Continued Fractions, Vol. 1. Convergence Theory, 2nd Edition

ATLANTIS STUDIES IN MATHEMATICS FOR ENGINEERING AND SCIENCE VOLUME 1 SERIES EDITOR: C.K. CHUI Atlantis Studies in Mat...

Author: Lisa Lorentzen | Haakon Waadeland

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ATLANTIS STUDIES IN MATHEMATICS FOR ENGINEERING AND SCIENCE VOLUME 1 SERIES EDITOR: C.K. CHUI

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) publishes high quality monographs in applied mathematics, computational mathematics, and statistics that have the potential to make a significant impact on the advancement of engineering and science on the one hand, and economics and commerce on the other. We welcome submission of book proposals and manuscripts from mathematical scientists worldwide who share our 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

AMSTERDAM – PARIS

© ATLANTIS PRESS / WORLD SCIENTIFIC

Continued Fractions Second edition

Volume 1: Convergence Theory

Lisa Lorentzen Haakon Waadeland Department of Mathematics Norwegian University of Science and Technology Trondheim Norway

AMSTERDAM – PARIS

Atlantis Press 29 avenue Laumière 75019 Paris, France For information on all Atlantis Press publications, visit our website at: www.atlantis-press.com. Copyright This book, or any parts thereof, may not be reproduced for commercial purposes in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system known or to be invented, without prior permission from the Publisher.

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

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

© 2008 ATLANTIS PRESS / WORLD SCIENTIFIC

Preface to the second edition 15 years have passed since the first edition of this book was written. A lot has happened since then – also in continued fraction theory. New ideas have emerged and some old results have gotten new proofs. It was therefore time to revise our book “Continued Fractions with Applications” which appeared in 1992 on Elsevier. The interest in using continued fractions to approximate special functions has also grown since then. Such fractions are easy to program, they have impressive convergence properties, and their convergence is often easy to accelerate. They even have good and reliable truncation error bounds which makes it possible to control the accuracy of the approximation. The bounds are both of the a posteriori type which tells the accuracy of a done calculation, and of the a priori type which can be used to determine the number of terms needed for a wanted accuracy. This important aspect is 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 analytic functions. There are several beautiful connections between analytic function theory and continued fraction expansions. We can for instance mention orthogonal polynomials, moment theory and Padé approximation. We have tried to give credit to people who have contributed to the continued fraction theory up through the ages. But some of the material we believe to be new, at least we have found no counterpart in the literature. In particular, we believe that tail sequences play a more fundamental role in this book than what is usual. This way of looking at continued fractions is very fruitful. Each chapter is still followed by a number of problems. This time we have marked the 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 received that we wanted it to stay as part of the book. Finally, we have kept the informality in the sense that the first chapter consists almost entirely of examples which show what 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 edition. Through the first year of its making, Haakon Waadeland was busy writing a handbook on continued fractions, together with an international group of people. This left Lisa Lorentzen with quite free hands to choose the contents and the way of presentation. Still, he has played an important part in the later phases of the work on volume 1. For volume 2 Lisa Lorentzen bears the blame alone. Trondheim, 14 February 2008 Lisa Lorentzen

Haakon Waadeland v

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 our way through the chapters on a last minute search for errors, unintended omissions and overlaps, or other unfortunate occurrences, we feel that this title might have been the right one even for the whole book. In most of the chapters, in particular in the applications, a lot of work has been put into the process of cutting, canceling and “nonwriting”. In many cases we are just left with a “taste”, or rather a glimpse of the role of the continued fractions within the topic of the chapter. We hope that we thereby can open 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”. The purpose 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 adjust the expectations to the right (moderate) level. Each chapter ends with a reference list, reflecting essentially literature used in preparing that particular chapter. As a result, 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 field will look in vain. To present such a bibliography has not been one of the purposes of the book. The authors The two authors are different in style and approach. We have not made an effort to hide this, but to a certain extent the creative process of tearing up each other’s drafts and 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 of cooperation leaves us with a joint responsibility for the whole book. The way we then distribute 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 also have lived with and loved the monographs by Perron, Wall and Jones/Thron. Actually the love for continued fractions most likely has been initiated by one or more of these books. This is at least the case for the authors of the present book, and more so: these three books have played an essential role in our lives. The present book is in no way an attempt to replace or compete with these books. To the contrary, we hope to urge the reader to go on to these sources for further information. vi

Preface

vii

For whom? We are aiming at two kinds of readers: On the one hand people in or near mathematics, who are curious about continued fractions; on the other hand senior-graduate level students who would like an introduction (and a little more) to the analytic theory of continued fractions. Some basic knowledge about functions of a complex variable, a little linear algebra, elementary differential equations and occasionally a little dash of measure theory is what is needed of mathematical background. Hopefully the students will appreciate the problems included and the examples. They may even appreciate that 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 inspiration and 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 also extends to Bill Jones, his closest coworker, to Arne Magnus, whose recent death struck us with sadness, and to all other members of the Colorado continued fraction community. Here in Trondheim Olav Njåstad has been a key person in the field, and we have on several occasions had a rewarding cooperation with him. Many people, who had received our Chapter I, responded by sending friendly and encouraging letters, often with valuable suggestions. We thank them all for their interest and kind help. The main person in the process of changing the hand-written drafts to a cameraready copy was Leiv Arild Andenes Jacobsen. His able mastering of LaTeX, in combination with hard work, often at times when most people were in bed, has left us with a great debt of gratitude. We also want to thank Arild Skjølsvold and Irene 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 Trondheim generously covered most of the typing expenses. The rest was covered by Elsevier Science Publishers. We are most grateful to Claude Brezinski and Luc Wuytack for urging us to write this book, and to Elsevier Science Publishers for publishing it. Trondheim, December 1991, Lisa Lorentzen

Haakon Waadeland

Contents Preface to the second edition Preface 1

2

v vi

Introductory examples 1.1 Basic concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Prelude to a definition . . . . . . . . . . . . . . . . . . . . . . . 1.1.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.3 Computation of approximants . . . . . . . . . . . . . . . . . 1.1.4 Approximating the value of K(an/bn) . . . . . . . . . . . . 1.2 Regular continued fractions . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2 Best rational approximation . . . . . . . . . . . . . . . . . . . 1.2.3 Solving linear diophantine equations . . . . . . . . . . . . . 1.2.4 Grandfather clocks . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.5 Musical scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Rational approximation to functions . . . . . . . . . . . . . . . . . . 1.3.1 Expansions of functions . . . . . . . . . . . . . . . . . . . . . . 1.3.2 Hypergeometric functions . . . . . . . . . . . . . . . . . . . . 1.4 Correspondence between power series and continued fractions 1.4.1 From power series to continued fractions . . . . . . . . . 1.4.2 From continued fractions to power series . . . . . . . . . 1.4.3 One fraction, two series; analytic continuation . . . . . . 1.4.4 Padé approximation . . . . . . . . . . . . . . . . . . . . . . . . 1.5 More examples of applications . . . . . . . . . . . . . . . . . . . . . . 1.5.1 A differential equation . . . . . . . . . . . . . . . . . . . . . . . 1.5.2 Moment problems and divergent series . . . . . . . . . . . 1.5.3 Orthogonal polynomials . . . . . . . . . . . . . . . . . . . . . . 1.5.4 Thiele interpolation . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.5 Stable polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1 . 2 . 2 . 5 . 10 . 11 . 14 . 14 . 17 . 21 . 22 . 23 . 25 . 25 . 27 . 30 . 30 . 33 . 33 . 35 . 38 . 38 . 39 . 42 . 43 . 45 . 46 . 48

Basics 2.1 Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Properties of linear fractional transformations . . . 2.1.2 Convergence of continued fractions . . . . . . . . . . . 2.1.3 Restrained sequences . . . . . . . . . . . . . . . . . . . . . 2.1.4 Tail sequences . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.5 Tail sequences and three term recurrence relations

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ix

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53 54 54 59 60 63 65

x

Contents . . . . . . . . . .

70 73 77 77 77 82 85 87 89 91

3

Convergence criteria 3.1 Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 The Stern-Stolz Divergence Theorem . . . . . . . . . . . . . . . . . . 3.1.2 The Lane-Wall Characterization . . . . . . . . . . . . . . . . . . . . . 3.1.3 Truncation error bounds. . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.4 Mapping with linear fractional transformations. . . . . . . . . . . 3.1.5 The Stieltjes-Vitali Theorem . . . . . . . . . . . . . . . . . . . . . . . . 3.1.6 A simple estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Classical convergence theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Positive continued fractions . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Alternating continued fractions . . . . . . . . . . . . . . . . . . . . . . 3.2.3 Stieltjes continued fractions . . . . . . . . . . . . . . . . . . . . . . . . ´ leszyn 3.2.4 The S ´ ski-Pringsheim Theorem . . . . . . . . . . . . . . . . . . 3.2.5 Worpitzky’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.6 Van Vleck’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.7 The Thron-Lange Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.8 The parabola theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Additional convergence theorems. . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Simple bounded circular value sets . . . . . . . . . . . . . . . . . . . 3.3.2 Simple unbounded circular value sets. . . . . . . . . . . . . . . . . . 3.4 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

99 100 100 103 106 108 114 115 116 116 122 124 129 135 142 148 151 160 160 163 165 166

4

Periodic and limit periodic continued fractions 4.1 Periodic continued fractions . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Iterations of linear fractional transformations . . . . . 4.1.3 Classification of linear fractional transformations . . 4.1.4 General convergence of periodic continued fractions 4.1.5 Convergence in the classical sense . . . . . . . . . . . . . 4.1.6 Approximants on closed form . . . . . . . . . . . . . . . . 4.1.7 A connection to the Parabola Theorem . . . . . . . . . 4.2 Limit periodic continued fractions . . . . . . . . . . . . . . . . . . 4.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Finite limits, loxodromic case . . . . . . . . . . . . . . . .

171 172 172 172 174 176 179 181 183 186 186 187

2.2

2.3 2.4

2.1.6 Value sets . . . . . . . . . . . . . . . 2.1.7 Element sets . . . . . . . . . . . . . . Transformations of continued fractions 2.2.1 Introduction . . . . . . . . . . . . . . 2.2.2 Equivalence transformations . . 2.2.3 The Bauer-Muir transformation 2.2.4 Contractions and extensions . . 2.2.5 Contractions and convergence . Remarks . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . .

