . . . . . . . . . . . . . . . . . . . . . . . . . 381.5.1 A differential equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381.5.2 Moment problems and divergent series . . . . . . . . . . . . . . . . 39 1.5.3 Orthogonal polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 421.5.4 Thiele interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431.5.5 Stable polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

1.6 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

2 Basics 53 2.1 Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.1.1 Properties of linear fractional transformations . . . . . . . . . . . 54 2.1.2 Convergence of continued fractions . . . . . . . . . . . . . . . . . . . 59 2.1.3 Restrained sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 602.1.4 Tail sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632.1.5 Tail sequences and three term recurrence relations . . . . . . . . 65

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2.1.6 Value sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702.1.7 Element sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

2.2 Transformations of continued fractions . . . . . . . . . . . . . . . . . . . . . 77 2.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 772.2.2 Equivalence transformations . . . . . . . . . . . . . . . . . . . . . . . 772.2.3 The Bauer-Muir transformation . . . . . . . . . . . . . . . . . . . . . 822.2.4 Contractions and extensions . . . . . . . . . . . . . . . . . . . . . . . 852.2.5 Contractions and convergence . . . . . . . . . . . . . . . . . . . . . . 87

2.3 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 892.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

3 Convergence criteria 993.1 Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

3.1.1 The Stern-Stolz Divergence Theorem. . . . . . . . . . . . . . . . . . 100 3.1.2 The Lane-Wall Characterization . . . . . . . . . . . . . . . . . . . . . 103 3.1.3 Truncation error bounds. . . . . . . . . . . . . . . . . . . . . . . . . . . 1063.1.4 Mapping with linear fractional transformations. . . . . . . . . . . 108 3.1.5 The Stieltjes-Vitali Theorem . . . . . . . . . . . . . . . . . . . . . . . . 114 3.1.6 A simple estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

3.2 Classical convergence theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 1163.2.1 Positive continued fractions . . . . . . . . . . . . . . . . . . . . . . . . 1163.2.2 Alternating continued fractions . . . . . . . . . . . . . . . . . . . . . . 122 3.2.3 Stieltjes continued fractions . . . . . . . . . . . . . . . . . . . . . . . . 1243.2.4 The Sleszy nski-Pringsheim Theorem . . . . . . . . . . . . . . . . . . 1293.2.5 Worpitzky’s Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1353.2.6 Van Vleck’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1423.2.7 The Thron-Lange Theorem. . . . . . . . . . . . . . . . . . . . . . . . . 1483.2.8 The parabola theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

3.3 Additional convergence theorems. . . . . . . . . . . . . . . . . . . . . . . . . . 1603.3.1 Simple bounded circular value sets . . . . . . . . . . . . . . . . . . . 160 3.3.2 Simple unbounded circular value sets. . . . . . . . . . . . . . . . . . 163

3.4 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1653.5 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

4 Periodic and limit periodic continued fractions 171 4.1 Periodic continued fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

4.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1724.1.2 Iterations of linear fractional transformations . . . . . . . . . . . . 172 4.1.3 Classification of linear fractional transformations . . . . . . . . . 174 4.1.4 General convergence of periodic continued fractions . . . . . . . 176 4.1.5 Convergence in the classical sense . . . . . . . . . . . . . . . . . . . . 179 4.1.6 Approximants on closed form . . . . . . . . . . . . . . . . . . . . . . . 1814.1.7 A connection to the Parabola Theorem . . . . . . . . . . . . . . . . 183

4.2 Limit periodic continued fractions . . . . . . . . . . . . . . . . . . . . . . . . . 186 4.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1864.2.2 Finite limits, loxodromic case . . . . . . . . . . . . . . . . . . . . . . . 187

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Contents xi

4.2.3 Finite limits, parabolic case . . . . . . . . . . . . . . . . . . . . . . . . 1924.2.4 Finite limits, elliptic case . . . . . . . . . . . . . . . . . . . . . . . . . . 1964.2.5 Infinite limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

4.3 Continued fractions with multiple limits . . . . . . . . . . . . . . . . . . . . 203 4.3.1 Periodic continued fractions with multiple limits . . . . . . . . . 203 4.3.2 Limit periodic continued fractions with multiple limits . . . . . 204

4.4 Fixed circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2054.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2054.4.2 Fixed circles for M . . . . . . . . . . . . . . . . . . . . . . . . . . . 2054.4.3 Fixed circles and periodic continued fractions . . . . . . . . . . . . 207

4.5 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2114.6 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

5 Numerical computation of continued fractions 217 5.1 Choice of approximants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

5.1.1 Fast convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2185.1.2 The fixed point method . . . . . . . . . . . . . . . . . . . . . . . . . . . 2195.1.3 Auxiliary continued fractions . . . . . . . . . . . . . . . . . . . . . . . 223 5.1.4 The improvement machine for the loxodromic case . . . . . . . . 227 5.1.5 Asymptotic expansion of tail values . . . . . . . . . . . . . . . . . . . 232 5.1.6 The square root modification . . . . . . . . . . . . . . . . . . . . . . . 235

5.2 Truncation error bounds. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2385.2.1 The ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2385.2.2 Truncation error bounds. . . . . . . . . . . . . . . . . . . . . . . . . . . 2405.2.3 The Oval Sequence Theorem. . . . . . . . . . . . . . . . . . . . . . . . 2435.2.4 An algorithm to find value sets for a given continued

fraction of form K(an/1) . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 5.2.5 Value sets and the fixed point method . . . . . . . . . . . . . . . . . 248 5.2.6 Value sets B(wn n ) for limit 1-periodic continued

fractions of loxodromic or parabolic type . . . . . . . . . . . . . . . 255 5.2.7 Error bounds based on idea 3 . . . . . . . . . . . . . . . . . . . . . . . 258

5.3 Stable computation of approximants . . . . . . . . . . . . . . . . . . . . . . . 260 5.3.1 Stability of the backward recurrence algorithm. . . . . . . . . . . 260

5.4 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2615.5 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262

A Some continued fraction expansions 265 A.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265

A.1.1 Notation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265A.1.2 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266

A.2 Elementary functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266A.2.1 Mathematical constants . . . . . . . . . . . . . . . . . . . . . . . . . . . 266A.2.2 The exponential function . . . . . . . . . . . . . . . . . . . . . . . . . . 268A.2.3 The general binomial function . . . . . . . . . . . . . . . . . . . . . . 269 A.2.4 The natural logarithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270A.2.5 Trigonometric and hyperbolic functions. . . . . . . . . . . . . . . . 272

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A.2.6 Inverse trigonometric and hyperbolic functions . . . . . . . . . . 273 A.2.7 Continued fractions with simple values . . . . . . . . . . . . . . . . 274

A.3 Hypergeometric functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275A.3.1 General expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275A.3.2 Special examples with 0F1 . . . . . . . . . . . . . . . . . . . . . . . . . 277A.3.3 Special examples with 2F0 . . . . . . . . . . . . . . . . . . . . . . . . . 277A.3.4 Special examples with 1F1 . . . . . . . . . . . . . . . . . . . . . . . . . 279A.3.5 Special examples with 2F1 . . . . . . . . . . . . . . . . . . . . . . . . . 281A.3.6 Some integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282A.3.7 Gamma function expressions by Ramanujan . . . . . . . . . . . . 284

A.4 Basic hypergeometric functions. . . . . . . . . . . . . . . . . . . . . . . . . . . 291A.4.1 General expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291A.4.2 Two general results by Andrews . . . . . . . . . . . . . . . . . . . . . 292 A.4.3 q -expressions by Ramanujan . . . . . . . . . . . . . . . . . . . . . . . 292

Bibliography 295Index 306

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Chapter 1

Introductory examplesWe have often been asked questions, by students as well as by established math-ematicians, about continued fractions: what they are and what they can be usedfor. Sometimes the questions have been raised under circumstances where a quickanswer is the only alternative to no answer: in the discussion after a talk or lecture,by a cup of coffee in a short break, in an airplane cabin or on a mountain hike. Inresponding to these questions we have often been pleased by the sparks of interestwe have seen, indicating that we had managed to transmit a glimpse of new andapparently appealing knowledge. In quite a few cases this led to further contactand follow-up activities”.

This introductory chapter is to a large extent inspired by the questions we havereceived and governed by the answers we have given. There is of course a greatdanger: A quick answer is often a wrong answer. It may tell the truth and nothingbut the truth, but it definitely does not tell the whole truth. This may lead tofalse guesses. This danger is in particular great in cases where observations andexperiments are used to create and support guesses. But we still wanted to keepthis often non-accepted, but highly useful aspect of mathematics as part of theintroductory chapter. We have tried to reduce the danger, partly by the way thingsare phrased, partly by indicating briefly how wrong such guesses can be, and finallyby referring to a more careful treatment later in the book. Still, we have chosen tointroduce some basic notation and definitions in the first section of the first chapter.

In this new edition it has been important not only to maintain the intention ofthe introductory chapter, but also to increase the number of examples. We havetherefore moved the convergence theorems to Chapter 3 to make some more space.In doing this we are violating rules and traditions for presentation of mathematics,namely to present the basic theory first, and then illustrate it by examples. This isdone on purpose, in the belief that what is lost in mathematical style and structureis gained in glimpses of what it is all about, and in wetting the appetite and curiosity.

L. Lorentzen and H. Waadeland, Continued Fractions, Atlantis Studies in Mathematics for Engineering and Science 1, DOI 10.1007/978-94-91216-37-4_1, © 2008 Atlantis Press/World Scientific

1

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2 Chapter 1: Introductory examples

1.1 Basic concepts

1.1.1 Prelude to a definition

Let {an} be a sequence of complex numbers. When we talk about the series∞∑

n=1

an = a1 + a2 + · · · + an + · · · , (1.1.1)

we have in mind the sequence {σn} of partial sums

σn :=n∑

k=1

ak .

(An empty sum∑n

k=m ak := 0 for n < m.) Convergence of the series (1.1.1) meansconvergence of {σn} to a complex number σ, in which case we write

∞∑n=1

an = σ . (1.1.2)

Similarly we are familiar with infinite products∞∏

n=1

an = a1 · a2 · · · an · · · . (1.1.3)

{pn} is the sequence of partial products

pn :=n∏

k=1

ak .

(An empty product∏n

k=m ak := 1 for n < m.) Convergence of the infinite product(1.1.3) means convergence of the sequence {pn} to a complex number p �= 0, inwhich case we write ∞∏

n=1

an = p . (1.1.4)

Let {fn} be the sequence

f1 := a1 , f2 :=a1

1 + a2, f3 :=

a1

1 +a2

1 + a3

,

and generally

fn :=a1

1 +a2

1 +a3

1 + ···+an

.

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1.1.1 Prelude to a definition 3

For simplicity we assume that all an �= 0, and we allow fn = ∞. Then {fn} is welldefined in C := C∪ {∞} where C is the set of complex numbers. Similarly to whatwe have for sums and products, this also leads to a concept, having to do with thenonterminating continuation of the process, in this case the concept of a continuedfraction

∞Kn=1

an

1:=

a1

1 +a2

1 +a3

1 + ···

. (1.1.5)

(The letter K is chosen since Kettenbruch is the German name for a continued frac-tion.) Convergence of (1.1.5) means convergence of the sequence {fn} of approxi-mants. We also accept convergence to ∞, as suggested by Pringsheim ([Prin99a]).The limit f := lim fn is the value of the convergent continued fraction when it exists,and then we adopt the tradition from series and infinite products and write

f =∞Kn=1

an

1= K∞

n=1(an/1) = K(an/1) . (1.1.6)

(An empty continued fraction Knk=m(ak/1) := 0 for m > n.) For simplicity we shall

write the continued fraction (1.1.5) as

∞Kn=1

an

1=:

a1

1 +a2

1 +a3

1 + · · · .

Note where we place the plus signs to indicate the fraction structure. This distin-guishes the continued fraction (1.1.5) from the series (1.1.1).

Example 1. For the continued fraction

∞Kn=1

61

=6

1 +6

1 +6

1 +6

1 + ···

=61 +

61 +

61 + · · ·

we findf1 = 6 , f2 =

67

, f3 =4213

, · · · .

By induction (see Problem 13 on page 49 with x = −2, y = 3) it follows thatgenerally

fn = 63n − (−2)n

3n+1 − (−2)n+1.

Therefore the continued fraction converges to 2. �

Page 17: Lisa Lorentzen, Haakon Waadeland Continued Fractions

4 Chapter 1: Introductory examples

Quite similarly we can construct, from any sequence {bn} of complex numbers, acontinued fraction

∞Kn=1

1bn

=1

b1 +1

b2 +1

b3 + ···

=1b1 +

1b2 +

1b3 + · · · , (1.1.7)

or from two sequences, {an} and {bn} of complex numbers, where all an �= 0, acontinued fraction

∞Kn=1

an

bn=

a1

b1 +a2

b2 +a3

b3 + ···

=a1

b1 +a2

b2 +a3

b3 + · · · . (1.1.8)

We write K(1/bn) and K(an/bn) for these structures.

In all the three cases the nth approximant fn is what we get by truncating thecontinued fraction after n fraction terms ak/bk, and convergence means convergenceof {fn}. (1.1.5) and (1.1.7) are obviously special cases of (1.1.8). In the particularcase when in (1.1.7) all bn are natural numbers, we get the regular continued fraction,well known in number theory.

Let us take a look at the common pattern in the three cases: series, products andcontinued fractions (and other constructions for that matter). In all three casesthe construction can be described in the following way: We have a sequence {φk}of mappings from C to C. By composition we construct a new sequence {Φn} ofmappings

Φ1 := φ1 , Φn := Φn−1 ◦ φn = φ1 ◦ φ2 ◦ · · · ◦ φn . (1.1.9)

For series we haveφk(w) := w + ak ,

andΦn(w) = φ1 ◦ φ2 ◦ · · · ◦ φn(w) = a1 + a2 + · · · + an + w .

For products we haveφk(w) := w · ak ,

andΦn(w) = φ1 ◦ φ2 ◦ · · · ◦ φn(w) = a1 · a2 · · · an · w .

For continued fractions (1.1.8) we have

φk(w) :=ak

bk + w,

andΦn(w) = φ1 ◦ φ2 ◦ · · · ◦ φn(w) =

a1

b1 +a2

b2 +· · ·+an

bn + w. (1.1.10)

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1.1.2 Definitions 5

These infinite structures can be regarded formally, but in applications the questionof convergence often comes up. Convergence of a series is defined as convergence of{Φn(0)}, convergence of an infinite product is defined as convergence of {Φn(1)},and convergence of a continued fraction is defined as convergence of {Φn(0)}.

1.1.2 Definitions

A linear fractional transformation (or Mobius transformation) τ is a mapping ofform

τ(w) =aw + b

cw + d; a, b, c, d ∈ C with ad − bc �= 0.

This representation of τ is not unique since we can multiply the coefficients a, b, cand d with any fixed non-zero constant without changing the mapping. However,we identify all these representations as one single transformation (mapping) τ .

We shall denote this family of mappings by M. Sometimes one emphasizes that ad−bc �= 0 by saying that τ is non-singular, as opposed to the singular transformationswith ad− bc = 0 which do not belong to M. A singular transformation is constantwherever it is meaningful, whereas τ ∈ M maps C univalently onto C.

We define a continued fraction in terms of linear fractional transformations:�

Definition 1.1. A continued fraction b0 + K(an/bn) is an ordered pair(({an}, {bn}), {Sn}) where {an} and {bn} are sequences of complex numberswith all an �= 0 and {Sn} is the sequence from M given by

Sn(w) := b0 +a1

b1 +a2

b2 +· · ·+an

bn + w. (1.2.1)

Remark: That {Sn} is a sequence from M is easily seen by the fact that

Sn = s0 ◦ s1 ◦ · · · ◦ sn where

s0(w) := b0 + w , sn(w) :=an

bn + wfor n ∈ N.

(1.2.2)

Obviously sn ∈ M when an �= 0, and thus Sn ∈ M since compositions of linearfractional transformations are again linear fractional transformations (which is eas-ily verified). We owe it to Weyl ([Weyl10]) for this very useful connection betweencontinued fractions and the class M. Normally we set b0 := 0, in which case φk = sk

and Φk = Sk in (1.1.10). Still, it is useful to have the possibility to set b0 �= 0.

The numbers an and bn are called the elements of b0 + K(an/bn), and an

bnis called

a fraction term of b0 + K(an/bn). Evaluations Sn(w) of Sn are called nth approxi-mants. The name convergents” has also been used in the literature. Historically,“

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6 Chapter 1: Introductory examples

the word approximant” always meant

fn := Sn(0) (1.2.3)

whereas Sn(wn) was referred to as a modified approximant” where wn was themodifying factor”. Classical approximants fn = Sn(0) play a special role in the

theory. In particular, the concept of convergence is based on {fn}:�

Definition 1.2. A continued fraction converges (in the classical sense) toa value f ∈ C if lim fn = f .

If K(an/bn) fails to converge, we say that it diverges.

A classical approximant fn is obtained by truncating the continued fraction after nfraction terms. The part we cut away,

f (n) :=an+1

bn+1 +an+2

bn+2 +an+3

bn+3 + · · · (1.2.4)

is called the nth tail of b0 + K(an/bn). This is also a continued fraction, and itconverges if and only if b0 + K(an/bn) converges. Indeed, (1.2.4) converges to f (n)

if and only if b0 + K(an/bn) converges to

f = Sn(f (n)). (1.2.5)

The sequence {f (n)} is then called the sequence of tail values for b0 + K(an/bn).This sequence will be important in our investigations.�

Lemma 1.1. Let Sn be given by (1.2.1). Then

Sn(w) =An−1w + An

Bn−1w + Bnfor n = 1, 2, 3, . . . (1.2.6)

where

An = bnAn−1 + anAn−2, Bn = bnBn−1 + anBn−2 (1.2.7)

with initial values A−1 = 1, A0 = b0, B−1 = 0 and B0 = 1.

Proof : It is clear that

S0(w) = b0 + w =b0 + w

1 + 0w, S1(w) = b0 +

a1

b1 + w=

b0b1 + a1 + b0w

b1 + w,

““

Page 20: Lisa Lorentzen, Haakon Waadeland Continued Fractions

1.1.2 Definitions 7

so (1.2.7) holds for n = 1. To see that it holds for general n ∈ N, we observe that

Sn(w) = Sn−1(sn(w)) =An−1 + An−2

anbn + w

Bn−1 + Bn−2an

bn + w

.

An and Bn are called the nth canonical numerator and denominator of b0+K(an/bn),or just its nth numerator and denominator for short. These names are quite naturalsince

fn = Sn(0) = An/Bn . (1.2.8)

The useful determinant formula

Δn := An−1Bn − AnBn−1 =n∏

k=1

(−ak) (1.2.9)

is a consequence of (1.2.7) (it follows by induction). Therefore also

An−1Bn+1 − An+1Bn−1 = bn+1Δn. (1.2.10)

Another simple, but important observation follows from the fact that

Sn(wn) = S0(w0) +n∑

k=1

(Sk(wk) − Sk−1(wk−1)). (1.2.11)

By the recurrence relations (1.2.7) and the determinant formula (1.2.9)

Sk(wk) − Sk−1(wk−1) =λk

∏k−1m=1(−am)

(Bk−1wk + Bk)(Bk−2wk−1 + Bk−1)where λk := ak − wk−1(bk + wk),

(1.2.12)

and thus

Sn(wn) = b0 + w0 +n∑

k=1

λk

∏k−1m=1(−am)

(Bk−1wk + Bk)(Bk−2wk−1 + Bk−1). (1.2.13)

For the case wk := 0 for all k, we get the well known Euler-Minding formula

fn =An

Bn= b0 −

n∑k=1

(−1)ka1a2 · · · ak

BkBk−1= b0 −

n∑k=1

Δk

BkBk−1, (1.2.14)

where Δk is given by (1.2.9). This formula was used by Euler ([Euler48]) andrediscovered by Minding ([Mind69]). Of course, Euler considered classical approxi-mants An/Bn, and the recurrence relations (1.2.7) are normally attributed to him([Euler48], Chapter 18).

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8 Chapter 1: Introductory examples

Remark: For a given continued fraction b0 + K(an/bn), we shall use the notationsn, Sn, An, Bn, fn, f (n) and Δn throughout the book.

In Lemma 1.1 we saw that the canonical numerators {An} and denominators {Bn}for a given continued fraction b0 + K(an/bn) are uniquely determined. In fact,Daniel Bernoulli ([Berno75]) proved that also the converse holds true if all Δn �= 0:�

Theorem 1.2. The given sequences {An}∞n=0 and {Bn}∞n=0 of complexnumbers are the canonical numerators and denominators of some contin-ued fraction b0 + K(an/bn) if and only if Δn �= 0 for n ≥ 1 and B0 = 1.Then b0 + K(an/bn) is uniquely determined by

b0 := A0 , b1 := B1 , a1 := A1 − A0B1 ,

an := − ΔnΔn−1

, bn := An−2Bn − Bn−2An

Δn−1for n ≥ 2 .

(1.2.15)

Proof : Let {An} and {Bn} be given with all Δn �= 0 and B0 = 1. Then theelements an and bn must be solutions of the system

bnAn−1 + anAn−2 = An ,bnBn−1 + anBn−2 = Bn

(1.2.16)

of linear equations. The determinant of this system is −Δn−1 �= 0. Hence thesolution is given by (1.2.15), and it is unique. �

This theorem allows us to construct as many continued fraction identities

f = K(an/bn)

as we may possibly want.

Example 2. We shall find the continued fraction b0+K(an/bn) for which An := n2

and Bn := n2 + 1 for n = 0, 1, 2, . . . . In this case

Δn = An−1Bn − Bn−1An = 1 − 2n

andAn−2Bn − Bn−2An = 4 − 4n ,

so by Theorem 1.2

b0 = 0 , b1 = 2 , a1 = 1 ,

an = −2n − 12n − 3

, bn =4n − 42n − 3

for n = 2, 3, 4, . . . .

Page 22: Lisa Lorentzen, Haakon Waadeland Continued Fractions

1.1.2 Definitions 9

Hence, the continued fraction

12+

−3/14/1 +

−5/38/3 +

−7/512/5 +

−9/716/7 +

−11/920/9 +· · ·

,

which also can be written

12 −

3/14/1 −

5/38/3 −

7/512/5 −

9/716/7 −

11/920/9 − · · · ,

is the one with An = n2 and Bn = n2 + 1. Since lim An/Bn = 1, this continuedfraction converges to 1, and we have proved the continued fraction identity

12 −

3/14/1 −

5/38/3 −

7/512/5 −

9/716/7 −

11/920/9 − · · · = 1.

Since fn = An/Bn, we also have, with N := {natural numbers}:

Theorem 1.3.A. The sequence {fn}∞n=0 from C is a sequence of classical approximantsfor some continued fraction K(an/bn) if and only if f0 = 0, f1 �= 0,∞ andfn �= fn−1 for all n ∈ N.B. If {fn} is a sequence of classical approximants for K(an/bn), then fn �=fn−2 if and only if bn �= 0.

Proof : A. Let first f0 = 0, f1 �= 0,∞ and fn �= fn−1 for all n. Let {An}∞n=0 and{Bn}∞n=0 be given by

An :=

{fn if fn �= ∞,

1 if fn = ∞,Bn :=

{1 if fn �= ∞,

0 if fn = ∞.

Then fn = An/Bn, Δn �= 0 for n ≥ 1 and B0 = 1, and thus the existence ofK(an/bn) follows from Theorem 1.2. Conversely, let {fn} be the classical approxi-mants for K(an/bn). Then f0 = 0, B0 = 1 and fn = Sn(0) �= Sn(∞) = fn−1 sinceSn is a univalent mapping of C to C.

B. This follows since fn = Sn(0) and fn−2 = Sn(−bn). �

As already mentioned, the nth tail K∞m=n+1(am/bm) of K(an/bn) is also a contin-

ued fraction. We shall let A(n)k and B

(n)k denote its kth canonical numerator and

denominator. Then the following follows by manipulating the recurrence relations:

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10 Chapter 1: Introductory examples

Lemma 1.4. A

(n)k = an+1B

(n+1)k−1 . (1.2.17)

B(n)k = bn+1B

(n+1)k−1 + an+2B

(n+2)k−2 . (1.2.18)

Proof : Let n ∈ N∪{0} be arbitrarily chosen. The first identity holds trivially fork = 0 and k = 1. Hence it holds for all k since {A(n)

k } and {B(n+1)k−1 } are solutions

of the same recurrence relation

Xk = bn+kXk−1 + an+kXk−2 . (1.2.19)

The second identity holds trivially for all n for k = 1 and k = 2 since B(n)−1 = 0,

B(n)0 = 1, B

(n)1 = bn+1 and B

(n)2 = bn+1bn+2 + an+2. Assume it holds for 2 ≤ k ≤

ν − 1. Then it also holds for k = ν since

B(n)k = bn+kB

(n)k−1 + an+kB

(n)k−2

= bn+k

(bn+1B

(n+1)k−2 + an+2B

(n+2)k−3

)+ an+k

(bn+1B

(n+1)k−3 + an+2B

(n+2)k−4

)= bn+1

(bn+kB

(n+1)k−2 + an+kB

(n+1)k−3

)+ an+2

(bn+kB

(n+2)k−3 + an+kB

(n+2)k−4

)which is equal to the right hand side of (1.2.18). �

Our definition of a continued fraction leads to an infinite structure. In some appli-cations one only needs a continued fraction-like structure which terminates after nterms,

b0 +a1

b1 +a2

b2 + · · ·+an

bn.

This will be called a terminating continued fraction even though it is not a continuedfraction by our definition. A related situation occurs if an = 0 for some or all n ∈ N.Let n0 be the first index for which an = 0. Then Sn is singular for n ≥ n0, andthe effect is essentially the same as if b0 + K(an/bn) was truncated after (n0 − 1)fraction terms. (The only thing that can complicate the situation is that the n0thtail converges to −bn0 . In such cases it is a matter of definition to determine theaction.) This is in particular relevant if the elements an and bn are functions of acomplex variable z.

1.1.3 Computation of approximants

There are several algorithms to compute approximants

Sn(wn) = b0 +a1

b1 +a2

b2 +· · ·+an

bn + wn=

An−1wn + An

Bn−1wn + Bn.

We shall only mention the most obvious ones:

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1.1.4 Approximating the value 11

1. The forward recurrence algorithm consists of computing An and Bn by therecurrence relation (1.2.7).

2. The backward recurrence algorithm starts by setting qn := wn and then workbackwards by setting

qk−1 :=ak

bk + qk

for k = n, n − 1, . . . , 1. Then Sn(wn) = b0 + q0. (Dirichlet [Diri63], p 46.)

3. Euler-Minding summation consists of computing Bn by the recurrence relation(1.2.7) and then finding fn or Sn(wn) by means of the Euler-Minding formula(1.2.14) or by (1.2.13).

The arithmetic complexity ω(fn) or ω(f1, f2, . . . , fn) of an algorithm is the mini-mal number of operations ( +, ×, : ) needed to compute fn or (f1, f2, . . . , fn).Straightforward counting shows:

1. The arithmetic complexity of the forward algorithm is

ω(f1) = 3, ω(fn) = 6n − 5, ω(f1, . . . , fn) = 7n − 5 for n ≥ 2.

2. The arithmetic complexity of the backward algorithm is

ω(f1) = 2, ω(fn) = 2n, ω(f1, . . . , fn) = n2 + n.

3. The arithmetic complexity of the Euler-Minding summation is ω(f1, . . . , fn) =6n − 3.

Note that the work involved is essentially independent of what we choose for wn.Method 1 and 3 has the advantage that if you have found Sn(wn), you can usethis to find Sn+1(wn+1), whereas you must start again from scratch in the secondmethod. On the other hand, the backward recurrence algorithm is in general morestable. We return to this claim in Section 5.3.1 on page 260. The computations inthis book are done by means of the backward recurrence algorithm. Approximantsare computed with high precision, and the value of a convergent continued fractionis estimated by high order approximants with reliable truncation error bounds.

1.1.4 Approximating the value of K(an/bn)

Example 3. The continued fraction

∞Kn=1

30 + 0.9n

1(1.4.1)

is known to converge. We want to find its value f . If we cannot find the exactvalue, we can use an approximant fn = Sn(0) as an approximation, since fn → f .However, the nth tail that we then cut away, looks more and more like K(30/1) asn increases. Since K(30/1) converges to 5 (Problem 13 on page 49 with x = −5,y = 6), it is reasonable to believe that Sn(5) is a better approximation than Sn(0)to the value of (1.4.1). The table below indicates strongly that this is true.

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12 Chapter 1: Introductory examples

In fact, we can do even better! The tail values

f (n) :=30 + 0.9n+1

1 +30 + 0.9n+2

1 +30 + 0.9n+3

1 + · · ·

satisfy the recurrence

f (n) =30 + 0.9n+1

1 + f (n+1), i.e., f (n)(1 + f (n+1)) = 30 + 0.9n+1.

Assume that f (n) = 5 + crn+1 + o(rn+1) for some c ∈ R and |r| < 1, where o(rn+1)is the usual little o-notation” defined by

un = o(rn+1) ⇐⇒ limn→∞

un

rn+1= 0.

That is, we assume that c is chosen such that

limn→∞

f (n) − (5 + crn+1)rn+1

= 0.

Since then

f (n)(1 + f (n+1)) = (5 + crn+1)(6 + crn+2) + o(rn+1)= 30 + c(6 + 5r)rn+1 + o(rn+1) ,

we must have r = 0.9 and c(6+5r) = 1; i.e., c = 2/21. Therefore the approximantsSn(wn) with wn = 5 + 2

21 · 0.9n+1 should be an even better choice than Sn(5), inparticular for large n. This is in fact true. We shall return to this idea in Section5.1.5 on page 232. The table below gives a very clear indication that this is so. Thevalue f is in this case f = 5.085066164924 correctly rounded to 12 decimal places.

n Sn(0) f − Sn(0) f − Sn(5) f − Sn(wn)1 30.9000 −25.8 −6.5 · 10−2 4.4 · 10−4

2 0.97139 4.1 4.8 · 10−2 −3.0 · 10−4

3 15.6770 −10.6 −3.7 · 10−2 2.0 · 10−4

4 1.85765 3.2 2.7 · 10−2 −1.4 · 10−4

35 5.10127 −1.6 · 10−2 −3.8 · 10−6 7.3 · 10−10

36 5.07160 1.3 · 10−2 2.8 · 10−6 −4.9 · 10−10

37 5.09631 −1.1 · 10−2 −2.1 · 10−6 3.3 · 10−10

38 5.07571 9.3 · 10−3 1.6 · 10−6 −2.2 · 10−10

85 5.08507 −1.8 · 10−6 −2.1 · 10−12 2.1 · 10−18

86 5.08506 1.5 · 10−6 1.6 · 10−12 −1.4 · 10−18

87 5.08507 −1.2 · 10−6 −1.2 · 10−12 9.7 · 10−19

88 5.08507 1.0 · 10−6 9.0 · 10−13 −6.5 · 10−19

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1.1.4 Approximating the value 13

Example 4. For the continued fraction

3 + 1/12

1 +4 + 3/22

1 +3 + 1/32

1 +4 + 3/42

1 +3 + 1/52

1 +· · ·(1.4.2)

the tails look more and more like”

31+

41+

31+

41+· · ·

(1.4.3)

and41+

31+

41+

31+· · ·

. (1.4.4)

We take convergence for granted, since it will be obvious later. The value of thecontinued fraction (1.4.3) is then clear: it is the positive root of the quadraticequation

u =3

1 +4

1 + u

,

which is 1. The value of the continued fraction (1.4.4) is therefore v = 4/(1+1) = 2,and it seems reasonable to choose approximants Sn(wn) for (1.4.2) with w2n := 1and w2n+1 := 2. It will be proved later that {Sn(wn)} converges faster than {Sn(0)}to the value of (1.4.2), something that is clearly indicated by the table below. Inthis example the value of the continued fraction is f = 1.1873788917921, correctlyrounded.

n Sn(0) f − Sn(0) Sn(wn) f − Sn(wn)1 4.00000000 −2.8 · 100 1.33333333 −1.5 · 10−1

2 0.69565217 4.9 · 10−1 1.18518519 2.2 · 10−3

3 1.85579937 −6.7 · 10−1 1.20054570 −1.3 · 10−2

4 1.00774785 1.8 · 10−1 1.18752336 −1.4 · 10−4

10 1.18032950 7.0 · 10−3 1.18738410 −5.2 · 10−6

11 1.19450818 −7.1 · 10−3 1.18739871 −2.0 · 10−5

12 1.18502153 2.4 · 10−3 1.18738030 −1.4 · 10−6

13 1.18975231 −2.4 · 10−3 1.18738379 −4.9 · 10−6

23 1.18738866 −9.8 · 10−6 1.18737890 −7.1 · 10−9

24 1.18737564 3.3 · 10−6 1.18737889 −7.1 · 10−10

25 1.18738215 −3.3 · 10−6 1.18737889 −2.0 · 10−9

This idea of carefully choosing the approximants goes way back, at least to Sylvester([Sylv69]) who in 1769 claimed that this possibility was one of the main advantagesof continued fractions. But it requires that {wn} is properly picked. We can easilymake improper choices”: Take any sequence {βn} from C and choose

wn := S−1n (βn) .

Then βn = Sn(wn). This shows that we can make {Sn(wn)} converge to anythingwe want, or diverge, regardless of the convergence behavior of the continued fraction

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14 Chapter 1: Introductory examples

itself. There is no reason for panic, though. This warning looks more serious thanit really is. It is one of the wonders of continued fractions that if Sn(0) → f , thenSn(wn) → f for almost every sequence {wn}. (Section 2.1.2.)

Example 3, and more so Example 4, may perhaps look artificial. But there is ahuge family of functions, including some hypergeometric functions and ratios ofhypergeometric functions, for which this method of convergence acceleration canbe used. (Section 1.3.2 on page 27.) Another matter is that such acceleration isnot much use for practical purposes unless one has fast converging, easy to usetruncation error bounds. This will be a recurrent theme in this book.

1.2 Regular continued fractions

1.2.1 Introduction

Regular continued fractions are continued fractions of the form b0 + K(1/bn) withbn ∈ N, although we also allow b0 = 0. They play an important role in numbertheory. In this section we shall give a few samples of their power. We first notethat An and Bn are positive integers and relatively prime since by the determinantformula (1.2.9)

Δn := An−1Bn − AnBn−1 = (−1)n. (2.1.1)

A regular continued fraction always converges, and we have some very useful trun-cation error bounds:

Theorem 1.5. A regular continued fraction K(1/bn) (with bn ∈ N forn ∈ N) converges to a value 0 < f < 1. Moreover,

f2 < f4 < f6 < · · · f5 < f3 < f1 , (2.1.2)

and|f − fn| < |fn+1 − fn| =

1BnBn+1

(2.1.3)

where the canonical denominators {Bn}∞n=1 are strictly increasing naturalnumbers.

Proof : The monotonicity of {Bk} follows from the recurrence formula Bn =bnBn−1 + Bn−2. The Euler-Minding formula (1.2.14) on page 7 therefore showsthat {fn} are the partial sums for an alternating series with monotonely decreasingterms 1/BkBk−1 → 0. The result follows therefore from standard properties ofalternating series. �

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1.2.1 Regular continued fractions 15

The alternating character of (fn+1 − fn) also suggests the rational approximation

f ≈ f∗n := 1

2(fn + fn+1) (2.1.4)

which guarantees that|f − f∗

n| < 12 |fn − fn+1| . (2.1.5)

This is also a good approximation, with a better truncation error bound.

We now turn to the question of how to find the regular continued fraction b0 +K(1/bn) which converges to a given positive number x; i.e., the regular continuedfraction expansion of x. The most natural method is probably the one we use forx := π in the following example.

Example 5. We shall expand π = 3.1415926535897932385 . . . in a regular contin-ued fraction:

π = 3.14159265358979323 . . . = 3 +11

0.14159265358979323 . . .

= 3 +1

7.062513305931045 . . .= 3 +

1

7 +11

0.062513305931045 . . .

= 3 +1

7 +1

15.996594406685 . . .

= 3 +17 +

1

15 +11

0.996594406685 . . .

= · · · = 3 +17+

115+

11+

1292+

11+· · ·

.

Of course, we only get the start of the continued fraction in this way since we inessence started with an approximation to π. �

The method applied in this example can be described as follows: for a positivenumber u let u� denote the largest integer ≤ u. Starting with a positive numberu0 we write

u0 = u0� + (u0 − u0�) = u0� +11

u0 − u0�.

We proceed with

u1 :=1

u0 − u0�= u1� + (u1 − u1�) = u1� +

11

u1 − u1�,

and repeat this process until it stops (if un ∈ N) or to get an infinite regularcontinued fraction.

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16 Chapter 1: Introductory examples�

�Theorem 1.6. To a given x > 0 there exists an essentially unique regu-lar continued fraction b0 + K(1/bn) which converges to x. This continuedfraction is terminating if and only if x > 0 is a rational number.

The convergence to x in the non-terminating case follows since

x = u0� +1

u1�+· · ·+1

un� + (un − un�)= Sn(un − un�),

so x − fn−1 = Sn(un − un�) − Sn−1(0) → 0 by (1.2.12). The only problem withthe uniqueness occurs in the terminating case: the two regular continued fractions

b0 +1b1 +

1b2 +· · ·+

1bn +

11

and b0 +1b1 +

1b2 +· · ·+

1bn + 1

are expansions of the same rational number. The next example demonstrates whyexpanding a rational number in a regular continued fraction by this method termi-nates, and why the process is equivalent to the euclidean algorithm.

Example 6. The euclidean algorithm ([Eucl56], book 7) attributed to the Greekmathematician Euclid (325 - 265 BC), is a tool to find the greatest common divisorof two integers. Applied to the pair (71, 47) it works as follows:

71 = 1 · 47 + 24,

47 = 1 · 24 + 23,

24 = 1 · 23 + 1,

23 = 23 · 1.

The greatest common divisor is therefore 1. That is, (71, 47) are relatively prime.A rewriting of these equations gives

7147

= 1 +1

47/24,

4724

= 1 +1

24/23,

2423

= 1 +1

23/1.

Linking these together we get the (terminating) regular continued fraction expansionof 71/47:

7147

= 1 +11+

11+

123

.

Its classical approximants are f1 =21, f2 =

32

and f3 =7147

. That the process hasto terminate follows since the left hand side in the euclidean algorithm consists ofdecreasing positive integers. �

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1.2.2 Best rational approximation 17

1.2.2 Best rational approximation

Example 7. In Example 5 we found the first terms of the regular continued fractionexpansion of π. This expansion can be used to produce rational approximations toπ. We compute the first canonical numerators and denominators:

n −1 0 1 2 3 4 5 · · ·bn — 3 7 15 1 292 1 · · ·An 1 3 22 333 355 103993 104348 · · ·Bn 0 1 7 106 113 33102 33215 · · ·

This gives the rational approximations

f0 = 3, f1 =227

= 3.142857 . . . , f2 =333106

= 3.14150943 . . . ,

f3 =355113

= 3.14159292 . . . , f4 =10399333102

= 3.1415926530119 . . .

to π. They are actually quite good. Indeed, even for such low order approximantswe have

|π − f3| < 3 · 10−7 and |π − f4| < 6 · 10−10 .

It is no coincidence that the approximants are good. Indeed, as the heading ofthis section indicates, the rational approximants we obtain from regular continuedfractions are best in a certain sense:�

Theorem 1.7. Let K(1/bn) be a regular continued fraction with value f ,and let p and q be two natural numbers such that∣∣∣f − p

q

∣∣∣ ≤ ∣∣∣f − An

Bn

∣∣∣ . (2.2.1)

Then q ≥ Bn. If q = Bn, then p = An.

Proof : Assume that q < Bn. We shall first prove that then

|qf − p| ≥ |Bn−1f − An−1| > |Bnf − An| . (2.2.2)

Let M and N be such that

AnM + An−1N = p ,BnM + Bn−1N = q .

(2.2.3)

Since the determinant of this 2×2-system of linear equations with unknowns M, Nis

AnBn−1 − BnAn−1 = (−1)n−1 ,

Page 31: Lisa Lorentzen, Haakon Waadeland Continued Fractions

18 Chapter 1: Introductory examples

such numbers M, N exist uniquely, and they are integers. If N = 0, then An/Bn =p/q, which means that An = p and Bn = q since An and Bn are relatively prime.Let N �= 0. Then M is either = 0 or has opposite sign of N , since otherwise q > Bn,contradicting our assumption.

With M and N as given above we get the identity

qf − p = M(Bnf − An) + N(Bn−1f − An−1) . (2.2.4)

Here the two expressions in parentheses have opposite signs by Theorem 1.5, andM, N also have opposite signs (unless M = 0). Hence

|qf − p| = |M(Bnf − An)| + |N(Bn−1f − An−1)| ,

and, since N is an integer �= 0, we have

|qf − p| ≥ |Bn−1f − An−1| . (2.2.5)

Since always |Bn−1f − An−1| > |Bnf − An| (see Problem 18 on page 50), (2.2.2)follows. Simultaneous division, left by q and right by Bn, gives∣∣∣f − p

q

∣∣∣ > ∣∣∣f − An

Bn

∣∣∣ . (2.2.6)

This contradicts (2.2.1), and thus q ≥ Bn.

Finally, assume that q = Bn. If N �= 0, then (2.2.4) still holds, and thus, by thesame argument as above (2.2.5) holds, which leads to the contradiction (2.2.6).Therefore N = 0, and thus An = p and Bn = q. �

0 1 2 3 4 5 6 70

1

2

3

4

5

6

7 This was first proved by H.J.S. Smith([Smith60]), but the present proofis due to Lagrange ([Lagr98]). Thetheorem shows that in order to finda better rational approximation tof than An/Bn, we have to increasethe denominator. Smith also gavea very nice geometric interpretationof this fact. A description of thisinterpretation can also be found ina paper by Felix Klein ([Klein95]):Let a be a positive irrational num-ber. We shall let this number a berepresented by the ray y = ax fromthe origin into the first quadrant ofa cartesian coordinate system. The

lattice point (n, m) with integer n and m represents the fraction m/n. Assume thereis a nail in every lattice point, and a rubber band fastened in the points (1, 0) and(0, 1) on the figure. Stretch the rubber band with a pencil following the ray y = ax.

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1.2.2 Best rational approximation 19

The ray y = ax does not pass through any lattice points, but the rubber band willcreate a polygon with corners at certain nails, say the nails (n2,m2), (n4,m4), . . .above the ray and the nails (n1,m1), (n3, m3), . . . below the ray, numbered withincreasing distance from the origin. Then

m2k−2

n2k−2<

m2k

n2k< a <

m2k+1

n2k+1<

m2k−1

n2k−1for all k.

Since there are no nails inside the polygon, these fractions are the best rational ap-proximations to a in our meaning of best”, and therefore they are the approximantsof the regular continued fraction expansion of a. That is,

mk/nk = Ak/Bk.

In our figure we have used a =√

2. The points (1, 1), (2, 3), (5, 7) on the figurecorresponds exactly to the first approximants 1/1, 3/2, 7/5 of the regular continuedfraction expansion 1 + K(2/1) of

√2 found in our next example.

Example 8. We shall find rational numbers which approximate the irrational num-ber

√2. From the equality

√2 − 1 =

12 + (

√2 − 1)

(2.2.7)

follow the equalities

√2 − 1 =

12 +

12 + (

√2 − 1)

=12 +

12 +

12 + (

√2 − 1)

=12 +

12 +

12 +

12 + (

√2 − 1)

,

and so on, as far out as we want. Since obviously

√2 = 1 +

12+

12+· · ·+

12+

12 + (

√2 − 1)

for any length of the row of dots, it seems to be a good idea to take a look at theapproximants of the regular continued fraction

1 +12+

12+

12+· · ·+

12+· · ·

.

Indeed, by Problem 13 with x = 1 −√

2 and y = 1 +√

2, this continued fractionconverges to

√2. To determine its first approximants we can use the recurrence

relations An = bnAn−1 + An−2, Bn = bnBn−1 + Bn−2 to obtain

n −1 0 1 2 3 4 5 · · ·bn — 1 2 2 2 2 2 · · ·An 1 1 3 7 17 41 99 · · ·Bn 0 1 2 5 12 29 70 · · ·

Page 33: Lisa Lorentzen, Haakon Waadeland Continued Fractions

20 Chapter 1: Introductory examples

This gives

1 +12

= 1.5,

1 +12+

12

=75

= 1.4,

1 +12+

12+

12

=1712

= 1.4166 . . . ,

1 +12+

12+

12+

12

=4129

= 1.41379 . . . ,

1 +12+

12+

12+

12+

12

=9970

= 1.4142857 . . . ,

which seems to approach√

2 pretty quickly. Already the fifth one, the last onelisted, has an error less than .00008. Indeed, by Theorem 1.5

f4 = 41/29 <√

2 < f5 = 99/70,

or alternatively

√2 ≈ 1

2(f4 + f5) =

57414060

≈ 1.41404,∣∣∣∣√2 − 57414060

∣∣∣∣ < 12(f5 − f4) =

14060

≈ 2.5 · 10−4

which agrees with the value√

2 = 1.41421356 . . . . �

This is a good place for a warning: using identities like (2.2.7) to derive a continuedfraction as in this example, must always be followed by a check: does the continuedfraction really converge to the expected value? The equality

−√

2 − 1 =1

2 + (−√

2 − 1)

will for instance lead to

−√

2 = 1 +12+

12+· · ·+

12 + (−

√2 − 1)

,

but the continued fraction (still) converges to√

2. More generally, if we start withany number p, the recurrence

p0 := p, pn−1 = an/(bn + pn) for n = 1, 2, 3, ... (2.2.8)

defines a sequence {pn}. So how do we know which p is the value of K(an/bn), ifthis continued fraction converges at all? This question will be resolved in the nextchapter. (Definition 2.9 on page 63, combined with Corollary 2.7 on page 68, givesone possible answer.)

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1.2.3 Solving linear diophantine equations 21

1.2.3 Solving linear diophantine equations

For regular continued fractions b0 + K(1/bn) the determinant in the determinantformula (1.2.9) on page 7 is

An−1Bn − AnBn−1 = (−1)n. (2.3.1)

This property is essential when terminating regular continued fractions are used tosolve linear diophantine equations of the form

Mx + Ny = P, (2.3.2)

where M, N, P are integers, and we seek integer solutions x, y. The Greek math-ematician Diophantos (ca 250 AD) studied such equations, but solving them bymeans of terminating regular continued fractions, as we shall do in our next ex-ample, seems to have been introduced by the Indian mathematician Aryabhata theelder (ca 475 - 550 AD).

Example 9. A shipowner wants to update and upgrade her fleet of ships. Afterhaving received bids from different shipyards, she made up her mind and placeda contract for her ships at Beakersheep Shipyard where they offered two types ofships, X-type for 71 · 108 (7.1 billion) US-dollars a piece and Y-type for 47 · 108

(4.7 billion) US-dollars a piece. The number and types of ships were meant to beconfidential until further notice. However, an industrial spy finds information onthe total price paid, 449 ·108 US-dollars. He also has a price list from the ship yard.How can he determine how many ships of the two types were contracted?

Let x and y be the number of X-ships and Y-ships, respectively. The problem isthen to solve the diophantine equation

71x + 47y = 449. (2.3.3)

We expand the rational number 71/47 in a terminating regular continued fraction,as done in Example 6, and find that the expansion is

7147

= 1 +11+

11+

123

,A3 = 71B3 = 47

A2 = 3B2 = 2

and thus, by (2.3.1) with n := 3

A3B2 − A2B3 = 71 · 2 + 47 · (−3) = 1.

To get 449 on the right hand side of this identity, we multiply by 449

71 · 898 + 47 · (−1347) = 449.

From this follows that for all integer values t, the pair

x = 898 + 47t, y = −1347 − 71t

Page 35: Lisa Lorentzen, Haakon Waadeland Continued Fractions

22 Chapter 1: Introductory examples

is a solution of (2.3.3). Actually, these pairs are the only solutions. Since we wantnon-negative solutions we must have

−19.10 · · · = −89847

≤ t ≤ −134771

= −18.97 . . . .

The only t-value for which this holds is t = −19, leading to the solution

x = 5, y = 2.

That is, the shipowner had placed a contract for 5 ships of X-type and 2 ships ofY-type. �

Of course, to find the integer solution x = 5, y = 2 in the example above wouldonly take a few seconds of trial and error. But that is only because we can expectthe solution to be relatively small. For large solutions the method illustrated byExample 9 is a clear winner.

1.2.4 Grandfather clocks

If cogwheel M with m teeth interlace with cogwheel N with n teeth, then N hasrotated m/n times around when M has rotated once. This is the idea of grandfatherclocks where cogwheels transmit the movement from the shaft of the second handto the shaft of the minute hand, and ditto from the minute hand to the hour hand.

But some of these old treasures also show the faces of the moon. The moon takesabout 29.53059 nights and days to go from full moon to full moon. The cogwheelfor the moon’s movement should therefore have 2 · 29.53059 times as many teeth asthe wheel for the hour hand. Therefore m = 5906118 teeth and n = 100000 teethcould be a solution if it was at all possible and practical to make durable cogwheelswith so many teeth. A slightly better solution would be to use an extra cogwheelin between, such that

5906118100000

=610

· 98435310000

describes ratios in the two transmissions. But the number of teeth is still unrealis-tically high.

A better solution still is to find a good rational approximation to the rationalnumber 59.06118 (which is also a rational approximation to the true number). And

Page 36: Lisa Lorentzen, Haakon Waadeland Continued Fractions

1.2.5 Musical scales 23

here the regular continued fractions come into play:

5906118100000

= 59 +1

1000006118

= 59 +1

16 +21126118

= 59 +1

16 +1

61882112

= 59 +1

16 +1

2 +18942112

= · · · = 59 +116+

12+

11+

18+

11+

12+

14+

11+

16

.

The choicem

n= 59 +

116

=94516

= 59.0625

gives a more practical solution to the problem. Huygens ([Huyg95]) used thismethod for designing cogwheels for his planetarium.

1.2.5 Musical scales

In a piano the fixed tones with their frequencies fn are essentially chosen such that(i) the ratio fn+1/fn between two neighboring tones is fixed, no matter where

they are chosen on the piano. This makes a melody independent of where itwas started on the piano.

(ii) the consonant ratios2 : 1 octave3 : 2 fifth4 : 3 fourth5 : 4 major third6 : 5 minor third

are present; i.e. there are frequencies such that their ratios give these fractions.This is done because such tones sound beautiful together.

The problem is that it is impossible to achieve both (i) and (ii) exactly, simulta-neously. In a piano the octave is divided into 12 intervals, so we get the twelvetones

C, #C, D, #D, E, F, #F, G, #G, A, #A, H

with corresponding frequencies f1, f2, . . . , f12, and f13 := 2f1 is the next C. Ifq := fn+1/fn is constant, this means that

f13

f1= 2 =

f13

f12· f12

f11· · · f2

f1= q12

Page 37: Lisa Lorentzen, Haakon Waadeland Continued Fractions

24 Chapter 1: Introductory examples

and thus q = 21/12 which is an irrational number! We can therefore not get theconsonant ratios we asked for exactly, except for the octave. We only get approxi-mations to the consonant ratios. For instance

f8

f1= q7 = 27/12 ≈ 1.498 · · · ≈ 3

2= 1.5

f6

f1= q5 = 25/12 ≈ 1.335 · · · ≈ 4

3= 1.333 . . .

f5

f1= q4 = 24/12 ≈ 1.260 · · · ≈ 5

4= 1.250

f4

f1= q3 = 23/12 ≈ 1.189 · · · ≈ 6

5= 1.200

For a mathematician, a natural question is then: given our two concerns (i) and(ii), what is an optimal number of tones in a octave? That is, for which N ∈ N willthere exist integers k such that (21/N )k are good approximations to the consonantratios? This is a question of rational approximation: we want to choose N suchthat 21/N is almost rational”. In particular we want, say,

(21/N )k = 2k/N ≈ 32; i.e.,

k

N≈ log2

32

=Ln 3/2Ln 2

for some k, N ∈ N where Ln x = loge x is the natural logarithm. In other words, k/Nshould be a good rational approximation to the irrational number log2

32 . Therefore

we expand log232 ≈ 0.5849625007 in a regular continued fraction ([Lore06]). Of

course, the regular continued fraction expansion of log232

never stops, whereas theregular continued fraction expansion of the rational number 0.5849625007 termi-nates. However, we only want to use the first few terms of the continued fractionanyway, since we do not want to have thousands of tones in one octave. We get

log2

32≈ 0.5846925007 =

11+

11+

12+

12+

13+

11+· · ·

.

The approximation

log2

32≈ 1

1+11+

12

=35

= 0.60000

says: 5 equally spaced tones in an octave with (f1, f4) as the fifth is reasonablygood. Taking one more term of the continued fraction gives

log2

32≈ 1

1+11+

12+

12

=712

≈ 0.58333 .

Hah!! That is exactly what we have on a piano. 12 tones in an octave, and forinstance C–G makes up a fifth. What is the next good choice? We let the continuedfraction come up with an answer:

log2

32≈ 1

1+11+

12+

12+

13

=2441

≈ 0.58537 .

Interesting! A piano with 41 equally spaced tones in an octave will give very goodapproximations to the consonant ratio 3:2, with (f1, f25) as the fifth.

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1.3.1 Expansions of functions 25

1.3 Rational approximation to functions

1.3.1 Expansions of functions

Example 10. In this book √. . . shall always denote the principal branch of the

square root; that is, −π2

< arg√. . . ≤ π2. For z ∈ D := {z ∈ C; | arg(1 + z)| < π},

the identity √1 + z − 1 =

z

2 + (√

1 + z − 1)

leads to the identities√

1 + z − 1 =z

2+z

2+· · ·+z

2+z

2 + (√

1 + z − 1),

just as in Example 8. This suggests the continued fraction

z

2+z

2+· · ·+z

2+· · ·. (3.1.1)

But the same warning still applies: we do not know off-hand that this continuedfraction converges to (

√1 + z − 1). However, by Problem 13 on page 49 with

x = 1 +√

1 + z and y = 1 −√

1 + z we get its classical approximants fn(z) onclosed form. They show that (3.1.1) converges to

√1 + z − 1 for all z ∈ D. The

approximants of (3.1.1) are rational functions,

f1(z) =z

2, f2(z) =

z

2+z

2=

2z

z + 4, f3(z) =

z

2+z

2+z

2=

z2 + 4z

4z + 8,

and so on. Hence fn(z) is a rational approximation to the function√

1 + z − 1 inD. Strictly speaking, the continued fraction (3.1.1) is not defined for z = 0, but wecan make the additional definition that it has the value 0 for z = 0. �

1�

� Continued fraction expansions areless known than power series ex-pansions. The Taylor series ex-pansion of

√1 + z − 1 at 0 is

√1 + z − 1 =

∞∑k=1

(12

k

)zk.

It converges for |z| < 1 and di-verges for |z| > 1. Its partialsums are

σ1(z) =12z, σ2(z) =

12z − 1

8z2,

σ3(z) =12z − 1

8z2 +

116

z3 , . . . ,

Page 39: Lisa Lorentzen, Haakon Waadeland Continued Fractions

26 Chapter 1: Introductory examples

and they are polynomial approximations to√

1 + z −1 for z in the unit disk D :={z ∈ C; |z| < 1}. The approximants fn(z) in Example 10 approximate

√1 + z − 1

in a much larger region, D := {z ∈ C; | arg(1 + z)| < π}. In our next example wecompare their degree of approximation.

Example 11. We compute the two types of approximants σn(z) and fn(z) for√1 + z − 1 for a given z–value to see what happens. We choose z := 0.96, in which

case the value of the function is exactly 0.4. The table below shows σn(0.96) andfn(0.96), all correctly rounded in the 4th decimal place:

n 1 2 3 4 5 6 7σn .4800 .3648 .4201 .3869 .4092 .3932 .4053fn .4800 .3871 .4022 .3996 .4001 .4000 .4000

Or, if we rather list the index n(k) which is the smallest index for which the ap-proximation is correct, rounded to k decimals, for all n ≥ n(k):

n(3) n(6) n(12) n(20)σn 23 129 424 849fn 4 8 16 26

This of course does not prove anything, but it certainly suggests that the continuedfraction is better (converges faster) than the power series expansion. (It is, however,only fair to say, that such a comparison, based merely upon the order n of theapproximant, does not always give the correct picture. Essential in the comparisonis the resources needed for the computation.)

Even more flattering for the continued fraction expansion is the choice z = 3 forwhich

√1 + z − 1 = 1. In this case it does not make sense to compute power series

approximants, since we know that the power series diverges. In the next table thefirst seven continued fraction approximants are listed, correctly rounded in the 4thdecimal place.

n 1 2 3 4 5 6 7fn 1.5000 .8571 1.0500 .9836 1.0055 .9982 1.0006�

Example 12. We use the notation Ln z or Ln(z) to denote the principal branch ofthe natural logarithm; that is, −π < Im Ln z ≤ π. The continued fraction

z

1+z/21 +

z/61 +

2z/61 +

2z/101 +

3z/101 +· · ·

(3.1.2)

converges to Ln(1 + z) for all z in the cut plane D := {z ∈ C; | arg(1 + z)| < π}when we set its value to 0 for z = 0. More precisely,

Ln(1 + z) =z

1+a2z

1 +a3z

1 +· · ·for z ∈ D ,

Page 40: Lisa Lorentzen, Haakon Waadeland Continued Fractions

1.3.2 Hypergeometric functions 27

where for all k ≥ 1

a2k :=k

2(2k − 1), a2k+1 :=

k

2(2k + 1).

Therefore its classical approximants fn(z) (which are rational functions) approxi-mate the value of Ln(1 + z) for z ∈ D. Right now we shall not worry about howone gets this continued fraction expansion or prove its convergence.

We shall compare this rational approximation to the polynomial approximationobtained from the power series expansion

Ln(1 + z) = z − z2

2+

z3

3− z4

4+ · · ·

which converges in the unit disk and diverges for |z| > 1. Let us take z = 1.The series then converges to Ln(2), but very slowly. The first seven continuedfraction approximants are listed in the table below, correctly rounded. The valueis Ln(2) = .69314718, correctly rounded.

n 1 2 3 4 5 6 7fn 1.000000 .666667 .700000 .692308 .693333 .693121 .693152

In order to get the polynomial approximation with the same accuracy we needn > 105 terms of the power series.

Let us also try a z-value where the series does not converge, for instance z = 3. Inthe next table the first 7 classical approximants fn are listed, correctly rounded.The value is now Ln(4) = 2Ln(2), so a better idea is to approximate Ln(2), andthen multiply by 2. But as an experiment we try Ln(1 + 3) ≈ 1.38629436.

Since an → 14 , the approximants Sn(w(z)) (probably) works better if w(z) is the

value of the periodic continued fraction K(z4/1); i.e., w(z) = 1

2 (√

1 + z − 1). Theapproximants are no longer rational, but they give good approximations. For z := 3,w(3) = 1

2, and f∗

n := Sn(12), correctly rounded, is shown in the next line of the table.

n 1 2 3 4 5 6 7fn 3.00000 1.20000 1.50000 1.36363 1.39726 1.38367 1.38744f∗

n 2.00000 1.50000 1.41176 1.39286 1.38806 1.38679 1.38644

Even better approximants Sn(wn(z)) will be suggested in Chapter 5. �

1.3.2 Hypergeometric functions

For n ∈ N and a ∈ C, the Pochhammer symbol (a)n is defined to be

(a)0 := 1, (a)n := (a)n−1 · (a + n − 1) =n−1∏k=0

(a + k) . (3.2.1)

Page 41: Lisa Lorentzen, Haakon Waadeland Continued Fractions

28 Chapter 1: Introductory examples

For given a, b, c ∈ C with c �= 0,−1,−2, . . . , the hypergeometric function 2F1(a, b; c; z)is the analytic function with power series expansion

∞∑n=0

(a)n(b)n

(c)n

zn

n!= 1 +

ab

c

z

1!+

a(a + 1)b(b + 1)c(c + 1)

z2

2!+ · · · (3.2.2)

at 0. Many useful special functions are special cases of 2F1(a, b; c; z). If we assumethat also a and b are different from 0 and the negative integers, then (3.2.2) isan infinite power series whose radius of convergence is 1. The following formalidentities can be established from (3.2.2) by comparing the power series on bothsides term by term:

2F1(a, b; c; z) = 2F1(a, b + 1; c + 1; z)

− a(c − b)c(c + 1)

z 2F1(a + 1, b + 1; c + 2; z)

2F1(a, b + 1; c + 1; z) = 2F1(a + 1, b + 1; c + 2; z)

− (b + 1)(c − a + 1)(c + 1)(c + 2)

z 2F1(a + 1, b + 2; c + 3; z).

Assuming that we avoid zeros in the denominators, this can be written

2F1(a, b; c; z)2F1(a, b + 1; c + 1; z)

= 1 +

−a(c − b)zc(c + 1)

2F1(a, b + 1; c + 1; z)

2F1(a + 1, b + 1; c + 2; z)

2F1(a, b + 1; c + 1; z)2F1(a + 1, b + 1; c + 2; z)

= 1 +

−(b + 1)(c − a + 1)z(c + 1)(c + 2)

2F1(a + 1, b + 1; c + 2; z)

2F1(a + 1, b + 2; c + 3; z)

.

Observe that the denominator on the right hand side of the first equality is equalto the left hand side of the second equality. Furthermore, the denominator on theright hand side of the second equality coincides with the left hand side of the firstone, if in the former a is replaced by a + 1, b is replaced by b + 1 and c by c + 2 inall places. Hence, we have

2F1(a, b; c; z)2F1(a, b + 1; c + 1; z)

= 1 +a1z

2F1(a, b + 1; c + 1; z)2F1(a + 1, b + 1; c + 2; z)

= 1 +a1z

1 +a2z

2F1(a + 1, b + 1; c + 2; z)2F1(a + 1, b + 2; c + 3; z)

= 1 +a1z

1 +a2z

1 +a3z

2F1(a + 1, b + 2; c + 3; z)2F1(a + 2, b + 2; c + 4; z)

Page 42: Lisa Lorentzen, Haakon Waadeland Continued Fractions

1.3.2 Hypergeometric functions 29

and so on, where

a2n+1 =−(a + n)(c − b + n)(c + 2n)(c + 2n + 1)

for n ≥ 0,

a2n =−(b + n)(c − a + n)(c + 2n − 1)(c + 2n)

for n ≥ 1,

(3.2.3)

and we get the continued fraction

1 +a1z

1 +a2z

1 +a3z

1 + · · · . (3.2.4)

Again we do not have any guarantee off-hand that this continued fraction converges,let alone that it converges to this ratio of hypergeometric functions. However, sincean → − 1

4 , it is a consequence of Theorem 4.13 on page 188 that (3.2.4) convergesin the cut plane D := {z ∈ C; 0 < arg(z − 1) < 2π}. That its value indeed is

2F1(a, b; c; z)2F1(a, b + 1; c + 1; z)

(3.2.5)

will be proved in volume 2. This means that its classical approximants provide ra-tional approximation to this ratio. This continued fraction expansion was developedby Gauss ([Gauss13]).

A particularly interesting example occurs for the choice b = 0. Since F (a, 0; c; z) ≡1, we get

2F1(a, 1; c + 1; z) =11 +

a1z

1 +a2z

1 +a3z

1 + · · · (3.2.6)

where

a2n+1 =−(a + n)(c + n)

(c + 2n)(c + 2n + 1), a2n =

−n(c − a + n)(c + 2n − 1)(c + 2n)

(3.2.7)

for n = 0, 1, 2, . . . . Examples of such functions are for instance

2F1(1, 1; 2; z) = −z−1Ln(1 − z)

2F1(12, 1;

32;−z2) = z−1Arctan z (principal part)

z · 2F1(1n

, 1; 1 +1n

;−zn) =∫ z

0

dt

1 + tn.

(3.2.8)

Since an → −14 , the nth tail of (3.2.4) looks more and more like the periodic

continued fraction−z/4

1 +−z/4

1 +−z/4

1 + · · ·which converges to w(z) := (

√1 − z − 1)/2 for z ∈ D. Therefore, we are not

surprised to find than Sn(w(z)) converges considerably faster to the value of (3.2.4)that fn = Sn(0). In Chapter 5 we also suggest faster converging approximantsSn(wn).

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30 Chapter 1: Introductory examples

1.4 Correspondence between power series and con-tinued fractions

1.4.1 From power series to continued fractions

A continued fraction of the form

b0 +a1z

1 +a2z

1 +· · ·+anz

1 +· · ·; 0 �= an ∈ C (4.1.1)

is called a regular C-fraction. (This concept has no particular connection to regularcontinued fractions.) Regular C-fractions often work better than power series asfar as speed of convergence and domain of convergence are concerned. Hence it isof interest to go from a power series to a continued fraction of this particular form.One way of doing this was demonstrated in the previous section. A more primitiveway is the method of successive substitutions ([Lamb61]) due to Lambert (1728 -1777), a colleague of Euler and Lagrange in Berlin. We illustrate this method byan example:

Example 13. We shall compute the circumference L of the ellipse

x2

a2+

y2

b2= 1 , a ≥ b ≥ 0 , a > 0 . (4.1.2)

The well known arc length formula leads to the elliptic integral

L = 4a

∫ π/2

0

√1 − ε2 sin2 θ dθ

where ε :=√

a2 − b2/a is the eccentricity of the ellipse, and thus b2 = a2(1 − ε2).By setting

t :=(a − b

a + b

)2

, (4.1.3)

and expanding the integrand in a series and integrate, we get

L = π(a + b)∞∑

n=0

(1/2n

)2

tn = π(a + b)(1 +

t

22+

t2

26+

t3

28+

25t4

214+ · · ·

), (4.1.4)

([Hutte55], [LoWa85]). One way of finding approximate values for L is to truncatethe series. But we can also transform the series into a continued fraction:

1 +t

22+

t2

26+

t3

28+

25 t4

214+ · · · = 1 +

t/22(1 + t

24 + t2

26 + 25 t3

212 + · · ·)−1

Page 44: Lisa Lorentzen, Haakon Waadeland Continued Fractions

1.4.1 From power series to continued fractions 31

= 1 +t/4

1 − t16 − 3 t2

256 − 9 t3

2048 + · · ·= 1 +

t/41 +

−t/16(1 + 3 t

16 + 9 t2

128 + · · ·)−1

= 1 +t/41 +

−t/161 − 3t

16 − 9t2

256 + · · ·= 1 +

t/41 +

−t/161 +

−3t/16(1 + 3t

16+ · · ·

)−1

= 1 +t/41 +

−t/161 +

−3t/161 − 3t/16 + · · · .

That is, we have obtained the first fraction terms of a continued fraction expansion

1 +t/41 +

−t/161 +

−3t/161 +

−3t/161 +· · ·

(4.1.5)

of L/(π(a + b)). We shall prove in volume 2 that this continued fraction actuallyconverges to L/(π(a + b)). Therefore its classical approximants provide rationalapproximations to L:

L ≈ π(a + b)fn(t) .

This approximation is surprisingly accurate, even for moderate n. For n = 2, simpleas it is, it has an error less than 3 mm for an ellipse with size and eccentricity as theorbit of the planet Mercury. For n = 3 it has for the same ellipse an error roughly= 1/10 of the wave length of blue light. In the flat” case b = 0, which is likely tobe the worst” case (t = 1), the exact value of L is 4a. The approximate formulasgive

L ≈

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

πa f0(1) = 3.1416 a

πa f1(1) = 3.9270 a

πa f2(1) = 3.9794 a

πa f3(1) = 3.9924 a

πa f4(1) = 3.9964 a ,

correctly rounded in the 4th decimal place.

However, we can do better. In the continued fraction (4.1.5) we pretend that allfraction terms from the second one are equal to

−t/161

,

i.e. we replace the continued fraction (4.1.5) by

1 +t/41 +

−t/161 +

−t/161 +

−t/161 +· · ·

.

(This is of course only an experiment, but one can prove that it will lead to a goodapproximation.) This periodic continued fraction has the value

S1(w) = 1 +t/4

1 + w, (4.1.6)

““

Page 45: Lisa Lorentzen, Haakon Waadeland Continued Fractions

32 Chapter 1: Introductory examples

where w is the value of its tail

w =−t/16

1 +−t/16

1 +−t/16

1 +· · ·=

12

(√1 − t/4 − 1

). (4.1.7)

Here we have used that the continued fraction K( z4/1) has the value (

√1 + z−1)/2,

and replaced z by z := −t/4 with 0 < t ≤ 1. Since therefore

S1(w) = 1 +t/4

1 + (√

1 − t/4 − 1)/2= 3 −

√4 − t ,

we get the formulaL ≈ π(a + b)(3 −

√4 − t) . (4.1.8)

This formula was first found by Ramanujan ([Bern78]). For an ellipse of size andeccentricity as the orbit of Mercury this formula has an error less than 2 mm. Forthe degenerate case ( flat” case) it gives the value 3.9834a.

The next formula uses the observation that the continued fraction (4.1.5) has twoequal fraction terms

−3t/161

.

If we pretend that all subsequent ones also are (−3t/16)/1 (again merely an exper-iment), the first tail of (4.1.5) would be

w =13

(−3t/16

1 +−3t/16

1 +−3t/16

1 +−3t/16

1 +· · ·

),

which by (4.1.7) has the value

w =16(√

1 − 3t/4 − 1) .

This suggests the approximate formula

L ≈ π(a + b)

(1 +

t/41 + (

√1 − 3t/4 − 1)/6

),

or, rewritten in a nicer form

L ≈ π(a + b)(

1 +3t

10 +√

4 − 3t

). (4.1.9)

For the flat” case it gives the value 3.9984a. This was published in [JaWa85]. Toour delight it was pointed out by Almquist and Berndt ([AlBe88]) that also thiswas one of Ramanujan’s formulas, although it was not known how he establishedit. It may therefore well be assumed that the method shown in the present sectionwas the one he used. �

Page 46: Lisa Lorentzen, Haakon Waadeland Continued Fractions

1.4.3 Analytic continuation 33

1.4.2 From continued fractions to power series

Example 14. Sometimes we want to go in the opposite direction, i.e. from con-tinued fraction to power series. We shall use our continued fraction expansion forLn(1 + z) as an example; i.e.,

Ln(1 + z) =z

1 +z/21 +

z/61 +

2z/61 +

2z/101 +

3z/101 + · · · .

We look at its first classical approximants and their power series expansions at 0:

f1(z) =z

1= z + 0z2 + 0z3 + 0z4 + · · ·

f2(z) =z

1 + z/2= z − z2

2+

z3

4− z4

8+ · · ·

f3(z) =z

1 + z/21 + z/6

= z − z2

2+

z3

3− 2

9z4 + · · ·

f4(z) =z

1 + z/2

1 +z/6

1 + z/3

= z − z2

2+

z3

3− z4

4+ · · ·

The underlined terms coincide with terms in the power series for Ln(1+z). Observethat the agreement increases with the order of the approximants. It can be proved(and it will be proved in volume 2) that this continues. It is called correspondencebetween the power series and the continued fraction expansion at 0. �

1.4.3 One fraction, two series; analytic continuation

Example 15. The identityz =

z

1 − z + z

used repeatedly leads to the identities

z =z

1 − z +z

1 − z + z, z =

z

1 − z +z

1 − z +z

1 − z + z,

and generallyz =

z

1 − z +z

1 − z +· · ·+z

1 − z + z.

Similarly, the identity−1 =

z

1 − z − 1, z �= 0 ,

Page 47: Lisa Lorentzen, Haakon Waadeland Continued Fractions

34 Chapter 1: Introductory examples

leads to−1 =

z

1 − z +z

1 − z +· · ·+z

1 − z − 1.

Inspired by these two identities we look at the continued fraction

z

1 − z +z

1 − z +· · ·+z

1 − z +· · ·. (4.3.1)

For simplicity we assume that z �= −1. Then its classical approximants fn(z) canbe written

f1(z) =z

1 − z=

z(1 + z)1 − z2

,

f2(z) =z

1 − z +z

1 − z=

z(1 − z)1 − z + z2

=z(1 − z2)1 + z3

,

f3(z) =z

1 − z +z

1 − z +z

1 − z=

z(1 + z3)1 − z4

,

and by Problem 13 on page 49 with x = −z and y = 1,

fn(z) =z(1 − (−z)n)1 − (−z)n+1

=z + (−z)n+1

1 − (−z)n+1.

We therefore distinguish between two cases:

0 < |z| < 1 : The continued fraction converges to z. Since

fn(z) = z + (−z)n+1 + higher powers of z

it corresponds at 0 to the series

z + 0z2 + 0z3 + · · ·

|z| > 1 : The continued fraction converges to −1. Since

fn(z) = −1 + (−z)−n + higher powers of z−1

it corresponds at ∞ to the series

−1 + 0z−1 + 0z−2 + 0z−3 + · · ·

This example shows that one and the same continued fraction expansion may con-verge to two different functions in two different regions and correspond to twodifferent series at two different points (here 0 and ∞). There also exist non-trivialanalogues, where one continued fraction simultaneously represents two different an-alytic functions by convergence and correspondence. But we can say more:

Page 48: Lisa Lorentzen, Haakon Waadeland Continued Fractions

1.4.4 Pade approximation 35

1. The approximants Sn(z) in Example 15 are all equal to z. This implies twothings: The convergence to z in |z| < 1 is accelerated (bull’s eye, the valueis hit right away), and the convergence to z is extended also to the region|z| ≥ 1, i.e. we have an analytic continuation of the limit function in |z| < 1to the whole plane.

2. The approximants Sn(−1) are all equal to −1. This implies two things: Accel-eration of convergence to −1 in |z| > 1 (again bull’s eye), and the convergenceto −1 is extended to 0 < |z| ≤ 1, i.e. we have an analytic continuation of thelimit function in |z| > 1 to the whole plane, minus the origin.

Also these properties have their non-trivial analogues. For instance, if

an(z) → z and bn(z) =→ 1 − z (4.3.2)

then the continued fraction K(an(z)/bn(z)) converges to one function for |z| < 1and to another function for |z| > 1. We have already seen that Sn(z) (probably)converges faster to its value than its classical approximants Sn(0). If the convergencein (4.3.2) is fast enough, then Sn(z) actually converges in a larger domain, andthus provides analytic continuation under proper conditions. So also for Sn(−1).This idea was presented and developed further in [Waad66], [Waad67], [ThWa80b],[Jaco88], [Lore93]. We return to this idea in volume 2.

1.4.4 Pade approximation

Problem: For a given (formal) power series

L(z) :=∞∑

k=0

ckzk

and given non-negative integers m and n, find the rational function

Rm/n(z) :=Pm/n(z)Qm/n(z)

whose Taylor series at z = 0 agrees with L(z) as far out as possible, whenPm/n and Qm/n are polynomials of degree ≤ m and ≤ n respectively.

Solutions to this problem are essentially what is called Pade approximants (in oneof the two possible definitions). The idea is due to Stirling ([Stir30]) who probablybased his work on a paper by Bernoulli ([Berno78]). For finer distinctions we referto the extensive literature on Pade approximation, such as for instance [BaGM96].Pade approximants for a given series L(z) are usually presented in an array calleda Pade table:

Page 49: Lisa Lorentzen, Haakon Waadeland Continued Fractions

36 Chapter 1: Introductory examples

m \n 0 1 2 3 4 · · ·0 R0/0 R0/1 R0/2 R0/3 R0/4 · · ·1 R1/0 R1/1 R1/2 R1/3 R1/4 · · ·2 R2/0 R2/1 R2/2 R2/3 R2/4 · · ·3 R3/0 R3/1 R3/2 R3/3 R3/4 · · ·4 R4/0 R4/1 R4/2 R4/3 R4/4 · · ·...

......

......

.... . .

The 0-column {Rn/0} is just the partial sums of L(z). (Some authors have chosento use the transposed table where the numerator-degree increases in each row andthe denominator-degree increases column-wise.) For practical computation of Padeapproximants different algorithms are available. We refer to ([CuWu87], sect 2.3)and references therein. Some key words deserve to be mentioned: the qd-algorithm,the Viskovatov algorithm, Gragg’s algorithm and the ε-algorithm.

Example 16. Let

L(z) := 1 +∞∑

n=1

(−1)n+1 zn

2n − 1= 1 + z − z2

3+

z3

5− z4

7+

z5

9− z6

11+ · · · .

The upper left corner of the Pade table is then

m \n 0 1 2 · · ·

0 11

1 − z

3

3 − 3z + 4z2· · ·

1 1 + z3 + 4z

3 + z

15 + 21z

15 + 6z − z2· · ·

23 + 3z − z2

3

15 + 24z + 4z2

15 + 9z

105 + 195z + 64z2

105 + 90z + 9z2· · ·

......

......

. . .

For instance

R2/1(z) =15 + 24z + 4z2

15 + 9z∼ 1 + z − z2

3+

z3

5− 3z4

25+

9z5

125+ · · ·

which agrees with L(z) up to and including the term of order 3. It is quite easy tosee that Rm/n(z) always agrees with L(z) up to and including at least the term oforder m + n.

Now, in our particular case L(z) can be written

L(z) = 1 + z 2F1( 12 , 1; 3

2 ;−z)

Page 50: Lisa Lorentzen, Haakon Waadeland Continued Fractions

1.4.4 Pade approximation 37

where the hypergeometric series 2F1( 12 , 1; 3

2 ;−z) has a continued fraction expansiongiven by (3.2.6) - (3.2.7) on page 29; that is,

2F1(12, 1; 3

2;−z) =

11+

12z

1 · 31 +

22z

3 · 51 +

32z

5 · 71 +

42z

7 · 91 +· · ·

.

Therefore L(z) has the continued fraction expansion

1 +∞Kn=1

anz

1where a1 = 1 and an+1 =

n2

4n2 − 1for n ≥ 1.

As in the previous example, the correspondence shows up when we compare L(z)to the Taylor expansions of the classical approximants

f1(z) = 1 + z

f2(z) = 1 +z

1+

13z

1=

3 + 4z

3 + z

f3(z) = 1 +z

1+z/31 +

4z/151

=15 + 24z + 4z2

15 + 9z

f4(z) = 1 +z

1+z/31 +

4z/151 +

9z/351

=105 + 195z + 64z2

105 + 90z + 9z2

and so on. We recognize these approximants from the Pade table. The approximantsof the regular C-fraction are indeed the staircase diagonal Pade approximants

R0/0, R1/0, R1/1, R2/1, R2/2, R3/2, . . .

as illustrated below.

m \n 0 1 2 3 4 5 6 · · ·0 R0/0 · · ·1 R1/0 R1/1 · · ·2 R2/1 R2/2 · · ·3 R3/2 R3/3 · · ·4 R4/3 R4/4 · · ·5 R5/4 R5/5 · · ·6 R6/5 R6/6 · · ·...

.

.....

.

.....

.

.....

.

... . .

This means that we can

• use regular C-fraction expansions to compute Pade approximants.

• use convergence theory for continued fractions to prove convergence of Padeapproximants.

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38 Chapter 1: Introductory examples

1.5 More examples of applications

1.5.1 A differential equation

One can solve certain differential equations by means of continued fractions. Weinclude a very simple example here. (More substantial examples can for instancebe found in [Khov63], [Steen73], [Ince26], [Waad83].)

Example 17. To solve the differential equation

y = 2y′ + y′′

we first differentiate the equation repeatedly to get

y′ = 2y′′ + y′′′ ,...

y(n) = 2y(n+1) + y(n+2) .

Assuming that we do not divide by 0, this gives

y

y′ = 2 +1

y′/y′′ ,

y′

y′′ = 2 +1

y′′/y′′′ ,

...y(n)

y(n+1)= 2 +

1y(n+1)/y(n+2)

.

From this it follows that

y

y′ = 2 +12+

12+· · ·+

12︸ ︷︷ ︸

n+1

+1

y(n+1)/y(n+2).

This suggests the continued fraction

2 +12+

12+· · ·+

12+· · ·

.

We know from Example 8 on page 19 that it converges to√

2 + 1, which suggeststhat the differential equation has a solution y satisfying

y

y′ =√

2 + 1 ,

Page 52: Lisa Lorentzen, Haakon Waadeland Continued Fractions

1.5.2 Moment problems and divergent series 39

that is,y′

y=

√2 − 1 ,

from which it would follow that

y = C exp[(√

2 − 1)x]

.

This is actually a solution of the given differential equation.

By the warning on page 20, the value 1 −√

2 is also associated with the continuedfraction above. Since the recursion more than the asymptotics is the important partof the investigation so far, we try

y

y′ = 1 −√

2 .

This leads toy′

y= −(

√2 + 1),

and thus y = C1 exp(−(√

2 + 1)x). A quick check shows that this also is a solutionof the differential equation. The method therefore gives the general solution

y = C1 exp[(√

2 − 1)x] + C2 exp[(−√

2 − 1)x] .

There is of course no good reason to use this method” for the present differentialequation, since there exists a perfectly good, simple method taught in elementarycalculus. But there are cases, where this idea leads to non-trivial results. �

1.5.2 Moment problems and divergent series

In Example 10 on page 25 and Example 12 on page 26 a series with convergenceradius 1 corresponded to a continued fraction which converged to the right valuein a much larger region D containing the convergence disk of the series. Therefore,transforming the series to the continued fraction in question worked as a method tosum the divergent series. The effect is even more spectacular in the next examplewhere the power series diverges for all z ∈ C.

Example 18. Take a look at the series

L(z) :=∞∑

n=0

n!(−z)−n = 1 − 1!z

+2!z2

− 3!z3

+ · · · .

Page 53: Lisa Lorentzen, Haakon Waadeland Continued Fractions

40 Chapter 1: Introductory examples

It diverges for all z ∈ C. A function closely associated with this series is the functionF , given as an integral

F (z) :=∫ ∞

0

ze−t

z + tdt =

∫ ∞

0

e−t · 11 − (−t/z)

dt

=∫ ∞

0

e−t

(1 − t

z+

t2

z2− · · · + (−t)n

zn+

(−t)n+1

zn(z + t)

)dt

= 1 − 1!z

+2!z2

− 3!z3

+ · · · + (−1)n n!zn

+∫ ∞

0

e−t (−t)n+1

zn(z + t)dt.

This function is holomorphic in the cut plane | arg(z)| < π, and

F (z) − σn(z) =∫ ∞

0

e−t (−t)n+1

zn(z + t)dt for σn(z) =

n∑m=0

(−1)m m!zm

.

For an arbitrary, but fixed value of α ∈ [0, π2) we assume in the following that z is

in a part of the plane where | arg(z)| ≤ α. Then

|F (z) − σn(z)| ≤ 1|z|n

∫ ∞

0

e−ttn+1

Re(z + t)dt

≤ 1|z|n

∫ ∞

0

e−ttn+1dt

|z| cos α=

(n + 1)!|z|n+1 cos α

.

Now, keep n fixed, and let z → ∞. Then the right hand side (and hence the lefthand side) tends to 0. This property is expressed as follows: The series L(z) is anasymptotic expansion of F (z) as z → ∞ in the angular opening where | arg(z)| ≤ α,and we write F (z) ∼ L(z) as z → ∞.