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Contents

4.3

4.4

4.5 4.6 5

A

xi

4.2.3 Finite limits, parabolic case . . . . . . . . . . . . . . . . . . . 4.2.4 Finite limits, elliptic case . . . . . . . . . . . . . . . . . . . . . 4.2.5 Infinite limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Continued fractions with multiple limits . . . . . . . . . . . . . . . 4.3.1 Periodic continued fractions with multiple limits . . . . 4.3.2 Limit periodic continued fractions with multiple limits Fixed circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 Fixed circles for U M . . . . . . . . . . . . . . . . . . . . . . 4.4.3 Fixed circles and periodic continued fractions . . . . . . . Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Numerical computation of continued fractions 5.1 Choice of approximants . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Fast convergence . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 The fixed point method . . . . . . . . . . . . . . . . . . . . . 5.1.3 Auxiliary continued fractions . . . . . . . . . . . . . . . . . 5.1.4 The improvement machine for the loxodromic case . . 5.1.5 Asymptotic expansion of tail values . . . . . . . . . . . . . 5.1.6 The square root modification . . . . . . . . . . . . . . . . . 5.2 Truncation error bounds . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 The ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Truncation error bounds. . . . . . . . . . . . . . . . . . . . . 5.2.3 The Oval Sequence Theorem. . . . . . . . . . . . . . . . . . 5.2.4 An algorithm to find value sets for a given continued fraction of form K(an /1) . . . . . . . . . . . . . . . . . . . . . 5.2.5 Value sets and the fixed point method . . . . . . . . . . . 5.2.6 Value sets B(wn Sn ) for limit 1-periodic continued fractions of loxodromic or parabolic type . . . . . . . . . 5.2.7 Error bounds based on idea 3 . . . . . . . . . . . . . . . . . 5.3 Stable computation of approximants . . . . . . . . . . . . . . . . . 5.3.1 Stability of the backward recurrence algorithm . . . . . 5.4 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Some continued fraction expansions A.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . A.1.1 Notation. . . . . . . . . . . . . . . . . . . . . . . A.1.2 Transformations . . . . . . . . . . . . . . . . . A.2 Elementary functions. . . . . . . . . . . . . . . . . . . A.2.1 Mathematical constants . . . . . . . . . . . . A.2.2 The exponential function . . . . . . . . . . . A.2.3 The general binomial function . . . . . . . A.2.4 The natural logarithm . . . . . . . . . . . . . A.2.5 Trigonometric and hyperbolic functions.

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xii

Contents A.2.6 Inverse trigonometric and hyperbolic functions A.2.7 Continued fractions with simple values . . . . . . A.3 Hypergeometric functions. . . . . . . . . . . . . . . . . . . . . A.3.1 General expressions . . . . . . . . . . . . . . . . . . . . A.3.2 Special examples with 0F1 . . . . . . . . . . . . . . . A.3.3 Special examples with 2F0 . . . . . . . . . . . . . . . A.3.4 Special examples with 1F1 . . . . . . . . . . . . . . . A.3.5 Special examples with 2F1 . . . . . . . . . . . . . . . A.3.6 Some integrals . . . . . . . . . . . . . . . . . . . . . . . A.3.7 Gamma function expressions by Ramanujan . . A.4 Basic hypergeometric functions. . . . . . . . . . . . . . . . . A.4.1 General expressions . . . . . . . . . . . . . . . . . . . . A.4.2 Two general results by Andrews . . . . . . . . . . . A.4.3 q -expressions by Ramanujan . . . . . . . . . . . . .

Bibliography Index

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Chapter 1

Introductory examples We have often been asked questions, by students as well as by established mathematicians, about continued fractions: what they are and what they can be used for. Sometimes the questions have been raised under circumstances where a quick answer 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. In responding to these questions we have often been pleased by the sparks of interest we have seen, indicating that we had managed to transmit a glimpse of new and apparently appealing knowledge. In quite a few cases this led to further contact and “ follow-up activities”. This introductory chapter is to a large extent inspired by the questions we have received and governed by the answers we have given. There is of course a great danger: A quick answer is often a wrong answer. It may tell the truth and nothing but the truth, but it deﬁnitely does not tell the whole truth. This may lead to false guesses. This danger is in particular great in cases where observations and experiments are used to create and support guesses. But we still wanted to keep this often non-accepted, but highly useful aspect of mathematics as part of the introductory chapter. We have tried to reduce the danger, partly by the way things are phrased, partly by indicating brieﬂy how wrong such guesses can be, and ﬁnally by referring to a more careful treatment later in the book. Still, we have chosen to introduce 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 of the introductory chapter, but also to increase the number of examples. We have therefore 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 is done on purpose, in the belief that what is lost in mathematical style and structure is 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

2

1.1 1.1.1

Chapter 1: Introductory examples

Basic concepts Prelude to a deﬁnition

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

an = a1 + a2 + · · · + an + · · · ,

(1.1.1)

n=1

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

n

ak .

k=1

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

an = σ .

(1.1.2)

n=1

Similarly we are familiar with inﬁnite products ∞

a n = a1 · a 2 · · · a n · · · .

(1.1.3)

n=1

{pn } is the sequence of partial products pn :=

n

ak .

k=1

n (An empty product 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, in which case we write ∞ an = p . (1.1.4) n=1

Let {fn } be the sequence f1 := a1 ,

f2 :=

a1 , 1 + a2

f3 :=

a1 , a2 1+ 1 + a3

and generally fn := 1+ 1+

a1 a2 a3 1 + ··

· +an

.

1.1.1 Prelude to a deﬁnition

3

For simplicity we assume that all an = 0, and we allow fn = ∞. Then {fn } is well := C ∪ {∞} where C is the set of complex numbers. Similarly to what deﬁned in C we have for sums and products, this also leads to a concept, having to do with the nonterminating continuation of the process, in this case the concept of a continued fraction ∞ a1 an := . (1.1.5) n=1 1 a2 1+ a3 1+ 1 + ·· · (The letter K is chosen since Kettenbruch is the German name for a continued fraction.) Convergence of (1.1.5) means convergence of the sequence {fn } of approximants. 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

K

∞

f=

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

K an =: a11 n=1 1

a2 a3 . + 1 + 1 +···

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

Example 1. For the continued fraction ∞

K6 =

n=1 1

6

=

6

1+

6 6 6 1 + 1 + 1 +···

6

1+ 1+

6 1 + ·· ·

we ﬁnd

6 42 , f3 = ,··· . 7 13 By induction (see Problem 13 on page 49 with x = −2, y = 3) it follows that generally 3n − (−2)n fn = 6 n+1 . 3 − (−2)n+1 f1 = 6 ,

f2 =

Therefore the continued fraction converges to 2. 3

4

Chapter 1: Introductory examples

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

K1

n=1 bn

1

=

1

b1 + b2 +

=

1 1 1 , b1 + b2 + b3 + · · ·

(1.1.7)

1 b 3 + ·· ·

or from two sequences, {an } and {bn } of complex numbers, where all an = 0, a continued fraction ∞

K an = n=1 bn

a1 a2

b1 + b2 +

=

a2 a3 a1 . b1 + b2 + b3 + · · ·

(1.1.8)

a3 b 3 + ·· ·

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 the continued fraction after n fraction terms ak /bk , and convergence means convergence of {fn }. (1.1.5) and (1.1.7) are obviously special cases of (1.1.8). In the particular case 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 and continued fractions (and other constructions for that matter). In all three cases the construction can be described in the following way: We have a sequence {φk } to C. By composition we construct a new sequence {Φn } of of mappings from C mappings Φn := Φn−1 ◦ φn = φ1 ◦ φ2 ◦ · · · ◦ φn . (1.1.9) Φ1 := φ1 , For series we have φk (w) := w + ak , and Φn (w) = φ1 ◦ φ2 ◦ · · · ◦ φn (w) = a1 + a2 + · · · + an + w . For products we have φk (w) := w · ak , and Φn (w) = φ1 ◦ φ2 ◦ · · · ◦ φn (w) = a1 · a2 · · · an · w . For continued fractions (1.1.8) we have φk (w) := and Φn (w) = φ1 ◦ φ2 ◦ · · · ◦ φn (w) =

ak , bk + w a1 a2 an . b1 + b2 +· · · + bn + w

(1.1.10)

1.1.2 Deﬁnitions

5

These inﬁnite structures can be regarded formally, but in applications the question of 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 M¨obius transformation) τ is a mapping of form aw + b ; a, b, c, d ∈ C with ad − bc = 0. τ (w) = cw + d This representation of τ is not unique since we can multiply the coeﬃcients a, b, c and 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 transformations with ad − bc = 0 which do not belong to M. A singular transformation is constant univalently onto C. wherever it is meaningful, whereas τ ∈ M maps 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 numbers with all an = 0 and {Sn } is the sequence from M given by Sn (w) := b0 +

a1 a2 an . b1 + b2 +· · · + bn + w

(1.2.1)

&

%

Remark: That {Sn } is a sequence from M is easily seen by the fact that S n = s 0 ◦ s 1 ◦ · · · ◦ sn s0 (w) := b0 + w ,

where an for n ∈ N. sn (w) := bn + w

(1.2.2)

Obviously sn ∈ M when an = 0, and thus Sn ∈ M since compositions of linear fractional transformations are again linear fractional transformations (which is easily veriﬁed). We owe it to Weyl ([Weyl10]) for this very useful connection between continued 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 abnn is called a fraction term of b0 + K(an /bn ). Evaluations Sn (w) of Sn are called nth approximants. The name “ convergents” has also been used in the literature. Historically,

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 the “ modifying 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) to if lim fn = f . a value f ∈ C

If K(an /bn ) fails to converge, we say that it diverges. A classical approximant fn is obtained by truncating the continued fraction after n fraction terms. The part we cut away, f (n) :=

an+2 an+3 an+1 bn+1 + bn+2 + bn+3 + · · ·

(1.2.4)

is called the nth tail of b0 + K(an /bn ). This is also a continued fraction, and it converges 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−1 w + An Bn−1 w + Bn

for n = 1, 2, 3, . . .

(1.2.6)

where An = bn An−1 + an An−2 ,

Bn = bn Bn−1 + an Bn−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 b0 b1 + a1 + b0 w = , b1 + w b1 + w

%

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 bn + w . Bn−1 + Bn−2 an bn + w An−1 + An−2

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 natural since (1.2.8) fn = Sn (0) = An /Bn . The useful determinant formula Δn := An−1 Bn − An Bn−1 =

n

(−ak )

(1.2.9)

k=1

is a consequence of (1.2.7) (it follows by induction). Therefore also An−1 Bn+1 − An+1 Bn−1 = bn+1 Δn .

(1.2.10)

Another simple, but important observation follows from the fact that Sn (wn ) = S0 (w0 ) +

n

(Sk (wk ) − Sk−1 (wk−1 )).