This is one bridge between the divergent series L(z) and the function F . Anotherbridge is by way of continued fractions. It can be shown that the continued fraction

11+

1/z

1 +1/z

1 +2/z

1 +2/z

1 +3/z

1 +3/z

1 +4/z

1 +4/z

1 +· · ·

corresponds at ∞ to the series L(z). It converges in the cut plane | arg(z)| < π to ananalytic function f (Theorem 3.21 on page 124). Indeed, in volume 2 we shall seethat f = F . So again we have given the divergent series a meaning by convertingit to a convergent continued fraction. �

What we have seen here is closely connected to the moment problem, or moreprecisely the Stieltjes moment problem. A real non-decreasing function Ψ on R

+

with infinitely many points of increase is called a distribution function, and

cn :=∫ ∞

0

tndΨ(t) (5.2.1)

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1.5.2 Moment problems and divergent series 41

is called the nth moment with respect to Ψ. With

F (z) :=∫ ∞

0

z dΨ(t)z + t

=∫ ∞

0

( ∞∑n=0

(−1)n

(t

z

)n)

dΨ(t) , (5.2.2)

termwise integration leads to the series

L(z) :=∞∑

n=0

(−1)n cn

zn. (5.2.3)

The Stieltjes moment problem is as follows: For a given sequence {cn}∞n=0 of realnumbers, find a distribution function Ψ on (0,∞) such that

cn =∫ ∞

0

tn dΨ(t) for n = 0, 1, 2, . . . .

Any such function Ψ is called a solution of the moment problem. Related tasks arenow:

1. Find necessary and sufficient conditions on {cn} for existence of a solution Ψ.

2. Find necessary and sufficient conditions for a solution Ψ to be unique up toan additive constant.

It is the beauty of this theory that the Stieltjes moment problem for {cn} has asolution if and only if the series (5.2.3) corresponds at z = ∞ to a continued fractionof the form

a1

1 +

∞Kn=2

an/z

1where all an > 0

called an S-fraction. The solution is unique if and only if this continued fractionconverges for z > 0. The value of the continued fraction is then the function F (z)in (5.2.2), from which the distribution function can be derived. This was proved byStieltjes in his famous 1894-paper ([Stie94]).

Example 19. As a consequence of Example 18, the moment problem with themoments cn := n! has a unique solution Ψ(t) with dΨ(t) = e−t dt = Ψ′(t)dt. Thesolution is usually written Ψ(t) := 1− e−t, since it then increases from 0 to 1 whent increases from 0 to ∞ . �

Terminology from measure theory is often used in this connection. In Example 19the measure Ψ is absolutely continuous with derivative e−t. Since∫ ∞

0

dΨ(t) =∫ ∞

0

e−tdt = 1,

it is an example of a probability measure.

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42 Chapter 1: Introductory examples

1.5.3 Orthogonal polynomials

Example 20. Let us define the inner product

〈f, g〉 :=∫ 1

−1

f(x)g(x) dx

in the space of real functions continuous on [−1, 1]. The Legendre polynomials{Pn}∞n=0 are the real polynomials Pn of degree n with the property that

〈Pm, Pn〉 =∫ 1

−1

Pm(x)Pn(x)dx =2

2n + 1δm,n for all m, n ≥ 0,

where δm,n is the Kronecker delta; i.e., δm,n = 1 if m = n and 0 otherwise. We saythat the Legendre polynomials are orthogonal on [−1, 1] with respect to this innerproduct.

The Legendre polynomials {Pn} are known to be the solution of the recurrencerelation

(n + 1)Pn+1(x) = (2n + 1)xPn(x) − nPn−1(x) for n = 1, 2, 3, . . .

with initial expressionsP0(x) := 1, P1(x) := x,

so for instance

P2(x) =32x2 − 1

2, P3(x) =

52x3 − 3

2x, P4(x) =

358

x4 − 154

x2 +38.

An alternative way of writing this recurrence relation is

Pn+1(x) =2n + 1n + 1

xPn(x) − n

n + 1Pn−1(x) .

Therefore Pn(x) are exactly the canonical denominators Bn(x) for the continuedfraction

1x+

−1/23x/2 +

−2/35x/3 +

−3/47x/4 +· · ·

.

Example 21. The Chebyshev polynomials Un(x) of the second kind are the solu-tions of the recurrence relation

Un+1(x) = 2xUn(x) − Un−1(x)

with initial values U0(x) := 1 and U1(x) := 2x. Therefore {Un(x)} are the canonicaldenominators of the continued fraction

−12x +

−12x +· · ·+

−12x +· · ·

.

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1.5.4 Thiele interpolation 43

One can prove that ∫ 1

−1

Um(x)Un(x)(1 − x2)1/2dx =π

2δm,n.

We therefore say that {Un(x)} are orthogonal on [−1, 1] with respect to the weightfunction W (x) := (1 − x2)1/2. �

These are merely two simple cases of a general theory. The connection betweenorthogonal sequences and continued fractions is the three term recurrence rela-tion: a famous theorem on orthogonal polynomials, commonly but incorrectly calledFavard’s theorem, says that {Pn} is a sequence of orthogonal polynomials with re-spect to some real measure dΨ(t) if and only if {Pn} is a solution of a three termlinear recurrence relation

Pn(x) = (x − cn)Pn−1(x) − λnPn−2(x) for n = 1, 2, 3, . . .

with initial values P−1(x) := 0, P0(x) := 1 (or some other positive constant), whereλn > 0 and cn ∈ R. On the other hand, a three term recurrence relation of thisform gives rise to a continued fraction. The continued fractions in question are theJacobi continued fractions, briefly called J-fractions

λ1

x − c1 −λ2

x − c2 − · · ·−λn

x − cn − · · · .

This connection can be exploited to find asymptotic properties and zero-free regionsfor {Pn(x)}.

1.5.4 Thiele interpolation

Let f be an unknown function with known values f(zn) at given distinct pointsz0, z1, z2 . . . in C. We want to find f . What we do is successively to find constantsφm ∈ C such that the functions

Fn(z) := φ0 +n

Km=1

z − zm−1

φmfor n = 0, 1, 2, . . .

satisfyFn(zk) = f(zk) for k = 0, 1, . . . n − 1. (5.4.1)

The idea is then: if the resulting Thiele continued fraction

φ0 +∞Kn=1

z − zn−1

φn

converges, then its value is such a function f(z). Normally we do not get the wholecontinued fraction — we have to stop after a finite number of terms, say at Fn(z).

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44 Chapter 1: Introductory examples

Then we have a rational function Fn(z) which interpolates f(z) at the first n pointszk.

There exists an algorithm for computing the numbers φm (inverse differencies). Itfails if not all operations are meaningful, but otherwise it works as follows:

φ0 := φ0[z0] := f(z0), φ1 := φ1[z0, z1] :=z1 − z0

f(z1) − f(z0),

φk := φk[z0, z1, . . . , zk] :=zk − zk−1

φk−1[z0, . . . , zk−2, zk] − φk−1[z0, . . . , zk−1]

for k = 2, 3, 4, . . . . Then (5.4.1) holds if

Fn(zk) = φ0 +zk − z0

φ1 +zk − z1

φ2 + · · ·+zk − zk−1

φk

and φk+1 +zk − zk+1

φk+2 + · · ·+zk − zn−1

φn�= 0 for k < n.

(5.4.2)

Example 22. Let

z0 := 0, z1 := 4, z2 := 9, z3 := 25,

f(z0) := 0, f(z1) := 2, f(z2) := 3, f(z3) := 5.

In this case we find φ0[z0] = 0 and

φ1[z0, z1] =4 − 02 − 0

= 2, φ1[z0, z2] =9 − 03 − 0

= 3, φ1[z0, z3] =25 − 05 − 0

= 5,

φ2[z0, z1, z2] =9 − 43 − 2

= 5, φ2[z0, z1, z3] =25 − 45 − 2

= 7.

φ3[z0, z1, z2, z3] =25 − 97 − 5

= 8.

0

1

2

3

4

5

5 10 15 20 25z

Therefore

F3(z) =z

2+z − 4

5 +z − 9

8

=z(31 + z)10(3 + z)

.

We verify that F3(0) = 0, F3(4) = 2, F3(9) =3, F3(25) = 5. This is therefore one function(of many possibilities) with the required val-ues at 0, 2, 9 and 25. (Another possibilityis√

z.) The figure shows the graph of F3(z)for real z > 0. �

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1.5.5 Stable polynomials 45

1.5.5 Stable polynomials

A polynomialQn(r) := rn + a1r

n−1 + · · · + an (5.5.1)

is called stable if all zeros lie in the left half plane Re(r) < 0. The concept isimportant in the theory of linear differential equations with constant coefficients,as illustrated in the example below.

Example 23. To solve the differential equation

y + 3y + 2y = 0

where y denotes the derivative with respect to the real variable t, one solves thecorresponding characteristic equation

Q2(r) := r2 + 3r + 2 = 0.

It has the roots r1 = −1 and r2 = −2. Hence the differential equation has thegeneral solution

y(t) = C1e−t + C2e

−2t,

which means that y(t) → 0 as t → + ∞. This corresponds to the fact that Q2 is astable polynomial.

Similarly, the characteristic equation for the differential equation...y + 4y + 6y + 4y = 0

isQ3(r) := r3 + 4r2 + 6r + 4 = 0

which has the solutions r1 = −2, r2,3 = −1 ± i, so Q3 is stable. Every solutiony(t) of the differential equation is therefore damped; i.e., y(t) → 0 as t → ∞. Thegeneral solution is of course

y(t) = C1e−2t + e−t(C2 cos t + C3 sin t).

We want to decide whether a polynomial is stable or not, without having to computeits zeros. This has been done for the case of real coefficients by Hurwitz ([Hurw95]).For a given polynomial (5.5.1) he introduced the auxiliary polynomial

Pn(r) := a1rn−1 + a3r

n−3 + a5rn−5 . . . ,

and proved that Qn is stable if and only if Pn(r)/Qn(r) can be written as a termi-nating continued fraction of the form

11 + d1r +

1d2r +· · ·+

1dnr

where all dk > 0.

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46 Chapter 1: Introductory examples

The proof can be found in [LoWa92], p 470, along with a result for the complexcase. Here we only check the result for Q2(r) and Q3(r) in Example 23.

Q2(r) = r2 + 3r + 2, P2(r) := 3r,P2(r)Q2(r)

=1

1 + 13 r +

132 r

,

Q3(r) = r3 + 4r2 + 6r + 4, P3(r) := 4r2 + 4P3(r)Q3(r)

=1

1 + 14

r +1

45

r +1

54

r.

1.6 Remarks

1. The birth of continued fractions. The euclidean algorithm, the basis for theregular continued fraction, goes back to around 300 B.C., but in a slightly differentform, as a subtraction algorithm rather than a division algorithm ([Eucl56]). Butat that time it did not lead to a continued fraction.

The birth of continued fractions, like many other things in the culture of mankind,took place in Italy in the renaissance, by Bombelli in 1572 ([Bomb72]) and Cataldi in1613 ([Cata13]), in both cases as approximate values for a square root. But alreadyFibonacci, around 1200, i.e., long before the renaissance, touched upon the idea of acontinued fraction, but an ascending one ([Fibo02]). Of the further development wemention, as we did in Section 1.2.4 of the present chapter, Huygens’ use of continuedfractions for designing cogwheels ([Huyg95]). In 1965 Lord Brouncker derived a non-terminating continued fraction for π/4 on request by Wallis who was working on hisbook ([Wallis85]) at that time. Lord Brouncker neither published his result nor hismethod, but Wallis writes enough about the ideas for Khrushchev to reconstructthem in his very interesting paper ([Khru06a]). In the same book Wallis also givesrecurrence relations for the approximants of a continued fraction.

With Euler ([Euler67]) a general theory for continued fractions began to develop, atheory where Gauss’ famous continued fractions for ratios of hypergeometric func-tions ([Gauss13]) was an early high point, still of great value in modern theory forcomputing special functions.

2. Additional literature. For those who want to go deeper into the analytic theory ofcontinued fractions we refer to the three standard monographs in the field: the clas-sical text-book by Perron ([Perr54], [Perr57] (in German)), Wall’s book ([Wall48])and the exposition by Jones and Thron ([JoTh80]). In Henrici’s 3 volume work onApplied and Computational Complex Analysis ([Henr77]) a large portion of volume2 is devoted to analytic theory of continued fraction. Khovanskii’s book ([Khov63])contains some interesting applications.

The history of continued fractions up to 1939 is described in Brezinski’s book([Brez91]). Also the texts mentioned above, in particular the book by Jones andThron, contain interesting comments on the historic development of concepts, meth-ods and applications. For the computational aspects of continued fractions we rec-ommend the handbook ([CJPVW7]).

To most people (meaning mathematicians) continued fractions are the regular con-tinued fractions in number theory. More information on this side of the continued

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Remarks 47

fraction theory can be found in [Perr54], [Ries85] and the recent book by Hensley([Hens06]).

3. The definition of a continued fraction. Our definition of a continued fractionb0 +K(an/bn) on page 5 is inspired by the classical definition by Henrici and Pfluger([HePf66]) who defined b0 +K(an/bn) as the ordered pair (({an}, {bn}), {fn}) where{fn} is the sequence of classical approximants.

Neither of these two definitions is an absolutely obvious choice. Approximants

fn := b0 +a1

b1 +

a2

b2 +· · ·+an−1

bn−1 + an

are for instance also known from the classical literature. And the sequence {Sn} inour definition could be replace by

sn := ϕ−1n−1 ◦ sn ◦ ϕn, Sn := s0 ◦ s1 ◦ · · · sn = Sn ◦ ϕn

for any sequence {ϕn} from M with ϕ−1(w) ≡ w.

Or we could follow Beardon ([Bear04]) and define a continued fraction as a sequence{S∗

n} from M with the properties that

S∗n := s∗1 ◦ s∗2 ◦ · · · ◦ s∗n where sk ∈ M with s∗k(a) = b

for two fixed (possibly coinciding) constants a, b ∈ C.

4. G-continued fractions. A continued fraction K(an/bn) is closely connected tothe recurrence relation

Xn = bnXn−1 + anXn−2 for n = 1, 2, 3, . . .

where an �= 0. For one thing, {An} and {Bn} are solutions of this recurrence.Moreover,

−Xn−1

Xn−2=

an

bn − Xn/Xn−1= sn(−Xn/Xn−1)

for every non-trivial solution. Hence {−Xn/Xn−1} is a tail sequence for K(an/bn).

A G-continued fraction of dimension N is connected to a longer recurrence relation

Xn + a(N)n Xn−1 + a(N−1)

n Xn−2 + · · · + a(1)n Xn−N = 0

where a(1)n �= 0. Let {Bn}∞n=1−N and {A(k)

n }∞n=1−N for k = 1, 2, . . . , N − 1 be thesolutions of this recurrence with initial values⎛⎜⎜⎜⎜⎝

A(1)1−N A

(1)2−N · · · A

(1)0

......

...

A(N−1)1−N A

(N−1)2−N · · · A

(N−1)0

B1−N B2−N · · · B0

⎞⎟⎟⎟⎟⎠ =

⎛⎜⎜⎜⎝1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

⎞⎟⎟⎟⎠Then (A

(1)n

Bn, A

(2)n

Bn, . . . , A

(N−1)nBn

) are the approximants of the G-continued fraction.These continued fractions are also called vector-valued continued fractions. Formore information and further references we refer to ([LoWa92], p 225).

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48 Chapter 1: Introductory examples

1.7 Problems

1. Continued fractions with given An, Bn. Construct the continued fraction with

(a) An = 2n, Bn = 3n + 1 for n ≥ 0,

(b) An = sin nπ2

, Bn = cos nπ2

for n ≥ 0,

(c) A2n−1 = n2, B2n−1 = n2, A2n = 2n2 + 1 and B2n = 2n2 for n ≥ 1.

2. Continued fraction identities. Construct a continued fraction K(1/bn) whichconverges to

(a) 1. (b) −1. (c) ∞.

3. Periodic continued fraction. Assume that the continued fraction

1

1+

1

1+

1

1+· · ·

converges. Prove that it its value is (√

5 − 1)/2.

4. Denominators = 1. Prove that Bn = 1 for all n ≥ 0 for K(an/bn) if and only ifb1 = 1 and an + bn = 1 for n ≥ 2.

5. Reversed terminating continued fraction. Let KNn=1(an/bn) for some N ∈ N

be a terminating continued fraction (with all an �= 0), and let

bN +aN

bN−1 +

aN−1

bN−2 +· · ·+a2

b1

be its reversed continued fraction. Prove that the (N − 1)th denominator BN−1 isthe same for the two terminating continued fractions.

6. Periodic continued fraction. The periodic continued fraction

Kn=1

z

2=

z

2+

z

2+

z

2+· · ·

converges for all z ∈ D := {z ∈ C; | arg(z + 1)| < π}.

(a) What possible values can it have?

(b) What is its value for z > 0? Use this to find a continued fraction expansionfor

√5.

(c) Let z := 1−2i. Compute the first classical approximants for K(z/2) and guessits value.

7. Approximants. Compute the first three classical approximants and the first threeapproximants Sn(1) of the continued fraction K(n/1) by means of

(a) the forward recurrence algorithm,

(b) the backwards recurrence algorithm,

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Problems 49

(c) the Euler-Minding formula and its generalization.

8. Euclidean algorithm. Use the Euclidean algorithm to find the greatest commondivisor

(a) gcd(119, 221), (b) gcd(3839, 1711), (c) gcd(49907, 22243).

9. Terminating regular continued fraction. Find the regular continued fractionexpansion and its first few approximants for the following numbers:

(a) 47/99, (b) 3839/1711, (c) 15015/7429.

10. Periodic regular continued fractions. For each of the given numbers, find itsregular continued fraction expansion and show that it is periodic (possibly after afew fraction terms). Further compute some of its first approximants.

(a)√

82, (b)√

51, (c)√

53.

11. ♠ Periodic regular continued fractions. Prove that the value f of a (non-

terminating) regular continued fraction K(1/bn) has the form f =M +

√Q

Nwith

M, N, Q are integers with N �= 0, Q > 0 and√

Q �∈ N, if and only if {bn} is periodic(from some n on). (Lagrange [Lagr70], [Perr54], p 66.)

12. ♠ The family M and matrices. Show that if we identify transformations

τn(w) :=anw + bn

cnw + dnfrom M with the matrix Tn :=

(an bn

cn dn

), then τ1 ◦ τ2 corre-

sponds to the matrix product T1T2.

13. ♠ Periodic continued fractions. Prove that

fn = −xyxn − yn

xn+1 − yn+1

for the continued fraction

− xy

x + y −xy

x + y −xy

x + y − · · · ; x, y �= 0, x �= y.

14. Best rational approximation. Find the first five classical approximants of theregular continued fraction expansion of e = 2.718281828459 . . . and compare themto other rational approximations with no larger denominators.

15. Diophantine equation. Find all positive integer solutions (x, y) with y < 20 ofthe diophantine equation 19x − 13y = 5.

16. Diophantine equations. Find the general solution of the linear diophantine equa-tion

(a) 11x − 3y = 5, (b) 99x − 47y = 3,

(c) 3839x − 1711y = 1, (d) 3839x + 1711y = 1.

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50 Chapter 1: Introductory examples

17. Fibonacci sequence.

(a) Use the identity √5 − 1

2=

1

1 +

√5 − 1

2

to produce a continued fraction by the procedure of Example 8. Compute thefirst 7 classical approximants fn and compare the values to (

√5 − 1)/2.

(b) Prove that fn = Fn−1/Fn where F0 = 1, F1 = 1, F2 = 2, F3 = 3, F4 = 5, andgenerally

Fn+1 = Fn + Fn−1 for n ≥ 1 .

(c) Prove that fn → (√

5− 1)/2. (The sequence {Fn} is the sequence of Fibonaccinumbers, and (

√5 − 1)/2 ≈ 0.61803 . . . is the golden ratio.)

18. ♠ Regular continued fractions. Let {An} and {Bn} be the canonical numeratorsand denominators for the regular continued fraction K(1/bn) as in Theorem 1.7 onpage 17.

(a) Show that Bnf − An = Bn(f − fn).

(b) Show that |Bn+1f −An+1| = −bn+1|Bnf −An|+ |Bn−1f −An−1| > 0. (Hint:Use the recurrence relations for {An} and {Bn} and the fact that f − fn

alternates in sign.)

(c) Show that |Bn−1f − An−1| > |Bnf − An|.

19. ♠ Divergence of periodic continued fraction. Prove that if the continuedfraction

a

1+

a

1+

a

1+· · ·converges, then it converges to one of the roots of the equation x = a/(1 + x). Usethis to prove that the continued fraction diverges for all a < −1/4.

20. ♠ The binomial series. Let α be real and not a positive integer, and let 2F1(a, b; c; z)be the hypergeometric series (3.2.2) on page 28. Prove that

2F1(−α, 1; 1;−z) = (1 + z)α for |z| < 1.

Use (3.2.5) - (3.2.6) on page 29 to find a continued fraction expansion of the formK(anz/1) for z(1 + z)α.

21. From power series to continued fraction. Use the procedure of Example 13 onpage 30 to find the first terms

a1z

1 +

a2z

1 +

a3z

1 +

a4z

1 + · · ·

of a continued fraction expansion of f(z) := ez − 1.

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Problems 51

22. From continued fraction to power series. Use the procedure of Example 14 onpage 33 to find the first terms of the power series expansion at 0 corresponding tothe continued fraction

z

1+

−z/2

1 +

z/6

1 +

−z/6

1 +· · ·

23. ♠ From approximants to continued fraction. Let {fn}∞n=0 with f0 := 0 andfn �= fn−1 for all n be a given sequence of complex numbers. Prove that thecontinued fraction K(an/bn) with b1 := 1, a1 := f1 and

an := − fn − fn−1

fn−1 − fn−2, bn :=

fn − fn−2

fn−1 − fn−2for n = 2, 3, 4, . . .

has classical approximants {fn}. (Bernoulli [Berno75].) (Hint: see Problem 4.)

24. From approximants to continued fraction. Prove that a given sequence {fn}of complex numbers is a sequence of classical approximants for a continued fractionb0 + K(an/1) if and only if b0 = f0, fn �= fn−1, fn �= fn−2 and

a1 = f1 − f0, a2 = −f1 − f2

f0 − f2and an = − (fn−3 − fn−2)(fn−1 − fn)

(fn−3 − fn−1)(fn−2 − fn)

for n > 2. (Bernoulli [Berno75].)

25. ♠ From series to continued fraction. Let σn :=∑n

k=1 ck be the partial sumsof the series

∑∞k=1 cn where all cn �= 0. Show that the continued fraction K(an/bn)

with

a1 := c1, an := − cn

cn−1, b1 := 1, bn :=

cn−1 + cn

cn−1

has canonical numerators An = σn and canonical denominators Bn = 1 and thusclassical approximants fn = σn. (Euler ([Euler48]), Stern ([Stern32]), Glaisher([Glai74]).)

26. ♠ From infinite product to power series. Let pn :=∏n

k=0 ak be the par-tial products of the infinite product

∏∞k=0 ak where all ak �= 0, 1. Show that the

continued fraction

a0 +a0(a1 − 1)

1 −a1(a2 − 1)/(a1 − 1)

(a1a2 − 1)/(a1 − 1) −a2(a3 − 1)/(a2 − 1)

(a2a3 − 1)/(a2 − 1) −

· · ·−an−1(an − 1)/(an−1 − 1)

(an−1an − 1)/(an−1 − 1) − · · ·has canonical numerators An = pn and canonical denominators Bn = 1 and thusclassical approximants fn = pn. (Stern [Stern32], Glaisher [Glai74].)

27. From continued fraction to continued fraction. Let {An}∞n=0 and {Bn}∞n=0

be the canonical numerators and denominators of 1 + K(an/1).

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52 Chapter 1: Introductory examples

(a) Show that

A1, A0, A3, A2, A5, A4, . . . and B1, B0, B3, B2, B5, B4, . . .

are the sequences of canonical numerators and denominators for the continuedfraction

1 + a1 −a1

1 +

1 + a3

a2 +

a2a31+a3

11+a3

+

1 + a4

a4 +

a4a51+a5

11+a5

+· · ·.

(Perron [Perr57], p 7.)

(b) Show that the sequence of classical approximants for

1 + a1−a1

1 +

1 + a3

a2 +

a2a3

1 +

(1 + a3)(1 + a5)

a4 +

+

a4a5

1 +

(1 + a5)(1 + a7)

a6 +· · ·is the same as for the continued fraction in (a).

(c) Find the continued fraction with canonical numerators and denominators

A0, A2, A1, A4, A3, A6, . . . and B0, B2, B1, B4, B3, B6, . . . .

28. Pade table. Given the power series∑∞

n=0 n!zn. Find the 3 × 3 upper left entriesin the Pade table for the series.

29. Differential equation. Solve the differential equation y = 2y′ + 3y′′ by using themethod” in Subsection 1.5.1.

30. Moments. Let a be an arbitrary positive number, and let Ψ be the distributionfunction given by Ψ(t) = 0 for 0 ≤ t < a and Ψ(t) = 1 for a ≤ t < ∞.

(a) Find the moments cn with respect to Ψ(t).

(b) Find the S-fraction corresponding to the series∑∞

m=0(−1)ncnz−n. (The seriescan be written in closed form, and the S-fraction is terminating.)

31. Moments. Let Ψ be the distribution function given by Ψ(t) = 0 for 0 ≤ t < 1 andΨ(t) = 1/2 for 1 ≤ t < 2 and Ψ(t) = 1 for 2 ≤ t < ∞. Try to answer questions (a)and (b) in Problem 30. Do you recognize anything from your everyday life?

32. Thiele fractions. Find the shortest Thiele fraction which has the values 1, 2 and3 at the points 0, 1 and 2 respectively.

33. ♠ Thiele fractions. Find the shortest Thiele fraction F (z) which has the values1, 2, 3, b at the points 0, 1, 2 and 3 respectively. What can be said about F (z) fordifferent values of b?

34. ♠ Stable polynomials.

(a) For which values of k is the polynomial Q3(x) := x3 + 3x2 + 3x + 1 + k stable?

(b) For which values of p and q is the polynomial Q2(x) := x2 + px + q stable?

(c) For which values of p q and r is the polynomial Q3(x) := x3 + px2 + qx + rstable?

Page 66: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Chapter 2

Basics

The transformations sn(w) := an/(bn + w) are linear fractional transformationswith sn(∞) = 0. Hence Sn := s1 ◦ s2 ◦ · · · ◦ sn are linear fractional transformationswith Sn(∞) = Sn−1(0), and a continued fraction is essentially a sequence of linearfractional transformations {Sn} with Sn(∞) = Sn−1(0) for all n. It is thereforenatural to define convergence of continued fractions as convergence of {Sn} in somesense. Traditionally what is required is that {Sn(0)} converges in C. This definitionis easy to grasp, but not quite what we are after. In this chapter we shall definean alternative concept, general convergence, which essentially requires that {Sn}converges to a constant function in C. The only snag is that we have to acceptexceptional sequences to be defined.

Two important tools in the convergence theory for continued fractions are tail se-quences and value sets which both will be defined in this chapter. The interplaybetween these two gives a deeper understanding of a continued fraction, but alsouseful methods for proving convergence and deriving truncation error bounds. In-deed, they are two of the main pillars on which the theory in this book is built.

The last part of this chapter is devoted to transformations of one continued fractionto another. The idea is that the new one shall have some desired properties lackingin the old one.

L. Lorentzen and H. Waadeland, Continued Fractions, Atlantis Studies in Mathematics for Engineering and Science 1, DOI 10.1007/978-94-91216-37-4_2, © 2008 Atlantis Press/World Scientific

53

Page 67: Lisa Lorentzen, Haakon Waadeland Continued Fractions

54 Chapter 2: Basics

2.1 Convergence

2.1.1 Properties of linear fractional transformations

As always, M denotes the family of linear fractional transformations

τ(w) :=aw + b

cw + d; ad − bc �= 0. (1.1.1)

These transformations have a number of nice properties. First of all, τ is a bijectivemeromorphic mapping of C onto C; that is, τ(C) = C and τ is univalent (one-to-one). Some additional properties are given below. Others will be described later.

Properties:

1. The cross ratio is invariant under linear fractional transformations; that is, ifu, v, w and z are four distinct points in C, then

τ(u) − τ(z)τ(u) − τ(w)

· τ(v) − τ(w)τ(v) − τ(z)

=u − z

u − w· v − w

v − z(1.1.2)

with the natural limiting forms if one or two of the quantities involved areinfinite. This property follows by straight forward checking.

2. From (1.1.2) it follows that whenever w1, w2, w3 are three distinct points inC and γ1, γ2, γ3 are three distinct points in C, then there exists a unique τfrom M such that

τ(w1) = γ1, τ(w2) = γ2 and τ(w3) = γ3.

This does not mean that the coefficients of τ are unique, but the mapping τis unique in the sense described on page 5.

3. From 2. it follows immediately that if a sequence {τn} from M converges(pointwise) at three distinct points to distinct values, then it converges tosome τ ∈ M in all of C.

4. If {τn} converges at n ≥ 3 distinct points, but not to a τ ∈ M, then the limitis the same at at least two of the three points.

The point ∞ ∈ C is not a special point for a linear fractional transformation.Indeed, for τ in (1.1.1)

τ(∞) = a/c and τ(−d/c) = ∞

where a/c = ∞ and d/c = ∞ if c = 0, and finite numbers otherwise. This callsfor a representation of C which makes no distinction between ∞ and the complexnumbers. The Riemann sphere is a perfect choice.

Page 68: Lisa Lorentzen, Haakon Waadeland Continued Fractions

2.1.1 Linear fractional transformations 55

In the complex plane, |w1 − w2| is the natural way to define the distance betweentwo points (the euclidean metric). On the Riemann sphere C one way to do it is tomeasure the euclidean length of the chord connecting the two points. This gives usthe chordal distance (or chordal metric) introduced by Ahlfors ([Ahlf53]):

m(w1, w2) :=

⎧⎪⎪⎪⎨⎪⎪⎪⎩2|w1 − w2|√

1 + |w1|2√

1 + |w2|2for w1, w2 ∈ C

2√1 + |w1|2

for w1 ∈ C, w2 = ∞

0 for w1 = w2 = ∞ .

(1.1.3)

For our purpose the main advantages of this metric are:

(i) it is bounded, since m(w1, w2) ≤ 2 for all w1, w2 ∈ C. (The radius of theRiemann sphere is taken to be 1.)

(ii) C equipped with this metric is compact; that is, every sequence {wn} on C

has a convergent subsequence.

(iii) wn → w ∈ C if and only if m(wn, w) → 0.

(iv) if τn → τ ∈ M, then the convergence is uniform on C with respect to thechordal metric.

(v) the cross ratio (1.1.2) implies that

m(τ(u), τ(z))m(τ(u), τ(w))

· m(τ(v), τ(w))m(τ(v), τ(z))

=m(u, z)m(u, w)

· m(v, w)m(v, z)

(1.1.4)

when u, v, w, z are four distinct points on C.

The euclidean distance is not bounded, and one has to distinguish between the casesw �= ∞ and w = ∞ to define the convergence wn → w in this metric. If w �= ∞,then wn → w if and only if |wn − w| → 0. If w = ∞, this characterization fails.

We use the chordal distance to define convergence in M, or rather the metric

σ(τ1, τ2) := supw∈C

m(τ1(w), τ2(w)). (1.1.5)

��

�Definition 2.1. A sequence {τn} from M converges to some τ ∈ M if and

only if limσ(τn, τ) = 0.

That is, τn(w) → τ(w) uniformly in C with respect to the chordal metric. The typeof convergence described in Property 4 requires some closer attention:

Page 69: Lisa Lorentzen, Haakon Waadeland Continued Fractions

56 Chapter 2: Basics

Example 1. We shall see three examples of convergence of a sequence {τn} fromM given by

τn(w) :=anw + bn

cnw + dn, Δn := andn − bncn �= 0 for n = 1, 2, 3, . . . .

Example 1A: Let the coefficients an, bn, cn and dn converge to finite values a, b, cand d respectively, where ad− bc �= 0. Then τn(w) → τ(w) := (aw + b)/(cw +d) forall w ∈ C. In the euclidean metric this convergence is uniform on compact subsetsof C \ {−d/c}. In the chordal metric it is uniform on C. In short, τn → τ .

Example 1B: The coefficients converge to finite values a, b, c and d respectively,where c �= 0 and ad − bc = 0. This time {τn} approaches a constant function

τ(w) :=aw + b

cw + d=

a

c− ad − bc

c(cw + d)=

a

cfor w �= −d

c.

So τn(w) → a/c for every w ∈ C except possibly at the point w† := −d/c. Theconvergence is locally uniform in C \ {−d/c} with respect to the chordal metric,and τn(wn) → a/c whenever lim inf m(wn,−d/c) > 0. For wn := −dn/cn we alwayshave τn(wn) = ∞ �= a/c for all n, so there exist sequences {wn} approaching −d/cfor which τn(wn) �→ a/c.

Example 1C: Let an → a �= ∞, cn → c �= 0,∞, and (andn − bncn) → 0. τn(w)still seems to converge to a/c, but we have a problem. This time {dn} may havea large number of limit points d, and w ought to stay away from −d/c for all ofthem. This can be very restrictive and even impossible since the limit points for{dn} may be dense in C. On the other hand, for every limit point d there existsa subsequence {dnk

} converging to d, and for this subsequence we are back to theformer case: {τnk

} converges (locally uniformly with respect to the chordal metric)to a/c in C \ {−d/c}. So there is some extensive converging to a/c going on. Toget hold of this convergence we look at

τn(wn) =anwn + bn

cnwn + dn=

an

cn− andn − bncn

c2n(wn + dn/cn)

.

Let w∗n := −dn/cn from some n on. Then τn(wn) → a/c whenever wn stays a

uniform distance away from w∗n; i.e., whenever

lim infn→∞ m(wn, w∗

n) > 0.

��

Definition 2.2. A sequence {τn} from M converges generally to a constantγ ∈ C if and only if there exists a sequence {w†

n} from C such that

limn→∞ τn(wn) = γ whenever lim inf

n→∞ m(wn, w†n) > 0 . (1.1.6)

Page 70: Lisa Lorentzen, Haakon Waadeland Continued Fractions

2.1.1 Linear fractional transformations 57

Remarks.

1. We write τn → γ to denote that {τn} converges generally to the constantγ ∈ C, and {w†

n} is called an exceptional sequence for{τn} in this case.

2. The exceptional sequence is not unique. If the sequence {τn} of transforma-tions from M converges generally to γ with exceptional sequence {w†

n}, thenevery sequence {w∗

n} with lim m(w†n, w∗

n) = 0 is exceptional. It is only itsasymptotic behavior that matters.

3. We say that two sequences {un} and {vn} from C are equivalent if and onlyif m(un, vn) → 0. Thereby the exceptional sequences for {τn} form an equiv-alence class, and {w∗

n} or {w†n} are just representatives for this equivalence

class.

4. If τn → γ with exceptional sequence {w†n}, it is not always clear what happens

to τn(w†n). It may converge to γ or to some other value, or it may diverge.

Its behavior depends on how {w†n} is chosen compared to {τn}.

5. A sequence {τn} from M can not converge uniformly in C to a constant γ, noteven in the chordal metric. One always has to accept exceptional sequences.This follows since τn(wn) = μ �= γ for all n whenever wn := τ−1

n (μ). Indeed,it is more of a mystery that we can get away with only one equivalence classof exceptional sequences. We shall return to this question in Section 2.1.3 onpage 60.

Condition (1.1.6) is not always so easy to check since it requires that we knowan exceptional sequence {w†

n}. The following equivalent definition makes checkingeasier:

Definition 2.3. A sequence {τn} from M converges generally to a constantγ ∈ C if and only if there exist two sequences {vn} and {wn} from C suchthat

lim infn→∞ m(vn, wn) > 0 (1.1.7)

andlim

n→∞ τn(vn) = limn→∞ τn(wn) = γ . (1.1.8)

In other words, we just require that τn(vn) → γ and τn(wn) → γ for two sufficientlydistinct sequences {vn} and {wn}. We shall prove that this definition is equivalentto the former one, and thus that the limit γ in Definition 2.3 is unique.

Page 71: Lisa Lorentzen, Haakon Waadeland Continued Fractions

58 Chapter 2: Basics

Proof of the equivalence of Definition 2.2 and 2.3: Let first {τn} satisfy (1.1.6).Then there clearly exist two sequences {vn} and {wn} bounded away from the ex-ceptional sequence {w†

n} so that (1.1.7) - (1.1.8) holds.

Next, let there exist two sequences {vn} and {wn} such that (1.1.7) - (1.1.8) holds.Let γ(n) := τ−1

n (γ) and

pn :=

{vn if m(vn, γ(n)) > m(wn, γ(n)),wn otherwise .

Then lim inf m(pn, γ(n)) > 0 and γ �= τn(pn) → γ. Let w†n := τ−1

n (μ) for someμ ∈ C, μ �= γ, and let {un} be any sequence from C with lim inf m(un, w†

n) > 0.We shall prove that lim τn(un) = γ. Since we already know that τn(pn) → γ andτn(γ(n)) = γ, we only have to consider indices n for which

un �= γ(n), pn. (1.1.9)

For these indices we can apply the invariance of the cross ratio (1.1.4) with τ := τn,z := un, u := γ(n), w := pn and v := w†

n = τ−1n (μ):

m(γ, τn(un))m(γ, τn(pn))

· m(μ, τn(pn))m(μ, τn(un))

=m(γ(n), un)m(γ(n), pn)

· m(w†n, pn)

m(w†n, un)

.

The right hand side of this equality stays bounded as n → ∞. Since τn(pn) → γ,this means that m(γ, τn(un)) → 0; i.e. lim τn(un) = γ. �

Summary. If a sequence {Fn} of univalent functions converges in a domain D,then it either converges to a univalent function or to a constant. So also for linearfractional transformations, but of course, for τn ∈ M we know more:

(i) if {τn(w)} converges at three distinct points to distinct values, then {τn}converges to a τ ∈ M on C, uniformly with respect to the chordal metric.

(ii) if {τn(w)} converges at two distinct points to the same value, then {τn(w)}converges to this constant whenever {τn(w)} converges, except possibly at onesingle point.

Here (ii) is fairly trivial, but it may also be very restrictive. General convergencegeneralizes (ii) by allowing w to vary with n. Thereby we only have to stay awayfrom one single point for each n, and we get convergence to the constant in all ofC except at one point for each n”. Therefore there are only two possibilities for aconvergent sequence {τn} from M: either

(i) {τn} converges to some (non-singular) τ ∈ M, and we write τn → τ as n → ∞,

(ii) or {τn} converges generally to some constant γ ∈ C with exceptional sequence{w†

n}. Then τn(wn) → γ whenever lim inf m(wn, w†n) > 0, and we write

τn → γ as n → ∞.

In all other cases we say that {τn} diverges.

Page 72: Lisa Lorentzen, Haakon Waadeland Continued Fractions

2.1.2 Convergence of continued fractions 59

2.1.2 Convergence of continued fractions

So far we have used the classical convergence concept for continued fractions; i.e.,K(an/bn) converges to f if {Sn(0)} converges to f . Since

Sn+1(∞) = Sn(0), (1.2.1)

classical convergence implies that {Sn} converges generally to the value f . Inciden-tally, this is the reason why the classical definition of convergence works so well.Still, it has some drawbacks, as illustrated by the following example:

Example 2. Let a3n+1 := 2, a3n+2 := 1 and a3n+3 := −1 for all n ≥ 0 in thecontinued fraction K(an/1); that is,

∞Kn=1

an

1:=

21 +

11 −

11 +

21 +

11 −

11 + · · · .

By the recurrence relations (1.2.7) on page 6 and induction, we find that

A3n−2 = 2n, A3n−1 = 2n, A3n = 0 ,B3n−2 = 2n+1 − 3 , B3n−1 = 2n+1 − 2 , B3n = 1 ,

and so lim S3n−2(0) = limS3n−1(0) = 12

and limS3n(0) = 0, and thus {Sn(0)}diverges. On the other hand

Sn(wn) =An−1wn + An

Bn−1wn + Bn,

so

limn→∞ S3n−2(w3n−2) = 12

whenever {w3n−2} is boundedlimn→∞ S3n−1(w3n−1) = 1

2 whenever {w3n−1} is boundedaway from − 1

limn→∞ S3n(w3n) = 12 whenever {w3n} is bounded

away from 0

Indeed, with w†3n−2 := ∞, w†

3n−1 := −1 and w†3n := 0 for all n ≥ 1, we have

limn→∞Sn(wn) =

12

whenever lim inf m(wn, w†n) > 0. (1.2.2)

Therefore, {Sn} converges generally to 12 with exceptional sequence {w†

n}. �

We therefore define general convergence of continued fractions ([Jaco86]):

Page 73: Lisa Lorentzen, Haakon Waadeland Continued Fractions

60 Chapter 2: Basics�

�Definition 2.4. The continued fraction K(an/bn) converges generally tof ∈ C with exceptional sequence {w†

n} if and only if {Sn} converges generallyto f with exceptional sequence {w†

n}.

Properties of general convergence:

1. The classical convergence to f implies general convergence to f . Hence thenew concept includes the classical one, and picks up additional cases, forinstance the one in Example 2.

2. The definitions of general convergence add some insight to the classical con-cept of convergence. In particular it makes sense to talk about exceptionalsequences {w†

n} for a (classically) convergent continued fraction.

3. If K(an/bn) fails to converge generally, we say that K(an/bn) diverges gener-ally.

General convergence has also another important advantage: it is sometimes mucheasier to prove! (This will be evident in Chapter 4.) We still use the classicaldefinition of convergence alongside with general convergence, mainly because theSn(0)–definition is so well established, but also because it has some merits of itsown. For instance: it is simple and uniquely defined, and it is important in manyapplications. To distinguish between the two, the Sn(0)–convergence will be referredto as just convergence or classical convergence, whereas the convergence in Definition2.4 always will be called general convergence.

2.1.3 Restrained sequences

Let K(an/bn) converge generally to f , and let q �= f . Then Sn(wn) = q for alln for the choice wn := S−1

n (q). That is, {S−1n (q)} is an exceptional sequence for

K(an/bn). This is true for every q �= f . Hence all these sequences {S−1n (q)} belong

to the same equivalence class. (See Remark 3 on page 57.)

Now, {S−1n } is also a sequence of linear fractional transformations. Since m(S−1

n (q1),S−1

n (q2)) → 0 whenever q1 �= f and q2 �= f , no subsequence of {S−1n } can converge

to a transformation from M. Following Jacobsen and Thron ([JaTh87]) we saythat {S−1

n } is a restrained sequence. Note that {S−1n } need not be generally con-

vergent since {S−1n (q)} with q �= f may have more than one limit point. But every

subsequence of {S−1n } must have a generally convergent subsequence.

Page 74: Lisa Lorentzen, Haakon Waadeland Continued Fractions

2.1.3 Restrained sequences 61�

Definition 2.5. A sequence {τn} from M is restrained if and only if thereexists a sequence {w†

n} from C such that whenever

lim inf m(vn, w†n) > 0 and lim inf m(wn, w†

n) > 0, (1.3.1)

limn→∞m(τn(vn), τn(wn)) = 0. (1.3.2)

�Definition 2.6. A sequence {τn} from M is restrained if and only if thereexist two sequences {vn} and {wn} from C with lim inf m(vn, wn) > 0 suchthat (1.3.2) holds.

Definition 2.7. A sequence {τn} from M is restrained if and only if nosubsequence of {τn} converges to some τ ∈ M.

You are asked to prove that these three definitions are equivalent in Problem 3on page 91. Definition 2.5 shows that {τn} is restrained if and only if the asymp-totic behavior of {τn(wn)} is independent of {wn} in the (1.3.2)-sense wheneverlim inf m(wn, w†

n) > 0. That is, all sequences {τn(wn)} with lim inf m(wn, w†n) > 0

are equivalent (in the sense of Remark 3 on page 57). A sequence from this equiv-alence class is called a generic sequence for {τn}. We evidently have:

�Theorem 2.1. Let the sequence {τn} ⊆ M be restrained with generic se-quence {zn} and exceptional sequence {w†

n}. Then {τ−1n } is restrained with

generic sequence {w†n} and exceptional sequence {zn}.

Remarks.

1. If {τn} converges generally to some constant, then {τn} is in particular re-strained.

2. If {τn} converges generally to f with exceptional sequence {w†n}, then the

constant sequence {f} is a generic sequence for {τn}, and {τ−1n (q)} is an

exceptional sequence for every q �= f . Therefore {τ−1n (q)} is generic for {τ−1

n }when q �= f and {f} is exceptional for {τ−1

n }.

Page 75: Lisa Lorentzen, Haakon Waadeland Continued Fractions

62 Chapter 2: Basics�

Definition 2.8. A continued fraction K(an/bn) is restrained if and only if{Sn} is restrained.

Example 3. The continued fraction in Example 2 on page 59 has the approximants

Sn(wn) =An−1wn + An

Bn−1wn + Bn

where A3n−2 = A3n−1 = 2n, A3n = 0, B3n−2 = 2n+1 − 3, B3n−1 = 2n+1 − 2 andB3n = 1 for all n. We showed that {Sn} converges generally to 1

2with exceptional

sequence {w†n} given by

w†3n−2 = ∞, w†

3n−1 = −1 and w†3n = 0 for all n.

Every sequence {zn} given by zn := Sn(wn) with lim inf m(wn, w†n) > 0 is a generic

sequence for this continued fraction. Indeed, every sequence {μn} converging to 12 is

generic for K(an/1), including the constant sequence { 12}. By Theorem 2.1, {S−1

n }should therefore be restrained with exceptional sequence {1

2} and generic sequence{w†

n}. Let us check this out. Simple computation shows that

S−1n (w) = − Bnw − An

Bn−1w − An−1,

so

S−13n−1(wn) = − (2n+1 − 2)wn − 2n

(2n+1 − 3)wn − 2n→ −1,

S−13n (wn) = − wn − 0

(2n+1 − 2)wn − 2n→ 0,

S−13n+1(wn) = − (2n+2 − 3)wn − 2n+1

wn − 0→ ∞

whenever lim inf d(wn, 12) > 0. �

A sequence {τn} from M is either restrained, or it has a subsequence {τnk} which

converges to a non-singular transformation. These are the only two possibilities. If{τn} has no restrained subsequence, we say that {τn} is totally non-restrained. Todetermine whether {τn} is restrained or not, the following observation may be ofhelp.�

�Theorem 2.2. Let the sequence {τn} from M converge to a transformationτ ∈ M. Then σn := τ−1

n−1 ◦ τn converges to the identity transformationI(w) ≡ w in C.

Page 76: Lisa Lorentzen, Haakon Waadeland Continued Fractions

2.1.4 Tail sequences 63

Proof : σn = τ−1n−1 ◦ τn where τn → τ and thus τ−1

n → τ−1. �

2.1.4 Tail sequences

Let us return to continued fractions K(an/bn); i.e., to sequences {Sn} of linearfractional transformations characterized by Sn+1(∞) = Sn(0) (although we mightas well work with general sequences {τn}; τn ∈ M).

Definition 2.9. For every t ∈ C, the sequence

tn := S−1n (t) for n = 0, 1, 2, . . . (1.4.1)

is called a tail sequence for K(an/bn) or for {Sn}.

Properties:

1. On page 6 we defined the sequence {f (n)} of tail values for a convergent con-tinued fraction K(an/bn). Evidently {f (n)} is an example of a tail sequence.

2. If K(an/bn) converges generally to f and t �= f , we recognize {tn} as arepresentative for the equivalence class of exceptional sequences for K(an/bn).

3. Since Sn = s1 ◦ s2 ◦ · · · ◦ sn, we have

tn−1 = sn(tn) for n = 1, 2, 3, . . . . (1.4.2)

4. If {tn}∞n=0 is a tail sequence for K(an/bn), then {tn}∞n=N is a tail sequencefor its Nth tail. More generally, if {tn} is a tail sequence for {Sn}, then thesubsequence {tnk

} is a tail sequence for {Snk}.

5. If {tn} and {tn} are two tail sequences for K(an/bn) with tk = tk for oneindex k, then tn = tn for all n by (1.4.2) since sn, s−1

n ∈ M are univalentfunctions.�

Theorem 2.3. Let {tn} be a tail sequence for K(an/bn). Then

tn = S−1n (t0) = s−1

n ◦ s−1n−1 ◦ · · · ◦ s−1

1 (t0)

= −{

bn +an

bn−1 +an−1

bn−2 +· · ·+a2

b1 +a1

(−t0)

}for n ≥ 1 .

(1.4.3)

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64 Chapter 2: Basics

Proof : The result follows since

s−1k (w) = −bk +

ak

w= −

{bk +

ak

(−w)

}for k ≥ 1 . (1.4.4)

��

Theorem 2.4. Let K(an/bn) have three distinct tail sequences {tn}, {tn}and {t∗n} where

lim m(tn, t∗n) = 0 and lim inf m(tn, t∗n) > 0.

Then K(an/bn) converges generally to t0 with exceptional sequence {tn}.

Proof : Since lim m(tn, t∗n) = lim m(S−1n (t0), S−1

n (t∗0)) = 0 where t0 �= t∗0, it fol-lows that {S−1

n } is restrained with generic sequence {tn}. On the other hand weknow that lim inf m(tn, t∗n) = lim inf m(S−1

n (t0), S−1n (t∗0)) > 0. Therefore the con-

stant sequence {t0} is exceptional for {S−1n }. Hence {Sn} is restrained with generic

sequence {t0} and exceptional sequence {tn} (Theorem 2.1 on page 61). That is,K(an/bn) converges generally to t0. �

The tail sequence {ζn} defined by

ζn := S−1n (∞) for n = 0, 1, 2, . . . , (1.4.5)

plays a special role in our theory. Here ζ0 = ∞, ζ1 = −b1, and by (1.4.3)

ζn = − Bn

Bn−1= −bn − an

bn−1 +an−1

bn−2 +· · ·+a2

b1(1.4.6)

for n ≥ 2. In earlier work one has rather used the notation

hn := −S−1n (∞), (1.4.7)

and the sequence {hn} is called the critical tail sequence for K(an/bn) (although,strictly speaking, it is {−hn} which is a tail sequence). The importance of {ζn} or{hn} is that if K(an/bn) converges generally to f �= ∞, then {−hn} is an exceptionalsequence which often shows up in useful formulas.

Example 4. In Example 2 on page 59 we saw that the 3-periodic continued fraction

∞Kn=1

an

1=

21 +

11 −

11 +

21 +

11 −

11 +

21 + . . .

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2.1.5 Tail sequences and recurrence relations 65

converges generally to f = 1/2 with exceptional sequence

w†n : 0, ∞, −1, 0, ∞, −1, 0, . . .

(starting with w†0 = 0). Every tail sequence {tn} with t0 �= 1

2 must have the sameasymptotic behavior; i.e.,

lim t3n = 0, lim t3n+1 = ∞, lim t3n+2 = −1.

The tail values {f (n)} is the tail sequence starting with t0 = 12:

f (0) =12

,

f (1) = s−11 (f (0)) = −1 +

21/2

= 3 ,

f (2) = s−12 (f (1)) = −1 +

13

= −23

,

f (3) = s−13 (f (2)) = −1 +

−1−2/3

=12

,

and so on. That is, {f (n)} is periodic with period 3, as it just has to be. In Problem4 on page 92 you are asked to prove that {w†

n} as given above and {f (n)} are theonly periodic tail sequences for K(an/1). �

2.1.5 Tail sequences and three term recurrence relations

There are strong connections between tail sequences for a continued fraction K(an/bn)and solutions of the corresponding recurrence relation

Xn = bnXn−1 + anXn−2 for n = 1, 2, 3, . . . . (1.5.1)

We have already seen that ζn = −Bn/Bn−1 forms a tail sequence, where Bn is asolution of (1.5.1). The following is therefore not so surprising:

�Theorem 2.5. Let {Xn}∞n=−1 be a non-trivial solution of (1.5.1) (i.e., notall Xn = 0). Then tn := −Xn/Xn−1 is well defined in C and {tn}∞n=0 is atail sequence for K(an/bn).

Proof : The sequence {tn} is well defined since Xn = Xn−1 = 0 for a fixed nimplies that all Xn = 0, something we have excluded. From (1.5.1) we find that

− Xn

Xn−1= −bn +

an

−Xn−1/Xn−2,

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66 Chapter 2: Basics

that is, tn = −bn +an/tn−1 = s−1n (tn−1) for all n. The result follows therefore from

(1.4.2). �

There is also an interesting connection the other way around: we can express so-lutions of (1.5.1) in terms of tail sequences. We shall only do so for the particularsolutions {An} and {Bn}, which are the solutions of (1.5.1) starting with

A−1 = 1, A0 = 0, A1 = a1, B−1 = 0, B0 = 1, B1 = b1, (1.5.2)

in other words: the canonical numerators and denominators of K(an/bn). But itis not difficult to extend the results to general solutions {Xn} of (1.5.1), since itturns out that every solution {Xn} is a linear combination of {An} and {Bn}. Theformulas may seem rather complicated, but they are indeed quite useful. As always:an empty sum is equal to 0 and an empty product is equal to 1. For some historyon this subject we refer to the remark section on page 89.

Theorem 2.6. Let {tn}∞n=0 be a tail sequence for K(an/bn) with all tn �=∞. Then all tn �= 0,−bn and

Bn + Bn−1tn =n∏

k=1

(bk + tk) , (1.5.3)

An − Bnt0 =n∏

k=0

(−tk) , (1.5.4)

An = t0

n∑k=1

k∏j=1

(bj + tj)n∏

j=k+1

(−tj) , (1.5.5)

Bn =n∑

k=0

k∏j=1

(bj + tj)n∏

j=k+1

(−tj) , (1.5.6)

t0 −An

Bn= t0/Σn (1.5.7)

where Σn :=n∑

k=0

Pk and Pk :=k∏

j=1

bj + tj−tj

, (1.5.8)

t0 − Sn(w) =t0(w − tn)

wΣn−1 − tnΣn. (1.5.9)

Proof : That tn �= 0,−bn follows from (1.4.2) which says that tn−1 = an/(bn+tn)for all n. Now, B0 + B−1t0 = 1 and B1 + B0t1 = b1 + t1. Since an = tn−1(bn + tn)

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2.1.5 Tail sequences and recurrence relations 67

by (1.4.2), and thus, by the recurrence relation,

Bn + Bn−1tn = bnBn−1 + anBn−2 + Bn−1tn

= (bn + tn)Bn−1 + tn−1(bn + tn)Bn−2 = (bn + tn)(Bn−1 + Bn−2tn−1) ,

equality (1.5.3) follows by induction.

To prove (1.5.4) we again use induction. Since A−1 − B−1t0 = 1, A0 − B0t0 = −t0and am = tm−1(bm + tm), we have

Am − Bmt0 = bmAm−1 + amAm−2 − t0(bmBm−1 + amBm−2)= bm(Am−1 − Bm−1t0) + tm−1(bm + tm)(Am−2 − Bm−2t0) .

Hence, if (1.5.4) holds for n := m − 1 and n := m − 2, then

Am − Bmt0 = bm

m−1∏k=0

(−tk) + tm−1(bm + tm)m−2∏k=0

(−tk)

= bm

m−1∏k=0

(−tk) − (bm + tm)m−1∏k=0

(−tk) =m∏

k=0

(−tk) .

The expression (1.5.6) for Bn follows from (1.5.3):

Bn =n∏

k=1

(bk + tk) − tnBn−1

=n∏

k=1

(bk + tk) − tn

n−1∏k=1

(bk + tk) + tntn−1Bn−2 = · · · =

=n∏

k=1

(bk + tk) − tn

n−1∏k=1

(bk + tk) + tntn−1

n−2∏k=1

(bk + tk) − · · · +n∏

k=1

(−tk)

which is exactly the expression for Bn. The expression (1.5.5) follows since byLemma 1.4 on page 10, An = a1B

(1)n−1 where B

(1)k denotes the canonical denomina-

tors of the first tail of K(an/bn). Finally, (1.5.7) and (1.5.9) follow from (1.5.4) and(1.5.6) (after some work). �

Formulas (1.5.7) and (1.5.9) are in particular useful for proving classical convergenceand estimate the speed of this convergence. For instance, the following corollaryfollows easily:

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68 Chapter 2: Basics�

Corollary 2.7. (Waadeland’s Tail Theorem.) Let {tn}∞n=0 be a tailsequence for K(an/bn) with all tn �= ∞. Then K(an/bn) converges in theclassical sense if and only if {Σn} given by (1.5.8) converges in C.

If {Σn} converges to Σ∞ ∈ C, then K(an/bn) converges to f := t0

(1− 1

Σ∞

),

and f − fn = t0

( 1Σn

− 1Σ∞

).

Example 5. For given b, t ∈ C with t �= 0,−b and b �= 0, the periodic continuedfraction

t(b + t)b +

t(b + t)b +

t(b + t)b +· · ·

has a tail sequence {tn} with all tn := t. Therefore, by (1.5.7)

t − fn = t − An

Bn=

tn∑

k=0

(b + t

−t

)k=

t

(1 +

b + t

t

)1 −(

b + t

−t

)n+1 (1.5.10)

which shows that fn → t if |t| < |b + t| and fn → −(b + t) if |t| > |b + t|.If |t| = |b + t|, then still t �= b + t under our conditions, and {t − fn} oscillates. Ifin particular (b + t)/t is an Nth root of unity; i.e., ((b + t)/(−t))N = 1 for someN ∈ N, then {t−fn} is periodic. Otherwise {((b+ t)/(−t))n+1} is a dense sequenceof distinct points on the unit circle. �

Formulas (1.5.3) and (1.5.6) imply that

ζn = − Bn

Bn−1= −Bn + Bn−1tn

Bn−1+ tn = tn −

∏nk=1(bk + tk)

Bn−1

= tn −∏n

k=1(bk + tk)n−1∑k=0

k∏j=1

(bj + tj)n−1∏

j=k+1

(−tj)

= tn +tnPn

Σn−1

(1.5.11)

where we have used the notation (1.5.8). That is, the critical tail sequence is givenin terms of {tn}. More generally, we can express any tail sequence {Tn} in terms of{tn} when all tn �= ∞:

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2.1.5 Tail sequences and recurrence relations 69�

Corollary 2.8. Let {tn} be a tail sequence for K(an/bn) with all tn �= ∞.Then the tail sequence {Tn} for K(an/bn) beginning with T0 is given by

Tn = tn(t0 − T0)Σn − t0

(t0 − T0)Σn−1 − t0where Σn :=

n∑k=0

k∏j=1

bj + tj−tj

.

Proof : SinceTn = S−1

n (T0) = − An − BnT0

An−1 − Bn−1T0

= − An − Bnt0 + Bn(t0 − T0)An−1 − Bn−1t0 + Bn−1(t0 − T0)

,

the result follows by use of (1.5.4) and (1.5.6). �

To any given sequence {tn} from C\{0} we can always construct continued fractionsK(an/bn) for which {tn} is a tail sequence. The only requirement is that

tn−1(bn + tn) = an.

If {bn} is chosen such that {Σn} given by (1.5.8) converges in C, then this continuedfraction converges. Its value is t0 if Σ∞ = ∞ and t0(1−1/Σ∞) otherwise (Corollary2.7). Thereby we have derived a continued fraction identity

f = K(an/bn).

Example 6. Let {tn} be given by

t2n := n + 1, t2n+1 := 1 for n ≥ 0,

and choose bn := 1 for all n. Then {tn} is a tail sequence for K(an/1) given by

a2n−1 := t2n−2(1 + t2n−1) = 2n, a2n := t2n−1(1 + t2n) = n + 2.

Since

P2k =k∏

j=1

1 + t2j−1

−t2j−1· 1 + t2j

−t2j=

k∏j=1

2 · j + 2j + 1

= 2k k + 22

= 2k−1(k + 2)

P2k−1 = P2k−2 ·1 + t2k−1

−t2k−1= −2 P2k−2 = −2k−1(k + 1) ,

it follows that

Σ2n = 1 +n∑

k=1

(P2k−1 + P2k) = 1 +n∑

k=1

2k−1 = 1 +1 − 2n

1 − 2= 2n → ∞,

Σ2n+1 = Σ2n + P2n+1 = 2n − 2n(n + 2) → −∞ .

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70 Chapter 2: Basics

Therefore ∞Kn=1

an

1:=

21+

31+

41+

41+

61+

51+

81+

61+

101 +

71+· · ·

= 1 .

2.1.6 Value sets

Tail sequences are useful to determine convergence properties of certain contin-ued fractions. We shall return to this later. We also have another potent tool todetermine such properties: value sets !

Definition 2.10. A sequence {Vn}∞n=0 of sets Vn ⊂ C is a sequence of valuesets for K(an/bn) if and only if

sn(Vn) =an

bn + Vn⊆ Vn−1 for n = 1, 2, 3, . . . . (1.6.1)

If {Vn} is 1-periodic, so that all Vn = V , we say that V is a simple value set forK(an/bn). If {Vn} is 2-periodic, so that all V2n = V0 and all V2n+1 = V1, we saythat (V0, V1) are twin value sets for K(an/bn).

In the literature {Vn} is often referred to as prevalue sets, and the term value setsis reserved for prevalue sets with the property an/bn ∈ Vn−1 for all n ≥ 1. This isa natural choice when convergence is based on the asymptotic behavior of classicalapproximants. Our wider view of approximants Sn(wn) and convergence calls forthe wider concept of value sets in Definition 2.10.

The importance of value sets follows from the nestedness

Kn := Sn(Vn) = Sn−1(sn(Vn)) ⊆ Sn−1(Vn−1) = Kn−1 (1.6.2)

which means that

Sn(wn) ∈ Kn ⊆ Kn−1 ⊆ V0 for wn ∈ Vn. (1.6.3)

Let us assume that Vn are closed, non-empty sets. Then Kn are closed, non-emptysets, and thus the nestedness (1.6.2) implies that the limit set

K := lim Kn :=∞⋂

n=1

Kn (1.6.4)

exists and is closed and non-empty. If diam(K) = 0, the limit point case, then wehave hit the jackpot, since then:

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2.1.6 Value sets 71

• K consists of just one point, say f ∈ C, and Sn(wn) → f whenever wn ∈ Vn

for all n.

• if 0 ∈ Vn for all n, then K(an/bn) converges to f in the classical sense.

• if lim inf diamm(Vn) > 0 for the chordal diameter

diamm(Vn) := sup{m(v, w); v, w ∈ Vn} (1.6.5)

of Vn, then K(an/bn) converges generally to f .

• the truncation error bound

|f − Sn(wn)| ≤ diam(Kn) for wn ∈ Vn (1.6.6)

approaches 0 as n → ∞. (Clearly, Sn(wn) ∈ Kn and f ∈ K ⊆ Kn.)

But also if the limit point case fails to occur, the existence of a sequence of valuesets for a continued fraction is useful. With the notation (1.6.4) we have:

Theorem 2.9. Let {Vn} be a sequence of closed value sets for K(an/bn).Then tn ∈ S−1

n (K) ⊆ Vn for every tail sequence {tn} starting with a t0 ∈ K,and every tail sequence {tn} with tk ∈ C \ Vk for some k ∈ N∪ {0} satisfiestn ∈ C \ Vn for all n ≥ k.

Proof : Let first t0 ∈ K. Then tn = S−1n (t0) ∈ S−1

n (K) ⊆ S−1n (Kn) = Vn. Next,

let tk ∈ C\Vk. Since tk = sk+1(tk+1) �∈ Vk whereas sk+1(Vk+1) ⊆ Vk, it follows thattk+1 ∈ C\Vk+1. By induction it follows therefore that tn ∈ C\Vn for all n ≥ k. �

Corollary 2.10. Let {Vn} be a sequence of closed value sets for the gener-ally convergent continued fraction K(an/bn) with lim supn→∞ diam m(Vn) >0. Then f (n) ∈ Vn for all n for the tail values {f (n)} of K(an/bn). If more-over Vk �= C for some k ∈ N ∪ {0}, then K(an/bn) has an exceptionalsequence {w†

n} with w†n ∈ C \ Vn for all n ≥ k.

Proof : We first prove that the value f of K(an/bn) belongs to K. Let {w†n}

be an exceptional sequence for K(an/bn). Under our conditions there exists asubsequence {Vnk

} of {Vn} with diamm(Vnk) ≥ d for some d > 0. Therefore there

exists a wnk∈ Vnk

with m(wnk, w†

nk) ≥ d/2 for each k ∈ N. Hence Snk

(wnk) → f ,

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72 Chapter 2: Basics

and the result follows since Snk(wnk

) ∈ Knk→ K Hence f (n) ∈ Vn for all n by

Theorem 2.9.

Let Vk �= C and tk ∈ C \ Vk. Then tk �= f (k), and the corresponding tail sequence{tn} is an exceptional sequence with tn ∈ C \ Vn for all n ≥ k by Theorem 2.9. �

Remark. In the particular case where all bn = b �= 0 and all Vn = V , then

sn(V ) :=an

b + V⊆ V ⇐⇒ s−1

n (−b − V ) = −b +an

−b − V⊆ −b − V,

so {w†n} has all its limit points in −b − V . A similar result holds if all b2n−1 = b1,

b2n = b2 and V2n = V0, V2n+1 = V1.

Corollary 2.10 generalizes a result from [Jaco86b]. Of course, it also holds if wereplace the condition on diamm(Vn) with the condition 0 ∈ Vn for infinitely manyn. Another question is: how can we find value sets for a given continued fraction?

Example 7. Let the real interval V := [−12, 1

2] be a simple value set for K(an/1);

that is,an

1 + V⊆ V for all n.

Since 1 + V = [12 , 32 ], we have 1/(1 + V ) = [23 , 2], and thus an/(1 + V ) ⊆ V if and

only if either an > 0 and [ 2an

3, 2an] ⊆ [− 1

2, 1

2] or an < 0 and [2an, 2an

3] ⊆ [−1

2, 1

2];

i.e., if and only if an ∈ [−14 , 1

4 ]. �

In this example we started with the value set, and found sufficient conditions on acontinued fraction for V to be its simple value set. This is the easy way. To findvalue sets for a given continued fraction is harder:

Example 8. By Example 4 on page 64 the 3-periodic continued fraction

∞Kn=1

an

1:=

21 +

11 −

11 +

21 +

11 −

11 +

21 + . . .

converges generally and has tail values

f (3n) =12, f (3n+1) = 3, f (3n+2) = −2

3.

We want to find useful value sets for K(an/1). Therefore we want Vn to containf (n) (Corollary 2.10). The easiest procedure seems to be to look for circular disks

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2.1.7 Element sets 73

Vn centered at f (n). This leads for instance to the three-periodic sequence

V3n ={

w ∈ C;∣∣∣w − 1

2

∣∣∣ < 16

}V3n+1 =

{w ∈ C;

∣∣∣w − 3∣∣∣ < 1

}V3n+2 =

{w ∈ C;

∣∣∣w +23

∣∣∣ < 112

}.

We shall return to this idea in Section 5.2 on page 238. �

Now, K(an/bn) does not necessarily converge generally, even if it is ε-contractivewith respect to V ; i.e., if there exists an ε > 0 such that sn(V ) ⊆ V , but omits achordal ε-disk

Bm(γn, ε) := {w ∈ C; m(w, γn) ≤ ε} (1.6.7)

with some center γn ∈ V for all n ∈ N. (The parabola theorem on page 151constitutes a counterexample.) But we can conclude that K(an/bn) is restrained:��

Theorem 2.11. Let V be a simple value set for K(an/bn). If K(an/bn) isε-contractive with respect to V , then K(an/bn) is restrained.

Proof : Assume that K(an/bn) is not restrained. Then there exists a subsequence{Snk

} of {Sn} converging to some S ∈ M. Therefore σk := S−1nk

◦ Snk+1 = snk+1 ◦· · · ◦ snk+1 converges uniformly to I(w) ≡ w (as in Theorem 2.2 on page 62). Thisis impossible when K(an/bn) is ε-contractive with respect to V . �

Remark. A similar result holds if {Vn} is a periodic sequence of value sets forK(an/bn).

2.1.7 Element sets

Families of continued fractions can be described by means of element sets {Ωn}∞n=1

where Ωn ⊆ C2. A continued fraction K(an/bn) belongs to the family if (an, bn) ∈Ωn for all n ∈ N. For short, we say that K(an/bn) is a continued fraction from{Ωn}.Similarly we say that K(an/1) is a continued fraction from the element sets {En}∞n=1

where En ⊆ C if an ∈ En for all n ∈ N. And K(1/bn) is a continued fraction fromthe element sets {Gn}∞n=1 where Gn ⊆ C if bn ∈ Gn for all n ∈ N.

If {Vn}∞n=0; Vn ⊆ C is a sequence of value sets for every continued fraction from asequence {Ωn} (or {En} or {Gn}) of element sets, we say that {Vn} is a sequence

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74 Chapter 2: Basics

of value sets for {Ωn} (or {En} or {Gn}). To a given sequence of value sets, thereexists an extreme sequence of element sets:

Definition 2.11. For a given sequence {Vn}∞n=0 with Vn ⊆ C, the sequence{Ωn} given by

Ωn := {(a, b) ∈ C2; a/(b + Vn) ⊆ Vn−1} (1.7.1)

is called the element sets (for continued fractions K(an/bn) ) correspondingto {Vn}.

These element sets characterize all continued fractions for which {Vn} is a sequenceof value sets (which may be an empty family). Similarly

En := {a ∈ C; a/(1 + Vn) ⊆ Vn−1} for n = 1, 2, 3, . . . (1.7.2)

andGn := {b ∈ C; 1/(b + Vn) ⊆ Vn−1} for n = 1, 2, 3, . . . (1.7.3)

are called element sets for continued fractions K(an/1) and K(1/bn) respectively,corresponding to {Vn}.If all Ωn = Ω, we say that Ω is a simple element set. Similarly, if {Ωn} is 2-periodic,we say that (Ω1,Ω2) are twin element sets for continued fractions K(an/bn). Ofcourse we use the same vocabulary to describe the element sets for K(an/1) andK(1/bn). For instance, the set E := [−1

4 , 14 ] in Example 7 on page 72 is a simple

element set corresponding to the simple value set V := [− 12 , 1

2 ].

Example 9. Let V := D, the closure of the unit disk D := {w ∈ C; |w| < 1}. Thena/(b + V ) ⊆ V if and only if |b| ≥ |a| + 1. Therefore

Ω := {(a, b) ∈ C2; |b| ≥ |a| + 1}

is a simple element set corresponding to V , and V is a simple value set for ev-ery continued fraction K(an/bn) from Ω. On page 129 we shall prove that everycontinued fraction from Ω converges. �

Example 10. Let

V0 := B(1, 3) and V1 := C \ B(0, 2)◦

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2.1.7 Element sets 75

where B(a, r) := {w ∈ C; |w − a| ≤ r} for a ∈ C and r > 0, and B(a, r)◦

is its interior. Then 1 + V0 = B(2, 3), 1/(1 + V0) = C \ B(−25 , 3

5)◦, and thusa/(1 + V0) = C \ B(−2a

5 , 3|a|5 )◦. Therefore a/(1 + V0) ⊆ V1 if and only if∣∣∣∣−2a

5− 0∣∣∣∣+ 2 ≤ 3|a|

5; i.e., |a| ≥ 10.

Similarly, 1+V1 = C\B(1, 2)◦, 1/(1+V1) = B(− 13 , 2

3), and a/(1+V1) = B(−a3 , 2|a|

3 )which is contained in V0 if and only if∣∣∣−a

3− 1∣∣∣+ 2|a|

3≤ 3; i.e., |a + 3| + 2|a| ≤ 9.

Therefore

E1 := {a ∈ C; |a + 3| + 2|a| ≤ 9}, E2 := {a ∈ C; |a| ≥ 10}

are the twin element sets for continued fractions K(an/1) corresponding to (V0, V1).�

To a given sequence of element sets there also exists an extreme sequence of valuesets:�

Theorem 2.12. For a given sequence {Ωn}∞n=1 with Ωn ⊆ C2, let for

each n ∈ N ∪ {0}, Vn ∈ C be the set of nth tail values for every generallyconvergent continued fraction from {Ωn}. Then {Vn} is a sequence of valuesets for {Ωn}.

Proof : Let (a, b) ∈ Ωn with a �= 0 and f (n) ∈ Vn be arbitrarily chosen. Then

f (n) =an+1

bn+1 +an+2

bn+2 +an+3

bn+3 +· · ·

for some generally convergent continued fraction K(an/bn) from {Ωn}, and thusalso

f (n−1) :=a

b + f (n)=

a

b +an+1

bn+1 +an+2

bn+2 +an+3

bn+3 +· · ·is the (n − 1)th tail value for some generally convergent continued fraction from{Ωn}. That is, f (n−1) ∈ Vn−1. �

This particular sequence of value sets for a sequence {Ωn} of element sets is calledthe sequence of limit sets for {Ωn}, and we use the terms simple limit set and twinlimit sets with obvious interpretation.

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76 Chapter 2: Basics

More importantly: when does K(an/bn) from {Ωn} converge? This will be the topicof Chapter 3. The following slight generalization of a result from [Jaco86b] gives asimple answer:

Theorem 2.13. Let (E1, E2) be twin element sets corresponding to thetwin value sets (V0, V1), and let E∗

1 , E∗2 be closed, bounded subsets of E◦

1

and E◦2 respectively. Then every continued fraction K(an/1) from (E∗

1 , E∗2 )

converges.

The proof is postponed to Chapter 4. The last example in this section indicateshow the knowledge of corresponding element and value sets can be used to estimatecontinued fraction values when we already know convergence.

Example 11. We shall see later (Worpitzky’s theorem on page 135) that the con-tinued fraction

∞Kn=1

an

1:=

−1/41 +

1/81 +

a3

1 +a4

1 +· · ·(1.7.4)

with −14 ≤ an ≤ 1

4 for all n converges. What can be said about the value f ofK(an/1)?

Since the interval V := [−12, 1

2] is a simple value set for K(an/1) (Example 7), it

follows that the value f (2) of the second tail

a3

1 +a4

1 +· · ·+an

1 +· · ·

lies in this interval. Hence f ∈ S2(V ) where

S2(w) =−1/4

1 + 1/81 + w

= −14· 1 + w

98 + w

= −14

(1 − 1/8

w + 98

).

Now, V + 98 = [58 , 13

8 ], so 1/(V + 98 ) = [ 8

13 , 85 ] and − 1

8/(V + 98) = [− 1

5 ,− 113 ], and

thus

S2(V ) = −14

[1 − 1

5, 1 − 1

13

]=[− 3

13,−1

5

].

That is,

−0.23 · · · < − 313

≤ f ≤ −15

= −0.20 . . . .

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2.2.2 Equivalence transformations 77

2.2 Transformations of continued fractions

2.2.1 Introduction

We have already seen a number of transformations:

• from power series to continued fractions (Section 1.4.1 on page 30)

• from continued fraction to power series (Section 1.4.2 on page 33)

• from An and Bn to b0 + K(an/bn) (Section 1.1.2 on page 5)

• from {fn} to b0 + K(an/bn) (Problem 23 on page 51)

• from an infinite product to a continued fraction (Problem 26 on page 51)

In this section we shall introduce some transformations between continued fractions.The idea is that the transformed continued fraction may be easier to work with thanthe original one.

2.2.2 Equivalence transformations

Definition 2.12. We say that two continued fractions are equivalent if theyhave the same sequence of classical approximants.

We write K(an/bn) ∼ K(cn/dn) to express that K(an/bn) and K(cn/dn) are equiv-alent. Let the canonical numerators and denominators be denoted by An and Bn

for K(an/bn) and by Cn and Dn for K(cn/dn). If we require that all Cn = An

and Dn = Bn, then it follows from Theorem 1.2 on page 8 that the two continuedfractions are identical; that is, cn = an and dn = bn for all n. So that has no point.What we have done is to require that An/Bn = Cn/Dn for all n.

The idea of equivalent continued fractions is due to Seidel ([Seid55]) who also proved:

Theorem 2.14. K(an/bn) ∼ K(cn/dn) if and only if there exists a se-quence {rn} of complex numbers with r0 = 1, rn �= 0 for all n ∈ N, suchthat

cn = rnrn−1an , dn = rnbn for all n ∈ N . (2.2.1)

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78 Chapter 2: Basics

Proof : Let An, Bn be the canonical numerators and denominators of K(an/bn).Then K(an/bn) ∼ K(cn/dn) if and only if there exist numbers rn �= 0 such that thecanonical numerators Cn and denominators Dn of K(cn/dn) can be written

C−1 = 1, D−1 = 0, Cn = An

n∏k=0

rk, Dn = Bn

n∏k=0

rk (2.2.2)

for all n. Since D0 = B0 = 1 we need r0 = 1. From Theorem 1.2 on page 8 itfollows then that K(cn/dn) is given by (2.2.1). �

Properties:

1. If (2.2.1) holds and K(an/bn) has canonical numerators An and denominatorsBn, then K(cn/dn) has canonical numerators Cn and denominators Dn givenby (2.2.2).

2. The concept of equivalence is tied to the classical approximants. If K(an/bn) ∼K(cn/dn) by the relations (2.2.1), then

Sn(w) = Tn(rnw) for n = 0, 1, 2, . . . , (2.2.3)

where Sn(w) are approximants of K(an/bn), and Tn(w) are approximants ofK(cn/dn).

3. If {tn} is a tail sequence for K(an/bn), then {tnrn} is a tail sequence forK(cn/dn).

Example 12. The continued fraction

∞Kn=1

anz

1:=

z

1+z/21 +

z/61 +

2z/61 +

2z/101 +

3z/101 +

3z/141 +

4z/141 +· · ·

,

wherea1 := 1, a2k :=

k

2(2k − 1), a2k+1 :=

k

2(2k + 1)for k ≥ 1,

converges to Ln(1 + z) for | arg(1 + z)| < π. An equivalence transformation with

r1 := 1, r2 := 2, r3 := 3, r4 := 2, r5 := 5, r6 := 2, r7 := 7, . . .

brings it over to the form

z

1+z

2+z

3+2z

2 +2z

5 +3z

2 +3z

7 +4z

2 +4z

9 +5z

2 +5z

11+6z

2 +· · ·

which therefore also converges to Ln(1 + z). �

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2.2.2 Equivalence transformations 79

Example 13. In Section 1.3.2 on page 27 we found that

1 +∞Kn=1

anz

1:= 1 −

a(c − b)c(c + 1)

z

1 −

(b + 1)(c − a + 1)(c + 1)(c + 2)

z

1 −

(a + 1)(c − b + 1)(c + 2)(c + 3)

z

1 −

(b + 2)(c − a + 2)(c + 3)(c + 4)

z

1 −· · ·

where a, b, c �∈ −N ∪ {0}, converges to the ratio F (a, b; c; z)/F (a, b + 1; c + 1; z) ofhypergeometric functions in the cut plane D := {z ∈ C; 0 < arg(1 + z) < 2π}. Anequivalence transformation with rn := c + n for n ≥ 1, and multiplication with c,gives the equivalent identity

cF (a, b; c; z)

F (a, b + 1; c + 1; z)

= c − a(c − b)zc + 1 −

(b + 1)(c − a + 1)zc + 2 −

(a + 1)(c − b + 1)zc + 3 −· · ·

for z ∈ D. �

Example 14. A regular C-fraction is a continued fraction of the form

∞Kn=1

anz

1=

a1z

1 +a2z

1 +a3z

1 +· · ·,

and if all an > 0, it is called an S-fraction. The equivalence transformation withr0 := 1, r2n−1 := w := 1/z and r2n := 1 for all n ∈ N brings it over to the form

a1

w +a2

1 +a3

w +a4

1 +· · ·; w :=

1z

.

If we instead use r0 := 1, rn := w := 1/√

z for all n ∈ N, we find that K(anz/1) isequivalent to

a1/w

w +a2

w +a3

w +a4

w +a5

w +· · ·; w :=

1√z

.

Example 15. The continued fraction in Problem 26 on page 51 has the form

a0 +a0(a1 − 1)

1 −a1(a2 − 1)/(a1 − 1)(a1a2 − 1)/(a1 − 1)−

a2(a3 − 1)/(a2 − 1)(a2a3 − 1)/(a2 − 1)−

· · ·−an−1(an − 1)/(an−1 − 1)(an−1an − 1)/(an−1 − 1)−· · ·

.

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80 Chapter 2: Basics

An equivalence transformation with r0 := 1, r1 := 1, rn := an−1 − 1 for n ≥ 2 leadsto

a0+a0(a1 − 1)

1 −a1(a2 − 1)a1a2 − 1 −

a2(a3 − 1)(a1 − 1)a2a3 − 1 −

−a3(a4 − 1)(a2 − 1)

a3a4 − 1 −a4(a5 − 1)(a3 − 1)

a4a5 − 1 −· · ·which therefore also has classical approximants fn =

∏nk=0 ak. �

Two equivalence transformations are of particular interest, namely the ones we getby using

rn :=n∏

k=1

a(−1)n−k+1

k and rn :=1bn

for all n ≥ 1 :

Corollary 2.15.

A. K(an/bn) ∼ K(1/dn) where

dn := bn

n∏k=1

a(−1)n+1−k

k for n = 1, 2, 3, . . . . (2.2.4)

B. If bn �= 0 for all n ≥ 1, then K(an/bn) ∼ K(cn/1), where

c1 :=a1

b1, cn :=

an

bnbn−1for n = 2, 3, 4, . . . . (2.2.5)

Remark: The transformation in A can always be performed. That is, every con-tinued fraction K(an/bn) can be brought to the equivalent form K(1/dn). Theelements dn have the structure dn = bn/andn−1, so

d1 = b1 ·1a1

, d2 = b2a1

a2, d3 = b3

a2

a1a3, d4 = b4

a1a3

a2a4, . . . .

The transformation in B can only be applied if all bn �= 0, since otherwise cn is nota well defined complex number.

The equivalence transformation is extremely useful in the continued fraction the-ory. Of course, if K(an/bn) ∼ K(cn/dn), then K(an/bn) converges if and only ifK(cn/dn) converges. This is no longer true for general convergence. If K(an/bn)converges generally, but not in the classical sense, then some of its equivalent formsK(cn/dn) may still diverge generally.

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2.2.2 Equivalence transformations 81

Example 16. Let K(an/1) be as in Example 2 on page 59, and let rn := −Bn−1/Bn

for all n ∈ N. Then K(an/1) ∼ K(cn/dn) where cn := rn−1rnan and dn := rnbn

where r0 := 1, so by (2.2.3) the approximants of K(cn/dn) are Tn(w) = Sn(w/rn).In particular

T3n(w) =A3n−1w/r3n + A3n

B3n−1w/r3n + B3n=

A3n − A3n−1wB3n/B3n−1

B3n(1 − w),

where we know that A3n = 0, A3n−1 = 2n, B3n = 1 and B3n−1 = 2n+1 − 2.Therefore

T3n(w) = − 2n

2n+1 − 2· w

1 − w→ 1

2w

w − 1,

a non-singular transformation from M. Hence {Tn} is not restrained, and K(cn/dn)diverges generally. �

Indeed, this is an example of a general phenomenon:�

�Theorem 2.16. Let K(an/bn) diverge in the classical sense, and let {tn}be a tail sequence for K(an/bn) with t0 not a limit point for {Sn(0)}. Then{Tn} given by Tn(w) := Sn(tnw) is not a restrained sequence.

Proof : Since {Sn(0)} diverges, there exists a subsequence {nk} of the naturalnumbers such that

fnk−1 = Snk−1(0) → g1 and fnk= Snk

(0) → g2 �= g1.

Assume first that t0 = ∞ and g1, g2 �= 0,∞. Then tn = ζn = −Bn/Bn−1, andfnk−1 and fnk

are �= 0,∞ from some k on. So, for n := nk with k sufficiently large,

Sn(tnw) =An−1tnw + An

Bn−1tnw + Bn=

−fn−1Bnw + An

−Bnw + Bn=

−fn−1w + fn

−w + 1

which converges to the non-singular transformation

limk→∞

Snk(tnk

w) =g2 − g1w

1 − w.

If t0 �= ∞ and/or gk ∈ {0,∞} for k = 1 or 2, there always exist complex numbersc1, c2, d1, d2 such that c1c2 �= 0 and

c1

d1 +c2

d2 + t0= ∞,

c1

d1 +c2

d2 + gk�= 0,∞ for k = 1, 2.