(1.2.11)

k=1

By the recurrence relations (1.2.7) and the determinant formula (1.2.9) Sk (wk ) − Sk−1 (wk−1 ) =

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

(1.2.12)

k−1 λk m=1 (−am ) . (Bk−1 wk + Bk )(Bk−2 wk−1 + Bk−1 )

(1.2.13)

and thus Sn (wn ) = b0 + w0 +

n k=1

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

(−1)k a1 a2 · · · ak Δk An = b0 − = b0 − , Bn Bk Bk−1 Bk Bk−1 n

n

k=1

k=1

(1.2.14)

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

8

Chapter 1: Introductory examples

Remark: For a given continued fraction b0 + K(an /bn ), we shall use the notation sn , 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 complex numbers are the canonical numerators and denominators of some continued 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 ,

an := − ΔΔn , n−1

bn :=

a1 := A1 − A0 B1 ,

An−2 Bn − Bn−2 An for n ≥ 2 . Δn−1

(1.2.15)

&

%

Proof : Let {An } and {Bn } be given with all Δn = 0 and B0 = 1. Then the elements an and bn must be solutions of the system bn An−1 + an An−2 = An , bn Bn−1 + an Bn−2 = Bn

(1.2.16)

of linear equations. The determinant of this system is −Δn−1 = 0. Hence the solution 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−1 Bn − Bn−1 An = 1 − 2n and An−2 Bn − Bn−2 An = 4 − 4n , so by Theorem 1.2 b0 = 0 ,

b1 = 2 , a1 = 1 , 2n − 1 4n − 4 , bn = an = − 2n − 3 2n − 3

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

1.1.2 Deﬁnitions

9

Hence, the continued fraction 1 −3/1 −5/3 −7/5 −9/7 −11/9 , 2 + 4/1 + 8/3 + 12/5 + 16/7 + 20/9 +· · · which also can be written 3/1 5/3 7/5 9/7 11/9 1 , 2 − 4/1 − 8/3 − 12/5 − 16/7 − 20/9 − · · · is the one with An = n2 and Bn = n2 + 1. Since lim An /Bn = 1, this continued fraction converges to 1, and we have proved the continued fraction identity 3/1 5/3 7/5 9/7 11/9 1 = 1. 2 − 4/1 − 8/3 − 12/5 − 16/7 − 20/9 − · · · 3 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 approximants for some continued fraction K(an /bn ) if and only if f0 = 0, f1 = 0, ∞ and fn = 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 }∞ be given by n=0 fn if fn = ∞, 1 if fn = ∞, Bn := An := 1 if fn = ∞, 0 if fn = ∞. Then fn = An /Bn , Δn = 0 for n ≥ 1 and B0 = 1, and thus the existence of K(an /bn ) follows from Theorem 1.2. Conversely, let {fn } be the classical approximants for K(an /bn ). Then f0 = 0, B0 = 1 and fn = Sn (0) = Sn (∞) = fn−1 since to C. Sn is a univalent mapping of 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(n) (n) ued fraction. We shall let Ak and Bk denote its kth canonical numerator and denominator. Then the following follows by manipulating the recurrence relations:

10

Chapter 1: Introductory examples

(n)

Ak

Lemma 1.4. (n)

Bk

(n+1)

= an+1 Bk−1 . (n+1)

(1.2.17) (n+2)

= bn+1 Bk−1 + an+2 Bk−2 .

(1.2.18)

Proof : Let n ∈ N ∪{0} be arbitrarily chosen. The ﬁrst identity holds trivially for (n) (n+1) k = 0 and k = 1. Hence it holds for all k since {Ak } and {Bk−1 } are solutions of the same recurrence relation Xk = bn+k Xk−1 + an+k Xk−2 .

(1.2.19) (n)

The second identity holds trivially for all n for k = 1 and k = 2 since B−1 = 0, (n) (n) (n) B0 = 1, B1 = bn+1 and B2 = bn+1 bn+2 + an+2 . Assume it holds for 2 ≤ k ≤ ν − 1. Then it also holds for k = ν since (n)

(n)

(n)

= bn+k Bk−1 + an+k Bk−2 (n+1) (n+2) (n+1) (n+2) + an+k bn+1 Bk−3 + an+2 Bk−4 = bn+k bn+1 Bk−2 + an+2 Bk−3 (n+1) (n+1) (n+2) (n+2) = bn+1 bn+k Bk−2 + an+k Bk−3 + an+2 bn+k Bk−3 + an+k Bk−4

Bk

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 applications one only needs a continued fraction-like structure which terminates after n terms, a2 an a1 . b0 + b1 + b2 + · · · + bn This will be called a terminating continued fraction even though it is not a continued fraction 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 , and the 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 n0 th tail converges to −bn0 . In such cases it is a matter of deﬁnition to determine the action.) This is in particular relevant if the elements an and bn are functions of a complex variable z.

1.1.3

Computation of approximants

There are several algorithms to compute approximants Sn (wn ) = b0 +

an a1 a2 An−1 wn + An = . b1 + b2 +· · · + bn + wn Bn−1 wn + Bn

We shall only mention the most obvious ones:

1.1.4 Approximating the value

11

1. The forward recurrence algorithm consists of computing An and Bn by the recurrence relation (1.2.7). 2. The backward recurrence algorithm starts by setting qn := wn and then work backwards by setting ak qk−1 := 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 minimal 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 use this to ﬁnd Sn+1 (wn+1 ), whereas you must start again from scratch in the second method. On the other hand, the backward recurrence algorithm is in general more stable. We return to this claim in Section 5.3.1 on page 260. The computations in this book are done by means of the backward recurrence algorithm. Approximants are computed with high precision, and the value of a convergent continued fraction is 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 ∞

K 30 +10.9 n=1

n

(1.4.1)

is known to converge. We want to ﬁnd its value f . If we cannot ﬁnd the exact value, 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) as n 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.

12

Chapter 1: Introductory examples

In fact, we can do even better! The tail values f (n) :=

30 + 0.9n+2 30 + 0.9n+3 30 + 0.9n+1 1 + 1 + 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(r n+1 ) for some c ∈ R and |r| < 1, where o(rn+1 ) is the usual “ little o-notation” deﬁned by un = o(rn+1 )

⇐⇒

lim

n→∞

un = 0. rn+1

That is, we assume that c is chosen such that f (n) − (5 + crn+1 ) = 0. n→∞ rn+1 lim

Since then f (n) (1 + f (n+1) ) = (5 + cr n+1 )(6 + crn+2 ) + o(rn+1 ) = 30 + c(6 + 5r)rn+1 + o(r n+1 ) , we must have r = 0.9 and c(6 + 5r) = 1; i.e., c = 2/21. Therefore the approximants 2 Sn (wn ) with wn = 5 + 21 · 0.9n+1 should be an even better choice than Sn (5), in particular for large n. This is in fact true. We shall return to this idea in Section 5.1.5 on page 232. The table below gives a very clear indication that this is so. The value f is in this case f = 5.085066164924 correctly rounded to 12 decimal places. n 1 2 3 4 35 36 37 38 85 86 87 88

3

Sn (0) 30.9000 0.97139 15.6770 1.85765 5.10127 5.07160 5.09631 5.07571 5.08507 5.08506 5.08507 5.08507

f − Sn (0) −25.8 4.1 −10.6 3.2 −1.6 · 10−2 1.3 · 10−2 −1.1 · 10−2 9.3 · 10−3 −1.8 · 10−6 1.5 · 10−6 −1.2 · 10−6 1.0 · 10−6

f − Sn (5) −6.5 · 10−2 4.8 · 10−2 −3.7 · 10−2 2.7 · 10−2 −3.8 · 10−6 2.8 · 10−6 −2.1 · 10−6 1.6 · 10−6 −2.1 · 10−12 1.6 · 10−12 −1.2 · 10−12 9.0 · 10−13

f − Sn (wn ) 4.4 · 10−4 −3.0 · 10−4 2.0 · 10−4 −1.4 · 10−4 7.3 · 10−10 −4.9 · 10−10 3.3 · 10−10 −2.2 · 10−10 2.1 · 10−18 −1.4 · 10−18 9.7 · 10−19 −6.5 · 10−19

1.1.4 Approximating the value

13

Example 4. For the continued fraction 3 + 1/12 4 + 3/22 3 + 1/32 4 + 3/42 3 + 1/52 1 1 1 1 1 + + + + +· · ·

(1.4.2)

the tails “ look more and more like” 3 4 3 4 1 + 1 + 1 + 1 +· · ·

(1.4.3)

and

4 3 4 3 . (1.4.4) 1 + 1 + 1 + 1 +· · · We take convergence for granted, since it will be obvious later. The value of the continued fraction (1.4.3) is then clear: it is the positive root of the quadratic equation 3 u= , 4 1+ 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 := 1 and 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. In this example the value of the continued fraction is f = 1.1873788917921, correctly rounded. n 1 2 3 4 10 11 12 13 23 24 25

Sn (0) 4.00000000 0.69565217 1.85579937 1.00774785 1.18032950 1.19450818 1.18502153 1.18975231 1.18738866 1.18737564 1.18738215

f − Sn (0) −2.8 · 100 4.9 · 10−1 −6.7 · 10−1 1.8 · 10−1 7.0 · 10−3 −7.1 · 10−3 2.4 · 10−3 −2.4 · 10−3 −9.8 · 10−6 3.3 · 10−6 −3.3 · 10−6

Sn (wn ) 1.33333333 1.18518519 1.20054570 1.18752336 1.18738410 1.18739871 1.18738030 1.18738379 1.18737890 1.18737889 1.18737889

f − Sn (wn ) −1.5 · 10−1 2.2 · 10−3 −1.3 · 10−2 −1.4 · 10−4 −5.2 · 10−6 −2.0 · 10−5 −1.4 · 10−6 −4.9 · 10−6 −7.1 · 10−9 −7.1 · 10−10 −2.0 · 10−9

3 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 advantages of continued fractions. But it requires that {wn } is properly picked. We can easily and choose make “ improper choices”: Take any sequence {βn } from C wn := Sn−1 (βn ) . Then βn = Sn (wn ). This shows that we can make {Sn (wn )} converge to anything we want, or diverge, regardless of the convergence behavior of the continued fraction

14

Chapter 1: Introductory examples

itself. There is no reason for panic, though. This warning looks more serious than it really is. It is one of the wonders of continued fractions that if Sn (0) → f , then Sn (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 a huge family of functions, including some hypergeometric functions and ratios of hypergeometric functions, for which this method of convergence acceleration can be used. (Section 1.3.2 on page 27.) Another matter is that such acceleration is not much use for practical purposes unless one has fast converging, easy to use truncation 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 ) with bn ∈ N, although we also allow b0 = 0. They play an important role in number theory. In this section we shall give a few samples of their power. We ﬁrst note that An and Bn are positive integers and relatively prime since by the determinant formula (1.2.9) (2.1.1) Δn := An−1 Bn − An Bn−1 = (−1)n . A regular continued fraction always converges, and we have some very useful truncation error bounds: $

' Theorem 1.5. A regular continued fraction K(1/bn ) (with bn ∈ N for n ∈ N) converges to a value 0 < f < 1. Moreover, f2 < f4 < f6 < · · · f5 < f3 < f1 ,

(2.1.2)

1 Bn Bn+1

(2.1.3)

and |f − fn | < |fn+1 − fn | =

where the canonical denominators {Bn }∞ n=1 are strictly increasing natural numbers. &

%

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

1.2.1 Regular continued fractions

15

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

which guarantees that

f ≈ fn∗ := 12 (fn + fn+1 )

(2.1.4)

|f − fn∗ | < 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 continued fraction expansion of x. The most natural method is probably the one we use for x := π in the following example.