Since thereforec1

d1 +c2

d2 + Sn(tnw)

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82 Chapter 2: Basics

is non-restrained, so is {Sn(tnw)}. �

This means in particular that if K(an/bn) converges generally with exceptionalsequence {w†

n}, but not in the classical sense, then one should stay away fromequivalence transformations (2.2.1) with lim inf m(rn, w†

n) = 0.

Typical in this situation is that {w†n} has limit points at 0 and ∞. Indeed, if {rn}

is bounded away from 0 and ∞, the equivalence transformation preserves generalconvergence:

�Theorem 2.17. Let K(an/bn) converge generally to f . If the sequence{rn} of complex numbers is bounded and bounded away from 0, thenK(rn−1rnan/rnbn) converges generally to r0f .

Proof : This follows from (2.2.3) since

lim infn→∞ m(un, vn) > 0 ⇐⇒ lim inf

n→∞ m(un/rn, vn/rn) > 0.

2.2.3 The Bauer-Muir transformation

For given continued fraction b0 + K(an/bn) and given sequence {wn} from C, theidea is to construct a new continued fraction d0 + K(cn/dn) whose sequence ofclassical approximants {Tn(0)} is exactly {Sn(wn)}. The new continued fraction isonly unique up to an equivalence transformation. The Bauer-Muir transform, theway we define it, is the canonical one in this equivalence class:

Definition 2.13. The Bauer-Muir transform of a continued fraction b0 +K(an/bn) with respect to a sequence {wn} from C is the continued fractiond0 + K(cn/dn) whose canonical numerators Cn and denominators Dn aregiven by

C−1 = 1 , D−1 = 0 ,Cn = An−1wn + An , Dn = Bn−1wn + Bn; n ≥ 0.

(2.3.1)

This transformation dates back to the 1870’s when Bauer ([Bauer72]) and Muir([Muir77]), independently of each other, proved what can be formulated as follows:

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2.2.3 The Bauer-Muir transformation 83�

Theorem 2.18. The Bauer-Muir transform of b0 + K(an/bn) with respectto {wn} from C exists if and only if

λn := an − wn−1(bn + wn) �= 0 for n = 1, 2, 3, . . . . (2.3.2)

If it exists, then it is given by

b0 + w0 +λ1

b1 + w1 +c2

d2 +c3

d3 + · · · where

cn := an−1λn

λn−1and dn := bn + wn − wn−2

λn

λn−1.

(2.3.3)

Proof : Let {Cn} and {Dn} be given by (2.3.1). Then {Cn} and {Dn} are canon-ical numerators and denominators of a continued fraction d0 +K(cn/dn) if and onlyif C−1 = D0 = 1, D−1 = 0 and Δn := Cn−1Dn − Dn−1Cn �= 0 for all n ≥ 1.(See Theorem 1.2 on page 8.) The initial conditions for Cn and Dn are satisfied.Moreover

Δn := Cn−1Dn − Dn−1Cn

= (An−2wn−1 + An−1)(Bn−1wn + Bn)− (Bn−2wn−1 + Bn−1)(An−1wn + An)

= (An−2wn−1 + An−1)(Bn−1wn + bnBn−1 + anBn−2)−(Bn−2wn−1 + Bn−1)(An−1wn + bnAn−1 + anAn−2)

= −(An−2Bn−1 − An−1Bn−2)(an − wn−1bn − wn−1wn)

where the first factor is different from 0 by the determinant formula on page 7,and the second factor is equal to λn in (2.3.2). This proves the existence part ofTheorem 2.18. The expressions for cn and dn follow from Theorem 1.2 on page 8.�

This transformation is in particular useful if b0+K(an/bn) and its Bauer-Muir trans-form converge to the same value. This was proved to be the case for positive contin-ued fractions and positive wn by Perron ([Perr57], p 27), but of course it holds when-ever K(an/bn) converges with exceptional sequence {w†

n} and lim inf m(wn, w†n) > 0,

([Lore94c]).

Example 17. Stable computation. We want to compute the approximantsSn(−6) of the continued fraction

K an

1:=

30 + 0.51 +

30 + (0.5)2

1 +30 + (0.5)3

1 + · · · .

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84 Chapter 2: Basics

It will be evident later that K(an/1) converges to a finite number f > 0 withexceptional sequence {−6}. Computation shows that f = 5.05859 correctly roundedto 6 digits. But what happens to Sn(−6)? As noted in Remark 4 on page 57 wedo not know off-hand whether {Sn(−6)} diverges or converges, and if it converges,we do not know its limit. If we try to compute Sn(−6) from the continued fraction,we have a problem. Small inaccuracies in the input or computation will have theeffect that {−6}∞n=0 is not recognized as an exceptional sequence. Our numericallycomputed value for Sn(−6) will approach f as n → ∞.

The Bauer-Muir transformation gives a more stable method to compute Sn(−6).Since

λn = 30 + (0.5)n − (−6)(1 − 6) = (0.5)n ,

the Bauer-Muir transform of K((30+ (0.5)n)/1) with respect to wn = −6 exists. Itis given by

n Sn(−6) Tn(0)1 −6.09999 −6.039602 −6.03962 −6.075823 −6.07580 −6.05404

12 −6.06215 −6.0622813 −6.06218 −6.0621514 −6.06229 −6.0622315 −6.06208 −6.0621818 −6.06247 −6.0622119 −6.06189 −6.0622020 −6.06257 −6.06220

161 5.05858 −6.06220162 5.05859 −6.06220163 5.05859 −6.06220

− 6 +0.5−5+

(30 + 0.5) · 0.5−5 − (−6) · 0.5+

(30 + (0.5)2) · 0.5−5 − (−6) · 0.5 +· · ·

= −6 +0.5−5+

15 + (0.5)2

−2 +15 + (0.5)3

−2 +· · ·,

and its classical approximants Tn(0) (which can becomputed stably) are exactly Sn(−6). In the ta-ble we have computed Sn(−6) and Tn(0). The twocolumns should have been identical. The compu-tation is done with only 6 digits to illustrate theinstability in the computation of Sn(−6). �

The Bauer-Muir transformation has many applications. Let us look at one more:

Example 18. Functional equation. The continued fraction∞Kn=1

z + n

n:=

z + 11 +

z + 22 +

z + 33 +· · ·

converges to some f(z) with exceptional sequence {w†n} where (w†

n/n) → −1 for allz ∈ C. (This is a consequence of Theorem 4.13 on page 188 after an equivalencetransformation.) Hence also Sn(1) converges to f(z). Therefore its Bauer-Muirtransform d0 + K(cn/dn) with respect to wn := 1 converges to f(z). Since

λn = z + n − 1(n + 1) = z − 1 for all n

with this choice for wn, we have for z �= 1

f(z) = d0 +∞Kn=1

cn

dn= 1 +

z − 12 +

z + 12 +

z + 23 +

z + 34 +· · ·

,

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2.2.5 Contractions and extensions 85

which looks similar to K((z + n)/n). Indeed, the first tail g(1)(z) of d0 + K(cn/dn)is such that

z

1 + g(1)(z)=

z

1+z + 1

2 +z + 2

3 +· · ·= f(z − 1) ,

that is, g(1)(z) = −1 + z/f(z − 1). This means that

f(z) = 1 +z − 1

2 + g(1)(z)= 1 +

z − 11 + z

f(z − 1)= z · f(z − 1) + 1

f(z − 1) + z

for z �= 1. This is a functional equation for f(z). For z = 1 the original continuedfraction is

f(1) =21 +

32 +

43 +

54 + · · · = 1

since the constant sequence {1} is a tail sequence for this continued fraction, and

∞∑n=0

n∏k=1

bk + tk−tk

=∞∑

n=0

n∏k=1

k + 1−1

=∞∑

n=0

(−1)n(n + 1)! = ∞

(corollary 2.7 on page 68). Hence,

f(2) = 2 · 1 + 11 + 2

=43

, f(3) = 3 ·43

+ 1

43

+ 3=

2113

,

f(4) = 4 ·2113

+ 1

2113

+ 4=

13673

, f(5) = 5 ·13673

+ 1

13673

+ 5=

1045501

,

and so on. �

2.2.4 Contractions and extensions

We say that d0 +K(cn/dn) is a contraction of b0 +K(an/bn) if its classical approx-imants {gn} is a subsequence of the classical approximants {fn} of b0 + K(an/bn).We call b0 + K(an/bn) an extension of d0 + K(cn/dn) in this case. We say inparticular that d0 + K(cn/dn) is a canonical contraction of b0 + K(an/bn) if

Ck = Ank, Dk = Bnk

for k = 0, 1, 2, . . . , (2.4.1)

where Cn, Dn, and An, Bn are the canonical numerators and denominators ofd0 +K(cn/dn) and b0 +K(an/bn) respectively. To derive a general expression for acanonical contraction we can use Theorem 1.2 on page 8. This idea is due to Seidel([Seid55]). Rather than considering the general case we shall look at two importantspecial cases:

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86 Chapter 2: Basics

• an even part of b0 + K(an/bn) is a contraction with classical approximantsgk := f2k. (That is, an even part is a continued fraction!) It is the canonicaleven part if Ck = A2k and Dk = B2k for all k.

• an odd part of b0 + K(an/bn) is a contraction with classical approximantsgk := f2k+1. It is the canonical odd part if Ck = A2k+1 and Dk = B2k+1 forall k.

Theorem 2.19. K(an/bn) has an even part if and only if all b2n �= 0. Itscanonical even part is then given by

b2a1

b2b1 + a2−a2a3b4/b2

a4 + b3b4 + a3b4/b2−a4a5b6/b4

a6 + b5b6 + a5b6/b4−· · ·(2.4.2)

If {tn}∞n=0 is a tail sequence for K(an/bn) with all tn �= ∞, thent0,−t1t2,−t3t4,−t5t6, . . . is a tail sequence for (2.4.2).

Proof : From Theorem 1.2 on page 8 it follows that K(an/bn) has an even partif and only if f2n �= f2n−2 for all n, that is, if and only if all b2n �= 0 (Theorem1.3 on page 9). From Theorem 1.2 on page 8 we find that the canonical even partd0 + K(cn/dn) has elements d0 := 0,

d1 := D1 = B2 = b2b1 + a2 ,

c1 := C1 − C0D1 = A2 − A0B2 = b2a1 ,

cn := −Δn/Δn−1 , dn := Δ(1)n /Δn−1 ,

whereΔn := Cn−1Dn − Dn−1Cn = A2n−2B2n − B2n−2A2n

Δ(1)n := Cn−2Dn − Dn−2Cn = A2n−4B2n − B2n−4A2n .

The recurrence relation for {An} and {Bn} (page 8) and the determinant formula(page 7) lead to

Δn = b2n

2n−1∏j=1

(−aj) ,

Δ(1)n =

(b2n(b2n−1b2n−2 + a2n−1) + a2nb2n−2

) 2n−3∏j=1

(−aj) .

This proves (2.4.2). Let {tn} be a tail sequence for K(an/bn) with all tn �= ∞.To see that t0,−t1t2,−t3t4, . . . is a tail sequence for (2.4.2) it suffices to prove that−t2n−3t2n−2 = cn/(dn − t2n−1t2n) for n ≥ 2; i.e.,

cn = −t2n−3t2n−2(dn − t2n−1t2n) and c1 = t0(d1 − t1t2) .

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2.2.5 Contractions and convergence 87

This follows by straight forward computation using that an = tn−1(bn + tn). �

The even part (2.4.2) is equivalent to

b0 +b2a1

b2b1 + a2−a2a3b4

b2(a4 + b3b4) + a3b4

−a4a5b6b2

b4(a6 + b5b6) + a5b6−a6a7b8b4

b6(a8 + b7b8) + a7b8−· · ·, (2.4.3)

which is more widely used (but which is no longer canonical). This even part wasestablished by Seidel ([Seid55]) although Lagrange had some special cases alreadyin 1774–76 ([Lagr76]).�

Theorem 2.20. K(an/bn) has an odd part if and only if all b2k+1 �= 0. Itscanonical odd part is then given by

a1

b1− a1a2b3/b1

b1(a3 + b2b3) + a2b3−a3a4b5b1/b3

a5 + b4b5 + a4b5/b3

−a5a6b7/b5

a7 + b6b7 + a6b7/b5−a7a8b9/b7

a9 + b8b9 + a8b9/b7−· · ·(2.4.4)

If {tn} is a tail sequence for b0 + K(an/bn) with all tn �= ∞, then−t0t1/b1, −b1t2t3, −t4t5, −t6t7, . . . is a tail sequence for (2.4.4).

The proof follows the same lines as the proof of Theorem 2.19 and is omitted. Theodd part (2.4.4) is equivalent to

a1

b1− a1a2b3/b1

b1(a3 + b2b3) + a2b3

−a3a4b5b1

b3(a5 + b4b5) + a4b5−a5a6b7b3

b5(a7 + b6b7) + a6b7−· · ·. (2.4.5)

2.2.5 Contractions and convergence

Example 19. For given 0 < q < 1 we study the continued fraction

∞Kn=1

(1/qn) =1q+

1q2 +

1q3 +· · ·

. (2.5.1)

Now, A0 = 0, A1 = 1, B0 = 1 and B1 = q, and the recurrence relations

An = qnAn−1 + An−2, Bn = qnBn−1 + Bn−2

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88 Chapter 2: Basics

show that {A2n}, {A2n+1}, {B2n} and {B2n+1} are strictly increasing, positivesequences, and that by induction

An <n∏

k=1

(1 + qk), Bn ≤n∏

k=1

(1 + qk) .

Hence the four sequences are bounded, and thus convergent, and the finite, positivelimits

A0 := lim A2n, A1 := lim A2n+1,

B0 := lim B2n, B1 := lim B2n+1

all exist. Moreover, the determinant formula shows that

|An−1Bn − Bn−1An| = 1; i.e., |A0B1 − B0A1| = 1 ,

and thus {S2n} and {S2n+1} converge to non-singular linear fractional transforma-tions

S(w) =:A1w + A0

B1w + B0, S∗(w) =:

A0w + A1

B0w + B1

respectively. In particular this means that both {S2n} and {S2n+1} are totallynon-restrained. Still

lim S2n(0) = A0/B0 and limS2n+1(0) = A1/B1 ,

so the even and odd parts of K(1/qn) converge to separate values. The canonicaleven part of K(1/qn) is

q2

1 + q1+2−q2

1 + q2 + q3+4−q2

1 + q2 + q5+6−· · ·(2.5.2)

and the canonical odd part is

1q− q2

q(1 + q2 + q2+3)−q3

1 + q2 + q4+5−q2

1 + q2 + q6+7−q2

1 + q2 + q8+9−· · ·−q2

1 + q2 + q4n+1−· · ·.

(2.5.3)

The sequences

Sn(w) :=An−1w + An

Bn−1w + Bn, Tn(w) :=

A2n−2w + A2n

B2n−2w + B2n,

Un(w) :=A2n−1w + A2n+1

B2n−1w + B2n+1

belonging to (2.5.1), (2.5.2) and (2.5.3) respectively, have different properties: {S2n}and {S2n+1} converge to distinct non-singular linear fractional transformations, so{Sn} is totally non-restrained. On the other hand, both (2.5.2) and (2.5.3) convergein the classical sense, so {Tn} and {Un} are restrained, even generally convergent.�

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Remarks 89

There is an important lesson to be learned from this example: contractions arecontinued fractions, so if they converge, they converge generally. But K(an/bn)itself does not even have to be restrained. This is no longer so if lim inf |bn| > 0,which in particular includes the continued fractions K(an/1):

Theorem 2.21. Let lim inf |bn| > 0 for K(an/bn). If {S2n+m(0)} convergesfor a fixed m ∈ {0, 1}, then {S2n+m} converges generally.

Proof : Let S2n+m(0) → f as n → ∞. The result follows since

S2n+m(−b2n+m) = S2n+m−1(∞) = S2n+m−2(0),

so lim S2n+m(0) = lim S2n+m(−b2n+m) = f where lim inf m(0,−b2n+m) > 0. �

Contractions are defined in terms of classical approximants. It is not much to begained from extending the concept to more general approximants, but it is inter-esting to study the connection between the linear fractional transformations

Sn(w) =An−1w + An

Bn−1w + Bn

for K(an/bn) and

Sen(w) =

A2n−2w + A2n

B2n−2w + B2n, So

n(w) =A2n−1w + A2n+1

B2n−1w + B2n+1

for its canonical even and odd parts when these continued fractions exist. Therecurrence relation (1.2.7) for {An} and {Bn} on page 6 then shows that

Sen(w) =

A2n−2w + b2nA2n−1 + a2nA2n−2

B2n−2w + b2nB2n−1 + a2nB2n−2

=A2n−2

w+a2n

b2n+ A2n−1

B2n−2w+a2n

b2n+ B2n−1

= S2n−1

(w + a2n

b2n

),

(2.5.4)

and similarly

Son(w) = S2n

(w + a2n+1

b2n+1

). (2.5.5)

2.3 Remarks

1. The convergence concept. The natural way to define convergence of a continuedfraction K(an/bn) is to truncate it after n terms to get either

fn :=a1

b1 +

a2

b2 +· · ·+an

bnor fn :=

a1

b1 +

a2

b2 +· · ·+an

bn + an+1

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90 Chapter 2: Basics

and to require convergence of {fn} or {fn}. This was also the idea for a long time.The first proper definition is due to Seidel ([Seid46]). However, as time passed on,one got a little uneasy with this definition, in particular since also approximants ofthe type Sn(w) were in use.

In 1918 Hamel ([Hamel18]) suggested that convergence should be defined in termsof Sn(w) instead of Sn(0). His rather vague ideas was made into a proper definitionof strong convergence by Thron and Waadeland in 1982 ([ThWa82] p 42).

But even this was not quite right. The definition of general convergence dates backto 1986 ([Jaco86]). It is really a convergence concept for sequences of functions fromM (or rather functions from quasi-normal families [Lore03a]).

Now, there exist continued fractions for which {fn} converges and {fn} diverges andvice versa. More generally, for any sequence {ϕn} from M with ϕ0(w) ≡ w

sn := ϕ−1n−1 ◦ sn ◦ ϕn and Sn := s1 ◦ · · · ◦ sn = Sn ◦ ϕn

could in principle have been used to define K(an/bn), and the convergence propertiesof {Sn} may differ from the ones of {Sn}.

2. Tail sequences. In early works on continued fractions the critical tail sequencehn := −S−1

n (∞) pops up in formulas of different types, mainly because hn =Bn/Bn−1. Otherwise tail sequences have not been recognized as objects of its ownright until Waadeland in a series of papers starting in 1966 ([Waad66]) used peri-odic tail sequences of periodic continued fractions. His aim was to continue analyticfunctions defined by limit periodic continued fractions to larger domains. Lateron, Thron and Waadeland ([ThWa80a]) applied sequences of tail values for peri-odic continued fractions to accelerate the convergence of limit periodic ones. Theseapplications were the inspiration for the names of these tail sequences: right tailsequence for the tail values, and wrong tail sequence for the other tail sequences ofa convergent continued fraction.

In this book the tail sequences are used for a number of purposes, such as provingconvergence, estimating the value and finding truncation error bounds for a givencontinued fraction. The vital point is the asymptotic behavior of the sequence oftail values on the one hand and the remaining tail sequences which all work asexceptional sequences for a restrained continued fraction on the other hand. Thisuse of tail sequences has been advocated by the authors for a long time.

Chihara ([Chih78]) has derived a number of results by use of what he called chainsequences {an}; i.e., sequences of positive numbers which can be written an =gn(1 − gn−1) with 0 < gn < 1. The sequence {gn} was called a parameter sequencefor {an}. The reason why this works so well is that {gn − 1} actually is a tailsequence for K(−an/1).

The first formulas in Theorem 2.6 on page 66 was published in [JaWa82], and thelast one by Waadeland ([Waad84]) for continued fractions K(an/1) and extendedby Jacobsen ([Jaco86a]).

3. Value sets. The early convergence results were obtained by manipulating therecurrence relation Xn = bnXn−1 +anXn−2. The idea of value sets possibly startedwith Weyl ([Weyl10]) and Hamel ([Hamel18]). The concept was applied with greatsuccess by a number of authors, such as Jones, Leighton, Paydon, Thron, Scott,

Page 104: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Problems 91

Wall. They required that either 0 ∈ Vn or an/bn ∈ Vn for all n so that fn = Sn(0) ∈Sn−1(Vn−1) ⊆ V0. The shift to our definition started with Jacobsen ([Jaco82]) andThron and Waadeland ([ThWa82], p 43). Indeed, the idea of general convergencecame as a result of struggling with these more general value sets.

4. Limit sets. The (unique) sequence of best value sets {Wn} for a given sequence{Ωn} of element sets was defined by Jones and Thron ([JoTh80], p 65) as

Wn := {sn+1 ◦ sn+2 ◦ · · · ◦ sn+m(0); m ∈ N, (ak, bk) ∈ Ωk for all k}

for n = 0, 1, 2, . . . . That is, Wn is the set of all classical approximants f(n)m for

the nth tail of every continued fraction from {Ωk}. It was best in the sense thatWn ⊆ Vn for all n for every sequence of value sets {Vn} of the traditional type for{Ωk}.Since we are no longer limited to look at classical approximants, these best valuesets are too large in most cases; we can do better. We want our value sets to besmall” closed sets. The concept of limit sets {Vn} was introduced in [Jaco86b].

With the present definition, they are also best” in the sense that Vn ⊆ Vn for all nfor every sequence {Vn} of closed value sets for {Ωn}.

5. Transformations of continued fractions. One can construct a number of dif-ferent types of transformations for continued fractions, for instance based on ap-proximants, classical or non-classical. But one must always keep in mind what thetransformation actually does, because it may change certain convergence properties.

Equivalence transformations are totally based on classical approximants. HenceK(an/bn) converges if and only if it is equivalent to a convergent continued fraction,whereas general convergence does not have this same property.

One can define some kind of general equivalence with respect to two sequences {un}and {vn} by demanding

Sn(un) = Tn(un), Sn(vn) = Tn(vn) and lim inf m(un, vn) > 0.

This would be the counterpart to traditional equivalence, there we actually haveSn(0) = Tn(0), and thus also Sn(∞) = Tn(∞), but it would not be half as nice . . . .

2.4 Problems

1. Linear fractional transformations. Determine the linear fractional transforma-tion τ with τ(0) = −1, τ(1) = 0 and τ(∞) = 1

2.

2. Chordal metric. Prove that lim an = γ ∈ C if and only if lim m(an, γ) = 0.

3. Restrained sequences. Prove that the three definitions 2.5, 2.6 and 2.7 of arestrained sequence beginning on page 61 are equivalent.

““

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92 Chapter 2: Basics

4. Periodic tail sequences. Prove that {w†n} in Example 4 on page 64 is a tail se-

quence for K(an/1). Prove that {w†n} and {f (n)} are the only periodic tail sequences

for K(an/1).

5. Computation of tail sequences. Given the periodic continued fraction

Kn=1

2

1=

2

1+

2

1+

2

1+· · ·.

(a) Prove that

Sn(w) =(1 + 2 · (− 1

2)n)w + 2(1 − (− 1

2)n)

(1 − (− 12)n)w + 2 + (− 1

2)n

.

(b) Prove that {Sn} converges generally to 1.

(c) For t0 := 1 + h ∈ C, find the tail sequence {tn} of K(2/1). Distinguish

between the three cases h = 0, h = −3 and h ∈ C \ {0,−3}. Which ones ofthese tail sequences are exceptional sequences for K(2/1)? What is the criticaltail sequence for K(2/1)?

(d) Find an explicit expression for S−1n (w), and prove that {S−1

n } converges gen-erally to −2 with exceptional sequence {1}.

(e) Use the results above to illustrate Theorem 2.1 on page 61.

6. Computation of tail sequences. Given the continued fraction

Kn=1

an

1:=

Kn=1

n(n + 2)

1=

1 · 31 +

2 · 41 +

3 · 51 +· · ·

.

(a) Prove that {n + 1} is a tail sequence for K(an/1).

(b) Prove that the continued fraction converges to 1.

(c) Use Corollary 2.8 on page 69 to find an expression for the terms in the tailsequence {tn} for K(an/1) starting with t0 := 1 + h �= 1.

7. ♠ Critical tail sequence. Let hn := −S−1n (∞) for K(an/bn). Show that

hn = bn + an/hn−1 for n = 1, 2, 3, . . . .

8. Computation of tail sequences and approximants. Given the periodic con-tinued fraction

Kn=1

an

bn=

2

4+

−4

1 +

2

4+

−4

1 +

2

4+

−4

1 +· · ·=

2

4−4

1+

2

4−4

1+

2

4−4

1+· · ·.

(a) Find the periodic tail sequences of K(an/bn).

(b) Find the first few terms of its critical tail sequence {hn}.

Page 106: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Problems 93

(c) Show that

1

3≤ h2n ≤ 1 and 6 ≤ h2n+1 ≤ 10 for all n ≥ 2 .

(d) Use the results from (a) and Corollary 2.7 on page 68 to prove that K(an/bn)converges to 1, and to find bounds for |fn − 1|.

(e) Use Corollary 2.8 on page 69 to determine the asymptotic behavior of {hn}.(f) Compute the first few approximants of the types Sn(0), Sn(tn) and Sn(tn) for

the continued fraction

2 + 0.5

4 −4 + (0.5)2

1 +

2 + (0.5)3

4 −4 + (0.5)4

1 +

2 + (0.5)5

4 −· · ·

where {tn} and {tn} are the periodic tail sequences ofK(an/bn) from (a). Compare these sequences of approximants.

(g) Construct the Bauer-Muir transform of the continued fraction in (f) with re-spect to {tn} and with respect to {tn}.

9. Tail sequences. In each of the following cases, find a tail sequence for the givencontinued fraction, and use this to find its value.

(a)

Kn=1

(x + n)(x + n + 2)

1.

(b)

Kn=1

(x + n)2

−1.

(c)

Kn=1

n2 − x2

2x − 1.

(d)

Kn=1

1

bnwhere bn =

(n + a)2 + n + a − 1

n + a + 1.

10. Tail sequences. Prove that if the continued fraction

1 +a1

1 +

(a1 − z)(a1 + 1)

z +

a1a2

1 +

(a2 − z)(a2 + 1)

z +

a2a3

1 +· · ·

converges, then its value is either 0 or 1 + z. (Perron [Perr57], p 9.)

11. Tail sequences. Prove that if the continued fraction

a − a(a + z)

a + b + z + 1−(b + 1)(b + z + 1)

a + b + z + 2 −(a + 1)(a + z + 1)

a + b + z + 3 −

−(b + 2)(b + z + 2)

a + b + z + 4 −(a + 2)(a + z + 2)

a + b + z + 5 −· · ·

converges, where a, b, z ∈ C are chosen such that the terms are �= 0, then it convergesto either 0 or −z. (Perron [Perr57], p 25.)

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94 Chapter 2: Basics

12. Tail sequence. Let t2n := 1 and t2n+1 := 22n+1 + 1 for n = 0, 1, 2, . . . , and let

Kn=1

an

1:=

4

1+

6

1+

10

1 +

18

1 +· · ·+2n + 2

1 +· · ·.

(a) Prove that {tn} is a tail sequence for K(an/1).

(b) Prove that K(an/1) diverges.

(c) Use (1.5.7) on page 66 to prove that the even part of K(an/1) converges to 1and the odd part to 2.

(Part (c) was proved by Angell and Hirschhorn ([AnHi05]) in a completely differentway.)

13. Continued fraction identity. Determine the continued fraction K(an/1) whichhas the tail sequence {1/(n + 1)}∞n=0, and prove that K(an/1) converges to 1.

14. Corresponding element sets. For which bn > 0 is V := {w ∈ C; |w − 1| < 1} avalue set for K(1/bn)? For which an > 0 is V a value set for K(an/1)?

15. Corresponding element sets. For given p ∈ R, let V be the half plane V = {w ∈C; Re(w) > p}. For which values of p do there exist continued fractions K(an/1)which has V as a value set? How about continued fractions K(1/bn)?

16. Restrained sequences. For given d ∈ C, let

τn(w) =1 + (2 + 1

n)w

2 + (d − 1n)w

for n = 1, 2, 3, . . . .

(a) For which pairs (n, d) are τn ∈ M? Assume in the following that all τn ∈ M.

(b) Find the d-values for which {τn} is a restrained sequence. Then find an excep-tional sequence {w†

n} and a generic sequence {zn} for {τn}.(c) Give an illustration of Theorem 2.1 on page 61 by explicit computation of τ−1

n .

(d) What can be said about {τn} for d-values different from the ones in (b)?

17. ♠ Totally non-restrained sequence. In Theorem 2.2 on page 62 we saw that ifτn → τ in M, then σn := τ−1

n−1 ◦ τn → I. For

τn :=anw + bn

cnw + dn

with an → a, bn → b, cn → c, dn → d and ad − bc �= 0 for a, b, c, d ∈ C, compute σn

explicitly and verify that σn → I.

18. Periodic continued fractions. For arbitrary p > 0 we are given the 3-periodiccontinued fraction ∞

Kn=1

an

1=

p

1+

1

1−1

1+

p

1+

1

1−1

1+

p

1+· · ·

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Problems 95

(a) Find its 3-periodic tail sequences.

(b) Find an expression for Sn. (Compute the coefficients for n = 3m, n = 3m + 1and n = 3m + 2 separately.)

(c) For which values of p > 0 is the continued fraction

(i) divergent ?

(ii) convergent ?

(iii) generally convergent ? Write down an exceptional sequence.

(d) Give in particular the set of p-values where the continued fraction convergesgenerally but not classically.

19. Periodic continued fraction. We return to the 3-periodic continued fraction

Kn=1

an

1=

2

1+

1

1−1

1+

2

1+

1

1−1

1+

2

1+· · ·from Example 2 on page 59.

(a) Write h9 := −S−19 (∞) as a terminating continued fraction by means of formula

(1.4.6) on page 64.

(b) Show that the canonical denominators Bn of the terminating continued fraction

in (b) satisfy B3n−1 = B3n−1 for n = 0, 1, 2.

(c) Compute Sn(wn) for the following values of wn. In which cases does {Sn(wn)}converge?

(i) wn = 1, (ii) wn = n,

(iii) wn = 2(n+2)/3, (iv) wn = 1/n,

(v) wn = 2−n/3.

20. The reversed periodic continued fraction. Let

a1

b1 +

a2

b2 +

a3

b3 +

a1

b1 +

a2

b2 +

a3

b3 +

a1

b1 +· · ·

be a 3-periodic continued fraction. The reversed (or dual) continued fraction is thenthe 3-periodic continued fraction

b3 +a3

b2 +

a2

b1 +

a1

b3 +

a3

b2 +

a2

b1 +

a1

b3 +

a3

b2 +· · ·.

Prove that the canonical denominators B3n−1 are the same for each n ∈ N for thetwo continued fractions for all n. (Hint: use formula (1.4.6) on page 64.)

21. ♠ The reversed periodic continued fraction. (Galois [Galo28].) Let K(an/bn)be a p-periodic continued fraction for a fixed p ∈ N. Prove that for each n ∈ N,the canonical denominators Bnp−1 are the same for this continued fraction and itsreversed continued fraction

bp +ap

bp−1 +

ap−1

bp−2 +· · ·+a2

b1 +

a1

bp +· · ·.

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96 Chapter 2: Basics

22. ♠ General convergence of periodic continued fraction. Give an exampleother than the one in Example 2 on page 59 of a periodic continued fraction K(an/1)which converges generally but not classically. Why do you need to have a period oflength ≥ 3 ?

23. ♠ From series to continued fraction. (Euler [Euler48].) Prove that the contin-ued fraction

ρ0 +ρ1

1 −ρ2

1 + ρ2 −ρ3

1 + ρ3−· · ·where all ρk �= 0

has canonical numerators and denominators given by

An = ρ0 +n∑

k=1

(k∏

j=1

ρj

), Bn = 1 for n = 0, 1, 2, . . . .

24. Equivalence transformation. Prove that K(an/(−bn)) ∼ −K(an/bn).

25. Equivalence transformation. (Stern [Stern32].) Prove that

b1

b1 +

b2

b2 +

b3

b3 +· · ·∼ b0

b0 +

b0

b1 +

b1

b2 +· · ·when all bn �= 0.

26. ♠ Shift transformation. Show that the canonical contraction of b0 + K(an/bn)with canonical numerators and denominators given by Cn := An+1 and Dn := Bn+1

for n ≥ 0 exists if and only if b1 �= 0, and that then it is given by

b0b1 + a1

b1− a1a2/b1

(b1b2 + a2)/b1 +

a3

b3 +

a4

b4 +· · ·. (∗)

Show further that if {tn}∞n=0 is a tail sequence for b0 + K(an/bn) with t0, t1 �= ∞,then −t0t1, t2, t3. . . . is a tail sequence for (∗).

27. ♠ Shift transformation. Let N > 1 be a fixed integer. Show that the canonicalcontraction of b0 + K(an/bn) with

Cn = An, Dn = Bn for n = 0, 1, 2, . . . , N − 1

andCn = An+1, Dn = Bn+1 for n = N, N + 1, N + 2, . . .

exists if and only if bN+1 �= 0, and show that then it is given by

b0 +a1

b1 +· · ·+aN−1

bN−1 +

bN+1aN

bN+1bN + aN+1

+

−aN+1aN+2/bN+1

(bN+2bN+1 + aN+2)/bN+1 +

aN+3

bN+3 +· · ·.

(Perron [Perr57], p 16.)

Show further that if {tn}∞n=0 is a tail sequence for b0 + K(an/bn) with tN �= ∞ andtN+1 �= ∞, then t0, t1, . . . , tN−1, −tN tN+1, tN+2, tN+3, . . . is a tail sequence for thiscontraction.

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Problems 97

28. ♠ The Wall transformations. Let {fn}∞n=0 be the sequence of classical approxi-mants for a given continued fraction K(an/1). Prove the following statements:

(a) f1, f2, f4, f5, f7, f8, . . . , f3n+1, f3n+2, . . . is the sequence of classical approxi-mants for the continued fraction

a1 −a1a2

1 + a2 +

a3

1 + a4 −a4a5

1 + a5 +

a6

1 + a7 −a7a8

1 + a8 +

a9

1 + a10−· · ·

(b) f0, f2, f3, f5, f6, f8, . . . , f3n, f3n+2, . . . is the sequence of classical approximantsfor the continued fraction

a1

1 + a2 −a2a3

1 + a3 +

a4

1 + a5−a5a6

1 + a6 +

a7

1 + a8−a8a9

1 + a9 +· · ·

(c) f0, f1, f3, f4, f6, f7, . . . , f3n, f3n+1, . . . is the sequence of classical approximantsfor the continued fraction

a1

1 +

a2

1 + a3−a3a4

1 + a4 +

a5

1 + a6−a6a7

1 + a7 +

a8

1 + a9 −· · ·

(d) If two of the continued fractions in (a)–(c) converge, then also the third oneconverges.

(e) K(an/1) converges if and only if the three continued fractions in (a)–(c) con-verge.

(Wall ([Wall57]), where also a number of similar results are presented.)

29. ♠ Extensions. Show that

a1

b1 − a2 +

a2

1 −1

1 + b2 − a3 +

a3

1 −1

1 + b3 − a4 +

a4

1 −· · ·

anda1

b1 − 1−1

1+

a2

1 + b2 − a2−1

1+

a3

1 + b3 − a3−1

1+· · ·are extensions of K(an/bn), (McLaughlin and Wyshinski [McWy07]).

30. ♠ Continued fractions with canonical denominators equal to 1. Let b0 +K(an/bn) have critical tail sequence {hn} (i.e., hn = −S−1

n (∞)) with hn �= ∞ forn ≥ 1. Prove that the continued fraction

b0 +a1/h1

1 +

a2/h1h2

b2/h2 +

a3/h2h3

b3/h3 +· · ·

is equivalent to b0 + K(an/bn) and has canonical denominators Bn = 1 for n ≥ 0.(Euler.)

31. ♠ Extension of continued fraction. (Perron [Perr57], p15.) Let b0 + K(an/bn)

have classical approximants Sn(0) = fn, let N ∈ N with N ≥ 2 and let g ∈ C bechosen such that

ρ := −S−1N (g) =

AN − BNg

AN−1 − BN−1g�= 0,∞.

Page 111: Lisa Lorentzen, Haakon Waadeland Continued Fractions

98 Chapter 2: Basics

Prove that

b0 +a1

b1 +· · ·+aN−1

bN−1 +

aN

bN − ρ+

ρ

1−aN+1/ρ

bN+1 + aN+1/ρ+

aN+2

bN+2 +· · ·

is an extension of b0 + K(an/bn) with classical approximants

f∗n =

⎧⎨⎩fn for n = 0, 1, . . . , N − 1 ,g for n = N ,fn−1 for n = N + 1, N + 2, . . . .

32. ♠ The Khovanskii transform. The Khovanskii transform ([Khov63], p 23) ofK(an/1) is given by

a1

1 + 2a2−a2

1 −a3

1 + 2a3 + 2a4 −a4

1 −a5

1 + 2a5 + 2a6

−· · ·−a2n

1 −a2n+1

1 + 2a2n+1 + 2a2n+2−· · ·.

Prove that if both K(an/1) and its Khovanskii transform converge (in the classicalsense), then they converge to the same value.

33. The Bauer-Muir transform.

(a) Find the Bauer-Muir transform of

z2 − 1 +22

1 +

22

z2 − 1+

42

1 +

42

z2 − 1+

62

1 +

62

z2 − 1+· · ·

with respect to {wn} given by

wn :=

{n + 1z + 1 − 1

2 if n is odd,

n(z + 1) + 12(3 + 2z − z2) if n is even.

(b) Assume that the continued fraction in (a) and its Bauer-Muir transform con-verge to the same value f(z) for z > 1. Find a functional equation for f(z).

34. The Bauer-Muir transform. Assume that the two continued fractions

a1 + h

1 +

a1

b +

a2 + h

1 +

a2

b +· · ·

h +a1

1 +

a1 + h

b +

a2

1 +

a2 + h

b +· · ·

converge. Prove that then they converge to the same value. (Ramanujan [Bern89],p 122), ([Jaco90]).

Page 112: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Chapter 3

Convergence criteria

The convergence theory for continued fractions relies on some basic concepts andideas. The first part of this chapter is devoted to some of these tools.

Next we go on to present some classical convergence theorems. They are classical intwo meanings of the word. For one thing they are old, well-proven results. But theyare also classical in the sense that they aim for classical convergence. The proofs weoffer are not always the classical ones, though. We have chosen to see the theoremsin a more modern setting whenever convenient. In particular we use value sets toprove uniform convergence of {Sn(w)} both with respect to w and with respect toa family of continued fractions. This allows us to produce some newer results aswell.

Our convergence criteria are mainly stated as conditions on the elements {an} and{bn} of K(an/bn). Even if these conditions are met only from some n on, K(an/bn)still converges, since a tail of K(an/bn) converges.

Let F be a family of continued fractions which satisfy a given convergence criterion.In applications it is often vital to have reliable truncation error bounds for continuedfractions from F ; i.e., bounds for the error |f − fn| in the approximation f ≈ fn.We have collected such bounds, valid for a family characterized by a convergencecriterion. They are useful because of their generality and their simplicity. Forbounds valid in more special (i.e., smaller) families, we refer to Chapter 5.

L. Lorentzen and H. Waadeland, Continued Fractions, Atlantis Studies in Mathematics for Engineering and Science 1, DOI 10.1007/978-94-91216-37-4_3, © 2008 Atlantis Press/World Scientific

99

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100 Chapter 3: Convergence criteria

3.1 Tools

3.1.1 The Stern-Stolz Divergence Theorem

Stern ([Stern48]) and Stolz ([Stolz86]) proved independently that a continued frac-tion K(1/bn) diverges in the classical sense if

∑|bn| < ∞. Later on, von Koch

([Koch95a], [Koch95b]) proved that {A2n+m}n and {B2n+m}n converge to finitevalues for which (1.1.1) holds in this case. From their proofs we can even extracta little more information. We say that a sequence {cn} converges absolutely if∑

|cn − cn−1| < ∞. Absolute convergence implies convergence to a finite valuesince

∑nk=1(ck − ck−1) = cn − c0. We also say that a continued fraction converges

absolutely if its sequence of classical approximants converges absolutely.�

Theorem 3.1. (The Stern-Stolz Divergence Theorem.) If∑

|bn| <∞, then the continued fraction K(1/bn) diverges generally, the sequences{A2n+m}n and {B2n+m}n converge absolutely to finite values Am and Bm

respectively (for m = 0, 1), and

A1B0 −A0B1 = 1. (1.1.1)

Proof : We shall use the classical proof of this theorem. It is related to Example19 on page 87. Let

∑|bn| < ∞. Now, {An} and {Bn} are solutions of the recurrence

relationXn = bnXn−1 + Xn−2 for n = 1, 2, 3, . . . . (1.1.2)

For every solution {Xn} of this relation, |Xn| ≤ |bn| · |Xn−1| + |Xn−2| ≤ |bn| ·|Xn−1| + (|bn−1| + 1)|Xn−2|, so if |Xm| ≤ λ(|b1| + 1) · · · (|bm| + 1) for some λ > 0for m := n − 1 and m := n − 2, then

|Xn| ≤ (|bn| + 1)λn−1∏k=1

(|bk| + 1) = λ

n∏k=1

(|bk| + 1).

It follows therefore by induction that

|Xn| ≤ max{|X−1|, |X0|} · (|b1| + 1)(|b2| + 1) · · · (|bn| + 1).

Since∏∞

n=1(1 + |bn|) < ∞ if and only if∑∞

n=1 |bn| < ∞, this means that {An} and{Bn} are bounded under our conditions. Therefore the two series

∑bnAn−1 and∑

bnBn−1 converge absolutely. Since solutions {Xn} of (1.1.2) satisfy Xn−Xn−2 =bnXn−1, this means that

∑|An − An−2| < ∞ and

∑|Bn − Bn−2| < ∞. In other

words, {A2n}, {A2n+1}, {B2n} and {B2n+1} converge absolutely to finite values.

The identity (1.1.1) follows since by the determinant formula on page 7,

A2n−1B2n − A2nB2n−1 = (−1)2n = 1 for all n .

Page 114: Lisa Lorentzen, Haakon Waadeland Continued Fractions

3.1.1 The Stern-Stolz Divergence Theorem 101

Finally, since

limn→∞S2n(w) =

A1w + A0

B1w + B0, lim

n→∞S2n+1(w) =A0w + A1

B0w + B1, (1.1.3)

the sequence {Sn} is totally non-restrained, and the general divergence follows. �

Remarks:

1. Since K(1/bn) diverges generally, it definitely diverges in the classical sense.Indeed, lim S2n(w) and limS2n+1(w) exist for every w ∈ C, but their limitsdepend totally on w. In particular,

limn→∞S2n(0) =

A0

B0and lim

n→∞S2n+1(0) = limn→∞S2n(∞) =

A1

B1

where A0/B0 �= A1/B1 by (1.1.1). (Am/Bm is well defined in C since Bm = 0implies that Am �= 0 by (1.1.1).)

2. If for m = 0 or m = 1 all B2n+m �= 0 and Bm �= 0, then also {S2n+m(0)}n

converges absolutely since by the determinant formula

|Sn(0) − Sn−2(0)| =∣∣∣AnBn−2 − An−2Bn

BnBn−2

∣∣∣ = ∣∣∣ bn

BnBn−2

∣∣∣ .Bm is in particular �= 0 if limn→∞ S2n+m(0) �= ∞ (see (1.1.3)).

3. An equivalence transformation does not change the classical approximants ofa continued fraction K(an/bn). The equivalence transformation in Corollary2.15 on page 77 brings K(an/bn) to the form K(1/dn). Hence, K(an/bn)diverges if

S :=∞∑

n=1

|dn| =∞∑

n=1

∣∣∣bn

n∏k=1

a(−1)n−k+1

k

∣∣∣ < ∞ , (1.1.4)

and its sequence of even and sequence of odd classical approximants both con-verge to distinct values. The series S in (1.1.4) is called the Stern-Stolz Seriesof K(an/bn). It is invariant under equivalence transformations of K(an/bn),and it can also be written

S =∞∑

n=1

∣∣∣b2na1a3 · · · a2n−1

a2a4 · · · a2n

∣∣∣+ ∞∑n=1

∣∣∣b2n+1a2a4 · · · a2n

a1a3 · · · a2n+1

∣∣∣ . (1.1.5)

4. In principle it may occur that K(an/bn) converges generally even if its Stern-Stolz Series (1.1.4) converges and thus K(an/bn) diverges in the classical sense(Theorem 2.16 on page 81). However, by Theorem 2.17 on page 82, thisproblem can not occur if

inf |rn| > 0 and sup |rn| < ∞ for rn := Πnk=1a

(−1)n−k+1

k . (1.1.6)

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102 Chapter 3: Convergence criteria

5. A continued fraction of the form K(an/1) diverges generally if its Stern-StolzSeries converges. This follows since Sn(−1) = Sn−2(0), which means that if{S2n+m(0)}n converges to f , then {S2n+m}n converges generally to f . But{S2n}n and {S2n+1}n converge generally to distinct values in this case.

6. If the limits A := lim An and B := lim Bn both exist and are finite, we say thatK(an/bn) converges separately. Hence, if all bn �= 0 and

∑|bn| < ∞, then the

even and odd parts of K(1/bn) converge separately. (The condition bn �= 0 isneeded for the existence of the even and odd parts.) Indeed, Theorem 3.1 isactually a convergence theorem in this sense.

Condition (1.1.4) can be difficult to check at times. Then the following theorem([Prin99b]) may come in handy:

Theorem 3.2. The Stern-Stolz Series of K(an/bn) has sum ∞ if at leastone of the following three conditions hold:

(i)∞∑

n=2

√|bnbn−1/an| = ∞ , (1.1.7)

(ii) lim infn→∞

∣∣∣∣ an

bnbn−1

∣∣∣∣ < ∞, (1.1.8)

(iii)∞∑

n=2

|bnbn−1|n|an|

= ∞. (1.1.9)

Proof : (i): The series in (1.1.7) and the Stern-Stolz Series are invariant underequivalence transformations, so it suffices to consider continued fractions K(1/bn)and to prove that

∞∑n=2

√|bnbn−1| = ∞ =⇒

∞∑n=1

|bn| = ∞ .

This implication follows easily since√

pq ≤ 12 (p + q) for positive numbers p and q,

and thus ∞∑n=2

√|bnbn−1| ≤

∞∑n=2

12(|bn| + |bn−1|) ≤

∞∑n=1

|bn|.

(ii): This follows immediately since (1.1.8) ⇒ (1.1.7).

(iii): It suffices to prove that (1.1.9) ⇒ (1.1.7), or rather that

∞∑n=2

qn

n= ∞ =⇒

∞∑n=2

√qn = ∞

Page 116: Lisa Lorentzen, Haakon Waadeland Continued Fractions

3.1.2 The Lane-Wall Characterization 103

for an arbitrary sequence {qn} with qn ≥ 0. If lim sup qn > 0, then clearly∑√

qn =∞. Let qn → 0. Then qn ≤ n2 from some n on, and thus qn/n ≤ qn/

√qn =

√qn,

and the implication follows. �

3.1.2 The Lane-Wall Characterization

The Stern-Stolz Divergence Theorem gives sufficient conditions for divergence ofclassical approximants. But how about convergence ? When is divergence of theStern-Stolz Series sufficient for convergence of K(an/bn)? The following theoremby Lane and Wall ([LaWa49]), gives a very useful answer:

Theorem 3.3. (The Lane-Wall Characterization.) Let K(an/bn) withclassical approximants fn = Sn(0) �= ∞ satisfy

∞∑n=1

|fn+1 − fn−1| < ∞. (1.2.1)

Then K(an/bn) converges if and only if

∞∑n=1

∣∣∣∣∣bn

n∏k=1

a(−1)n−k+1

k

∣∣∣∣∣ = ∞. (1.2.2)

Proof : Since the approximants fn and the Stern-Stolz Series (1.2.2) are invariantunder equivalence transformations, we may assume that K(an/bn) has the formK(1/bn). By the Stern-Stolz Divergence Theorem we know that K(1/bn) divergesif∑

|bn| < ∞. Assume that∑

|bn| = ∞. Since {f2n} and {f2n+1} convergeabsolutely, the limit L0 of {f2n} and the limit L1 of {f2n+1} exist and are finite.We want to prove that they coincide.

Assume that L0 �= L1. Then |fn+1 − fn| → |L0 − L1| > 0, and thus

∞∑n=1

|δn| < ∞ for δn := −fn+1 − fn−1

fn+1 − fn

since fn+1 �= fn by Theorem 1.3A on page 9. In particular δn �= ∞,−1. Thedeterminant formula and the recurrence relation (1.1.2) on page 100 for {An} and

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104 Chapter 3: Convergence criteria

{Bn} show that

δn = −

An+1

Bn+1− An−1

Bn−1

An+1

Bn+1− An

Bn

= −An+1Bn−1 − An−1Bn+1

An+1Bn − AnBn+1· Bn

Bn−1

= −bn+1AnBn−1 − An−1Bn

An+1Bn − AnBn+1· Bn

Bn−1= bn+1

Bn

Bn−1

(1.2.3)

where all Bn �= 0 since all fn �= ∞. That is,

δnBn−1 = bn+1Bn for all n ∈ N. (1.2.4)

We shall prove that∑

|δn| < ∞ implies∑

|bn| < ∞, a contradiction, and thusL0 = L1. We first observe that

Bn−1

Bn=

Bn−1

bnBn−1 + Bn−2=

Bn−1

δn−1Bn−2 + Bn−2=

1δn−1 + 1

· 1Bn−2/Bn−1

. (1.2.5)

Therefore it follows from (1.2.4) that

bn+1 = δnBn−1

Bn=

δn

δn−1 + 1· 1Bn−2/Bn−1

=δn(δn−2 + 1)

δn−1 + 1· Bn−3

Bn−2

=δn(δn−2 + 1)

(δn−1 + 1)(δn−3 + 1)· 1Bn−4/Bn−3

=δn(δn−2 + 1)(δn−4 + 1)(δn−1 + 1)(δn−3 + 1)

· Bn−5

Bn−4

and so on. That is,

b2n+1 =δ2n(δ2n−2 + 1)(δ2n−4 + 1) · · · (δ2 + 1)

(δ2n−1 + 1)(δ2n−3 + 1) · · · (δ3 + 1)· B1

B2(1.2.6)

and

b2n+2 =δ2n+1(δ2n−1 + 1)(δ2n−3 + 1) · · · (δ1 + 1)

(δ2n + 1)(δ2n−2 + 1) · · · (δ2 + 1)· B0

B1. (1.2.7)

Since all δn �= −1,∞ and all Bn �= 0,∞, this shows that if∑

|δn| < ∞, then∑|bn| < ∞. �

Remark. If fn �= ∞ for n ≥ m ∈ N only, and∑∞

n=m+1 |fn+1 − fn| < ∞, then theconclusion of Theorem 3.3 still holds. The proof only needs the minor modificationthat the process of using (1.2.5) to produce (1.2.6)-(1.2.7) stops at Bm/Bm+1 andBm+1/Bm+2.

The absolute convergence in (1.2.1) is vital. As Wall proved in [Wall56], there existsa divergent continued fraction for which the even and odd parts converge to distinctvalues and (1.2.2) holds:

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3.1.2 The Lane-Wall Characterization 105

Example 1. Let K(an/1) be given by

an := (−1)n−1n2 for n = 1, 2, 3, . . . .

Then K(an/1) ∼ K(1/bn) where 1/b1 := a1 = 1 and 1/bn−1bn := an for n ≥ 2.Since the geometric mean

√pq of two non-negative real numbers is less that or equal

to their arithmetic mean 12(p + q), it follows that√|bn−1bn| = 1

n≤ 1

2(|bn−1| + |bn|),

and thus ∞∑n=1

|bn| =|b1|2

+∞∑

n=2

12(|bn−1| + |bn|) ≥

12

+∞∑

n=2

1n

= ∞;

i.e., (1.2.2) holds for K(an/1). The even part of K(an/1) is the continued fraction

a1

1 + a2−a2a3

1 + a3 + a4−a4a5

1 + a5 + a6−· · ·−a2n−2a2n−1

1 + a2n−1 + a2n−· · ·∼

∞Kn=1

cn

1

where

c1 :=a1

1 + a2= −1

3, c2 :=

−a2a3

(1 + a2)(1 + a3 + a4)= 2,

cn :=−a2n−2a2n−1

(1 + a2n−3 + a2n−2)(1 + a2n−1 + a2n)=

(n − 1)2(2n − 1)2n − 3

for n ≥ 3. Similarly, the odd part of K(an/1) is

a1 −a1a2

1 + a2 + a3−a3a4

1 + a4 + a5−a5a6

1 + a6 + a7−· · ·∼ a1 +

∞Kn=1

dn

1

whered1 :=

−a1a2

1 + a2 + a3=

23,

dn :=−a2n−1a2n

(1 + a2n−2 + a2n−1)(1 + a2n + a2n+1)=

n2(2n − 1)2n + 1

.

Both K(cn/1) and K(dn/1) converge, as you are asked to prove in Problem 3 onpage 166. Indeed, since all dn > 0, it follows that K(dn/1) converges to a positivenumber. Hence the odd part of K(an/1) converges to a number > a1 = 1. However,K(cn/1) converges to a number f := c1/(1 + f (1)) where f (1) ≥ 0, and thus f < 0.Hence the even and odd parts of K(an/1) converge to distinct values. �

On the other hand, Wall ([Wall56]) has proved the following useful result:

Theorem 3.4. Let the even and odd approximants Sn(0) for K(an/1) con-verge. If lim inf |an| < ∞, then K(an/1) converges.

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106 Chapter 3: Convergence criteria

Proof : Let L0 := lim S2n(0) and L1 := limS2n+1(0), and assume that L0 �= L1.Without loss of generality we assume that L1 �= ∞ and L2 �= ∞. (Otherwisewe consider a continued fraction a/(1 + K(an/1)) (or a/(1 + a/(1 + K(an/1))) ifnecessary) for appropriately chosen a, a �= 0.) Since Sn(−1) = Sn−2(0), the twosequences {S2n+m}n for m = 0, 1 converge generally to Lm, respectively, withexceptional sequences {ζ2n+m}n. Let δn be as defined in the previous proof. Thenδn → 0, and by (1.2.3)

δn := −fn+1 − fn−1

fn+1 − fn=

1an+1

· Bn

Bn−1= − ζn

an+1→ 0.

Since δn → 0 and lim inf |an| < ∞, there therefore exists a subsequence {ζnk}

converging to 0. Without loss of generality we may assume that all nk are eithereven or odd numbers, say nk are even. Then limSnk

(∞) = L0. However, Snk(∞) =

Snk−1(0) which converges to L1 as k → ∞. This is a contradiction. Hence L0 = L1.�

3.1.3 Truncation error bounds

Let K(an/bn) converge generally to some finite value f ∈ C. The approximantsSn(wn) can be used to approximate f . But we need to control the truncation error(f − Sn(wn)). This is done by deriving reliable truncation error bounds λn; i.e.,positive numbers λn such that

|f − Sn(wn)| ≤ λn. (1.3.1)

We have already seen three methods to establish such bounds:

1. In formula (1.6.6) on page 71 we observed that

|f − Sn(wn)| ≤ diam(Kn) for wn ∈ Vn (1.3.2)

whenever K(an/bn) converges generally to f and wn ∈ Vn with diamm(Vn) ≥ε > 0 for all n. Here {Vn} are value sets for K(an/bn) and Kn := Sn(Vn).Therefore, if we find bounds for diam(Kn), then we have quite general boundsfor the absolute error |f − Sn(wn)|. More generally, (1.3.2) implies that

|Sn+m(wn+m) − Sn(wn)| ≤ diam(Kn) for m,n ∈ N (1.3.3)

whenever wk ∈ Vk for k = n, m+n, since then Sn+m(wn+m) ∈ Sn+m(Vn+m) ⊆Sn(Vn) = Kn. This still holds if K(an/bn) diverges.

2. By Corollary 2.7 on page 68

f − fn = t0

( 1Σn

− 1Σ∞

)(1.3.4)

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3.1.3 Truncation error bounds 107

when K(an/bn) converges to f , where

Σn :=n∑

k=0

Pk → Σ∞ ∈ C for Pk :=k∏

j=1

bj + tj−tj

(1.3.5)

for an arbitrary tail sequence {tn} for K(an/bn) with all tn �= ∞. Of course,fn+m − fn = t0(Σ−1

n − Σ−1n+m) holds even if K(an/bn) diverges.

3. From the invariance of the cross ratioSn(u) − Sn(z)Sn(u) − Sn(w)

· Sn(v) − Sn(w)Sn(v) − Sn(z)

=u − z

u − w· v − w

v − z(1.3.6)

with u := 0, z := f(n)k (the kth classical approximant of the nth tail), w := ∞

and v := ζn := S−1n (∞), we obtain

fn − fn+k

fn − fn−1· 1 =

−f(n)k

ζn − f(n)k

when f(n)k �= 0,∞ and ζn �= 0,∞, and thus fn, fn−1 �= ∞. I.e.,

fn+k − fn =−f

(n)k

f(n)k − ζn

(fn − fn−1) . (1.3.7)

Similarly, if K(an/bn) converges to a finite value f and ζn, f (n) �= 0,∞, then

f − fn =−f (n)

f (n) − ζn(fn − fn−1). (1.3.8)

If K(an/bn) converges to f = ∞, it is natural to use

1f− 1

fn= − 1

fnor

1f− 1

Sn(wn)= − 1

Sn(wn)

as a measure for the truncation error. Bounds for this error can be found from thetruncation error of a continued fraction a/(b + K(an/bn)) which then converges to0, regardless of the choice of a �= 0 and b.

Now, a lot of the truncation error bounds derived up through the ages concernclassical approximants. On the other hand we have seen that a clever choice ofapproximants Sn(wn) can give faster convergence, so we really want bounds for|f−Sn(wn)|. Fortunately we have the following method to adjust existing truncationerror bounds to this newer situation:�

Theorem 3.5. Let K(an/bn) converge generally to f �= ∞. If both f (n) �=∞ and ζn �= ∞, then

f − Sn(wn) =f (n) − wn

ζn − wn(f − fn−1). (1.3.9)

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108 Chapter 3: Convergence criteria

Proof : Also this follows from (1.3.6), but this time we use u := f (n), z := wn,w := ∞ and v := ζn if these are distinct points. Now, f (n) �= ζn since f =Sn(f (n)) �= ∞ and Sn(ζn) = ∞. If wn = ζn or wn = f (n), the identity (1.3.9)reduces to ∞ = ∞ or 0 = 0. Finally, if wn = ∞, the identity reduces to f − fn−1 =f − fn−1. �

Equation (1.3.9) also shows that to get fast convergence, a sensible choice for wn isa choice that makes |κ(wn)| small, preferably κ(wn) → 0 for

κ(wn) :=f (n) − wn

ζn − wn. (1.3.10)

That is, we want wn to be close to the tail value f (n) and far away from the criticaltail ζn.

We distinguish between a priori bounds which can be computed in advance, beforewe compute the approximants, and a posteriori bounds which are expressed in termsof the approximants. The bounds (1.3.2) and (1.3.4) are normally a priori bounds,whereas (1.3.8) is a typical a posteriori bound. A posteriori bounds are normallytighter than a priori bounds. A priori bounds on the other hand make it possibleto estimate in advance the order of the approximant and thus the number of digitsneeded in the computation to reach a wanted degree of approximation.

Example 2. Let 0 < g < 12 be given. Then the real interval V := [−g, g] is a simple

value set for every continued fraction K(an/1) from E := [−g(1− g), g(1− g)]. LetK(an/1) be a convergent continued fraction from E. By (1.4.6) on page 64 we knowthat

−ζn = hn = 1 +an

1 +an−1

1 + · · ·+a2

1.

Since 0 ∈ V and ak/(1 + V ) ⊆ V for all k, it follows that hn ∈ 1 + V ; i.e.,ζn ∈ −1−V = [−1− g, −1 + g]. It is also clear that f (n) ∈ V . Therefore |f (n)| ≤ gand |ζn − f (n)| ≥ 1 − 2g, and thus by (1.3.8)

|f − fn| ≤g

1 − 2g|fn − fn−1| for n = 1, 2, 3, . . . .

3.1.4 Mapping with linear fractional transformations

A circle C on the Riemann sphere C is also a circle in the complex plane C if ∞ �∈ C,but if ∞ ∈ C, then C \ {∞} is a straight line in C. Therefore circles and lines onC are also called generalized circles as a common name.

Linear fractional transformations τ ∈ M have beautiful mapping properties. In-deed, they

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3.1.4 Mapping with linear fractional transformations 109

1. map circles on C onto circles on C,

2. map points symmetric with respect to a generalized circle C in C onto pointssymmetric with respect to τ(C). (Two points u and v are symmetric withrespect to the circle with center Γ ∈ C and radius ρ > 0 if (u−Γ)(v−Γ) = ρ2.(Two points u and v are symmetric with respect to the line L if L is theperpendicular bisector of the line segment connecting u and v.)

We shall focus on the family V of closed sets V on C where ∂V (the boundary ofV ) is a circle on C. We distinguish between the cases where

∞ �∈ V : We say that V is a closed (circular) disk, and we write V = B(Γ, ρ) :={w ∈ C; |w − Γ| ≤ ρ} where Γ ∈ C is the center and ρ > 0 is the (euclidean)radius of V .

∞ ∈ ∂V : We say that V is a closed half plane, and we write V = ξ + eiαH whereξ ∈ C, α ∈ R (the set of real numbers), H := {w ∈ C; Re w > 0}, and H is itsclosure in C.

∞ ∈ V ◦ (the interior of V ): We say that V is the complement of (or the exteriorof) an open disk, and we write V = B(Γ,−ρ) := {w ∈ C; |w − Γ| ≥ ρ} withρ > 0, when V is the exterior of B(Γ, ρ)◦.

Let τ ∈ M be given by

τ(w) :=aw + b

cw + dwith Δ := ad − bc �= 0. (1.4.1)

We want to describe τ(V ) for V ∈ V. The easy case is the case where c = 0. Thenτ can be written τ(w) = a

dw + bd , and therefore

c = 0 =⇒{

τ(B(Γ, μ)) = B(adΓ + b

d, |a

d|μ) ,

τ(ξ + eiαH) = τ(ξ) + ei(α+β)H where β := arg ad .

(1.4.2)

Otherwise we have:

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110 Chapter 3: Convergence criteria�

Lemma 3.6. Let τ be given by (1.4.1) with c �= 0, and let V ∈ V. If−d

c�∈ ∂V , then τ(V ) = B(Γ1, μ1) where

Γ1 :=a

c− Δ(cΓ + d)/c

|cΓ + d|2 − |cμ|2and μ1 :=

μ|Δ||cΓ + d|2 − |cμ|2

if V = B(Γ, μ) with Γ ∈ C and μ ∈ R \ {0}(1.4.3)

and

Γ1 :=a

c−

12Δ e−iα/c2

Re[(ξ + dc )e−iα]

and |μ1| :=∣∣∣ 1

2Δ/c2

Re[(ξ + dc )e−iα]

∣∣∣if V = ξ + eiα

H with ξ ∈ C and α ∈ R.

(1.4.4)

Proof: This time τ can be written

τ(w) =a

c− Δ/c2

w + βwhere β :=

d

c. (1.4.5)

Since −β �∈ ∂V ; i.e., 0 �∈ ∂(β + V ), the case where τ(V ) is a half plane is ruled out.

β + Γ

|μ|

v− = β + Γ − β+Γ|β+Γ| |μ|

Re

Im

v+ = β + Γ + β+Γ|β+Γ| |μ|

Let first V := B(Γ, μ). If β+Γ = 0,then the boundary of (β + V ) is acircle with center at the origin andradius |μ|. Therefore 1/∂(β + V ) isa circle with center at the origin andradius 1/|μ|. Therefore Δ

c2/∂(β+V )

is a circle with center at the ori-gin and radius |Δ

c2 /μ|. Therefore V

is mapped onto B(ac ,−|Δc2 |/μ), as

claimed in (1.4.3).

Let β + Γ �= 0. Since

v± := (β + Γ)(1 ± |μ|/|β + Γ|)are the two points on ∂(β + V ) fur-thest away from 0 and closest to 0respectively (see the picture), it fol-lows that the circle 1/(β + ∂V ) hascenter Γ and radius r given by

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3.1.4 Mapping with linear fractional transformations 111

∂(β + V )

v

ξ + β

Re

Im

α

α

β + V

Γ :=12

( 1v−

+1v+

)=

β + Γ|β + Γ|2 − μ2

r :=12

∣∣∣ 1v−

− 1v+

∣∣∣ = ∣∣∣ μ

|β + Γ|2 − μ2

∣∣∣.Since τ(V ) is bounded if and only if0 �∈ β + V , this proves (1.4.3).

Next, let V := ξ+eiαH. Then V +β =ξ + β + eiαH. The point on ∂(V + β)closest to the origin is v := eiαRe[(ξ +β)e−iα]. Therefore 1/∂(V + β) is thecircle through the origin with center12v and radius 1

2|v| . Hence the expres-sion for τ(V ) follows from (1.4.5). �

We conclude this section with some useful observations ([Lore94a]), strongly inspiredby work of Jones and Thron. Here rad V denotes the radius of a disk V .

Lemma 3.7. Let Tn := τ1 ◦ τ2 ◦ · · · ◦ τn for all n ∈ N where all τn ∈ Mmap the unit disk D into itself with

lim supn→∞

rad τn(D) < 1. (1.4.6)

Then {Tn} is restrained with exceptional sequence {T −1n (∞)} from C \ D.

If moreover lim rad Tn(D) > 0, then∑

m(|T −1n (∞)|, 1) < ∞, and {Tn} has

an exceptional sequence from ∂D.

Proof : Let Γn and Rn be the center and radius of the disk Tn(D). The nestednessTn+1(D) = Tn(τn+1(D)) ⊆ Tn(D) shows that {Rn} is a non-increasing sequence ofpositive numbers. Hence Rn → R ≥ 0. If R = 0, the limit point case, then thesequence {Tn} is clearly restrained. Indeed, it converges generally to the limit pointγ ∈ lim Tn(D) ⊆ D.

Assume that R > 0, the limit circle case. A linear fractional transformation mappingD onto D can always be written on the form eiα(w−q)/(1−qw) for some real constantα and complex constant q ∈ D. Therefore Tn can be written

Tn(w) = Γn + Pnw − Qn

1 − Qnw=

(Γn − PnQn) + (Pn − QnΓn)w1 − Qnw

(1.4.7)

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112 Chapter 3: Convergence criteria

where |Pn| = Rn and |Qn| < 1. Similarly, τn can be written

τn(w) = γn + pnw − qn

1 − qnwfor n = 1, 2, 3, . . . ,

where γn and rn := |pn| are the center and radius of τn(D) and |qn| < 1. By (1.4.6)lim sup rn < 1; i.e., rn ≤ r from some n on for some r < 1. Since τn(D) ⊆ D, we have|γn| ≤ 1−rn. From (1.4.3) it follows that the radius Rn+1 of Tn+1(D) = Tn(τn+1(D))is equal to

Rn+1 =|(Γn − PnQn)(−Qn) − (Pn − QnΓn) · 1|

|1 − Qnγn+1|2 − |Qn|2r2n+1

rn+1

=Rn(1 − |Qn|2)rn+1

|1 − Qnγn+1|2 − |Qn|2r2n+1

.

Therefore

Rn+1

Rn≤ (1 − |Qn|2)rn+1

(1 − |Qn|(1 − rn+1))2 − |Qn|2r2n+1

=(1 + |Qn|)rn+1

1 − |Qn| + 2|Qn|rn+1= 1 − (1 − rn+1)(1 − |Qn|)

1 − |Qn| + 2|Qn|rn+1

≤ 1 − (1 − r)(1 − |Qn|)1 − |Qn| + 2r|Qn|

=: 1 − δn

where δn > 0. Since∏∞

n=1(Rn+1/Rn) = R/R1 > 0, it follows that∑

δn < ∞;i.e.,

∑(1 − |Qn|) < ∞. Since T −1

n (∞) = 1/Qn where |Qn| → 1, this proves thatT −1

n (∞) �= ∞ for n ≥ n0 for some n0 ∈ N and that∑∞

n=n0(|T −1

n (∞)| − 1) < ∞,or, equivalently,

∑∞n=1 m(|T −1

n (∞)|, 1) < ∞. From the expression (1.4.7) for Tn itfollows that its derivative is

T ′n(w) = Pn

1 − |Qn|2

(1 − Qnw)2(1.4.8)

where |Pn| = Rn → R < 1 and |Qn| → 1. Hence T ′n(w) → 0 for every w ∈ D, and

{Tn} is restrained.

It remains to prove that {T −1n (∞)} is an exceptional sequence. But this fol-

lows trivially from (1.4.8) since w := 1/Qn = T −1n (∞) gives T ′

n(w) = ∞. Since|T −1

n (∞)| → 1, there also exist exceptional sequences {w†n} with |w†

n| = 1 for all n;i.e., w†

n ∈ ∂D. �

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3.1.4 Mapping with linear fractional transformations 113�

Lemma 3.8. Let Tn and τn be as in Lemma 3.7, and assume there existsa sequence {wn} ⊆ C such that

lim inf∣∣|wn| − 1

∣∣ > 0 and lim inf∣∣|τn(wn)| − 1

∣∣ > 0. (1.4.9)

Then {Tn} converges generally to some constant γ ∈ D with an exceptionalsequence {w†

n} with w†n ∈ C \ D. If the limit point case fails to occur for

{Tn(D)}, then {Tn(w)} converges absolutely to γ for every w ∈ D.

Proof : We use the notation from the proof of Lemma 3.7. If the limit pointcase occurs for Tn(D), then the general convergence is clear. Assume that R > 0.Assume there exists such a sequence {wn} bounded away from the unit circle ∂D,with the property that also {τn(wn)} is bounded away from ∂D, as required in(1.4.9). From (1.4.7)

|Tn(u) − Tn(v)| =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩Rn

(1 − |Qn|2)|u − v||1 − Qnu| · |1 − Qnv|

if u, v �= ∞,

Rn1 − |Qn|2

|Qn| · |1 − Qnv|if u = ∞,

0 if u = v = ∞.

(1.4.10)

Since |Qn| → 1 in this case, there therefore exists a constant A > 0 such that

|Tn+1(wn+1) − Tn(wn)| = |Tn(τn+1(wn+1)) − Tn(wn)| < A(1 − |Qn|)

from some n on, say n ≥ n0. Since∑

(1 − |Qn|) < ∞ under our conditions,this means that

∑∞n=n0

|Tn+1(wn+1) − Tn(wn)| < ∞; i.e., {Tn(wn)}∞n0converges

absolutely to some γ ∈ D.

Let w ∈ D be arbitrarily chosen. Then (1 − Qnw) is bounded away from 0. Hencealso |Tn(wn) − Tn(w)| ≤ A(1 − |Qn|) for some A > 0 from some n on. There-fore {Tn(w)} converges absolutely to γ for every w ∈ D, and thus {Tn} convergesgenerally to γ. �

Remark: The conditions in Lemma 3.8 are in particular satisfied if all τn mapD into itself and |τn(w0)| ≤ r < 1 for all n for some given point w0 �∈ D, sincethen rad(τn(D)) ≤ 1+r

2 for all n. This generalizes a result by Hillam and Thron([HiTh65]) who required that τn(w) = z for all n for some w ∈ C \ D and z ∈ D.

One application of Lemma 3.8 is to derive convergence criteria for continued frac-tions K(an/bn) with value sets {Vn}∞n=0 where Vn ∈ V (i.e., ∂Vn is a circle on C).The idea is to define linear fractional transformations ϕn such that ϕn(D) = Vn.Then

τn := ϕ−1n−1 ◦ sn ◦ ϕn

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114 Chapter 3: Convergence criteria

maps D into D. Moreover, sn = ϕn−1 ◦ τn ◦ ϕ−1, so

Sn := s1 ◦ s2 ◦ · · · ◦ sn = ϕ−10 ◦ τ1 ◦ τ2 ◦ · · · ◦ τn ◦ ϕn = ϕ−1

0 ◦ Tn ◦ ϕn.

Therefore, if {ϕn} is totally non-restrained, then {Sn} converges generally if andonly if {Tn} converges generally. Hence the following result is for instance a conse-quence of Lemma 3.8:

Corollary 3.9. Let {ϕn}∞n=0 be a totally non-restrained sequence of linearfractional transformations, and let {Vn}∞n=0 given by Vn := ϕn(D) be asequence of value sets for the continued fraction K(an/bn). If

lim supn→∞

radϕ−1n−1 ◦ sn ◦ ϕn(D) < 1 (1.4.11)

and there exists a sequence {wn} from C such that

lim infn→∞ distm(wn, ∂Vn) > 0 and lim inf

n→∞ distm(sn(wn), ∂Vn−1) > 0,

(1.4.12)then K(an/bn) converges generally to some value f ∈ V0 with exceptionalsequence {w†

n} with w†n ∈ C \ Vn for all n.

Here distm(w, V ) := inf{m(w, v); v ∈ V } denotes the chordal distance between apoint w ∈ C and a set V ⊆ C.

3.1.5 The Stieltjes-Vitali Theorem

Convergence of a continued fraction is defined as convergence of the sequence {Sn}from M, either evaluated at the origin or in the sense of general convergence.There is a useful theorem on convergence of sequences of holomorphic functions,generally known as Vitali’s Theorem, ([Vita35]). However, Stieltjes presented theidea earlier in his important 1894-paper ([Stie94]). Therefore we follow Jones andThron ([JoTh80], p 89) in naming the theorem:

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3.1.6 A simple estimate 115�

Theorem 3.10. (The Stieltjes-Vitali Theorem.) Let {Fn} be a se-quence of holomorphic functions in a region D, such that

(i) there exist two distinct values p, q ∈ C for which Fn(z) �= p andFn(z) �= q for all n ∈ N and all z ∈ D, and

(ii) {Fn(z)} converges to a finite value for each z in a set D∗ ⊆ D whichhas at least one accumulation point in D.

Then {Fn(z)} converges locally uniformly in D to a holomorphic function.

Here region is meant in the strict sense; i.e., an open, connected set ⊆ C. Moreover,D∗ needs to contain infinitely many points since it shall have an accumulation point.For the proof of this theorem we refer to text books on functions of a complexvariable, for instance ([Hille62], p 248–251).

3.1.6 A simple estimate

In [Lange99b] the following estimate was proved:

Lemma 3.11. For given positive constants a, b with a < b + 1,

n∏k=1

b + k − a

b + k<( b + 1

b + n + 1

)a

for n > 0. (1.6.1)

Proof : For u := (b + k + 1)/(b + k) we consider the integral

I :=∫ u

1

t−a−2(u − t) dt =[ t−a−1

−a − 1u − t−a

−a

]u1

= − 1a+1 (u−a − u) + 1

a (u−a − 1)

=1

a(a + 1)

[( b + k

b + k + 1

)a

− b + k − a

b + k

].

It is clear from the definition of I that I > 0. Hence

b + k − a

b + k<( b + k

b + k + 1

)a

,

and (1.6.1) follows by taking products on both sides. �

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116 Chapter 3: Convergence criteria

3.2 Classical convergence theorems

3.2.1 Positive continued fractions

A continued fraction K(an/bn) with all an > 0 and bn ≥ 0 is called a positivecontinued fraction. (Note that we allow bn = 0.) It has the following useful property:�

Theorem 3.12. Let K(an/bn) be a positive continued fraction. Then

S2(0) ≤ S4(0) ≤ S6(0) ≤ · · · ≤ S5(0) ≤ S3(0) ≤ S1(0) . (2.1.1)

If moreover b1 > 0, then {S2n(0)} and {S2n+1(0)} converge absolutely tofinite, non-negative values. If all bn > 0, then (2.1.1) holds with strictinequalities.

Proof : Let first b1 > 0. Then Bn > 0 for n ≥ 1 by the recurrence relation.Therefore the Euler-Minding series

∞∑n=1

(fn − fn−1) where fn − fn−1 = (−1)n+1

∏nk=1 ak

BnBn−1(2.1.2)

is an alternating series. Since∣∣∣∣ fn − fn−1

fn−1 − fn−2

∣∣∣∣ = anBn−2

Bn=

Bn − bnBn−1

Bn≤ 1, (2.1.3)

we find that {|fn − fn−1|} is non-increasing, and thus (2.1.1) follows since Sn(0) =fn =

∑nm=1(fm − fm−1). The convergence of {f2n} and {f2n+1} is therefore clear.

Indeed, since (f2n+2 − f2n) ≥ 0 and (f2n+1 − f2n−1) ≤ 0 for all n, both {f2n} and{f2n+1} converge absolutely.

Next let b1 = 0. If all bn = 0, then S2n(0) = 0 and S2n+1(0) = ∞ for all n, and(2.1.1) holds trivially. Otherwise, let bn0+1 be the first non-zero bn. Then (2.1.1)holds for the n0th tail K∞

n=n0+1(an/bn) of K(an/bn). Since

K(an/bn) = Sn0(K∞n=n0+1(an/bn)) (2.1.4)

where

Sn0(w) =a1

0 +a2

0 + · · ·+an0

w=

⎧⎪⎨⎪⎩a1a3 · · · an0

a2a4 · · · an0−1w if n0 is odd

a1a3 · · · an0−1a2a4 · · · an0

w if n0 is even

(2.1.1) holds also now. �

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3.2.1 Positive continued fractions 117

The following classical result due to Seidel ([Seid46]) and Stern ([Stern48]) is a directconsequence of Theorem 3.12 combined with the Stern-Stolz Divergence Theoremon page 100 and the Lane-Wall Characterization on page 103:

Theorem 3.13. A positive continued fraction K(1/bn) converges if andonly if

∑bn = ∞. If

∑bn < ∞, then K(1/bn) diverges generally.

Proof : Let first∑

bn = ∞. Without loss of generality we assume that b1 > 0.(Otherwise we consider a tail of K(1/bn).) Then {f2n} and {f2n+1} converge abso-lutely (Theorem 3.12), and thus K(1/bn) converges (the Lane-Wall Characteriza-tion).

Next let∑

bn < ∞. The general divergence follows then from the Stern-StolzDivergence Theorem. �

Seidel and Stern required that all bn > 0. That bn ≥ 0 suffices was proved byBroman ([Brom77]). An equivalence transformation gives the formulation:

Theorem 3.14. (The Seidel-Stern Theorem.) A positive continuedfraction K(an/bn) converges if and only if its Stern-Stolz Series diverges to∞; i.e., if and only if

S :=∞∑

n=1

bn

n∏k=1

a(−1)n+k+1

k = ∞. (2.1.5)

Positive continued fractions and tail sequences.

It is clear that if the positive continued fraction K(an/bn) with all bn > 0 convergesgenerally, then its sequence {f (n)} of tail values is a positive sequence. What is moreinteresting is that a kind of converse result is true. The following was published byKhrushchev ([Khru06b]) under the name of Markov’s Theorem:

�Theorem 3.15. Let {tn} be a positive tail sequence for the generally con-vergent positive continued fraction K(an/bn). Then K(an/bn) converges tot0 and {tn} is its sequence of tail values.

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118 Chapter 3: Convergence criteria

Proof : Since K(an/bn) converges and all tn �= 0,−bn, it follows from Corollary2.7 on page 68 that {Σn} given by

Σn :=n∑

k=0

Pk where Pk :=k∏

m=1

bm + tm−tm

converges to some Σ∞ ∈ C as n → ∞. Since |Pk| ≥ 1 for all n, the possibility thatΣ∞ is finite is ruled out. Hence, by Corollary 2.7 t0 is the value of K(an/bn). �

Truncation error bounds.

Let R+ be the set of positive numbers, and R+ be its closure in C. Both R+ and R+

are simple value sets for the family of positive continued fractions. Indeed, R+ isthe limit set for this family. We can use this to construct truncation error bounds:

Theorem 3.16. If K(an/bn) is a positive continued fraction, then

diamSn(R+) = |fn − fn−1|. (2.1.6)

Proof : Sn(R+) is an interval ⊆ R+ with end points Sn(0) = fn and Sn(∞) =fn−1. �

This means that if K(an/bn) converges, then (fn − fn−1) → 0 and

|f − Sn(wn)| ≤ |fn − fn−1| for wn ∈ R+. (2.1.7)

In particular the choice wn := ∞ leads to

|f − fn−1| ≤ |fn − fn−1| (2.1.8)

which also follows directly from (2.1.1). Indeed, from (2.1.1) we also know that

(f − fn−1) is

{≤ 0 if n odd,

≥ 0 if n even.(2.1.9)

We can even exploit this oscillatory character further to get

|f − f∗n| ≤

12(f2n+1 − f2n) for f∗

n :=12(f2n+1 + f2n). (2.1.10)

These simple a posteriori bounds secure reliable computation of positive continuedfractions. We shall see later (Theorem 3.24 on page 128 and Corollary 3.41 on page148) that the following a priori bounds hold:

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3.2.1 Positive continued fractions 119�

Theorem 3.17. If K(an/bn) is a positive continued fraction, then

diamSn(R+) ≤ 2a1

n∏k=2

√1 + 4ak − 1√1 + 4ak + 1

if all bn = 1, (2.1.11)

and

diamSn(R+) ≤ 1/b1∏nk=2(1 + bkbk−1)

if all an = 1. (2.1.12)

Example 3. In Example 12 on page 26 we used the continued fraction

Ln 2 =11+

1/21 +

1/(2 · 3)1 +

2/(2 · 3)1 +

2/(2 · 5)1 +

3/(2 · 5)1 +

3/(2 · 7)1 +· · ·

to estimate the value of Ln 2. The first seven approximants fn = Sn(0) were givenin a table. The oscillatory character of {Sn(0)} in this table is consistent with(2.1.1). By (2.1.8)-(2.1.9) the approximation Ln(2) ≈ f6 ≈ 0.693121 satisfies

0 < Ln 2 − f6 ≤ (0.693152 − 0.693121) ≈ 3.1 · 10−5,

and by (2.1.11)

|Ln 2 − f6| ≤ 2 ·√

1 + 2 − 1√1 + 2 + 1

·

√1 + 2

3− 1√

1 + 23 + 1

·

√1 + 4

3− 1√

1 + 43 + 1

·

√1 + 6

5− 1√

1 + 65 + 1

(2.1.13)

which is ≈ 2.8 · 10−3. Now,

a2k =k/2

2k − 1=

14

+1/4

2k − 1, a2k+1 =

k/22k + 1

=14− 1/4

2k + 1

for k ≥ 1. Hence an ≤ 14

+ 1/45

= 0.3 for n ≥ 6, and so

|Ln 2 − fn| ≤ 2 ·√

3 − 1√3 + 1

·√

5 −√

3√5 +

√3·√

7 −√

3√7 +

√3×

×√

9 −√

5√9 +

√5·(√

2.2 − 1√2.2 + 1

)n−5 (2.1.14)

for n ≥ 4, which is an easier bound to compute. In the table below we compare theactual value of |Ln 2 − fn| to the three bounds given above.

n fn |Ln 2 − fn| (2.1.8) (2.1.11) (2.1.14)10 .693147157853 . . . 2 · 10−8 2 · 10−7 5 · 10−8 1 · 10−7

11 .693147184962 . . . 4 · 10−9 3 · 10−8 1 · 10−8 2 · 10−8

12 .693147179886 . . . 7 · 10−10 5 · 10−9 2 · 10−9 4 · 10−9

13 .693147180688 . . . 1 · 10−10 8 · 10−10 3 · 10−10 8 · 10−10

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120 Chapter 3: Convergence criteria

The improved approximant f∗3 := 1

2(f7 + f6) gives for instance

Ln 2 ≈ 0.693136 with |Ln 2 − f∗3 | ≤ 1.5 · 10−5.