Example 5. We shall expand π = 3.1415926535897932385 . . . in a regular continued fraction: 1 1 0.14159265358979323 . . . 1 1 =3+ =3+ 7.062513305931045 . . . 1 7+ 1 0.062513305931045 . . . 1 1 1 =3+ =3+ 1 1 7+ 7+ 15 + 15.996594406685 . . . 1 0.996594406685 . . . 1 1 1 1 1 = ··· = 3 + . 7 + 15 + 1 + 292 + 1 +· · ·

π = 3.14159265358979323 . . . = 3 +

Of course, we only get the start of the continued fraction in this way since we in essence started with an approximation to π. 3 The method applied in this example can be described as follows: for a positive number u let u denote the largest integer ≤ u. Starting with a positive number u0 we write 1 u0 = u0 + (u0 − u0 ) = u0 + . 1 u0 − u0 We proceed with u1 :=

1 = u1 + (u1 − u1 ) = u1 + u0 − u0

1 , 1 u1 − u1

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

16

Chapter 1: Introductory examples

Theorem 1.6. To a given x > 0 there exists an essentially unique regular continued fraction b0 + K(1/bn ) which converges to x. This continued fraction 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 1 = Sn (un − un ),

u1 +· · · + un + (un − un )

so x − fn−1 = Sn (un − un ) − Sn−1 (0) → 0 by (1.2.12). The only problem with the uniqueness occurs in the terminating case: the two regular continued fractions b0 +

1 1 1 1 b1 + b2 +· · · + bn + 1

and b0 +

1 1 1 b1 + b2 +· · · + bn + 1

are expansions of the same rational number. The next example demonstrates why expanding a rational number in a regular continued fraction by this method terminates, and why the process is equivalent to the euclidean algorithm. Example 6. The euclidean algorithm ([Eucl56], book 7) attributed to the Greek mathematician Euclid (325 - 265 BC), is a tool to ﬁnd the greatest common divisor of 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 71 1 =1+ , 47 47/24 1 47 =1+ , 24 24/23 1 24 =1+ . 23 23/1 Linking these together we get the (terminating) regular continued fraction expansion of 71/47: 71 1 1 1 =1+ . 47 1 + 1 + 23 2 3 71 . That the process has Its classical approximants are f1 = , f2 = and f3 = 1 2 47 to terminate follows since the left hand side in the euclidean algorithm consists of decreasing positive integers. 3

1.2.2 Best rational approximation

1.2.2

17

Best rational approximation

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

−1 — 1 0

0 3 3 1

1 7 22 7

2 15 333 106

3 1 355 113

4 292 103993 33102

5 1 104348 33215

··· ··· ··· ···

This gives the rational approximations f0 = 3, f3 =

f1 =

22 = 3.142857 . . . , 7

355 = 3.14159292 . . . , 113

333 = 3.14150943 . . . , 106 103993 f4 = = 3.1415926530119 . . . 33102 f2 =

to π. They are actually quite good. Indeed, even for such low order approximants we have |π − f3 | < 3 · 10−7 and |π − f4 | < 6 · 10−10 . 3 It is no coincidence that the approximants are good. Indeed, as the heading of this section indicates, the rational approximants we obtain from regular continued fractions 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

An

p

(2.2.1)

.

f − ≤ f − q Bn Then q ≥ Bn . If q = Bn , then p = An . & Proof :

% Assume that q < Bn . We shall ﬁrst prove that then |qf − p| ≥ |Bn−1 f − An−1 | > |Bn f − An | .

(2.2.2)

Let M and N be such that An M + An−1 N = p , Bn M + Bn−1 N = q .

(2.2.3)

Since the determinant of this 2×2-system of linear equations with unknowns M, N is An Bn−1 − Bn An−1 = (−1)n−1 ,

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 (Bn f − An ) + N (Bn−1 f − An−1 ) .

(2.2.4)

Here the two expressions in parentheses have opposite signs by Theorem 1.5, and M, N also have opposite signs (unless M = 0). Hence |qf − p| = |M (Bn f − An )| + |N (Bn−1 f − An−1 )| , and, since N is an integer = 0, we have |qf − p| ≥ |Bn−1 f − An−1 | .

(2.2.5)

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

An

p

(2.2.6)

.

f − > f − q Bn 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 the same 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. This was ﬁrst proved by H.J.S. Smith ([Smith60]), but the present proof is due to Lagrange ([Lagr98]). The 6 theorem shows that in order to ﬁnd 5 a better rational approximation to f than An /Bn , we have to increase 4 the denominator. Smith also gave a very nice geometric interpretation 3 of this fact. A description of this 2 interpretation can also be found in a paper by Felix Klein ([Klein95]): 1 Let a be a positive irrational number. We shall let this number a be 0 0 1 2 3 4 5 6 7 represented by the ray y = ax from the origin into the ﬁrst quadrant of a cartesian coordinate system. The lattice point (n, m) with integer n and m represents the fraction m/n. Assume there is 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. 7

1.2.2 Best rational approximation

19

The ray y = ax does not pass through any lattice points, but the rubber band will create 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 with increasing distance from the origin. Then m2k−2 m2k m2k+1 m2k−1 <
for all k.

Since there are no nails inside the polygon, these fractions are the best rational approximations to a in our meaning of “ best”, and therefore they are the approximants of 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 ﬁgure corresponds exactly to the ﬁrst approximants 1/1, 3/2, 7/5 of the regular continued √ fraction 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=

1 √ 2 + ( 2 − 1)

(2.2.7)

follow the equalities √

2−1 = =

1 1 1 1 1 √ √ = 2 + 2 + ( 2 − 1) 2 + 2 + 2 + ( 2 − 1) 1 1 1 1 √ , 2 + 2 + 2 + 2 + ( 2 − 1)

and so on, as far out as we want. Since obviously √ 1 1 1 1 √ 2=1+ 2 + 2 +· · · + 2 + 2 + ( 2 − 1) for any length of the row of dots, it seems to be a good idea to take a look at the approximants of the regular continued fraction 1 1 1 1 . 2 + 2 + 2 +· · · + 2 +· · · √ √ Indeed, by Problem 13 with x = 1 − 2 and y = 1 + 2, this continued fraction √ converges to 2. To determine its ﬁrst approximants we can use the recurrence relations An = bn An−1 + An−2 , Bn = bn Bn−1 + Bn−2 to obtain 1+

n bn An Bn

−1 — 1 0

0 1 1 1

1 2 3 2

2 2 7 5

3 2 17 12

4 2 41 29

5 2 99 70

··· ··· ··· ···

20

Chapter 1: Introductory examples

This gives 1 = 1.5, 2 1 1 7 1+ = = 1.4, 2+2 5 1 1 1 17 1+ = = 1.4166 . . . , 2+2+2 12 41 1 1 1 1 = = 1.41379 . . . , 1+ 2+2+2+2 29 1 1 1 1 1 99 1+ = = 1.4142857 . . . , 2+2+2+2+2 70 √ which seems to approach 2 pretty quickly. Already the ﬁfth one, the last one listed, has an error less than .00008. Indeed, by Theorem 1.5 √ f4 = 41/29 < 2 < f5 = 99/70, 1+

or alternatively √ 1 5741 2 ≈ (f4 + f5 ) = ≈ 1.41404, 2 4060

√

2 − 5741 < 1 (f5 − f4 ) = 1 ≈ 2.5 · 10−4

4060 2 4060 √ which agrees with the value 2 = 1.41421356 . . . . 3 This is a good place for a warning: using identities like (2.2.7) to derive a continued fraction as in this example, must always be followed by a check: does the continued fraction really converge to the expected value? The equality √ − 2−1=

1 √ 2 + (− 2 − 1)

will for instance lead to √ 1 1 1 √ , − 2=1+ 2 + 2 +· · · + 2 + (− 2 − 1) √ but the continued fraction (still) converges to 2. More generally, if we start with any 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 ), if this continued fraction converges at all? This question will be resolved in the next chapter. (Deﬁnition 2.9 on page 63, combined with Corollary 2.7 on page 68, gives one possible answer.)

1.2.3 Solving linear diophantine equations

1.2.3

21

Solving linear diophantine equations

For regular continued fractions b0 + K(1/bn ) the determinant in the determinant formula (1.2.9) on page 7 is An−1 Bn − An Bn−1 = (−1)n .

(2.3.1)

This property is essential when terminating regular continued fractions are used to solve linear diophantine equations of the form M x + N y = P,

(2.3.2)

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

Example 9. A shipowner wants to update and upgrade her ﬂeet of ships. After having received bids from diﬀerent shipyards, she made up her mind and placed a contract for her ships at Beakersheep Shipyard where they oﬀered two types of ships, 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 be conﬁdential until further notice. However, an industrial spy ﬁnds information on the 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 is then 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 71 1 1 1 =1+ , 47 1 + 1 + 23

A3 = 71 B3 = 47

A2 = 3 B2 = 2

and thus, by (2.3.1) with n := 3 A3 B2 − A2 B3 = 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

22

Chapter 1: Introductory examples

is a solution of (2.3.3). Actually, these pairs are the only solutions. Since we want non-negative solutions we must have −19.10 · · · = −

1347 898 ≤t≤− = −18.97 . . . . 47 71

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 of Y-type. 3

Of course, to ﬁnd the integer solution x = 5, y = 2 in the example above would only take a few seconds of trial and error. But that is only because we can expect the solution to be relatively small. For large solutions the method illustrated by Example 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 has rotated m/n times around when M has rotated once. This is the idea of grandfather clocks where cogwheels transmit the movement from the shaft of the second hand to 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 takes about 29.53059 nights and days to go from full moon to full moon. The cogwheel for the moon’s movement should therefore have 2 · 29.53059 times as many teeth as the wheel for the hour hand. Therefore m = 5906118 teeth and n = 100000 teeth could be a solution if it was at all possible and practical to make durable cogwheels with so many teeth. A slightly better solution would be to use an extra cogwheel in between, such that 6 984353 5906118 = · 100000 10 10000 describes ratios in the two transmissions. But the number of teeth is still unrealistically high. A better solution still is to ﬁnd a good rational approximation to the rational number 59.06118 (which is also a rational approximation to the true number). And

1.2.5 Musical scales

23

here the regular continued fractions come into play: 1 5906118 1 = 59 + = 59 + 2112 100000 100000 16 + 6118 6118 = 59 +

1 1 16 + 6188 2112

= · · · = 59 +

1

= 59 + 16 +

1 1894 2+ 2112

1 1 1 1 1 1 1 1 1 . 16 + 2 + 1 + 8 + 1 + 2 + 4 + 1 + 6

The choice

1 945 m = 59 + = = 59.0625 n 16 16 gives a more practical solution to the problem. Huygens ([Huyg95]) used this method 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 it was started on the piano. (ii) the consonant ratios 2 : 1 octave 3:2

ﬁfth

4:3

fourth

5:4

major third

6:5

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, simultaneously. In a piano the octave is divided into 12 intervals, so we get the twelve tones 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. If q := fn+1 /fn is constant, this means that f13 f12 f2 f13 =2= · ··· = q 12 f1 f12 f11 f1

24

Chapter 1: Introductory examples

and thus q = 21/12 which is an irrational number! We can therefore not get the consonant ratios we asked for exactly, except for the octave. We only get approximations to the consonant ratios. For instance f8 3 = q 7 = 27/12 ≈ 1.498 · · · ≈ = 1.5 f1 2 f6 4 = q 5 = 25/12 ≈ 1.335 · · · ≈ = 1.333 . . . f1 3 f5 5 = q 4 = 24/12 ≈ 1.260 · · · ≈ = 1.250 f1 4 6 f4 3 3/12 =q =2 ≈ 1.189 · · · ≈ = 1.200 f1 5 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 will there exist integers k such that (21/N )k are good approximations to the consonant ratios? This is a question of rational approximation: we want to choose N such that 21/N is “ almost rational”. In particular we want, say, (21/N )k = 2k/N ≈

3 ; 2

i.e.,

3 k Ln 3/2 ≈ log2 = N 2 Ln 2

for some k, N ∈ N where Ln x = loge x is the natural logarithm. In other words, k/N should be a good rational approximation to the irrational number log2 32 . Therefore we expand log2 32 ≈ 0.5849625007 in a regular continued fraction ([Lore06]). Of course, the regular continued fraction expansion of log2 32 never stops, whereas the regular continued fraction expansion of the rational number 0.5849625007 terminates. However, we only want to use the ﬁrst few terms of the continued fraction anyway, since we do not want to have thousands of tones in one octave. We get log2

3 1 1 1 1 1 1 ≈ 0.5846925007 = . 2 1 + 1 + 2 + 2 + 3 + 1 +· · ·

The approximation 3 1 1 1 3 ≈ = = 0.60000 2 1+1+2 5 says: 5 equally spaced tones in an octave with (f1 , f4 ) as the ﬁfth is reasonably good. Taking one more term of the continued fraction gives log2

log2

1 1 1 1 7 3 ≈ = ≈ 0.58333 . 2 1+1+2+2 12

Hah!! That is exactly what we have on a piano. 12 tones in an octave, and for instance C–G makes up a ﬁfth. What is the next good choice? We let the continued fraction come up with an answer: log2

3 1 1 1 1 1 24 ≈ = ≈ 0.58537 . 2 1+1+2+2+3 41

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

1.3.1 Expansions of functions

1.3 1.3.1

25

Rational approximation to functions Expansions of functions

√ Example 10. In this book . . . shall always denote the principal branch of the √ π square root; that is, − 2 < arg . . . ≤ π2 . For z ∈ D := {z ∈ C; | arg(1 + z)| < π}, the identity √ z √ 1+z−1= 2 + ( 1 + z − 1) leads to the identities √ z z z z √ 1+z−1= , 2 + 2 +· · · + 2 + 2 + ( 1 + z − 1) just as in Example 8. This suggests the continued fraction z z z . 2 + 2 +· · · + 2 +· · ·

(3.1.1)

But the same warning √ still applies: we do not know oﬀ-hand that this continued 1+z − fraction converges to ( √ √ 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) on √ closed 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, z 2 + 4z z z z = , 2+2+2 4z + 8 √ and so on. Hence fn (z) is a rational approximation to the function 1 + z − 1 in D. Strictly speaking, the continued fraction (3.1.1) is not deﬁned for z = 0, but we can make the additional deﬁnition that it has the value 0 for z = 0. 3 f1 (z) =

z , 2

f2 (z) =

2z z z = , 2+2 z+4

1

f3 (z) =

Continued fraction expansions are less known than power series expansions. √ The Taylor series expansion of 1 + z − 1 at 0 is ∞ 1 √ 2 zk . 1+z−1= k k=1

It converges for |z| < 1 and diverges for |z| > 1. Its partial sums are 1 1 1 z, σ2 (z) = z − z 2 , 2 2 8 1 1 3 1 2 σ3 (z) = z − z + z , . . . , 2 8 16 σ1 (z) =

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 we compare 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) and fn (0.96), all correctly rounded in the 4th decimal place: n σn fn

1 .4800 .4800

2 .3648 .3871

3 .4201 .4022

4 .3869 .3996

5 .4092 .4001

6 .3932 .4000

7 .4053 .4000

Or, if we rather list the index n(k) which is the smallest index for which the approximation is correct, rounded to k decimals, for all n ≥ n(k): σn fn

n(3) 23 4

n(6) 129 8

n(12) 424 16

n(20) 849 26

This of course does not prove anything, but it certainly suggests that the continued fraction 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 the approximant, does not always give the correct picture. Essential in the comparison is the resources needed for the computation.) Even more √ ﬂattering for the continued fraction expansion is the choice z = 3 for which 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 the ﬁrst seven continued fraction approximants are listed, correctly rounded in the 4th decimal place. 3

n fn

1 1.5000

2 .8571

3 1.0500

4 .9836

5 1.0055

6 .9982

7 1.0006

Example 12. We use the notation Ln z or Ln(z) to denote the principal branch of the natural logarithm; that is, −π < Im Ln z ≤ π. The continued fraction z z/2 z/6 2z/6 2z/10 3z/10 1 + 1 + 1 + 1 + 1 + 1 +· · ·

(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, z a 2 z a3 z for z ∈ D , Ln(1 + z) = 1 + 1 + 1 +· · ·

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) approximate the value of Ln(1 + z) for z ∈ D. Right now we shall not worry about how one gets this continued fraction expansion or prove its convergence. We shall compare this rational approximation to the polynomial approximation obtained from the power series expansion Ln(1 + z) = z −

z3 z4 z2 + − + ··· 2 3 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 continued fraction approximants are listed in the table below, correctly rounded. The value is Ln(2) = .69314718, correctly rounded. n fn

1 1.000000

2 .666667

3 .700000

4 .692308

5 .693333

6 .693121

7 .693152

In order to get the polynomial approximation with the same accuracy we need n > 105 terms of the power series. Let us also try a z-value where the series does not converge, for instance z = 3. In the 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), and then 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) = 12 ( 1 + z − 1). The approximants are no longer rational, but they give good approximations. For z := 3, w(3) = 12 , and fn∗ := Sn ( 12 ), correctly rounded, is shown in the next line of the table. n fn fn∗

1 3.00000 2.00000

2 1.20000 1.50000

3 1.50000 1.41176

4 1.36363 1.39286

5 1.39726 1.38806

6 1.38367 1.38679

7 1.38744 1.38644

Even better approximants Sn (wn (z)) will be suggested in Chapter 5. 3

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

(a + k) .

k=0

(3.2.1)

28

Chapter 1: Introductory examples

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

(3.2.2)

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

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

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

−

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

Assuming that we avoid zeros in the denominators, this can be written −a(c − b)z c(c + 1) =1+ 2 F1 (a, b + 1; c + 1; z) 2 F1 (a, b + 1; c + 1; z) 2 F1 (a + 1, b + 1; c + 2; z) −(b + 1)(c − a + 1)z (c + 1)(c + 2) 2 F1 (a, b + 1; c + 1; z) =1+ . F (a + 1, b + 1; c + 2; z) 2 F1 (a + 1, b + 1; c + 2; z) 2 1 2 F1 (a + 1, b + 2; c + 3; z) 2 F1 (a, b; c; z)

Observe that the denominator on the right hand side of the ﬁrst equality is equal to the left hand side of the second equality. Furthermore, the denominator on the right hand side of the second equality coincides with the left hand side of the ﬁrst one, if in the former a is replaced by a + 1, b is replaced by b + 1 and c by c + 2 in all places. Hence, we have 2 F1 (a, b; c; z) =1+ F (a, b + 1; c + 1; z) 2 1

a1 z F (a, b + 1; c + 1; z) 2 1 2 F1 (a + 1, b + 1; c + 2; z) a1 z =1+ a2 z 1+ F (a + 1, b + 1; c + 2; z) 2 1 2 F1 (a + 1, b + 2; c + 3; z) a3 z a1 z a2 z =1+ 1 + 1 + 2 F1 (a + 1, b + 2; c + 3; z) 2 F1 (a + 2, b + 2; c + 4; z)

1.3.2 Hypergeometric functions

29

and so on, where −(a + n)(c − b + n) (c + 2n)(c + 2n + 1) −(b + n)(c − a + n) = (c + 2n − 1)(c + 2n)

for n ≥ 0,

a2n+1 = a2n

(3.2.3) for n ≥ 1,

and we get the continued fraction 1+

a2 z a3 z a1 z . 1 + 1 + 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, since an → − 14 , it is a consequence of Theorem 4.13 on page 188 that (3.2.4) converges in the cut plane D := {z ∈ C; 0 < arg(z − 1) < 2π}. That its value indeed is 2 F1 (a, b; c; z) 2 F1 (a, b

+ 1; c + 1; z)

(3.2.5)

will be proved in volume 2. This means that its classical approximants provide rational approximation to this ratio. This continued fraction expansion was developed by Gauss ([Gauss13]). A particularly interesting example occurs for the choice b = 0. Since F (a, 0; c; z) ≡ 1, we get a1 z a2 z a3 z 1 (3.2.6) 2 F1 (a, 1; c + 1; z) = 1 + 1 + 1 + 1 +··· 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 2 F1 (1, 1; 2; z)

= −z −1 Ln(1 − z)

3 1 , 1; ; −z 2 ) = z −1 Arctan z 2 2

z dt 1 1 . z · 2 F1 ( , 1; 1 + ; −z n ) = n n n 0 1+t 2 F1 (

(principal part)

(3.2.8)

Since an → − 14 , the nth tail of (3.2.4) looks more and more like the periodic continued fraction −z/4 −z/4 −z/4 1 + 1 + 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 approximants Sn (wn ).

30

1.4

1.4.1

Chapter 1: Introductory examples

Correspondence between power series and continued fractions From power series to continued fractions

A continued fraction of the form an z a1 z a2 z ; 1 + 1 +· · · + 1 +· · ·

b0 +

0 = an ∈ C

(4.1.1)

is called a regular C-fraction. (This concept has no particular connection to regular continued fractions.) Regular C-fractions often work better than power series as far as speed of convergence and domain of convergence are concerned. Hence it is of 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 primitive way is the method of successive substitutions ([Lamb61]) due to Lambert (1728 1777), a colleague of Euler and Lagrange in Berlin. We illustrate this method by an example:

Example 13. We shall compute the circumference L of the ellipse y2 x2 + = 1, a2 b2

a ≥ b ≥ 0,

a > 0.