The correct value is of course Ln 2 = 0.69314718055 . . . . �

If we use approximants Sn(wn), the useful oscillation property (2.1.1) may get lost.It can be saved, though, if we are a little careful:�

Theorem 3.18. Let K(an/bn) be a positive continued fraction with b1 > 0,and let wn ≥ 0 satisfy

wn <an+1

bn+1 + wn+1and wn <

an+1

bn+1 +an+2

bn+2 + wn+2

(2.1.15)

for all n ∈ N. Then

S2(w2) < S4(w4) < S6(w6) < · · · < S5(w5) < S3(w3) < S1(w1). (2.1.16)

Proof : Since K(an/bn) is positive with b1 > 0, it follows from the recurrencerelation for {Bn} that Bn > 0 for n ≥ 0. Hence ζn := S−1

n (∞) = −Bn/Bn−1 < 0.For w > ζn, the derivative of Sn satisfies

S′n(w) =

An−1Bn − Bn−1An

(Bn−1w + Bn)2=

∏nk=1(−ak)

(Bn−1w + Bn)2

{< 0 for n odd> 0 for n even

and so, for fixed n, S2n(w) increases and S2n+1(w) decreases as w > ζn increases.In particular this holds for w ≥ 0. Since

Sn+1(wn+1) = Sn

(an+1

bn+1 + wn+1

),

the first condition in (2.1.15) implies that S2n(w2n) < S2n+1(w2n+1) and S2n+1(w2n+1) >S2n+2(w2n+2). Moreover, since

Sn+2(wn+2) = Sn(vn) where vn :=an+1

bn+1 +an+2

bn+2 + wn+2

,

the second condition in (2.1.15) implies that the sequence {S2n(w2n)} increases and{S2n+1(w2n+1)} decreases. This proves (2.1.16). �

Remark. It is also clear from the sign of S′n(w) that

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3.2.1 Positive continued fractions 121

• if both <” are replaced by ≤” in (2.1.15), then (2.1.16) holds with all <”replaced by ≤”.

• if both <” are replaced by >” (or ≥”) in (2.1.15), then (2.1.16) holds withall <” replaced by >” (or ≥”).

Corollary 3.19. Let K(an/1) be a positive continued fraction with an →a < ∞, and let x be the non-negative solution of the equation x = a/(1+x).If (an − a) > 0 and (an − a) > x(an+1 − an) for all n, then

S2(x) < S4(x) < S6(x) < · · · < S5(x) < S3(x) < S1(x) . (2.1.17)

Proof : Condition (2.1.15) holds since

an+1

1 + x− x =

an+1 − a

1 + x> 0

andan+1

1 + an+21 + x

− x =an+1 − a + x(an+1 − an+2)

1 + x + an+2> 0.

Remark: As in the previous remark, the inequalities in (2.1.17) are for instancereversed if (an − a) < 0 and (an − a) < x(an+1 − an) for all n.

Example 4. The positive continued fraction

∞Kn=1

an

1:=

2 + 11

1 +2 + 1

2

1 +2 + 1

3

1 +· · ·+2 + 1

n

1 +· · ·

has elements 0 < an → 2, and x = 1 is the positive root of the equation x =2/(1+x). Here (an − 2) = 1

n > 0 and (an − 2) = 1n > x(an+1 −an) = 1 · ( 1

n+1 −1n ) .

Hence (2.1.17) holds. Since K(an/1) converges by the Seidel-Stern Theorem (seeRemark 3 on page 101 and Theorem 3.2(ii) on page 102), this means that

|f − Sn(x)| < |Sn+1(x) − Sn(x)| (2.1.18)

and

|f − f∗n| ≤ 1

2(S2n+1(x) − S2n(x)) for f∗n := 1

2(S2n+1(x) + S2n(x)).

“ ““

“ “ ““ “ “

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122 Chapter 3: Convergence criteria

In the table below we have listed the approximants fn := Sn(0) and the truncationerrors (f − fn) and the expression (fn+1 − fn) from the truncation error bound in(2.1.8) for 5 ≤ n ≤ 10. Moreover, (f − Sn(x)) and (Sn+1(x)− Sn(x)) from (2.1.18)are listed. The result is consistent with the Seidel-Stern Theorem, Corollary 3.19and the idea of convergence acceleration from Example 3 on page 11. The value ofK(an/1) is 1.37587055 correctly rounded to 8 decimals.

n fn f − fn fn+1 − fn f − Sn(x) Sn+1(x) − Sn(x)5 1.4599... −0.084... −0.125... −0.0030... −0.0043...6 1.3342... 0.041... 0.063... 0.0013... 0.0019...7 1.3974... −0.021... −0.032... −0.00059... −0.00085...8 1.3649... 0.010... 0.016... 0.00026... 0.00039...9 1.3814... −0.005... −0.008... −0.00012... −0.00017...

10 1.3730... 0.002... 0.004... 0.000056... 0.00008...

3.2.2 Alternating continued fractions

A continued fraction b0 + K(an/bn) with real elements an and bn is called a realcontinued fraction. And if all bn ≥ 0 and an alternates in sign, we say that K(an/bn)is an alternating continued fraction. We include the following result due to Perron([Perr57], p 53):

�Theorem 3.20. (Alternating continued fraction.) Let the even partof the alternating continued fraction K(an/1) with a2 > −1 be a positivecontinued fraction. Then K(an/1) converges if and only if its even partconverges.

The canonical even part of K(an/1) is given by

∞Kn=1

cn

dn=:

a1

1 + a2 +−a2a3

1 + a3 + a4 +−a4a5

1 + a5 + a6 +· · ·. (2.2.1)

Since {an} is alternating in sign, this is a positive continued fraction if and only ifa2n−1 > 0 and a2n < 0 for all n and

a2n−1 + a2n ≥ −1 for n ≥ 2. (2.2.2)

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3.2.2 Alternating continued fractions 123

Proof of Theorem 3.20: Clearly, K(an/1) diverges if its even part diverges, solet its even part converge. That is, f2n = (A2n/B2n) → f where 0 ≤ f < ∞ by theSeidel-Stern Theorem. Now,∣∣∣∣f2n+1 − f2n

f2n+2 − f2n

∣∣∣∣ =∣∣∣∣A2n+1B2n − A2nB2n+1

A2n+2B2n − A2nB2n+2· B2n+2

B2n+1

∣∣∣∣=

∣∣∣∣B2n+2

B2n+1

∣∣∣∣ = ∣∣∣∣ B2n+2

B2n+2 − a2n+2B2n

∣∣∣∣by the recurrence relation for {An} and {Bn}. Since a2n+2 < 0 and all B2n > 0({B2n} are the canonical denominators of the positive continued fraction (2.2.1)with 1 + a2 > 0), this ratio is < 1. Hence also f2n+1 → f . �

Example 5. The alternating continued fraction

11 −

11 +

21 −

31 +

41 −

51 +

61 − · · ·+

2n1 −

2n + 11 + · · · (2.2.3)

has even part∞Kn=1

cn

dn:=

10+

1 · 20 +

3 · 40 +

5 · 60 +

7 · 80 +· · ·

which diverges by the Seidel-Stern Theorem. Hence the alternating continued frac-tion (2.2.3) diverges.

The alternating continued fraction

11−

1/21 +

11−

3/21 +

21−

5/21 +

31−

7/21 +· · ·+

n

1−n + 1/2

1 +· · ·(2.2.4)

has even part1

1/2+

12· 1

1/2 +

32· 2

1/2 +

52· 3

1/2 +

72· 4

1/2 +· · ·which is equivalent to

∞Kn=1

cn

1:=

21+

1 · 21 +

3 · 41 +

5 · 61 +

7 · 81 +· · ·

.

Now, K(cn/1) converges by the Seidel-Stern Theorem and Theorem 3.2(i) on page102 since ∞∑

n=1

1√

cn+1>

∞∑n=1

12n

= ∞.

Hence the convergence of (2.2.4) follows by Theorem 3.20. �

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124 Chapter 3: Convergence criteria

3.2.3 Stieltjes continued fractions

A continued fraction of the form

b0 +∞Kn=1

anz

1= b0 +

a1z

1 +a2z

1 +a3z

1 +· · ·; b0 ≥ 0, an > 0 for all n (2.3.1)

is called a Stieltjes continued fraction, or an S-fraction for short. A large numberof continued fraction expansions of useful functions have this form. S-fractions arealso essential in the Stieltjes moment theory (Section 1.5.2 on page 39). Luckilythey have nice convergence properties ([Stie94]):�

Theorem 3.21. (Stieltjes continued fractions.) The even and oddparts of an S-fraction converge locally uniformly with respect to z in the cutplane D := {z ∈ C; | arg(z)| < π} to holomorphic functions. The S-fractionitself converges for z ∈ D if and only if

∞∑n=1

n∏k=1

a(−1)n−k+1

k = ∞ . (2.3.2)

Proof : Without loss of generality we set b0 := 0 and assume that arg z =: 2α ∈[0, π). (Otherwise we just consider the complex conjugate of K(anz/1).) We shallfirst see that for fixed z, the closed half plane

V (z) := eiαH where H := {w ∈ C; Re w > 0} (2.3.3)

is a simple value set for K(anz/1). Let w ∈ V (z). Then −π2 + α ≤ arg w ≤ π

2 + α,so −π

2 + α < arg(1 + w) < π2 + α, and thus

2α − (π2 + α) < arg

anz

1 + w< 2α + (π

2 − α),

i.e., anz/(1 + V (z)) ⊆ V (z). This means that the classical approximants f(1)n (z) of

its first tail satisfy

f (1)n (z) :=

a2z

1 +a3z

1 +· · ·+an+1z

1∈ V (z).

In particular f(1)n (z) �= −1. Therefore fn(z) := Sn(0) = a1z/(1 + f

(1)n−1(z)) �= ∞

for all n. That is, {fn(z)} is a sequence of rational, holomorphic functions in D.Moreover, fn(z) never takes a negative value for n ∈ N and z ∈ D.

For z > 0 the result follows from the Seidel-Stern Theorem. Therefore {f2n} and{f2n+1} converge locally uniformly in D to holomorphic functions by the Stieltjes-Vitali Theorem on page 115, and {fn} converges locally uniformly in D if and onlyif (2.3.2) holds. �

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3.2.3 Stieltjes continued fractions 125

Remarks.

1. A very nice consequence of Theorem 3.21 is that an S-fraction converges to aholomorphic function in D if and only if it converges at a single point z ∈ D.

2. If an S-fraction diverges for a z ∈ D, then it diverges generally for all z ∈ D.(See Remark 5 on page 102.)

Limit set.

Simple computation shows that for given z ∈ D with arg z =: 2α ∈ (−π, π), theangular opening (or ray if α = 0)

Vα :=

{(−i e2iα

H) ∩ (iH) if α ≥ 0,

(i e2iαH) ∩ (−iH) if α ≤ 0(2.3.4)

between the rays R+ and e2iαR+ is also a value set for every S-fraction K(anz/1). If

α = 0, then K(anz/1) is a positive continued fraction which was studied in Section3.2.1. Let α �= 0. Our next theorem shows that then every point in the interior ofVα, and no other point, is the value of an S-fraction with b0 = 0 and arg z = 2α:

Theorem 3.22. For given α ∈ R with 0 < |α| < π

2 , V ◦α is the limit set for

the family of continued fractions K(an/1) from the element set E := e2iαR+.

Proof : Let K(an/1) be a convergent continued fraction from E with value f .Since Vα is a closed, simple value set for K(an/1), we know that f ∈ Vα (Corollary2.10 on page 71). Since also the first tail value f (1) for K(an/1) is the value of acontinued fraction from E, also f (1) ∈ Vα. Since f = a1/(1 + f (1)) and −1 �∈ Vα,it follows that f �= ∞. Similarly, f (1) �= ∞, and thus f �= 0. Clearly f > 0 only ifarg(1+ f (1)) = 2α which is impossible. Therefore f �∈ R+, and similarly f (1) �∈ R+.Finally, arg f = 2α only if (1 + f (1)) > 0 which already is ruled out. This provesthat f ∈ V ◦

α .

Next, let f := r eiθ ∈ V ◦α be arbitrarily chosen. We need to prove that f is the

value of a continued fraction from E. We shall actually prove that f is the value ofa 2-periodic continued fraction

r1e2iα

1 +r2e

2iα

1 +r1e

2iα

1 +r2e

2iα

1 +r1e

2iα

1 +· · ·(2.3.5)

for some r1 > 0 and r2 > 0. Such a continued fraction clearly converges to a valuein V ◦

α by the arguments above. A necessary condition for f to be its value is that

f = r eiθ =r1e

2iα

1 +r2e

2iα

1 + r eiθ= r1e

2iα 1 + r eiθ

1 + r2e2iα + r eiθ. (2.3.6)

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126 Chapter 3: Convergence criteria

Without loss of generality (complex conjugation) we assume that α > 0. Then0 < θ < 2α, and 1 + r eiθ = ρ eiψ for some ρ > 0 and 0 < ψ < θ. Hence we need

r eiθ =r1ρ ei(2α+ψ)

ρ eiψ + r2e2iα. (2.3.7)

Let r2 > 0 be chosen such that arg(ρ eiψ +r2e2iα) = 2α+ψ−θ. This can always be

done since ψ < 2α + ψ − θ < 2α. Then (2.3.7) holds for the right choice of r1 > 0.With this choice, (2.3.5) converges to reiθ. �

Truncation error bounds.

Henrici and Pfluger ([sical approximants fn(z) := Sn(0) of Stieltjes fractions. These bounds are basedthe value set Vα given by (2.3.4). What they actually proved though, is in fact

bound for the diameter of Sn(Vα):

Theorem 3.23. (The Henrici-Pfluger Bounds.) For given z �= 0 withα := 1

2 arg z ∈ (−π2 , π

2 ), let Vα be given by (2.3.4). Then

diamSn(Vα) ≤

⎧⎪⎪⎨⎪⎪⎩|fn(z) − fn−1(z)| if |α| ≤ π

4,

|fn(z) − fn−1(z)|| sin 2α| if |α| > π

4

(2.3.8)

for the S-fraction K(anz/1).

Proof : Again we may without loss of generality assume that b0 = 0. If α = 0,then the bound follows from Theorem 3.16 on page 118. Let

fn

fn−1

Sn(R)

Sn(Re2iα)

α �= 0. Without loss of generality (complex con-jugation) we assume that α > 0. The set Kn :=Sn(Vα) is bounded by the circular arcs Sn(R+) andSn(e2iα

R+). These two arcs meet at Sn(0) = fn

= fn(z) and Sn(∞) = fn−1 = fn−1(z) at the an-gle 2α since Sn is a conformal map. Indeed, 2α isthe inner angle since ζn �∈ Vα because Sn(ζn) = ∞whereas Sn(Vα) ⊆ s1(Vα) which is a bounded set.Therefore Kn is a closed, bounded lens-shaped re-gion, for instance as illustrated in the present fig-ure. Now, if Kn is not too fat”, then diam(Kn) =|fn − fn−1|. Otherwise one measures the diameteralong the perpendicular bisector between fn−1 and

HePf66]) have proved a posteriori truncation error bounds forclasona

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3.2.3 Stieltjes continued fractions 127

fn. The worst case” occurs if Kn is convex and the two circles Sn(R) and Sn(Re2iα)have the same radius. Therefore

fn

fn−1

A C B

Rn

θ/2

θ/2

π − θ

α

diam(Kn) is less than or equal to the di-ameter 2Rn of two equally large circleswhich intersect at fn and fn−1 under anangle 2α. The figure illustrates the sit-uation. Here C is the center of the leftcircle, and B is the midpoint betweenthe two centers.

Let θ2 denote the angle ∠CAfn. Then

also ∠AfnC = θ2 and the angle ∠AfnB

is equal to π2 −

θ2 which again is less than

α; i.e., π − θ < 2α. Therefore, sinceπ2

< π−θ < 2α < π, the common radiusRn of these two circles is given by

Rn = |fnC| =|fnB|sin θ

=|fn − fn−1|

2 sin θ

=|fn − fn−1|2 sin(π − θ)

<|fn − fn−1|2| sin 2α| .

Of course, this means that if wn ∈ Vα for all n, then

|Sn+m(wn+m) − Sn(wn)| ≤ λ|fn(z) − fn−1(z)| (2.3.9)

where λ := 1 if |α| ≤ π4 and λ := 1/| sin 2α| otherwise. If in particular K(anz/1)

converges to f(z), then

|f(z) − Sn(wn)| ≤ λ|fn(z) − fn−1(z)|, (2.3.10)

and the choice wn := ∞ gives

|f(z) − fn−1(z)| ≤ λ|fn(z) − fn−1(z)|. (2.3.11)

The diameter of Sn(V(z)).

The Henrici-Pfluger Bounds are a posteriori bounds based on the value set Vα in(2.3.4). To prove a priori bounds we follow Thron ([Thron81]) and Gragg andWarner ([GrWa83]), and use the half plane V := V (z) = eiα

H which also is a valueset for K(anz/1) for given z, even though it is normally quite a lot larger.

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128 Chapter 3: Convergence criteria�

Theorem 3.24. (The Thron-Gragg-Warner Bounds.) For given z :=rei2α with −π

2< α < π

2and r > 0,

diam(Sn(V )) ≤ 2a1r

cos α

n∏k=2

√1 + 4akr/ cos2 α − 1√1 + 4akr/ cos2 α + 1

(2.3.12)

for the S-fraction K(anz/1) and V := eiαH.

Since ∞ ∈ V , it follows in particular that

|fn+m − fn−1| ≤ 2a1r

cos α

n∏k=2

√1 + 4akr/ cos2 α − 1√1 + 4akr/ cos2 α + 1

(2.3.13)

for m,n ∈ N, which essentially was the original result by Thron and Gragg-Warner.If K(anz/1) converges to f(z), then |f(z)− fn−1(z)| and |f(z)− Sn(w)| for w ∈ Vare bounded by the same expression.

We postpone the proof of Theorem 3.24 to page 158.

Example 6. Let us once again turn to the continued fraction expansion

K anz

1=

z

1+z/21 +

z/61 +

2z/61 +

2z/101 +

3z/101 +· · ·

of Ln(1+z). This is an S-fraction, and it converges for | arg(z)| < π (Theorem 3.21).Its classical approximants fn(z) are rational functions in z with rational coefficients.In Example 3 on page 119 we used this continued fraction to approximate Ln(2).The bounds we obtained for |Ln 2−fn| are consistent with (2.3.8) and (2.3.12). Letnow z := 1 + i =

√2 eiπ/4. Then still

|Ln(1 + i) − fn(1 + i)| ≤ |fn(1 + i) − fn+1(1 + i)| (2.3.14)

by (2.3.8), and by (2.3.13) we get the a priori bounds

|Ln(1 + i) − fn(1 + i)| ≤ 2√

2cos π

8

n+1∏k=2

√1 + 4ak

√2/ cos2 π

8− 1√

1 + 4ak

√2/ cos2 π

8+ 1

≤ 2√

2cos π

8

·√

1 + 2w − 1√1 + 2w + 1

·

√1 + 2

3w − 1√

1 + 23w + 1

·

√1 + 4

3w − 1√

1 + 43w + 1

×

×

√1 + 4

5w − 1√1 + 4

5w + 1

√1 + 6

5w − 1√1 + 6

5w + 1

)n−5

(2.3.15)

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3.2.4 The Sleszynski-Pringsheim Theorem 129

where w := |1+i|/ cos2 π8 = 4(

√2−1). The true value of Ln(1+i) is 0.804718956217+

0.463647609001 i, correctly rounded to 12 decimals. The table below shows the or-der of the true truncation errors compared to the one in (2.3.14) and the two in(2.3.15):

n |Ln(1 + i) − fn| |fn+1 − fn| (2.3.15)1 (2.3.15)210 5.7 · 10−7 7.1 · 10−7 1.1 · 10−5 1.7 · 10−5

11 1.5 · 10−7 1.8 · 10−7 2.4 · 10−6 4.5 · 10−6

12 3.1 · 10−8 3.7 · 10−8 6.0 · 10−7 1.2 · 10−6

13 7.8 · 10−9 9.3 · 10−9 1.4 · 10−7 3.2 · 10−7

3.2.4 The Sleszynski-Pringsheim Theorem

A continued fraction K(an/bn) with

|bn| ≥ |an| + 1 for n ≥ 1 (2.4.1)

is called a Sleszynski-Pringsheim continued fraction. The unit disk D := {z ∈C; |z| < 1} is a simple value set for K(an/bn) if and only if (2.4.1) holds. Now,0 ∈ D, so fn = Sn(0) ∈ D for all n. In 1889 Sleszynski ([Sles89]) proved that(2.4.1) was sufficient for the (classical) convergence of K(an/bn) (although he neverused the term value set”). This result was rediscovered by Pringsheim in 1899([Prin99a]). Today it is a simple consequence of Corollary 3.9 on page 114 withϕn(w) ≡ w and wn := ∞. But the classical proof gives some additional informationon the convergence:�

�Theorem 3.25. (The Sleszynski-Pringsheim Theorem.) ASleszynski-Pringsheim continued fraction converges absolutely to some valuef with 0 < |f | ≤ 1. Moreover, {|An|} and {|Bn|} are strictly increasing se-quences, and 0 < |Sn(w)| ≤ 1 for all w ∈ D.

Proof : Solutions {Xn} of the recurrence relation Xn = bnXn−1+anXn−2 satisfy

|Xn| ≥ |bn||Xn−1| − |an||Xn−2| ≥ |bn|Xn−1| − (|bn| − 1)|Xn−2| , (2.4.2)

and hence, if |Xm| − |Xm−1| > 0, then

|Xn| − |Xn−1| ≥ (|bn| − 1)(|Xn−1| − |Xn−2|)

≥ · · · ≥ (|Xm| − |Xm−1|)n∏

k=m+1

(|bk| − 1) > 0(2.4.3)

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130 Chapter 3: Convergence criteria

for n ≥ m. Now, {An} and {Bn} are such solutions with |A1| − |A0| = |a1| > 0and |B0| − |B−1| = 1 > 0. Therefore {|An|}∞0 and {|Bn|}∞0 are strictly increasingsequences. Moreover,

|Bn| − |Bn−1| ≥n∏

k=1

(|bk| − 1) ≥n∏

k=1

|ak| (2.4.4)

by (2.4.3). The determinant formula on page 7 therefore gives that∣∣∣∣An

Bn− An−1

Bn−1

∣∣∣∣ = ∏nk=1 |ak|

|BnBn−1|≤ |Bn| − |Bn−1|

|BnBn−1|=

1|Bn−1|

− 1|Bn|

(2.4.5)

where∑∞

n=1(1/|Bn−1|−1/|Bn|) = 1−1/ lim |Bn|. Hence the absolute convergence offn = Sn(0) = An/Bn is established. That |Sn(w)| ≤ 1 for w ∈ D and thus |f | ≤ 1follows from the nestedness of {Sn(D)}. This also holds true for the first tail ofK(an/bn); i.e., |S(1)

n−1(w)| ≤ 1 and |f (1)| ≤ 1. Since |a1/(b1+q)| ≥ |a1|/(|b1|+1) > 0for |q| ≤ 1, it therefore follows that |Sn(w)| > 0 and |f | > 0. �

Example 7. We consider the continued fraction

b0 +∞Kn=1

an

bn:= 1 +

2z

2 − z+z2

6 +z2

10+· · ·+z2

4n − 2+· · ·. (2.4.6)

For given z ∈ C there always exists an n0 ∈ N such that |bn| ≥ |an| + 1 for n ≥ n0.Therefore the n0th tail of b0 +K(an/bn) converges, and thus the continued fractionitself converges. Hence (2.4.6) converges for every z ∈ C. Indeed, its value is ez

((2.2.1) on page 268). �

Remark: If K(an/bn) is equivalent to a Sleszynski-Pringsheim continued fraction,then also K(an/bn) converges absolutely, and its classical approximants are stillcontained in the open unit disk D. Therefore K(an/bn) converges if

|b1| ≥ |a1| +√|a2|, |bn| ≥

√|an| +

√|an+1| for n ≥ 2, (2.4.7)

since then∞Kn=1

an

bn∼ a1/

√|a2|

b1/√|a2| +

a2/√|a2a3|

b2/√

|a3| +a3/√

|a3a4|b3/√|a4| + · · ·

which is a Sleszynski-Pringsheim continued fraction. Similarly, if

|b1| ≥ |a1| + |a2|, |bn| ≥ |an+1| + 1 for n ≥ 2, (2.4.8)

then K(an/bn) converges since∞Kn=1

an

bn∼ a1/a2

b1/a2 +1/a3

b2/a3 +1/a4

b3/a4 + · · ·

Page 144: Lisa Lorentzen, Haakon Waadeland Continued Fractions

3.2.4 The Sleszynski-Pringsheim Theorem 131

is a Sleszynski-Pringsheim continued fraction under condition (2.4.8), ([Prin05],[Prin18]). These are just two examples of infinitely many possibilities.

Limit set.

Every f ∈ D\{0} is actually the value of a Sleszynski-Pringsheim continued fraction:

Theorem 3.26. D \ {0} is the limit set for the family of Sleszynski-Pringsheim continued fractions.

Proof : It follows from Theorem 3.25 that the value of a Sleszynski-Pringsheimcontinued fraction is contained in D \ {0}. We need to prove that every f ∈ D \ {0}is the value of a Sleszynski-Pringsheim continued fraction. For a fixed b > 2 leta := 1 − b. Then the periodic continued fraction

∞Kn=1

a

b=

a

b +a

b +a

b +· · ·

is a Sleszynski-Pringsheim continued fraction which converges to a root of the equa-tion a/(b + x) = x. Since this value must be ∈ D, it follows that K(a/b) convergesto −1. Let f ∈ D\{0,−1} and b1 > 1 be arbitrarily chosen, and let a1 := f ·(b1−1).Then |a1| + 1 ≤ b1 − 1 + 1 = |b1|, and

a1

b1 +a

b +a

b +a

b +· · ·converges to a1/(b1 − 1) = f . �

Truncation error bounds.

In [BeLo01] we studied the speed of convergence of Sleszynski-Pringsheim continuedfractions. From (2.4.4) it follows immediately that

|Bn| ≥ Σn :=n∑

k=0

Pk where Pk :=k∏

m=1

(|bm| − 1). (2.4.9)

Therefore, by (2.4.5)

|f − fn| ≤∞∑

k=n+1

|fk − fk−1| ≤∞∑

k=n+1

∏kj=1 |aj |

|BkBk−1|≤

∞∑k=n+1

∏kj=1 |aj |

ΣkΣk−1, (2.4.10)

and since |aj | ≤ |bj | − 1 and Pk = Σk − Σk−1,

|f − fn| ≤∞∑

k=n+1

Σk − Σk−1

ΣkΣk−1=

∞∑k=n+1

( 1Σk−1

− 1Σk

)=

1Σn

− 1Σ∞

. (2.4.11)

Page 145: Lisa Lorentzen, Haakon Waadeland Continued Fractions

132 Chapter 3: Convergence criteria

Alternatively, (2.4.4) and (2.4.9) lead to

|f − fn| ≤1

Σ∗n

− 1Σ∗∞

where Σ∗m :=

m∑k=0

P ∗k and P ∗

k :=k∏

j=1

|aj | . (2.4.12)

But we can say more:

Case 1: an < 0, bn = 1 − an for all n.

In this case the constant sequence {−1} is a tail sequence for K(an/bn), so we havefull control. From Theorem 2.6 on page 66 with tn := −1 we immediately find that

Bn − Bn−1 = Pn, An + Bn = 1 and 1 + fn = 1/Σn

where Pk and Σn are given by (2.4.9). In particular

Bn = Σn and f − fn =1

Σ∞− 1

Σn. (2.4.13)

Case 2: an < 0, bn ≥ 1 − an for all n.

It follows by induction that all Bn > 0 since B0 = B0 − B−1 = 1 and

Bn − Bn−1 = (bn − 1)Bn−1 − |an|Bn−2 ≥ (bn − 1)(Bn−1 − Bn−2)

if Bn−2 ≥ 0. Therefore Bn+1 > Bn ≥ Σn and

fn − fn−1 = −∏n

k=1(−ak)BnBn−1

< 0, (2.4.14)

so {fn} is a decreasing sequence. Since f1 = a1/b1 < 0, this means that

−1 < fn < fn−1 < 0 and Bn+1 > Bn ≥ Σn. (2.4.15)

Case 3: The general case |an| + 1 ≤ |bn| for all n.

Let K(an/bn) and K(an/bn) be the two continued fractions given by

an := an := −|an|, bn := |bn| and bn := |an| + 1. (2.4.16)

Then K(an/bn) is as described in case 1 and K(an/bn) as described in case 2. Weshall prove (with obvious notation):�

Theorem 3.27. Let K(an/bn) be a Sleszynski-Pringsheim continued frac-tion and let K(an/bn) and K(an/bn) be given by (2.4.16). Then, with Σn

given by (2.4.9),

|f − fn| ≤ fn − f ≤ fn − f =1

Σn− 1

Σ∞. (2.4.17)

Page 146: Lisa Lorentzen, Haakon Waadeland Continued Fractions

3.2.4 The Sleszynski-Pringsheim Theorem 133

Proof : The equality follows from (2.4.13) and the second inequality from (2.4.11).It therefore just remains to prove the first inequality. By (2.4.14),

fn − f = −n∑

k=1

( k∏m=1

|am|)/(

BkBk−1

)with obvious notation. The first inequality in (2.4.17) follows therefore from (2.4.10)if we can prove that |Bn| ≥ Bn, and thus

|Bn| ≥ Bn ≥ Bn = Σn for n ≥ 1. (2.4.18)

Let Xn := |Bn| − Bn. Then X−1 = X0 = 0 and

Xn ≥ |bnBn−1| − |anBn−2| − Bn = |bn|Xn−1 − |an|Xn−2,

so if Xn−2 ≥ 0, then

Xn − Xn−1 ≥ (|bn| − 1)(Xn−1 − Xn−2).

It follows therefore by induction that Xn−Xn−1 ≥ 0 and Xn ≥ 0 for all n. Therefore(2.4.18) and (2.4.17) follow. �

Remark. The equality (fn − f) = (1/Σn −1/Σ∞) shows that there are Sleszynski-Pringsheim continued fractions which converge arbitrarily slowly. (Just choose tn :=−1 and bn > 1 so that Σn =

∑nk=0

∏kj=1(bj − 1) converges as slowly as you want,

and set an := −(bn − 1) for all n.) Hence the convergence is not uniform withrespect to the family of Sleszynski-Pringsheim continued fractions.

The radius of Sn(D).

The quantity Σn is still given by (2.4.9).�

Theorem 3.28. Let K(an/bn) be a Sleszynski-Pringsheim continued frac-tion. Then

radSn(D) =∏n

k=1 |ak|(|Bn| − |Bn−1|)(|Bn| + |Bn−1|)

≤∏n

k=1 |ak|/(|bk| − 1)|Bn| + |Bn−1|

≤ 1|Bn| + |Bn−1|

≤ 1Σn + Σn−1

.

(2.4.19)Let further K(an/bn) and K(an/bn) be given by (2.4.16). Then

rad Sn(D) ≤ rad Sn(D) ≤ rad Sn(D) =1

Σn + Σn−1, (2.4.20)

Page 147: Lisa Lorentzen, Haakon Waadeland Continued Fractions

134 Chapter 3: Convergence criteria

Proof : The equality in (2.4.19) follows from (1.4.3) on page 110. The first in-equality follows from (2.4.4). The second inequality is a consequence of (2.4.1), andthe last inequality follows from (2.4.18). The bounds (2.4.20) is a consequence of(2.4.18) and (2.4.19) if we can prove that

|Bn| − |Bn−1| ≥ |Bn| − |Bn−1| ≥ |Bn| − |Bn−1| = Σn − Σn−1.

Of course, |Bn| = Bn = Σn, so the equality is clear. Moreover,

|Bn| − |Bn−1| = Bn − Bn−1 = (|bn| − 1)Bn−1 − |an|Bn−2

≥ (|bn| − 1)(Bn−1 − Bn−2) ≥n∏

k=1

(|bk| − 1) = Bn − Bn−1,

and finallyQn := (|Bn| − |Bn−1|) − (Bn − Bn−1) ≥ 0

sinceQn ≥ (|bn| − 1)(|Bn−1| − Bn−1) − |an|(|Bn−2| − Bn−2)

≥ (|bn| − 1)Qn−1 ≥ · · · ≥ Q1

n∏k=2

(|bk| − 1) ≥ 0.

Remark. There exist Sleszynski-Pringsheim continued fractions for which the limitpoint case for Sn(D) fails to occur. For instance, for K(an/bn) the radius Rn

converges to a positive value if Σ∞ < ∞.

Example 8. For given q ∈ C with |q| = 1, the continued fraction∞Kn=1

z

2qn=

z

2q +z

2q2 +z

2q3 + · · ·

is a Sleszynski-Pringsheim continued fraction for |z| ≤ 1. So let |z| ≤ 1. ThenK(z/2qn) converges to some value f(z) with 0 < |f(z)| ≤ 1. Moreover Pk and Σn

given by (2.4.9) have values Pk = 1 and Σn = n + 1. Therefore, by (2.4.10) and(2.4.18)

|f(z) − fn(z)| ≤∞∑

k=n+1

|z|kΣkΣk−1

≤∞∑

k=n+1

|z|k(k + 1)k

≤∞∑

k=n+1

1(k + 1)k

=∞∑

k=n+1

(1k− 1

k + 1

)=

1n + 1

.

For |z| < 1 we also have the bound

|f(z) − fn(z)| <∞∑

k=n+1

|z|k(n + 2)(n + 1)

=|z|n+1/(1 − |z|)(n + 2)(n + 1)

.

Page 148: Lisa Lorentzen, Haakon Waadeland Continued Fractions

3.2.5 Worpitzky’s Theorem 135

We can also majorize the expression by a telescoping series to get

|f(z) − fn(z)| ≤∞∑

k=n+1

( |z|kk

− |z|kk + 1

)<

∞∑k=n+1

( |z|kk

− |z|k+1

k + 1

)=

|z|n+1

n + 1.

3.2.5 Worpitzky’s Theorem

Worpitzky was a high school teacher who published deep results in the yearly reportfrom his school. His convergence result from 1865 ([Worp65]) can be stated asfollows:

Theorem 3.29. (Worpitzky’s Theorem.) Let

0 < |an| ≤ 14 for all n ≥ 1 . (2.5.1)

Then K(an/1) converges absolutely to some value f with 0 < |f | ≤ 12 , and

0 < |Sn(w)| ≤ 12

for all n ∈ N and |w| ≤ 12.

This was quite remarkable in 1865. Today we observe that the result follows directlyfrom the Sleszynski-Pringsheim Theorem on page 129 since 2K(an/1) is equivalentto the Sleszynski-Pringsheim continued fraction

4a1

2 +4a2

2 +4a3

2 +· · ·+4an

2 +· · ·. (2.5.2)

This equivalence also means that V := 12D is a simple value set for the continued

fractions K(an/1) from E := 14D, and the truncation error bounds for Sleszynski-

Pringsheim continued fractions can easily be adjusted to K(an/1) from E. We stillchose to include Worpitzky’s Theorem since it is so easy to apply and so widelyknown and used. The following version of Worpitzky’s Theorem is more general. Itcan be derived from the work of Pringsheim ([Prin05]) and Wall ([Wall48], p 45-50):

Page 149: Lisa Lorentzen, Haakon Waadeland Continued Fractions

136 Chapter 3: Convergence criteria�

Theorem 3.30. For a given sequence {gn}∞n=0 of positive numbers gn < 1,let {En}∞n=1 be given by

En := gn−1(1 − gn)D = {a ∈ C; |a| ≤ gn−1(1 − gn)}, (2.5.3)

and let K(an/1) be a continued fraction from {En}. Then K(an/1) con-verges to some value f with 0 < |f | ≤ g0, and {gnD}∞n=0 is a sequence ofvalue sets for K(an/1).

This follows since g−10 K(an/1) is equivalent to the Sleszynski-Pringsheim continued

fraction

K cn

dn:=

a1/g0g1

1/g1 +a2/g1g2

1/g2 +a3/g2g3

1/g3 +· · ·+an/gn−1gn

1/gn +· · ·. (2.5.4)

The choice gn := 12 for all n gives back the original Worpitzy Theorem.

The sequence {−gn}.The sequence {−gn}∞n=0 is a tail sequence for the continued fraction K(an/1) givenby

an := −gn−1(1 − gn) for all n. (2.5.5)

This continued fraction is special since

• it defines En as {a ∈ C; |a| ≤ |an|} = |an|D,

• it is equivalent to g0K(cn/dn) where cn := −(1 − gn)/gn and dn := 1/gn, soK(cn/dn) is a Sleszynski-Pringsheim continued fraction as described in case1 on page 132.

It follows from Theorem 2.6 on page 66 (and also from Property 1 on page 78) thatK(an/1) has approximants fn and denominators Bn given by

Bn = Σn, fn =g0

Σn

− g0, f =g0

Σ∞− g0, f − fn =

g0

Σ∞− g0

Σn

where Σn :=Σn∏nk=1 gk

=n∑

k=0

Pk with Pk :=k∏

j=1

1 − gj

gj.

(2.5.6)

If Σ∞ = ∞, then {−gn} is the sequence of tail values for K(an/1) (Corollary 2.7on page 68). Otherwise K(an/1) still converges, but it has another sequence {f (n)}of tail values. Also f (n) ∈ Vn according to Theorem 3.30, and f (n) �= −gn. That is,−gn < f (n), and thus also

f (n−1) =an

1 + f (n)= −gn−1(1 − gn)

1 + f (n)< −gn−1(1 − gn) < 0.

Page 150: Lisa Lorentzen, Haakon Waadeland Continued Fractions

3.2.5 Worpitzky’s Theorem 137

Therefore we can replace {gn} by {−f (n)} without changing {En}:

Theorem 3.31. For given sequence {gn} of positive numbers < 1, let

Σ∞ :=∞∑

k=0

Pk < ∞, where Pk :=k∏

j=1

1 − gj

gj.

Then there exists a sequence {g∗n} with gn(1−gn+1) < g∗n < gn for all n ≥ 0such that g∗n−1(1 − g∗n) = gn−1(1 − gn) and

Σ∗∞ = ∞ where Σ∗

n :=n∑

k=0

P ∗k and P ∗

k :=k∏

j=1

1 − g∗jg∗j

.

Remark: Corollary 2.8 on page 69 shows that

g∗n = −f (n) = gn(g0 − g∗0)Σn − g0

(g0 − g∗0)Σn−1 − g0

for n ≥ 1

where g∗0 = −f = −g0/Σ∞ + g0 (Corollary 2.7 on page 68). That is,

g∗n = gnΣ∞ − Σn

Σ∞ − Σn−1

for n ≥ 1. (2.5.7)

Some choices of {gn}.To make En large, we want to choose {gn} such that |an| is large, where an is givenby (2.5.5). If all gn = g, then g := 1

2 gives the maximal value for |an| = |a|, andwe are back to the original Worpitzky Theorem. If {gn} is allowed to vary, we canactually obtain |an| > 1

4for all n. For instance, if

gn :=12

+1

4n + 2for all n, (2.5.8)

then

|an| =(

12

+1

4n − 2

)(12− 1

4n + 2

)=

14

+1/4

4n2 − 1. (2.5.9)

This is not an optimal choice, though. Indeed, there exists no optimal choice of thistype:

Page 151: Lisa Lorentzen, Haakon Waadeland Continued Fractions

138 Chapter 3: Convergence criteria�

�Lemma 3.32. Let 1

2< gn < 1 be chosen such that |an| := gn−1(1−gn) > 1

4for all n. Then there exists a sequence {gn} with gn < gn < 1 such that|an| > |an| for all n for an := −gn−1(1 − gn).

Proof : Let Σn and Pn be given by (2.5.6), and set

gn := gn(1 + εn) where εn :=PnPn−1

2Σn−1

.

Then gn < gn. Now, 0 < (1− gn)/gn < 1, so 0 < Pn < Pn−1 < 1 and Pn−1 < Σn−1

for all n ≥ 2. Therefore

gn = gn + gnPnPn−1

2Σn−1

= gn + (1 − gn)P 2

n−1

2Σn−1

< gn + (1 − gn) = 1.

Moreover

|an| − |an| = gn−1(1 − gn)(

(1 + εn−1)(

1 − gnεn

1 − gn

)− 1)

= |an|(

εn−1 −gnεn

1 − gn− gnεnεn−1

1 − gn

)where

εn−1 −gnεn

1 − gn− gnεnεn−1

1 − gn=

Pn−1Pn−2

2Σn−2

−P 2

n−1

2Σn−1

−P 3

n−1Pn−2

4Σn−1Σn−2

=Pn−1

4Σn−1Σn−2

(2Pn−2Σn−1 − 2Pn−1Σn−2 − P 2

n−1Pn−2

)>

Pn−1

4Σn−1Σn−2

(2Pn−1(Σn−1 − Σn−2) − P 2

n−1Pn−2

)=

Pn−1

4Σn−1Σn−2

(2Pn−1Pn−2 − P 2

n−1Pn−2

)=

P 2n−1Pn−2

4Σn−1Σn−2

(2 − Pn−1

)> 0.

Indeed, it was proved in [JaMa90] that there is a hierarchy of sequences {g(k)n }∞n=0

for k = −1, 0, 1, 2, . . . given by

g(k)nk+n :=

12

+k∑

m=0

14Lm(n)

Page 152: Lisa Lorentzen, Haakon Waadeland Continued Fractions

3.2.5 Worpitzky’s Theorem 139

where Lm(n) is defined recursively by

L0(n) := n, Lm(n) := Ln(Lm−1(n)) for m ≥ 1

for n ≥ nm sufficiently large. Every continued fraction from {En} converges when

En := {w; |w| ≤ rn} where rn :=14

+k∑

m=0

1(4Lm(n))2

(2.5.10)

from some n on, and the continued fraction K(an/1) diverges when

an := −14−

k−1∑m=0

1(4Lm(n))2

− 1 + ε

(4Lk(n))2(2.5.11)

for all n sufficiently large, where k ∈ N and ε > 0 are arbitrarily chosen. Inparticular K(an/1) with an := − 1

4− 1+ε

16n2 diverges for ε > 0.

Limit sets.

If Σ∞ = ∞, then every f ∈ V0\{0} is the value of a continued fraction K(an/1) from{En}. This follows by the following arguments: K(an/1) with an given by (2.5.5)converges to −g0, and thus its first tail converges to −g1. Let f ∈ V0 \ {0,−g1} bearbitrarily chosen, and set a := f · (1 − g1). Then |a| ≤ g0(1 − g1) and

a

1+a2

1 +a3

1 +a4

1 +· · ·

converges to f . If Σ∞ < ∞, this is no longer true. Then V0 is too large”. Indeed,no value f ∈ V0 \ V ∗

0 with V ∗0 as in Theorem 3.31 can be the value of a continued

fraction from {En}. Therefore:�

�Theorem 3.33. Let {gn} and {En} be as in Theorem 3.30. If Σ∞ = ∞,then {gnD \{0}}∞n=0 is the sequence of limit sets for the family of continuedfractions K(an/1) from {En}.

Truncation error bounds.

We have already (in (2.5.6)) established an expression for the truncation error f−fn

for K(an/1) with an := −gn−1(1 − gn).

K(an/1) with an ≤ an < 0 is equivalent to g0K(cn/dn) where cn := an/gngn−1 anddn := 1/gn, and thus K(cn/dn) is a Sleszynski-Pringsheim continued fraction asdescribed in case 2 on page 132. Therefore the denominators Bn and approximantsfn for K(an/1) satisfy

Bn+1 ≥ Bn ≥ Σn and − gn < fn < fn−1 < 0 (2.5.12)

Page 153: Lisa Lorentzen, Haakon Waadeland Continued Fractions

140 Chapter 3: Convergence criteria

where Σn still is given by (2.5.6). For the general case we get from Theorem 3.27(again with obvious notation):

Theorem 3.34. Let {gn} and K(an/1) be as in Theorem 3.30, and letK(an/1) and K(an/1) be given by an := −|an| and an := −gn−1(1 − gn)for all n. Then

|f − fn| ≤ fn − f ≤ fn − f =g0

Σn

− g0

Σ∞. (2.5.13)

Corollary 3.35. For given positive r ≤ 1, let κ :=1 +

√1 − r

1 −√

1 − r. If all

|an| ≤ r4 for K(an/1), then

|f − fn| ≤

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩1/2

n + 1if r = 1,

√1 − r

κn+1 − 1if r < 1.

(2.5.14)

Proof : Let gn := g := (1 −√

1 − r)/2 for all n. Then |an| ≤ g(1 − g) = r4 and

κ = (1 − g)/g. Therefore Pk = κk and Σn = n + 1 if κ = 1 and

Σn =n∑

k=0

κn =κn+1 − 1

κ − 1

otherwise. In both cases Σn → Σ∞ = ∞. If r = 1, then κ = 1 and g = 12 . If r < 1,

then κ > 1 and g < 12 and

g0

Σn

=(κ − 1)gκn+1 − 1

=1 − 2g

κn+1 − 1=

√1 − r

κn+1 − 1.

The result follows therefore from Theorem 3.34. �

Page 154: Lisa Lorentzen, Haakon Waadeland Continued Fractions

3.2.5 Worpitzky’s Theorem 141

The radius of Sn(gnD).�

Theorem 3.36. Let {gn} and {En} be as in Theorem 3.30, and let K(an/1)be a continued fraction from {En}. Then

radSn(gnD) =∏n

k=1 |ak||Bn|2/gn − gn|Bn−1|2

≤ g0

Σn + Σn−1

n∏k=1

|ak|gk−1(1 − gk)

≤ g0

Σn + Σn−1

.

(2.5.15)

Let further K(an/1) and K(an/1) be given by an := −|an| and an :=−gn−1(1 − gn). Then

radSn(gnD) ≤ rad Sn(gnD) ≤ rad Sn(gnD) =g0

Σn + Σn−1

. (2.5.16)

Proof : The equivalence (2.5.4) shows that the radius Rn of Sn(gnD) is equal tog0 · (radius of Tn(D)) where

Tn(w) =c1

d1 +c2

d2 +c3

d3 +· · ·+cn

dn + w=:

Cn−1w + Cn

Dn−1w + Dn

with cn :=an

gngn−1and dn :=

1gn

. Here K(cn/dn) is a Sleszynski-Pringsheim

continued fraction, and the approximants of K(an/1) satisfy

Sn(w) :=An−1w + An

Bn−1w + Bn; An := Cn

n∏k=0

gk, Bn := Dn

n∏k=1

gk

for n ≥ 2, so by Theorem 3.28 on page 133

Rn =g0

∏nk=1 |ck|

|Dn|2 − |Dn−1|2≤ g0

|Dn| + |Dn−1|

n∏k=1

|ck||dk| − 1

≤ g0

|Dn| + |Dn−1|

where by (2.4.9)

|Dn| ≥n∑

k=0

k∏m=1

(|dm| − 1) =n∑

k=0

k∏m=1

(1

gm− 1)

= Σn .

Therefore (2.5.15) follows since |ak| ≤ gk−1(1−gk). Similarly we have g0K(an/1) ∼K(cn/dn) and g0K(an/1) ∼ K(cn/dn) where cn := −|an|/gngn−1, dn := 1/gn,cn := −(1 − gn)/gn and dn := 1/gn = 1 + |cn|. Hence also (2.5.16) follows fromTheorem 3.28. �

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142 Chapter 3: Convergence criteria

Remark: The limit point case Sn(gnD) → {f} occurs for every continued fractionK(an/1) from {En} only if Σ∞ = ∞. Indeed, if Σ∞ < ∞, then Vn := gnD

contains both −gn and −g∗n (as given in Theorem 3.31), and thus Sn(Vn) containsboth −g0 and −g∗0 for all n for the continued fraction K(an/1) from {En} withan := −gn−1(1 − gn), and the limit point case can not occur.

3.2.6 Van Vleck’s Theorem

This is a convergence theorem for continued fractions K(1/bn) with

bn ∈ Gε := {w ∈ C; | arg w| ≤ π2 − ε} ∪ {0} (2.6.1)

for some arbitrarily given 0 < ε < π/2. Let

Vε := Gε ∪ {∞} for ε ≥ 0. (2.6.2)

Thensn(Vε) =

1bn + Vε

⊆ Vε for bn ∈ Gε. (2.6.3)

Therefore Vε is a simple value set for K(1/bn) from Gε. This even holds for ε := 0.Moreover, Vδ is a value set for Gε for every 0 ≤ δ ≤ ε. We know by Theorem3.13 on page 117 that a positive continued fraction K(1/bn) converges if and onlyif∑

bn = ∞, where the case∑

bn < ∞ is described by the Stern-Stolz Theoremon page 100. Van Vleck ([VanV01]) showed that this is true, not only for positivecontinued fractions, but for every continued fraction from Gε:�

Theorem 3.37. (Van Vleck’s Theorem.) Let bn ∈ Gε for all n ∈ N

for a given 0 < ε < π2 with b1 �= 0. Then the even and odd approximants

of K(1/bn) converge to finite values. Moreover, K(1/bn) itself converges ifand only if

∑|bn| = ∞.

If K(1/bn) diverges, then it diverges generally, and its even and odd classicalapproximants converge absolutely.

Proof : The nestedness

fn ∈ Sn(Vε) ⊆ Sn−1(Vε) ⊆ s1(Vε) =1

b1 + Vε(2.6.4)

shows that {Sn(w)} is uniformly bounded with respect to n ∈ N and w ∈ Vε

since 0 �∈ b1 + Vε, and thus 1/(b1 + Vε) is bounded. In particular all Bn �= 0. If∑|bn| < ∞, then K(1/bn) diverges generally, but its even and odd approximants

converge absolutely (the Stern-Stolz Theorem and the subsequent remark on page101).

Page 156: Lisa Lorentzen, Haakon Waadeland Continued Fractions

3.2.6 Van Vleck’s Theorem 143

Let∑

|bn| = ∞. Following Jones and Thron ([JoTh80], p 89) we shall use theStieltjes-Vitali Theorem to prove that then K(1/bn) converges. Let

βn := arg(bn) and dn(z) := |bn|eiβnz for all n . (2.6.5)

(If bn = 0, we just set βn := 0.) Then |βn| ≤ π2 −ε and dn(z) ∈ Gε/2 if | arg(dn(z))| ≤

π2− ε

2; i.e.,

dn(z) ∈ Gε/2 if z ∈ D :={

z ∈ C; |Re(z)| ≤ π − ε

π − 2ε

}. (2.6.6)

Now,dn(z) ≥ 0 if z ∈ D∗ := {z ∈ D; Re(z) = 0} . (2.6.7)

Since for every fixed z ∈ D∗∑|dn(z)| =

∑|bn|e−βnIm(z) > e−|Im(z)|π/2

∑|bn| = ∞ ,

it follows from the Seidel-Stern Theorem on page 117 that K(1/dn(z)) converges toa finite value for every z ∈ D∗. We have proved: the classical approximants fn(z)of K(1/dn(z)) form a sequence of uniformly bounded holomorphic functions in Dwhich converges for z ∈ D∗. By the Stieltjes-Vitali Theorem on page 115 it followstherefore that K(1/dn(z)) converges for all z ∈ D◦. In particular it converges forz = 1. Hence K(1/bn) converges. �

Remark. Equivalence transformations show for instance that the conclusions ofVan Vleck’s Theorem also hold for K(1/(−bn)) and K(−1/ibn) when all bn ∈ Gε.

Example 9. The two-periodic continued fraction

−1−1 + i +

−11 + i +

−1−1 + i +

−11 + i +

−1−1 + i +

−11 + i + · · ·

is equivalent to the continued fraction

i

1 + i +1

1 − i +1

1 + i +1

1 − i +1

1 + i +1

1 − i + · · · .

(Use rn := −i in the equivalence transformation.) This new continued fraction hasthe form i · K(1/bn) where all bn ∈ Gε for ε := π

4. Since

∑|bn| = ∞, the original

continued fraction converges. It is rather easy to find its value f . It must be f = ixwhere x satisfies the equation

x =1

1 + i +1

1 − i + x

.

This quadratic equation has the two roots 12(−1±

√3)(1− i). Since x ∈ Vε, the real

part of x can not be negative, so x = 12(√

3−1)(1−i) and thus f = 12(√

3−1)(1+i).�

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144 Chapter 3: Convergence criteria

Limit set.�

Theorem 3.38. For given 0 ≤ ε ≤ π/2, Vε \ {0,∞} is the limit set for thefamily of continued fractions K(1/bn) from Gε \ {0}.

Proof : Let K(1/bn) be a convergent continued fraction from Gε with all bn �= 0.It is then a consequence of Van Vleck’s Theorem that its value f and its first tailvalue f (1) are finite and belong to Vε. Hence also f = 1/(b1 + f (1)) �= 0, and thusf ∈ Vε \ {0,∞}.We need to prove that every f ∈ Vε \ {0,∞} is the value of a continued fractionK(1/bn) from Gε with all bn �= 0. Let first f be a point on the boundary; i.e.,f ∈ ∂Vε\{0,∞}. It suffices to study the case where arg f > 0; i.e., f = r ei(π/2−ε) =i r e−iε (complex conjugation). Let b1 := r1 e−i(π/2−ε) = −i r1 eiε and b2 := i r2 e−iε

where r1 > 0, r2 > 0 and r1 + 1/(r2 + r) = 1/r. Then b1, b2 ∈ ∂Gε and

1b2 + f

=1

i(r2 + r)e−iε=

−i

r2 + reiε =: f ∈ ∂Vε

and1

b1 + f=

1−i(r1 + 1

r2+r )eiε= i r e−iε = f .

That is, the 2-periodic sequence f, f , f, f , f, . . . is a tail sequence for the 2-periodiccontinued fraction

∞Kn=1

1bn

:=1b1 +

1b2 +

1b1 +

1b2 +

1b1 +

1b2 +· · ·

.

Now, K(1/bn) converges (the Van Vleck Theorem), and its value must belong toVε. Moreover, its value must be a solution of the quadratic equation

1

b1 +1

b2 + x

= x; i.e., x = 12(−r1r2 ±

√r21r

22 + 4ir2e−iε).

Page 158: Lisa Lorentzen, Haakon Waadeland Continued Fractions

3.2.6 Van Vleck’s Theorem 145

Re

Im

f (1)

1f

= b + f (1)

ε

ε

b

L1

L2

Only one of the two solutions belongs toVε. Hence K(1/bn) converges to f . Thisproves that ∂Vε \ {0,∞} belongs to thelimit set. If ε = π/2 we are finished, sincethen V ◦

ε = ∅. Let ε < π/2, and let f :=r eiα ∈ V ◦

ε with −(π2 −ε) < α < π

2 −ε and0 < r < ∞ be arbitrarily chosen. Then1f = 1

r e−iα ∈ V ◦ε , and it suffices to find

a b ∈ Gε and an f (1) ∈ ∂Vε such thatf = r eiα = 1

b+f(1) ; i.e., b + f (1) = 1f .

This can for instance be done by draw-ing lines L1 and L2 through 1

f parallelto the two boundary lines (see the pic-ture). The points where L1 and L2 inter-sects the boundary of Vε can then be ourpoints f (1) and b.

Remark. Clearly, K(1/bn) from Gε converges to f = ∞ only if b1 = 0 and f (1) = 0.And f (1) = 0 only if f (2) = ∞, and so on. Hence, if K(1/bn) from Gε converges to∞, then all b2n−1 = 0, and if K(1/bn) converges to 0, then all b2n = 0. For instance

10+

11+

10+

11+

10+· · ·

and11+

10+

11+

10+

11+· · ·

converge to ∞ and 0 respectively.

Truncation error bounds.

We have the following a posteriori bounds for K(1/bn) from Gε:�

Theorem 3.39. For given 0 < ε < π2, let K(1/bn) with b1 �= 0 be a

continued fraction from Gε. Then

diamSn(Vε) ≤{|fn − fn−1| if ε ≥ π/4,

|fn − fn−1|/ sin 2ε if ε < π/4.

Proof : As always, H denotes the right half plane, so Vε can be written

Vε = (eiεH) ∩ (e−iε

H).

Now, Sn(eiεH) and Sn(e−iε

H) are two circular disks since −b1 �∈ e±iεH, and thus

Sn(e±iεH) is bounded. The boundaries of these two disks intersect at Sn(0) = fn

Page 159: Lisa Lorentzen, Haakon Waadeland Continued Fractions

146 Chapter 3: Convergence criteria

and Sn(∞) = fn−1 under the inner angle 2α := π − 2ε. That is, Sn(Vε) is a lens-shaped set as illustrated in the figure on page 126. Therefore the result followsjust as in the proof of the Henrici-Pfluger Bounds. (We have used that sin 2α =sin(π − 2ε) = sin 2ε.) �

The diameter of Sn(H).

Our proof of Van Vleck’s Theorem was based on the Stieltjes-Vitali Theorem whichgives no information on the speed of convergence. We therefore give an alterna-tive proof which also opens up for an extension of Van Vleck’s classical result andfor truncation error bounds. It is due to Jensen ([Jens09]), and was later found,independently, by Lange ([Lange99a]).

Proof 2 of Van Vleck’s Theorem: Also the half plane V0 := H is a simplevalue set for K(1/bn). Therefore

Kn := Sn(V0) ⊆ Sn−1(V0) ⊆ · · · ⊆ S1(V0) ⊆ V0. (2.6.8)

Now, Re(b1) �= 0, so K1 is a circular disk, and the nestedness (2.6.8) makes allKn circular disks. Since 0 ∈ V0, we have fn ∈ Kn, so all Bn �= 0, and thusζn := S−1

n (∞) = −Bn/Bn−1 �= ∞ for all n ∈ N.

ζn is symmetric to (−ζn) with respect to ∂V0 = i R+. The center Cn of Kn issymmetric to ∞ with respect to ∂Kn = ∂Sn(V0). Therefore, by Property 2 on page109, Cn = Sn(−ζn). Hence, the radius of Kn is

Rn = |Sn(0) − Cn| =∣∣∣∣An

Bn− An − An−1ζn

Bn − Bn−1ζn

∣∣∣∣=

|(AnBn−1 − An−1Bn)ζn||Bn(Bn − Bn−1ζn)|

=1 · |Bn/Bn−1|

|BnBn−1( Bn

Bn−1+ Bn

Bn−1)|

=|Bn/Bn|

|BnBn−1 + BnBn−1|=

12|Re(BnBn−1)|

.

Now, setQn := Re(BnBn−1).

Then the recurrence relation for {Bn} gives

Qn = Re(bn|Bn−1|2 + Bn−2Bn−1) = Qn−1 + |Bn−1|2Re(bn). (2.6.9)

Since Q1 = Re(b1) > 0 and Re(bn) ≥ 0, this means that Qn > 0 and non-decreasingas n increases. Let Q := lim Qn. If Q = ∞, then Rn → 0, and the convergence isclear. Therefore also

∑|bn| = ∞ (the Stern-Stolz Theorem).

Page 160: Lisa Lorentzen, Haakon Waadeland Continued Fractions

3.2.6 Van Vleck’s Theorem 147

Let Q < ∞. Then

∞∑n=2

∣∣∣∣ 1Qn−1

− 1Qn

∣∣∣∣ = ∞∑n=2

(1

Qn−1− 1

Qn

)=

1Q1

− 1Q

< ∞. (2.6.10)

So far we have only used that Re(b1) > 0 and that V0 is a value set for K(1/bn);i.e., bn ∈ V0 for n ≥ 2. At this point we assume that all bn ∈ Gε with a given ε > 0.Then λ := sin ε > 0 and |bn| ≤ Re(bn)

cos(π/2−ε) = Re(bn)/λ, so

|fn − fn−2| =∣∣∣∣AnBn−2 − BnAn−2

BnBn−2

∣∣∣∣ = ∣∣∣∣ bn

BnBn−2

∣∣∣∣ ≤ Re(bn)λ|BnBn−2|

=|Bn−1|2Re(bn)

λ · |BnBn−1| · |Bn−1Bn−2|=

Qn − Qn−1

λ|BnBn−1| · |Bn−1Bn−2|

≤ 1λ· Qn − Qn−1

QnQn−1=

1λ·(

1Qn−1

− 1Qn

) (2.6.11)

where we have used the equality (2.6.9). Hence∑

|fn − fn−2| < ∞, and the evenand odd approximants of K(1/bn) converge absolutely. Van Vleck’s Theorem followstherefore from the Lane-Wall Characterization on page 103. �

Actually, the limit point case always occurs when K(1/bn) converges:

Corollary 3.40. Let K(1/bn) be as in Van Vleck’s Theorem. ThendiamSn(H) → 0 if and only if

∑|bn| = ∞.

Proof : If∑

|bn| < ∞, then K(1/bn) diverges, and diamSn(H) can not vanishas n → ∞. Let

∑|bn| = ∞. We want to prove that now Rn → 0; i.e., Qn → ∞.

From (2.6.9)

Qn = Qn−1 + Re(bn) · |ζn−1| · |Bn−1Bn−2|

≥ Qn−1(1 + Re(bn)|ζn−1|) ≥ Q1

n∏k=2

(1 + Re(bk)|ζk−1|)(2.6.12)

where Q1 = Re(b1) > 0 and Re(bk) ≥ λ|bk|. Therefore it suffices to prove that∑|bkζk−1| = ∞. For convenience we set δk−1 := −bkζk−1 = bkBk−1/Bk−2. Then

{δk} is exactly as in (1.2.3) on page 104, and thus, if∑

|δn| < ∞, then∑

|bn| < ∞as in the proof the Lane-Wall Characterization. Therefore

∑|δn| =

∑|bkζk−1| =

∞. �

Page 161: Lisa Lorentzen, Haakon Waadeland Continued Fractions

148 Chapter 3: Convergence criteria�

Corollary 3.41. For given K(1/bn) with Re(b1) > 0 and Re(bn) ≥ 0 forn > 2,

diamSn(H) =1

Re(BnBn−1)≤ 1/Re(b1)∏n

k=2(1 + |ζk−1|Re(bk))

≤ 1/Re(b1)∏nk=2(1 + Re(bk−1)Re(bk))

for n ∈ N. (2.6.13)

Proof : The first inequality follows from (2.6.12). The second one follows since−ζk ∈ bk + H (Theorem 2.3 on page 63). �

Remark. This proves that if Re(b1) > 0 and Re(bn) ≥ 0 for n ≥ 2, then K(1/bn)converges if Re(BnBn−1) → ∞, which happens if

∑|ζn−1| ·Re(bn) = ∞. Although

this extends Van Vleck’s Theorem, the criterion is more difficult to check. Beardonand Short ([BeSh07]) have proved that the limit point case may fail to occur for aconvergent continued fraction in this case.

3.2.7 The Thron-Lange Theorem

The Thron-Lange Theorem concerns a family of continued fractions K(1/bn) forwhich {bn} stays away from a circular disk B(−2γ, 2r)◦ symmetric about the realaxis. The unifying factor for this family is that the disk B(γ, r) is a simple valueset for all its continued fractions.�

Theorem 3.42. (The Thron-Lange Theorem). For given γ ∈ R, letr :=

√1 + γ2, V := B(γ, r) and G := B(−2γ,−2r). Then G is the element

set for continued fractions K(1/bn) corresponding to the value set V , andevery continued fraction K(1/bn) from G converges with

diamSn(V ) ≤ 2r

n∏k=2

2k − σ

2k + σ≤ 2r

( 4 + σ

2n + 2 + σ

; σ := 1 − |γ|r

. (2.7.1)

The convergence of K(1/bn) was proved by Thron. Actually, it is a corollary of amuch more general result from [Thron49]. Lange ([Lange99b]) has proved a numberof corollaries of Thron’s general theorem, but our particular version can be foundin Thron’s paper. However, Lange is responsible for the truncation error bounds([Lange99b]).

To prove the convergence in Theorem 3.42 is not difficult. It follows from Corollary3.9 on page 114 with ϕn(w) := ϕ(w) := γ + rw and wn := ∞. However, it is

Page 162: Lisa Lorentzen, Haakon Waadeland Continued Fractions

3.2.7 The Thron-Lange Theorem 149

also a consequence of Lange’s truncation error bound (2.7.1) since this vanishes asn → ∞, and thus the limit point case occurs. We shall prove this bound:

Proof : We shall first prove that the inclusion 1/(b + V ) ⊆ V holds if and only ifb ∈ G. Evidently, 1/(b + V ) ⊆ V if and only if

b + V = B(b + γ, r) ⊆ 1V

= B( γ

γ2 − r2,

r

γ2 − r2

)= B(−γ,−r)

(Lemma 3.6 on page 110); that is, if and only if

|b + γ − (−γ)| ≥ r + r

which proves the assertion.

In the following we assume that γ ≥ 0. We can do so without loss of generalitysince K(1/(−bn)) is equivalent to −K(1/bn). Let K(1/bn) be a continued fractionfrom G. Since 0 ∈ V , we know that fn ∈ V , and thus Bn �= 0 since V is bounded.Hence it follows from Lemma 3.6 on page 110 that the radius Rn of Sn(V ) is givenby

Rn =r

|Bn−1|2(|γ − ζn|2 − r2),

and thusRn+1

Rn=

1|ζn|2

· |γ − ζn|2 − r2

|γ − ζn+1|2 − r2(2.7.2)

whereζn+1 = −bn+1 +

1ζn

. (2.7.3)

Now, bn+1 ∈ B(−2γ,−2r), and we shall first prove that

ζn ∈ B(n + 1

nγ, −n + 1

nr)

for n ≥ 1 (2.7.4)

which by Lemma 3.6 is equivalent to

1ζn

∈ B(− nγ

n + 1,

nr

n + 1

)for n ≥ 1. (2.7.5)

(2.7.4) holds trivially for n = 1 since ζ1 = −b1 ∈ (−G) = B(2γ,−2r). Assume itholds for a given n ∈ N. Then we can write

1ζn

= − nγ

n + 1+

nr

n + 1(1 − μ)eiθ, bn+1 = −2γ + 2(1 + λ)r eiϕ (2.7.6)

for some 0 ≤ μ ≤ 1, λ ≥ 0 and θ, ϕ ∈ R, and thus

ζn+1 = 2γ − 2(1 + λ)r eiϕ − nγ

n + 1+

nr

n + 1eiθ =

(2 − n

n + 1

)γ + ρ eiψ

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150 Chapter 3: Convergence criteria

where ρ ≥ 2(1 + λ)r − nr

n + 1=

r(n + 2)n + 1

. That is, ζn+1 ∈ B(n + 2

n + 1γ, −n + 2

n + 1r).

Hence (2.7.4) holds for all n by induction.

With the notation (2.7.6), the ratio (2.7.2) takes the form

Rn+1

Rn=

1|ζn|2

· |γ − ζn|2 − r2

|γ + bn+1 − 1ζn|2 − r2

=| γζn

− 1|2 − r2

|ζn|2|γ + bn+1 − 1

ζn|2 − r2

=

∣∣− nγ2

n+1 + nγrn+1 (1 − μ)eiθ − 1

∣∣2 − r2∣∣− nγ

n+1 + nrn+1 (1 − μ)eiθ

∣∣2∣∣− γ + 2(1 + λ)r eiϕ + nγn+1 − nr

n+1 (1 − μ)eiθ∣∣2 − r2

.

(2.7.7)

Letun := γ − nγ

n + 1+

nr

n + 1(1 − μ)eiθ =

γ

n + 1+

nr

n + 1(1 − μ)eiθ.

Then|un| ≤

γ

n + 1+

nr

n + 1<

r + nr

n + 1= r,

so the denominator of (2.7.7) is bounded below by

(2r − |un|)2 − r2 = (3r − |un|)(r − |un|) > 0.

The numerator of (2.7.7) can be written∣∣∣γ(− nγ

n + 1+ un − γ

n + 1

)− 1∣∣∣2 − ∣∣∣− nγ

n + 1+ un − γ

n + 1

∣∣∣2r2

= |1 + γ2 − γun|2 − |un − γ|2r2 = |r2 − γ un|2 − |un − γ|2r2

= r4 + γ2|un|2 − 2r2γ Re(un) − (γ2 + |un|2 − 2γ Re(un))r2

= r4 + γ2|un|2 − r2γ2 − r2|un|2 = (r2 − |un|2)(r2 − γ2)

where r2 − γ2 = 1. Therefore

Rn+1

Rn≤ r2 − |un|2

(3r − |un|)(r − |un)=

r + |un|3r − |un|

≤ r + max |un|3r − max |un|

where |un| ≤ γ/(n + 1) + nr/(n + 1). That is,

Rn+1

Rn≤ (n + 1)r + γ + nr

3(n + 1)r − γ − nr=

(2n + 1)r + γ

(2n + 3)r − γ=

2n + 2 − σ

2n + 2 + σ.

Therefore

Rn+1 = R1

n∏k=1

Rk+1

Rk≤ r

n∏k=1

2k + 2 − σ

2k + 2 + σ= r

n∏k=1

(1 + σ/2) + k − σ

(1 + σ/2) + k

which proves the first inequality in (2.7.1). The second one follows from Lemma3.11 on page 115. �

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3.2.8 The parabola theorems 151

3.2.8 The parabola theorems

The parabola theorems are convergence theorems for continued fractions K(an/1)with half planes as value sets. The version with the simple value set

Vα := − 12 + eiα

H = {w ∈ C; Re(we−iα) ≥ − 12 cos α} ∪ {∞} (2.8.1)

for some fixed α ∈ R with |α| < π2 , has got the prominent name the Parabola The-

orem. The convergence theorem for S-fractions on page 124 implies that K(an/1)from the ray arg an = 2α converges if and only if its Stern-Stolz Series

S :=∞∑

n=1

∣∣∣∣a1a3 · · · a2n−1

a2a4 · · · a2n

∣∣∣∣+ ∞∑n=1

∣∣∣∣ a2a4 · · · a2n

a1a3 · · · a2n+1

∣∣∣∣ (2.8.2)

diverges to ∞. The Parabola Theorem extends this to continued fractions K(an/1)from a closed parabolic neighborhood

Eα := {a ∈ C; |a| − Re(ae−i2α) ≤ 12 cos2 α} (2.8.3)

of this ray. More precisely, with the notation above we have:

Theorem 3.43. (The Parabola Theorem.) For fixed α ∈ R with|α| < π

2, the set Eα is the element set for continued fractions K(an/1)

corresponding to the value set Vα.Let K(an/1) be a continued fraction from Eα. If S = ∞, then K(an/1)converges to a finite value. If S < ∞, then {f2n} and {f2n+1} converge ab-solutely to distinct finite values, and {S2n} and {S2n+1} converge generallyto these values.

Re

Im

−12

Both V ◦α and E◦

α contain the origin. Theboundary of the half plane Vα is a line pass-ing through −1

2. The boundary of Eα is a

parabola with axis along the ray arg z = 2α,focus at the origin and vertex at the point−1

4ei2α cos2 α. It intersects the real axisat z = − 1

4 . The half plane Vα is illus-trated to the left and the parabolic regionEα on the next page. Since 0 is an inte-rior point in Vα, it follows that the approx-imants Sn(0) = An/Bn of a continued frac-tion K(an/1) from Eα are interior pointsin Vα. In particular Sn(0) �= ∞, and thus

Bn �= 0 and ζn �= ∞ for n ≥ 1.

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152 Chapter 3: Convergence criteria

1

2

3

y

1 2 3

x

For α := 0, the set E0 contains the Worpitzky disk E := {a ∈ C; |a| ≤ 14}. The

Parabola Theorem therefore generalizes the classical Worpitzky Theorem consider-ably. It also generalizes the Seidel-Stern Theorem on page 117 which says that apositive continued fraction of the form K(an/1) converges if and only if S = ∞.The Parabola Theorem implies that this holds for real continued fractions K(an/1)with an ≥ −1

4 .

Proof of the Parabola Theorem: We shall first prove that s(Vα) :=

Re

Im

γa

0

α

−12

∂Vα

a/(1 +Vα) ⊆ Vα if and only if a ∈Eα. The inclusion is clear for a = 0,so let a �= 0. Then s maps Vα ontothe closed circular disk with cen-ter at γa := (ae−iα)/ cos α and ra-dius ρa := |a|/ cos α (Theorem 3.6on page 110). This disk is containedin Vα if and only if γa ∈ Vα and γa

has a distance δa to ∂Vα such thatδa ≥ ρa. Now

δa =12

cos α + Re(γae−iα) ,

where δa > 0 if and only if γa ∈ Vα.(See the figure.)

So s(Vα) ⊆ Vα if and only if

12

cos α+Re(

ae−iα

cos αe−iα

)≥ |a|

cos α,

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3.2.8 The parabola theorems 153

i.e. if and only if a ∈ Eα.

Let K(an/1) be a continued fraction from Eα. Since ∞ �∈ S1(Vα), the nested setsKn := Sn(Vα) are bounded disks. If diam(Kn) → 0, the limit point case, thenthe convergence is clear, and it follows from the Stern-Stolz Theorem that S = ∞.Assume that Kn → K where diam(K) > 0. The linear fractional transformation

ϕ(w) :=−1 + eiα cos α − w

1 + w(2.8.4)

maps the closed unit disk D onto Vα with ϕ(∞) = −1 and ϕ(−1) = ∞. Let

τn := ϕ−1 ◦ s2n−1 ◦ s2n ◦ ϕ for n = 1, 2, 3, . . . . (2.8.5)

Then, τn(D) = ϕ−1 ◦ s2n−1 ◦ s2n(Vα) ⊆ ϕ−1 ◦ s2n−1(Vα) ⊆ ϕ−1(Vα) = D, andτn(∞) = ϕ−1 ◦ s2n−1 ◦ s2n(−1) = ϕ−1 ◦ s2n−1(∞) = ϕ−1(0) = −1 + eiα cos α =: kwhere |k e−iα| = |i sinα| < 1. It follows therefore by Lemma 3.8 on page 113 withwn := ∞ that the sequence {Tn} given by Tn := τ1 ◦ τ2 ◦ · · · ◦ τn converges generallyto some value γ ∈ D and that

∑|Tn(∞)−Tn−1(∞)| < ∞. Since Tn = ϕ−1 ◦S2n ◦ϕ,

this means that Tn(∞) = ϕ−1(S2n(−1)) = ϕ−1(S2n−2(0)) = ϕ−1(f2n−2) where ϕ−1

is a fixed linear fractional transformation with pole at −1. Since all fn ∈ s1(Vα),and {fn} thus is bounded away from ∞ and −1, it follows that

∑|f2n−f2n−2| < ∞.

Similarly, also∑

|f2n+1 − f2n−1| < ∞. (Just use τn := ϕ−1 ◦ s2n ◦ s2n+1 ◦ ϕ.)It is therefore a consequence of the Lane-Wall Characterization on page 103 thatK(an/1) converges if and only if S = ∞. That {S2n} and {S2n+1} converge gener-ally, also when K(an/1) diverges, follows since Sn+2(−1) = Sn(0). �

The Parabola Theorem is in many ways the queen among the convergence theoremsfor continued fractions K(an/1). It is best in several respects. For instance, one cannot enlarge the set Eα, not even by adding just one point, without destroying theproperty that every continued fraction K(an/1) from Eα with divergent Stern-StolzSeries converges, ([Lore92]).

The Parabola Theorem has an extension of type similar to the extension of Wor-pitzky’s Theorem on page 136. It is due to Thron, ([Thron58]). This time we havea sequence {Vα,n}∞n=0 of half planes given by

Vα,n := −gn + eiαH = {w ∈ C; Re(we−iα) ≥ −gn cos α} ∪ {∞} (2.8.6)

where

−π2 < α < π

2 , g0 > 0 and 0 < ε ≤ gn ≤ 1 − ε for n ≥ 1 (2.8.7)

are given constants.

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154 Chapter 3: Convergence criteria�

Theorem 3.44. (The Parabola Sequence Theorem.) For given con-stants (2.8.7), the sequence {Eα,n}∞n=1 given by

Eα,n := {a ∈ C; |a| − Re(ae−i2α) ≤ 2gn−1(1 − gn) cos2 α} (2.8.8)

is the sequence of element sets corresponding to the value sets {Vα,n} givenby (2.8.6).Let K(an/1) be a continued fraction from {Eα,n}, and let S be the sum ofits Stern-Stolz Series (2.8.2). If S = ∞, then K(an/1) converges to a finitevalue. If S < ∞, then {f2n} and {f2n+1} converge absolutely to distinctfinite values, and {S2n} and {S2n+1} converge generally to these values.

Proof : The proof is similar to the proof of the Parabola Theorem. This time thedisk sn(Vα,n) has center at γα,n := (an e−iα)/(2(1 − gn) cos α) and radius ρα,n :=|an|/(2(1 − gn) cos α). Hence sn(Vα,n) ⊆ Vα,n−1 if and only if γα,n ∈ Vα,n−1 andγα,n has a distance δα,n ≥ ρα,n to ∂Vα,n−1; i.e.,

δα,n = gn−1 cos α + Re(

ane−iα

2(1 − gn) cos αe−iα

)≥ |an|

2(1 − gn) cos α;

i.e., if and only if an ∈ Eα,n. If the limit point case occurs for Sn(Vα,n), thenK(an/1) converges and S = ∞.

Assume that the limit circle case occurs. The linear fractional transformation

ϕn(w) :=−1 + 2(1 − gn)eiα cos α − w

1 + w

maps D onto Vα,n with ϕn(∞) = −1 and ϕn(−1) = ∞. Hence τn+1 := ϕ−12n ◦s2n+1 ◦

s2n+2 ◦ϕ2n+2 maps D into D with τn+1(∞) = ϕ−12n (0) = −1+2(1− g2n)eiα cos α =:

kn.

Now, |kn| is bounded away from 1 since kne−iα = (1−2g2n) cos α+i sin α where |1−2g2n| ≤ 1−2ε, and thus |kne−iα|2 ≤ (1−2ε)2 cos2 α+sin2 α = 1−4ε(1−ε) cos2 α < 1.Therefore we also now get

∑|Tn(∞)−Tn−1(∞)| < ∞. Since Tn = ϕ−1

0 ◦ S2n ◦ϕ2n

where ϕ−10 is a fixed linear fractional transformation with pole at −1 and {ϕ2n} is

totally non-restrained, we still have∑

|f2n − f2n−2| < ∞ in the limit circle case.Similarly, also

∑|f2n+1 − f2n−1| < ∞. Finally, Sn+2(−1) = Sn(0) as before. �

The diameter of Sn(Vα,n).

As in the Worpitzky situation, the quantities

Σn :=n∑

m=0

Pm where Pm :=m∏

k=1

1 − gk

gk(2.8.9)

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3.2.8 The parabola theorems 155

are important. The following truncation error bound is due to Thron ([Thron58]):�

Theorem 3.45. Let α, {gn} and K(an/1) be as in the Parabola SequenceTheorem, and let Pm and Σn be given by (2.8.9). Then

diamSn(Vα,n) ≤ |a1|/((1 − g1) cos α)n∏

j=2

(1 +

Pj−1gj−1(1 − gj) cos2 α

|aj |Σj−2

) . (2.8.10)

Proof : Since a1/(1 + Vα,1) is bounded, we know that Sn(Vα,n) is a sequence ofnested circular disks. Therefore fn−1 = An−1/Bn−1 �= ∞ which means that Bn �= 0and ζn = −Bn/Bn−1 �= ∞ for n ≥ 1. We can therefore write

Sn(w) = fn−1 +AnBn−1 − BnAn−1

B2n−1(w − ζn)

,

and thus the radius Rn of Sn(Vα,n) is given by

Rn =12

supw∈∂Vα,n

|Sn(w) − Sn(∞)| =12

supw∈∂Vα,n

∣∣∣∣AnBn−1 − BnAn−1

B2n−1(w − ζn)

∣∣∣∣where ζn �∈ Vα,n since ζ1 = −1 �∈ Vα,1. Now, points on ∂Vα,n can be writtengneiα(−1 + i vn) cos α with vn ∈ R. Therefore ζn can be written

ζn = gneiα(−1 − un + ivn) cos α where un > 0, vn ∈ R, (2.8.11)

and so

Rn =12

∣∣∣∣AnBn−1 − BnAn−1

B2n−1gnun cos α

∣∣∣∣ .

By means of the determinant formula on page 7 we therefore get that

Rn

Rn−1=

|an||ζn−1|2

gn−1un−1

gnun. (2.8.12)

Since an ∈ Eα,n, we may also write

an = gn−1(1 − gn)e2iα(xn + iyn) cos2 α where y2n ≤ 4xn + 4.

Then gn(1 + un) cos α = −Re(ζne−iα) where

−Re(ζne−iα) = −Re((

− 1 +an

ζn−1

)e−iα

)= −Re

(−e−iα +

ane−2iα

ζn−1e−iα

)= cos α − Re

(gn−1(1 − gn)(xn + iyn) cos2 α

gn−1(−1 − un−1 + i vn−1) cos α

).

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156 Chapter 3: Convergence criteria

This means that

gnun = gn(1 + un) − gn = (1 − gn)(1 + Re

xn + iyn

1 + un−1 − ivn−1

)= (1 − gn)

(1 +

xn(1 + un−1) − ynvn−1

(1 + un−1)2 + v2n−1

).

(2.8.13)

Therefore

Rn

Rn−1=

gn−1(1 − gn)√

x2n + y2

n

g2n−1{(1 + un−1)2 + v2

n−1}· gn−1un−1/(1 − gn)

1 +xn(1 + un−1) − ynvn−1

(1 + un−1)2 + v2n−1

=

√x2

n + y2n un−1

(1 + un−1)2 + v2n−1 + xn(1 + un−1) − ynvn−1

.

(2.8.14)

Since y2n ≤ 4xn + 4, the last denominator can be written

(1 + un−1)2 + (vn−1 − yn/2)2 − y2n/4 + xn(1 + un−1)

≥ (1 + un−1)2 − 14 (4xn + 4) + xn(1 + un−1) = (2 + un−1 + xn)un−1 .

Therefore, since x2n + y2

n ≤ x2n + 4xn + 4 = (xn + 2)2,

Rn

Rn−1≤

√x2

n + y2n

xn + 2 + un−1≤

√x2

n + y2n√

x2n + y2

n + un−1

=1

1 + un−1√x2

n+y2n

. (2.8.15)

Equality (2.8.13) also gives a lower bound for un: we first observe that |yn| ≤2√

xn + 1 implies that

xn(1 + un−1) − ynvn−1

(1 + un−1)2 + v2n−1

≥ xn(1 + un−1) − 2√

xn + 1 · |vn−1|(1 + un−1)2 + v2

n−1

,

where a standard minimalization procedure shows that the right hand side attainsits minimum for vn−1 := (1 + un−1)

√xn + 1; that is,

xn(1 + un−1) − 2√

xn + 1 · |vn−1|(1 + un−1)2 + v2

n−1

≥ xn(1 + un−1) − 2(1 + un−1)(xn + 1)(1 + un−1)2 + (1 + un−1)2(xn + 1)

=−1

1 + un−1.

With this inequality we get from (2.8.13) that

gnun ≥ (1 − gn)(

1 − 11 + un−1

)=

(1 − gn)un−1

1 + un−1,

i.e.,1un

≤ gn

1 − gn

(1 +

1un−1

). (2.8.16)

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3.2.8 The parabola theorems 157

Now, ζ1 = −1, and thus, by (2.8.11),1u1

=g1

1 − g1. Hence it follows by induction,

using (2.8.16), that

1un

≤ gn

1 − gn+

gn

1 − gn

gn−1

1 − gn−1+ · · · +

n∏k=1

gk

1 − gk

= (Pn−1 + Pn−2 + · · · + 1)/Pn = Σn−1/Pn .

Therefore the bound (2.8.15) is again bounded by

Rn

Rn−1≤ 1

1 +Pn−1/Σn−2√

x2n + y2

n

=1

1 +Pn−1gn−1(1 − gn) cos2 α

|an|Σn−2

.

Now, diamSn(Vα,n) = 2Rn = 2R1

∏nk=2 Rk/Rk−1 where R1 is the radius of a1/(1+

Vα,1); i.e., 2R1 = |a1|/δ1 where δ1 is the euclidean distance between −1 and ∂Vα,1;i.e., δ1 = (1 − g1) cos α. This proves (2.8.10). �

The choice gn := 12

for all n in the Parabola Sequence Theorem gives back theParabola Theorem. The truncation error bound (2.8.10) then takes the form:

Corollary 3.46. Let α and K(an/1) be as in the Parabola Theorem. Then

diamSn(Vα) ≤ 2|a1|/ cos αn∏

j=2

(1 +

cos2 α

4(j − 1)|aj |) . (2.8.17)

Proof : For all gn := 12 we get Pk = 1 and Σn = n + 1. �

Remarks.

1. The bound (2.8.17) depends on {an}. If also |an| ≤ M for all n, then thisbound is again bounded by

2M/ cos α∏n−1j=1 (1 + (cos2 α)/(4Mj))

→ 0 as n → ∞. (2.8.18)

That is, every continued fraction K(an/1) from Eα ∩B(0,M) converges, andthe convergence is uniform with respect to K(an/1) from Eα∩B(0,M). More-over, the limit point case for Sn(Vα) occurs for all these continued fractions.

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158 Chapter 3: Convergence criteria

2. Also the truncation error bound (2.8.10) depends on K(an/1). If we add thecondition that |an| ≤ Mn for all n, then

diamSn(Vα,n) ≤ M1/((1 − g1) cos α)n−1∏j=1

(1 +

Pjgj(1 − gj+1) cos2 α

Σj−1Mj+1

) . (2.8.19)

If∑

Pj/(Σj−1Mj+1) = ∞, then every continued fraction K(an/1) from {Eα,n∩B(0,Mn)} converges to some finite value f ∈ Vα,0, uniformly with respect toK(an/1) from {Eα,n ∩ B(0,Mn)}, and Sn(Vα,n) approaches a limit point.

3. Let K(an/1) be from {Eα,n ∩ B(0,M)} for some M > 0. Then the bound(2.8.19) vanishes as n → ∞ if and only if

∑Pj/Σj−1 = ∞, which happens if

and only if Σ∞ = ∞ (standard property of positive series).

4. If Σn → Σ∞ < ∞, then the limit point case fails to occur for the continuedfraction K(an/1) with an := −gn−1(1 − gn) from {Eα,n}, since the limit setmust contain both −g0 and f �= −g0. As in Theorem 3.31 on page 137, wecan replace {gn} by some smaller constants {g∗n} to get smaller value sets

V ∗α,n := −g∗n + eiα

H

for {Eα,n} where Σ∗∞ :=

∑∞n=0

∏nk=1(1 + g∗k)/g∗k = ∞. If 0 < ε ≤ gn ≤ 1 − ε

for all n, then 0 < ε2 ≤ g∗n ≤ 1 − ε < 1 − ε2 for all n.

Proof of Theorem 3.24 on page 128: For fixed z := r e2iα with |α| < π/2, theS-fraction K(anz/1) is a continued fraction from Eα given by (2.8.3), and the halfplane Vα in (2.8.1) is a simple value set for K(anz/1). By (2.8.14) (with all gn = 1

2)

it follows therefore that the radius Rn of Sn(Vα) satisfies

Rn

Rn−1=

√x2

n + y2n un−1

(1 + un−1)2 + v2n−1 + xn(1 + un−1) − ynvn−1

.

Now, with this notation, anz = anr e2iα = 14 e2iαxn cos2 α and ζn = 1

2 eiα(−1 −un + i vn) with un > 0 by (2.8.11). That is, xn = 4anr/ cos2 α > 0, yn = 0 and

Rn

Rn−1≤ xnun−1

(1 + un−1)2 + xn(1 + un−1)

which obtains its maximum with respect to un−1 for un−1 :=√

1 + xn. That is,

Rn

Rn−1≤ xn

√1 + xn

(1 +√

1 + xn)2 + xn(1 +√

1 + xn)

=xn

√1 + xn

2(1 + xn) + (2 + xn)√

1 + xn

=xn

2√

1 + xn + 2 + xn=

xn

(1 +√

1 + xn)2=

1(1√xn

+√

1 + 1xn

)2

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3.2.8 The parabola theorems 159

where xn = 4anr/ cos2 α. Since R1 = a1r/( 12

cos α), this proves that

diamSn(Vα) = 2Rn = 2R1

n∏k=2

Rk

Rk−1≤ 4a1r

cos α

n∏k=2

(1√xk

+√

1 +1xk

)−2

which actually was the way Thron presented his error bound. To get our formulation(which agrees with the formulation by Gragg and Warner), we just observe that

√1 + xn − 1√1 + xn + 1

=xn

(√

1 + xn + 1)2=

1(1√xn

+√

1 + 1xn

)2 .

Limit sets.

Also for the Parabola Theorem it turns out that every point in Vα \ {0,∞} is thevalue of a continued fraction K(an/1) from Eα:�

�Theorem 3.47. For fixed α ∈ R with |α| < π

2 let Eα and Vα be givenby (2.8.3) and (2.8.1). Then Vα \ {0,∞} is the limit set for the family ofcontinued fractions K(an/1) from Eα.

Proof : Every convergent continued fraction K(an/1) from Eα has its value inVα \ {0,∞} (the Parabola Theorem). Let w0 ∈ Vα \ {0,∞} be arbitrarily chosen,and let w ∈ ∂Vα and μ ≥ 0 be the (uniquely determined) numbers which makew0 = w + μeiα. Let further

w1 := −1 − w + μ eiα, a1 := −w2 + μ2e2iα, a2 := −(1 + w)2 + μ2e2iα.

Then also w1 ∈ Vα \ {0,∞}. Moreover, −w2 ∈ ∂Eα since w can be written w =− 1

2 + it eiα (see Problem 19 on page 169). Similarly, −(1 + w)2 ∈ ∂Eα. Thereforea1, a2 ∈ Eα and the 2-periodic continued fraction

∞Kn=1

an

1:=

a1

1 +a2

1 +a1

1 +a2

1 +a1

1 +· · ·is a continued fraction from Eα. Therefore K(an/1) converges to some f ∈ Vα.Indeed, by Remark 1 on page 157 Sn(wn) → f whenever all wn ∈ Vα. Now,

a1

1 + w1=

−w2 + μ2e2iα

1 − 1 − w + μ eiα=

w2 − μ2e2iα

w − μ eiα= w + μ eiα = w0,

a2

1 + w0=

−(1 + w)2 + μ2e2iα

1 + w + μ eiα= −(1 + w) + μ eiα = w1.

That is, S2n(w0) = S2n+1(w1) = w0 for all n, and therefore K(an/1) converges tow0. �

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160 Chapter 3: Convergence criteria

3.3 Additional convergence theorems

3.3.1 Simple bounded circular value sets

The Sleszynski - Pringsheim Theorem required that V = D. This easily generalizesto cases where V is a disk B(Γ, ρ) with 0 ∈ V ◦; i.e., ρ > |Γ|.

Theorem 3.48. For given Γ ∈ C and ρ > 0 with ρ > |Γ|, let

Ω := {(a, b) ∈ C2; |a(b+Γ)−Γd|+ρ|a| ≤ ρd for d := |b+Γ|2−ρ2}. (3.1.1)

Then every continued fraction K(an/bn) from Ω converges to some valuef ∈ V := B(Γ, ρ).

Proof : Let a, b ∈ C with a �= 0. Then s(V ) := a/(b+V ) ⊆ V only if 0 �∈ (b+V );i.e., |b + Γ| > ρ. So assume that d > 0. Then s(V ) = B(Γ1, ρ1) given by

Γ1 :=a

d(b + Γ) and ρ1 :=

|a|d

ρ (Lemma 3.6 on page 110.) (3.1.2)

Hence s(V ) ⊆ V if and only if |Γ − Γ1| + ρ1 ≤ ρ; i.e., if and only if (a, b) ∈ Ω. Inother words, Ω is the element set corresponding to the value set V . This was firstproved by Lane ([Lane45]). Since moreover s(∞) = 0 ∈ V ◦ for all (a, b) ∈ Ω witha �= 0, the convergence follows from Corollary 3.9 on page 114. (See the remark onpage 113.) �

If we set Γ := 0 and ρ := 1, we get back the Sleszynski-Pringsheim criterion on page135.

If 0 �∈ V ◦, we still get general convergence under mild conditions:

Theorem 3.49. For given Γ ∈ C, and 0 < ρ �= |Γ| and 0 < ε < 1, let

Ω∗ := {(a, b) ∈ Ω; |a| ≤ (1 − ε)d} (3.1.3)

where Ω and d are given by (3.1.1). Then every continued fraction K(an/bn)from Ω∗ converges generally to some value f ∈ V := B(Γ, ρ) with excep-tional sequence {w†

n} from C \ V .

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3.3.1 Simple bounded circular value sets 161

Proof : With Γ1 and ρ1 as in the previous proof, the extra condition |a| ≤ (1−ε)dmakes ρ1 = |a|ρ/d ≤ (1 − ε)ρ, and thus (1.4.11) on page 114 holds with ϕn(w) :=Γ + ρw for K(an/bn) from Ω∗.

If 0 ∈ V ◦, the convergence follows from Theorem 3.48. If 0 �∈ V , the generalconvergence follows from Corollary 3.9 on page 114 with wn := ∞. Since V containsmore than two points, the value f of K(an/1) belongs to V . In particular f �= ∞,so {S−1

n (∞)} is an exceptional sequence. Finally, w†n := S−1

n (∞) ∈ C \ V sinceSn(w†

n) = ∞ �∈ V whereas Sn(V ) ⊆ V . �

For continued fractions of the form K(an/1), the element set Ω simplifies to

E := {a ∈ C; |a(1 + Γ) − Γd| + ρ|a| ≤ ρd for d := |1 + Γ|2 − ρ2}. (3.1.4)

If Γ = 0, then E is the circular disk |a| ≤ ρ(1 − ρ) where 0 < ρ < 1, a situation werecognize from the generalized version of the Worpitzky Theorem. If Γ �= 0, then

E = a∗E∗ where a∗ := Γ(1 + Γ)(1 − ρ2

|1 + Γ|2)

1 + Γd

and E∗ :={

ξ ∈ C; |ξ − 1| + ρ

|1 + Γ| |ξ| ≤ρ

|Γ|

}.

(3.1.5)

Clearly E∗ �= ∅ if and only if 1 ∈ E∗; i.e., if and only if |Γ| ≤ |1 + Γ|. And E◦ �= ∅if and only if |Γ| < |1 + Γ|. The boundary ∂E of E is then called a cartesian oval.

Lemma 3.50. The cartesian oval O := {ξ ∈ C; |ξ − 1| + |ξ|k = �} with0 < k < 1 and k < � is a closed convex curve, symmetric about the real axiswith O ∩ R = {μ, ν} given by

μ :=1 − �

1 − kif � ≤ 1, μ :=

1 − �

1 + kif � ≥ 1 and ν :=

� + 1k + 1

. (3.1.6)

Moreover, μ ≤ |ξ| ≤ ν for all ξ ∈ O.

Proof : Let t ∈ (0, 1) and ξ1, ξ2 ∈ O. Then ξ := tξ1 + (1 − t)ξ2 satisfies ξ − 1 =t(ξ1 − 1) + (1 − t)(ξ2 − 1), so

|ξ − 1| + |ξ|k ≤ t|ξ1 − 1| + (1 − t)|ξ2 − 1| + t|ξ1|k + (1 − t)|ξ2|k= t � + (1 − t)� = �,

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162 Chapter 3: Convergence criteria

and thus O is convex. The symmetry follows since ξ ∈ O implies that its complexconjugate ξ also is an element in O. For ξ ∈ O ∩ R,

ξ ≥ 1 =⇒ ξ − 1 + ξk = � =⇒ ξ =� + 1k + 1

,

0 < ξ < 1 =⇒ 1 − ξ + ξk = � =⇒ ξ =1 − �

1 − k=⇒ � < 1,

ξ ≤ 0 =⇒ 1 − ξ − ξk = � =⇒ ξ =1 − �

1 + k=⇒ � ≥ 1.

That is, O ∩ R = {μ, ν}. Finally, if ξ ∈ O, then

� = |ξ − 1| + |ξ|k ≤ |ξ| + 1 + |ξ|k =⇒ |ξ| ≥ (� − 1)/(k + 1),� = |ξ − 1| + |ξ|k ≥ |ξ| − 1 + |ξ|k =⇒ |ξ| ≤ (� + 1)/(k + 1),� = |ξ − 1| + |ξ|k ≥ 1 − |ξ| + |ξ|k =⇒ |ξ| ≥ (1 − �)/(1 − k)

which proves that μ ≤ |ξ| ≤ ν. �

Hence E is a convex set, symmetric about the line a∗R. For more information onthe shape of E we refer to Section 5.2.4 on page 244 in Chapter 5.

Theorem 3.51. (The Oval Theorem.) Let E be given by (3.1.4) with|Γ| < |1 + Γ| and |1 + Γ| > ρ. Then every continued fraction K(an/1) fromE converges to some value f ∈ V := B(Γ, ρ) with exceptional sequence from−1 − V , and V is a simple value set for K(an/1).

Proof : Let K(an/1) be a continued fraction from E. Assume first that |a| = d.Then E consists of the single point a = a∗. But this is impossible since we haveassumed that |Γ| �= |1+Γ|. Since E is a compact set, there therefore exists an ε > 0such that |a| ≤ (1 − ε)d for all a ∈ E. Hence it follows from Theorem 3.49 thatK(an/1) converges generally to some value f ∈ V := B(Γ, ρ). It has an exceptionalsequence {S−1

n (−1 − w)} for some w ∈ V with −1 − w �= f , where

S−1n (−1 − w) = −1 − an

1 +an−1

1 + · +a1

1 + w∈ −1 − V for all n.

Since V is bounded, this means that also Sn(∞) = Sn−1(0) → f . �

Truncation error bounds.

As a corollary to Theorem 5.9 which we prove on page 241, we get the followingtruncation error bound:

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3.3.2 Simple unbounded circular value sets 163�

Theorem 3.52. Let Ω be given by (3.1.1) with Γ ∈ C and 0 < ρ �= |Γ|.Then

|f − Sn(Γ)| ≤ ρ|Γ| + ρ

|bn + Γ| − ρ

n−1∏k=1

Mk (3.1.7)

for every convergent continued fraction K(an/bn) from Ω, where Mk :=max{|w/(bk + w)|; w ∈ V }.

For continued fractions K(an/1) from E the error bound simplifies:�

Theorem 3.53. Let E be given by (3.1.4) with |1 + Γ| > |Γ| �= ρ. Then

|f − Sn(Γ)| ≤ ρ|Γ| + ρ

|1 + Γ| − ρMn−1 (3.1.8)

for every continued fraction K(an/1) from E, where M := max{|w/(1 +w)|; w ∈ V }.

Remarks.

1. Since s(w) := w/(1 + w) maps B(Γ, ρ) onto B(Γ1, ρ1) given by

Γ1 := 1 − 1 + Γ|1 + Γ|2 − ρ2

=Γ(1 + Γ) − ρ2

|1 + Γ|2 − ρ2, ρ1 :=

ρ

|1 + Γ|2 − ρ2(3.1.9)

(Lemma 3.6 on page 110), it follows that

M = |Γ1| + ρ1 =|Γ(1 + Γ) − ρ2| + ρ

|1 + Γ|2 − ρ2. (3.1.10)

2. The convergence of K(an/1) in the Oval Theorem follows from the ParabolaTheorem since Lange ([Lange95]) proved that E is a subset of the parabolicregion Eα with α := arg(Γ + 1

2 ). The importance of the Oval Theorem liestherefore with its better truncation error bound for this smaller family ofcontinued fractions K(an/1). We shall return to this idea in Chapter 5.

3.3.2 Simple unbounded circular value sets

An unbounded, closed circular set V is either a half plane or the exterior of a disk. Inparticular ∞ ∈ V , so if V is a simple value set for K(an/bn), then an/(bn + ∞) =

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164 Chapter 3: Convergence criteria

0 ∈ V . Hence we are back to the classical type of value sets which contain theclassical approximants.

If we stick to continued fractions of the form K(an/1) or K(1/bn), we have thefollowing interesting observation:

Theorem 3.54.A. Let V be a simple value set for K(an/1). Then W := C \ (−1 − V ) andW ∩ V are also simple value sets for K(an/1).B. Let V be a simple value set for K(1/bn). Then W := C \ (−1/V ) andW ∩ V are also simple value sets for K(1/bn).

Proof : A. Since sn(V ) := an/(1 + V ) ⊆ V , we have V ⊆ s−1n (V ) = −1 + an/V ;

i.e., (−1−V ) ⊆ an/(−V ) = sn(−1−V ), and thus W ⊇ sn(W ). Clearly, sn(V ) ⊆ Vand sn(W ) ⊆ W implies that sn(V ∩ W ) ⊆ V ∩ W .

B. Since sn(V ) := 1/(bn + V ) ⊆ V , we have (bn + V ) ⊆ 1/V , and thus −V ⊆bn − 1/V , which means that

− 1V

⊆ 1bn − 1/V

= sn(−1/V )

and the result follows. �

From this follows immediately:

1. If V := B(Γ, μ) with μ < 0 is a simple value set for K(an/1), then also thedisk B(−1−Γ, |μ|) is a simple value set for K(an/1), and K(an/1) convergesby virtue of the Oval Theorem.

2. If V := B(Γ, μ) with μ < 0 is a simple value set for K(1/bn), then also−1/V is the exterior of a disk since ∞ ∈ V ◦ implies that 0 ∈ V ◦. ThereforeW := C \ (−1/V ) is a circular disk, and we can apply Theorem 3.49.

3. If V := w0 + eiαH is a simple value set for K(an/1), then ∞ ∈ ∂V , and thus

0 ∈ V is necessary. With the extra condition 0 ∈ V ◦ and −1 �∈ V , we areback in the Parabola Theorem.

4. If V := w0 + eiαH is a simple value set for K(1/bn), then 0 ∈ V is still

necessary. With the extra condition 0 ∈ V ◦, it follows that (−1/V ) is theexterior of a disk, and thus W := C \ (−1/V ) is a disk, and we can againapply Theorem 3.49.

Page 178: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Remarks 165

3.4 Remarks

1. Lemma 3.7. This crucial lemma is inspired by the work of W.J.Thron. In 1965 heand his student K.L.Hillam ([HiTh65]) published the result that {Tn(0)} converges

if there exist two points, w1 ∈ D and w2 ∈ C \ D, such that τn(w2) = w1 for all n.The present version dates from 1994 ([Lore94a]) where it also was proved that theconclusions still hold if we replace (1.4.6) by lim inf rad(τ−1

n (D)) > 1. (If τ−1n (D)

is unbounded, we set rad τ−1n (D) := ∞.) Also Lemma 3.8 still holds under this

condition. Lemma 3.7 has later been further generalized in [Lore07].

2. Truncation error bounds. The demand for truncation error estimates becamemore prominent in the 1960s with the growing use of computers. W. J. Thron([Thron58]) realized that value sets could also be used for the purpose of deriv-ing such estimates. This started a series of useful publications in this area, suchas [HePf66], [JoTh76], [CrJT94], [BaJo85], [FiJo72] and [CJPVW7]. For furtherreferences we refer to ([JoTh76], p 298).

3. The parabola theorems. The first parabola theorem was published by Scott andWall in 1940, ([ScWa40]). It was valid for the special case α = 0. It was proved byexploiting what they called the fundamental inequalities. The result was generalizedalmost immediately ([PaWa42] and [LeTh42]). The most general parabola theoremis due to Jones and Thron, ([JoTh68]) who also proved convergence for cases ofcontinued fractions K(an/bn) with half planes as value sets.

4. Classical convergence theorems. The Worpitzky Theorem was proved already in1865, but remained unknown to workers in the field until Pringsheim rediscovered itmore than 30 years later. It was not until 1905, through Van Vleck, that Worpitzkygot the credit he deserved. Part of the reason may be the way it was published,(in an annual report from the school where Worpitzky was teaching, [Worp65]), butthere may be other more significant reasons, see [JaTW89]. Beardon ([Bear01a])has given a generalization of Worpitzky’s Theorem.

Theorem 3.25 on page 129 usually carries the name of Pringsheim only. However,as pointed out to us by W. J. Thron, J. Sleszynski is the right one to credit, sincehe already proved the theorem in 1888, see [Sles89].

5. Boundary versions of convergence theorems. The limit regions in Worpitzky’sTheorem and the Parabola Theorem are created by using the whole element set inboth cases. A natural question to ask is: what happens to the limit regions if werestrict the elements to the boundary of the element sets? The answer is given in[Waad89]: the Worpitzky limit set is reduced to the annulus 1

6≤ |f | ≤ 1

2whereas

the half plane minus {0,∞} remains unchanged in the parabola case.

In [Waad92] an investigation of the Oval Theorem is carried out. The process andthe result are somewhat more complicated, but the limit region is a circular disk witha hole (not centered). Proper limits of parameters make the oval region approach theparabolic region of the Parabola Theorem and the corresponding limit sets approachthe half plane limit set in the parabola case.

A probabilistic aspect also turned up in the Worpitzky case. When we computedvalues (or rather high order approximants) of continued fractions from the boundaryof the Worpitzky disk, they all seemed to lie in a more narrow annulus that expectedfrom the theoretic results. The reason for this turned out to be that the probabilitydensity for the values is very small close to the boundary of the annulus, ([Waad98]).

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166 Chapter 3: Convergence criteria

6. More convergence theorems. It is clear that with the help of simple valuesets one can produce large numbers of convergence theorems. This has been doneby a number of authors, in particular by Jones and Thron ([Thron48], [Thron74],[JoTh68]). Let us also mention Lange’s strip convergence regions ([Lange94]) andOverholts polygons ([Over82]) which are based on non-circular value sets, and thegeneral work by Cordova Yevenes ([Cord92]).

7. Twin convergence theorems. The situation where {Vn} is a periodic sequenceof value sets has also been studied in detail. In particular the 2-periodic case is veryinteresting. Here Thron has given a large number of convergence results based oncircular value sets, in later years in collaboration with Jones and Lange, ([Thron43],[Thron49], [Thron59], [LaTh60], [Lange66], [JoTh68], [JoTh70]). In [Lore08a] thecases where 0 �∈ Vn are also included.

3.5 Problems

1. Convergence to ∞. Prove that if the continued fraction K(an/bn) with b2n−1 = 0for all n converges, then it converges to ∞. (Broman [Brom77].)

2. Convergence of continued fractions. Determine whether K(an/bn) convergesor diverges and whether its even and odd parts converge or diverge in each of thefollowing cases.

(a) All an = 1 and bn = z/n2 for z ∈ C.

(b) All an = z/n2 and bn = 1 for z ∈ C \ {0}.(c) All an = zn2 and bn = 1 for z ∈ C \ R.

3. Convergence of positive continued fractions. Prove that the two positivecontinued fractions K(cn/1) and K(dn/1) given by

c1 :=1

3, c2 := 2, cn :=

(n − 1)2(2n − 1)

2n − 3for n ≥ 3,

d1 :=2

3, dn :=

n2(2n − 1)

2n + 1for n ≥ 2

converge to finite values. (Hint: Theorem 3.2(i) on page 102 and the Seidel-SternTheorem on page 117 may be of help.)

4. Convergence of continued fractions. Determine whether K(1/bn) converges ordiverges when

(a) bn := (−1)nn. (b) bn := 1 + (−1)n. (c) bn := in.

(d) bn := 1/n. (e) bn := (2i)n. (f) bn = (sin n)/n2 .

5. Convergence of continued fractions. Determine whether K(an/1) converges ordiverges when

(a) an := (−1)nn. (b) an := 2 + (−1)n. (c) an := (−2)n.

(d) an := 1/n. (e) an := 1/n2. (f) an = (sin n)/n2 .

Page 180: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Problems 167

6. Mapping of disks. Find the center and radius of the disk

(a) 3/(1 + 2D).

(b) τ(B(1,−3)) where τ(w) :=w + 2

w − 2.

7. Positive continued fraction. Given the continued fraction

b +

Kn=1

a

2b= b +

a

2b+

a

2b+

a

2b+· · ·

Prove that if a > 0 and b > 0 then b + K(a/2b)converges to√

a + b2. Use this tofind a rational approximation to

√13 with an error less than 10−4.

8. Positive continued fractions. Let b > 0 and p > 0. Prove that K(np/b) convergesif and only if p ≤ 2. (Perron [Perr57], p 48.)

9. Oscillations of approximants. The continued fraction

Kn=1

an

1=

1

1+

1/2

1 +

1/(2 · 3)

1 +

2/(2 · 3)

1 +

2/(2 · 5)

1 +

3/(2 · 5)

1 +· · ·

+

n/(2(2n − 1))

1 +

n/(2(2n + 1))

1 +· · ·

converges to Ln(2). Suggest three sequences {Sn(wn)} of approximants convergingto Ln(2). Does any of these sequences have an oscillating character which can beused to obtain upper bounds for the truncation error |Ln(2) − Sn(wn)|?

10. Truncation error bounds. Let K(1/bn) be the continued fraction where bn =4 + (0.9)n for all n.

(a) Find a connected value set V for K(1/bn). (Try to make V small.)

(b) Does K(1/bn) converge to a finite value f?

(c) Are the classical approximants fn of K(1/bn) all distinct; i.e. fn �= fm ifn �= m?

(d) Use the value set V found in (a) to derive upper bounds for the truncationerror |f − Sn(w)| for suitably chosen w ∈ C.

11. Truncation error bounds. Use Theorem 3.46 on page 157 to derive a prioritruncation error bounds for

Ln(1 + i) =

Kn=1

ani

1=

i

1+

i/2

1 +

i/(2 · 3)

1 +

2i/(2 · 3)

1 +

2i/(2 · 5)

1 +· · ·

where a2n = n/(2(2n− 1)) and a2n+1 = n/(2(2n + 1)) for all n ≥ 1. Compare thesewith the Gragg-Warner truncation error bounds in Theorem 3.24 on page 128.

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168 Chapter 3: Convergence criteria

12. ♠ Alternating continued fraction. Let K(an/1) have real elements an such that(−1)nan > 0 and

|a2n−1| < 1 + a2n, |a2n+1| < 1 + a2n for all n .

Prove that {S4n+p(0)}∞n=1 converges for p = 1, 2, 3 and 4.

13. Oscillating approximants. Suggest expressions for wn such that the sequenceSn(wn) of approximants for

Kn=1

an

1=

12

1 +

5 · 22

1 +

32

1 +

5 · 42

1 +

52

1 +

5 · 62

1 +

72

1 +· · ·(hopefully) converges faster to the value of K(an/1) than Sn(0). Compute the first6 approximants of Sn(0) and Sn(wn). Use the oscillating character to determine anerror bound for S6(0), and if possible for S6(w6).

14. Convergence of continued fractions. Let α be a positive number. For whichvalues of α does the continued fraction K∞

n=1(1/n−α) converge, and for which valuesdoes it diverge? (Hint: Use Van Vleck’s Theorem.)

15. ♠ Convergence neighborhoods. Use the Parabola Theorem to prove:

(a) For given a ∈ C with |a| < 14, every continued fraction K(an/1) from B(a, |a+

14|) converges.

(b) For given a := r e2iα with r ≥ 14

and −π2

< α < π2, every continued fraction

K(an/1) from B(a,√

r/2 cos α) converges. (Jones and Thron [JoTh80], p108.)

16. From Van Vleck to Stieltjes. Prove the following implications:

(a) Van Vleck’s Theorem on page 142 =⇒ Theorem 3.21 on page 124.

(b) Theorem 3.39 on page 145 =⇒ the Henrici-Pfluger Bounds on page 126.

17. Convergence of even and odd parts. Prove that if the even and odd parts ofK(an/1) with an ∈ C \ {0} converge to distinct values, then an → ∞. (Bowmanand McLaughlin [BoML04].)

18. The Cardioid Theorem. Use the Parabola Sequence Theorem to prove that if

|an| − Re(an) ≤ k gn−1(1 − gn) for n = 1, 2, . . .

and ∞∑n=1

n∏k=1

|ak|n−k+1 = ∞,

then K(anz/1) converges locally uniformly to a holomorphic function in the cardioiddomain

D := {z = r e2iθ ∈ C; r < 1k

cos θ}.(Jones and Thron [JoTh80], p 134.)

Page 182: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Problems 169

19. ♠ The Parabola Theorem. Let Vα and Eα be as in the Parabola Theorem. Provethat

(a) a ∈ Eα if and only if a = c2 for some c ∈ C with |Im(c e−iα)| ≤ 12

cos α.

(b) a ∈ Eα if and only if a = −c2 for some c ∈ C with |Re(c e−iα)| ≤ 12

cos α.

20. ♠ A convergence theorem. Let {ρn} be a sequence of positive numbers. Provethat

(a) {Gn}∞n=1 given by

Gn := B(0,−ρn − 1/ρn−1) = {b ∈ C; |b| ≥ ρn + 1/ρn−1}

is the sequence of element sets for continued fractions K(1/bn) correspondingto the value sets Vn := B(0, ρn) for n ≥ 0. (Jones and Thron [JoTh80], p 75.)

(b) {−ρn} is a tail sequence for K(−1/bn) with bn := ρn + 1/ρn−1.

(c) K(−1/bn) in (b) converges. What is its value?

(d) K(−1/bn) with

bn :=

√b + n + 1

b + n+

√b + n − 1

b + n

for a fixed constant 0 ≤ b ≤ 1 converges to −√

1 + 1/b.

(e) Prove that

diamSn(Vn) ≤ 2ρ0

n∏j=2

ρjρj−1(1 + 3ρj−1ρj−2)

2ρjρ2j−1ρj−2 + 2ρjρj−1 − ρj−1ρj−2 + 1

for every K(1/bn) from {Gn}. (Craviotto et al [CrJT94].)

21. ♠ Comment to the Stern-Stolz Theorem. Let K(1/bn) be given by

b1 := 1, b2n+1 := 2(−1)n/√

n + 1 and b2n := (−1)n−1/(√

n + 1 +√

n).

(a) Prove that∑

bn converges to a finite value.

(b) Prove that K(1/bn) converges (and thus that convergence of∑

bn to a finitevalue is not sufficient for divergence in general).

22. ♠ An extension of Thron-Lange’s Theorem. For given γ ∈ C, let r :=√1 + |γ|2, V := B(γ, r) and G := B(−2γ,−2r). Let K(1/bn) be a continued

fraction from G.

(a) Let ϕ(w) := (w − γ)/r. Prove that τn := ϕ ◦ sn ◦ ϕ−1 maps D into D withτn(∞) = −γ/r ∈ D.

(b) Use Lemma 3.8 on page 113 to prove that {Tn}∞n=1 given by Tn := τ1◦τ2◦· · ·◦τn

converges generally.

(c) Prove that K(1/bn) converges (in the classical sense).

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170 Chapter 3: Convergence criteria

23. ♠ Element sets. Show that the sequence {En}∞n=1 of element sets (possibly empty)for continued fractions K(an/1) corresponding to a given sequence {Vn}∞n=0 of valuesets, is given by

En = ∩{(1 + w)Vn−1; w ∈ Vn \ {−1,∞}}.(Jones and Thron, [JoTh80] p 77.)

24. ♠ Element sets. Show that the sequence {Gn}∞n=1 of element sets (possibly empty)for continued fractions K(1/bn) corresponding to a given sequence {Vn}∞n=0 of valuesets, is given by

Gn = ∩{−w + 1/Vn−1; w ∈ Vn \ {∞}}.(Jones and Thron, [JoTh80] p 78.)

25. Location of the value of a continued fraction. Use Worpitzky’s Theorem toprove the following:

(a) The value of any continued fraction

1/4

1 +

−1/4

1 +

a3

1 +

a4

1 +· · ·,

with |an| ≤ 1/4 for all n must lie in the disk∣∣∣∣w − 2

5

∣∣∣∣ ≤ 1

10.

(b) The value of any continued fraction

i/4

1 +

a2

1 +

a3

1 +· · ·,

with |an| ≤ 1/4 for all n must lie in the disk∣∣∣∣w − i

3

∣∣∣∣ ≤ 1

6.

(c) Show that the disk in (b) is a limit set for the family of continued fractionsfrom (a).

26. ♠ The Parabola Theorem. Prove that Eα in the Parabola Theorem on page 151also is given by

Eα :=

{r ei(θ+2α); 0 ≤ r ≤

12

cos2 α

1 − cos θ

}.

Page 184: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Chapter 4

Periodic and limit periodiccontinued fractions

Periodic continued fractions are well understood. We can determine when they con-verge, when they diverge, their values if they converge and the asymptotic behaviorof their tail sequences. Indeed, it is all a matter of iterations of linear fractionaltransformations, and we have closed expressions for their approximants Sn(w).

Only few continued fraction expansions of interesting functions are periodic, butlarge numbers are limit periodic, where the tail of K(an/bn) looks more and morelike a periodic continued fraction the further out it starts. Then K(an/bn) essen-tially inherits the asymptotic properties from the periodic one in most cases. Thismeans that also limit periodic continued fractions are reasonably well understood.

Equivalence transformations are sometimes useful to bring a continued fraction toa wanted form. Naturally, there is no point in transforming a given continuedfraction to some K(an/bn) with lim an = lim bn = 0 or with lim an = lim bn = ∞.But sometimes one has to live with outcomes like lim an = ∞, lim bn = b ∈ C \ {0}or lim an = a ∈ C \ {0}, lim bn = 0. Also some of these continued fractions areanalyzed in this chapter.

L. Lorentzen and H. Waadeland, Continued Fractions, Atlantis Studies in Mathematics for Engineering and Science 1, DOI 10.1007/978-94-91216-37-4_4, © 2008 Atlantis Press/World Scientific

171

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172 Chapter 4: Periodic and limit periodic continued fractions

4.1 Periodic continued fractions

4.1.1 Introduction

A continued fraction K(an/bn) is called strictly periodic with period length p ∈ N,or just p-periodic or periodic for short, if the sequences {an}∞n=1 and {bn}∞n=1 arep-periodic; i.e. if

an+p = an, bn+p = bn for all n ∈ N , (1.1.1)

and p is the smallest positive integer for which this holds. For such continuedfractions the approximants can be written

Snp+m(w) = S[n]p ◦ Sm(w) = Sm ◦ (S(m)

p )[n](w) (1.1.2)

where F [n] := F ◦ F ◦ · · · ◦ F is the nth iterate of a function F , and

S(m)p (w) := S−1

m ◦ Sp ◦ Sm(w) =am+1

bm+1 +am+2

bm+2 +· · ·+am+p

bm+p + w. (1.1.3)

The convergence behavior of K(an/bn) therefore depends on how sequences of it-erates of linear fractional transformations S

(m)p behave asymptotically. For conve-

nience we also use the notation

S0(w) := w and S(0)m := Sm(w). (1.1.4)

4.1.2 Iterations of linear fractional transformations

We consider linear fractional transformations τ ∈ M; i.e.,

τ(w) =aw + b

cw + d, a, b, c, d ∈ C with Δ := ad − bc �= 0. (1.2.1)

If {τ [n]} converges generally to some x ∈ C, then x must be a fixed point for τ ; i.e.τ(x) = x. Unless τ is the identity transformation I(w) ≡ w; i.e. a = d �= 0, b = c =0, τ has only two (possibly coinciding) fixed points x and y. From (1.2.1) we seethat ∞ is a fixed point for τ if and only if c = 0. For c �= 0 and a + d �= 0, the fixedpoints are

x, y =a − d ±

√(a − d)2 + 4bc

2c=

a − d ±√

(a + d)2 − 4Δ2c

=a − d ± (a + d)u

2cwhere u :=

√1 − 4Δ/(a + d)2

(1.2.2)

(with Re√. . . ≥ 0 as always), and for c �= 0 and a = −d

x, y =

{ac (1 ±

√−Δ/a2) if a �= 0,

±√

b/c if a = 0.(1.2.3)

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4.1.2 Iterations of linear fractional transformations 173

Thereby we obtain the standard identity

(cx + d)(cy + d) = Δ if c �= 0. (1.2.4)

If c = 0 and τ �= I, then the fixed points are at ∞ and b/(d − a).

Case 1: τ has only one fixed point.

Let first c �= 0. Then it follows from (1.2.2) that

(a + d)2 = 4Δ and x = y = (a − d)/2c.

In particular a + d �= 0. In this case

1τ(w) − x

=1

τ(w) − τ(x)=

(cx + d)(cw + d)(ad − bc)(w − x)

=cx + d

ad − bc

(c +

cx + d

w − x

)=

2c

a + d+

1w − x

.

(1.2.5)

That is, ϕ ◦ τ(w) = q + ϕ(w) where q := 2c/(a + d) and ϕ(w) := 1w−x

, and so

τ(w) := ϕ ◦ τ ◦ ϕ−1(w) = q + w. (1.2.6)

Next, let c = 0. Then x = y = ∞ and τ(w) := τ(w) = w+ b/d already has the formq + w, this time with q := b/d. The advantage of this form is that iterations of τjust gives τ [n](w) = w + nq, and thus we know how iterations of τ = ϕ−1 ◦ τ ◦ ϕbehave. It is also clear that τ [n] = ϕ−1 ◦ τ [n] ◦ ϕ, so the following result can bestated:�

Theorem 4.1. Let τ given by (1.2.1) have only one fixed point x. Then

τ [n](w) =

{x + w−x

nq(w−x)+1where x = a−d

2c, q := 2c

a+dif c �= 0

nq + w where q := b/d if c = 0.

Case 2: τ has exactly two distinct fixed points.

Let first c �= 0. Then x and y are finite, and

τ(w) − x

τ(w) − y=

τ(w) − τ(x)τ(w) − τ(y)

=cy + d

cx + d· w − x

w − y(1.2.7)

where cx + d �= 0 and cy + d �= 0 by (1.2.4). That is, ϕ ◦ τ(w) = � · ϕ(w) where

ϕ(w) :=w − x

w − yand � := cy+d

cx+d , so

τ(w) := ϕ ◦ τ ◦ ϕ−1(w) = �w. (1.2.8)

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174 Chapter 4: Periodic and limit periodic continued fractions

Next, let c = 0. Then ad �= 0 since Δ �= 0. Moreover, τ has the two distinct fixedpoints ∞ and b/(d − a), where a �= d. Since τ(w) = (a/d)w + (b/d), we have

τ(w) − b

d − a=

a

dw +

b

d− b

d − a=

a

d

(w − b

d − a

)ϕ ◦ τ = � · ϕ(w) for � :=

a

dand ϕ(w) := w − b

d − a,

(1.2.9)

so τ(w) := ϕ ◦ τ ◦ ϕ−1(w) = �w, and thus τ [n](w) = �nw, which tells the wholestory about iterations of τ = ϕ−1 ◦ τ ◦ ϕ:

Theorem 4.2. Let τ given by (1.2.1) have exactly two fixed points x �= y.Then

τ [n](w) =

⎧⎨⎩(x −�ny)w − xy(1 −�n)

(1 −�n)w + �nx − yfor � :=

cy + d

cx + dif c �= 0,

(ad)nw + (1 − (a

d)n) b

d−aif c = 0.

4.1.3 Classification of linear fractional transformations

We classify τ ∈ M according to the behavior of iterations τ [n] as n → ∞. A linearfractional transformation τ is conjugate (or similar) to τ if there exists a linearfractional transformation ϕ such that τ = ϕ ◦ τ ◦ ϕ−1. As seen in the previoussection, we may therefore look at the conjugates τ(w) = �w or τ(w) = w + q of τ ,or equivalently, use Theorem 4.2 or Theorem 4.1.

Definition 4.1. For τ given by (1.2.1) with fixed points x and y (coincidingif necessary), the ratio � := �(τ) of τ is a complex number with 0 < |�| ≤ 1given by

� :=

⎧⎪⎨⎪⎩(cy + d)/(cx + d) if c �= 0,

a/d if c = 0 and |a| ≤ |d|,d/a if c = 0 and |a| > |d|.

(1.3.1)

(If |cy + d| > |cx + d|, we just interchange the names of the two fixed points.)Then � is uniquely defined for given τ , except if � �= 1 with |�| = 1. This lack ofuniqueness will not be a problem for our applications. With u as given by (1.2.2),

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4.1.3 Classification of linear fractional transformations 175

� can be written

� =

⎧⎨⎩1 − u

1 + uif c �= 0, a + d �= 0,

−1 if c �= 0, a + d = 0.(1.3.2)

This ratio determines to a large degree the asymptotic behavior of {τ [n]}:

• τ ∈ M is a loxodromic transformation if |�| < 1. Then τ has exactly twodistinct fixed points x, y ∈ C, and τ is conjugate to τ(w) := �w which hasfixed points at 0 and ∞. Since τ [n](w) = �nw → 0 for w �= ∞, the sequence{τ [n]} converges generally to x with exceptional sequence {y}∞n=1. We saythat x is the attracting fixed point and y is the repelling fixed point for τ .

• τ ∈ M is an elliptic transformation if |�| = 1 with � �= 1. Also now τ hasexactly two distinct fixed points x, y ∈ C, and τ is conjugate to τ(w) = �w,but {τ [n]} is totally non-restrained and diverges for every w �= 0,∞. Therefore{τ [n]} is totally non-restrained, and {τ [n](w)} diverges for every w �= x, y. Wesay that x and y are indifferent fixed points for τ .

• τ ∈ M is the identity transformation if τ(w) = I(w) ≡ w. Then every pointin C is a fixed point for τ , and � = 1. τ = I is only conjugate to itself, andτ [n](w) = w.

• τ ∈ M is a parabolic transformation if τ �= I and � = 1. Then τ has only onefixed point x = y ∈ C, and τ is conjugate to τ(w) := w + q with q �= 0. Sinceτ [n](w) = w + nq → ∞ for every fixed w ∈ C, it follows that τ [n](w) → x forevery w ∈ C. Also in this case we say that x is an attracting fixed point.

Remarks:

1. This classification is invariant under conjugation.

2. This classification is invariant under inversion since w is a fixed point for τ ifand only if w is a fixed point for τ−1, and straight forward computation showsthat � is invariant under inversion. The roles of x and y must be interchanged,though. (See Remark 4 below.)

3. This classification is essentially invariant under iteration. This follows sincefor τ with ratio � and fixed points x, y, τ [n] has ratio �n and fixed pointsx, y. The only exception occurs when τ is elliptic with ratio �, where �N = 1for some N ∈ N. Then τ [N ] is the identity transformation. Indeed, τ [nN ] = Ifor every n ∈ N.

4. If τ is loxodromic with attracting fixed point x and repelling fixed point y,then τ−1 is loxodromic with attracting fixed point y and repelling fixed pointx.

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176 Chapter 4: Periodic and limit periodic continued fractions

5. If w �= ∞ is a fixed point for τ , then the derivative τ ′(w) = � if w is theattracting fixed point and τ ′(w) = 1/� if w is the repelling fixed point for τ(Problem 2 on page 212).

Example 1. The linear fractional transformation

τ(w) :=4w + 32w + 5

has fixed points 1 and −3/2. Since |2 · 1 + 5| = 7 and |2 · (−3/2) + 5| = 2, we setx := 1, y := −3

2, and the ratio for τ is � = 2/7. Hence {τ [n]} converges generally

to 1 with exceptional sequence {−32}. Indeed, by Theorem 4.2,

τ [n](w) =(1 + 3

2 ( 27)n)w + 3

2(1 − ( 27 )n)

(1 − ( 27)n)w + (2

7)n + 3

2

→ 1 for w �= −32

= y.

Even though τ [n] has ratio �n when τ has ratio �, the ratio is not multiplicative ingeneral:

Example 2. The transformation s1(w) := (−1/4)/(1 + w) is parabolic with fixedpoint x = −1

2 and ratio 1, and the transformation s2(w) := 2/(1+w) is loxodromicwith fixed points 1 and −2 and ratio −1/2. Their composition is

s1 ◦ s2(w) =−1/4

1 + 2/(1 + w)= −1

4· w + 1w + 3

which is loxodromic with fixed points x, y = 18(−13 ±

√153) and ratio � = (3 +

18(−13 +

√153))/(3 + 1

8(−13 −

√153)). �

4.1.4 General convergence of periodic continued fractions

We say that a p-periodic continued fraction K(an/bn) is of parabolic or loxodromicor elliptic or identity type according to the classification of Sp. We further say that� is the ratio for the p-periodic continued fraction if � is the ratio for Sp. Theconvergence properties for iterations of Sp are described in the previous section. By(1.1.3) we see that S

(m)p is conjugate to Sp, and that x (or y) is a fixed point for Sp

if and only ifx(m) := S−1

m (x) (or y(m) := S−1m (y)) (1.4.1)

is a fixed point for S(m)p .

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4.1.4 Convergence of periodic continued fractions 177

If Sp is loxodromic, then x and x(m) shall always denote the attracting fixed pointof Sp and S

(m)p respectively, and y and y(m) the repelling ones. By (1.1.2) we then

have:

Theorem 4.3. Let K(an/bn) be a p-periodic continued fraction.

A. If Sp is parabolic or loxodromic, then K(an/bn) converges generally tothe attracting fixed point of Sp.

B. If Sp is elliptic or the identity transformation, then {Sn} is totallynon-restrained, and K(an/bn) diverges generally.

Example 3. Straight forward checking shows that the 1-periodic continued fractionK(a/b) with a, b ∈ C, a �= 0 is of

• elliptic type if b = 0 or ab2 < − 1

4 ,

• parabolic type if ab2

= −14

,

• loxodromic type otherwise.

Hence K(a/b) converges generally if and only if b �= 0 and (a/b2) lies in the cutplane C \ (−∞,−1

4 ). Its value is then

x = b2(−1 + u) where u :=

√1 + 4a/b2 (with Re u ≥ 0). (1.4.2)

If b = 0, then � = −1, and {Sn} is the 2-periodic sequence aw

, w, aw

, w, . . . of linearfractional transformations. �

The picture does not change much if we consider tail sequences {S−1n (t0)} instead

of approximants {Sn(w)}. Our classification is independent of inversion, and

tnp+m = S−1np+m(t0) = (S(m)−1

p )[n](S−1m (t0)) = (S(m)−1

p )[n](tm). (1.4.3)

Hence {tnp+m}n is the sequence of iterates of S(m)−1

p evaluated at tm.

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178 Chapter 4: Periodic and limit periodic continued fractions�

Theorem 4.4. Let K(an/bn) be a p-periodic continued fraction, and let{tn} ⊆ C be a tail sequence for K(an/bn).

A. If Sp is parabolic with fixed point x, then

limn→∞ tpn+m = S−1

m (x) =: x(m) for 0 ≤ m ≤ p − 1. (1.4.4)

B. If Sp is loxodromic with attracting fixed point x and repelling fixedpoint y, and t0 �= x, then

limn→∞ tpn+m = S−1

m (y) =: y(m) for 0 ≤ m ≤ p − 1. (1.4.5)

Of course, if t0 = x, then the p-periodic sequence {x(m)}∞m=0 is the sequence of tailvalues for K(an/bn).

Example 4. We consider the 1-periodic continued fraction K(a/b) from Example3. In the parabolic case, K(a/b) has one and only one periodic tail sequence {x}∞n=0,where x = − b

2 is the value of K(a/b). Indeed, {− b2}∞n=0 is its sequence of tail values,

and every tail sequence for K(a/b) converges to − b2 .

In the loxodromic case K(a/b) has two periodic tail sequences, {x}∞n=0 and {y}∞n=0

where x and y are the attracting and the repelling fixed point for S1(w) = s1(w) =a/(b+w); that is, x = b(−1+u)/2 and y = b(−1−u)/2, where u is given by (1.4.2).Hence {x}∞n=0 is the sequence of tail values for K(a/b), and tn → y for every othertail sequence {tn} for K(a/b). In particular {y}∞n=0 is an exceptional sequence. �

Example 5. The 4-periodic continued fraction

∞Kn=1

an

1:=

11 +

11 −

31 −

21 +

11 +

11 −

31 −

21 +

11 + · · · (1.4.6)

converges generally to the attracting fixed point x = 1 of the loxodromic transfor-mation

S4(w) =11 +

11 −

31 −

21 + w

=4 + 2w

5 + w.

The repelling fixed point of S4 is y = −4, so

y(0) = −4, y(1) = s−11 (−4) = −5

4 ,

y(2) = s−12 (−5

4) = − 95 , y(3) = s−1

3 (−95 ) = 2

3 ,

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4.1.5 Convergence in the classical sense 179

and y(4) = s−14 (y(3)) = −4 = y(0), and so on. Therefore, every tail sequence {tn}

with t0 �= 1 satisfies

limn→∞ t4n+m =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−4 if m = 0 ,

−54 if m = 1 ,

−95

if m = 2 ,

23 if m = 3 .

4.1.5 Convergence in the classical sense

The parabolic case.

Let K(an/bn) be p-periodic of parabolic type. Iterations {τ [n]} of a parabolictransformation τ converge to the fixed point for τ for every w ∈ C. Thereforelimn→∞ S

(m)np (w) = x(m) for every w ∈ C and m ∈ {0, 1, . . . , p − 1}, and thus

limSn(w) = x for every w ∈ C. In particular Sn(0) → x. Therefore:

Theorem 4.5. A periodic continued fraction of parabolic type converges tox in the classical sense and Sn(w) → x for every w ∈ C.

Still, {Sn(w)} does not converge uniformly to x in C, not even with respect tothe chordal metric. We always have exceptional sequences when {Sn} convergesgenerally. For example, every tail sequence {tn} with t0 �= x is an exceptionalsequence.

The loxodromic case.

This is a case where K(an/1) converges generally, but may diverge in the classicalsense:

Example 6. We have seen in Example 2 on page 59 that the three-periodic con-tinued fraction

∞Kn=1

an

1:=

21 +

11 −

11 +

21 +

11 −

11 +

21 + · · ·

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180 Chapter 4: Periodic and limit periodic continued fractions

converges generally to 12, but diverges in the classical sense. The linear fractional

transformation

S3(w) :=21 +

11 −

11 + w

=2w

1 + 2w

is loxodromic with attracting fixed point 1/2. But S3(0) = 0, and thus S3n(0) = 0for all n whereas limn→∞ S3n+m(0) = 1/2 for m := 1, 2. �

�Theorem 4.6. Let K(an/bn) be a p-periodic continued fraction of loxo-dromic type. If Sm(0) = y for some m ∈ {1, 2, . . . , p}, then K(an/bn)diverges in the classical sense. Otherwise, K(an/bn) converges to x in theclassical sense.

Proof : K(an/bn) converges generally to x (Theorem 4.3). Hence K(an/bn) con-verges to x in the classical sense if its exceptional sequences have no limit point at0. Now, {S−1

n (y)} is an exceptional sequence. It is p-periodic, so if S−1m (y) �= 0; i.e.,

Sm(0) �= y for m = 1, 2, . . . , p, then 0 is no limit point for this sequence.

Let Sm(y) = 0 for some m ∈ {1, 2, . . . , p}. Evidently Sm(0) = y can not happen forall m ∈ {1, 2, . . . , p} since Sp(0) = y implies that y = 0, whereas S1(0) = a1/b1 �= 0.Therefore Snp+m(0) = y for all n, whereas Snp+k(0) → x as n → ∞ for all k forwhich Sk(0) �= y. Hence K(an/bn) diverges in this case. �

Remark: This type of divergence of periodic continued fractions was first describedby Thiele ([Thie79]), and is therefore called Thiele oscillation. If Thiele oscillationoccurs for a p-periodic continued fraction of loxodromic type, then

limn→∞Snp+m(0) =

{x whenever Sm(0) �= y,

y whenever Sm(0) = y.

Since all y(m) �= 0 in Example 5, the continued fraction in this example convergesin the classical sense.

Corollary 4.7. A 1- or 2-periodic continued fraction K(an/bn) of loxo-dromic type with b1b2 �= 0, converges to x in the classical sense.

Page 194: Lisa Lorentzen, Haakon Waadeland Continued Fractions

4.1.6 Approximants on closed form 181

Proof : Let first K(an/bn) be 1-periodic. Since S1(w) = a1/(b1+w) is loxodromic,we know from Example 3 that

a1/b21 ∈ C \ (−∞,− 1

4],

x =b1

2(−1 + u1), y =

b1

2(−1 − u1) where u1 :=

√1 + 4a1/b2

1.(1.5.1)

Hence y �= 0.

Next, let K(an/bn) be 2-periodic with b1b2 �= 0. The fixed points of S2(w) are

x =2a1 − λ + λu2

2b1, y =

2a1 − λ − λu2

2b1

where λ := a1 + a2 + b1b2 and u2 :=√

1 − 4a1a2/λ2.

(1.5.2)

Since x �= y, this means that u2 �= 0 and λ �= 0. Therefore y = 0 only if (2a1−λ)2 =λ2u2

2; i.e., if b1b2 = 0 which we have excluded. Moreover, also y(1) := a2/(b2+y) �= 0,and the result follows. �

Example 7. The continued fractionz

2+z

2+z

2+z

2+· · ·converges to f(z) =

√1 + z−1 for all z in the cut plane D := {z ∈ C; | arg(1+z)| <

π} since S1(w) = z/(2 + w) is loxodromic with attracting fixed point x = f(z)for z ∈ D. Every tail sequence {tn} for K(z/2) with t0 �= x converges to y =−√

1 + z − 1. The continued fraction also converges to f(z) for z := −1, sincethen S1 is parabolic. But for z on the cut z < −1, S1 is elliptic and the continuedfraction diverges generally (Theorem 4.3B). �

4.1.6 Approximants on closed form

The expressions for τ [n](w) in Section 4.1.2 can be used to find closed expressionsfor Sn(w) = (An−1w + An)/(Bn−1w + Bn) when K(an/bn) is p-periodic. But aperiodic tail sequence also simplifies the formulas in Theorem 2.6 on page 66. Forinstance:�

Corollary 4.8. Let K(a/b) be 1-periodic, and let x and y be the two pos-sibly coinciding fixed points for s(w) := a/(b + w), where x is attracting orindifferent. Then x �= 0,∞,−b, � = (b + y)/(b + x) = x/y, y = −b− x and

Bn + Bn−1x = (b + x)n, An − Bnx = (−x)n+1,

An = x(−x)nn∑

k=1

�−k, Bn = (−x)nn∑

k=0

�−k , x − An

Bn= x/ n∑

k=0

�−k.

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182 Chapter 4: Periodic and limit periodic continued fractions

Proof : We know that a �= 0. If b �= 0, it follows from (1.5.1) that y = −b − xwhere x �= 0,∞,−b. If b = 0, then S1 has the two fixed points ±

√a �= 0,∞. The

expression for � follows since a/(b + x) = x and a/(b + y) = y. �

Remarks:

1. If s is parabolic, then a = −b2/4, x = −b/2, � = 1 and the formulas inCorollary 4.8 imply that

An = −n( b2)n+1, Bn = (n + 1)( b

2)n,

x − An

Bn= − b/2

n + 1,

An

Bn= − nb/2

n + 1.

If |b| < 2, then {An} and {Bn} therefore converge to 0, and if |b| ≥ 2, theyconverge to ∞.

2. For � �= 1 we can sum the geometric sums in the formulas in Corollary 4.8 toget

An = x(−x)n�−n − 11 −� = xy

(−y)n − (−x)n

y − x,

Bn = (−x)n�−n −�1 −� =

(−x)n+1 − (−y)n+1

y − x,

x − An

Bn=

x(1 −�)�−n −� ,

An

Bn=

x(�−n − 1)�−n −� .

Hence, for |�| < 1, the sequences {An} and {Bn} both converge to ∞ if|y| > 1, and to 0 if |y| < 1.

Our next result gives an alternative expression for the ratio � of a transformationSp:

Theorem 4.9. Let K(an/bn) be p-periodic with a p-periodic tail sequencet := {tn} for which all tn �= ∞, and let

X := X(t) :=p−1∏m=0

(−tm) and Y := Y (t) :=p∏

m=1

(bm + tm) .

Then either X/Y = � or X/Y = �−1. If t := {tn} is a second p-periodictail sequence for K(an/bn) with all tn �= ∞, then

X(t) = Y ( t ) and X( t ) = Y (t). (1.6.1)

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4.1.7 Connection to the Parabola Theorem 183

Remark. We recognize the expression

Pp :=Y

X=

p∏m=1

bm + tm−tm

(since t0 = tp)

from Theorem 2.6 on page 66.

Proof of Theorem 4.9: Evidently tnp+m is a fixed point for S(m)p , say tnp+m =

x(m). The derivative of Sp at x(p) is (using the chain rule)

S′p(x

(0)) = S′p(x

(p)) = s′1(x(1)) · s′2(x(2)) · · · s′p(x(p)) (1.6.2)

where

s′m(x(m)) =−am

(bm + x(m))2=

−x(m−1)

bm + x(m). (1.6.3)

This proves that S′p(x

(0)) = X/Y . That X/Y = � follows now from Remark 5on page 176. Next, assume that {tn} is a second p-periodic tail sequence with alltn �= ∞. Then tnp+m must be a second fixed point for S

(m)p ; i.e., tnp+m = y(m).

Writing X := X( t ) and Y := Y ( t ) we similarly find that X/Y = 1/�. That is,� = X/Y = Y /X. Moreover,

Bp + Bp−1tp =p∏

m=1

(bm + tm) (1.6.4)

by Theorem 2.6 on page 66. Hence by Definition 4.1 of � on page 174

� =Bp + Bp−1y

(p)

Bp + Bp−1x(p)=

p∏m=1

bm + y(m)

bm + x(m)=

Y

Y.

This proves that X = Y and Y = X. �

4.1.7 A connection to the Parabola Theorem

In the Parabola Theorem on page 151, the half plane Vα := − 12

+ eiαH is a simplevalue set for every continued fraction K(an/1) from

Eα := {a ∈ C; |a| − Re(a e−2iα) ≤ 12 cos2 α} (1.7.1)

for given α ∈ R with |α| < π/2. It follows from this Parabola Theorem that everyperiodic continued fraction from Eα converges. Hence:

• there are no periodic continued fractions of elliptic or identity type from Eα.• every periodic continued fraction from Eα converges also in the classical sense.

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184 Chapter 4: Periodic and limit periodic continued fractions

With this notation we have:

Theorem 4.10. Let x ∈ ∂Vα \ {∞}. Then a1 := −x2 ∈ ∂Eα and a2 :=−(1 + x)2 ∈ ∂Eα, and

S2 := s1 ◦ s2 for s1(w) :=a1

1 + w, s2(w) :=

a2

1 + w(1.7.2)

is a parabolic transformation.

Proof : That S2 is parabolic follows since

S2(w) =−x2

1 −(1 + x)2

1 + w=

−x2(1 + w)w − 2x − x2

has only one fixed point, namely x. Since ∞ �= x ∈ ∂Vα, it can be written x =− 1

2 + it eiα for some t ∈ R, and thus

|a1| − Re(a1e−2iα) = |x|2 − Re[(−1

4 + it eiα + t2e2iα)e−2iα]

= 14 + t2 + t sin α + 1

4 cos 2α − t sin α − t2 = 12 cos2 α

which proves that a1 ∈ ∂Eα. Since −(1 + x) = − 12 − it eiα, also a2 ∈ ∂Eα. �

Remarks.

1. This means that∂Eα = −(∂Vα)2, (1.7.3)

and that ∂Eα is made up of parabolic pairs (a1, a2) in the sense that S2 in(1.7.2) is parabolic with fixed point x. Similarly, S

(1)2 := s2 ◦ s1 is parabolic

with fixed point −(1 + x).

2. Theorem 4.10 implies that the boundary of Eα is a kind of natural boundary:for every a1 = −x2 ∈ ∂Eα there exist points a†

2 arbitrarily close to a2 :=−(1 + x)2 ∈ ∂Eα such that s1 ◦ s†2 is elliptic.

3. To every 0 �= a1 = −x2 ∈ C \ (−∞,−14 ] there exist exactly two points a2 such

that S2 in (1.7.2) is parabolic. They are given by −(1 + x)2 and −(1 − x)2.

4. To every 0 �= a1 ∈ C \ (−∞,− 14 ] there exist exactly two parabolas of the type

∂Eα through a1. They are the parabola through a1 = −x2 and a2 := −(1+x)2

and the parabola through a1 = −x2 and a2 := −(1 − x)2.

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4.1.7 Connection to the Parabola Theorem 185�

Theorem 4.11. K(an/1) is a p-periodic continued fraction of parabolictype from Eα if and only if p = 1 or p = 2 and

∞Kn=1

an

1=

−x2

1 −(1 + x)2

1 −x2

1 −· · ·where x = Sp(x) ∈ ∂Vα. (1.7.4)

Proof : That (1.7.4) is a parabolic continued fraction from Eα follows from The-orem 4.10. Let K(an/1) be a p-periodic continued fraction of parabolic type fromEα. It is clear that am ∈ ∂Eα for all m since there are no p-periodic continuedfractions of elliptic type from Eα. Since K(an/1) is from Eα, it converges to a finitevalue, and so do also all its tails (the Parabola Theorem). Hence, the attractingfixed point x(m) of S

(m)p is �= ∞ for all m. Moreover, all x(m) ∈ ∂Vα since all

am = x(m−1)(1 + x(m)) ∈ ∂Eα. That is,

x(m) = − 12 + iμmeiα with μm ∈ R for m = 1, 2, . . . , p

am = −(− 12 + iλmeiα)2 with λm ∈ R,

and since am = x(m−1)(1 + x(m)),

−( 14 − iλmeiα − λ2

me2iα) = (− 12 + iμm−1e

iα)(12 + iμmeiα)

iλmeiα + λ2m(cos α + i sin α)eiα = i

2(μm−1 − μm)eiα − μm−1μme2iα,

which gives the two equations

λm + λ2m sin α = 1

2(μm−1 − μm) − μm−1μm sinα , λ2m cos α = −μm−1μm cos α

where cos α �= 0. That is,

λm = 12(μm−1 − μm) and λ2

m = 14(μm−1 − μm)2 = −μmμm−1

which only holds for μm−1 = −μm. Hence {μm} is either the constant sequence{0} or a 2-periodic sequence μ0,−μ0, μ0,−μ0, . . . . Therefore x(2n) = x(0) =: x andx(2n+1) = −1

2− iμ0e

iα = −(1 + x), and K(an/1) has the form (1.7.4). �

For later reference we note:

Lemma 4.12. The tangent of ∂Eα at a finite point a = −x2 ∈ ∂Eα is theline a + xieiα

R.

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186 Chapter 4: Periodic and limit periodic continued fractions

Proof : The boundary of Vα is the line

∂Vα : w = − 12 + ti eiα for t ∈ R ∪ {∞},

and thus the curve ∂Eα = −(∂Vα)2 is given by

w = −(− 12 + ti eiα)2. (1.7.5)

The result follows therefore since

∂w

∂t= −2

(−1

2 + tieiα)· ieiα = −2xieiα for x := − 1

2 + tieiα.

4.2 Limit periodic continued fractions

4.2.1 Definition

A continued fraction K(an/bn) is limit periodic with period length p, or just limitp-periodic or limit periodic for short, if the limits

limn→∞ anp+m =: am , lim

n→∞ bnp+m =: bm for m = 1, 2, . . . , p (2.1.1)

exist in C, and p is the smallest positive integer for which (2.1.1) holds (although wesometimes find it convenient to relax this last condition). An arbitrary continuedfraction K(an/bn) can always be made limit periodic by an equivalence transfor-mation. But if the result is K(cn/dn) where cn → 0, dn → 0 or where cn → ∞,dn → ∞, this just hides the structure of K(an/bn). We are interested in continuedfractions for which limn→∞ anp+m/(bnp+m−1bnp+m) exists in C for m = 1, 2, . . . , p.

Let us assume that the limits in (2.1.1) are finite with all am �= 0. (Other cases willbe treated later on.) Then the (np)th tail of K(an/bn) looks more and more likethe corresponding p-periodic continued fraction

∞Kn=1

an

bn

=a1

b1 +a2

b2 +· · ·+ap

bp +a1

b1 +a2

b2 +· · ·+ap

bp +a1

b1 +a2

b2 +· · ·(2.1.2)

as n increases. The convergence of K(an/bn) is determined by the classification ofthe linear fractional transformation

Sp(w) :=a1

b1 +a2

b2 +· · ·+ap

bp + w. (2.1.3)

So we expect the convergence properties of K(an/bn) to depend on this classificationtoo. We say that K(an/bn) is a limit periodic continued fraction of parabolic,loxodromic, elliptic or identity type depending on the classification of Sp. We further

Page 200: Lisa Lorentzen, Haakon Waadeland Continued Fractions

4.2.2 Finite limits, loxodromic case 187

say that � is the ratio for K(an/bn) if � is the ratio for Sp. The important pointin our investigations is actually that

limn→∞S(np+m)

p = S(m)p for m = 1, 2, . . . , p, (2.1.4)

but this follows in general from (2.1.1).

Notation. We let x(m) be an attracting or indifferent fixed point of

S(m)p (w) =

am+1

bm+1 +am+2

bm+2 +· · ·+am+p

bm+p + w(2.1.5)

for all m ≥ 0, and y(m) be a second fixed point of S(m)p if it exists, and

x(m−1) = sm(x(m)), y(m−1) = sm(y(m)) (2.1.6)

for all m with obvious notation. We shall also let x(m)n , y

(m)n and �(m)

n denotetwo fixed points (coinciding if necessary) and the ratio for S

(np+m)p , where x

(m)n is

attracting or indifferent. Here, as always,

S(n)k (w) := S−1

n ◦ Sn+k(w) =an+1

bn+1 +· · ·+an+k

bn+k + w. (2.1.7)

4.2.2 Finite limits, loxodromic case

Let K(an/bn) with approximants Sn(w) satisfy (2.1.1) with finite limits am, bm,where Sp is a loxodromic transformation from M. Then

limn→∞x(m)

n = x(m), limn→∞ y(m)

n = y(m) and limn→∞�(m)

n = � (2.2.1)

where � is the ratio for Sp and thus for S(m)p . In particular

x(m) �= y(m) for all m and |�| < 1. (2.2.2)

We shall also allow that one or more am = 0, as long as the limits (2.1.1) still existin C and (2.2.2) holds. Then S

(m)p is a singular transformation, but we still say that

K(an/bn) is a limit p-periodic continued fraction of loxodromic type.

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188 Chapter 4: Periodic and limit periodic continued fractions�

Theorem 4.13. Let K(an/bn) be a limit p-periodic continued fraction ofloxodromic type with finite limits (2.1.1).

A. Then K(an/bn) converges generally to some value f ∈ C with thep-periodic exceptional sequence {y(m)}∞m=0.

B. K(an/bn) converges in the classical sense if all y(m) �= 0.

C. limn→∞S−1

np+m(t0) ={

x(m) if t0 = fy(m) if t0 �= f

for m = 1, . . . , p .

To prove this theorem we shall use a couple of lemmas on transformations {τn}from M. In the first one we consider compositions

Tn := τ1 ◦ τ2 ◦ · · · ◦ τn and Tn := τn ◦ τn−1 ◦ · · · ◦ τ1

and show that {Tn} and {Tn} converge generally:

�Lemma 4.14. Let τn ∈ M for all n ∈ N have ratios �n → � and fixedpoints xn → x and yn → y, where |�| < 1, x �= y and yn is the repellingfixed point of τn from some n on. Then Tn → f for some f ∈ C, andTn → x.

Proof : Without loss of generality we assume that x = 0 and y = ∞. (Otherwisewe choose a ψ ∈ M with ψ(x) = 0 and ψ(y) = ∞, and consider transformationsψ ◦ τn ◦ ψ−1.) Then xn �= ∞ from some n on, and thus τ ′

n(xn) = �n → �. Sincexn → 0 and |�| < 1, there exist constants n0 ∈ N and μ, ρ > 0 such that |τ ′

n(w)| ≤μ < 1 for all n ≥ n0 and |w| ≤ ρ. That is, τn(ρD) ⊆ ρD for n ≥ n0. Without loss ofgenerality we assume that n0 = 1. (Otherwise we look at T (k)

n := τk+1◦· · ·◦τk+n andT (k)

n := τk+n ◦ · · · ◦ τk+1 for sufficiently large k ∈ N, and use that Tk+n = Tk ◦ T (k)n

and Tk+n = T (k)n ◦ Tk.) The chain rule leads to

T ′n(wn) = τ ′

1(w1) · τ ′2(w2) · · · τ ′

n(wn),

T ′n(w0) = τ ′

n(wn−1) · τ ′n−1(wn−2) · · · τ ′

1(w0)

where wk−1 := τk(wk) and wk := τk(wk−1) for all k.

Now, wn ∈ ρD implies that wk ∈ ρD for all k ≤ n, and w0 ∈ ρD implies that allwk ∈ ρD. Hence |T ′

n(w)| ≤ μn → 0 and |T ′n(w)| ≤ μn → 0 for |w| ≤ ρ. Therefore

rad Tn(ρD) → 0 and rad Tn(ρD) → 0.

Page 202: Lisa Lorentzen, Haakon Waadeland Continued Fractions

4.2.2 Finite limits, loxodromic case 189

The nestedness Tn+1(ρD) ⊆ Tn(ρD) implies that {Tn} converges generally to somef ∈ ρD. The fact that for each ρ > 0 sufficiently small, |Tn(w)| ≤ ρ from some n

on at (more than) two points w = w1 and w = w2, implies that Tn → 0. �

Remark. From this proof it also follows that there exists a neighborhood V of xsuch that τn(V ) ⊆ V for all n from some n on.

Lemma 4.15. Let τn be as in Lemma 4.14. Then {Tn} converges generallywith exceptional sequence {y}.

Proof : That {Tn} converges generally follows from Lemma 4.14. Now, also τ−1k ∈

M with ratio �n → � and fixed points xn → x and yn → y, only this time xn is therepelling one for large n. Since T −1

n = τ−1n ◦ τ−1

n−1 ◦ · · · ◦ τ−11 , it follows from Lemma

4.14 that {T −1n } converges generally to y. Hence {y} is an exceptional sequence for

{Tn}. �

Proof of Theorem 4.13: A. By Lemma 4.15 it follows that for each given m ≥ 0,{S(m)

np }∞n=0 converges generally to some f (m) ∈ C with constant exceptional sequence{y(m)}. We need to prove that Sm(f (m)) = f (0) for all m ∈ N. This clearly holds if{f (np+m)}∞n=0 has no limit point at y(m) for m = 1, 2, . . . , p. Since by Lemma 4.14limn→∞ f (np+m) = x(m) �= y(m), the conclusion in A follows with f := f (0).

B. This is a straight forward consequence of the result in A.

C. Since {S(m)np }∞n=1 converges generally to f (m) with constant exceptional sequence

{y(m)}, the sequence {(S(m)np )−1}∞n=1 converges generally to y(m) with constant ex-

ceptional sequence {f (m)}. Since f (np+m) → x(m) as n → ∞, the result follows.��

�Corollary 4.16. A limit 1- or limit 2-periodic continued fraction K(an/bn)of loxodromic type with limits am �= ∞ and bm �= 0, converges in the classicalsense.

Proof : Let first K(an/bn) be limit 1-periodic with an → a �= ∞ and bn → b �= 0.For sufficiently large n, bn �= 0, and the fixed points of sn(w) are

xn :=bn

2(−1 + un), yn :=

bn

2(−1 − un)

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190 Chapter 4: Periodic and limit periodic continued fractions

where un :=√

1 + 4an/b2n with Re un ≥ 0. Hence, yn → y �= 0, and the result

follows from Theorem 4.13B.

Next, let K(an/bn) be limit 2-periodic with a1, a2 �= ∞ and b1, b2 �= 0. By (1.5.2)on page 181 and the subsequent arguments, the fixed points of S

(2n+m)2 are

x(m)n =

2a2n+m+1 − λ(m)n + λ

(m)n u

(m)n

2b2n+m+1,

y(m)n =

2a2n+m+1 − λ(m)n − λ

(m)n u

(m)n

2b2n+m+1

for sufficiently large n, where λ(m)n → a1 + a2 + b1b2 �= 0 and y

(m)n → y(m) �= 0 for

m = 0, 1. �

Example 8. The continued fraction K(an/1) where an → 0 is limit 1-periodic ofloxodromic type, since sn(w) := an/(1 + w) has fixed points xn, yn and ratio �n

given by

xn =−1 + un

2, yn =

−1 − un

2, �n =

1 − un

1 + un

where un :=√

1 + 4an → 1. Hence K(an/1) converges to some value f ∈ C. Itstail values f (n) → 0, and every tail sequence {tn} with t0 �= f converges to −1. �

Example 9. The continued fraction

∞Kn=1

an

1:=

3 + 1/12

1 +4 + 3/22

1 +3 + 1/32

1 +4 + 3/42

1 +3 + 1/52

1 +· · ·

in Example 4 on page 13 is limit 2-periodic with corresponding 2-periodic continuedfraction ∞

Kn=1

an

1:=

31+

41+

31+

41+

31+· · ·

.

Since

S2(w) =3

1 +4

1 + w

=3(1 + w)5 + w

has fixed points x(0) = 1 and y(0) = −3 where |5 + y(0)| < |5 + x(0)|, it follows thatK(an/1) is of loxodromic type, and that x(0) is the attracting fixed point for S2.Therefore: K(an/1) converges to some value f ∈ C. Since

x(1) :=4

1 + x(0)= 2 and y(1) :=

41 + y(0)

= −2,

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4.2.2 Finite limits, loxodromic case 191

every tail sequence {tn} for K(an/1) satisfies

limn→∞ t2n =

{1 if t0 = f,

−3 if t0 �= f,lim

n→∞ t2n−1 =

{2 if t0 = f,

−2 if t0 �= f.

Example 10. Let K(an/1) be a limit 3-periodic continued fraction with

lim a3n+1 = 2, lim a3n+1 = 1, lim a3n = −1.

We recognize the corresponding periodic continued fraction∞Kn=1

an

1=

21+

11−

11+

21+

11−

11+· · ·

from Example 6 on page 179 and Example 2 on page 59. S(m)3 is loxodromic with

attracting fixed point x(m) and repelling fixed point y(m) given by

x(0) = 12 , x(1) = 3, x(2) = −2

3, y(0) = 0, y(1) = ∞, y(2) = −1.

Hence K(an/1) converges generally to some f ∈ C, and its tail sequences satisfy

limn→∞ t3n+m =

{x(m) if t0 = f,

y(m) if t0 �= f,

just as the tail sequences for K(an/1).

Now, K(an/1) diverges by Thiele oscillation. This does not necessarily imply thatK(an/1) diverges. It depends on how {a3n+m}∞n=1 approaches its limit am form = 1, 2, 3. It may actually be very complicated to determine whether K(an/1)converges or diverges in the classical sense. �

Example 11. Let K(an/bn) satisfy an → 0, bn → 0, but an/bnbn−1 → 2. Thenbn �= 0 from some n on, so K(an/bn) is equivalent to K(cn/dn) where cn :=an/bnbn−1 and dn := 1 from some n on. That is, K(cn/dn) is limit 1-periodicof loxodromic type, and thus converges in the classical sense to some f ∈ C. There-fore also K(an/bn) converges to f . Since s(w) := 2/(1 + w) has attracting fixedpoint 1 and repelling fixed point −2, the tail sequences {tn} for K(cn/dn) satisfy

limn→∞ tn =

{1 if t0 = f

−2 if t0 �= f.

The tail sequences {tn} for K(an/bn) satisfy tn = bntn from some n on. Hencetn → 0, or more precisely,

tn ∼{

bn if t0 = f

−2bn if t0 �= f.

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192 Chapter 4: Periodic and limit periodic continued fractions

4.2.3 Finite limits, parabolic case

Also in this section we consider limit p-periodic continued fractions K(an/bn) wherethe limits am and bm in (2.1.1) are finite, but now we assume that Sp given by (2.1.3)is a parabolic transformation from M. We shall not allow am = 0 in this section.

The situation is significantly different from the loxodromic case. Periodic continuedfractions of parabolic type converge (Theorem 4.5 on page 179), but among theirclosest neighbors are the elliptic ones which diverge generally. This is reflected inthe convergence behavior of limit p-periodic continued fractions of parabolic type.We shall see that they converge or diverge generally depending on how the elements(an, bn) approach their limit points (2.1.1).

K(an/bn) limit 1-periodic.

K(an/bn) is limit 1-periodic of parabolic type if an → a and bn → b where a/b2 =− 1

4. We can therefore assume that all bn �= 0, without loss of generality. Then

K(an/bn) ∼ K(cn/1) where cn → −14 . This is of course also so for continued

fractions K(an/bn) with an/bn−1bn → − 14 more generally.

So let K(an/1) be a continued fraction with an → − 14 . What can we say about

the convergence of K(an/1)? It turns out that the Parabola Sequence Theorem onpage 154 is very useful in this situation. It concerns continued fractions K(an/1)from {Eα,n} given by

Eα,n :={a ∈ C; |a| − Re(a e−2iα) ≤ 2gn−1(1 − gn) cos2 α}where − π

2 < α < π2 and 0 < ε ≤ gn ≤ 1 − ε < 1.

(2.3.1)

Without loss of generality we assume that

Σ∞ :=∞∑

n=0

Pn = ∞ where Pn :=n∏

k=1

1 − gk

gk. (2.3.2)

(Otherwise we replace {gn} by a sequence {g∗n} as explained in the remarks onpage 136.) Since an → − 1

4 , the limit point case then occurs for Sn(Vα,n) whereVα,n := −gn + eiαH. Hence K(an/1) converges. We can say even more (the proofis postponed to page 209):�

Theorem 4.17. Let K(an/1) with an → −14 be a continued fraction from

{Eα,n}∞n=1 given by (2.3.1). If gn → 12 , then {Sn(w)} converges to some

finite value f ∈ −1 + ε + eiαH for all w ∈ C with w �= − 1

2 , and every tailsequence for K(an/1) converges to − 1

2. If gn−1(1 − gn) ≥ 1

4from some n

on, then also Sn(−12 ) → f .

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4.2.3 Finite limits, parabolic case 193

Example 12. Let |an| − Re an ≤ 12

for all n for K(an/1) with an → −14. Then

an ∈ E0,n with gn := 12 for all n. Hence every tail sequence for K(an/1) converges

to − 12, so naturally {Sn(w)} converges to the value f of K(an/1) for every w �= − 1

2.

But according to Theorem 4.17B also Sn(−12 ) → f .

Let K(cn/dn) with approximants Tn(wn) be equivalent to K(an/1). Then every tailsequence {tn} of K(cn/dn) behaves like tn ∼ −dn/2 and Tn(dnw) → f for everyw ∈ C. �

We have a mixture of two kinds of restrictions on an in Theorem 4.17: K(an/1)converges if either

(i) {an} approaches −14 from the right direction”; i.e., from within a sequence

{Eα,n} of parabolic regions for a fixed |α| < π2. By Lemma 4.12 on page 185,

the tangent of ∂Eα at a = − 14 = −(−1

2 )2 is the line −14 + i eiαR. Hence

K(an/1) converges in particular if an → −14 through a compact subset of the

open half plane − 14 + eiαH,

(ii) or {an} approaches −14

fast enough. That is, if

|an + 14 | ≤ gn−1(1 − gn) − 1

4 for all n, (2.3.3)

since then an ∈ Eα,n with α = 0. (Then {an} also satisfies condition (2.5.3)in the extended Worpitzky Theorem on page 136.)

Condition (2.3.3) is best in the following sense:�

Theorem 4.18. Let En := {a ∈ C; |a + 14 | ≤ ρn} with

ρn := (gn−1(1 − gn) − 14)(1 + δ) > 0 for n = 1, 2, 3, . . . (2.3.4)

for some fixed numbers 0 < gn < 1 and δ > 0. Then there exist divergentcontinued fractions K(an/1) from {En}.

To prove this, we first observe that gn → 12 :

Lemma 4.19. Let

ρn := gn−1(1 − gn) − 14 > 0 for n = 1, 2, 3, . . . . (2.3.5)

where 0 < gn < 1 for all n. Then 12 < gn → 1

2 strictly monotonely.

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194 Chapter 4: Periodic and limit periodic continued fractions

Proof : With such a choice of {gn}, we have − 14∈ E◦

0,n for all n, where E0,n isgiven by (2.3.1). Since K((−1

4 )/1) converges, its tail values −12 ∈ V ◦

0,n = −gn + H

for all n. Hence gn > 12

for n ≥ 0.

It also follows from (2.3.5) that (1 − gn) > (14 )/gn−1, and thus

gn < 1 − 1/4gn−1

< 1 − 1/41 −

1/4gn−2

< · · · < 1 + Sn(g0)

where Sn(g0) → − 12 (Theorem 4.5). Hence lim sup gn ≤ 1

2 . This shows that gn → 12

The monotonicity follows since

gn < 1 − 1/4gn−1

< gn−1 ⇔ gn−1 −14

< g2n−1 ⇔ 0 < (gn−1 − 1

2)2.

Remark. It follows similarly that if ρn ≥ 0 for all n, then 12≤ gn → 1

2monotonely.

Proof of Theorem 4.18: Without loss of generality we assume that (2.3.2) holds,so that {−gn} is the sequence of tail values for K(an/1) from {En} with an :=−gn−1(1 − gn) for all n. Since gn > 1

2 , the sequence {Pn} is a monotonely de-creasing sequence of positive numbers with

∑Pn = ∞. Let z �= 1 be some Nth

root of unity; i.e., z �= 1 with zN = 1 for some N ∈ N. Then∑

Pnzn diverges byoscillation. Let further

tn :=−gn

gn + z − gnz, so that

1 + tn−tn

=1 − gn

gnz,

and thus∑∞

n=0

∏nk=1

1+tk

−tkdiverges by oscillation. Since {tn} is a tail sequence for

the continued fraction K(an/1) given by

an := tn−1(1 + tn) =−gn−1

gn−1 + z − gn−1z· (1 − gn)zgn + z − gnz

,

this continued fraction diverges (Corollary 2.7 on page 68). On the other hand,limz→1 an = gn−1(1− gn), so K(an/1) is a continued fraction from {En} for z closeenough to 1, which can be achieved for N large enough. �

K(an/bn) limit 2-periodic.

Also here we assume that all bn �= 0, so that an equivalence transformation bringsthe continued fraction to the form K(cn/1). We therefore study continued fractionsK(an/1). Our analysis thereby covers continued fractions K(an/bn) where

limn→∞

a2n+1

b2nb2n+1= a1 and lim

n→∞a2n

b2n−1b2n= a2,

with S2 := s1 ◦ s2 parabolic when sm(w) := am/(1 + w) .(2.3.6)

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4.2.3 Finite limits, parabolic case 195

K(an/1) is limit 2-periodic of parabolic type with finite limits a1, a2 if

a2n−1 → a1 = −x2 and a2n → a2 = −(1 + x)2 (2.3.7)

for some x �= −1, 0, ∞; that is, if (1+a1+a2)2 = 4a1a2 �= 0. The transformations S2

and S(1)2 are then parabolic with attracting fixed points x and −(1+x) respectively.

In this situation, Theorem 3.4 on page 105 is very useful: if the even and odd partsof K(an/1) converge, then K(an/1) itself converges. And the even and odd partsof K(an/1) are limit 1-periodic continued fractions of parabolic type.

But we can also use the Parabola Sequence Theorem directly. As noted in Remark1 on page 184 the boundary of Eα is made up of parabolic pairs. Indeed, to everyparabolic pair (−x2, −(1 + x)2) with x ∈ C \ (−∞,− 1

2 ], there exists a parabolicregion Eα from the Parabola Theorem, with boundary passing through both −x2

and −(1+x)2 (Remark 2 and 3 on page 184). Moreover, we can always find {Eα,n}such that gn−1(1 − gn) > 1

4 and thus Eα ⊆ Eα,n for all n.