(4.1.2)

The well known arc length formula leads to the elliptic integral

π/2

L = 4a 0

where ε := By setting

1 − ε2 sin2 θ dθ

√ a2 − b2 /a is the eccentricity of the ellipse, and thus b2 = a2 (1 − ε2 ). t :=

a − b 2 a+b

,

(4.1.3)

and expanding the integrand in a series and integrate, we get L = π(a + b)

2 ∞ 1/2 n=0

n

t t2 t3 25t4 tn = π(a + b) 1 + 2 + 6 + 8 + 14 + · · · , (4.1.4) 2 2 2 2

([H¨ utte55], [LoWa85]). One way of ﬁnding approximate values for L is to truncate the series. But we can also transform the series into a continued fraction: 1+

t t2 t3 25 t4 + + + + ··· = 1 + 22 26 28 214

t/22 1+

t 24

+

t2 26

+

25 t3 212

+ ···

−1

1.4.1 From power series to continued fractions =1+

=1+

1−

t 16

−

t/4 1 +1 −

t/4 −

3 t2 256

9 t3 2048

+ ···

=1+

31 t/4 1 +

−t/16 1+

3t 16

+

9 t2 128

+ ···

−1

−t/16 −3t/16 t/4 −t/16 =1+ −1 9t2 1 1 + + − 256 + · · · 1 + 3t + · · ·

3t 16

16

−3t/16 t/4 −t/16 . =1+ 1 + 1 + 1 − 3t/16 + · · · That is, we have obtained the ﬁrst fraction terms of a continued fraction expansion 1+

t/4 −t/16 −3t/16 −3t/16 1 + 1 + 1 + 1 +· · ·

(4.1.5)

of L/(π(a + b)). We shall prove in volume 2 that this continued fraction actually converges to L/(π(a + b)). Therefore its classical approximants provide rational approximations to L: L ≈ π(a + b)fn (t) . This approximation is surprisingly accurate, even for moderate n. For n = 2, simple as it is, it has an error less than 3 mm for an ellipse with size and eccentricity as the orbit 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 to be the “ worst” case (t = 1), the exact value of L is 4a. The approximate formulas give ⎧ ⎪ πa f0 (1) = 3.1416 a ⎪ ⎪ ⎪ ⎪ ⎪ ⎨πa f1 (1) = 3.9270 a L ≈ πa f2 (1) = 3.9794 a ⎪ ⎪ ⎪πa f3 (1) = 3.9924 a ⎪ ⎪ ⎪ ⎩πa f (1) = 3.9964 a , 4 correctly rounded in the 4th decimal place. However, we can do better. In the continued fraction (4.1.5) we pretend that all fraction terms from the second one are equal to −t/16 , 1 i.e. we replace the continued fraction (4.1.5) by 1+

t/4 −t/16 −t/16 −t/16 . 1 + 1 + 1 + 1 +· · ·

(This is of course only an experiment, but one can prove that it will lead to a good approximation.) This periodic continued fraction has the value S1 (w) = 1 +

t/4 , 1+w

(4.1.6)

32

Chapter 1: Introductory examples

where w is the value of its tail −t/16 −t/16 −t/16 1 = 1 − t/4 − 1 . (4.1.7) 1 + 1 + 1 +· · · 2 √ 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 w=

S1 (w) = 1 +

1+(

√ t/4 = 3 − 4 − t, 1 − t/4 − 1)/2

we get the formula L ≈ π(a + b)(3 −

√ 4 − t) .

(4.1.8)

This formula was ﬁrst found by Ramanujan ([Bern78]). For an ellipse of size and eccentricity as the orbit of Mercury this formula has an error less than 2 mm. For the degenerate case (“ ﬂat” case) it gives the value 3.9834a. The next formula uses the observation that the continued fraction (4.1.5) has two equal fraction terms −3t/16 . 1 If we pretend that all subsequent ones also are (−3t/16)/1 (again merely an experiment), the ﬁrst tail of (4.1.5) would be 1 −3t/16 −3t/16 −3t/16 −3t/16 w= , 3 1 + 1 + 1 + 1 +· · · which by (4.1.7) has the value w=

1 ( 1 − 3t/4 − 1) . 6

This suggests the approximate formula

t/4 L ≈ π(a + b) 1 + 1 + ( 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]. To our delight it was pointed out by Almquist and Berndt ([AlBe88]) that also this was one of Ramanujan’s formulas, although it was not known how he established it. It may therefore well be assumed that the method shown in the present section was the one he used. 3

1.4.3 Analytic continuation

1.4.2

33

From continued fractions to power series

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

z/2 z/6 2z/6 2z/10 3z/10 z . 1 + 1 + 1 + 1 + 1 + 1 +···

We look at its ﬁrst classical approximants and their power series expansions at 0: f1 (z) = f2 (z) = f3 (z) =

f4 (z) =

z = z + 0z 2 + 0z 3 + 0z 4 + · · · 1

z z3 z2 z4 + =z− − + ··· 1 + z/2 2 4 8

z3 z 2 z2 + − z4 + · · · =z− z/2 2 3 9 1+ 1 + z/6

z3 z4 z2 z + − =z− + ··· z/2 2 3 4 1+ z/6 1+ 1 + z/3

The underlined terms coincide with terms in the power series for Ln(1+z). Observe that 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 correspondence between the power series and the continued fraction expansion at 0. 3

1.4.3

One fraction, two series; analytic continuation

Example 15. The identity z=

z 1−z+z

used repeatedly leads to the identities z=

z z , 1−z +1−z+z

and generally z=

z=

z z z , 1−z +1−z +1−z+z

z z z . 1 − z + 1 − z +· · · + 1 − z + z

Similarly, the identity −1 =

z , 1−z−1

z = 0 ,

34

Chapter 1: Introductory examples

leads to

z z z . 1 − z + 1 − z +· · · + 1 − z − 1 Inspired by these two identities we look at the continued fraction −1 =

z z z . 1 − z + 1 − z +· · · + 1 − z +· · ·

(4.3.1)

For simplicity we assume that z = −1. Then its classical approximants fn (z) can be written f1 (z)

=

f2 (z)

=

f3 (z)

=

z(1 + z) z = , 1−z 1 − z2 z z(1 − z) z z(1 − z 2 ) = = , 1−z +1−z 1 − z + z2 1 + z3 z z z z(1 + z 3 ) , = 1−z +1−z +1−z 1 − z4

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

z(1 − (−z)n ) z + (−z)n+1 = . 1 − (−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 + 0z 2 + 0z 3 + · · · |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 + · · · 3 This example shows that one and the same continued fraction expansion may converge to two diﬀerent functions in two diﬀerent regions and correspond to two diﬀerent series at two diﬀerent points (here 0 and ∞). There also exist non-trivial analogues, where one continued fraction simultaneously represents two diﬀerent analytic functions by convergence and correspondence. But we can say more:

1.4.4 Pad´e approximation

35

1. The approximants Sn (z) in Example 15 are all equal to z. This implies two things: The convergence to z in |z| < 1 is accelerated (bull’s eye, the value is 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| < 1 to the whole plane. 2. The approximants Sn (−1) are all equal to −1. This implies two things: Acceleration of convergence to −1 in |z| > 1 (again bull’s eye), and the convergence to −1 is extended to 0 < |z| ≤ 1, i.e. we have an analytic continuation of the limit 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| < 1 and 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 convergence in (4.3.2) is fast enough, then Sn (z) actually converges in a larger domain, and thus 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

Pad´ e approximation

Problem: For a given (formal) power series L(z) :=

∞

ck z k

k=0

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, when Pm/n and Qm/n are polynomials of degree ≤ m and ≤ n respectively. Solutions to this problem are essentially what is called Pad´e approximants (in one of the two possible deﬁnitions). The idea is due to Stirling ([Stir30]) who probably based his work on a paper by Bernoulli ([Berno78]). For ﬁner distinctions we refer to the extensive literature on Pad´e approximation, such as for instance [BaGM96]. Pad´e approximants for a given series L(z) are usually presented in an array called a Pad´e table:

36

Chapter 1: Introductory examples m \n 0 1 2 3 4 .. .

0 R0/0 R1/0 R2/0 R3/0 R4/0 .. .

1 R0/1 R1/1 R2/1 R3/1 R4/1 .. .

2 R0/2 R1/2 R2/2 R3/2 R4/2 .. .

3 R0/3 R1/3 R2/3 R3/3 R4/3 .. .

4 R0/4 R1/4 R2/4 R3/4 R4/4 .. .

··· ··· ··· ··· ··· ··· .. .

The 0-column {Rn/0 } is just the partial sums of L(z). (Some authors have chosen to use the transposed table where the numerator-degree increases in each row and the denominator-degree increases column-wise.) For practical computation of Pad´e approximants 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 +

∞

(−1)n+1

n=1

z2 z3 z4 z5 z6 zn =1+z− + − + − + ··· . 2n − 1 3 5 7 9 11

The upper left corner of the Pad´e table is then m \n

0

1

2

···

0

1

1 1−z

3 3 − 3z + 4z 2

···

1

1+z

3 + 4z 3+z

15 + 21z 15 + 6z − z 2

···

2

3 + 3z − z 2 3

15 + 24z + 4z 2 15 + 9z

105 + 195z + 64z 2 105 + 90z + 9z 2

···

.. .

.. .

.. .

.. .

..

.

For instance R2/1 (z) =

z2 z3 3z 4 9z 5 15 + 24z + 4z 2 ∼1+z− + − + + ··· 15 + 9z 3 5 25 125

which agrees with L(z) up to and including the term of order 3. It is quite easy to see that Rm/n (z) always agrees with L(z) up to and including at least the term of order m + n. Now, in our particular case L(z) can be written L(z) = 1 + z 2 F1 ( 12 , 1; 32 ; −z)

1.4.4 Pad´e approximation

37

where the hypergeometric series 2 F1 ( 12 , 1; 32 ; −z) has a continued fraction expansion given by (3.2.6) - (3.2.7) on page 29; that is, 12 z 22 z 32 z 42 z 1 1·3 3·5 5·7 7·9 1 3 . 2 F1 ( 2 , 1; 2 ; −z) = 1 + 1 + 1 + 1 + 1 +· · · Therefore L(z) has the continued fraction expansion ∞

1+

K anz n=1 1

where a1 = 1 and an+1 =

n2 −1

for n ≥ 1.

4n2

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 3 + 4z z 13 z = 1+ 1 3+z 15 + 24z + 4z 2 z z/3 4z/15 f3 (z) = 1 + = 1+ 1 + 1 15 + 9z z z/3 4z/15 9z/35 105 + 195z + 64z 2 f4 (z) = 1 + = 1+ 1 + 1 + 1 105 + 90z + 9z 2 f2 (z) = 1 +

and so on. We recognize these approximants from the Pad´e table. The approximants of the regular C-fraction are indeed the staircase diagonal Pad´e 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 R1/0

. ..

1 R1/1 R2/1

. ..

2

R2/2 R3/2

. ..

3

R3/3 R4/3

. ..

4

R4/4 R5/4 . ..