Theorem 4.20. Let K(an/1) be a limit 2-periodic continued fraction ofparabolic type (with finite limits) from {Eα,n} given by (2.3.1). If gn → 1

2,

then K(an/1) converges to some value f ∈ −1 + ε + eiαH, and every tailsequence {tn} for K(an/1) satisfies lim t2n = x, lim t2n+1 = −1 − x wherex is given by (2.3.7). If gn−1(1− gn) ≥ 1

4from some n on, then Sn(w) → f

for every w ∈ C.

The proof of this theorem is similar to the proof of Theorem 4.17 and will beomitted.

More general cases.

For longer periods than p = 2, p-contractions can often give some answers, forexample in combination with the following lemma:

�Lemma 4.21. Let K(an/1) be limit p-periodic with finite limits am :=limn→∞ anp+m �= 0 for m = 1, 2, . . . , p. If {S(m)

np }∞n=0 converges generallyfor m = 0, 1, . . . , p − 1, then K(an/1) converges generally to f (0).

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196 Chapter 4: Periodic and limit periodic continued fractions

Proof : Since {am} is bounded and bounded away from 0, there always existsequences {un} and {vn} from C such that

f = limn→∞Snp+m(wnp+m)

= limn→∞Snp+m

( anp+m+1

1 + wnp+m+1

)= lim

n→∞Snp+m+1(wnp+m+1)

for both {wn} := {un} and {wn} := {vn}, where

lim inf m(un, vn) > 0 and lim inf m( an+1

1 + un+1,

an+1

1 + vn+1

)> 0.

4.2.4 Finite limits, elliptic case

Since periodic continued fractions of elliptic type diverge, it is easy to believe thatalso limit periodic ones diverge. However, as pointed out by Gill ([Gill73]) therealso exist convergent ones. Here is a rather simple example (compared to otherexamples in the literature) of this phenomenon:

Example 13. A limit 1-periodic continued fraction K(an/1) is of elliptic type if

an → a < −14 . (2.4.1)

Then S1(w) = a/(1 + w) has the two indifferent fixed points

x = −12 + iq, y = − 1

2 − iq where q :=√|a| − 1

4 > 0. (2.4.2)

In particular y = x, 1 + x = −x and a = −|x|2. Let tn := x − 2|x|2/(n + 1) forall n ≥ 0. Then {tn} is a tail sequence for the limit 1-periodic continued fractionK(an/1) of elliptic type given by an := tn−1(1 + tn). Now,

t0 − fn =t0Σn

where Σn :=n∑

k=0

k∏j=1

1 + tj−tj

. (2.4.3)

(See formula (1.5.7) on page 66.) We have

1 + tj−1

−tj−1=

1 + x − 2|x|2/j

−x + 2|x|2/j

=−x − 2|x|2/j

−x + 2|x|2/j=

1 + 2x/j

1 − 2x/j· x

x=

j − 1 + 2iq

j + 1 + 2iq· x

x

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4.2.4 Finite limits, elliptic case 197

and thusk∏

j=1

1 + tj−tj

=(1 + 2iq)(2 + 2iq)

(k + 1 + 2iq)(k + 2 + 2iq)·(

x

x

)k

which means that

∞∑k=0

k∏j=1

∣∣∣∣1 + tj−tj

∣∣∣∣ = ∞∑k=0

k+1∏j=2

∣∣∣∣1 + tj−1

−tj−1

∣∣∣∣ = 1 + |1 + 2iq| · |2 + 2iq|∞∑

k=1

1k2 + O(k)

.

That is, Σn converges to a finite number, and K(an/1) converges to some f �= t0.�

Still, K(an/bn) diverges if {S(np)p } converges fast enough to an elliptic transforma-

tion.�

Lemma 4.22. Let τn ∈ M have ratio �n and fixed points xn, yn for alln ∈ N, where∑

|xn| < ∞,∑

1/|yn| < ∞ and∑

|�n −�| < ∞ (2.4.4)

for some � �= 1 with |�| = 1, and let Tn := τ1 ◦ · · · ◦ τn for all n ∈ N. Then{Tn} is totally non-restrained, and {Tn(�−nw)} converges to T (w) for allw ∈ C for a T ∈ M.

To prove this we shall use the following auxiliary result:�

Lemma 4.23. For a given sequence {rn} of positive numbers with r :=∑∞n=1 rn < ∞, let {(Xn, Yn)}∞n=0 satisfy

0 ≤ Xn ≤ (1 + rn)Xn−1 + rnYn−1,

0 ≤ Yn ≤ rnXn−1 + (1 + rn)Yn−1 for n = 1, 2, 3, . . .(2.4.5)

with X0 ≥ 0 and Y0 ≥ 0. Then {(Xn, Yn)} is bounded.

Proof : The worst case occurs when we have equality for all n in the two inequal-ities in (2.4.5); i.e., when(

Xn

Yn

)= Mn

(Xn−1

Yn−1

)where Mn :=

(1 + rn rn

rn 1 + rn

);

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198 Chapter 4: Periodic and limit periodic continued fractions

i.e., when (Xn

Yn

)= Mn,1

(X0

Y0

)where Mn,1 := MnMn−1 · · ·M1.

Now, the matrix product Mk+1Mk is

Mk+1Mk =(

1 + rk+1,k rk+1,k

rk+1,k 1 + rk+1,k

)where rk+1,k := rk + rk+1 + 2rkrk+1,

so all Mn,1 have the same form with some rn,1 for which r1,1 = r1 and

rk+1,1 = rk,1 + rk+1 + 2rk,1rk+1 = rk,1(1 + 2rk+1) + rk+1.

We shall prove by induction that

rn,1 =12

n∏j=1

(1 + 2rj) −12. (2.4.6)

Evidently r1,1 = r1 = 12(1 + 2r1) − 1

2 , and if (2.4.6) holds for n = k, then

rk+1,1 = rk,1(1 + 2rk+1) + rk+1

=12

k+1∏j=1

(1 + 2rj) −12(1 + 2rk+1) + rk+1 =

12

k+1∏j=1

(1 + 2rj) −12

which proves (2.4.6). Since∑

rk < ∞, it follows that {rn,1} is bounded. �

Proof of Lemma 4.22: Let τn ∈ M be given, with ratio �n and fixed pointsxn, yn (coinciding if necessary), where xn → x, yn → y and �n → � with |�| = 1,� �= 1. Without loss of generality we assume that x = 0, y = ∞, and that xn �= ∞,yn �= 0 and �n �= 1 for all n. (Otherwise we consider τn := ϕ ◦ τn ◦ ϕ−1 whereϕ(x) = 0 and ϕ(y) = ∞, and look at Tn := τm+1 ◦ · · · ◦ τm+n = T −1

m ◦ Tm+n forsufficiently large m.)

Since all �n �= 1, we have all xn �= yn, and by Theorem 4.2 on page 174, τn can bewritten

τn(w) =(xn −�nyn)w − (1 −�n)xnyn

(1 −�n)w + �nxn − yn=

(�n − xn

yn)w + (1 −�n)xn

− 1yn

(1 −�n)w −�nxn

yn+ 1

with the natural limit form if yn = ∞. That is,

τn(w) =(1 + δ

(1)n )�w + δ

(2)n

δ(3)n w + 1 + δ

(4)n

for n = 1, 2, 3, . . . (2.4.7)

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4.2.4 Finite limits, elliptic case 199

where∑

|δ(k)n | < ∞ for k = 1, 2, 3, 4 under our conditions. Let {rn} be a sequence

of positive numbers such that

|δ(k)n | ≤ rn for all n and k = 1, 2, 3, 4 and r :=

∑rn < ∞.

We want to prove that

Tn(w) =Anw + Bn

Cnw + Dnwhere

limAn�−n = C(1), limBn = C(2)

lim Cn�−n = C(3), limDn = C(4) (2.4.8)

with C(1)C(4) − C(2)C(3) �= 0 for r sufficiently small. Since Tn = Tn−1 ◦ τn, we findthat

An = (1 + δ(1)n )�An−1 + δ(3)

n �Bn−1, Bn = δ(2)n An−1 + (1 + δ(4)

n )Bn−1,

Cn = (1 + δ(1)n )�Cn−1 + δ(3)

n �Dn−1, Dn = δ(2)n Cn−1 + (1 + δ(4)

n )Dn−1

(2.4.9)

for n ≥ 2. It follows therefore from Lemma 4.23 that {An}, {Bn}, {Cn} and {Dn}are bounded sequences.

From the recurrence relations (2.4.9) we find that

|An�−n −An−1�−(n−1)| = |δ(1)n �−(n−1)An−1 + δ(3)

n �−(n−1)Bn−1|≤ rn|An−1| + rn|Bn−1|

since |�| = 1, where {An} and {Bn} are bounded. Hence {An�−n} convergesabsolutely to some finite constant C(1) as n → ∞. Similarly Cn�−n → C(3) �= ∞.Moreover, |Bn−Bn−1| ≤ rn(|An−1|+|Bn−1|) and |Dn−Dn−1| ≤ rn(|Cn−1|+|Dn−1|)which proves that the finite limits C(2) := limBn and C(4) := limDn also exist.Finally, by (2.4.9)

Δn := An�−nDn − BnCn�−n

= [(1 + δ(1)n )An−1 + δ(3)

n Bn−1]�−(n−1) · [δ(2)n Cn−1 + (1 + δ(4)

n )Dn−1]

− [(1 + δ(1)n )Cn−1 + δ(3)

n Dn−1]�−(n−1) · [δ(2)n An−1 + (1 + δ(4)

n )Bn−1]

= [(1 + δ(1)n )(1 + δ(4)

n ) − δ(2)n δ(3)

n ]Δn−1

= · · · = Δ1

n∏k=2

[(1 + δ(1)k )(1 + δ

(4)k ) − δ

(2)k δ

(3)k ]

which stays bounded away from 0 when∑

rn < ∞ and all rn < 1 with (1− rn)2 −r2n = 1 − 2rn > 0; i.e., all rn < 1

2. �

Our main theorem follows immediately:

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200 Chapter 4: Periodic and limit periodic continued fractions�

Theorem 4.24. Let τn ∈ M have ratio �n and fixed points xn, yn, and letTn := τ1 ◦ τ2 ◦ · · · ◦ τn for all n ∈ N. If τn → τ ∈ M where τ is elliptic withratio � and fixed points x, y, and∑

|�n −�| < ∞,∑

m(xn, x) < ∞ and∑

m(yn, y) < ∞,

then Tn(�−nw) → T (w) for a T ∈ M. In particular {Tn} is totally non-restrained.

Therefore, a limit periodic continued fraction of elliptic type diverges if {S(np)p }

approaches the elliptic transformation fast enough. For instance:�

�Corollary 4.25. Let K(an/bn) be limit 1-periodic of elliptic type with an →a �= 0,∞ and bn → b �= ∞. If

∑|an − a| < ∞ and

∑|bn − b| < ∞, then

K(an/bn) diverges generally.

Proof : The linear fractional transformations

sn(w) :=an

bn + wand s(w) :=

a

b + w

have finite fixed points

xn, yn = − 12(bn ±

√b2n + 4an), x, y = −1

2 (b ±√

b2 + 4a).

Therefore∑

|xn − x| < ∞ and∑

|yn − y| < ∞. Moreover, the ratios of sn and sare given by

�n =bn + yn

bn + xn=

xn

yn, � =

x

y,

where yn, y �= 0 since an, a �= 0, so also∑

|�n −�| < ∞. �

Example 14. Let 0 �= an → a < − 14

so fast that∑

|an − a| < ∞. Then K(an/1)is a generally divergent limit periodic continued fraction of elliptic type. �

4.2.5 Infinite limits

Let K(an/bn) be limit p-periodic where one or more of the limits (2.1.1) are equalto ∞. Then it may be a good idea to use an equivalence transformation to bring it

Page 214: Lisa Lorentzen, Haakon Waadeland Continued Fractions

4.2.5: Infinite limits 201

over to the form K(cn/1) or K(1/dn) if possible. The cases cn → 0 and dn → ∞are very simple: the continued fraction converges. The cases cn → ∞ and dn → 0can sometimes be resolved by use of a Parabola Theorem or Van Vleck’s Theorem,but far from always. In this section we shall consider the case

K(an/1) where an → ∞ . (2.5.1)

We know that K(an/1) diverges if its Stern-Stolz series

S :=∞∑

n=0

n∏k=1

|ak|(−1)n+k+1(2.5.2)

converges to a finite value (the Stern-Stolz Theorem on page 100). Hence we con-centrate on the case S = ∞. We follow an idea from [JaWa86], and analyze theeven and odd parts of (2.5.1). They are given by

a1

1 + a2−a2a3

1 + a3 + a4−a4a5

1 + a5 + a6−· · ·(even part)

a1 −a1a2

1 + a2 + a3−a3a4

1 + a4 + a5−· · ·(odd part).

If they both converge to the same value, then K(an/1) converges. This is in par-ticular the case in the following situation:�

�Theorem 4.26. K(an/1) converges if an → ∞ with

limn→∞

a2n−1

a2n=: α ∈ C, lim

n→∞a2n

a2n+1=: β ∈ C where |α|, |β| �= 1.

Proof : Let an → ∞ with (a2n−1/a2n) → α and (a2n/a2n+1) → β, and let Sn(w)denote the approximants of K(an/1). Assume first that |α| < 1. The even part ofK(an/1) is equivalent to

a1/a2

1/a2 + 1−a3/a4

1/a4 + a3/a4 + 1−a5/a6

1/a6 + a5/a6 + 1−· · ·. (2.5.3)

The transformation s(w) :=−α

α + 1 + whas fixed points x = −α, y = −1 and

ratio � =α + 1 + y

α + 1 + x= α. Hence s is loxodromic with repelling fixed point −1.

Therefore (2.5.3) converges generally to some f (e) ∈ C with constant exceptionalsequence {−1}. Hence it also converges to f (e) in the classical sense.

Next, let |α| > 1. The even part of K(an/1) is also equivalent to

11/a1 + a2/a1−

a2/a1

1/a3 + 1 + a4/a3−a4/a3

1/a5 + 1 + a6/a5−· · ·. (2.5.4)

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202 Chapter 4: Periodic and limit periodic continued fractions

The transformation s(w) :=−1/α

1/α + 1 + wis also loxodromic with repelling fixed

point y = −1. Hence {S2n} still converges generally and in the classical sense tosome f (e) ∈ C with exceptional sequence {−1}.

Similarly, {S2n+1} converges generally to some f (o) ∈ C with exceptional sequence{−1} if |β| �= 1. It remains to prove that f (o) = f (e). By (2.5.4) on page 89

S(e)n (−a2n) = S2n−1(0).

Since (−a2n) → ∞, and thus stays asymptotically away from the exceptional se-quence {−1}, it follows that S

(e)n (−a2n) → f (e). Since S2n−1(0) → f (o), the result

follows. �

If |α| = 1, then (2.5.3) and (2.5.4) are limit 1-periodic of parabolic (α = 1) orelliptic (α �= 1) type, and the question of convergence is more subtle.

Example 15. The continued fraction

Rη(a, b) :=a

η+12b2

η +22a2

η +32b2

η +42a2

η +52b2

η +· · ·where a, b, η are complex numbers �= 0, is called Ramanujan’s AGM-fraction. Thereason for this is the surprising identity

Rη(a+b2 ,

√ab) = 1

2 (Rη(a, b) + Rη(b, a))

for a, b, η > 0. That is, AGM is an abbreviation for the arithmetic-geometric mean.

Since Rη(a, b) ∼ R1(a/η, b/η), we may without loss of generality assume that η =1. It follows from the Parabola Theorem on page 151 that R1(a, b) converges ifa2 = b2 ∈ C \ [−∞, 0]. Let a2 �= b2. Then R1(a, b) has the form K(an/1) wherean → ∞ and

limn→∞

a2n

a2n−1=

b2

a2and lim

n→∞a2n−1

a2n=

a2

b2.

Therefore Theorem 4.26 shows that R1(a, b) also converges whenever |a| �= |b|. Thiswas first proved by J. Borwein et al ([BoCF04], [BoCR04]) and later on by Lorentzen([Lore08b]).

Let |a| = |b|, but a2 �= b2. Then (2.5.3) and (2.5.4) are limit periodic continuedfractions of elliptic type. Since∑∣∣a2n+2

a2n+1− b2

a2

∣∣ =∑∣∣ ba

∣∣2(2n + 12n

− 1)2 =

∑ 14n2

< ∞,

it follows from Corollary 4.25 that (2.5.4) diverges generally. Hence R1(a, b) divergesgenerally in this case. This was also first proved by J. Borwein et al ([BBCM07])and later on by Lorentzen ([Lore08b]). �

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4.3.1 Periodic continued fractions with multiple limits 203

4.3 Continued fractions with multiple limits

4.3.1 Periodic continued fractions with multiple limits

A p-periodic continued fraction of elliptic type diverges, of course. Still, the asymp-totic behavior of {Sn} is quite interesting.

Example 16. The 1-periodic continued fraction K(−1/1) is of elliptic type sinceS1(w) = s1(w) = −1/(1 + w) has the two fixed points (−1 ± i

√3)/2, and thus the

ratio

� =1 − i

√3

1 + i√

3= e−2iπ/3.

Since �3 = 1, the sequence {Sn} of linear fractional transformations is 3-periodic.Indeed,

S1(w) = s1(w) =−1

1 + w,

S2(w) = s1 ◦ s1(w) = −1 + w

w,

S3(w) = s1 ◦ s1 ◦ s1(w) = S2 ◦ s1(w) = w,

S4(w) = S3 ◦ s1(w) = S1(w) and so on.

Therefore, even though {Sn} is totally non-restrained, {Sn(0)} is the 3-periodicsequence

{fn} : −1, ∞, 0, −1, ∞, 0, −1, . . . .

We follow Bowman and Mc Laughlin ([BoML06]) and say that K(−1/1) has 3limits. Also {S−1

n } is 3-periodic, so every tail sequence for K(−1/1) is 3-periodic.For instance, starting with t0 = 1 we get

{tn} : 1, −2, − 12 , 1, −2, − 1

2 , 1, −2, −12 , . . . .

Indeed, the continued fraction K(an/1) can be regarded as Np-periodic of identitytype. �

This is part of a general picture. The essential thing is that �N = 1; i.e., the ratios{�n} for {τ [n]} are N -periodic. Since τ [n] has the same fixed points as τ itself, thismeans that {τ [n]} is an N -periodic sequence of linear fractional transformations.(See Property 2 on page 54.)�

Lemma 4.27. Let K(an/bn) be a p-periodic continued fraction with ratio� �= 1. If �N = 1 for some N ∈ N, then {Sn} is an (Np)-periodic sequenceof linear fractional transformations. If moreover K(an/bn) has a tail se-quence {tn} with all tn �= ∞ and

∏pn=1(−tn)N = 1, then also {An} and

{Bn} are (Np)-periodic.

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204 Chapter 4: Periodic and limit periodic continued fractions

Proof : Since {S(m)np }n is N -periodic for m = 0, 1, . . . , p − 1, it follows that {Sn}

is (Np)-periodic. (In principle it may also have a shorter period length, of course.)Indeed, SNp(w) = I(w) ≡ w is the identity transformation, and thus SnNp = I forall n ∈ N, so there exists a γ �= 0,∞ such that

ANp = 0, ANp−1 = γ, , BNp = γ and BNp−1 = 0.

On the other hand it follows from Theorem 2.6 on page 66 that

ANp − BNpt0 =Np∏n=0

(−tn); i.e., − γ t0 = −t0

Np∏n=1

(−tn).

Hence, if∏Np

n=1(−tn) = 1, then γ = 1, and thus also {An} and {Bn} are (Np)-periodic. �

This was proved by Bowman and Mc Laughlin who studied such continued fractionsin a series of papers ([BoML06], [BoML07], [BoML08]).

Example 17. The constant sequence {x} with x := (−1 + i√

3)/2 = e2iπ/3 is atail sequence for K(−1/1) from Example 16. Since �3 = 1 and p = 1, this meansthat {An/Bn} is 3-periodic. Moreover, (−x)6 = 1 and �6 = 1. Therefore {An} and{Bn} are 6-periodic. Straight forward computation confirms this:

n −1 0 1 2 3 4 5 6 7 8 9 · · ·an −1 −1 −1 −1 −1 −1 −1 −1 −1 · · ·An 1 0 −1 −1 0 1 1 0 −1 −1 0 · · ·Bn 0 1 1 0 −1 −1 0 1 1 0 −1 · · ·

An/Bn −1 ∞ 0 −1 ∞ 0 −1 ∞ 0 · · ·

4.3.2 Limit periodic continued fractions with multiple limits

A similar situation occurs if τn → τ ∈ M fast enough, where τ has ratio � �= 1 with�N = 1. Then {Tn} given by Tn := τ1 ◦ τ2 ◦ · · · ◦ τn is a limit N -periodic (totallynon-restrained) sequence. To see this, we let τ have fixed points x = 0 and y = ∞,without loss of generality. Then we can use Lemma 4.22 on page 197 to prove:�

�Lemma 4.28. Let {τn} and {Tn} be as in Lemma 4.22. If � �= 1, but�N = 1 for some N ∈ N, then {TnN+m}n converges to some Tm ∈ M form = 1, 2, . . . , N .

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4.4.2 Fixed circles for τ ∈ M 205

Proof : From Lemma 4.22 we know that Tn(�−nw) → T (w) for a T ∈ M. Since�N = �−N = 1, this means that limn→∞ TnN+m(w) = T (�mw) =: Tm(w) for eachfixed 1 ≤ m ≤ N . �

Remark: This means that {TnN+m(w)}n converges for each m ∈ N, but to valuesdepending on w.

Example 18. The limit 1-periodic continued fraction K(1/bn) where bn → bi isof elliptic type if b ∈ R with |b| < 2. In particular this holds for b = 0 whenS1(w) := 1/(b + w) = 1/w, and thus has the ratio � = −1. That is, �2 = 1.Therefore K(1/bn) with

∑|bn| < ∞ is totally non-restrained and {S2n−1} and

{S2n} converge to functions T1 ∈ M and T2 ∈ M respectively in this case. Thiswas exactly what we proved in Van Vleck’s Theorem on page 142. �

4.4 Fixed circles

4.4.1 Introduction

In order to get a more geometric idea of the asymptotic behavior of {Sn(w)} and{S−1

n (w)}, we shall look at fixed circles for Sp for p-periodic continued fractions.This will also give us a better idea of what happens in the limit periodic case.

A circle C in C is called a fixed circle for τ ∈ M if it is invariant under the mappingτ ; i.e., if τ(C) = C. If ∞ ∈ C, it may also be called a fixed line for τ . The crucialpoint is that if w is a point on such a circle C, then iterations of τ will move thispoint along C in the sense that τ [n](w) ∈ C for all n.

Not every τ ∈ M has fixed circles, but if τ has one, then it has infinitely many.Indeed, we shall see that if τ has fixed circles, and τ �= I and � �= −1, then for eachw ∈ C not a fixed point for τ , there is a unique fixed circle for τ passing throughw. We let Cw denote this fixed circle.

4.4.2 Fixed circles for τ ∈ M

If C is a fixed circle for τ , then ϕ(C) is a fixed circle for τ := ϕ ◦ τ ◦ ϕ−1. Indeed,

ϕ(Cw) = Cϕ(w) (4.2.1)

where Cv denotes the fixed circle for τ through v. We shall use this fact to describethe fixed circles for τ .

If τ = I; i.e., τ(w) ≡ w, then the case is clear: every circle C on C is a fixed circlefor τ . But this fact is of minor interest. Nothing happens to w under iterations of

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206 Chapter 4: Periodic and limit periodic continued fractions

I. Not much happens if � = −1 either. Then τ is conjugate to τ(w) = −w whichmoves a point from w to −w, and back again to w in the sense that τ [2n−1](w) = −wand τ [2n](w) = w for all n. In the following we assume that τ �= I and � �= −1; i.e.,τ [2] �= I.

The elliptic case: Let τ ∈ M be elliptic with fixed points x, y and ratio � �= −1.Then |�| = 1 with � �= ±1; i.e., � = eiθ for some 0 < θ < 2π with θ �= π, and τis conjugate to τ(w) := �w which has fixed points at 0 and ∞. τ moves a pointw �= 0,∞ the angle θ around on the circle Cw through w centered at the origin.Hence every circle C with center at the origin is a fixed circle for τ , and no othercircle has this property. A circle with center at 0 can also be seen as a circle withcenter at ∞.

Let ϕ ∈ M be the transformation which makes τ = ϕ−1 ◦ τ ◦ ϕ. Then the circlesC := ϕ−1(C) on the Riemann sphere are fixed circles for τ . They are no longernecessarily concentric, but they are disjoint. For

τ(w) :=aw + b

cw + dwith Δ := ad − bc �= 0, c �= 0,

ϕ(w) := (w−x)/(w− y), and thus ϕ−1(w) = (x− yw)/(1−w). The fixed circle forτ through ϕ(∞) = 1 is the unit circle ∂D. Therefore the fixed line L for τ is givenby L = ϕ−1(∂D). That is, L is given by

ϕ−1(eit) =x − y eit

1 − eit=

x + y

2+

x − y

21 + eit

1 − eitfor 0 ≤ t < 2π,

where

1 + eit

1 − eit=

e−it/2 + eit/2

e−it/2 − eit/2=

cos t/2−i sin t/2

= it1 for t1 ∈ R ∪ {∞} =: R.

That is, L is the perpendicular bisector of the line segment connecting

x

y

L

C

x and y. The other fixed circles C = ϕ−1(C)for τ have their centers on the line through xand y, and each one of them winds around xor y, and none of them intersects or touchesL or each other. Moreover, since 0 and ∞are symmetric points with respect to everycircle C centered at the origin, it follows thatthe fixed points x and y are symmetric withrespect to each fixed circle (remarks on page109). We distinguish between the followingtwo cases:

Case 1: �N �= 1 for all N ∈ N. Then {�n}is a sequence of distinct points dense on ∂D,

and thus {τ [n]} is a sequence of distinct elliptic transformations �nw from M.

Page 220: Lisa Lorentzen, Haakon Waadeland Continued Fractions

4.4.3 Fixed circles and periodic continued fractions 207

Hence also {τ [n]} is a sequence of distinct elliptic transformations from M. Itera-tions of τ move a point w �= x, y around and around on Cw, indefinitely, withoutever hitting the same point twice. The limit points for {τ [n](w)} are dense on Cw.

Case 2: �N = 1 where N > 2 is the smallest positive integer for which this holds.Then {�n} is an N -periodic sequence of equidistant points on ∂D with �nN = 1 forall n. Hence {τ [n]} is an N -periodic sequence from M with τnN = I for all n. Allother transformations τ [n] are elliptic. For w �= x, y, {τ [n](ϕ(w))} is an N -periodicsequence of equidistant points on the circle Cϕ(w) with center at the origin, and so{τ [n](w)} is an

w

τ(w)τ [2](w)τ [3](w)x

Cw

L

N -periodic sequence of (not necessarily equidistant)points on the fixed circle ϕ−1(Cϕ(w)) = Cw for τ .

The parabolic case: Let τ be parabolic with fixedpoint x. If x = ∞, then τ(w) = w+q for some q �= 0.Since τ(qt + λ) = q(t + 1) + λ, it follows that everyline λ+qR with λ ∈ C is invariant under τ . No othercircle on C has this property. If x �= ∞, then τ isconjugate to τ(w) := w + q, and the line L throughx and the point ζ := τ−1(∞) = −d/c is a fixed linefor τ . This means that the fixed circles for τ are

exactly L plus all circles tangent to L at x. Iterations of τ move a point w �= xmonotonely along Cw towards x without ever passing x.

The loxodromic case: Let τ ∈ M be loxodromic. Then τ is conjugate to τ(w) =�w with 0 < |�| < 1 and attracting fixed point x = 0 and repelling fixed pointy = ∞.

Let first � ∈ R. Then −1 < � < 1, and we say that τ is hyperbolic. In this case R

is a fixed line for τ . Indeed, every line eiαR through the origin is invariant under τ ,

and these are the only fixed circles for τ . If � �∈ R, then there are no fixed circlesfor τ . Roughly speaking, τ [n](w) spirals in towards the attracting fixed point at theorigin.

This means that a loxodromic transformation τ has fixed circles if and only if it ishyperbolic. In the hyperbolic case, its fixed circles are exactly all the circles passingthrough both its fixed points x, y. If � > 0, then iterations of τ move a point wmonotonely along the fixed circle Cw towards x. If � < 0, then iterations of τ alsomove a point w along Cw towards x, but this time in an alternating manner.

4.4.3 Fixed circles and periodic continued fractions

A p-periodic continued fraction K(an/bn) has approximants

Snp+m(w) = S[n]p ◦ Sm(w) = Sm ◦ (S(m)

p )[n](w). (4.3.1)

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208 Chapter 4: Periodic and limit periodic continued fractions

If Sp has fixed circles C, then S(m)p has the fixed circles S−1

m (C). We let C(m)w denote

the fixed circle of S(m)p through w, and x(m), y(m) be the fixed points of S

(m)p and

x := x(0).

The parabolic case.

The fixed circles for τ(w) := w + q are the lines λ + qR for λ ∈ C. Iterationsτ [n](w0) = w0 + nq of τ move a point w0 monotonely along this line towards ∞.Iterations of τ−1 also move w0 monotonely along this line towards ∞, but in theopposite direction since (τ−1)[n](w0) = w0 − nq. Therefore, iterations S

(m)np (w) of

x

p = 3L

Cw

CS1(w)

CS2(w)

the parabolic transformation S(m)p moves a

point w �= x(m) monotonely towards x(m)

along the fixed circle C(m)w of S

(m)p . Similarly,

S(m)−1

np moves w monotonely along C(m)w to-

wards x(m) from the other direction.

Now, Snp+m = Sm ◦ S(m)np , and CSm(w) =

Sm(C(m)w ) is the corresponding fixed circle for

Sp for each m ∈ {1, 2, . . . , p}. Hence, Sn(w)jumps periodically between the p fixed circlesCSm(w) for Sp, gradually moving towards x,and a tail sequence {tn} with t0 �= x jumpsperiodically between the p circles C

(m)tm

, since

tnp+m := S−1np+m(t0) = S

(m)−1

np ◦ S−1m (t0) =

S(m)−1

np (tm) ∈ C(m)tm

. We have proved:

Theorem 4.29. Let K(an/bn) be p-periodic of parabolic type. For eachm ∈ {1, 2, . . . , p} and w ∈ C\{x}, {Snp+m(w)}n is a monotone sequence onCSm(w) and {S−1

np+m(w)}n is a monotone sequence on C(m)

S−1m (w)

. {S(m)np (w)}n

and {S(m)−1

np (w)}n approach x(m) monotonely along C(m)w from opposite

sides.

Example 19. The 1-periodic continued fraction K((− 14 )/1) is of parabolic type,

and S1 has the fixed point x = − 12 . The fixed circles for S1 are the line R and every

circle tangent to the real line at x = −12 . Iterations of S1 move every point w �= −1

2monotonely along Cw towards −1

2 , and K((−14 )/1) converges to − 1

2 . {− 12} is the

sequence of tail values for K((− 14)/1). Hence it follows from Remark 2 on page 182

Page 222: Lisa Lorentzen, Haakon Waadeland Continued Fractions

4.4.3 Fixed circles and periodic continued fractions 209

that An = −n( 12)n+1 and Bn = (n + 1)(1

2)n, so that

Sn(w) =−(n − 1)( 1

2)nw − n( 12)n+1

n(12)n−1w + (n + 1)(1

2)n

= −12

+12n

− 1/22n2w + n2 + n

.

Hence Sn(w) approaches −12 from the right, and a tail sequence {tn} with t0 �= x

approaches −12

monotonely along Ct0 from the left. �

This observation, combined with the following lemma, makes it easy to prove The-orem 4.17 on page 192.�

�Lemma 4.30. Let K(an/bn) be a p-periodic continued fraction of parabolictype with value x, and let V ⊆ C be a closed set �= C, containing at leasttwo points. If Sp(V ) ⊆ V , then x ∈ ∂V .

Proof : x ∈ V since K(an/bn) converges to x. Assume that x ∈ V ◦. Then everyfixed circle C of Sp contains an arc A ⊆ V with x an inner point in A. For everytail sequence {tn} for K(an/bn), the points tnp lies on such an arc from some n

on. Since Sp(V ) ⊆ V , this means that t0 = Snp(tnp) ∈ V for all t0 ∈ C which isimpossible. Hence x ∈ ∂V . �

Proof of Theorem 4.17: Since an → − 14, the convergence of Sn(wn) with

wn ∈ Vα,n := −gn + eiαH for n = 0, 1, 2, . . . (4.3.2)

follows from the Parabola Sequence Theorem on page 154. Therefore {Sn} and thusalso {S−1

n } are restrained. Let gn → 12. Then Vα,n → Vα := − 1

2+ eiαH. Assume

that K(an/1) has a tail sequence {tn} with a limit point x �= − 12 . The fixed circle

Cx of s(w) := −14/(1 + w) through x is a circle tangent to the real line at − 1

2or

the line itself. And every term in the two sequences {s[n](x)} and {(s−1)[n](x)} onCx is also a limit point for {tn} since

tnk→ x =⇒ tnk−1 = snk

(tnk) → s(x), tnk−2 → s[2](x), · · ·

tnk→ x =⇒ tnk+1 = s−1

nk(tnk

) → s−1(x), tnk+2 → (s−1)[2](x), · · ·

Since s[n](x) → − 12 from one side and (s−1)[n](x) → −1

2 from the other side of − 12

along Cx, a subarc of Cx with − 12 as an inner point must be contained in Vα. This

is a contradiction, and tn → − 12 .

This also means that Sn(w) → f for every w �= − 12 since the exceptional sequences

for {Sn} converge to − 12 .

Page 223: Lisa Lorentzen, Haakon Waadeland Continued Fractions

210 Chapter 4: Periodic and limit periodic continued fractions

Let ρn := gn−1(1 − gn) − 14≥ 0 from some n on. Without loss of generality we

assume that

Σ∞ :=∞∑

n=0

n∏k=1

1 − gk

gk= ∞ (remarks on page 136). (4.3.3)

Then the limit point case occurs for Sn(Vα,n) (the Parabola Sequence Theorem).We know from the remark on page 194 that 1

2 ≤ gn → 12 . Therefore − 1

2 ∈ Vα,n

from some n on, so also Sn(−12) → f . �

The loxodromic case.

If Sp is hyperbolic, then {Snp+m(0)}n approaches x along CSm(0), and {tnp+m}n

with t0 �= x, y approaches y(m) along C(m)tm

. If � > 0, then this convergence ismonotone on CSm(0) and C

(m)tm

respectively. If � < 0, they converge in an alternatingmanner.

If Sp is loxodromic with � �∈ R, then {Snp+m(0)}n and {tnp+m}n with t0 �= x spiralin towards their limits x and y(m) respectively.

The elliptic case.

A p-periodic continued fraction of elliptic type diverges, of course, and {Sn} istotally non-restrained (Theorem 4.3). But elliptic transformations have fixed circles,so:

• the points Snp+m(w) ∈ CSm(w) and S−1np+m(w) ∈ C

(m)

S−1m (w)

for all n, and theymove around and around indefinitely as n increases.

• If the ratio � of Sp is an Nth root of unity, then {Snp+m}∞n=1 and {S−1np+m}∞n=1

are N -periodic for each m ∈ {1, 2, . . . , p}. Otherwise the points {Snp+m(w)}∞n=1

and {S−1np+m(w)}∞n=1 are dense on CSm(w) and C

(m)

S−1m (w)

respectively.

Example 20. The continued fraction K(−2/1) is 1-periodic of elliptic type sinceS1(w) = s1(w) = −2/(1 + w) has the two fixed points x := (−1 + i

√7)/2 and

y = (−1− i√

7)/2 which gives the ratio � = (1 + y)/(1 + x) = (1− i√

7)/(1 + i√

7).Now, � seems to be no Nth root of unity. If this is as it seems, then the sequence{Sn(w)} of approximants and the tail sequence {S−1

n (w)} with w �= x, y consist ofdistinct points dense on the fixed circle Cw for S1. For instance, {fn} and {tn}with t0 ∈ R are dense on R. Below we have recorded the first few terms of {fn}and {tn} with t0 := 1:

{fn} : −2, 2, −23, −6,

25, −10

7,

143

, − 617

, −3411

,2223

, . . .

{tn} : 1, −3, −13, 5, −7

5,

37, −17

3, −11

17,

2311

, −4523

, . . . .

Page 224: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Remarks 211

The identity case.

Let K(an/bn) be a p-periodic continued fraction of identity type. That is, Sp(w) ≡w. Then Snp+m = Sm for all n ∈ N and m ∈ {1, 2, . . . , p}, and both {Sn(w)} and{S−1

n (w)} are p-periodic. This can not happen with p = 1 since S1(w) = a1/(b1+w).Indeed, since S2(w) = a1(b2 + w)/(b1b2 + a2 + b1w) ≡ w only if b1 = b2 = 0 anda1 = a2, and thus K(an/bn) is 1-periodic of elliptic type, we need p ≥ 3.

Example 21. For the 3-periodic continued fraction∞Kn=1

an

bn= −2

2 −21 −

11 −

22 −

21 −

11 −

22 − · · · ,

S3(w) = w, so K(an/bn) is of identity type with

S3n(w) = w, S3n+1(w) =−2

2 + w, S3n+2(w) = −1 + w

w

for all n. Hence {Sn(0)} is the 3-periodic sequence −1, ∞, 0, −1, ∞, 0, −1, . . . ,and the tail sequence starting with t0 := 1 is the 3-periodic sequence given by1, −4, −1

2 , 1, −4, . . . . �

4.5 Remarks

1. Periodic continued fractions. According to Jones and Thron ([JoTh80], p 1),the first continued fractions in the literature were, as far as we know, terminatingperiodic continued fractions for approximating square roots. But the first study ofinfinite periodic continued fractions seems to be due to Daniel Bernoulli ([Berno75])who studied the 1-periodic continued fraction K(1/b). Periodic continued fractionswas also the topic of E. Galois’ first published paper ([Galo28]). Here he studiedreversed regular periodic continued fractions. The first comprehensive study of pe-riodic continued fractions in general seems to be due to Stolz ([Stolz86]). Lateron, a number of authors have added to the theory, in particular Thiele ([Thie79]),Pringsheim ([Prin00]) and Perron ([Perr05]).

2. Ratio and trace for τ ∈ M. We have chosen to use the ratio � to classify thetransformations τ from M. An equivalent criterion is based on the trace tr(τ) :=a + d of

τ(w) :=aw + b

cw + dwhere Δ := ad − bc = 1.

The connection between � and tr(τ) when Δ = 1 is

� =1 − u

1 + uwhere u :=

√1 − Δ

tr(τ)2=√

1 − tr(τ)−2 .

3. Limit periodic continued fractions. The systematic work on limit periodiccontinued fractions started with Van Vleck ([VanV04]). He studied S-fractionsK(anz/1) where 0 < an → a. Later on Pringsheim ([Prin10]) and Perron ([Perr13])made substantial contributions to the theory. This was later simplified in ([Perr29]).

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212 Chapter 4: Periodic and limit periodic continued fractions

4. Tail sequences and the linear recurrence relation. A sequence {tn} is a tailsequence for K(an/bn) if and only if tn = −Xn/Xn−1 for all n where {Xn} is anon-trivial solution of the recurrence relation

Xn = bnXn−1 + anXn−2 for n = 1, 2, 3, . . . .

Theorem 4.13 on page 188 can therefore also be seen as a consequence of Poincare’swork on recurrence relations ([Poin85]) if all am �= 0.

5. Periodic sequences of element sets. The remark on page 189 shows that ifK(an/bn) is a p-periodic continued fraction of loxodromic type, then there exists ap-periodic sequence {Ωn} of element sets such that

• K(an/bn) is from {Ω◦n}, and

• every K(an/bn) from {Ωn} converges generally.

For the case p = 2, several such sequences {Ωn} have been derived by Thronand coauthors in a number of papers ([Thron43], [Thron59], [LaTh60], [Lange66],[JoTh68], [JoTh70], [Lore08a]. Results for larger p can be found in [Jaco82] and[Jaco87].

6. Continued fractions and conjugation. In Remark 3 on page 47 we indicatedthat our definition of continued fractions, as a sequence {Sn} of linear fractionaltransformations with Sn(∞) = Sn−1(0), could be generalized to give definitionsinvariant under conjugation.

4.6 Problems

1. ♠ Diagonalization of matrices. In Problem 12 on page 49 we saw that if

τn :=anw + bn

cnw + dnwas identified by Tn :=

(an bn

cn dn

),

then τ1 ◦ τ2 was identified by the matrix product T1T2.

(a) Prove that if τ1 is parabolic, then the composition ϕ ◦ τ ◦ ϕ−1(w) = q + w in

(1.2.6) on page 173 correspond to diagonalization of the matrix T1 :=

(a bc d

).

(b) Prove that if τ is elliptic or loxodromic, then the composition ϕ◦ τ ◦ϕ−1(w) =�w in (1.2.8) on page 173 also corresponds to diagonalization of T1.

2. ♠ Ratio and derivative. Let x be a finite fixed point for τ ∈ M with ratio �.Prove that then τ ′(x) = � or τ ′(x) = 1/�.

3. Convergence of periodic continued fractions. Prove that K(an/bn) convergesif an = 2 and bn = 4eiθ for all n.

4. Convergence of periodic continued fractions. Given the continued fraction

b +

Kn=1

a

2b= b +

a

2b+

a

2b+

a

2b+· · ·

Page 226: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Problems 213

(a) Prove that if a > 0 and b > 0 then b + K(a/2b)converges to√

a + b2. Use thisto find a rational approximation to

√13 with an error less than 10−4.

(b) For which values of (a, b) ∈ C × C does b + K(a/2b) converge/diverge?

5. ♠ Convergence neighborhood of loxodromic transformation. For given 0 �=a ∈ C, let A and B be the two statements:

A: There exists a region (open, connected set) E with a ∈ E such that everycontinued fraction K(an/1) from E converges.

B: s(w) := a/(1 + w) is a loxodromic transformation.

Show that A and B are equivalent.

6. ♠ Reversed continued fraction. The reversed continued fraction of a p-periodiccontinued fraction

Kn=1

an

bn=

a1

b1 +

a2

b2 +· · ·+ap

bp +

a1

b1 +

a2

b2 +· · ·+ap

bp +

a1

b1 +· · ·

is the p-periodic continued fraction

bp +ap

bp−1 +

ap−1

bp−2 +· · ·+a1

bp +

ap

bp−1 +

ap−1

bp−2 +· · ·+a1

bp +

ap

bp−1 +· · ·.

Let {Sn(w)} and {Sn(w)} be the approximants of K(an/bn) and its reverse, respec-tively. Prove that K(an/bn) converges generally if and only if its reversed continuedfraction converges generally.

7. ♠ Approximants of periodic continued fractions.

(a) Show that the approximants Sn(w) for the 1-periodic continued fraction K(− 14/1)

can be written

Sn(w) = −1

2+

w + 1/2

2nw + n + 1for n = 1, 2, 3, . . . .

Draw a picture of the fixed line for Sn and the fixed circle for Sn throughw0 := − 1

2+ 2i, and mark the approximants Sn(0) and Sn(w0) for n = 1, 2, 3.

(b) Find a closed expression for the nth term of a tail sequence {tn} for K(− 14/1),

expressed in terms of t0 and n, and verify that {tn} approaches − 12

from theleft when t0 := −1.

8. ♠ Approximants for 2-periodic continued fractions K(an/1). For givenx �= −1, 0,∞, find general formulas for the classical approximants of the 2-periodiccontinued fraction

−x2

1 −(1 + x)2

1 −x2

1 −(1 + x)2

1 −x2

1 −· · ·.

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214 Chapter 4: Periodic and limit periodic continued fractions

9. ♠ Divergence of limit periodic continued fractions of parabolic type. LetK(an/1) have a tail sequence {tn} given by

tn := −1

2− 1/4 + i μ

n + 1for n = 0, 1, 2, . . .

for some μ > 0. Prove that then K(an/1) is a divergent limit 1-periodic continued

fraction of parabolic type with an = − 14− 1/16+μ2

n(n+1).

10. General and classical convergence.

(a) Show that if K(a/b) is a 1-periodic continued fraction which converges gener-ally to f , then it converges to f also in the classical sense.

(b) Show that if K(an/1) is a 2-periodic continued fraction which converges gen-erally to f , then it converges to f also in the classical sense.

(c) Give an example of a periodic continued fraction which converges in the generalsense but not in the classical sense.

11. Thiele oscillation. For which values of z does the 4-periodic continued fraction

Kn=1

an

1=

1

1+

1

1−2

1−2z

1 +

1

1+

1

1−2

1−2z

1 +· · ·(a) converge/diverge generally?

(b) oscillate by Thiele oscillation?

(c) converge in the classical sense?

What is the value of K(an/1) when it converges generally? Determine the limitingbehavior of its tail sequences in this case.

12. 2-periodic continued fractions and the Parabola Theorem. Prove thatK(an/1) with all a2n−1 := a and a2n := a converges if and only if |a| − Re a ≤ 1

2.

(Here a is the complex conjugate of a.)

13. Convergence of (limit) periodic continued fractions.

(a) For which values of z ∈ C with |z| = 1 does the periodic continued fractionK( 5z

4/1) (i) converge? (ii) diverge?

(b) Find the periodic tail sequences for K( 5z4

/1).

(c) For which values of z ∈ C with |z| = 1 does the continued fraction K(anz/1)converge when an := (5n2 + 1)/(4(n + 1)2) for all n?

14. ♠ Convergence of limit periodic continued fractions. For given non-negativeintegers p, q and r, let

a2n−1 :=

p∑k=0

αknk, a2n :=

p∑k=0

γknk,

b2n−1 :=

q∑k=0

βknk, b2n :=r∑

k=0

δknk

Page 228: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Problems 215

be polynomials in n for n = 1, 2, 3, . . . , where all αk, γk, βk and δk are complexnumbers with αpγpβqδr �= 0. Prove that K(an/bn) converges if either (a), (b), (c)or (d) holds.

(a) q + r > p.

(b) q + r = p andαpγp

(αp + γp + βqδr)2�∈[

14,∞]

.

(c) q + r = p − 1, αp = γp and βqδr/αp �∈ [−∞, 0] .

(d) q + r = p − 2, αp = γp and arg αp = arg(αp−1 − αp

(βq−1

βq+

δr−1δr

))and

βqδr/αp �∈ [−∞, 0].

Hint: You may use the Parabola Theorem on page 151 and Theorem 3.4 on page105.

15. Convergence of a limit periodic continued fraction. Prove that K(an/1) with

a2n := 1 − 4n2, a2n+1 := −4n2

converges. (Hint: tn := (−1)nn is a tail sequence for K(an/1).) What is the valueof K(an/1)? Does K(−an/1) converge?

16. Convergence of a limit periodic continued fraction. Prove that K(anz/1)with

a2n−1 := a + n, a2n := b + n for n = 1, 2, 3, . . .

converges for a, b, z ∈ C with | arg z| < π (Wall [Wall45]).

17. ♠ Fixed circles for parabolic transformations. Let τ(w) := (aw + b)/(cw + d)be a parabolic transformation from M with fixed point x �= ∞. Prove that x+((a+

d)/2c)R is the fixed line for τ .

18. ♠ Fixed circles for parabolic transformations. Let τ(w) := (aw + b)/(cw + d)be a parabolic transformation from M with finite fixed point x and pole ζ. Provethat the circle with center Γ ∈ C and radius ρ > 0 is a fixed circle for τ if and onlyif there exists a real number t �= 0 such that

Γ = x + i(ζ − x)t, ρ = |(ζ − x)t| .

19. ♠ Fixed circles for parabolic transformations. Let {tn} be a tail sequence forK((− 1

4)/1).

(a) Prove that if t0 = − 12, then tn = − 1

2for all n, and that if t0 �= − 1

2, then all tn

are distinct points on the fixed circle Ct0 , approaching − 12

monotonely fromthe left.

(b) Let x ∈ C \ {0,−1} be arbitrarily chosen. Prove that the 2-periodic continuedfraction

K an

1=

−x2

1 +

−(1 + x)2

1 +

−x2

1 +

−(1 + x)2

1 +· · ·is of parabolic type with tail values f (2n) = x and f (2n+1) = −(1 + x).

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216 Chapter 4: Periodic and limit periodic continued fractions

(c) Let Sn(w) be the approximants of K(an/1) in (b), and let Cw and C(1)w de-

note the fixed circles through w for S2 and S(1)2 respectively. Explain why all

S2n(w) ∈ Cw and S2n+1(w) ∈ Cs1(w) = C−x2/(1+w).

(d) Let {tn} be a tail sequence for K(an/1). Explain why t2n → x, t2n+1 →−(1 + x), and t2n ∈ Ct0 , t2n+1 ∈ C

(1)t1

for t0 �= x.

(e) Prove that the fixed line for S2 in (b) is the line through x and S−12 (∞) =

x(2 + x).

(f) Sketch some fixed circles for S2 in (b) with x := i.

20. Fixed circle for elliptic transformation. Prove that the circle with center atγ and radius ρ is a fixed circle for the elliptic transformation s(w) := −2/(1 + w)when

(a) γ := − 12

+ 2i, ρ := 32. (b) γ := − 1

2+ 2

3

√7 i, ρ := 7

6.

21. Fixed circle for elliptic transformation. Let C be the circle with center at− 1

2+ iq and radius r where 0 < r < q. Prove that C is a fixed circle for s(w) :=

(− 14− t2)/(1 + w) with 0 �= t ∈ R if and only if q2 − t2 = r2.

Page 230: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Chapter 5

Numerical computation ofcontinued fractions

If K(an/bn) converges generally to f , then its approximants {Sn(wn)} converge to fas long as {wn} stays asymptotically away from its exceptional sequence. Thereforewe can use Sn(wn) as an approximation to f . However, the choice of {wn} makesquite a difference. For one thing, if one is looking for, say, a rational approximant,then this limits the choice for {wn}. But often one just wants fast convergence tof . In this chapter we suggest a number of ideas for how to choose {wn} to this aim.

If we still insist on Sn(wn) being rational, we chose the elements an and bn ofK(an/bn) to be polynomials or rational functions, and use a rational approximationwn to wn to get the rational approximants Sn(wn).

Even more important from a practical point of view is the control of the truncationerror |f − Sn(wn)|. One needs to have explicit bounds for this error in terms of n,in order to take full advantage of this faster convergence. The second part of thischapter deals with truncation error analysis, and we show how one can derive good(i.e., tight) and reliable truncation error bounds.

We also need a method to compute Sn(wn) in a stable manner. Luckily the back-ward algorithm is stable in most cases, and good choices for {wn} even stabilize itfurther. In the last part of this chapter we indicate why and how this works.

L. Lorentzen and H. Waadeland, Continued Fractions, Atlantis Studies in Mathematics for Engineering and Science 1, DOI 10.1007/978-94-91216-37-4_5, © 2008 Atlantis Press/World Scientific

217

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218 Chapter 5: Computation of continued fractions

5.1 Choice of approximants

5.1.1 Fast convergence

Let K(an/bn) converge generally to f and let {f (n)} be its sequence of tail values.For simplicity we assume that f �= ∞. We can do so without loss of generality,since if f = ∞, then f (1) �= ∞, and∣∣∣ 1

Sn(w)

∣∣∣ = ∣∣∣ 1f− 1

Sn(w)

∣∣∣ = ∣∣∣f (1) − S(1)n−1(w)

a1

∣∣∣ (1.1.1)

is a measure for the truncation error. Here, as always, S(k)n (w) denotes the nth

approximant of the kth tail of K(an/bn). For f �= ∞ we want bounds for the trun-cation error |f −Sn(wn)|, and the sequence {ζn} with ζn := S−1

n (∞) = −Bn/Bn−1

is an exceptional sequence for K(an/bn).

We know that Sn(wn) → f whenever {wn} stays asymptotically away from anexceptional sequence {w†

n}. But some choices of {wn} give faster convergence thanothers. Since f = Sn(f (n)), it seems reasonable to expect that Sn(wn) convergesfaster the better wn approximates f (n). Since by Lemma 1.1 on page 6

f − Sn(w)f − Sn(v)

=Sn(f (n)) − Sn(w)Sn(f (n)) − Sn(v)

=Bn−1v + Bn

Bn−1w + Bn· w − f (n)

v − f (n)=

v − ζn

w − ζn· w − f (n)

v − f (n)

(1.1.2)

(with the natural limit forms if f (n) = ∞ or ζn = ∞), we have

f − Sn(wn)f − Sn(vn)

→ 0 ⇐⇒ κn(wn)κn(vn)

→ 0 where κn(w) :=w − f (n)

w − ζn. (1.1.3)

We say that {Sn(wn)} accelerates the convergence of K(an/bn) when

limn→∞

f − Sn(wn)f − Sn(0)

= 0. (1.1.4)

The quantity κn(wn) is a measure for how well Sn(wn) approximates f for a givencontinued fraction. It is essentially the ratio of wn’s distances to the “good tailsequence” {f (n)} and the bad tail sequence” {ζn}. Hence even a rather lousyapproximation to f (n) may work well. In particular we do not need high precisionin the computation of wn to obtain a reasonably good approximation Sn(wn) tof . But {f (n)} is in general unknown. Still, if we have some information on itsasymptotic behavior, as we for example often have for limit periodic continuedfractions, then this can be used to find favorable wn. To determine such asymptoticbehavior one may need to transform the continued fraction. The good thing abouttransformations like equivalence transformations or contractions, is that we knowexactly what happens to tail sequences. It is also clear that if {wn} accelerates the

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5.1.2 The fixed point method 219

convergence of the kth tail of K(an/bn), then it also accelerates the convergence ofK(an/bn).

5.1.2 The fixed point method

This method gives the obvious” approximants for limit periodic continued frac-tions.

The loxodromic case.

Let K(an/bn) be limit p-periodic of loxodromic type with attracting fixed points(x(0), . . . , x(p−1)) and repelling fixed points (y(0), . . . , y(p−1)). Then we know fromTheorem 4.13 on page 188 that

• K(an/bn) converges generally to some f ∈ C,

• limn→∞ f (np+m) = x(m) for m = 1, 2, . . . , p

• limn→∞ w†np+m = y(m) �= x(m) for m = 1, 2, . . . , p for the exceptional se-

quences {w†n} for K(an/bn).

Therefore it is a very good idea to use the approximants Snp+m(x(m)). Indeed, by(1.1.3)

limn→∞

f − Snp+m(x(m))f − Snp+m(wnp+m)

= 0 for f �= ∞ (1.2.1)

for every other sequence {wn} with lim infn→∞ m(wnp+m, x(m)) > 0. In particular,

limn→∞

f − Snp+m(x(m))f − Snp+m(0)

= 0 for x(m) �= 0, f �= ∞. (1.2.2)

(If x(m) = 0, then Snp+m(0) can already be seen as a result of the fixed pointmethod.) The effect of this choice of approximants has been demonstrated in anumber of examples in Chapter 1. Of course, it works better the faster f (np+m)

approaches x(m), but there is always a positive effect in the long run. Since there isno extra work to speak of to compute Snp+m(x(m)) instead of Snp+m(0), we stronglyrecommend the use of Snp+m(x(m)) (or even faster converging approximants to besuggested later in this chapter) whenever convenient.

The parabolic case.

In the parabolic case the question is much more subtle, since {f (np+m)}n and{ζnp+m}n normally both converge to the attracting fixed point x(m) when K(an/bn)

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220 Chapter 5: Computation of continued fractions

converges. To get an impression of what to expect, we shall investigate some specialcases of the situation

K(an/1) with an → −14 and wn := − 1

2 , and

an ∈ Eα,n := {a ∈ C; |a| − Re(a e−2iα) ≤ 2gn−1(1 − gn) cos2 α}(1.2.3)

for some fixed α ∈ R with |α| < π/2 and 0 < ε ≤ gn ≤ 1−ε < 1 with gn−1(1−gn) ≥14

for all n. It follows then from the Parabola Sequence Theorem on page 154 thatK(an/1) converges to a finite value f ∈ −g0 + eiαH. Moreover Sn(−1

2) → f

(Theorem 4.17 on page 192). The question is now whether

limn→∞

f − Sn(−12 )

f − Sn(0)→ 0.

By (1.1.2),

f − Sn(−12)

f − Sn(0)=

κn(−12)

κn(0)=

f (n) + 12

f (n)· ζn

ζn + 12

∼f (n) + 1

2

ζn + 12

. (1.2.4)

To control this ratio, we assume that

|an + 14 | ≤ ρn := gn−1(1 − gn) − 1

4 for n = 1, 2, 3, . . . . (1.2.5)

Then either ρn = 0 for all n, or 0 < ρn → 0 monotonely (Lemma 4.19 on page 193).�

�Lemma 5.1. Let K(an/1) satisfy (1.2.5). Then {Vn}∞n=0 given by Vn :=B(−1

2 , gn − 12 ) is a sequence of value sets for K(an/1), and f (n) ∈ Vn for

all n.

Proof : We first prove that {Vn} is a sequence of value sets for K(an/1): letw := −1

2 + R eiψ ∈ Vn; that is, R ≤ gn − 12 . Since an = −1

4 + r eiθ for some0 ≤ r ≤ ρn, we get∣∣∣∣ an

1 + w+

12

∣∣∣∣ =∣∣∣∣∣ − 1

4 + r eiθ

1 − 12

+ R eiϕ+

12

∣∣∣∣∣ =∣∣∣∣∣r eiθ + 1

2R eiϕ

12

+ R eiϕ

∣∣∣∣∣≤

ρn + 12(gn − 1

2)12− (gn − 1

2)

=gn−1(1 − gn) − 1

4 + 12gn − 1

4

1 − gn= gn−1 − 1

2 .

Hence an/(1 + w) ∈ Vn−1.

Without loss of generality we assume that

∞∑n=0

Pn = ∞ for Pn :=n∏

k=1

1 − gk

gk(1.2.6)

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5.1.2 The fixed point method 221

(Theorem 3.31 on page 137). Then {Sn(V0,n)} with V0,n := −gn + H convergesto a limit point f (Theorem 3.45 on page 155). In particular Sn(wn) → f forwn ∈ Vn ⊆ V0,n. Since Sn(wn) ∈ V0 for all n, also f ∈ V0. Similarly f (n) =limk→∞ S

(n)k (wn+k) ∈ Vn for all n. �

Therefore, if an ∈ B(− 14 , ρn) for all n, then |f (n)+ 1

2 | ≤ gn− 12 and |ζn + 1

2 | > gn− 12

(ζn �∈ Vn since Sn(ζn) = ∞ �∈ V0), and we can guarantee that

lim supn→∞

∣∣∣f − Sn(−12)

f − Sn(0)

∣∣∣ ≤ 1. (1.2.7)

So, at least we do not loose much in the long run by using Sn(−12 ) instead of Sn(0)

if |an + 14 | ≤ ρn from some n on. But can we gain? The answer is yes: let {gn} be

a second sequence of positive numbers < 1 such that

0 < ρn := gn−1(1 − gn) − 14 ≤ ρn and

gn − 12

gn − 12

→ 0. (1.2.8)

Then the following holds:

Theorem 5.2. Let {gn} and {gn} with 12 < gn, gn < 1 be given such that

(1.2.8) holds, where ρn is given by (1.2.5). Then K(an/1) with |an+ 14| ≤ ρn

for all n ∈ N, converges to a finite value f and

limn→∞

∣∣∣f − Sn(−12)

f − Sn(0)

∣∣∣ = 0. (1.2.9)

Proof : This time |f (n) + 12| ≤ gn− 1

2since {B(−1

2, gn− 1

2)} is a sequence of value

sets for K(an/1). But also {B(− 12 , gn − 1

2 )} is a sequence of value sets for K(an/1)since |an + 1

4| ≤ ρn ≤ ρn. Therefore |ζn + 1

2| > gn − 1

2, and the result follows since

∣∣∣f − Sn(−12)

f − Sn(0)

∣∣∣ ∼ ∣∣∣f (n) + 12

ζn + 12

∣∣∣ ≤ gn − 12

gn − 12

→ 0.

The conditions in this theorem are really quite restrictive. Indeed, since gn →12 monotonely, we know from (2.5.10)-(2.5.11) on page 139 that gn−1(1 − gn) ∼1/(16n2) is the best we can do. With the standard little o - notation; i.e., mn =o(kn) means that limn→∞(mn/kn) = 0, we get:

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222 Chapter 5: Computation of continued fractions

Corollary 5.3. Let K(an/1) with |an + 1

4| = o(n−2) have a finite value f .

Then (1.2.9) holds.

Proof : Let gn := 12 + 1/(4n + 2) for n ≥ 0. Then

ρn := gn−1(1 − gn) − 14 =

116n2 − 4

= (gn−1 − 12)(gn − 1

2 )

and gn−1 − gn =4

16n2 − 4= 4(gn−1 − 1

2)(gn − 12 ) = 4ρn.

(1.2.10)

Then B(− 14, ρn) ⊆ E0,n given by (1.2.3) for all n. Since ρn := |an + 1

4| = o(ρn), we

have an ∈ B(−14 , ρn) from some n on. Without loss of generality we assume that

this holds for all n ≥ 1. Let an := −14 − |an + 1

4 |. Then also an ∈ B(− 14 , ρn) for all

n, so K(an/1) converges, and its nth tail value, which we denote by −gn, belongsto B(− 1

2 , gn− 12 ) for all n (Lemma 5.1). This means that 1−gn ≤ gn ≤ gn. Indeed,

since all an = −gn−1(1 − gn) ≤ − 14, it follows from Lemma 4.19 on page 193 and

the subsequent remark that 12 ≤ gn → 1

2 monotonely, so 12 ≤ gn ≤ gn.

Now, an ∈ B(−14, ρn) where (ρn/ρn) → 0. We want to prove that

δn :=gn − 1

2

gn − 12

→ 0.

Straight forward computation, using (1.2.10), shows that

ρn

ρn=

gn−1(1 − gn) − 14

ρn=

12(gn−1 − gn) − (gn−1 − 1

2 )(gn − 12)

(gn−1 − 12)(gn − 1

2)

=12(gn−1 − gn)

(gn−1 − 12)(gn − 1

2)− δn−1δn =

12

(gn−1 − 12 ) − (gn − 1

2)(gn−1 − 1

2)(gn − 1

2)

− δn−1δn

=12

δn−1

gn−1 − 12

gn − 12

− δn

gn−1 − 12

− δnδn−1

where 1 < (gn−1 − 12)/(gn − 1

2) → 1. Assume that a subsequence δnk−1 → δ > 0.Since (gnk−1 − 1

2) → 0, this means that also δnk

→ δ. Therefore δn → δ. However,this implies that

ρn = 12(gn−1 − gn) − (gn−1 − 1

2)(gn − 1

2)

∼ δ · 12 (gn−1 − gn) − δ2(gn−1 − 1

2 )(gn − 12) = δ(2ρn − δρn) = δ(2 − δ)ρn

which is impossible since ρn = o(ρn) and 0 < δ ≤ 1. Hence δn → 0. �

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5.1.3 Auxiliary continued fractions 223

Remark. If in particular all |an + 14| ≤ C n−d for some C > 0 and d > 2, or if

|an + 14 | ≤ C rn for some C > 0 and 0 < r < 1, then we can use

gn :=12

+1

4n + 2and gn :=

12

+1

nδ+1or gn :=

12

+ qn

from some n on in Theorem 5.2, where δ > d− 2 and r < q < 1 are fixed constants.Then |ζn + 1

2 | > gn and |f (n) + 12 | < gn from some n on, and thus it follows from

Corollary 5.3 and (1.2.8) that∣∣∣f − Sn(−12 )

f − Sn(0)

∣∣∣ = {o(n−δ) when |an + 14 | ≤ C n−d,

o(ρn) when |an + 14| ≤ C rn

(1.2.11)

when f �= ∞, as first proved by Thron and Waadeland ([ThWa80a]).

But what if K(an/bn) has a more general form? If K(an/bn) is limit 1-periodic withan/bn−1bn → −1

4 , it is equivalent to a continued fraction K(cn/dn) where cn → − 14

and dn = 1 from some n on.

If K(an/bn) is limit p-periodic, we can for instance use a p-contraction to approxi-mate the value of K(an/bn). This contraction is limit 1-periodic, and can possiblybe accelerated as described above.

5.1.3 Auxiliary continued fractions

The fixed point method can also be interpreted as follows: let K(an/bn) be limitp-periodic with finite limits

limn→∞ anp+m = am �= 0, lim

n→∞ bnp+m = bm for m = 1, 2, . . . , p.

Then its Nth tail looks more and more like the Nth tail of the correspondingperiodic continued fraction K(an/bn) when N increases. We have proved (Theorem4.13 on page 188) that if K(an/bn) is of loxodromic type, then the sequence of tailvalues for K(an/bn) has the same asymptotic behavior as the sequence {f (n)} oftail values for K(an/bn). Hence the approximants Snp+m(f (np+m)) = Snp+m(x(m))accelerate the convergence of K(an/bn) when all x(m) �= 0.

There is no reason why this should only work for pairs of periodic – limit periodiccontinued fractions. More generally, let K(an/bn) and K(an/bn) be two generallyconvergent continued fractions where

m(an, an) → 0 and m(bn, bn) → 0. (1.3.1)

Then their tail values f (n) and f (n) satisfy m(f (n), f (n)) → 0 under proper con-ditions. We say that K(an/bn) is an auxiliary continued fraction for K(an/bn),and {Sn(wn)} with wn := f (n) ought to be a good approximation to the value ofK(an/bn).

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224 Chapter 5: Computation of continued fractions

Example 1. The continued fraction K(n2/1) converges to some f �= ∞ (theParabola Theorem). Its convergence is slow, so it is a good idea to look for {wn}such that Sn(wn) → f faster.

Now, K(an/1) is close to the auxiliary continued fraction K((n2 − 1)/1) whichconverges with tail values f (n) := n. (This follows since f (n−1)(1 + f (n)) = n2 −1 and

∑∞n=0

∏nk=1(1 + f (k))/(−f (k)) = ∞.) It seems reasonable to believe that

approximants Sn(wn) with wn := n is a good choice for K(an/1). The table belowindicates that Sn(n) approximates the value f = 0.4426950409 (correctly roundedto 10 decimals) reasonably well, and far better than Sn(0). For instance, in orderto get an absolute error less than 0.002, we need n to be ≈ 500 for Sn(0), but only≈ 10 for Sn(n).

n Sn(0) Sn(n) |f − Sn(0)| |f − Sn(n)|1 1.00000000 0.50000000 −5.6 · 10−1 −5.7 · 10−2

2 0.20000000 0.42857143 2.4 · 10−1 1.4 · 10−2

3 0.71428571 0.44827586 −2.7 · 10−1 −5.6 · 10−3

23 0.48644179 0.44271504 −4.4 · 10−2 −2.0 · 10−5

24 0.40302141 0.44267739 4.0 · 10−2 1.8 · 10−5

25 0.48305030 0.44271070 −4.0 · 10−2 −1.6 · 10−5

501 0.44476903 0.44269504 −2.1 · 10−3 −2.1 · 10−9

502 0.44063101 0.44269504 2.1 · 10−3 2.0 · 10−9

503 0.44476080 0.44269504 −2.1 · 10−3 −2.0 · 10−9

Example 2. The limit periodic continued fraction K(an/1) with an := 110+210/nfor n ≥ 1 is of loxodromic type. It converges with tail values f (n) = 10(n+2)/(n+1)for n ≥ 0. (This can be seen as in the previous example.) To compute the value ofK((110 + 215/n)/1) we use approximants Sn(f (n)). The fixed point method givesapproximants Sn(10). The table below shows the values of the approximants Sn(0),Sn(10) and Sn(f (n)) for K((110 + 215/n)/1) and the corresponding truncation er-rors. The true value of K((110 + 215/n)/1) is f = 20.18548937, correctly roundedto 8 decimals.

n Sn(0) Sn(10) Sn(10n+2n+1

)

1 325.00000000 29.54545455 20.312500002 1.48741419 15.64551422 20.093457943 148.35485214 24.22152021 20.255741544 3.11185304 17.57667608 20.1306584197 20.19073608 20.18551791 20.1854900398 20.18071637 20.18546367 20.1854887899 20.18983340 20.18551253 20.18548991100 20.18153746 20.18546851 20.18548889

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5.1.3 Auxiliary continued fractions 225

n f − Sn(0) f − Sn(10) f − Sn(10 n+2n+1

)

1 −304.8 −9.4 −0.132 18.7 4.5 0.0923 −128.2 −4.0 −0.0704 17.1 2.6 0.05597 −5.2 · 10−3 −2.9 · 10−5 −6.6 · 10−7

98 4.8 · 10−3 2.6 · 10−5 5.9 · 10−7

99 −4.3 · 10−3 −2.3 · 10−5 −5.4 · 10−7

100 4.9 · 10−3 2.1 · 10−5 4.8 · 10−7

This continued fraction converges faster than the one in the previous example. Still,the improvement is more than worth the small effort it takes to achieve it. To getan absolute error ≤ 0.005, we need n ≈ 100 for Sn(0), n ≈ 50 for Sn(10) and n ≈ 10for Sn(10n+2

n+1 ). �

But when does (f (n)−f (n)) → 0 hold more generally? Let {Ωn}∞n=1 be a sequence ofelement sets for continued fractions K(an/bn). Let further {Vn}∞n=0 be a sequenceof value sets for {Ωn} and let dist(x, V ) denote the euclidean distance between apoint x ∈ C and a set V ⊆ C. We then follow ([Jaco83], [Jaco87]) and define:�

Definition 5.1. The sequence {Ωn}∞n=1 of element sets is a tusc with re-spect to its sequence {Vn}∞n=0 of value sets if there exists a sequence {λn};0 < λn → 0, such that

diamS(m)n (Vm+n) ≤ λn for all m,n ∈ N (1.3.2)

for every continued fraction K(an/bn) from {Ωn}.

Remarks.

1. The name tusc is an abbreviation for totally uniform sequence of convergencesets”.

2. Let {Ωn} be a tusc with respect to {Vn} where lim inf diamm(Vn) > 0. (Asalways, diamm(Vn) is the chordal diameter of Vn.) Then every continuedfraction K(an/bn) from {Ωn} converges generally (the limit point case oc-curs). The convergence of {Sn(wn)} is uniform with respect to wn ∈ Vn andK(an/bn) from {Ωn}.

3. If {En} is a tusc with respect to {Vn} for continued fractions K(an/1), wherelim inf diammVn > 0, then {En} is bounded; i.e., M := sup{|an|; an ∈ En, n ∈N} < ∞, since the euclidean diameter

diamam+1

1 + Vm+1≤ λ1 for all m ∈ N ∪ {0}.

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226 Chapter 5: Computation of continued fractions

If moreover dist(−1, Vn) ≥ ε > 0 for all n, then |S(m)n (w)| ≤ M/ε for all

w ∈ Vm+n and |f (m)| ≤ M/ε for every continued fraction from {En}. We cantherefore assume without loss of generality that also {Vn} is bounded.

4. Let {Eα,n} and {Vα,n} be the element sets and value sets in the ParabolaSequence Theorem, where we assume that

Σ(m)n :=

n∑k=0

P(m)k → ∞ as n → ∞, uniformly with respect to m

for P(m)k :=

m+n∏j=m+1

1 − gj

gj.

Then {Eα,n ∩ B(0, M)} is a tusc with respect to {Vα,n} for every fixed 0 <M < ∞. This follows since by Remark 2 on page 158

diamS(m)n (Vα,m+n) ≤ M/(ε cos α

n−1∏j=1

(1 +

P(m)j

Σ(m)j−1

· ε2 cos2 α

M

)where

n−1∏j=1

(1 +

P(m)j

Σ(m)j−1

)=

n−1∏j=1

Σ(m)j

Σ(m)j−1

= Σ(m)j → ∞

uniformly with respect to m.

This holds in particular if all gn = g.

Theorem 5.4. Let K(an/bn) and K(an/bn) be two continued fractionsfrom {Ωn}, where {Ωn} is a tusc with respect to {Vn}, dist(dn, Vn) ≥ ε > 0for all n and all K(cn/dn) from {Ωn}, and lim inf diamm(Vn) > 0. If {an}is bounded, (an − an) → 0 and (anbn − anbn) → 0, then (f (n) − f (n)) → 0.

Proof : Let {an} be bounded, (an − an) → 0 and (anbn − anbn) → 0. We shallfirst prove that for each given n ∈ N,

D(m)n := |S(m)

n (wm+n) − S(m)n (wm+n)| → 0 as m → ∞ (1.3.3)

for wm+n ∈ Vm+n.

Without loss of generality, {Vn} is bounded since

|S(m)n (w)| =

∣∣∣ am+1

bm+1 + S(m+1)n−1 (w)

∣∣∣ ≤ M/ε for w ∈ Vm+n where M := sup |am|.

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5.1.4 The improvement machine 227

For n := 1 and wm+1, wm+1 ∈ Vm+1,

D(m)1 = |S(m)

1 (wm+1) − S(m)1 (wm+1)| =

∣∣∣ am+1

bm+1 + wm+1− am+1

bm+1 + wm+1

∣∣∣=∣∣∣am+1bm+1 − am+1bm+1 + am+1wm+1 − am+1wm+1

(bm+1 + wm+1)(bm+1 + wm+1)

∣∣∣, (1.3.4)

where the denominator is ≥ ε2 > 0 and

am+1wm+1 − am+1wm+1 = am+1(wm+1 − wm+1) + wm+1(am+1 − am+1).

Hence D(m)1 → 0 if also (wm+1 − wm+1) → 0. In particular, D

(m)1 → 0 as m → ∞

for wm+1 := wm+1. We shall prove (1.3.3) by induction on n, so assume that itholds for all 1 ≤ n ≤ ν − 1. For n := ν we then get

D(m)n =

∣∣∣ am+1

bm+1 + S(m+1)n−1 (wn+m)

− am+1

bm+1 + S(m+1)n−1 (wn+m)

∣∣∣where vm+1 := S

(m+1)n−1 (wm+n) ∈ Vm and vm+1 := S

(m+1)n−1 (wm+n) ∈ Vm with

(vm+1 − vm+1) → 0 as m → ∞ by our induction hypothesis. Hence, also D(m)n → 0

as m → ∞ by (1.3.4). This proves (1.3.3).

Now, |f (m) − f (m)| ≤ |f (m) −S(m)n (wm+n)|+ |f (m) − S

(m)n (wm+n)|+ D

(m)n ≤ 2λn +

D(m)n for every n ∈ N. Hence, by (1.3.3), |f (m) − f (m)| → 0 as m → ∞. �

5.1.4 The improvement machine for the loxodromic case

Let K(an/bn) be a limit p-periodic continued fraction of loxodromic type with finitelimits, and assume that we have accelerated its convergence by using Sn(wn) insteadof Sn(0), where 0 �= (wn − f (n)) → 0, for instance by using the fixed point method.If the finite limits

limn→∞

λnp+m+1

λnp+m=: rm where λn := an − wn−1(bn + wn) (1.4.1)

exist for m = 1, 2, . . . , p, then we can improve the speed of convergence further.The idea was published by the authors and generalized by Levrie ([Levr89]). Wedemonstrate how it works for the case p = 1 where

an → a �= ∞, bn → b �= 0,∞ and |x| < |b + x|,

where x is the attracting fixed point of the loxodromic transformation S1(w) :=a/(b + w) if a �= 0 and x := 0 if a = 0. (The case p > 1 can be found in ([JaWa88],[JaWa90]).)

Page 241: Lisa Lorentzen, Haakon Waadeland Continued Fractions

228 Chapter 5: Computation of continued fractions�

Theorem 5.5. Let K(an/bn) be a limit 1-periodic continued fraction ofloxodromic type, with finite limits b �= 0 and a, attracting fixed point x, andfinite value f . Let further {wn} be a sequence from C with wn �= 0 and0 �= (wn − f (n)) → 0. If limn→∞ λn+1/λn = r for {λn} given by (1.4.1),then

limn→∞

f − Sn(w(1)n )

f − Sn(wn)= 0 for w(1)

n := wn +λn+1

(b + x + rx).

Proof : Let λn+1/λn → r. Since f (n) → x = (−b +√

b2 + 4a )/2 (Theorem 4.13on page 188), we may without loss of generality assume that all f (n) �= ∞, andbn + f (n) �= 0. Since 0 �= εn := (wn − f (n)) → 0, it follows that λn → 0, and thusthat |r| ≤ 1. Without loss of generality we may therefore assume that bn + wn �= 0and λn �= 0 for all n (since λn+1/λn → r). That is,

εn := wn−f (n) �= 0, f (n) �= ∞, λn �= 0, bn+f (n) �= 0, bn+wn �= 0 for all n.

We shall first prove that εn+1/εn → r. Since

λn = an − wn−1(bn + wn)

= f (n−1)(bn + f (n)) − (f (n−1) + εn−1)(bn + f (n) + εn)

= −εn−1(bn + f (n)) − f (n−1)εn − εnεn−1,

(1.4.2)

it follows thatλn+1

λn=

εn

εn−1· bn+1 + f (n+1) + wnεn+1/εn

bn + f (n) + wn−1εn/εn−1, (1.4.3)

soεn

εn−1=

λn+1λn

(bn + f (n)) + λn+1λn

wn−1εn

εn−1

bn+1 + f (n+1) + wnεn+1/εn.

We multiply this identity by wn−1 and solve for (wn−1εn/εn−1):

wn−1εn

εn−1=

λn+1λn

(bn + f (n))wn−1

bn+1 + f (n+1) − wn−1λn+1λn

+ wnεn+1εn

.

This shows that {wnεn+1/εn} is a tail sequence for the continued fraction K(cn/dn)given by

cn :=λn+1

λn(bn + f (n))wn−1 → r(b + x)x = ra =: c,

dn := bn+1 + f (n+1) − wn−1λn+1

λn→ b + x − xr =: d.

Page 242: Lisa Lorentzen, Haakon Waadeland Continued Fractions

5.1.4 The improvement machine 229

Since K(cn/dn) is limit 1-periodic of loxodromic type, where S1(w) := c/(d + w)has attracting fixed point rx and repelling fixed point y := −(b + x) with |b + x| >|rx|, it follows from Theorem 4.13 on page 188 that either wn−1εn/εn−1 → rx orwn−1εn/εn−1 → −(b + x). However, if wn−1εn/εn−1 → −(b + x), then εn/εn−1 →−(b + x)/x which is impossible since 0 < εn → 0 and |b + x|/|x| > 1. Hencewn−1εn/εn−1 → rx. This proves that εn/εn−1 → r when x �= 0.

If x = 0, then it follows from (1.4.3) that

λn+1

λn∼ εn

εn−1

b

b=

εn

εn−1as n → ∞.

Hence, also now εn/εn−1 → r.

Next, we divide (1.4.2) by εn−1 and let n → ∞. Then we find that limn→∞ λn/εn−1 =−(b + x + rx) �= 0. Since by (1.1.2) on page 218

f − Sn(w(1)n )

f − Sn(wn)=

f (n) − w(1)n

f (n) − wn· ζn − wn

ζn − w(1)n

,

where lim(ζn −wn) = lim(ζn −w(1)n ) = (y−x) �= 0,∞ ( y := −b−x is the repelling

fixed point for K(an/bn)) and

f (n) − w(1)n

f (n) − wn=

−εn − λn+1/(b + x + rx)−εn

→ 0,

the result follows. �

Remarks.

1. Indeed,lim

n→∞λn+1/λn = r ⇐⇒ limn→∞ εn+1/εn = r (1.4.4)

under the conditions of Theorem 5.5 since ⇐= follows from (1.4.3).

2. If wn = f (n) for infinitely many indices, then {w(1)n } does not lead to conver-

gence acceleration, since f −Sn(wn) = 0 for these indices. This is a drawbacksince {f (n)} is not known exactly. On the other hand, this will not disturbthe computations too much, since {Sn(w(1)

n )} still converges reasonably fastto the correct value.

3. The expression for λn is also found in the description of the Bauer-Muir trans-formation. In fact, our improvement machine is closely related to the followingidea:

◦ Find {wn} such that (f (n) − wn) → 0.

◦ Construct the Bauer-Muir transform b(1)0 +K(a(1)

n /b(1)n ) of K(an/bn) with

respect to {wn}.

Page 243: Lisa Lorentzen, Haakon Waadeland Continued Fractions

230 Chapter 5: Computation of continued fractions

◦ Find numbers {w(1)n } which accelerate the convergence of this new con-

tinued fraction b(1)0 +K(a(1)

n /b(1)n ), and repeat the process, etc. The main

difference is that by using the improvement machine, we do not have tocompute the Bauer-Muir transform K(a(1)

n /b(1)n ).

4. A simple, but important special case is the case where all bn = 1 and an → awith | arg(a+ 1

4 )| < π, and all wn = x, the attractive fixed point for K(an/1).Then λn = an − a, and

limn→∞

an+1 − a

an − a= r =⇒ lim

n→∞f − Sn(w(1)

n )f − Sn(x)

= 0

for w(1)n := x +

an+1 − a

1 + x + rx

when the value f of K(an/1) is finite.

Example 3. The Stieltjes fraction

∞Kn=1

anz

1:=

12z

1 · 31 +

22z

3 · 51 +

32z

5 · 71 +

42z

7 · 91 +

52z

9 · 111 +· · ·

converges to a holomorphic function f(z) in D := {z ∈ C; | arg(z + 1)| < π} sincean → 1/4 =: a. Clearly f(0) = 0, so let 0 �= z ∈ D. Then K(anz/1) is limit 1-periodic of loxodromic type. Indeed, s(w) := az/(1 + w) has attracting fixed pointx := (u− 1)/2 and repelling fixed point y := (−u− 1)/2 where u :=

√1 + z. Hence

the approximants

Sn(w(0)n ) where w(0)

n := x = (u − 1)/2; u :=√

1 + z (1.4.5)

can be a starting point for the improvement machine. Since

λ(0)n = anz − az =

z/44n2 − 1

=z/4

(2n − 1)(2n + 1)

in this case, we find that λ(0)n+1/λ

(0)n → 1, and thus the approximants Sn(w(1)

n ) with

w(1)n := w(0)

n +λ

(0)n+1

1 + 2x= x +

an+1z − az

u= x +

z/(4u)(2n + 1)(2n + 3)

converge faster to f(z) than Sn(x). Now, also (w(1)n − f (n)) → 0, and

λ(1)n := anz − w

(1)n−1(1 + w(1)

n )

= anz −(x +

z/(4u)4n2 − 1

)(1 + x +

z/(4u)(2n + 1)(2n + 3)

)= anz − x(1 + x) − xz/(4u)

(2n + 1)(2n + 3)− (1 + x)z/(4u)

4n2 − 1

− z2/(16u2)(4n2 − 1)(2n + 1)(2n + 3)

Page 244: Lisa Lorentzen, Haakon Waadeland Continued Fractions

5.1.4 The improvement machine 231

where anz − x(1 + x) = λ(0)n , so

anz − x(1 + x) − z/(4u)4n2 − 1

=(1 − 1

u

) z/44n2 − 1

=xz/(2u)4n2 − 1

.

Hence

λ(1)n =

xz/u

(4n2 − 1)(2n + 3)− z2/(16u2)

(4n2 − 1)(2n + 1)(2n + 3),

so also λ(1)n+1/λ

(1)n → 1. Hence {Sn(w(2)

n )} where

w(2)n := w(1)

n +λ

(1)n+1

1 + 2x= w(1)

n +λ

(1)n+1

u

converges even faster. We can continue the process, but for this example we preferto stop here.

From (2.6.1) in the appendix we know that the principal value of arctan z has thecontinued fraction expansion

Arctan z = z/(1 + K(anz2/1)) for z2 ∈ D.

That is, K(anz/1) has the value f(z) = −1 +√

z/Arctan√

z. For z := 1, thisis f(1) = −1 + 4/π = 0.273239544735, correctly rounded to 12 decimals. Forcomparison, the table below shows the first approximants Sn(0), Sn(x), Sn(w(1)

n )and Sn(w(2)

n ) for K(anz/1) for z := 1; that is,

x :=√

2 − 12

, w(1)n := x +

1/√

24(2n + 1)(2n + 3)

,

w(2)n := w(1)

n +2 −

√2

4(2n + 1)(2n + 3)(2n + 5)− 1/32

(2n + 1)(2n + 3)2(2n + 5).

n Sn(0) Sn(x) Sn(w(1)n ) Sn(w

(2)n )

5 .2732919254 .2732398207 .2732395555 .27323954116 .2732305259 .2732395100 .2732395435 .27323954517 .2732410964 .2732395493 .2732395449 .27323954478 .2732392779 .2732395441 .27323954479 .2732395906 .273239544810 .2732395368 .273239544711 .273239546112 .273239544813 .273239544814 .2732395447

Each column stops when we have reached the correctly rounded value with 10decimals, and this value is repeated for all larger indices. We already see theimprovement. Now, the continued fraction converges quite fast for z := 1. For valuescloser to the boundary of D the convergence is slower, and thus the improvementmore valuable. For instance, for z := −2 + i/100 we get:

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232 Chapter 5: Computation of continued fractions

n |f − Sn(0)| |f − Sn(x)| |f − Sn(w(1)n )| |f − Sn(w

(2)n )|

100 ca 1 1.0 · 10−5 1.5 · 10−6 2.9 · 10−8

1000 9.2 · 10−3 1.2 · 10−8 1.6 · 10−11 3.5 · 10−14

5000 1.9 · 10−10 9.5 · 10−19 2.7 · 10−22 1.1 · 10−26

This indicates that to obtain a bound for |f − Sn(wn)| of the order 10−8, we needmaybe n ≈ 4000 for wn := 0, n ≈ 1000 for wn := x, n ≈ 500 for wn := w

(1)n and

n ≈ 100 for wn := w(2)n . �

5.1.5 Asymptotic expansion of tail values

If we can use the improvement machine repeatedly, as we did in the previous exam-ple, we actually develop the first terms in some asymptotic expansion of f (n), wherethe method chooses the terms in the expansion. But we can also take more controlover the expansion, and choose its basis functions uj(n), such that (hopefully)

f (n) ∼∞∑

j=σ

kjuj(n) as n → ∞ where limn→∞

uj+1(n)uj(n)

= 0 for all j. (1.5.1)

We demonstrate the idea in the following situation: let K(an/bn) be a convergentcontinued fraction where an = a(n) and bn = b(n) are polynomials in n. Thatis, K(an/bn) is limit periodic. Assume that its tail values f (n) have asymptoticexpansions

f (n) = f(n) ∼∞∑

j=σ

kj n−j as n → ∞ (1.5.2)

for some σ ∈ N and kj ∈ C. That is, we choose an expansion in terms of uj(n) :=n−j . The functional equation

f(n − 1)(b(n) + f(n)) = a(n) (1.5.3)

allows us to determine σ and the first coefficients kj formally. The idea is then touse the approximants Sn(w(N)

n ) for K(an/bn) where

w(N)n :=

σ+N∑j=σ

kj n−j . (1.5.4)

This idea was suggested by Wynn ([Wynn59]).

That f (n) really has an asymptotic expansion of the form (1.5.1) is not always thecase. But by the Birkhoff-Trjzinzki theory it follows that if the process of findingσ and kj works, then K(an/bn) has a tail sequence {tn} with tn ∼

∑∞σ kjuj(n) as

n → ∞ where uj(n) is as specified for σ ≤ j ≤ σ + N . We just have to make surethat it is the sequence of tail values. (More information on this can be found in theremark section on page 262.)

Page 246: Lisa Lorentzen, Haakon Waadeland Continued Fractions

5.1.5 Asymptotic expansion of f (n) 233

Example 4. The continued fraction

∞Kn=1

an

bn=

2z

1 −12z2

3 −22z2

5 −32z2

7 −· · ·

converges to f(z) := Ln 1+z1−z

in the cut plane D where 0 < arg(1−z2) < 2π (formula(2.4.9) in the appendix on page 271). We assume that (1.5.2) holds. Then (1.5.3)can be written

( ∞∑j=σ

kj(n − 1)−j)(

2n − 1 +∞∑

j=σ

kj n−j)

= −(n2 − 2n + 1)z2 (1.5.5)

for n ≥ 2 where

(n − 1)−j = n−j (1 − 1/n)−j = n−j∞∑

m=0

(−j

m

)(− 1

n

)m

. (1.5.6)

Hence σ := −1 is necessary. Comparing the coefficients for n−j on each side of(1.5.5) gives (after some computation) that

k−1 = u − 1, k0 =1 − u

2, k1 =

−z2

8u, . . . where u2 = 1 − z2.

That is, we get two possibilities for w(2)n := k−1n + k0 + k1/n, depending on our

choice for u. To make the right choice we observe that K(an/1) is equivalent toK(cn/1) where

cn+1 =an+1

bnbn+1=

−n2z2

4n2 − 1→ −z2

4,

so the sequence {f (n)} of tail values for K(cn/1) converges to the attracting fixedpoint x of the loxodromic transformation s(w) := − z2

4/(1+w); i.e., to x = (u−1)/2

with Re(u) > 0. Since f (n) = bnf (n), we let this be our choice for u. Since K(an/bn)is limit periodic of loxodromic type for z ∈ D, it follows that Sn(w(N)

n ) convergesfaster to Ln 1+z

1−zthan Sn(0).

As an illustration, let z := 3/4. Then the continued fraction converges to Ln(7) =1.945910149055 correctly rounded to 12 decimals. We compare some low orderapproximants Sn(0) and Sn(w(N)

n ) for N = 0, 1, 2, where u :=√

1 − z2 and

w(0)n := (u − 1)n, w(1)

n := w(0)n +

1 − u

2, w(2)

n := w(1)n − z2

8un.

Each column stops when the value, correctly rounded to 8 decimals, is equal to therounded value of Ln 1+z

1−z for all indices ≥ n.

Page 247: Lisa Lorentzen, Haakon Waadeland Continued Fractions

234 Chapter 5: Computation of continued fractions

n Sn(0) Sn(w(0)n ) Sn(w

(1)n ) Sn(w

(2)n )

5 1.94498141 1.94602817 1.94589653 1.945912646 1.94571873 1.94593035 1.94590818 1.945910457 1.94587082 1.94591370 1.94590985 1.945910198 1.94590209 1.94591078 1.94591010 1.945910159 1.94590850 1.94591026 1.94591014

10 1.94590981 1.94591017 1.9459101511 1.94591008 1.9459101512 1.9459101313 1.94591015

The truncation errors are then for instance

n f − Sn(0) f − Sn(w(0)n ) f − Sn(w

(1)n ) f − Sn(w

(2)n )

8 8.1 · 10−6 −6.4 · 10−7 4.7 · 10−8 −5.6 · 10−9

9 1.7 · 10−6 −1.2 · 10−7 7.7 · 10−9 −8.1 · 10−10

10 3.4 · 10−7 −2.1 · 10−8 1.3 · 10−9 −1.2 · 10−10

11 6.9 · 10−8 −4.0 · 10−9 2.2 · 10−10 −1.9 · 10−11

12 1.4 · 10−8 −7.4 · 10−10 3.7 · 10−11 −3.0 · 10−12

It is more interesting to see what happens for z-values where K(an/bn) convergesmore slowly. For z := −2 + i/100 the continued fraction has the value

f = −1.098567846693 + 3.134926307891 i,

correctly rounded to 12 decimals. The truncation errors are in this case:

n |f − Sn(0)| |f − Sn(w(0)n )| |f − Sn(w

(1)n )| |f − Sn(w

(2)n )|

10 4.3 1.7 · 10−1 5.4 · 10−3 3.8 · 10−4

20 11.3 8.0 · 10−2 1.2 · 10−3 4.5 · 10−5

50 9.1 2.7 · 10−2 1.6 · 10−4 2.4 · 10−6

100 3.1 1.0 · 10−2 3.0 · 10−5 2.3 · 10−7

200 2.6 2.9 · 10−3 4.2 · 10−6 1.6 · 10−8

300 9.6 · 10−1 1.1 · 10−3 1.0 · 10−6 2.6 · 10−9

400 6.2 · 10−1 4.5 · 10−4 3.3 · 10−7 6.2 · 10−10

500 3.7 · 10−1 2.0 · 10−4 1.2 · 10−7 1.8 · 10−10

1000 1.9 · 10−2 5.6 · 10−6 1.6 · 10−8 1.2 · 10−11

To get an error less than ca 5 · 10−5, one actually needs n to be more than 2000 forSn(0), ca 750 for Sn(w(0)

n ), ca 100 for Sn(w(1)n ) and ca n = 20 for Sn(w(2)

n ). �

The method also works in situations where

anp+m = am(n), bnp+m = bm(n) for m = 1, 2, . . . , p.

We then try to find the first few terms of expansions f (np+m) ∼∑

k(m)j n−j for each

m ∈ {1, 2, . . . , p}. For further examples we refer to ([Wynn59]).

Page 248: Lisa Lorentzen, Haakon Waadeland Continued Fractions

5.1.6 The square root modification 235

5.1.6 The square root modification

Let K(an/bn) be a generally convergent continued fraction with tail values {f (n)},where {an} and {bn} have a monotonic character which makes it natural to replaceits nth tail by a periodic continued fraction:

f (n) =an+1

bn+1 +an+2

bn+2 +an+3

bn+3 +· · ·≈ an+1

bn+1 +an+1

bn+1 +an+1

bn+1 +· · ·(1.6.1)

(or a p-periodic continued fraction). If this last continued fraction converges to somevalue wn, then hopefully f (n) is close to this value. Hence we use the approximantsSn(wn).

This idea was suggested by Gill ([Gill80]). He used it to accelerate continued frac-tions K(an/1) where an → − 1

4. In [JaJW87] it was used to accelerate the con-

vergence of continued fractions K(an/1) with an → ∞. In both these cases theclassical convergence may be rather slow, so a method for convergence accelerationis useful.

Example 5. The continued fraction K(an/1) with

an = −14

+z

8nfor n = 1, 2, 3, . . .

is limit 1-periodic of parabolic type. For z in the cut plane D := C \ (−∞, 0] where| arg(z)| < π, it converges to a finite value f by virtue of the Parabola Theoremon page 151, since all an ∈ Eα for α := 1

2arg z. Therefore also {f (n)} and {ζn}

converge to −12 , so we do not have high expectations to the speed of convergence

of Sn(−12) as compared to Sn(0). However, an → −1

4monotonely, and its nth tail

looks similar to the periodic continued fraction

an+1

1 +an+1

1 +an+1

1 +· · ·

which converges to

wn :=√

1 + 4an+1 − 12

= −12

+12

√z

2(n + 1). (1.6.2)

We therefore expect Sn(wn) to converge faster, something which is confirmed by thetable below. For z := 1, the correct value of K(an/1) is −0.172160228791, roundedto 12 decimals.

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236 Chapter 5: Computation of continued fractions

n Sn(0) Sn(− 12) Sn(wn)

1 −0.12500000 −0.25000000 −0.166666672 −0.15384615 −0.20000000 −0.170366643 −0.16379310 −0.18421053 −0.171441404 −0.16793893 −0.17801047 −0.171833615 −0.16987898 −0.17523168 −0.171998426 −0.17086247 −0.17386837 −0.172074767 −0.17139164 −0.17315351 −0.172112798 −0.17168993 −0.17275888 −0.172132829 −0.17186450 −0.17253187 −0.1721438610 −0.17196993 −0.17239677 −0.1721501711 −0.17203530 −0.17231404 −0.1721539012 −0.17207677 −0.17226212 −0.17215617

For z := −2 + i/10 the continued fraction has the value

f = −0.634775601464989 + 0.578831734368452 i,

correctly rounded to 15 decimals. The truncation errors for this slower convergingcontinued fractions are for instance:

n |f − Sn(0)| |f − Sn(− 12)| |f − Sn(wn)|

100 4.03 · 10−1 5.90 · 10−1 6.15 · 10−3

500 1.28 · 10−1 1.64 · 10−1 8.03 · 10−4

1000 5.47 · 10−2 5.95 · 10−2 2.25 · 10−4

5000 1.15 · 10−3 1.15 · 10−3 2.03 · 10−6

10000 6.15 · 10−5 6.16 · 10−5 7.69 · 10−8

100000 2.52 · 10−14 2.52 · 10−14 9.98 · 10−18

In this case there is almost no difference between |f −Sn(0)| and |f −Sn(−12)|, but

|f − Sn(wn)| is considerably smaller. For instance is |f − Sn(wn)| ≤ 2 · 10−6 forn = 5000, whereas we need considerably more than n = 10 000 for |f −Sn(0)| to beso small. �

Example 6. Let an := n for all n ∈ N. Then K(an/1) converges to a finitevalue f (the Seidel-Stern Theorem on page 117). But the convergence of {Sn(0)}is relatively slow. The square root modification gives

Sn(wn) with wn :=(√

1 + 4(n + 1) − 1)

/2. (1.6.3)

The table below shows the approximants Sn(0) and Sn(wn) and the correspondingtruncation errors. Here f = 0.525135276161 is the value of K(an/1), correctlyrounded to 12 decimals.

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5.1.6 The square root modification 237

n Sn(0) Sn(wn) f − Sn(0) f − Sn(wn)1 1.00000000 0.50000000 −4.75 · 10−1 2.51 · 10−2

2 0.33333333 0.53518376 1.92 · 10−1 −1.00 · 10−2

3 0.66666667 0.52051760 −1.42 · 10−1 4.61 · 10−3

4 0.44444444 0.52752523 8.07 · 10−2 −2.39 · 10−3

5 0.58333333 0.52380952 −5.82 · 10−2 1.33 · 10−3

11 0.53312330 0.52504318 −7.99 · 10−3 9.21 · 10−5

12 0.51912183 0.52519962 6.01 · 10−3 −6.43 · 10−5

51 0.52514011 0.52513526 −4.83 · 10−6 1.24 · 10−8

52 0.52513106 0.52513529 4.21 · 10−6 −1.06 · 10−8

101 0.52513529 0.52513528 −1.53 · 10−8 1.97 · 10−11

102 0.52513526 0.52513528 1.39 · 10−8 −1.76 · 10−11

An alternative method is to rather use the even (or odd) part of K(an/1), ([JaWa86]).The even part of K(n/1) is

11 + 2−

2 · 31 + 3 + 4−

4 · 51 + 5 + 6−· · ·

which is equivalent to K(cn/1) with c1 := 1/3, c2 := −1/4 and

cn :=−(2n − 2)(2n − 1)

(1 + 2n − 3 + 2n − 2)(1 + 2n − 1 + 2n)

= −n − 1/24n

= −14

+18n

→ −14

as n →∞ .

We recognize this continued fraction from the previous example where we also usedthe square root modification. �

Of course, the square root modification also works for limit periodic continuedfractions of loxodromic type. In our next example we show a method to computeπ:

Example 7. The beautiful continued fraction∞Kn=1

an

1where a1 := 4, an+1 :=

n2

4n2 − 1for n ≥ 1 (1.6.4)

converges to π (appendix, page 267). K(an/1) converges rather fast, but it iseasy to accelerate its convergence. The fixed point method gives for instance theapproximants

Sn(x) where x := 12 (√

1 + 4 · 14 − 1) = 1

2 (√

2 − 1),

and the square root modification

Sn(wn) where wn := 12 (√

1 + 4an+1 − 1) =12

(√8n2 − 14n2 − 1

− 1)

ought to approximate π even better. The table below shows that this is indeed thecase:

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238 Chapter 5: Computation of continued fractions

n Sn(x) Sn(wn) π − Sn(x) π − Sn(wn)1 3.3137084990 3.1651513899 −1.7 · 10−1 −2.4 · 10−2

2 3.1344464996 3.1409646632 7.1 · 10−3 6.3 · 10−4

3 3.1421356273 3.1416290537 −5.4 · 10−4 −3.6 · 10−5

4 3.1415404008 3.1415898074 5.2 · 10−5 2.8 · 10−6

5 3.1415983790 3.1415929165 −5.7 · 10−6 −2.6 · 10−7

6 3.1415919726 3.1415926266 6.8 · 10−7 2.7 · 10−8

7 3.1415927393 3.1415926566 −8.6 · 10−8 −3.0 · 10−9

8 3.1415926423 3.1415926532 1.1 · 10−8 3.5 · 10−10

9 3.1415926551 3.1415926536 −1.5 · 10−9 −4.3 · 10−11

10 3.1415926534 3.1415926536 2.1 · 10−10 5.4 · 10−12

Both {Sn(x)} and {Sn(wn)} are alternating sequences – a fact that gives simple aposteriori truncation error bounds. �

5.2 Truncation error bounds

5.2.1 The ideas

Let K(an/bn) converge generally to a finite value f . We want to find bounds for thetruncation error |f −Sn(wn)|, where {wn} is a given sequence of complex numbers,chosen to make {Sn(wn)} converge fast to f . In particular we want the bounds toreflect the fast convergence due to the choice of {wn}. We shall therefore look fora sequence {Vn}∞n=0 of closed value sets for K(an/bn) with f (n) ∈ Vn and wn ∈ Vn

for all n. Then f = Sn(f (n)) ∈ Sn(Vn) and Sn(wn) ∈ Sn(Vn).

Idea 1: We can use {Vn} as in Chapter 3 to derive a priori bounds

|f − Sn(wn)| ≤ diamSn(Vn). (2.1.1)

These bounds are valid for every wn ∈ Vn.

Idea 2: We can use {Vn} to derive a posteriori bounds based on the followinglemma:�

Lemma 5.6. Let ∞ �= f∗n := Sn(wn) → f �= ∞ for the generally convergent

continued fraction K(an/bn) with value f , tail values {f (n)} and critical tailsequence {ζn} with ζn �= ∞. Then

f − f∗n =

an(f (n) − wn)−λn

· 1 − wn−1/ζn−1

f (n) − ζn(f∗

n − f∗n−1)

when λn := an − wn−1(bn + wn) �= 0 and wn �= 0,∞ for all n.

Page 252: Lisa Lorentzen, Haakon Waadeland Continued Fractions

5.2.1 The ideas 239

Proof : Let w∗n := s−1

n (wn−1) = −bn + an/wn−1. Then w∗n �= ∞ and

f∗n − f∗

n−1 = Sn(wn) − Sn−1(wn−1) = Sn(wn) − Sn(w∗n)

=An−1wn + An

Bn−1wn + Bn− An−1w

∗n + An

Bn−1w∗n + Bn

=(An−1Bn − AnBn−1)(wn − w∗

n)(Bn−1wn + Bn)(Bn−1w∗

n + Bn),

and thus, since f = Sn(f (n)),

f − f∗n

f∗n − f∗

n−1

=Sn(f (n)) − Sn(wn)Sn(wn) − Sn(w∗

n)=

f (n) − wn

wn − w∗n

· Bn−1w∗n + Bn

Bn−1f (n) + Bn.

Now, ζn = −Bn/Bn−1 = −bn + an/ζn−1, and both wn �= ζn and f (n) �= ζn sincef∗

n �= ∞ and f �= ∞. Since also Bn−1 �= 0 (because ζn �= ∞), we get

Bn−1w∗n + Bn

Bn−1f (n) + Bn=

−bn + an/wn−1 − ζn

f (n) − ζn= an

w−1n−1 − ζ−1

n−1

f (n) − ζn

and the result follows. �

If f (n) ∈ Vn, wn ∈ Vn and ζn �∈ Vn for all n, then we can hopefully find a bound for|(wn − f (n))(1 − wn−1/ζn−1)|/|f (n) − ζn|, and thus we have an a posteriori boundfor |f − Sn(wn)|.

Idea 3: We can combine an already known truncation error bound |f −Sn(wn)| ≤λn for K(an/bn) with the identity

|f − Sn(wn)| =∣∣∣ wn − ζn

wn − ζn· wn − f (n)

wn − f (n)

∣∣∣|f − Sn(wn)| (2.1.2)

( (1.1.2) on page 218). If wn ∈ Vn and f (n) ∈ Vn, then |wn − f (n)| ≤ diamVn. Ifwe can establish a bound ∣∣∣ wn − ζn

(wn − ζn)(wn − f (n))

∣∣∣ ≤ kn,

then |f − Sn(wn)| ≤ knλn diamVn.

In Chapter 3 we started with the value sets, and the purpose was to prove con-vergence for a large family of continued fractions. The truncation error boundswe obtained, were then valid for every continued fraction from this family and forevery wn ∈ Vn. What we now want, is to find small” value sets {Vn} for a givencontinued fraction K(an/bn) to get small” error bounds. It takes considerablymore effort to establish such bounds, so the truncation error bounds in Chapter 3are still useful because of their simplicity.

Now, {wn} is a known sequence, so it is easy to make sure that wn ∈ Vn for all n.The tail values f (n) on the other hand, are unknown. (That is in fact the wholepoint! If f (n) was known, then f = Sn(f (n)) makes the need for {wn} and truncation

““

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240 Chapter 5: Computation of continued fractions

error bounds quite obsolete.) So, how can we make sure that also f (n) ∈ Vn for alln? Well, for one thing, if f (n) ∈ Vn from some n on, then f (n) ∈ Vn for all n. Butmore importantly, the following holds:

�Lemma 5.7. Let {Vn} be a sequence of closed value sets for K(an/bn). Ifthere exists a sequence {wn} with wn ∈ Vn for all n such that Sn(wn) → f ,then f (n) := S−1

n (f) ∈ Vn for all n.

Proof : Since Sn(wn) ∈ V0 for all n, we have f ∈ V0, and similarly

S(n)k (wn+k) :=

an+1

bn+1 +· · ·+an+k

bn+k + wn+k∈ Vn for all n and k (2.1.3)

implies that

f (n) = S−1n (f) = lim

k→∞S−1

n ◦ Sn+k(wn+k) = limk→∞

S(n)k (wn+k) ∈ Vn.

This means that if K(an/bn) actually converges generally to this value f , then itstail values f (n) ∈ Vn. This gives us the tool we need to develop a priori truncationerror bounds.

5.2.2 Truncation error bounds

Let K(an/bn) be a generally convergent continued fraction, and let {wn} be asequence of complex numbers which approximate its tail values. We want to find asequence {Vn}∞n=0 of small” value sets for K(an/bn) where wn ∈ Vn, at least fromsome n on.

Since circular disks behave so nicely under linear fractional transformations, weshall let Vn be closed circular disks. As earlier, we use the notation

B(Γ, ρ) := {w ∈ C; |w − Γ| ≤ ρ} for Γ ∈ C and ρ > 0. (2.2.1)

We let Γn := wn be the center of Vn (or Γn ≈ wn). It then remains to determineρn > 0 such that an/(bn + B(Γn, ρn)) ⊆ B(Γn−1, ρn−1) for all n. We first note thefollowing result due to Lane ([Lane45]):

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5.2.2 Truncation error bounds 241�

Lemma 5.8. Let Vn := B(Γn, ρn) for Γn ∈ C and ρn > 0 for n ≥ 0, andlet Ωn be the set of all pairs (a, b) ∈ C2 for which |b + Γn| > ρn and∣∣∣ a(b + Γn)

|b + Γn|2 − ρ2n

− Γn−1

∣∣∣+ |a|ρn

|b + Γn|2 − ρ2n

≤ ρn−1 (2.2.2)

for n ≥ 1. Then {Vn} is a sequence of value sets for every K(an/bn) from{Ωn}.

Proof : From Lemma 3.6 on page 110 it follows that

a

b + B(Γn, ρn)= B(Γ∗

n, ρ∗n) where

Γ∗n :=

a(b + Γn)|b + Γn|2 − ρ2

n

, ρ∗n :=

ρn|a||b + Γn|2 − ρ2

n

.

(2.2.3)

Since B(Γ∗n, ρ∗n) ⊆ B(Γn−1, ρn−1) if and only if |b+Γn| > ρn and |Γ∗

n−Γn−1|+ρ∗n ≤ρn−1, the result follows. �

Theorem 5.9. Let K(an/bn) be a continued fraction from {Ωn} as givenin Lemma 5.8. Then

|Sn+m(w) − Sn(Γn)| ≤ ρn|Γ0| + ρ0

|bn + Γn| − ρn

n−1∏k=1

Mk for w ∈ Vn+m, (2.2.4)

where Mk := max{∣∣∣ w

bk + w

∣∣∣; w ∈ Vk

}and Vn := B(Γn, ρn) for all n.

Proof : Since Sn = s1 ◦ s2 ◦ · · · ◦ sn where sk(w) := ak/(1 + w), the chain ruleshows that its derivative at w ∈ Vn is given by

S′n(w) = s′1(wn,1) · s′2(wn,2) · · · s′n(wn,n) (2.2.5)

where wn,n := w ∈ Vn, wn,k := sk+1(wn,k+1) ∈ Vk for k := n− 1, n− 2, . . . , 1, and

s′k(wn,k) =−ak

(bk + wn,k)2=

−wn,k−1

bk + wn,k. (2.2.6)

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242 Chapter 5: Computation of continued fractions

Therefore

|S′n(w)| =

n∏k=1

|wn,k−1||bk + wn,k|

=|wn,0|

|bn + wn,n|

n−1∏k=1

∣∣∣∣ wn,k

bk + wn,k

∣∣∣∣≤ |Γ0| + ρ0

|bn + Γn| − ρn

n−1∏k=1

Mk,

and thus, for w∗ ∈ Vn,

|Sn(w∗) − Sn(Γn)| ≤ |w∗ − Γn| · supw∈Vn

|S′n(w)| ≤ ρn · |Γ0| + ρ0

|bn + Γn| − ρn

n−1∏k=1

Mk .

Since Sn+m(w) = Sn(w∗) for a w∗ ∈ Vn when w ∈ Vn+m, this proves (2.2.4). �

Remarks:

1. If Sn(Γn) → f �= ∞, then (2.2.4) implies that

|f − Sn(Γn)| ≤ ρn|Γ0| + ρ0

|bn + Γn| − ρn

n−1∏k=1

Mk.

2. It follows from the proof of (2.2.4) that

diamSn(B(Γn, ρn)) ≤ 2ρn|Γ0| + ρ0

|bn + Γn| − ρn

n−1∏k=1

Mk.

3. For given {ρ∗n} with ρn ≤ ρ∗n < |bn + Γn| for all n,

ρn|Γ0| + ρ0

|bn + Γn| − ρn

n−1∏k=1

Mk ≤ ρ∗n|Γ0| + ρ∗0

|bn + Γn| − ρ∗n

n−1∏k=1

M∗k

whereM∗

k := max{∣∣∣ w

bk + w

∣∣∣; |w − Γk| ≤ ρ∗k}

.

4. Just as in Remark 1 on page 163, we find that

Mk =|bk + Γk|2 − (bk + Γk) − ρ2

k| + ρk

|bk + Γk|2 − ρ2k

.

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5.2.3 The Oval Sequence Theorem 243

5.2.3 The Oval Sequence Theorem

The idea is now to let Γn := wn and choose ρn such that (an, bn) ∈ Ωn given inLemma 5.8. Now, Lemma 5.8 offers no guarantee for Ωn �= ∅. Indeed, it would behelpful to have an idea of what Ωn looks like. Its shape was investigated in [JaTh86]for the special case where all bn = 1. To keep things reasonably simple, we staywith this case. Then we need ρn < |1 + Γn|, and we replace Ωn by En × {1} where

En :={

a ∈ C;∣∣∣ a(1 + Γn)|1 + Γn|2 − ρ2

n

− Γn−1

∣∣∣+ |a|ρn

|1 + Γn|2 − ρ2n

≤ ρn−1

}. (2.3.1)

If Γn−1 = 0, then a ∈ En if

0 ≤ |a| |1 + Γn| + ρn

|1 + Γn|2 − ρ2n

=|a|

|1 + Γn| − ρn≤ ρn−1.

That is, En is the circular disk

En = B(0, rn) where rn := ρn−1(|1 + Γn| − ρn) > 0. (2.3.2)

Let Γn−1 �= 0. Then straight forward checking shows that En is bounded by thecurve ∂En = a∗

nOn where

a∗n :=

Γn−1

1 + Γn

dn with dn := |1 + Γn|2 − ρ2n (2.3.3)

and

On := {ξ ∈ C; |ξ−1|+kn|ξ| = �n} where kn :=ρn

|1 + Γn|, �n :=

ρn−1

|Γn−1|. (2.3.4)

Let On be the closed set bounded by On. Since kn < 1, we find that On = ∅ if1 �∈ On: i.e., if kn > �n. If kn = �n, then On consists of the point ξ = 1 only. Hencewe assume that

kn < 1 and kn < �n. (2.3.5)

Then 1 ∈ O◦, and we recognize On = ∂On as a cartesian oval. As a corollary toTheorem 5.9 we now get:�

Corollary 5.10. (The Oval Sequence Theorem.) Let K(an/1) be acontinued fraction from {En} given by (2.3.1). Then

|Sn+m(w) − Sn(Γn)| ≤ ρn|Γ0| + ρ0

|1 + Γn| − ρn

n−1∏k=1

Mk for w ∈ Vn+m

where Mk := max{∣∣∣ w

1 + w

∣∣∣; w ∈ Vk

}and V : n := B(Γn, ρn) for all n.

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244 Chapter 5: Computation of continued fractions

Remark. From Remark 4 on page 242, we find that

Mk =|Γk(1 + Γk) − ρ2

k| + ρk

|1 + Γk|2 − ρ2k

. (2.3.6)

If Γk > 0 and ρ2k ≤ Γk(1 + Γk), this reduces to

Mk =Γk + ρk

1 + Γk + ρk. (2.3.7)

5.2.4 An algorithm to find value sets for a given continuedfraction of form K(an/1)

Let K(an/1) be a given generally convergent continued fraction with (unknown)sequence {f (n)} of tail values. Let {wn} be a given sequence from C\{−1} approx-imating {f (n)}. We are looking for circular disks in C centered at Γn := wn (or ata point Γn close to wn) such that {Vn}∞n=0 with

wn ∈ Vn := B(Γn, ρn) for n ≥ n0 (2.4.1)

for some n0 ∈ N, is a sequence of value sets for K(an/1). That is, we are lookingfor radii ρn > 0 such that an ∈ En given by (2.3.1).

The shape of the ovals.

If Γn−1 �= 0, then En = a∗nO given by (2.3.3)–(2.3.4) has the following properties:

1. En is convex and symmetric about the line e2iαnR where

αn := 12 arg a∗

n = 12 arg(Γn−1(1 + Γn)). (2.4.2)

(Lemma 3.50 on page 161.)

2. The intersection En∩(e2iαnR) is called the axis of En. Lemma 3.50 shows thatEn ∩ (e2iαnR) = e2iαnIn where In := [un, vn] with un := a∗

nμ and vn := a∗nν;

that is,

un :=

{(|Γn−1| − ρn−1)(|1 + Γn| − ρn) if |Γn−1| ≤ ρn−1,

(|Γn−1| − ρn−1)(|1 + Γn| + ρn) if |Γn−1| ≥ ρn−1,

vn := (|Γn−1| + ρn−1)(|1 + Γn| − ρn),

(2.4.3)

as essentially pointed out in [JaTh86, p100].

3. Straight forward checking shows that

B(a∗n, r∗n) ⊆ En when Γn−1 �= 0, (2.4.4)

wherer∗n := (ρn−1|1 + Γn| − ρn|Γn−1|)(1 − ρn/|1 + Γn|). (2.4.5)

The boundary of this disk is tangent to ∂En at vne2iα.

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5.2.4 An algorithm to find value sets for a given continued fraction 245

4. It is often easier to work with a disk ⊆ En centered at an := Γn−1(1 + Γn).This disk must also be tangent to ∂En at vne2iα if it exists. This happens ifand only if

rn := vn − |Γn−1(1 + Γn)|= ρn−1|1 + Γn| − ρn|Γn−1| − ρnρn−1 > 0.

(2.4.6)

ThenB(Γn−1(1 + Γn), rn) ⊆ En. (2.4.7)

Strategies.

We can derive several strategies for our algorithm, strategies based on the oval sets{En}, strategies based on the interval In = [un, vn] given by (2.4.3), strategies basedon the disk B(a∗

n, r∗n) in (2.4.4), and strategies based on the disk

B(an, rn) where an := Γn−1(1 + Γn)and rn := ρn−1|1 + Γn| − ρn|Γn−1| − ρnρn−1 > 0

(2.4.8)

from (2.4.7). We choose the latter alternative, since it gives an easier algorithm(but not always the best result).

The algorithm.

The algorithm is due to Jacobsen ([Jaco82], [Jaco87], ([Lore03b]).

1. Let {σn} be a sequence of positive numbers such that {Ψn} given by

Ψn := σn|1 + Γn| − σn−1|Γn−1| > 0; n ∈ N (2.4.9)

is non-decreasing. Such a sequence always exists when Γn �= −1 for n ≥ 1.For instance, for arbitrary Ψ > 0, the choice

σn :=Ψ

|1 + Γn|

(1 +∣∣∣ Γn−1

1 + Γn−1

∣∣∣+∣∣∣ Γn−1

1 + Γn−1

∣∣∣∣∣∣ Γn−2

1 + Γn−2

∣∣∣+ · · · +n−1∏j=0

∣∣∣ Γj

1 + Γj

∣∣∣) (2.4.10)

for n = 0, 1, 2, . . . gives Ψn = Ψ for all n.

2. Let

ρn :=1

2σn

(Ψn+1 −

√Ψ2

n+1 − 4δn+1

)where

δn+1 := supm≥n+1

{σm−1σm|am − am|; am := Γm−1(1 + Γm)}(2.4.11)

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246 Chapter 5: Computation of continued fractions

for n ≥ n0, where n0 ∈ N is chosen such that 4δn0+1 ≤ Ψ2n0+1, and thus

ρn ≥ 0 for n ≥ n0. Then {ρnσn}∞n=n0is non-increasing since

2ρnσn = Ψn+1 −√

Ψ2n+1 − 4δn+1 =

4δn+1

Ψn+1 +√

Ψ2n+1 − 4δn+1

where {Ψn} is non-decreasing and {δn} is non-increasing..

We want |an − an| ≤ rn where an and rn are given by (2.4.8). We have

rnσnσn−1

= (ρn−1σn−1)σn|1 + Γn| − (ρnσn)σn−1|Γn−1| − (ρnσn)(ρn−1σn−1)≥ (ρn−1σn−1)(Ψn − ρn−1σn−1) = δn.

That is, |an − an|σnσn−1 ≤ δn ≤ rnσnσn−1; i.e., |an − an| ≤ rn for n ≥ n0 + 1.

Theorem 5.11. For a given continued fraction K(an/1) and a given se-quence {Γn} of complex numbers �= −1, let {σn} be a sequence of positivenumbers such that {Ψn} given by (2.4.9) is non-decreasing. Let further ρn

and δn be given by (2.4.11). If 4δn0+1 ≤ Ψ2n0+1 for some n0 ≥ 0, then there

exists a sequence {Vn} of value sets for K(an/1) with Vn := B(Γn, ρn) forn ≥ n0.

Remarks.

1. The expression (2.4.11) for ρn−1 satisfies

ρn−1 =2δn/σn−1

Ψn +√

Ψ2n − 4δn

≤ 2δn

Ψnσn−1=: ρ∗n−1. (2.4.12)

2. If n0 = 0; i.e., 4δ1 ≤ Ψ21, then {B(Γn, ρn)}∞n=0 is a sequence of value sets for

K(an/1).

3. If n0 > 0, then {Vn}∞n=0 given by Vn := B(Γn, ρn) for n ≥ n0 and Vn−1 :=an/(1 + Vn) for n = n0, n0 − 1, . . . , 1, is a sequence of value sets for K(an/1).

Error bounds.

If the sequence {Vn} of value sets derived for K(an/1) in this way consists ofbounded sets, then the Oval Sequence Theorem still applies for n ≥ n0 > 0 ifwe replace |Γ0| + ρ0 by sup{|w|; w ∈ V0} in the error bound.

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5.2.4 An algorithm to find value sets for a given continued fraction 247

Another method to obtain error bounds when n0 > 0, is to use

Sn0+k(w) = Sn0 ◦ S(n0)k (w) =

An0−1S(n0)k (w) + An0

Bn0−1S(n0)k (w) + Bn0

=fn0−1S

(n0)k (w) − fn0ζn0

S(n0)k (w) − ζn0

(Lemma 1.1 on page 6). Then

Sn+m(w) − Sn(v) =ζn0(fn0 − fn0−1)(S

(n0)n+m−n0

(w) − S(n0)n−n0

(v))

(S(n0)n+m−n0

(w) − ζn0)(S(n0)n−n0

(v) − ζn0),

where |S(n0)n+m−n0

(w) − ζn0 | > |Γn0 − ζn0 | − ρn0 if ζn0 �∈ Vn0 , and

fm =a1

1 +a2

1 +· · ·+am

1and ζm = −1 − am

1 +am−1

1 +· · ·+a2

1

are easy to compute for low values of m. Hence, for w ∈ Vn+m, it follows from theOval Sequence Theorem that

|Sn+m(w) − Sn(Γn)| <|ζn0(fn0 − fn0−1)| · ρn(|Γn0 | + ρn0)(|Γn0 − ζn0 | − ρn0)2(|1 + Γn| − ρn)

n−1∏j=n0+1

Mj (2.4.13)

for w ∈ Vn+m and n > n0.

Remarks.

1. For n0 = 1 we have ζ1 = −1, f1 = a1 and f0 = 0, and thus

|Sn+m(w) − Sn(Γn)| <|a1|ρn(|Γ1| + ρ1)

(|1 + Γ1| − ρ1)2(|1 + Γn| − ρn)

n−1∏j=2

Mj (2.4.14)

for w ∈ Vn+m and n ≥ 2.

2. For n0 = 1 we can also replace a1 by some a1 �= 0 which allows n0 = 0. Then

|Sn+m(w) − Sn(Γn)| ≤∣∣∣a1

a1

∣∣∣ ρn|Γ0| + ρ0

|1 + Γn| − ρn

n−1∏k=1

Mk (2.4.15)

where V0 = B(Γ0, ρ0) := a1/(1 + B(Γ1 + ρ1)).

3. For n0 = 2 we note that ζ2 = −1 − a2, f2 = a1/(1 + a2) and f1 = a1, so

|Sn+m(w) − Sn(Γn)| <

|a1a2|ρn(|Γ2| + ρ2)|1 + a2 + Γ2| − ρ2)2(|1 + Γn| − ρn)

n−1∏j=3

Mj

for w ∈ Vn+m for n ≥ 3.

(2.4.16)

4. These formulas easily generalize to continued fractions K(an/bn).

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248 Chapter 5: Computation of continued fractions

5.2.5 Value sets and the fixed point method

Let K(an/1) be a limit 1-periodic continued fraction of loxodromic type, such thatan → a with | arg(a + 1

4 )| < π. To find bounds for the truncation error |f − Sn(x)|where x is the attracting fixed point

x :=u − 1

2where u :=

√1 + 4a, (2.5.1)

we shall find a sequence {B(x, ρn)} of value sets for K(an/1) by means of ouralgorithm.

With σn := 1 for all n, we have Ψn = Ψ = |1 + x| − |x| for all n, so by (2.4.11)

ρn := 12(Ψ −

√Ψ2 − 4dn+1) where dn+1 := sup

m≥n+1|am − a|. (2.5.2)

That is, by Theorem 5.11, we get the following value sets {Vn} for K(an/1):�

Corollary 5.12. Let K(an/1) be limit 1-periodic of loxodromic type withtail values f (n) and attracting fixed point x ∈ C. Then there exists a se-quence {Vn}∞n=0 of value sets for K(an/1) with Vn := B(x, ρn) for n ≥ n0,where ρn is given by (2.5.2) and n0 ∈ N is chosen such that 4dn0+1 ≤ Ψ2.In particular, |f (n) − x| ≤ ρn for n ≥ n0.

Proof : It remains to prove that the tail values f (n) ∈ Vn, but this follows sincethe constant sequence {y} with y := −1 − x �= x is an exceptional sequence forK(an/1), and thus f (n) = limm→∞ S

(n)m (x) ∈ Vn. �

If a ≥ 0, this result takes a simpler form, since then x ≥ 0 and Ψ = 1.�

Corollary 5.13. Let K(an/1) be limit 1-periodic with a fixed point x ≥ 0.Then there exists a sequence {Vn}∞n=0 of value sets for K(an/1) with Vn :=B(x, ρn) for n ≥ n0, where ρn := 1

2(1−

√1 − 4dn+1), dn := supm≥n |am−a|

and n0 ∈ N is determined by dn0+1 ≤ 14 . In particular |f (n) − x| ≤ ρn for

n ≥ n0.

Remarks.

1. Since y := −1 − x is the second fixed point for S1(w), it follows immedi-ately that K(an/1) in Corollary 5.13 is of loxodromic type, and that x is itsattracting fixed point.

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5.2.5 Value sets and the fixed point method 249

2. If an → 0, and thus Vn = B(0, ρn) in our situation, then the oval regionEn reduces to the disk En = B(0, ρn−1(1 − ρn)) (see (2.3.2)), a situation werecognize from the extension of Worpitzky’s Theorem on page 136.

3. The radius ρn in (2.5.2) can be written

ρn =2dn+1

Ψ +√

Ψ2 − 4dn+1

<2dn+1

Ψ=: ρ∗n.

ρ∗n is therefore a simpler bound for |f (n) − x| than ρn, but ρ∗n ∼ 2ρn.

4. If an → − 14, then K(an/1) is limit 1-periodic of parabolic type with fixed

point − 12 . Then Lemma 5.1 on page 220 gives value sets for K(an/1) when

an → −14

fast enough.

Example 8. We shall find a sequence {Vn} of value sets for the continued fraction

∞Kn=1

an

1where a1 := 4, an+1 :=

n2

4n2 − 1for n ≥ 1

which converges to π (appendix, page 267). K(an/1) is limit 1-periodic of loxo-dromic type with a := lim an = 1

4and attracting fixed point x := 1

2(√

1 + 4a− 1) =12 (√

2 − 1). Now, d1 = a1 − 14 = 15

4 and

dn+1 := supm≥n+1

|am − a| = an+1 − a =n2

4n2 − 1− 1

4=

1/44n2 − 1

for n ≥ 1,

so dn+1 ≤ 14 for n ≥ 1. Hence it follows from Corollary 5.13 that K(an/1) has a

sequence {Vn}∞n=0 of value sets given by

Vn := B(x, ρn) where x := 12(√

2 − 1) and ρn := 12(1 −

√1 − 1/(4n2 − 1) )

for n ≥ 1 and

V0 := 4/(1 + V1) = B(Γ0, ρ0) with Γ0 :=4(1 + x)

|1 + x|2 − ρ21

, ρ0 :=4ρ1

|1 + x|2 − ρ21

.

In particular

∣∣∣f (n) −√

2 − 12

∣∣∣ ≤ ρn =12−

√n2 − 1

2

4n2 − 1< 2 dn+1 =

1/24n2 − 1

for n ≥ 1.

From the Oval Sequence Theorem it follows for instance that

|π − Sn(x)| ≤ 1/24n2 − 1

· |Γ0| + ρ0

|1 + x| − ρn

n−1∏k=1

|x(1 + x) − ρ2k| + ρk

|1 + x|2 − ρ2k

.

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250 Chapter 5: Computation of continued fractions

That is, ρ1 < 16, Γ0 ≈ 3.333, ρ0 < 0.47 and

Mk =|x(1 + x) − ρ2

k| + ρk

|1 + x|2 − ρ2k

< 0.2 for k ≥ 2,

so

|π − Sn(x)| ≤ 0.44 · (0.2)n−2

4n2 − 1for n ≥ 2.

The positive case.

Of course, Theorem 5.12 holds with Ψ = 1 when all an > 0. But it is a restrictionto require that an ∈ B(a, rn), as we have done in Corollary 5.12. (See Property 4on page 245.) We can do better if we use the oval sets En directly:�

Theorem 5.14. Let K(an/1) with 0 < an → a �= ∞ have attracting fixedpoint x. Then {B(x, ρn)} is a sequence of value sets for K(an/1) if for eachn ∈ N either

ρn−1 ≤ x and |an − a + ρn−1ρn| ≤ ρn−1(1 + x) − ρnx

orρn−1 ≥ x and an − a + ρn−1ρn ≤ ρn−1(1 + x) − ρnx.

(2.5.3)

Proof : Since an > 0 and x = 12(√

1 + 4a − 1) ≥ 0, it follows from Property 2on page 244 that {B(x, ρn)} is a sequence of value sets for K(an/1) if un ≤ an ≤vn for all n, where un and vn are given by (2.4.3) with all Γn := x. That is, ifeither

ρn−1 ≤ x, (x − ρn−1)(1 + x + ρn) ≤ an ≤ (x + ρn−1)(1 + x − ρn)or if ρn−1 ≥ x, an ≤ (x + ρn−1)(1 + x − ρn),

(2.5.4)

which is equivalent to (2.5.3) since x(1 + x) = a. �If we are willing to accept a little stronger conditions than (2.5.3), we have thefollowing simpler result:�

Corollary 5.15. Let K(an/1) with 0 < an → a �= ∞ have attracting fixedpoint x. Then {B(x, ρn)} is a sequence of value sets for K(an/1) if

|an − a + ρn−1ρn| ≤ ρn−1(1 + x) − ρnx for n ≥ 1. (2.5.5)

Page 264: Lisa Lorentzen, Haakon Waadeland Continued Fractions

5.2.5 Value sets and the fixed point method 251

Both in Theorem 5.14 and Corollary 5.15 it is up to us to choose {ρn}. In view ofRemark 3 on page 249, on could for instance try ρn := 2dn+1.

Example 9. Let K(an/1) be given by

a1 := 1, a2k :=k

4k − 2, a2k+1 :=

k

4k + 2for k ≥ 1.

Then an → 14

=: a, x = 12(√

2 − 1) and

a1 − a =34, a2k − a =

1/42k − 1

and a2k+1 − a =−1/42k + 1

for k ≥ 1. The choice ρn := 2 dn+1 gives

ρ0 :=32, ρ1 :=

12, ρ2k := ρ2k+1 :=

1/22k + 1

for k ≥ 1.

Condition (2.5.5) therefore looks like

|34 + 32 · 1

2 | < 32 · 1

2(√

2 + 1) − 12 · 1

2 (√

2 − 1) for n := 1,∣∣∣ 1/42k − 1

+1/4

4k2 − 1

∣∣∣ ≤ (√

2 + 1)/42k − 1

− (√

2 − 1)/42k + 1

for n := 2k,

∣∣∣ −1/42k + 1

+1/4

(2k + 1)2

∣∣∣ ≤ (√

2 + 1)/42k + 1

− (√

2 − 1)/42k + 1

for n := 2k + 1

for k ≥ 1, which actually is satisfied. Hence K(an/1) has a sequence {Vn} of valuesets given by

V0 := B(x, 32 ), V1 := B(x, 1

2 ), V2k := V2k+1 := B(x,

1/22k + 1

); k ≥ 1.

Hence it follows for instance from the Oval Sequence Theorem that

|f − Sn(x)| ≤ ρn

12(√

2 − 1) + 32

12(√

2 + 1) − ρn

n−1∏k=1

Mk (2.5.6)

where by (2.3.7)

M2m = M2m+1 =12(√

2 − 1) + 1/(4m + 2)12(√

2 + 1) + 1/(4m + 2)=

(√

2 − 1)(2m + 1) + 1(√

2 + 1)(2m + 1) + 1.

This bound is compared to the actual truncation error in the table below. Of course,since {Mk} is non-increasing, we can replace Mk by some Mk0 for all k > k0 tosimplify the computation. This is done in the last column with k0 := 4. This boundis easier to handle if we want to determine the order n of Sn(x) needed to reach a

Page 265: Lisa Lorentzen, Haakon Waadeland Continued Fractions

252 Chapter 5: Computation of continued fractions

given accuracy. A slight improvement is obtained if we do not insist on having thecenter Γ0 of V0 at x; i.e., if we take V0 := 1/(1 + V1) = B(1,

√2 − 1). Then the

Oval Sequence Theorem gives the bound

|f − Sn(x)| ≤ ρn

√2

12 (√

2 + 1) − ρn

n−1∏k=1

Mk. (2.5.7)

n Sn(x) |f − Sn(x)| (2.5.6) (2.5.6)modified (2.5.7)5 .6931772536 3.0 · 10−5 1.1 · 10−3 1.1 · 10−3 9.2 · 10−4

6 .6931515697 4.4 · 10−6 1.8 · 10−4 1.8 · 10−4 1.5 · 10−4

7 .6931478287 6.5 · 10−7 4.0 · 10−5 4.3 · 10−5 3.3 · 10−5

8 .6931472790 9.8 · 10−7 6.6 · 10−6 7.1 · 10−6 5.5 · 10−6

9 .6931471956 1.5 · 10−8 1.4 · 10−6 1.6 · 10−6 1.1 · 10−6

10 .6931471829 2.3 · 10−9 2.3 · 10−7 2.6 · 10−7 1.9 · 10−7

11 .6931471809 3.7 · 10−10 4.7 · 10−8 5.5 · 10−8 3.9 · 10−8

12 .6931471806 5.8 · 10−11 7.9 · 10−9 9.6 · 10−9 6.6 · 10−9

13 .6931471806 9.2 · 10−12 1.6 · 10−9 1.9 · 10−9 1.3 · 10−9

14 .6931471806 1.5 · 10−12 2.7 · 10−10 3.2 · 10−10 2.2 · 10−10

15 .6931471806 2.4 · 10−13 5.2 · 10−11 6.3 · 10−11 4.3 · 10−11

Limit 1-periodic S-fractions.

Let z �= 0 with | arg z| < π be kept fixed in the S-fraction K(anz/1) where 0 <an → a ≥ 0. Then K(anz/1) is limit 1-periodic of loxodromic type with attractingfixed point

x = x(z) = 12 (u(z) − 1) where u(z) :=

√1 + 4az.

The element sets {En} for the value sets {B(x, ρn)}∞n=0 are bounded by cartesianovals with axis along the line z + zR since arg a∗

n = arg(x(1 + x) = arg z in (2.3.3).By (2.4.3) with all Γn := x, the end points of the axis of En are unz/|z| and vnz/|z|where un < 0 if |x| < ρn−1 and

un = a|z| + |x|ρn − |1 + x|ρn−1 − ρnρn−1 if |x| ≥ ρn−1,

vn = a|z| − |x|ρn + |1 + x|ρn−1 − ρnρn−1.

Now, anz ∈ En if un ≤ an|z| ≤ vn. Hence, just as for the positive case we get:�

Theorem 5.16. Let K(anz/1) with 0 < an → a �= ∞ and fixed z with| arg z| < π have attracting fixed point x(z). Then {B(x(z), ρn(z))} is asequence of value sets for K(anz/1) if for each n ∈ N either

ρn−1 ≤ |x| and |(an − a)|z| + ρn−1ρn| ≤ ρn−1|1 + x| − ρn|x|orρn−1 > |x| and (an − a)|z| + ρn−1ρn ≤ ρn−1|1 + x| − ρn|x|.

(2.5.8)

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5.2.5 Value sets and the fixed point method 253

Example 10. The S-fraction K(anz/1) given by

a1 := 1, a2k :=k

4k − 2, a2k+1 :=

k

4k + 2for k ≥ 1,

converges to Ln(1 + z) for | arg(z + 1)| < π (see page 270). For z = 0 its value isclearly 0. In the following we assume that z �= 0 and | arg z| < π. Since an → 1

4 =: aand dn := supm≥n |amz − az| = |anz − az| in this case, we find that x := x(z) =12 (√

1 + z − 1),

d1 =3|z|4

, d2k =|z|/4

2k − 1and d2k+1 =

|z|/42k + 1

for k ≥ 1.

As in the previous example we choose ρn := 2 dn+1; i.e.,

ρ0 :=3|z|2

, ρ1 :=|z|2

, ρ2k := ρ2k+1 :=|z|/2

2k + 1for k ≥ 1.

Then condition (2.5.8) looks like

3|z|4

+3|z|2

4≤ 3|z|

2|1 + x| − |z|

2|x| for n := 1,

|z|/42k − 1

+|z|/2

2k − 1· |z|/22k + 1

≤ |z|/22k − 1

|1 + x| − |z|/22k + 1

|x| for n := 2k,

and for n := 2k + 1 ≥ 3 it only requires that∣∣∣−|z|/42k + 1

+( |z|/2

2k + 1

)2∣∣∣ ≤ |z|/22k + 1

(|1 + x| − |x|) if|z|/2

2k + 1< |x|.

Of course,

Ψ := |1 + x| − |x| = 12(|√

1 + z + 1| − |√

1 + z − 1|) > 0,

so these conditions hold

for n = 1 if 1 + |z| ≤ 2Ψ,

for n = 2k ≥ 2 if 1 +|z|

2k + 1≤ 2Ψ,

for n = 2k + 1 ≥ 3 if 2k + 1 ≤ |z|2|x| or

∣∣∣− 1 +|z|

2k + 1

∣∣∣ ≤ 2Ψ.

Hence there exists a sequence {Vn} of value sets for K(anz/1) with Vn := B(x, ρn)for n ≥ n0 where

n0 =

{0 if 1 + |z| ≤ 2|1 + x| − 2

3|x|,

2k − 1 if 1 + |z|/(2k + 1) ≤ 2Ψ otherwise.

If z > 0, then |1 + x| − |x| = 1. In particular n0 = 0 for z = 1, which is exactlywhat we obtained in Example 9. �

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254 Chapter 5: Computation of continued fractions

Limit 2-periodic continued fractions of loxodromic type.

If K(an/1) is limit 2-periodic of loxodromic type with attracting fixed points (x(0), x(1)),then Γ2n → x(0) and Γ2n+1 → x(1) for any natural choice of {Γn}. In the particularcase where we choose Γ2n := x(0) and Γ2n+1 := x(1) for all n, (2.4.9) holds withequality for all n with the choice

σ2n := |1 + x(1)| + |x(1)|, σ2n+1 := |1 + x(0)| + |x(0)| (2.5.9)

and Ψn := Ψ = |(1 + x(0))(1 + x(1))| − |x(0)x(1)| > 0 (Theorem 4.9 on page 182).That is,

ρn−1 :=1

2σn−1

(Ψ −

√Ψ2 − 4σ1σ2dn

)where dn := sup

m≥n|am − am| (2.5.10)

is a sensible choice, where

a2n−1 := a1 := lim a2m−1 and a2n := a2 := lim a2m. (2.5.11)

With this notation we get:�

Corollary 5.17. Let K(an/1) be limit 2-periodic of loxodromic type withtail values f (n) and finite attracting fixed points (x(0), x(1)). Let furthern0 ∈ N be such that 4σ1σ2dn0+1 ≤ Ψ2. Then there exists a sequence {Vn} ofvalue sets for K(an/1) with V2k := B(x(0), ρ2k) and V2k+1 := B(x(1), ρ2k+1)for n ≥ n0, and f (n) ∈ Vn for all n.

Example 11. We want to find small” value sets for∞Kn=1

an

1:=

3 + 1/12

1 +4 + 3/22

1 +3 + 1/32

1 +4 + 3/42

1 +3 + 1/52

1 +· · ·from Example 4 on page 13. K(an/1) is limit 2-periodic of loxodromic type withattracting fixed points (1, 2). Therefore σn given by (2.5.9) is σ2n := 3 + 2 = 5and σ2n+1 := 2 + 1 = 3, and Ψn = Ψ = 2 · 3 − 1 · 2 = 4. Hence σ1σ2 = 15 anddn := supm≥n |am − am| gives d1 = 1, d2 = 3

4and d2n−1 = d2n = 3/(4n2) for n ≥ 2.

Therefore 4σ1σ2dn < Ψ2 for n ≥ 3, and thus K(an/1) has a sequence {Vn} of valuesets with

V2n := B(2, ρ2n), V2n+1 := B(1, ρ2n+1) for n ≥ 1

where

ρ2n :=1

2σ2

(Ψ −

√Ψ2 − 4σ1σ2d2n+1

)=

25− 2

5

√1 − 45/16

(n + 1)2

ρ2n+1 :=1

2σ1

(Ψ −

√Ψ2 − 4σ1σ2d2n+2

)=

23− 2

3

√1 − 45/16

(n + 1)2.

Page 268: Lisa Lorentzen, Haakon Waadeland Continued Fractions

5.2.6 Value sets B(wn, ρn) for 1-periodic continued fractions 255

In particular |f (2n) − 2| ≤ ρ2n and |f (2n+1) − 1| ≤ ρ2n+1 for n ≥ 1. �

Limit p-periodic continued fractions of loxodromic type.

If K(an/1) is limit p-periodic of loxodromic type with attracting fixed points (x(0),x(1), . . . , x(p−1)), we can choose Γnp+m := Γm := x(m) for all m and n. Then wecan choose

σnp+p−1 := σp−1 :=p−2∑

n=−1

n∏j=0

|1 + Γj |p−2∏

j=n+1

|Γj | (2.5.12)

and the other σns are determined from σp−1 by cyclic shifts. For instance, for k = 3the three σn given by

σ0 = |Γ1Γ2| + |1 + Γ1||Γ2| + |1 + Γ1||1 + Γ2| ,σ1 = |Γ2Γ0| + |1 + Γ2||Γ0| + |1 + Γ2||1 + Γ0| ,σ2 = |Γ0Γ1| + |1 + Γ0||Γ1| + |1 + Γ0||1 + Γ1| .

With this choice we get Ψn = Ψ :=∏k−1

j=0 |1 + Γj | −∏k−1

j=0 |Γj | > 0. Also now The-orem 5.11 implies that f (np+m) ∈ Vnp+m := B(x(m), ρnp+m) for n ≥ n0 sufficientlylarge. We refer to [Jaco87] for value sets for more general continued fractions.

5.2.6 Value sets B(wnρn) for limit 1-periodic continued frac-tions of loxodromic or parabolic type

Let K(an/1) be limit 1-periodic of loxodromic type, and let wn ≈ f (n) be chosento make Sn(wn) → f �= ∞ fast, where f is the value of K(an/1). To take fulladvantage of this choice, we want Vn = B(wn, ρn). If wn → x, as it normally does,then also f (n) ∈ Vn in this case (Lemma 5.7 on page 240), so |f (n) − wn| ≤ ρn.

Example 12. We return to K(an/1) with a1 := 4, an+1 := n2/(4n2 − 1) for n ≥ 1from Example 7 on page 237. The square root modification gave approximantsSn(wn) with

wn :=12

(√8n2 − 14n2 − 1

− 1)

=12

(√2 ·√

1 +1/2

4n2 − 1− 1)

≈ 12

(√2(1 +

1/44n2 − 1

)− 1)

=√

2 − 12

+√

2/84n2 − 1

.

We therefore choose

Γn := wn :=√

2 − 12

+√

2/84n2 − 1

for n ≥ 1.

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256 Chapter 5: Computation of continued fractions

We first want to find positive σns such that Ψn := σn(1 + wn) − σn−1wn−1 > 0 isnon-decreasing. Since

1 + wn+1 − wn = 1 +√

28

( 1(2n + 1)(2n + 3)

− 1(2n − 1)(2n + 1)

)= 1 −

√2

8(2n + 3) − (2n − 1)

(2n − 1)(2n + 1)(2n + 3)= 1 −

√2/2

(2n − 1)(2n + 1)(2n + 3)

is monotonely increasing, we can take all σn := 1 to get

Ψn+1 = 1 + wn+1 − wn = 1 −√

2/2(4n2 − 1)(2n + 3)

for n ≥ 1.

Now,an+1 := Γn(1 + Γn+1)

=(√2 − 1

2+

√2/8

4n2 − 1

)(√2 + 12

+√

2/8(2n + 1)(2n + 3)

)=

14

+1/4

(2n − 1)(2n + 3)+

√2/4

(2n − 1)(2n + 1)(2n + 3)

+1/32

(2n − 1)(2n + 1)2(2n + 3),

so

an+1 − an+1 =(2 −

√2)/4

(2n − 1)(2n + 1)(2n + 3)− 1/32

(2n − 1)(2n + 1)2(2n + 3)

<(2 −

√2)/4

(2n − 1)(2n + 1)(2n + 3),

and we choose

ρn := 12 (Ψn+1 −

√Ψ2

n+1 − 4δn+1 ) where δn+1 :=(2 −

√2)/4

(2n − 1)(2n + 1)(2n + 3)

for n ≥ 1. Then 4δn+1 ≤ Ψ2n+1 if and only if

2 −√

2(2n − 1)(2n + 1)(2n + 3)

≤(1 −

√2/2

(2n − 1)(2n + 1)(2n + 3)

)2

which holds for n ≥ 1. Hence K(an/1) has a sequence {Vn} of value sets with

Vn := B(√2 − 1

2+

√2/8

4n2 − 1, ρn

)for n ≥ 1,

and

|f (n) − wn| ≤ ρn < ρ∗n := 2δn+1

Ψn+1=

(2 −√

2)/2(4n2 − 1)(2n + 3) −

√2/2

for n ≥ 1.

Page 270: Lisa Lorentzen, Haakon Waadeland Continued Fractions

5.2.6 Value sets B(wn, ρn) for 1-periodic continued fractions 257

To obtain a priori truncation error bounds, we set B(Γ0, ρ∗0) := 1/(1 + B(Γ1, ρ

∗1)).

Then (Lemma 3.6 on page 110)

Γ0 :=4(1 + Γ1)

|1 + Γ1|2 − ρ∗21

≤4(

√2+12 +

√2

8·3 )

(√

2+12

+√

28·3 )2 − ( (2−√

2)/2

3·5−√2/2

)2≈ 3.160305,

ρ∗0 :=4ρ∗1

|1 + Γ1|2 − ρ∗21

≈ 0.051153.

The Oval Sequence Theorem and Remark 3 on page 242 thereby imply that

|f − Sn(Γn)| ≤ ρ∗n|Γ0| + ρ∗0

|1 + Γn| − ρ∗n

n−1∏k=1

M∗k ≤

(2 −√

2) · (3.160305 + 0.051153)∏n−1

k=1 M∗k

(√

2 + 1)[(4n2 − 1)(2n + 3) −√

22

] +√

24

(2n + 3) − 1/44n2−1

− 2 +√

2

(2.6.1)

where

M∗k :=

Γk + ρ∗k1 + Γk + ρ∗

k

<

√2−12 +

√2/8

4k2−1 + (2−√2)/2

(4k2−1)(2k+3)−√2/2

√2+12 +

√2/8

4k2−1 + (2−√2)/2

(4k2−1)(2k+3)−√2/2

.

The table below compares this error bound to the true truncation errors. In thelast column we have replaced Mk by M5 for k > 5. This bound is well suited todetermine n in Sn(Γn) to achieve a wanted accuracy. The true value of K(an/1) isπ, and the approximants Sn(Γn) converge very fast to this value.

n |f − Sn(Γn)| (2.6.1) (2.6.1)modified5 2.6 · 10−7 7.4 · 10−7 7.4 · 10−7

6 2.7 · 10−8 7.6 · 10−8 7.6 · 10−8

7 3.0 · 10−9 8.5 · 10−9 8.5 · 10−9

8 3.5 · 10−10 1.0 · 10−9 1.0 · 10−9

9 4.3 · 10−11 1.2 · 10−10 1.2 · 10−10

10 5.4 · 10−12 1.6 · 10−11 1.6 · 10−11

11 7.1 · 10−13 2.0 · 10−12 2.1 · 10−12

12 9.5 · 10−14 2.7 · 10−13 2.8 · 10−13

13 1.3 · 10−14 3.7 · 10−14 3.9 · 10−14

14 1.8 · 10−15 5.2 · 10−15 5.4 · 10−15

15 2.5 · 10−16 7.2 · 10−16 7.6 · 10−16

Our next example demonstrates how difficult it can be to find useful value sets forlimit periodic continued fractions of parabolic type.

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258 Chapter 5: Computation of continued fractions

Example 13. The continued fraction K(an/1) with

an := −14

+(−1/2)n

4n + 2for n = 1, 2, 3, . . .

is limit 1-periodic of parabolic type. Let all Γn := − 12 . Then |1+Γn| = |Γn−1| = 1

2 ,and we can for instance use

σn := 2n + 1 for n ≥ 0.

Then Ψn := σn|1 + Γn| − σn−1|Γn−1| = 1 for all n. The choice (2.4.11) for {ρn}then gives

ρn−1 =1/2

2n − 1

(1 −√

1 − 4δn

)where δn := sup

m≥n

{(4m2 − 1)

(1/2)m

4m + 2

}i.e., δn = sup

m≥n{( 1

2)m+1(2m − 1)} = ( 1

2)n+1(2n − 1) for n ≥ 3

and δ1 = δ2 = 38 . In particular 4δn < 1 for n ≥ 4, so K(an/1) has a sequence

{Vn}∞n=0 of value sets with Vn := B(− 12 , ρn) for n ≥ 3. Of course, this means that

K(an/1) has a tail sequence {tn} with tn ∈ B(−12, ρn) for all n. But we do not

know that this is a sequence of tail values. We can not even say whether K(an/1)converges generally or not at this point. �

5.2.7 Error bounds based on idea 3

Idea 3 on page 239 was to combine the bound |f (n)−wn| ≤ ρn with known truncationerror bounds |f − Sn(wn)| ≤ λn to get

|f − Sn(wn)| ≤ |wn − ζn||wn − ζn|

λnρn

|wn − f (n)| . (2.7.1)

For instance, if {Vn} with Vn := (−gn+eiαH) for 0 < gn < 1 and fixed −π2 < α < π

2is a sequence of value sets for K(an/1), then we know from Theorem 3.45 on page155 that

diamSn(Vn) ≤ |a1|/((1 − g1) cos α)n∏

j=2

(1 +

Pj−1gj−1(1 − gj) cos2 α

|aj |Σj−2

) (2.7.2)

where

Σn :=n∑

m=0

Pm and Pm :=m∏

k=1

1 − gk

gk. (2.7.3)

Another typical example is valid for S-fractions K(anz/1) with an > 0 and z :=r e2iα for some fixed r > 0 and −π

2 < α < π2 . Then the Thron-Gragg-Warner bound

|f − Sn(w)| ≤ 2a1r

cos α

n∏k=2

√1 + 4akr/ cos2 α − 1√1 + 4akr/ cos2 α + 1

=: λn (2.7.4)

Page 272: Lisa Lorentzen, Haakon Waadeland Continued Fractions

5.2.7 Error bounds based on idea 3 259

from page 128 gives a very good truncation error bound in this case. A natural ideais therefore to use this error in (2.7.1).

Example 14. Again we consider K(an/1) with a1 := 4 and an+1 := n2/(4n2 − 1).A truncation error bound for

|f − Sn(wn)| for wn := Γn :=√

2 − 12

+√

2/84n2 − 1

was established by means of the Oval Sequence Theorem in Example 12. We shallnow derive bounds for |f − Sn(wn) by means of (2.7.1). If we base our analysis on(2.7.2) with α := 0 and all gn := 1

2 , then

|f − Sn(∞)| ≤ 8n∏

j=2

(1 +

1/4(j − 1)|aj |

)−1

= 8n−1∏j=1

(1 +

4j2 − 14j3

)−1

.

In Example

|f (n) − wn| ≤ ρ∗n =(2 −

√2)/2

(4n2 − 1)(2n + 3) −√

2/2for n ≥ 1.

Since all an > 0, we definitely have

ζn = −1 − an

1 +an−1

1 +· · ·+a2

1< −1.

Hence |wn − ζn| > |1+wn|−ρ∗n, and thus it follows from (2.7.1) with wn := ∞ that

|f − Sn(wn)| <ρ∗n

|1 + wn| − ρ∗n8

n−1∏j=1

(1 +

4j2 − 14j3

)−1

. (2.7.5)

Now, K(an/1) can also be considered as an S-fractions with z := 1. The Thron-Gragg-Warner bound (2.7.4) can therefore also be used; i.e.,

|f − Sn(wn)| <ρ∗n

|1 + wn| − ρ∗n8

n−1∏k=1

√1 + 4k2/(4k2 − 1) − 1√1 + 4k2/(4k2 − 1) + 1

. (2.7.6)

The table below shows the true truncation error |f − Sn(wn)| and the three trun-cation error bounds (2.6.1), (2.7.5) and (2.7.6).

n |f − Sn(Γn)| (2.6.1) (2.7.5) (2.7.6)5 2.6 · 10−7 7.4 · 10−7 3.6 · 10−4 1.7 · 10−6

6 2.7 · 10−8 7.6 · 10−8 1.8 · 10−4 1.8 · 10−7

7 3.0 · 10−9 8.5 · 10−9 9.9 · 10−5 2.0 · 10−8

8 3.5 · 10−10 1.0 · 10−9 5.9 · 10−5 2.3 · 10−9

9 4.3 · 10−11 1.2 · 10−10 3.8 · 10−5 2.9 · 10−10

10 5.4 · 10−12 1.6 · 10−11 2.5 · 10−5 3.7 · 10−11

11 7.1 · 10−13 2.0 · 10−12 1.7 · 10−5 4.8 · 10−12

12 9.5 · 10−14 2.7 · 10−13 1.2 · 10−5 6.4 · 10−13

13 1.3 · 10−14 3.7 · 10−14 9.0 · 10−6 8.7 · 10−14

14 1.8 · 10−15 5.2 · 10−15 6.8 · 10−6 1.2 · 10−14

15 2.5 · 10−16 7.2 · 10−16 5.2 · 10−6 1.7 · 10−15

12 we found that

Page 273: Lisa Lorentzen, Haakon Waadeland Continued Fractions

260 Chapter 5: Computation of continued fractions

5.3 Stable computation of approximants

5.3.1 Stability of the backward recurrence algorithm

In section 1.1.3 on page 10 we suggested three algorithms for computing approxi-mants

Sn(wn) =a1

b1 +a2

b2 +· · ·+an

bn + wn=

An−1wn + An

Bn−1wn + Bn. (3.1.1)

1. The forward recurrence algorithm where An and Bn are computed by meansof the recurrence relation

Ak = bkAk−1 + akAk−2, Bk = bkBk−1 + akBk−2, (3.1.2)

starting with A−1 = 1, A0 = 0 and B−1 = 0, B0 = 1.

2. The backward recurrence algorithm where we use the recursion

qk−1 = sk(qk) :=ak

bk + qkfor k = n, n − 1, . . . , 1 (3.1.3)

starting with qn := wn.

3. The Euler-Minding summation where we compute {Bk}nk=1 by means of (3.1.2)

and

Sn(0) =a1

B0B1− a1a2

B1B2+

a1a2a3

B2B3− a1a2a3a4

B3B4+ · · · −

∏nk=1(−ak)Bn−1Bn

(3.1.4)

which normally only is suited to find classical approximants for continued frac-tions K(1/bn), even though it can be modified to give general approximantsSn(wn).

The emphasis in section 1.1.3 was on the complexity of the algorithm; i.e., thennumber of operations needed to reach Sn(wn). Even more important is the sta-bility of the computation, in particular if we want to compute high order ap-proximants Sn(wn). The forward algorithm is clearly unstable if for instancebkAk−1 ≈ −akAk−2 or bkBk−1 ≈ −akBk−2 for some indices. One can also runinto very large or very small values for |An| or |Bn| fairly soon, which complicatesthe computation. The same problems may ruin the Euler-Minding summation.

In section 1.1.3 we claimed that the backward algorithm is usually more stable thanthe other two. So, what are we actually doing in this algorithm? Well, let us assumethat K(an/bn) converges generally to f , and that the purpose of computing Sn(w)is to approximate f . We are therefore not worried if the computed value Sn(w) ofSn(w) is closer to f than it should be. What we do, is: we start with a given valuew ∈ C and compute

wn,n := w, wn,k−1 := sk(wn,k) for k = n, n − 1, . . . , 1. (3.1.5)

Page 274: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Remarks 261

That is, we compute the first (n + 1) terms of the tail sequence {tn,k}k where

tn,k = wn,k = sk+1 ◦ sk+2 ◦ · · · ◦ sn(w) = S(k)n−k(w). (3.1.6)

We know a lot about tail sequences for a generally convergent continued fractionK(an/bn):

• there are two kinds of tail sequences for K(an/bn); the sequence {tn} of tailvalues where t0 = f , and all the other tail sequences which in the literatureoften are called the wrong tail sequences.

• all the wrong tail sequences have the same asymptotic behavior in the sensethat limm(tn, tn) = 0 when t0 �= f and t0 �= f .

• if we start with a given value for t0 and compute the tail sequence by therecurrence

tn+1 := s−1n (tn) for n = 0, 1, 2, . . . , (3.1.7)

then the computed sequence {tn} will always behave like a wrong tail sequenceasymptotically, even if we started with t0 = f . The reason is that even minorinaccuracies makes tn �= f (n), at least if

lim infn→∞ m(f (n), tn) > 0 for t0 �= f. (3.1.8)

Therefore, the computation of a wrong tail sequence is stable when (3.1.8) holds,whereas the computation of the tail values is highly unstable with this algorithm.In the backward recurrence algorithm the recursion (3.1.7) is inverted. We startwith tn and compute tn−1 = sn(tn). Therefore, the computation of the sequenceof tail values by this algorithm is highly stable, whereas computation of wrong tailsequences are unstable: they will be pulled in towards f , exactly what we want.

We can also argue in terms of value sets. Let for instance V be a simple boundedclosed value set for K(an/1) with V ◦ �= ∅ and V ∩ (−1−V ) = ∅. Let w ∈ V0. Thenthe part {tn,k}n

k=0 of the tail sequence {tn,k} given by (3.1.6) is contained in V0.As n increases, this tail sequence will be more and more like the sequence of tailvalues from V0. The other tail sequences have all their limit points in −1 − V0.

5.4 Remarks

1. Convergence acceleration. The idea of choosing {wn} carefully to make {Sn(wn)}converge faster is an old idea. Indeed, already in 1869 Sylvester ([Sylv69]) claimed:I think a substantial difference does arise in favor of the continued fraction form,

inasmuch as it indicates a certain obvious correction to be applied in order that theconvergence may become more exact.” The correction he had in mind was exactlythe idea of using Sn(w) instead of Sn(0). He was comparing the continued fractionto the series

∑(fn − fn−1) where fn := Sn(0).

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262 Chapter 5: Computation of continued fractions

This idea has been applied by a number of authors, but the first systematic investi-gation for complex continued fractions is due to Thron and Waadeland ([ThWa80a])who used the attracting fixed point of limit 1-periodic continued fractions K(an/1)to accelerate the convergence. Their investigations were inspired by previous workby Gill ([Gill75]). Later on, Gill and the authors of this book have extended thetheory in a number of papers.

2. Birkhoff-Trjzinski theory. In order to accelerate the convergence of a continuedfraction K(an/bn), we use approximants Sn(wn) where wn approximates its tailvalue f (n). Now, a tail sequence {tn} can always be written as {−Pn/Pn−1} where{Pn} is a non-trivial solution of the recurrence relation

Pn = bnPn−1 + anPn−2 for n = 1, 2, 3, . . . .

The Birkhoff-Trjzinski theory gives a method to approximate {Pn} when an = a(n)and bn = b(n) are functions of n with asymptotic expansions of the form

a(n) ∼ nλ1/ω∞∑

m=0

αm n−m/ω, b(n) ∼ nλ2/ω∞∑

m=0

βm n−m/ω as n → ∞,

where λi are integers, ω ∈ N and α0β0 �= 0. They proved that every solutionof the recurrence relation can be written as a linear combination of two solutionsPn := P (n) which have asymptotic expansions of the form

eQ(ρ,n)s(ρ, n) where

Q(ρ, n) := δ0 n ln(n) +

ρ∑j=1

δjn(ρ+1−j)/ρ,

s(ρ, n) := nθt∑

j=0

(ln n)jnrt−1/ρqj(ρ, n)

qj(ρ, n) :=

∞∑m=0

cj,mn−m/ρ

where the integers ρ ≥ 1, rj and δ0ρ and the complex numbers δj , θ and cj,m withcj,0 �= 0, r0 := 0 and −π ≤ Im δ1 < π can be determined by substituting this intothe recurrence relation and comparing coefficients, ([Birk30], [BiTr32]). For moreinformation we refer to [Wimp84].

5.5 Problems

1. Limit periodic continued fraction. Let K(1/bn) be the continued fraction wherebn = 4 + (0.9)n for all n.

(a) Find a value set V := B(Γ, ρ) for K(1/bn). (Try to make V small.)

(b) Does K(1/bn) converge to a finite value f?

(c) Are the classical approximants fn of K(1/bn) all distinct; i.e. fn �= fm ifn �= m?

(d) Use the value set V found in (a) to derive upper bounds for the truncationerror |f − Sn(w)| for suitably chosen w ∈ C.

Page 276: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Problems 263

2. Simple element set. Let K(an/1) have all elements an ∈ E := {w ∈ C; |w−3−i| ≤0.4}. Find a Γ ∈ C and a 0 < ρ < |1 + Γ| such that E is a subset of the cartesianoval in the Oval Sequence Theorem on page 243 with all Γn := Γ and ρn := ρ. Usethis to prove that K(an/1) converges to a finite value and to find truncation errorbounds for suitably chosen approximants of K(an/1).

3. The logarithm of (1 + i). Use the Oval Sequence Theorem on page 243 with allΓn := Γ and ρn := ρ to find bounds for |f − Sn(w)| for the continued fraction

K ani

1:=

i

1+

i/2

1 +

i/6

1 +

2i/6

1 +

2i/10

1 +

3i/10

1 +· · ·

when w = (√

1 + i − 1)/2.

4. Limit periodic continued fraction. We want to improve the speed of convergenceof the continued fraction

K an

1=

6 + (0.9)

1 +

6 + (0.9)2

1 +

6 + (0.9)3

1 +

6 + (0.9)4

1 +· · ·.

(a) Prove that K(an/1) converges to a finite value f .

(b) Suggest a value wn = w for all n such that

limn→∞

f − Sn(w)

f − Sn(0)= 0.

Does the sequence {Sn(w)} have an oscillating character which can be used toobtain upper bounds for the truncation error |f − Sn(w)|?

(c) Suggest values for wn such that

limn→∞

f − Sn(wn)

f − Sn(w)= 0

where w is the value from a). Does {Sn(wn)} have such an oscillating charac-ter?

5. The case an → ∞. Suggest expressions for wn such that the approximants Sn(wn)of

K an

1=

12

1 +

5 · 22

1 +

32

1 +

5 · 42

1 +

52

1 +

5 · 62

1 +

72

1 +· · ·(hopefully) converge faster to the value f of K(an/1) than Sn(0). Compute the first6 approximants of Sn(0) and Sn(wn) and use the oscillating character to determinean error bound for S6(0). Find an upper bound for |f − Sn(wn)|.

6. Choice of approximations.

(a) Suggest some favorable approximants Sn(wn) to compute K((−0.2+(0.4)n)/1).

(b) Compute the value of K((−0.2 + (0.4)n)/1) with an absolute error less than0.05.

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264 Chapter 5: Computation of continued fractions

7. The improvement machine. Let K(an/1) be given by

an = x(1 + x) + rn where 0 < |x| < |1 + x| and 0 < |r| < 1 .

Use the improvement machine on page 227 to derive approximants Sn(w(m)n ) for

K(an/1) such that

limn → ∞

f − Sn(w(m)n )

f − Sn(w(m−1)n )

= 0 for m = 1, 2, 3, . . . .

8. ♠ The Oval Theorem. Set all Γn := Γ and ρn := ρ in the oval sequence theoremon page 243, where ReΓ > − 1

2and ρ < |1 + Γ|.

(a) Prove that

E :={

a ∈ C;∣∣∣ a(1 + Γ)

|1 + Γ|2 − ρ2− Γ∣∣∣+ |a|ρ

|1 + Γ|2 − ρ2≤ ρ}

is the simple element set for continued fractions K(an/1) corresponding to thesimple value set V := B(Γ, ρ).

(b) Prove that every continued fraction K(an/1) from E converges to some valuein V . (Hint: the Parabola Theorem on page 151 or the lemma on page 113may be of help.)

(c) Prove that

|f − Sn(wn)| ≤ 2ρ|Γ| + ρ

|1 + Γ| − ρMn−1 where M := max

v∈V

∣∣∣ v

1 + v

∣∣∣for every continued fraction K(an/1) from E and wn ∈ V .

(d) Prove that M < 1 if ρ < 12(|1 + Γ| − |Γ|).

(e) Prove that M < 1 if Γ > 0 and ρ < Γ + 12.

9. ♠ A bridge between the Worpitzky disk and the Parabola Theorem. LetE be as in Problem 8 with ρ := Re Γ + 1

2> 0.

(a) Prove that − 14∈ ∂E.

(b) Prove that E reduces to the Worpitzky disk if Γ := 0.

(c) Prove that the Worpitzky disk from (b) is contained in E for all Γ with ReΓ >0.

(d) Prove that E is contained in the parabolic region Eα from the Parabola The-orem on page 151 when arg Γ(1 + Γ) = 2α.

(e) Prove that lim E = Eα from the Parabola Theorem when arg Γ(1 + Γ) = 2α iskept fixed and Γ → ∞.

Page 278: Lisa Lorentzen, Haakon Waadeland Continued Fractions

Appendix A

Some continued fractionexpansions

This is a catalogue of some of the known continued fraction expansions. The list is in noway complete. Still it can be useful, both to find a continued fraction expansion of somegiven function and to “sum” a given continued fraction.

We have not attempted to find the origin of each result. The references we give aretherefore just pointing to books or papers where the expansion also can be found.

A.1 Introduction

A.1.1 Notation

We write

f(· · · ) = K(an(· · · )/bn(· · · )); (· · · ) ∈ D (1.1.1)

to say that the continued fraction converges in the classical sense to f(· · · ) for the pa-rameters (· · · ) in the set D. In the literature the set D is often far too restrictive, if it isgiven at all. We have determined a (possibly larger) set Dc where the continued fractionconverges. This is done by methods presented in this book. However, it may well happenthat the equality (1.1.1) fails in a subset of Dc, even if f(· · · ) is interpreted as an analyticcontinuation of the expression in question. In some cases we therefore give expressions forsets Dc and Df such that K(an(· · · )/bn(· · · )) converges in Dc and the equality holds inDf ⊆ Dc. What happens outside these sets is not checked. The identities normally holdsalso at points where an(· · · ) = 0 unless otherwise stated.

The elements an(· · · ) and bn(· · · ) of almost all continued fractions in this appendix arepolynomials in the parameters. The classical approximants are then rational functions ofthe parameters, and can therefore not converge to multivalued functions. If the left handside of (1.1.1) is a multivalued function, we always take the principal part unless otherwise

L. Lorentzen and H. Waadeland, Continued Fractions, Atlantis Studies in Mathematics for Engineering and Science 1, DOI 10.1007/978-94-91216-37-4, © 2008 Atlantis Press/World Scientific

265

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266 Appendix A: Some continued fraction expansions

stated. The principal part is often written with a capital first letter, such as Ln z, Arctan zetc.

A.1.2 Transformations

It is evident that not every continued fraction expansion can find room in a book like this.On the other hand, quite a number of the known continued fraction expansions can bederived from one another by simple transformations. We have for instance

f = b0 + K(an/bn) ⇐⇒ c · 1

f=

c

b0 +

a1

b1 +

a2

b2 +· · ·; c �= 0. (1.2.1)

Similarly, if f = b0 + K(an/bn), then g = (f − 1)/(f + 1) = 1 − 2/(1 + f); i.e.,

f = b0 + K(an/bn) ⇐⇒ f − 1

f + 1= 1 − 2

1 + b0 +

a1

b1 +

a2

b2 +· · ·. (1.2.2)

Another simple transformation is maybe most easily described for S-fractions. Assume thatf(z) = b0 + K(anz/1). Then f(z−1) = b0 + K(anz−1/1). Equivalence transformationslead to

f(

1z

)= b0 +

a1

z +

a2

1 +

a3

z +

a4

1 +

a5

z +· · ·

= b0 +a1/z

1 +

a2

z +

a3

1 +

a4

z +

a5

1 +· · ·

= b0 +a1/ξ

ξ +

a2

ξ +

a3

ξ +

a4

ξ +

a5

ξ +· · ·,

where ξ2 = z. We shall normally not list equivalent continued fractions like this separately.