5

6

R5/5 R6/5 . ..

R6/6 . ..

··· ··· ··· ··· ··· ··· ··· ··· .. .

This means that we can • use regular C-fraction expansions to compute Pad´e approximants. • use convergence theory for continued fractions to prove convergence of Pad´e approximants. 3

38

1.5 1.5.1

Chapter 1: Introductory examples

More examples of applications A diﬀerential equation

One can solve certain diﬀerential equations by means of continued fractions. We include a very simple example here. (More substantial examples can for instance be 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 y y

1

=

2+

=

2+

1 , y /y

2+

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

y /y

,

.. . y (n) y (n+1)

=

From this it follows that 1 1 y 1 1 =2+ . (n+1) y 2 + 2 +· · · + 2 + y /y (n+2) n+1

This suggests the continued fraction 2+

1 1 1 . 2 + 2 +· · · + 2 +· · ·

We know from Example 8 on page 19 that it converges to that the diﬀerential equation has a solution y satisfying √ y = 2 + 1, y

√ 2 + 1, which suggests

1.5.2 Moment problems and divergent series

39

that is, √ y = 2 − 1, y 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 continued fraction above. Since the recursion more than the asymptotics is the important part of the investigation so far, we try √ y = 1 − 2. y This leads to

√ y = −( 2 + 1), y

√ and thus y = C1 exp(−( 2 + 1)x). A quick check shows that this also is a solution of 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 diﬀerential equation, since there exists a perfectly good, simple method taught in elementary calculus. But there are cases, where this idea leads to non-trivial results. 3

1.5.2

Moment problems and divergent series

In Example 10 on page 25 and Example 12 on page 26 a series with convergence radius 1 corresponded to a continued fraction which converged to the right value in 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 to sum the divergent series. The eﬀect is even more spectacular in the next example where the power series diverges for all z ∈ C.

Example 18. Take a look at the series L(z) :=

∞ n=0

n!(−z)−n = 1 −

2! 1! 3! + 2 − 3 + ··· . z z z

40

Chapter 1: Introductory examples

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

∞ −t

∞ ze 1 F (z) := dt = dt e−t · z + t 1 − (−t/z) 0 0

∞ (−t)n (−t)n+1 t t2 e−t 1 − + 2 − · · · + + = dt z z zn z n (z + t) 0

∞ 3! (−t)n+1 2! 1! n n! e−t n dt. = 1 − + 2 − 3 + · · · + (−1) n + z z z z z (z + t) 0 This function is holomorphic in the cut plane | arg(z)| < π, and

F (z) − σn (z) =

∞

−t

e 0

(−t)n+1 dt z n (z + t)

for σn (z) =

n

(−1)m

m=0

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

∞ 1 |z|n 0

∞ 1 ≤ n |z| 0

|F (z) − σn (z)| ≤

e−t tn+1 dt Re(z + t) e−t tn+1 dt (n + 1)! = n+1 . |z| cos α |z| cos α

Now, keep n ﬁxed, and let z → ∞. Then the right hand side (and hence the left hand side) tends to 0. This property is expressed as follows: The series L(z) is an asymptotic 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 . Another bridge is by way of continued fractions. It can be shown that the continued fraction 1 1/z 1/z 2/z 2/z 3/z 3/z 4/z 4/z 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +· · · corresponds at ∞ to the series L(z). It converges in the cut plane | arg(z)| < π to an analytic function f (Theorem 3.21 on page 124). Indeed, in volume 2 we shall see that f = F . So again we have given the divergent series a meaning by converting it to a convergent continued fraction. 3

What we have seen here is closely connected to the moment problem, or more precisely 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 := tn dΨ(t) (5.2.1) 0

1.5.2 Moment problems and divergent series

41

is called the nth moment with respect to Ψ. With n

∞

∞ ∞ z dΨ(t) t F (z) := dΨ(t) , = (−1)n z + t z 0 0 n=0

(5.2.2)

termwise integration leads to the series L(z) :=

∞

(−1)n

n=0

cn . zn

(5.2.3)

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

∞ tn dΨ(t) for n = 0, 1, 2, . . . . cn = 0

Any such function Ψ is called a solution of the moment problem. Related tasks are now: 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 to an additive constant. It is the beauty of this theory that the Stieltjes moment problem for {cn } has a solution if and only if the series (5.2.3) corresponds at z = ∞ to a continued fraction of the form ∞ an /z a1 where all an > 0 1 + n=2 1

K

called an S-fraction. The solution is unique if and only if this continued fraction converges 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 by Stieltjes in his famous 1894-paper ([Stie94]).

Example 19. As a consequence of Example 18, the moment problem with the moments cn := n! has a unique solution Ψ(t) with dΨ(t) = e−t dt = Ψ (t)dt. The solution is usually written Ψ(t) := 1 − e−t , since it then increases from 0 to 1 when t increases from 0 to ∞ . 3 Terminology from measure theory is often used in this connection. In Example 19 the measure Ψ is absolutely continuous with derivative e−t . Since

∞

∞ dΨ(t) = e−t dt = 1, 0

it is an example of a probability measure.

0

42

1.5.3

Chapter 1: Introductory examples

Orthogonal polynomials

Example 20. Let us deﬁne the inner product

1 f (x)g(x) dx f, g := −1

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

1 2 δm,n for all m, n ≥ 0, Pm , Pn = Pm (x)Pn (x)dx = 2n + 1 −1 where δm,n is the Kronecker delta; i.e., δm,n = 1 if m = n and 0 otherwise. We say that the Legendre polynomials are orthogonal on [−1, 1] with respect to this inner product. The Legendre polynomials {Pn } are known to be the solution of the recurrence relation (n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x) for n = 1, 2, 3, . . . with initial expressions P0 (x) := 1,

P1 (x) := x,

so for instance P2 (x) =

3 2 1 x − , 2 2

P3 (x) =

5 3 3 x − x, 2 2

P4 (x) =

35 4 15 2 3 x − x + . 8 4 8

An alternative way of writing this recurrence relation is Pn+1 (x) =

2n + 1 n xPn (x) − Pn−1 (x) . n+1 n+1

Therefore Pn (x) are exactly the canonical denominators Bn (x) for the continued fraction 1 −1/2 −2/3 −3/4 . x + 3x/2 + 5x/3 + 7x/4 +· · · 3 Example 21. The Chebyshev polynomials Un (x) of the second kind are the solutions 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 canonical denominators of the continued fraction −1 −1 −1 . 2x + 2x +· · · + 2x +· · ·

1.5.4 Thiele interpolation

43

One can prove that

1 −1

Um (x)Un (x)(1 − x2 )1/2 dx =

π δm,n . 2

We therefore say that {Un (x)} are orthogonal on [−1, 1] with respect to the weight function W (x) := (1 − x2 )1/2 . 3 These are merely two simple cases of a general theory. The connection between orthogonal sequences and continued fractions is the three term recurrence relation: a famous theorem on orthogonal polynomials, commonly but incorrectly called Favard’s theorem, says that {Pn } is a sequence of orthogonal polynomials with respect to some real measure dΨ(t) if and only if {Pn } is a solution of a three term linear recurrence relation Pn (x) = (x − cn )Pn−1 (x) − λn Pn−2 (x) for n = 1, 2, 3, . . . with initial values P−1 (x) := 0, P0 (x) := 1 (or some other positive constant), where λn > 0 and cn ∈ R. On the other hand, a three term recurrence relation of this form gives rise to a continued fraction. The continued fractions in question are the Jacobi continued fractions, brieﬂy called J-fractions λ1 λ2 λn . x − c 1 − x − c2 − · · · − x − cn − · · · This connection can be exploited to ﬁnd asymptotic properties and zero-free regions for {Pn (x)}.

1.5.4

Thiele interpolation

Let f be an unknown function with known values f (zn ) at given distinct points z0 , z1 , z2 . . . in C. We want to ﬁnd f . What we do is successively to ﬁnd constants φm ∈ C such that the functions n

Fn (z) := φ0 +

K z −φzmm−1 m=1

for n = 0, 1, 2, . . .

satisfy Fn (zk ) = f (zk ) for k = 0, 1, . . . n − 1.

(5.4.1)

The idea is then: if the resulting Thiele continued fraction ∞

φ0 +

K z −φznn−1

n=1

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

44

Chapter 1: Introductory examples

Then we have a rational function Fn (z) which interpolates f (z) at the ﬁrst n points zk . There exists an algorithm for computing the numbers φm (inverse diﬀerencies). It fails if not all operations are meaningful, but otherwise it works as follows: φ0 := φ0 [z0 ] := f (z0 ), φk := φk [z0 , z1 , . . . , zk ] :=

φ1 := φ1 [z0 , z1 ] :=

z 1 − z0 , f (z1 ) − f (z0 )

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 z k − z0 zk − z1 zk − zk−1 φ1 + φ2 + · · · + φk zk − zk+1 zk − zn−1 and φk+1 + = 0 for k < n. φk+2 + · · · + φn

Fn (zk ) = φ0 +

(5.4.2)

Example 22. Let z0 := 0, f (z0 ) := 0,

z1 := 4, f (z1 ) := 2,

z2 := 9,

z3 := 25,

f (z2 ) := 3,

f (z3 ) := 5.

In this case we ﬁnd φ0 [z0 ] = 0 and φ1 [z0 , z1 ] =

4−0 = 2, 2−0

φ2 [z0 , z1 , z2 ] =

φ1 [z0 , z2 ] =

9−0 = 3, 3−0

9−4 = 5, 3−2

φ2 [z0 , z1 , z3 ] =

φ3 [z0 , z1 , z2 , z3 ] =

φ1 [z0 , z3 ] =

25 − 0 = 5, 5−0

25 − 4 = 7. 5−2

25 − 9 = 8. 7−5

Therefore 5

z z−4 z−9 2+ 5 + 8 z(31 + z) . = 10(3 + z)

F3 (z) =

4 3 2 1

0

5

10

z

15

20

25

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 values√at 0, 2, 9 and 25. (Another possibility is z.) The ﬁgure shows the graph of F3 (z) for real z > 0. 3

1.5.5 Stable polynomials

1.5.5

45

Stable polynomials

A polynomial Qn (r) := rn + a1 rn−1 + · · · + an

(5.5.1)

is called stable if all zeros lie in the left half plane Re(r) < 0. The concept is important 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 the corresponding characteristic equation Q2 (r) := r2 + 3r + 2 = 0. It has the roots r1 = −1 and r2 = −2. Hence the diﬀerential equation has the general solution y(t) = C1 e−t + C2 e−2t , which means that y(t) → 0 as t → + ∞. This corresponds to the fact that Q2 is a stable polynomial. Similarly, the characteristic equation for the diﬀerential equation ... y + 4¨ y + 6y˙ + 4y = 0 is

Q3 (r) := r3 + 4r2 + 6r + 4 = 0

which has the solutions r1 = −2, r2,3 = −1 ± i, so Q3 is stable. Every solution y(t) of the diﬀerential equation is therefore damped; i.e., y(t) → 0 as t → ∞. The general solution is of course y(t) = C1 e−2t + e−t (C2 cos t + C3 sin t). 3 We want to decide whether a polynomial is stable or not, without having to compute its 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) := a1 r n−1 + a3 r n−3 + a5 r n−5 . . . , and proved that Qn is stable if and only if Pn (r)/Qn (r) can be written as a terminating continued fraction of the form 1 1 1 1 + d1 r + d2 r +· · · + dn r

where all dk > 0.