Another situation that often arises is the following: We have

f(z) = b0 +a1z

2

1 +

a2z2

1 +

a3z2

1 +· · ·. (1.2.3a)

Then

f(iz) = b0 −a1z

2

1 −a2z

2

1 −a3z

2

1 −· · ·. (1.2.4b)

Of course, every time we have a continued fraction expansion f = b0 + K(an/bn) withall an, bn �= 0, we can take its even or odd part and obtain a “new” continued fractionconverging to the same value f . Some of these variations will be listed, in particular ifthey turn out to be nice and simple.

A.2 Elementary functions

A.2.1 Mathematical constants

π = 3 +1

7+

1

15+

1

1+

1

292+

1

1+

1

1+

1

1+

1

2+

1

1+

1

3+

1

1+

1

14+· · ·, (2.1.1)

Page 280: Lisa Lorentzen, Haakon Waadeland Continued Fractions

A.2.1 Mathematical constants 267

([JoTh80], p 23). This is the regular continued fraction expansion of π.

π =4

1+

12

3 +

22

5 +

32

7 +

42

9 +· · ·, (2.1.2)

([JoTh80], p 25), (see also (3.6.1)).

π

2= 1 +

1

1+

1 · 21 +

2 · 31 +

3 · 41 +· · ·

, ([Khru06a]). (2.1.3)

For the Riemann zeta function we have

12ζ(2) =

π2

12=

1

1+

14

3 +

24

5 +

34

7 +· · ·, (2.1.4)

([Bern89], p 150).

ζ(2) =π2

6= 1 +

1

1+

12

1 +

1 · 21 +

22

1 +

2 · 31 +

32

1 +

3 · 41 +

42

1 +· · ·, (2.1.5)

([Bern89], p 153).

Apery’s constant:

ζ(3) = 1 +1

4+

13

1 +

13

12+

23

1 +

23

20+

33

1 +

33

28+

43

1 +

43

36+· · ·, (2.1.6)

([Bern89], p 155), (see also (4.7.37)).

Euler’s number:

e =1

1−1

1+

1

2−1

3+

1

2−1

5+

1

2−1

7+· · ·, (2.1.7)

([JoTh80], p 25), (see also (3.2.1)).

e = 2 +1

1+

1

2+

1

1+

1

1+

1

4+

1

1+

1

1+

1

6+· · ·, (2.1.8)

([JoTh80], p 23).

e = 1 +2

1+

1

6+

1

10+

1

14+

1

18+· · ·, (2.1.9)

([Khov63], p 114). (See also (3.2.2)).

e = 2 +2

2+

3

3+

4

4+

5

5+· · ·, (2.1.10)

([Perr57], p 57).

e =1

1−2

3+

1

6+

1

10+

1

14+

1

18+· · ·, (2.1.11)

([Khov63], p 114). (See also (3.2.2) for z = −1).

√e = 1 +

1

1+

1

1+

1

5+

1

1+

1

1+

1

9+

1

1+

1

1+

1

13+· · ·, (2.1.12)

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268 Appendix A: Some continued fraction expansions

([Euler37]).

3√

e = 1 +1

2+

1

1+

1

1+

1

8+

1

1+

1

1+

1

14+

1

1+

1

1+

1

20+· · ·, (2.1.13)

([Euler37]). (See also (2.2.4).)

coth 12

=e + 1

e − 1= 2 +

1

6+

1

10+

1

14+

1

18+· · ·, (2.1.14)

([Khru06b]).

The golden ratio: √5 − 1

2=

1

1+

1

1+

1

1+· · ·, (2.1.15)

([JoTh80], p 23).

Catalan’s constant: G :=∑∞

k=0(−1)k/(2k + 1)2 :

2G = 2 − 12

3 +

22

1 +

22

3 +

42

1 +

42

3 +

62

1 +

62

3 +· · ·, (2.1.16)

([Bern89], p 151). (See also (4.7.30) with z := 2.)

2G = 1 +1

1/2+

12

1/2+

1 · 21/2 +

22

1/2+

2 · 31/2 +

32

1/2+

3 · 41/2 +· · ·

, (2.1.17)

([Bern89], p 153). (See also (4.7.32) with z := 12.)

A.2.2 The exponential function

ez = 1F1(1; 1; z) =1

1−z

1+

z

2−z

3+

z

2−z

5+

z

2−z

7+· · ·

=1

1−z

1+

1z

2 −1z

3 +

2z

4 −2z

5 +

3z

6 −3z

7 +

4z

8 −4z

9 +· · ·; z ∈ C,

([JoTh80], p 207). (See also (4.1.4).) The odd part of this continued fraction is

ez = 1 +2z

2 − z +

z2

6 +

z2

10+

z2

14+

z2

18+· · ·; z ∈ C, (2.2.1)

([Khov63], p 114).

ez = 1 +z

1 − z +

1z

2 − z +

2z

3 − z +

3z

4 − z +· · ·; z ∈ C, (2.2.2)

([JoTh80], p 272).

Since ez = 1/e−z, we can find three more expansions from (2.2.1)–(2.2.2) by use of (1.2.1).For instance, (2.2.2) transforms into

ez =1

1−z

1 + z−1z

2 + z−2z

3 + z−3z

4 + z +· · ·; z ∈ C, (2.2.3)

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A.2.3 The general binomial function 269

([Khov63], p 113).

e1/z = 1 +1

z − 1+

1

1+

1

1+

1

3z − 1+

1

1+· · ·1

1+

1

5z − 1+

1

1+· · ·z ∈ C, (2.2.4)

([Khru06b]).

Lambert’s continued fraction

ez − e−z

ez + e−z=

z

1+

z2

3 +

z2

5 +

z2

7 +· · ·; z ∈ C, (2.2.5)

([Wall48], p 349), is easily obtained from (2.2.1) by use of (1.2.2).

A.2.3 The general binomial function

The general binomial function (1 + z)α is a multivalued function. As already mentionedin the introduction, we shall always let (1 + z)α mean the principal part of this function;i.e., as always,

(1 + z)α := exp(α Ln(1 + z)) where − π < Im(Ln(1 + z)) ≤ π.

We then have the following expansions:

(1 + z)α = 2F1(−α, 1; 1;−z)

=1

1−αz

1 +

(1 + α)z

2 +

(1 − α)z

3 +

(2 + α)z

2 +

(2 − α)z

5 +

(3 + α)z

2 +

(3 − α)z

7 +· · ·(2.3.1)

for α ∈ C and | arg(z + 1)| < π, ([JoTh80], p 202). (See also (3.1.6).) The odd part of thiscontinued fraction is

(1 + z)α =

1 +2αz

2 + (1 − α)z−(12 − α2)z2

3(z + 2) −(22 − α2)z2

5(z + 2) −(32 − α2)z2

7(z + 2) −· · ·(2.3.2)

for α ∈ C and | arg(z + 1)| < π, ([Khov63], p 105). (2.3.2) is also the odd part of

(1 + z)α =

1

1−αz

1(1 + z)−(1 − α)z

2 −(1 + α)z

3(1 + z)−(2 − α)z

2 −(2 + α)z

5(1 + z)−(3 − α)z

2 −· · ·(2.3.3)

for α ∈ C and | arg(z + 1)| < π, ([Khov63], p 101).

(1 + z)α =1

1−αz

1 + (1 + α)z−1(1 + α)z(1 + z)

2 + (3 + α)z −2(2 + α)z(1 + z)

3 + (5 + α)z −· · ·(2.3.4)

([Khov63], p 101). Dc := {(α, z) ∈ C2; Re(z) �= − 1

2, z �= −1}, Df := {(α, z) ∈

C2; Re(z) > − 1

2}.

(1 + z)α =1

1−αz

1 + αz +

1(1 − α)z

2 − (1 − α)z +

2(2 − α)z

3 − (2 − α)z +

3(3 − α)z

4 − (3 − α)z +· · ·, (2.3.5)

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270 Appendix A: Some continued fraction expansions

([Khov63], p 102). Dc := {(α, z) ∈ C2; |z| �= 1}, Dc := {(α, z) ∈ C

2; |z| < 1}.The general binomial function satisfies (1 + z)α = 1/(1 + z)−α. Hence the equality (1.2.1)applied to these 5 expansions gives us 5 new ones. To find a continued fraction expansionfor (

z + 1

z − 1

=

(1 +

2

z − 1

(2.3.6)

we can use any of the 5 expansions (2.3.1)–(2.3.5) with z replaced by 2/(z − 1).

Laguerre’s continued fraction

(z + 1

z − 1

= 1 +2α

z − α+

α2 − 12

3z +

α2 − 22

5z +

α2 − 32

7z +· · ·, (2.3.7)

for α ∈ C and z ∈ C \ [−1, 1], ([Perr57], p 153).

z1/z = 1 +z − 1

1z +

(1z − 1)(z − 1)

2 +

(1z + 1)(z − 1)

3z +

(2z − 1)(z − 1)

2 +

(2z + 1)(z − 1)

5z +

(3z − 1)(z − 1)

2 +

(3z + 1)(z − 1)

7z +· · ·

(2.3.8)

for | arg z| < π, ([Khov63], p 109). The even part of (2.3.8) is

z1/z = 1+2(z − 1)

z2 + 1 −(12z2 − 1)(z − 1)2

3z(z + 1) −(22z2 − 1)(z − 1)2

5z(z + 1) −(32z2 − 1)(z − 1)2

7z(z + 1) −· · ·

(2.3.9)

for | arg z| < π, ([Khov63], p 110).

(1 + az

1 + bz

= 1 +2α(a − b)z

2 + (a + b − α(a − b))z−(a − b)2(12 − α2)z2

3(2 + (a + b)z) −(a − b)2(22 − α2)z2

5(2 + (a + b)z) −(a − b)2(32 − α2)z2

7(2 + (a + b)z) −· · ·

(2.3.10)

for α ∈ C and 2+(a+b)z(a−b)z

∈ C \ [−1, 1], ([Perr57], p 264). (Note that 2+(a+b)z(a−b)z

= x+yx−y

for

x := 1 + az, y := 1 + bz.)

A.2.4 The natural logarithm

Ln(1 + z) = z 2F1(1, 1; 2;−z) = z

∫ 1

0

dt

1 + zt

=z

1+

1z

2 +

1z

3 +

2z

2 +

2z

5 +

3z

2 +

3z

7 +

4z

2 +

4z

9 +· · ·

=z

1+

12z

2 +

12z

3 +

22z

4 +

22z

5 +

32z

6 +

32z

7 +

42z

8 +

42z

9 +· · ·

(2.4.1)

for | arg(1 + z)| < π, ([JoTh80], p 203). (See also (3.1.6).)

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A.2.4 The natural logarithm 271

Ln(1 + z) =z

1 + z−z

1+

1

1 + z−z

1+

1/2

1 + z−z

1+

1

1 + z−z

1+

2/3

1 + z−· · ·(2.4.2)

for | arg(1 + z)| < π, ([JoTh80], p 319). Here the continued fraction has the formK(an(z)/bn(z)) where all a2n(z) = −z, a4n−1(z) = 1 and a4n+1(z) = n/(n + 1). (2.4.1) isthe even part of (2.4.2). The odd part of (2.4.2) can be written

Ln(1 + z) =z

1 + z

{1 +

z

2+

2z

3 +

z

2+

3z

5 +

2z

2 +

4z

7 +

3z

2 +

5z

9 +

4z

2 +· · ·

}=

z

1 + z

{1 +

z

2+

1 · 2z

3 +

1 · 2z

4 +

2 · 3z

5 +

2 · 3z

6 +

3 · 4z

7 +

3 · 4z

8 +· · ·

} (2.4.3)

for | arg(1 + z)| < π. The even part of (2.4.1) is

Ln(1 + z) =2z

1(2 + z)−12z2

3(2 + z)−22z2

5(2 + z)−32z2

7(2 + z)−· · ·(2.4.4)

for | arg(1 + z)| < π, ([Khov63], p 111).

Ln(1 + z) =z

1+

12z

2 − z +

22z

3 − 2z +

32z

4 − 3z +

42z

5 − 4z +

52z

6 − 5z +· · ·, (2.4.5)

([Khov63], p 111). Dc := {z ∈ C; |z| �= 1}, Df := {z ∈ C; |z| < 1}.The connection

Ln(1 + z) = −Ln

(1

1 + z

)= −Ln

(1 − z

1 + z

)(2.4.6)

can be applied to (2.4.1)–(2.4.5) to get 5 new continued fraction expansions. For instance,from (2.4.1) we get

Ln(1 + z) =z

1 + z−1z

2 −1z

3(1 + z)−2z

2 −2z

5(1 + z)−3z

2 −3z

7(1 + z)−· · ·(2.4.7)

for | arg(1 + z)| < π, ([Khov63], p 110), and from (2.4.5)

Ln(1 + z) =z

1 + z−12z(1 + z)

2 + 3z −22z(1 + z)

3 + 5z −32z(1 + z)

4 + 7z −· · ·, (2.4.8)

([Khov63], p 111). Dc := {z ∈ C; Re(z) �= − 12}, Df := {z ∈ C; Re(z) > − 1

2}.

Ln(1 + z

1 − z

)= 2z 2F1(

12, 1; 3

2; z2) = Ln

(1 +

2z

1 − z

)= z

∫ 1

−1

dt

1 + tz

=2z

1 −12z2

3 −22z2

5 −32z2

7 −42z2

9 −· · ·

(2.4.9)

for | arg(1 − z2)| < π, ([JoTh80], p 203). (See also (3.1.6).) From this we also get

Ln( z + 1

z − 1

)= Ln

(1 + 1/z

1 − 1/z

)=

2

z−12

3z−22

5z−32

7z−42

9z−· · ·(2.4.10)

for z ∈ C \ [−1, 1], ([Perr57], p 155). Of course, also other continued fraction expansionsfor Ln(1 + z) can be used to derive expressions for Ln((1 + z)/(1 − z)).

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272 Appendix A: Some continued fraction expansions

A.2.5 Trigonometric and hyperbolic functions

tan z :=sin z

cos z= z

0F1(3/2;−z2/4)

0F1(1/2;−z2/4)=

z

1−z2

3 −z2

5 −z2

7 −z2

9 −· · ·; z ∈ C (2.5.1)

([JoTh80], p 211), (See also (3.1.1).) The odd part of (2.5.1) is

tan z = z +5z3

1 · 3 · 5 − 6z2−1 · 9z4

5 · 7 · 9 − 14z2−5 · 13z4

9 · 11 · 13 − 22z2−9 · 17z4

13 · 15 · 17 − 30z2−· · ·; z ∈ C,

(2.5.2)

Another type of expansion is

tanzπ

4=

z

1+

12 − z2

2 +

32 − z2

2 +

52 − z2

2 +

72 − z2

2 +· · ·; z ∈ C, (2.5.3)

([Perr57], p 35). From these expansions one also gets continued fractions for cot z =1/ tan z, tanh z = −i tan(iz) and coth z = i/ tan(iz).

Quite another type of expansion for tan z follows from the identity

tan αz = −i(1 + i tan z)α − (1 − i tan z)α

(1 + i tan z)α + (1 − i tan z)α= −i

y − 1

y + 1, (2.5.4)

where y := ((1 + i tan z)/(1 − i tan z))α can be expanded according to (2.3.7). Combinedwith (1.2.2) we get

tan αz =α tan z

1 −(α2 − 12) tan2 z

3 −(α2 − 22) tan2 z

5 −(α2 − 32) tan2 z

7 −· · ·, (2.5.5)

([Khov63], p 108). Dc := {(α, z) ∈ C2; Re(cos z) �= 0}, Df := {(α, z) ∈ C

2; |Re(z)| <π/2}.

coth 1z

= 1 +1

3z +

1

5z +

1

7z +

1

9z +· · ·; z ∈ C, (2.5.6)

([

πz

2coth

πz

2= 1 +

z2

1 +

12(z2 + 12)

3 +

22(z2 + 22)

5 +

32(z2 + 32)

7 +· · ·(2.5.7)

for all z ∈ C, ([ABJL92], entry 44).

a tanh(πb/2) − b tanh(πa/2)

a tanh(πa/2) − b tanh(πb/2)=

ab

1 +

(a2 + 12)(b2 + 12)

3 +

(a2 + 22)(b2 + 22)

5 +· · ·(2.5.8)

for all a, b ∈ C, ([ABJL92], entry 47).

sinh(πz) − sin(πz)

sinh(πz) + sin(πz)=

2z2

1 +

4z4 + 14

3 +

4z4 + 24

5 +

4z4 + 34

7 +· · ·(2.5.9)

for all z ∈ C, ([ABJL92], entry 49).

Khru06b]).

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A.2.6 Inverse trigonometric and hyperbolic functions 273

A.2.6 Inverse trigonometric and hyperbolic functions

Arctan z = z 2F1(12, 1; 3

2;−z2) = − i

2Ln(1 + iz

1 − iz

)=

z

1+

12z2

3 +

22z2

5 +

32z2

7 +

42z2

9 +· · ·

(2.6.1)

for | arg(1 + z2)| < π; i.e. z ∈ C \ i((−∞,−1] ∪ [1,∞)), ([JoTh80], p 202). (See also(3.1.6).) This continued fraction can also be written

Arctan z = z − z3

3 +

32z2

5 +

22z2

7 +

52z2

9 +

42z2

11 +

72z2

13 +

62z2

15 +· · ·(2.6.2)

for z ∈ C \ i((−∞,−1] ∪ [1,∞)), ([Khov63], p 117).

Arctan z =z

1(1 + z2)−1 · 2z2

3 −1 · 2z2

5(1 + z2)−3 · 4z2

7 −3 · 4z2

9(1 + z2)−5 · 6z2

11 −· · ·(2.6.3)

for z ∈ C \ i((−∞,−1] ∪ [1,∞)), ([Khov63], p 121). (This follows from (2.6.6) with zreplaced by z(1+ z2)−1/2.) Since Artanh z = iArctan(−iz), we also get continued fractionexpansions for Artanh z from (2.6.1)– (2.6.3).

Also expressions for

Arcsin z = Arctan( z√

1 − z2

), Arccos z = Arctan

(√1 − z2

z

)can be obtained. For instance, from (2.6.1) we get

Arcsin z√1 − z2

=z

1(1 − z2)+

12z2

3 +

22z2

5(1 − z2)+

32z2

7 +

42z2

9(1 − z2)+· · ·(2.6.4)

for | arg(1 − z2)| < π, ([Khov63], p 118) and

Arccos z√1 − z2

=1

z +

12(1 − z2)

3z +

22(1 − z2)

5z +

32(1 − z2)

7z +· · ·, (2.6.5)

([Khov63], p 119). Here Dc := {z ∈ C; Re z �= 0} and Df := {z ∈ C; Re z > 0}. We alsohave

Arcsin z√1 − z2

= z2F1(

12, 1

2; 3

2; z2)

2F1(12,− 1

2; 1

2; z2)

=z

1−1 · 2z2

3 −1 · 2z2

5 −3 · 4z2

7 −3 · 4z2

9 −5 · 6z2

11 −5 · 6z2

13 −· · ·

(2.6.6)

for | arg(1 − z2)| < π, ([JoTh80], p 203), and thus, since Arccos z = Arcsin√

1 − z2 for0 ≤ z ≤ π

2

Arccos z√1 − z2

=z

1−1 · 2(1 − z2)

3 −1 · 2(1 − z2)

5 −3 · 4(1 − z2)

7 −3 · 4(1 − z2)

9 −· · ·, (2.6.7)

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274 Appendix A: Some continued fraction expansions

([Khov63], p 121) where Dc := {z ∈ C; Re(z) �= 0}, Df := {z ∈ C; Re(z) > 0}.Similar expressions for inverse hyperbolic functions can be derived, since Arsinh z =iArcsin(−iz) and (Arcosh z)/

√z2 − 1 = (Arccos z)/

√1 − z2 for 0 ≤ z ≤ π

2.

A neat formula can be obtained from (3.2.6) in the following way

(iz + 1

iz − 1

)iα

= exp(iαLn

( iz + 1

iz − 1

))= exp(2αArctan(1/z))

= 1 +2α

z − α+

α2 + 12

3z +

α2 + 22

5z +

α2 + 32

7z +· · ·

(2.6.8)

for | arg(1 + 1/z2)| < π, i.e. z �∈ i[−1, 1], ([Wall57], p 346).

Arsinh z

(1 + z2)1/2= z 2F1(1, 1; 3

2;−z2) =

z

1+

2z2

1 +

2(1 + z2)

1 +

4z2

1 +

4(1 + z2)

1 +· · ·(2.6.9)

for z ∈ R, ([ABJL92], entry 37).

Arctan z = z 2F1(12, 1; 3

2;−z2) =

z

1+

1z2

1 +

2(1 + z2)

1 +

3z2

1 +

4(1 + z2)

1 +· · ·(2.6.10)

for z ∈ R, ([ABJL92], entry 38).

A.2.7 Continued fractions with simple values

The continued fractions in this section all have easy to find tail sequences ([Lore08c]). Thefirst one is taken from ([Perr57], p 279). It converges for all pairs (a, z) ∈ C

2, but theequality

0 = −a − z +z

1 − a − z +

2z

2 − a − z +

3z

3 − a − z +· · ·(2.7.1)

given by Perron is only claimed for z �= 0 and a ∈ N ∪ {0}. In our next continued fractionthe constant sequence {1} is always a tail sequence, so

1 =z + 1

z +

z + 2

z + 1+

z + 3

z + 2+

z + 4

z + 3+· · ·; z ∈ C \ (−N), (2.7.2)

([Bern89], p 112). (See also (3.1.5) with z := 1, a := z + 1 and c := z + 1.) The sequenceof tail values for the next continued fraction is also known ([Lore08c]); it is in fact t0 := 1,tn+1 := z + na for n ≥ 0. Hence

1 =z + a

a +

(z + a)2 − a2

a +

(z + 2a)2 − a2

a +

(z + 3a)2 − a2

a +· · ·(2.7.3)

for a �= 0 and z/a �∈ (−N), ([Bern89], p 118). (See also (3.1.8) with z := −1, a := 0,b := (z/a) − 2 and c := z/a.)

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A.3.1 Hypergeometric functions 275

a =ab

a + b + d−(a + d)(b + d)

a + b + 3d −(a + 2d)(b + 2d)

a + b + 5d −· · ·, (2.7.4)

([Bern89], p 119). Here Dc := {(a, b, d) ∈ C3; Re((a−b)/d) �= 0 or a = b}. Both {−a−nd}

and {−b − nd} are tail sequences, so the value of the continued fraction is either a or b.The value is a in Df := {(a, b, d) ∈ C

3; Re((a− b)/d) < 0 or a = b}. (See also (3.1.6) withz = 1, a replaced by (a + d)/2d, b replaced by a/2d and c replaced by (a + b + d)/2d.) Forb = a replaced by a + 1 and d := 1, (3.7.4) can be transformed into

a = 2a + 1 − (a + 1)2

2a + 3 −(a + 2)2

2a + 5 −(a + 3)2

2a + 7 −· · ·; a ∈ C, (2.7.5)

([Perr57], p 105).

az =abz

b − (a + 1)z +

(a + 1)(b + 1)z

b + 1 − (a + 2)z +

(a + 2)(b + 2)z

b + 2 − (a + 3)z +· · ·, (2.7.6)

([Perr57], p 290). Dc := {(a, b, z) ∈ C3; |z| �= 1 and b �= 0,−1,−2, . . . }, Df := {(a, b, z) ∈

Dc; |z| < 1}. (See also (3.1.8) with z replaced by −z, a replaced by b − a, and b = creplaced by b − 1.)

z + a + 1

z + 1=

z + a

z − 1+

z + 2a

z + a − 1+

z + 3a

z + 2a − 1+· · ·, (2.7.7)

([Bern89], p 115). Dc := {(a, z) ∈ C2; a �= 0 and z/a �= 0,−1,−2, . . .} ∪ {(a, z) ∈ C

2; a =0 and |z| �= 1}, Df := {(a, z) ∈ Dc; if a = 0, then |z| > 1}. (See also (3.1.5) with zreplaced by 1/a, a replaced by z/a +1 and c replaced by z/a.) If we instead let z = 1 andreplace a by z − 1, c by z − 3 in (3.1.5) we get

z2 + z + 1

z2 − z + 1=

z

z − 3+

z + 1

z − 2+

z + 2

z − 1+

z + 3

z +

z + 4

z + 1+· · ·, (2.7.8)

([Bern89], p 118). Dc := C, Df := {z ∈ C; z �= 0,−1,−2, . . .}. For z := 1, a := z − 1 andc := z − 4 in (3.1.5) we get

z3 + 2z + 1

(z − 1)3 + 2(z − 1) + 1=

z

z − 4+

z + 1

z − 3+

z + 2

z − 2+

z + 3

z − 1+

z + 4

z +· · ·, (2.7.9)

([Bern89], p 118). Dc := C, Df := {z ∈ C; z �= 0,−1,−2, . . .}.

A.3 Hypergeometric functions

A.3.1 General expressions

c0F1(c; z)

0F1(c + 1; z)= c +

z

c + 1+

z

c + 2+

z

c + 3+· · ·; (c, z) ∈ C

2, (3.1.1)

([JoTh80], p 210).

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276 Appendix A: Some continued fraction expansions

2F0(a, b; z)

2F0(a, b + 1; z)= 1 − az

1 −(b + 1)z

1 −(a + 1)z

1 −(b + 2)z

1 −(a + 2)z

1 −· · ·(3.1.2)

for (a, b, z) ∈ C3 with | arg(−z)| < π, ([JoTh80], p 213). The even part of this one is

2F0(a, b; z)

2F0(a, b + 1; z)= 1 +

az

(b + 1)z − 1−(a + 1)(b + 1)z2

(a + b + 3)z − 1−(a + 2)(b + 2)z2

(a + b + 5)z − 1−· · ·(3.1.3)

for (a, b, z) ∈ C3 with | arg(−z)| < π.

c · 1F1(a; c; z)

1F1(a + 1; c + 1; z)= c − (c − a)z

c + 1 +

(a + 1)z

c + 2

−(c − a + 1)z

c + 3 +

(a + 2)z

c + 4 −(c − a + 2)z

c + 5 +· · ·; (a, c, z) ∈ C

3,

(3.1.4)

([JoTh80], p 206).

1F1(a + 1; c + 1; z)

1F1(a; c; z)=

c

c − z +

(a + 1)z

c + 1 − z +

(a + 2)z

c + 2 − z +

(a + 3)z

c + 3 − z +· · ·(3.1.5)

for (a, c, z) ∈ C3, ([JoTh80], p 278).

c · 2F1(a, b; c; z)

2F1(a, b + 1; c + 1; z)= c − a(c − b)z

c + 1 −(b + 1)(c − a + 1)z

c + 2

−(a + 1)(c − b + 1)z

c + 3 −(b + 2)(c − a + 2)z

c + 4 −(a + 2)(c − b + 2)z

c + 5 − · · ·

(3.1.6)

for (a, b, c, z) ∈ C4 with | arg(1− z)| < π, ([JoTh80], p 199). The Norlund fraction has the

form

c · 2F1(a, b; c; z)

2F1(a + 1, b + 1; c + 1; z)= c − (a + b + 1)z +

(a + 1)(b + 1)(z − z2)

c + 1 − (a + b + 3)z +

(a + 2)(b + 2)(z − z2)

c + 2 − (a + b + 5)z +

(a + 3)(b + 3)(z − z2)

c + 3 − (a + b + 7)z +· · ·

(3.1.7)

([LoWa92], p 304). Dc := {(a, b, c, z) ∈ C4; Re(z) �= 1

2}, Df := {(a, b, c, z) ∈ C

4; Re(z) <12}. The Euler fraction has the form

c · 2F1(a, b; c; z)

2F1(a, b + 1; c + 1; z)= c + (b − a + 1)z − (c − a + 1)(b + 1)z

c + 1 + (b − a + 2)z−(c − a + 2)(b + 2)z

c + 2 + (b − a + 3)z−(c − a + 3)(b + 3)z

c + 3 + (b − a + 4)z−· · ·,

(3.1.8)

([LoWa92], p 308). Dc := {(a, b, c, z) ∈ C4; |z| �= 1}, Dc := {(a, b, c, z) ∈ C

4; |z| <1, (c − a) �= −1,−2,−3, . . . }.By setting b := 0 in (3.1.2), (3.1.5), (3.1.6) or (3.1.7) and using (1.2.1) we get continuedfraction expansions for 2F0(a, 1; z) and 2F1(a, 1; c + 1; z). Similarly, a := 0 in (3.1.3) or(3.1.4) gives continued fraction expansions for 1F1(1; c + 1; z). A different expansion is

Page 290: Lisa Lorentzen, Haakon Waadeland Continued Fractions

A.3.3 Special examples with 2F0 277

2F1(a, 1; c + 1; z) =Γ(1 − a)Γ(c + 1)

Γ(c − a + 1)

(1 − z)c−a

(−z)c−

c

1 − c + (a − 1)z +

1(1 − c)(z − 1)

3 − c + (a − 2)z +

2(2 − c)(z − 1)

5 − c + (a − 3)z +· · ·,

(3.1.9)

([Bern89], p 164). Dc := {(a, c, z) ∈ C3; |z − 1| �= 1}, Df := {(a, c, z) ∈ C

3; |z − 1| <1 and c �= 0, 1, 2, . . . }. From this follows after some computation, ([Bern89], p 165) that

1F1(1; c + 1; z) =ezΓ(c + 1)

zc− c

z +

1 − c

1 +

1

z +

2 − c

1 +

2

z +

3 − c

1 +· · ·

=ezΓ(c + 1)

zc− c

z + 1 − c−1(1 − c)

z + 3 − c−2(2 − c)

z + 5 − c−3(3 − c)

z + 7 − c−· · ·

(3.1.10)

for | arg z| < π, ([Bern89], p 165) (the second continued fraction is the even part of thefirst one).

A.3.2 Special examples with 0F1

The Bessel function of the first kind and order ν is

Jν(z) :=( z

2

)ν∞∑

k=0

(−1)k(z/2)2k

k! Γ(ν + k + 1)=

(z/2)ν

Γ(ν + 1)0F1(ν + 1;− z2

4), (3.2.1)

so that by (3.1.1)

Jν+1(z)

Jν(z)=

z

2(ν + 1)· 0F1(ν + 2;−z2/4)

0F1(ν + 1;−z2/4)

=z

2(ν + 1)−z2

2(ν + 2)−z2

2(ν + 3)−z2

2(ν + 4)−· · ·

(3.2.2)

for z ∈ C, ν �= −1,−2,−3, . . . , ([JoTh80], p 211).

A.3.3 Special examples with 2F0

The connection (see for instance ([Wall48], p 352, p 355))

2F0(a, b;−z) ∼ 1

Γ(a)

∫ ∞

0

e−tta−1

(1 + tz)bdt =

1

Γ(b)

∫ ∞

0

e−ttb−1

(1 + tz)adt (3.3.1)

implies that (3.1.2) – (3.1.3) lead to continued fraction expansions for ratios of such inte-grals. In particular, the incomplete gamma function Γ(a, z) satisfies

Γ(a, z) :=

∫ ∞

z

e−tta−1dt ∼ e−zza−12F0(1 − a, 1;−1/z) , (3.3.2)

([EMOT53], p 266). Hence, by (3.1.2)

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278 Appendix A: Some continued fraction expansions

Γ(a, z) =e−zza

z +

1 − a

1 +

1

z +

2 − a

1 +

2

z +

3 − a

1 +

3

z +· · ·

=e−zza

1 + z − a−1(1 − a)

3 + z − a−2(2 − a)

5 + z − a−3(3 − a)

7 + z − a−· · ·

(3.3.3)

for (a, z) ∈ C2 with | arg z| < π, ([AbSt65], p 260, p 263), ([Khov63], p 144), where the

second continued fraction is the even part of the first one.

This (and the expressions to come) are to be interpreted in the following way: The integralin (3.3.2) is taken for real z. Then Γ(a, z) is the analytic continuation of this function tothe given domain.

The complementary error function erfc z satisfies

erfc z :=2√π

∫ ∞

z

e−t2dt =1√π

Γ( 12, z2) ∼ 1√

πe−z2

z−12F0(

12, 1;−1/z2) (3.3.4)

([EMOT53], p 266), which means that by (3.1.2)

erfcz =2√π

e−z2{ 1

2z +

2

2z +

4

2z +

6

2z +

8

2z +· · ·

}=

2√π

e−z2{ z

1 + 2z2−1 · 2

5 + 2z2−3 · 4

9 + 2z2 −5 · 6

13 + 2z2−7 · 8

17 + 2z2−· · ·

},

(3.3.5)

([JoTh80], p 219). (There is a slightly different notation in ([JoTh80]).) Dc := {z ∈C; Re z �= 0}, Df := {z ∈ C; Re z > 0}. Again the second continued fraction is the evenpart of the first one. If we integrate this complementary error function we get similarexpressions:

i−1erfc z =2√π

e−z2, i0erfc z = erfc z , inerfc z =

∫ ∞

z

in−1erfc t dt (3.3.6)

for n = 1, 2, 3, . . . . Therefore

in−1erfc z

inerfc z= 2z

2F0(n+1

2, n

2;−1/z2)

2F0(n+1

2, n

2+ 1;−1/z2)

= 2z +2(n + 1)

2z +

2(n + 2)

2z +

2(n + 3)

2z +· · ·

(3.3.7)

([JoTh80], p 219). Dc := {(n, z) ∈ C2; Re z �= 0}, Df := {(n, z) ∈ C

2; Re z > 0}.For the exponential integral

−Ei(−z) :=

∫ ∞

z

e−t

tdt ∼ e−z

z2F0(1, 1;− 1

z) , (3.3.8)

([EMOT53], p 267), we get by (3.1.2) and its even part

Ei(−z) = −e−z

z +

1

1+

1

z +

2

1+

2

z +

3

1+

3

z +

4

1+· · ·

= − e−z

1 + z−12

3 + z−22

5 + z−32

7 + z−42

9 + z−· · ·

(3.3.9)

for | arg z| < π, ([Khov63], p 145).

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A.3.2 Special examples with 1F1 279

Similarly, for the logarithmic integral

li z :=

∫ z

0

dt

Ln t= Ei(Ln z) =

z

Ln z−1

1−1

Ln z−2

1−2

Ln z−· · ·

= − z

1 − Ln z−12

3 − Ln z−22

5 − Ln z−32

7 − Ln z−42

9 − Ln z−· · ·

(3.3.10)

for | arg(−Ln z)| < π.

The plasma dispersion function is

P (z) :=1√π

∫ ∞

−∞

e−t2

t − zdt = i

√π e−z2

erfc(−iz)

=2z

1 − 2z2 −1 · 2

5 − 2z2 −3 · 4

9 − 2z2−5 · 6

13 − 2z2 −7 · 8

17 − 2z2 −· · ·,

(3.3.11)

([JoTh80], p 219). Dc := {z ∈ C; Im(z) �= 0}, Df := {z ∈ C; Im(z) > 0}.

∫∞0

tae−bt−t2/2dt∫∞0

ta−1e−bt−t2/2dt=

a

b· 2F0

(a2, a+1

2;− 1

b2

)2F0

(a2, a−1

2;− 1

b2

) =a

b +

a + 1

b +

a + 2

b +

a + 3

b +· · ·, (3.3.12)

([Perr57], p 297). Dc := {(a, b) ∈ C2; Re b �= 0}, Df := {(a, b) ∈ C

2; Re b > 0}.

A.3.4 Special examples with 1F1

From ([EMOT53], p 255) it follows that

1F1(a; c; z) =Γ(c)

Γ(a)Γ(c − a)

∫ 1

0

etzta−1(1 − t)c−a−1dt (3.4.1)

for Re(c) > 0, Re(a) > 0. Hence (3.1.4), (3.1.5) and (3.1.10) lead to continued fractionexpansions of ratios of such integrals.

The error function is given by

erf (z) :=2√π

∫ z

0

e−t2dt =2√π

z1F1(12; 3

2;−z2), ([EMOT53], p 266)

=2√π

ze−z2

1F1(1; 32; z2), ([JoTh80], p 282).

(3.4.2)

Hence,

erf (z) =2 e−z2

√π

z

1−2z2

3 +

4z2

5 −6z2

7 +

8z2

9 −· · ·

=2 e−z2

√π

z

1 − 2z2 +

4z2

3 − 2z2 +

8z2

5 − 2z2 +

12z2

7 − 2z2 +· · ·

(3.4.3)

for z ∈ C, ([JoTh80], p 208 and 282).

The error function is related to Dawson’s integral∫ z

0

et2dt =i√

π

2erf(−iz), ([JoTh80], p 208) (3.4.4)

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280 Appendix A: Some continued fraction expansions

and to the Fresnel integrals

C(z) :=

∫ z

0

cos(π

2t2)

dt, S(z) :=

∫ z

0

sin(π

2t2)

dt (3.4.5)

by

C(z) + iS(z) =

∫ z

0

eit2π/2 dt =

√−2

∫ √−iπ/2 · z

0

e−u2du

=1 + i

2erf

(√π

2(1 − i)z

).

(3.4.6)

The incomplete gamma function

γ(a, z) :=

∫ z

0

e−tta−1dt =za

ae−z

1F1(1; a + 1; z)

=zae−z

a −az

a + 1+

1z

a + 2−(a + 1)z

a + 3 +

2z

a + 4−(a + 2)z

a + 5 +· · ·

=zae−z

a −az

1 + a + z−(1 + a)z

2 + a + z−(2 + a)z

3 + a + z−(3 + a)z

4 + a + z−· · ·

(3.4.7)

for all (a, z) ∈ C2, ([JoTh80], p 209), ([Khov63], p 149–150).

The Coulomb wave function

FL(η, ρ) = ρL+1e−iρCL(η)1F1(L + 1 − iη; 2L + 2; 2iρ) (3.4.8)

where CL(η) = 2L exp(−πη/2)|Γ(L + 1 + iη)|/(2L + 1)! for η ∈ R, ρ > 0 and L ∈ N ∪ {0}satisfies

FL(η, ρ)

FL−1(η, ρ)=

(L + 1)(L2 + η2)1/2

(2L + 1)(η + L(L + 1)/ρ)−L(L + 2)((L + 1)2 + η2)

(2L + 3)(η + (L + 1)(L + 2)/ρ)−(L + 1)(L + 3)((L + 2)2 + η2)

(2L + 5)(η + (L + 2)(L + 3)/ρ)−· · ·,

(3.4.9)

([JoTh80], p 216). Dc := {(L, η, ρ) ∈ C3; ρ �= 0}, Df := {(L, η, ρ) ∈ (N∪{0})×C

2; ρ �= 0}.It is well known that

∞∑k=0

(−z)k

k! (a + k)= e−z

∞∑k=0

zk

(a)k+1=

e−z

a1F1(1; a + 1; z) , (3.4.10)

([Bern89], p 166). This means for instance that

∞∑k=0

(−z)k

k! (a + k)=

Γ(a)

za− e−z

z + 1 − a−1(1 − a)

z + 3 − a−2(2 − a)

z + 5 − a−· · ·(3.4.11)

for (a, z) ∈ C2 with | arg z| < π, and

∞∑k=0

zk

1 · 3 · · · (2k + 1)=

√π

2zez/2 − 1

z + 1−1 · 2z + 5−

3 · 4z + 9−

5 · 6z + 13−· · ·

(3.4.12)

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A.3.5 Special examples with 2F1 281

for | arg z| < π, ([Bern89], p 166), and

ze−z2∞∑

k=0

(2z2)k

1 · 3 · · · (2k + 1)=

∫ z

0

e−t2dt

=

√π

2− e−z2

2z2 + 1−1 · 2

2z2 + 5−3 · 4

2z2 + 9−5 · 6

2z2 + 13−· · ·,

(3.4.13)

([Bern89], p 166). Dc := {z ∈ C; Re z �= 0}, Df := {z ∈ C; Re z > 0}.

A.3.5 Special examples with 2F1

∫ z

0

tpdt

1 + tq=

zp+1

q2F1

(p + 1

q, 1; 1 +

p + 1

q;−zq

)=

zp+1

0q + p + 1+

(0q + p + 1)2zq

1q + p + 1 +

(1q)2zq

2q + p + 1+

(1q + p + 1)2zq

3q + p + 1 +

(2q)2zq

4q + p + 1+· · ·

(3.5.1)

for p, q > 0 with | arg(1 + zq)| < π, ([Khov63], p 126).

Incomplete beta functions are given by

Bx(p, q) :=

∫ x

0

tp−1(1 − t)q−1dt =xp

p2F1(p, 1 − q; p + 1; x) (3.5.2)

for p > 0, q > 0 and 0 ≤ x ≤ 1, ([EMOT53], p 87). Hence, by (3.1.6) and (3.1.8)

Bx(p + 1, q)

Bx(p, q)=

px

p + 12F1(p + 1, 1 − q; p + 2; x)

2F1(p, 1 − q; p + 1; x)

=px

p + 1−1(1 − q)x

p + 2 −(p + 1)(p + q + 1)x

p + 3 −2(2 − q)x

p + 4 −(p + 2)(p + q + 2)x

p + 5 −· · ·; | arg(1 − x)| < π

(3.5.3)

for p > 0, q > 0, ([JoTh80], p 217), and

Bx(p + 1, q)

Bx(p, q)=

px

p + 1 + (p + q)x−(p + q + 1)(p + 1)x

p + 2 + (p + q + 1)x

−(p + q + 2)(p + 2)x

p + 3 + (p + q + 2)x−· · ·,

(3.5.4)

([JoTh80], p 217). Dc := {p > 0, q > 0, x ∈ C; |x| �= 1}, Df := {(p, q, x) ∈ Dc; |x| < 1}.Legendre functions of the first kind of degree α ∈ R and order m ∈ N ∪ {0} are givenby

P mα (z) :=

1

π

Γ(α + m + 1)

Γ(α + 1)

∫ π

0

(z + (z2 − 1)1/2 cos t)α cos mt dt

=1

Γ(1 − m)

(z + 1

z − 1

)m/2

2F1

(−α, α + 1; 1 − m;

1 − z

2

).

(3.5.5)

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282 Appendix A: Some continued fraction expansions

From ([Gaut67], formula (6.1) on p 55) it follows that

P mα (z)

P m−1α (z)

=(m + α)(m − α − 1)

− 2mz

(z2 − 1)1/2

−(m + 1 + α)(m − α)

− 2(m + 1)z

(z2 − 1)1/2

(m + 2 + α)(m + 1 − α)

− 2(m + 2)z

(z2 − 1)1/2

− · · ·

∼ − (m + α)(m − α − 1)√

z2 − 1

2mz −(m + 1 + α)(m − α)(z2 − 1)

2(m + 1)z −(m + 2 + α)(m + 1 − α)(z2 − 1)

2(m + 2)z −(m + 3 + α)(m + 2 − α)(z2 − 1)

2(m + 3)z − · · ·

(3.5.6)

Dc := {(α, m, z) ∈ C3; Re(z) �= 0}, Df := {(α, m, z) ∈ R × (N ∪ {0}) × C; Re(z) > 0}.

Legendre functions of the second kind of degree α ∈ R and order m ∈ N∪{0} are givenby

Qmα (z) := (−1)m Γ(α + 1)

Γ(α − m + 1)

∫ ∞

0

cosh mt

(z + (z2 − 1)1/2 cosh t)α+1dt

=

√π eimπ

(2z)α+1

(1 − 1

z2

)m/2 Γ(α + m + 1)

Γ(α + m + 32)· 2F1

(α+m+2

2, α+m+1

2; α + 3

2; 1/z2) .

(3.5.7)

In ([JoTh80], p 205) it is proved that

Qmα (z)

Qmα+1(z)

=1

α + m + 1

{(2α + 3)z − (α + m + 2)2

(2α + 5)z −(α + m + 3)2

(2α + 7)z −(α + m + 4)2

(2α + 9)z −(α + m + 5)2

(2α + 11)z −(α + m + 6)2

(2α + 13)z −· · ·

}.

(3.5.8)

Dc := {(α, m, z) ∈ C3; z �∈ [−1, 1]}, Df := {(α, m, z) ∈ R × (N ∪ {0}) × C; z �∈ [−1, 1]}.

A.3.6 Some integrals

Hypergeometric functions can be written in terms of integrals. This has already been usedto some extent in the preceding subsections, and we refer to ([AbSt65]) and ([EMOT53])for further details. Here we shall just list some simple examples without bringing in thehypergeometric functions themselves.

∫ 1

0

xse1−xdx =1

s+

1

s + 1+

1

s + 1+

1

s + 1+· · ·=

∞∑n=1

1

(s + 1)n; s ∈ C, (3.6.1)

([Khru06b]).

∫ 1

0

xs

1 + x2dx =

1

s+

12

s +

22

s +

32

s +· · ·=

∞∑k=0

2 · (−1)k

s + 2k + 1=

∞∑k=1

4

(s + 2k)2 − 1, (3.6.2)

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A.3.6 Some simple integrals 283

([Khru06b]). Dc := {s ∈ C; Re s �= 0}, Df := {s ∈ C; Re s > 0}.

∫ ∞

0

e−tdt

t + z=

1

z + 1−12

z + 3−22

z + 5−32

z + 7−· · ·; | arg z| < π, (3.6.3)

([BoSh89], p 20).

∫ ∞

0

e−t/z

(1 + t)ndt

=z

1+

nz

1 +

1z

1 +

(n + 1)z

1 +

2z

1 +

(n + 2)z

1 +

3z

1 +· · ·

=z

1 + nz−nz2

1 + (n + 2)z−2(n + 1)z2

1 + (n + 4)z−3(n + 2)z2

1 + (n + 6)z−· · ·

(3.6.4)

for (n, z) ∈ R × C with | arg z| < π, ([BoSh89], p 157). The second continued fraction isthe even part of the first one.

∫ ∞

0

e−tz

cosh2 tdt =

1

z +

1 · 2z +

2 · 3z +

3 · 4z +· · ·

= 2z

∞∑k=0

(−1)k

(z + 2k)(z + 2k + 2), (3.6.5)

([Khru06b]). Dc := {z ∈ C; Re z �= 0}, Df := {z ∈ C; Re z > 0}.For Jacobi’s elliptic functions sn t, cn t and dn t with modulus k we have

∫ ∞

0

e−tzsn t dt =1

12(1 + k2) + z2−1 · 22 · 3k2

32(1 + k2) + z2 −3 · 42 · 5k2

52(1 + k2) + z2−· · ·, (3.6.6)

([Wall48], p 374). Dc := {(k, z) ∈ C2; |k| �= 1}, Df := {(k, z) ∈ C

2; |k| < 1},

∫ ∞

0

e−tzsn2t dt =2

22(1 + k2) + z2−2 · 32 · 4k2

42(1 + k2) + z2 −4 · 52 · 6k2

62(1 + k2) + z2−· · ·, (3.6.7)

([Wall48], p 375). Dc := {(k, z) ∈ C2; |k| �= 1}, Df := {(k, z) ∈ C

2; |k| < 1},

∫ ∞

0

e−tzcn t dt =1

z +

12

z +

22k2

z +

32

z +

42k2

z +

52

z +· · ·, (3.6.8)

([Perr57], p 220). Dc := {(k, z) ∈ R × C; Re z �= 0}, Df := {(k, z) ∈ R × C; Re z > 0},

∫ ∞

0

e−tzdn t dt =1

z +

12k2

z +

22

z +

32k2

z +

42

z +

52k2

z · · ·, (3.6.9)

([Wall48], p 374). Dc := {(k, z) ∈ R × C; Re z �= 0}, Df := {(k, z) ∈ R × C; Re z > 0},and

∫ ∞

0

sn t cn t

dn te−tzdt =

1

2 · 12(2 − k2) + z2 −1 · 22 · 3k4

2 · 32(2 − k2) + z2 −3 · 42 · 5k4

2 · 52(2 − k2) + z2 −· · ·

(3.6.10)

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284 Appendix A: Some continued fraction expansions

for (k, z) ∈ C2 with |1 − k2| < 1, ([Wall48], p 375).

∫ ∞

0

(1 − c

et(1−c) − cb

)a

e−tzdt =ra

z +

ar

1 +

rcb

z +

(a + 1)r

1 +

2rcb

z +

(a + 2)r

1 +· · ·(3.6.11)

where r := (1 − c)/(1 − cb), for (a, b, c, z) ∈ C4 with a > 0, cb > 0 and | arg(r/z)| < π,

([Wall48], p 359).

∫ ∞

0

te−tz

sinh tdt =

1

z +

14

3z +

24

5z +

34

7z +· · ·, (3.6.12)

([Wall48], p 371). Dc := {z ∈ C; Re z �= 0}, Df := {z ∈ C; Re z > 0}.

∫ ∞

0

2te−tz

et + e−tdt =

∫ ∞

0

te−tz

cosh tdt = 2

∞∑n=0

(−1)n

(z + 1 + 2n)2

=1

z2 − 1+

4 · 12

1 +

4 · 12

z2 − 1+

4 · 22

1 +

4 · 22

z2 − 1+

4 · 32

1 +· · ·,

(3.6.13)

([Perr57], p 30). Dc := {z ∈ C; | arg(z2 − 1)| < π}, Df := {z ∈ C; Re z > 0 and z �∈(0, 1]}. For instance, for z =:

√5 we get

∫ ∞

0

4te−√

5t

cosh tdt =

1

1+

12

1 +

12

1 +

22

1 +

22

1 +

32

1 +

32

1 +· · ·, ([Perr57], p 30). (3.6.14)

A.3.7 Gamma function expressions by Ramanujan

Ramanujan produced quite a number of continued fraction expansions of ratios of gammafunctions. These ratios have all proved to be connected to hypergeometric functions,([Rama57], [Bern89]). We use Ramanujan’s notation∏

ε

Γ(a + εb + c) := Γ(a + b + c)Γ(a − b + c),∏ε

Γ(a + εb + εc + d) := Γ(a + b + c + d)Γ(a − b + c + d)×

× Γ(a + b − c + d)Γ(a − b − c + d)

(3.7.1)

and so on. That is, ε = ±1, and the product is taken over all different combinations ofthe εs.

1 − R

1 + R=

p

z +

12 − q2

z +

22 − p2

z +

32 − q2

z +

42 − p2

z +· · ·

where R =∏

ε

Γ

(z + p + εq + 1

4

(z + p + εq + 3

4

) ·∏

ε

Γ

(z − p + εq + 3

4

(z − p + εq + 1

4

) ,

(3.7.2)

([Bern89], p 156). Dc := {(p, q, z) ∈ C3; Re z �= 0}, Df := {(p, q, z) ∈ C

3; Re z > 0}.

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A.3.7 Gamma function expressions by Ramanujan 285

From this it follows that

∞∑k=1

{(−1)k+1

z + q + 2k − 1+

(−1)k+1

z − q + 2k − 1

}

=

∫ ∞

0

cosh(qt)e−tz

cosh tdt =

1

z +

12 − q2

z +

22

z +

32 − q2

z +

42

z +· · ·,

(3.7.3)

([Bern89], p 148), Dc := {(q, z) ∈ C2; Re z �= 0}, Df := {(q, z) ∈ C

2; Re z > 0}, and

tanh

{∫ ∞

0

sinh(at)e−tz

t cosh tdt

}=

a

z +

12

z +

22 − a2

z +

32

z +

42 − a2

z +· · ·, (3.7.4)

([Wall48], p 372), Dc := {(a, z) ∈ C2; Re z �= 0}, Df := {(a, z) ∈ C

2; Re z > 0}, and

tanh{1

2

∫ ∞

0

sinh(2at)e−tz

t cosh tdt}

=

a

z +

12 − a2

z +

22 − a2

z +

32 − a2

z +

42 − a2

z +· · ·,

(3.7.5)

([Wall48], p 371), Dc := {(a, z) ∈ C2; Re z �= 0}, Df := {(a, z) ∈ C

2; Re z > 0}.Solving (3.7.2) for 1/R gives

1

R= 1 +

2p

z − p+

12 − q2

z +

22 − p2

z +

32 − q2

z +

42 − p2

z +· · ·, (3.7.6)

([Perr57], p 34). Dc := {(p, z) ∈ C2; Re z �= 0}, Df := {(p, z) ∈ C

2; Re z > 0}. Thevalues p := q := 1/2 lead to

z

4

Γ2( z

4

)Γ2

(z + 2

4

) = 1 +2

2z − 1+

1 · 32z +

3 · 52z +

5 · 72z +· · ·

for Re(z) > 0 (3.7.7)

and thus, for z := 4n or z := 4n − 2 where n ∈ N, we have

1

(2 · 4 · · · · (2n)

1 · 3 · · · · (2n − 1)

)2

= 1 +2

8n − 1+

1 · 38n +

3 · 58n +

5 · 78n +· · ·

, (3.7.8)

([Perr57], p 34),

2n2π

2n − 1

(1 · 3 · · · · (2n − 1)

2 · 4 · · · · (2n)

)2

= 1 +2

8n − 5+

1 · 38n − 4+

3 · 58n − 4+

5 · 78n − 4+· · ·

(3.7.9)

for n ∈ N, ([Perr57], p 34).

a + 1

a

∫ 1

0

ta

(1 − t

1 + t

)b

dt∫ 1

0

ta−1

(1 − t

1 + t

)b

dt

=a + 1

2b +

(a + 1)(a + 2)

2b +

(a + 2)(a + 3)

2b +· · ·, (3.7.10)

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286 Appendix A: Some continued fraction expansions

([Perr57], p 299). Dc := {(a, b) ∈ C2; Re(b) �= 0}, Df := {(a, b) ∈ C

2; Re(b) > 0}. Fromthis follows directly that also

∫ 1

0

ta

(1 − t

1 + t

)bdt

1 − t∫ 1

0

ta

(1 − t

1 + t

)bdt

1 − t2

= 1 +a + 1

2b +

(a + 1)(a + 2)

2b +

(a + 2)(a + 3)

2b +· · ·, (3.7.11)

([Perr57], p 300). Dc := {(a, b) ∈ C2; Re(b) �= 0}, Df := {(a, b) ∈ C

2; Re(b) > 0}. Aformula of the same character as (3.7.2) is

∏ε

(Γ( z + εp + εq + 1

4

))/(Γ( z + εp + εq + 3

4

))=

812(z2 − p2 + q2 − 1)+

12 − q2

1 +

12 − p2

z2 − 1 +

32 − q2

1 +

32 − p2

z2 − 1 +· · ·,

(3.7.12)

([Bern89], p 159). Dc := {(p, q, z) ∈ C3; | arg(z2 − 1)| < π}, Df := {(p, q, z) ∈ C

3; Re z >0} \ {(p, q, z) ∈ C

3; 0 < z ≤ 1}.

∏ε

Γ

(z + εq + 1

4

(z + εq + 3

4

) =4

z +

12 − q2

2z +

32 − q2

2z +

52 − q2

2z +· · ·, (3.7.13)

([Bern89], p 140). Dc := {(q, z) ∈ C2; Re z �= 0}, Df := {(q, z) ∈ C

2; Re z > 0}. Forq := 0 and z := 4n − 1 or z := 4n + 1 for an n ∈ N, this reduces to

4πn2

(1 · 3 · · · · (2n − 1)

2 · 4 · · · · (2n)

)2

= 4n − 1 +12

8n − 2+

32

8n − 2+

52

8n − 2+· · ·, (3.7.14)

([Perr57], p 36), or

1

π

(2n + 1

n + 1

)2 (2 · 4 · · · · (2n + 2)

1 · 3 · · · · (2n + 1)

)2

= 4n + 1 +12

8n + 2+

32

8n + 2+

52

8n + 2+· · ·, (3.7.15)

([Perr57], p 36).

A formula closely related to (3.7.13) is

exp{ ∫ ∞

0

(1−cosh 2at

cosh 2t

)e−tz dt

t

}=

1 +2(12 − a2)

z2 +

32 − a2

1 +

52 − a2

z2 +· · ·,

(3.7.16)

([Wall48], p 371). Dc := {(a, z) ∈ C2; Re z �= 0}, Df := {(a, z) ∈ C

2; Re z > 0}.The most involved of Ramanujan’s formulas of this type is

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A.3.7 Gamma function expressions by Ramanujan 287

R − Q

R + Q=

8abcdh

1{2S4 − (S2 − 2 · 0 · 1)2 − 4(02 + 0 + 1)2}+

64(a2 − 12)(b2 − 12)(c2 − 12)(d2 − 12)(h2 − 12)

3{2S4 − (S2 − 2 · 1 · 2)2 − 4(12 + 1 + 1)2} +

64(a2 − 22)(b2 − 22)(c2 − 22)(d2 − 22)(h2 − 22)

5{2S4 − (S2 − 2 · 2 · 3)2 − 4(22 + 2 + 1)2} +· · ·,

(3.7.17)

where S4 := a4 + b4 + c4 + d4 + h4 + 1, S2 := a2 + b2 + c2 + d2 + h2 − 1, and

R :=∏

ε

Γ

(a + ε(b + c) + ε(d + h) + 1

2

)·∏

ε

Γ

(a + ε(b + d) + ε(c + h) + 1

2

),

Q :=∏

ε

Γ

(a + ε(b − c) + ε(d + h) + 1

2

)·∏

ε

Γ

(a + ε(b + c) + ε(d − h) + 1

2

),

(3.7.18)([Bern89], p 163). The expansion (3.7.17) only holds if the continued fraction terminates.

1 − R

1 + R=

2abc

z2 − a2 − b2 − c2 + 1+

4(a2 − 12)(b2 − 12)(c2 − 12)

3(z2 − a2 − b2 − c2 + 5) +

4(a2 − 22)(b2 − 22)(c2 − 22)

5(z2 − a2 − b2 − c2 + 13) +· · ·for (a, b, c, z) ∈ C

4 where

R :=∏

ε

Γ

(z + a + ε(b + c) + 1

2

(z − a + ε(b + c) + 1

2

) ·∏

ε

Γ

(z − a + ε(b − c) + 1

2

(z + a + ε(b − c) + 1

2

) ,

(3.7.19)

([Bern89], p 157). The last number in each partial denominator of the continued fraction(i.e., 1, 5, 13, . . . ) is the number 2n2 + 2n + 1 for n = 0, 1, 2, . . . .)

1 − R

1 + R=

ab

z +

(a2 − 12)(b2 − 12)

3z +

(a2 − 22)(b2 − 22)

5z +

(a2 − 32)(b2 − 32)

7z +· · ·

where R =∏

ε

Γ( z + ε(a − b) + 1

2

)/∏ε

Γ( z + ε(a + b) + 1

2

),

(3.7.20)

([Bern89], p 155). Dc := {(a, b, z) ∈ C3; Re z �= 0}, Df := {(a, b, z) ∈ C

3; Re z > 0}. Inparticular

∞∑k=0

{1

z − a + 2k + 1− 1

z + a + 2k + 1

}= lim

b→0

1

b

1 − R

1 + R

=a

z +

12(12 − a2)

3z +

22(22 − a2)

5z +

32(32 − a2)

7z +· · ·

(3.7.21)

for Re(z) > 0, ([Bern89], p 149).

∞∑k=1

(−1)k+1

(a + k)(b + k)=

1

(a + 1)(b + 1)+

(a + 1)2(b + 1)2

a + b + 3 +

(a + 2)2(b + 2)2

a + b + 5 +· · ·(3.7.22)

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288 Appendix A: Some continued fraction expansions

for (a, b) ∈ C2 with b �= −1 if a ∈ (−N) and a �= −1 if b ∈ (−N), ([Bern89], p 123).

1 − R

1 + R=

ab

z2 − 1 − a2 +

22 − b2

1 +

22 − a2

z2 − 1 +

42 − b2

1 +

42 − a2

z2 − 1 +· · ·;

R :=∏

ε

Γ( z + ε(a + b) + 3

4

)Γ( z + ε(a + b) + 1

4

)/∏

ε

Γ( z + ε(a − b) + 3

4

)Γ( z + ε(a − b) + 1

4

) ,

(3.7.23)

([Bern89], p 158). Dc := {(a, b, z) ∈ C3; | arg(z2 − 1)| < π}, Df := {(a, b, z) ∈ C

3; Re z >0} \ {(a, b, z) ∈ C

3; 0 < z ≤ 1}. Dividing (3.7.23) by a and letting a → 0 in this equalitygives

∞∑k=0

{(−1)k

z − b + 2k + 1− (−1)k

z + b + 2k + 1

}=

∫ ∞

0

e−tz sinh(bt)

cosh tdt

=b

z2 − 1+

22 − b2

1 +

22

z2 − 1+

42 − b2

1 +· · ·,

(3.7.24)

([Bern89], p 150). Dc := {(b, z) ∈ C2; | arg(z2 − 1)| < π}, Df := {(b, z) ∈ C

2; Re z >0} \ {(b, z) ∈ C

2; 0 < z ≤ 1}. Of course, dividing this again by b and letting b → 0 gives

2∞∑

k=0

(−1)k

(z + 2k + 1)2=

1

z2 − 1+

22

1 +

22

z2 − 1+

42

1 +

42

z2 − 1+· · ·, (3.7.25)

([Bern89], p 151). Dc := {z ∈ C; | arg(z2 − 1)| < π}, Df := {z ∈ C; Re z > 0} \ {z ∈C; 0 < z ≤ 1}.

1 + 2z

∞∑k=1

(−1)k

z + 2k=

1

z +

1 · 2z +

2 · 3z +

3 · 4z +· · ·

, (3.7.26)

([Bern89], p 151). Dc := {z ∈ C; Re z �= 0}, Df := {z ∈ C; Re z > 0}.

1 + 2z2∞∑

k=1

(−1)k

(z + k)2=

1

z +

12

z +

1 · 2z +

22

z +

2 · 3z +

32

z +· · ·, (3.7.27)

([Bern89], p 152). Dc := {z ∈ C; Re z �= 0}, Df := {z ∈ C; Re z > 0}.

c

∫ ∞

0

sinh at sinh bt

sinh cte−tzdt =

ab

1(z2 + c2 − a2 − b2)−4 · 12(12c2 − a2)(12c2 − b2)

3(z2 + 5c2 − a2 − b2) −4 · 22(22c2 − a2)(22c2 − b2)

5(z2 + 13c2 − a2 − b2) −· · ·,

(3.7.28)

where the coefficients for c2 are 2k2 +2k +1 in the denominators ([Wall48], p 370). Dc :={(a, b, c, z) ∈ C

4; Re zc�= 0}, Df := {(a, b, c, z) ∈ R

4; Re zc

> 0 and Re(z+c−a−b) > 0}.

c

∫ ∞

0

sinh at

sinh cte−tzdt =

a

z +

12(12c2 − a2)

3z +

22(22c2 − a2)

5z +· · ·, (3.7.29)

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A.3.7 Gamma function expressions by Ramanujan 289

([Wall48], p 370). Dc := {(a, c, z) ∈ C3; Re(z/c) �= 0}, Df := {(a, c, z) ∈ C

3; Re(z/c) >0}.

∫ ∞

0

e−tzdt

(cosh t + a sinh t)b=

1

z + ab+

1 · b(1 − a2)

z + a(b + 2)+

2(b + 1)(1 − a2)

z + a(b + 4) +

3(b + 2)(1 − a2)

z + a(b + 6) +· · ·,

(3.7.30)

([Wall48], p 369). Dc := {(a, b, z) ∈ C3; Re a �= 0}, Df := {(a, b, z) ∈ C

3; Re a >0 and Re(b + z) > 0}.

∫ ∞

02F1

(a, b;

a + b + 1

2;− sinh2 t

)e−tzdt =

1

z +

4 · 1ab

(a + b + 1)z +

+

4 · 2(a + 1)(b + 1)(a + b)

(a + b + 3)z +

4 · 3(a + 2)(b + 2)(a + b + 1)

(a + b + 5)z +· · ·

(3.7.31)

([Wall48], p 370). Dc := {(a, b, z) ∈ C3; Re z �= 0}, Df := {(a, b, z) ∈ C

3; Re z > 0}.

ζ(3, z + 1) :=

∞∑k=1

1

(z + k)3

=1

2(z2 + z)+

13

1 +

13

6(z2 + z)+

23

1 +

23

10(z2 + z)+· · ·,

(3.7.32)

([Bern89], p. 153). Dc := {z ∈ C; | arg(z2 + z)| < π} = {z ∈ C; Re z �= − 12} \ [−1, 0],

Df := {z ∈ C; Re z > − 12} \ [− 1

2, 0]. The even part of this continued fraction is

ζ(3, z + 1) =1

1(2z2 + 2z + 1)−16

3(2z2 + 2z + 3)−26

5(2z2 + 2z + 7)−36

7(2z2 + 2z + 13)−· · ·.

(3.7.33)

Dc := {z ∈ C; Re(z2 + 12) �= 0}, Df := {z ∈ C; Re(z2 + 1

2) > 0}. (The numbers 1, 3, 7,

13, . . . in the denominators are n2 + n + 1 for n = 0, 1, 2, . . . .)

∞∑k=0

{1

z + a + b + 2k + 1+

1

z − a − b + 2k + 1−

1

z + a − b + 2k + 1− 1

z − a + b + 2k + 1

}=

∞∑k=0

8ab(z + 2k + 1)

{(z + 2k + 1)2 − a2 − b2}2 − 4a2b2

=2ab

1(z2 − 1) + b2 − a2 +

2(12 − b2)

1 +

2(12 − a2)

3(z2 − 1) + b2 − a2 +

4(22 − b2)

1 +

4(22 − a2)

5(z2 − 1) + b2 − a2 +· · ·,

(3.7.34)

([Bern89], p 158). Dc := {(a, b, z) ∈ C3; | arg(z2 − 1)| < π} = {(a, b, z) ∈ C

3; Re z �=0 and z �∈ (−1, 1)}, Df := {(a, b, z) ∈ C

3; Re z > 0 and z �∈ (0, 1)}. Dividing by 2a andletting a → 0 in (3.7.34) leads to

Page 303: Lisa Lorentzen, Haakon Waadeland Continued Fractions

290 Appendix A: Some continued fraction expansions

∞∑k=0

{1

(z − b + 2k + 1)2− 1

(z + b + 2k + 1)2

}=

∞∑k=0

4b(z + 2k + 1)

{(z + 2k + 1)2 − b2}2

=b

1(z2 − 1) + b2 +

2(12 − b2)

1 +

2 · 12

3(z2 − 1) + b2 +

4(22 − b2)

1 +

4 · 22

5(z2 − 1) + b2 +· · ·,

(3.7.35)([Bern89], p 158). Dc := {(b, z) ∈ C

2; Re z �= 0 and z �∈ (−1, 1)}, Df := {(b, z) ∈C

2; Re z > 0 and z �∈ (0, 1)}. The even part of this continued fraction is

∞∑k=0

{1

(z − b + 2k + 1)2− 1

(z + b + 2k + 1)2

}=

∞∑k=0

4b(z + 2k + 1)

{(z + 2k + 1)2 − b2}2

=b

1(z2 − b2 + 1)−4(12 − b2)14

3(z2 − b2 + 5)−4(22 − b2)24

5(z2 − b2 + 13)−4(32 − b2)34

7(z2 − b2 + 25)−· · ·.

(3.7.36)

Dc := {(b, z) ∈ C2; Re z �= 0}, Df := {(b, z) ∈ C

2; Re z > 0}. (The numbers 1, 5, 13, 25, . . .in the denominators have the form 2n2 + 2n + 1 for n = 0, 1, 2, 3, . . . .)

u − v

u + v=

2a2

1z +

4a4 + 14

3z +

4a4 + 24

5z +

4a4 + 34

7z +· · ·where

u :=∞∏

k=0

{1 +

( 2a

z + 2k + 1

)2}, v :=

Γ2( z + 1

2

)Γ( z + 2a + 1

2

)Γ( z − 2a + 1

2

) (3.7.37)

([ABJL92], entry 48). Dc := {(a, z) ∈ C; Re z �= 0}, Df := {(a, z) ∈ C; Re z > 0}.

u − v

u + v=

a3

1(2z2 + 2z + 1)+

a6 − 16

3(2z2 + 2z + 3)+

a6 − 26

5(2z2 + 2z + 7)+· · ·where

u :=

∞∏k=1

{1 +

(a

z + k

)3}

, v :=

∞∏k=1

{1 −

(a

z + k

)3}

,

(3.7.38)

([ABJL92], entry 50). Dc := {(a, z) ∈ C2; Re z �= − 1

2}, Df := {(a, z) ∈ C

2; Re z > − 12}.

(The numbers 1, 3, 7, . . . in the denominators are the numbers n2+n+1 for n = 0, 1, 2, . . . .)

2∞∑

k=0

(−1)ky2k+1

r + 2k + 1=

z

1 + a+

12z2

3 + a+

22z2

5 + a+

32z2

7 + a+· · ·

where y := (√

1 + z2 − 1)/z and r := a/√

1 + z2

(3.7.39)

for (a, z) ∈ C2 with | arg(z2 + 1)| < π; i.e., z ∈ C \ i((−∞,−1] ∪ [1,∞)), ([ABJL92], entry

14). With the same notation and same region for (a, z), also

y + r

(y +

1

y

) ∞∑k=1

(−1)ky2k

r + 2k=

z

2 + a+

1 · 2z2

4 + a +

2 · 3z2

6 + a +

3 · 4z2

8 + a +· · ·, (3.7.40)

([ABJL92], entry 15) and

Page 304: Lisa Lorentzen, Haakon Waadeland Continued Fractions

A.4.1 Basic hypergeometric functions 291

(1 +

1

z2

)(b−1)/2

(2y)b∞∑

k=0

(−1)k(b)ky2k

k!(r + b + 2k)=

z

a + b+

1 · bz2

a + b + 2+

2(b + 1)z2

a + b + 4 +

3(b + 2)z2

a + b + 6 +· · ·,

(3.7.41)

([ABJL92], entry 17), where b ∈ C.

A.4 Basic hypergeometric functions

In this chapter we use the standard notation

2ϕ1(a, b; c; q; z) :=∞∑

n=0

(a; q)n(b; q)n

(c; q)n(q; q)nzn

where (d; q)0 := 1, (d; q)n := (1 − d)(1 − dq) · · · (1 − dqn−1) for n ∈ N.

For convenience we always assume that q ∈ C with |q| < 1, although the continued fractionmay well converge, even to the right value, for other values of q ∈ C.

A.4.1 General expressions

(1 − c)2ϕ1(a, b; c; q; z)

2ϕ1(a, bq; cq; q; z)=

1 − c +(1 − a)(c − b)z

1 − cq +

(1 − bq)(cq − a)z

1 − cq2 +

(1 − aq)(cq − b)qz

1 − cq3 +

(1 − bq2)(cq2 − a)qz

1 − cq4 +

(1 − aq2)(cq2 − b)q2z

1 − cq5 +· · ·

(4.1.1)

for (a, b, c, z) ∈ C4, ([ABBW85], p 14).

(1 − c)2ϕ1(a, b; c; q; z)

2ϕ1(aq, bq; cq; q; z)= b0 + K(an/bn)

where an := (1 − aqn)(1 − bqn)cqn−1(1 − zabqn/c)z

bn := 1 − cqn − (a + b − abqn − abqn+1)qnz

(4.1.2)

for (a, b, c, z) ∈ C4.

q(1 − c)2ϕ1(a, b; c; q; z)

2ϕ1(a, bq; cq; q; z)= (1 − c)q + (a − bq)z − (a − cq)(1 − bq)qz

(1 − cq)q + (a − bq2)z−(a − cq2)(1 − bq2)qz

(1 − cq2)q + (a − bq3)z −(a − cq3)(1 − bq3)qz

(1 − cq3)q + (a − bq4)z − · · ·

(4.1.3)

for (a, b, c, z) ∈ C4, ([ABBW85], p 18).

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292 Appendix A: Some continued fraction expansions

If we choose b = 1 in (4.1.1), (4.1.2) or (4.1.3) we obtain continued fraction expansions for

2ϕ1(a, q; cq; q; z) or 2ϕ1(aq, q; cq; q; z).

A.4.2 Two general results by Andrews

G(a, b, c; q)

G(aq, b, cq; q)= 1 +

aq + cq

1 +

bq + cq2

1 +

aq2 + cq3

1 +

bq2 + cq4

1 +· · ·

where G(a, b, c; q) :=

∞∑k=0

(− ca; q)kqk(k+1)/2ak

(q; q)k(−bq; q)k

(4.2.1)

for (a, b, c) ∈ C3, ([ABJL89], p 80).

H(a1, a2; z; q)

H(a1, a2; qz; q)= 1 + bqz +

(1 + aq2z)qz

1 + bq2z +

(1 + aq3z)q2z

1 + bq3z +· · ·where a := −1/a1a2 and b := −1/a1 − 1/a2 and

H(a1, a2; z; q) :=

(qz

a1; q

)∞

(qz

a2; q

)∞

(qz; q)∞(1 − z)×

×∞∑

k=0

(1 − zq2k)(z; q)k(a1; q)k(a2; q)kqk(3k+1)/2(az2)k

(q; q)k

(qz

a1; q

)k

(qz

a2; q

)k

(4.2.2)

for (1/a1, 1/a2, z) ∈ C3, ([ABJL89], p 79).

A.4.3 q-expressions by Ramanujan

The formula (4.2.1) can also be found in Ramanujan’s lost notebook ([Andr79], p 90).Quite a number of Ramanujan’s expressions are special cases of (4.2.1) and (4.2.2). Werefer in particular to ([ABJL92]) for more details. From (4.1.1) we find that

(−a; q)∞(b; q)∞ − (a; q)∞(−b; q)∞(−a; q)∞(b; q)∞ + (a; q)∞(−b; q)∞

=a − b

1 − q·

2ϕ1

(bq

a,bq2

a; q3; q2; a2

)2ϕ1

(bq

a,b

a; q; q2; a2

)=

a − b

1 − q +

(a − bq)(aq − b)

1 − q3 +

(a − bq2)(aq2 − b)q

1 − q5 +

(a − bq3)(aq3 − b)q2

1 − q7 +· · ·

(4.3.1)

for (a, b) ∈ C2, ([ABBW85], p 14).

(a2q3; q4)∞(b2q3; q4)∞(a2q; q4)∞(b2q; q4)∞

=1

1 − ab+

(a − bq)(b − aq)

(1 − ab)(q2 + 1)+

(a − bq3)(b − aq3)

(1 − ab)(q4 + 1) +· · ·(4.3.2)

Page 306: Lisa Lorentzen, Haakon Waadeland Continued Fractions

A.4.3 q-expressions by Ramanujan 293

for (a, b) ∈ C2, ([ABBW85], entry 12).

F (b; a)

F (b; aq)= 1 +

aq

1 + bq +

aq2

1 + bq2 +

aq3

1 + bq3 +· · ·

where F (b; a) :=

∞∑k=0

akqk2

(−bq; q)k(q; q)k

(4.3.3)

for (a, b) ∈ C2, ([ABBW85], entry 15).

If we set a := 0 in (4.2.1) we get

ϕ(c)

ϕ(cq)= 1 +

cq

1 +

bq + cq2

1 +

cq3

1 +

bq2 + cq4

1 +

cq5

1 + · · ·

where ϕ(c) :=

∞∑k=0

qk2ck

(q; q)k(−bq; q)k),

(4.3.4)

([ABJL92], entry 56). If we moreover set b := −c, this reduces to

∞∑k=0

(−c)kqk(k+1)/2 =1

1+

cq

1 +

c(q2 − q)

1 +

cq3

1 +

c(q4 − q2)

1 +· · ·, (4.3.5)

([ABBW85], p 22).

G(z)

G(qz)= 1 − qz

1 + q +

q3z

1 + q2−q2z

1 + q3 +

q6z

1 + q4 −q3z

1 + q5 +

q9z

1 + q6−· · ·

where G(z) :=

∞∑k=0

(−z)kqk(k+1)/2

(q2; q2)k

(4.3.6)

for z ∈ C, ([ABJL92], formula 9.1).

(q2; q3)∞(q; q3)∞

=1

1−q

1 + q−q3

1 + q2−q5

1 + q3−q7

1 + q4 −· · ·, (4.3.7)

([ABJL92], entry 10).

(q3; q4)∞(q; q4)∞

=1

1−q

1 + q2 −q3

1 + q4 −q5

1 + q6−· · ·, (4.3.8)

([ABJL92], entry 11).

(−q2; q2)∞(−q; q2)∞

=1

1+

q

1+

q2 + q

1 +

q3

1 +

q4 + q2

1 +

q5

1 +· · ·, (4.3.9)

([ABJL92], entry 12).

(q; q2)∞{(q3; q6)∞}3

=1

1+

q + q2

1 +

q2 + q4

1 +

q3 + q6

1 +· · ·, (4.3.10)

([ABJL89], thm 7).

Page 307: Lisa Lorentzen, Haakon Waadeland Continued Fractions

294

(q; q5)∞(q4; q5)∞(q2; q5)∞(q3; q5)∞

=1

1+

q

1+

q2

1 +

q3

1 +

q4

1 +· · ·, (4.3.11)

([ABJL89], (5)).

(q; q8)∞(q7; q8)∞(q3; q8)∞(q5; q8)∞

=1

1+

q + q2

1 +

q4

1 +

q3 + q6

1 +

q8

1 +

q5 + q10

1 +· · ·, (4.3.12)

([ABJL89], thm 6).

∞∑k=1

(a; q)∞ak

(q; q)k(1 + qkz)=

a

1+

(1 − a)qz

1 +

(1 − q)aqz

1 +

(1 − aq)q2z

1 +

(1 − q2)aq2z

1 +

(1 − aq2)q3z

1 +· · ·

(4.3.13)

for (a, z) ∈ C2, ([Wall48], p 376).

Appendix A: Some continued fraction expansions

Page 308: Lisa Lorentzen, Haakon Waadeland Continued Fractions

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Index

(a)n Pochhammer symbol, 27

A(n)k , 9

An canonical numerator, 6B

(n)k , 9

Bn canonical denominator, 6F [n], 172I(w), 172Nth root of unity, 194Pk, 66Sn, 5S

(m)p , 172

Δn, 7�, 174Σ∞, 68Σn, 66C, 3�u�, 15C, 3D unit disk, 26H, 109N, 9R real numbers, 109B(a, r), 75Bm(γn, ε), 73V, 109Ln z, 26diam(V ), 71dist(x, V ), 225distm(w, V ) chordal distance, 114rad(D), radius of D, 111R+, 118∼ equivalence between continued fractions,

77√. . ., 25

{ζn} critical tail sequence, 64{hn} critical tail sequence, 64{tn} tail sequence, 63{w†

n} exceptional sequence, 56f (n) tail value, 6fn classical approximant, 6

o(kn), 221o(rn+1), 12p-periodic continued fraction, 177sn, 5Tn, 111M family of Mobius transformations, 5S Stern-Stolz Series, 101R

+, 118B(C,−r), 109B(C, r), 109ε-contractive, 73

2F1(a, b; c; z) hypergeometric function, 28Sleszynski-Pringsheim Theorem, 129Sleszynski-Pringsheim continued fraction, 129

a posteriori bounds, 108a priori bounds, 108absolute convergence of continued fraction,

100absolute convergence of sequence, 100absolutely continuous measure, 41alternating continued fraction, 122analytic continuation, 35approximant, 3, 5arithmetic complexity, 11asymptotic expansion, 40attracting fixed point, 175auxiliary continued fraction, 223axis of cartesian oval, 244

backward recurrence algorithm, 11Bauer-Muir transform, 82best rational approximation, 17binomial series, 50Birkhoff-Trjzinski theory, 262

canonical contraction, 85canonical denominator, 7canonical numerator, 7cartestian oval, 161

306

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307

chain sequence, 90Chebyshev polynomials, 42chordal diameter, 71chordal distance, 55chordal metric, 55classical approximant, 6classical convergence, 60conjugate transformation, 174continued fraction, 3continued fraction expansion, 25continued fraction of elliptic type, 176continued fraction of identity type, 176continued fraction of loxodromic type, 176continued fraction of parabolic type, 176continued fraction, definition, 5contraction, 85convergence acceleration, 218convergence neighborhood, 213convergence of continued fraction, 6convergents, 5correspondence, 33corresponding periodic continued fraction,

186critical tail sequence, 64cross ratio, 54

determinant formula, 7diagonalization of matrix, 212diamm , 71differential equation, 38diophantine equation, 21distribution function, 40divergent continued fraction, 6dual continued fraction, 95

element sets, 73elements of a continued fraction, 5ellipse, arc length of, 30elliptic transformation, 175empty product, 2empty sum, 2equivalence transformation, 77equivalent continued fractions, 77equivalent sequences, 57euclidean algorithm, 16Euler-Minding formula, 7Euler-Minding summation, 11even part, 86exceptional sequence, 56, 57extension, 85

Favard’s Theorem, 43Fibonacci numbers, 50fixed point, 172fixed circle for τ , 205fixed line for τ , 205fixed point method, 219forward recurrence algorithm, 11fraction term, 5functional equation, 85fundamental inequalities, 165

general convergence, 56, 60general divergence, 60generalized circle, 108generic sequence, 61golden ratio, 50greatest common divisor, 16

Henrici-Pfluger Bounds, 126history of continued fractions, 46hyperbolic transformation, 207hypergeometric functions, 27

identity transformation, 62improvement machine, 227indifferent fixed points, 175interpolation, 44inverse differencies, 44iterate, 172

J-fractions, 43Jacobi continued fraction, 43

Khovanskii transform, 98Kronecker delta, 42

Legendre polynomials, 42limit p-periodic continued fraction of loxo-

dromic type, 187limit periodic continued fraction, 186limit point case, 70limit sets, 75linear fractional transformation, 5loxodromic transformation, 175

Mobius transformation, 5measure, 41modified approximant, 6modifying factor, 6moment, 41

Index

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308

moment problem, 40, 41

odd part, 86orthogonal polynomials, 42oval, 161Oval Sequence Theorem, 243

Pade approximant, 35Pade approximants, diagonal, 37Pade table, 35Parabola Sequence Theorem, 154Parabola Theorem, 151parabolic pair, 184parabolic transformation, 175period length, 186periodic continued fraction, 172Pochhammer symbol, 27positive continued fraction, 116prevalue set, 70probability measure, 41

Ramanujan’s AGM-fraction, 202ratio for continued fraction, 176ratio of τ , 174rational approximation, 17real continued fraction, 122recurrence relations for An, Bn, 6regular C-fraction, 30, 79regular continued fraction, 4, 14regular continued fraction expansion, 15repelling fixed point, 175restrained continued fraction, 62restrained sequence, 61reversed continued fraction, 213reversed periodic continued fraction, 95reversed terminating continued fraction, 48right tail sequence, 90root of unity, 194

S-fraction, 41, 124Seidel-Stern Theorem, 117separate convergence, 102sequence of tail values, 6similar transformation, 174simple element set, 74simple value set, 70singular transformation, 5square root modification, 235stable polynomials, 45Stern-Stolz Divergence Theorem, 100

Stern-Stolz Series, 101Stieltjes continued fraction, 124Stieltjes moment problem, 41Stieltjes-Vitali Theorem, 115strong convergence, 90successive substitutions, 30sum of divergent series, 39symmetric points with respect to circle, 109

tail, 6tail sequence, 63tail values, 6terminating continued fraction, 10Thiele continued fraction, 43Thiele oscillation, 180Thron-Gragg-Warner Bounds, 128Thron-Lange Theorem, 148totally non-restrained, 62transformations of continued fractions, 77truncation error, 106truncation error estimate, 165truncation error bounds, 106tusc, 225twin element sets, 74twin value sets, 70

unit disk, 26

value set, 70Van Vleck’s Theorem, 142Vitali’s Theorem, 114

Worpitzky’s Theorem, 135wrong tail sequence, 90

Index