46

Chapter 1: Introductory examples

The proof can be found in [LoWa92], p 470, along with a result for the complex case. Here we only check the result for Q2 (r) and Q3 (r) in Example 23. Q2 (r) = r2 + 3r + 2,

P2 (r) := 3r,

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

1 1 P2 (r) = , Q2 (r) 1 + 13 r + 32 r

P3 (r) := 4r 2 + 4

1 1 P3 (r) 1 . = Q3 (r) 1 + 14 r + 45 r + 54 r

1.6

Remarks

1. The birth of continued fractions. The euclidean algorithm, the basis for the regular continued fraction, goes back to around 300 B.C., but in a slightly diﬀerent form, as a subtraction algorithm rather than a division algorithm ([Eucl56]). But at 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 in 1613 ([Cata13]), in both cases as approximate values for a square root. But already Fibonacci, around 1200, i.e., long before the renaissance, touched upon the idea of a continued fraction, but an ascending one ([Fibo02]). Of the further development we mention, as we did in Section 1.2.4 of the present chapter, Huygens’ use of continued fractions for designing cogwheels ([Huyg95]). In 1965 Lord Brouncker derived a nonterminating continued fraction for π/4 on request by Wallis who was working on his book ([Wallis85]) at that time. Lord Brouncker neither published his result nor his method, but Wallis writes enough about the ideas for Khrushchev to reconstruct them in his very interesting paper ([Khru06a]). In the same book Wallis also gives recurrence relations for the approximants of a continued fraction. With Euler ([Euler67]) a general theory for continued fractions began to develop, a theory where Gauss’ famous continued fractions for ratios of hypergeometric functions ([Gauss13]) was an early high point, still of great value in modern theory for computing special functions. 2. Additional literature. For those who want to go deeper into the analytic theory of continued fractions we refer to the three standard monographs in the ﬁeld: the classical 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 on Applied and Computational Complex Analysis ([Henr77]) a large portion of volume 2 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 and Thron, contain interesting comments on the historic development of concepts, methods and applications. For the computational aspects of continued fractions we recommend the handbook ([CJPVW7]). To most people (meaning mathematicians) continued fractions are the regular continued fractions in number theory. More information on this side of the continued

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 fraction b0 + 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 a1 a 2 an−1 f˜n := b0 + b1 + b2 +· · · + bn−1 + an are for instance also known from the classical literature. And the sequence {Sn } in our deﬁnition could be replace by s˜n := ϕ−1 n−1 ◦ sn ◦ ϕn ,

S˜n := s˜0 ◦ s˜1 ◦ · · · s˜n = 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 {Sn∗ } from M with the properties that Sn∗ := 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 to the recurrence relation Xn = bn Xn−1 + an Xn−2

for n = 1, 2, 3, . . .

where an = 0. For one thing, {An } and {Bn } are solutions of this recurrence. Moreover, Xn−1 an − = = sn (−Xn /Xn−1 ) Xn−2 bn − 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 ) (N −1) Xn + a(N Xn−2 + · · · + a(1) n Xn−1 + an n Xn−N = 0 (1)

(k)

∞ where an = 0. Let {Bn }∞ n=1−N and {An }n=1−N for k = solutions of this recurrence with initial values ⎞ ⎛ ⎛ (1) (1) (1) A2−N ··· A0 1 0 A1−N ⎜ .. .. .. ⎟ 0 1 ⎟ ⎜ ⎜ ⎜ . . . ⎟=⎜ ⎜. . ⎟ ⎜ (N −1) (N −1) (N −1) ⎠ ⎝ .. .. ⎝ A1−N A2−N · · · A0 0 0 B1−N B2−N ··· B0 (1)

(2)

(N −1)

1, 2, . . . , N − 1 be the ··· ··· ···

⎞ 0 0⎟ ⎟ .. ⎟ .⎠ 1

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

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

nπ , 2 2

Bn = cos

nπ 2 2

for n ≥ 0,

(c) A2n−1 = n , B2n−1 = n , A2n = 2n2 + 1 and B2n = 2n2 for n ≥ 1. 2. Continued fraction identities. Construct a continued fraction K(1/bn ) which converges 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 if b1 = 1 and an + bn = 1 for n ≥ 2. 5. Reversed terminating continued fraction. Let KN n=1 (an /bn ) for some N ∈ N be a terminating continued fraction (with all an = 0), and let bN +

aN −1 a2 aN bN −1 + bN −2 +· · · + b1

be its reversed continued fraction. Prove that the (N − 1)th denominator BN −1 is the same for the two terminating continued fractions. 6. Periodic continued fraction. The periodic continued fraction ∞

K z2 = z2 + z2 + z2 +· · ·

n=1

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 expansion for 5. (c) Let z := 1 − 2i. Compute the ﬁrst classical approximants for K(z/2) and guess its value. 7. Approximants. Compute the ﬁrst three classical approximants and the ﬁrst three approximants Sn (1) of the continued fraction K(n/1) by means of (a) the forward recurrence algorithm, (b) the backwards recurrence algorithm,

Problems

49

(c) the Euler-Minding formula and its generalization. 8. Euclidean algorithm. Use the Euclidean algorithm to ﬁnd the greatest common divisor (a) gcd(119, 221), (b) gcd(3839, 1711), (c) gcd(49907, 22243). 9. Terminating regular continued fraction. Find the regular continued fraction expansion 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 its regular continued fraction expansion and show that it is periodic (possibly after a few 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 (nonM+ Q terminating) regular continued fraction K(1/bn ) has the form f = with N √ 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 an w + bn an b n from M with the matrix Tn := τn (w) := , then τ1 ◦ τ2 correcn w + d n cn dn sponds to the matrix product T1 T2 . 13. ♠ Periodic continued fractions. Prove that fn = −xy

xn − y n − y n+1

xn+1

for the continued fraction −

xy xy xy ; x + y − x + y − x + y −···

x, y = 0,

x = y.

14. Best rational approximation. Find the ﬁrst ﬁve classical approximants of the regular continued fraction expansion of e = 2.718281828459 . . . and compare them to other rational approximations with no larger denominators. 15. Diophantine equation. Find all positive integer solutions (x, y) with y < 20 of the diophantine equation 19x − 13y = 5. 16. Diophantine equations. Find the general solution of the linear diophantine equation (a) 11x − 3y = 5, (b) 99x − 47y = 3, (c) 3839x − 1711y = 1,

(d) 3839x + 1711y = 1.

50

Chapter 1: Introductory examples

17. Fibonacci sequence. (a) Use the identity

√

5−1 = 2

1 √ 5−1 1+ 2 to produce a continued fraction by the procedure of Example√8. Compute the ﬁrst 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, and generally Fn+1 = Fn + Fn−1 for n ≥ 1 . √ (c) Prove that fn → √ ( 5 − 1)/2. (The sequence {Fn } is the sequence of Fibonacci numbers, and ( 5 − 1)/2 ≈ 0.61803 . . . is the golden ratio.) 18. ♠ Regular continued fractions. Let {An } and {Bn } be the canonical numerators and denominators for the regular continued fraction K(1/bn ) as in Theorem 1.7 on page 17. (a) Show that Bn f − An = Bn (f − fn ). (b) Show that |Bn+1 f − An+1 | = −bn+1 |Bn f − An | + |Bn−1 f − 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−1 f − An−1 | > |Bn f − An |. 19. ♠ Divergence of periodic continued fraction. Prove that if the continued fraction a a a 1 + 1 + 1 +· · · converges, then it converges to one of the roots of the equation x = a/(1 + x). Use this 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 2 F1 (a, b; c; z) be the hypergeometric series (3.2.2) on page 28. Prove that 2 F1 (−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 form K(an z/1) for z(1 + z)α . 21. From power series to continued fraction. Use the procedure of Example 13 on page 30 to ﬁnd the ﬁrst terms a1 z a2 z a3 z a4 z 1 + 1 + 1 + 1 +··· of a continued fraction expansion of f (z) := ez − 1.

Problems

51

22. From continued fraction to power series. Use the procedure of Example 14 on page 33 to ﬁnd the ﬁrst terms of the power series expansion at 0 corresponding to the continued fraction z −z/2 z/6 −z/6 1 + 1 + 1 + 1 +· · · 23. ♠ From approximants to continued fraction. Let {fn }∞ n=0 with f0 := 0 and fn = fn−1 for all n be a given sequence of complex numbers. Prove that the continued 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−2

for 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 fraction b0 + K(an /1) if and only if b0 = f0 , fn = fn−1 , fn = fn−2 and a1 = f1 − f0 ,

a2 = −

(fn−3 − fn−2 )(fn−1 − fn ) f1 − f2 and an = − f0 − f2 (fn−3 − fn−1 )(fn−2 − fn )

for n > 2. (Bernoulli [Berno75].) n 25. ♠ From series ∞ to continued fraction. Let σn := k=1 ck be the partial sums of the series k=1 cn where all cn = 0. Show that the continued fraction K(an /bn ) with cn cn−1 + cn a1 := c1 , an := − , b1 := 1, bn := cn−1 cn−1 has canonical numerators An = σn and canonical denominators Bn = 1 and thus classical approximants fn = σn . (Euler ([Euler48]), Stern ([Stern32]), Glaisher ([Glai74]).) n 26. ♠ From inﬁnite product to power ∞series. Let pn := k=0 ak be the partial products of the inﬁnite product k=0 ak where all ak = 0, 1. Show that the continued fraction a0 +

a0 (a1 − 1) a1 (a2 − 1)/(a1 − 1) a2 (a3 − 1)/(a2 − 1) 1 − (a1 a2 − 1)/(a1 − 1) − (a2 a3 − 1)/(a2 − 1) − an−1 (an − 1)/(an−1 − 1) · · · − (an−1 an − 1)/(an−1 − 1) − · · ·

has canonical numerators An = pn and canonical denominators Bn = 1 and thus classical 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).

52

Chapter 1: Introductory examples (a) Show that A 1 , A0 , A3 , A 2 , A5 , A4 , . . .

and

B1 , B0 , B3 , B2 , B5 , B4 , . . .

are the sequences of canonical numerators and denominators for the continued fraction a2 a3 a4 a5 a1 1 + a3 1+a3 1 + a4 1+a5 . 1 + a1 − 1 1 1 + a2 + 1+a + a4 + 1+a +· · · 3

5

(Perron [Perr57], p 7.) (b) Show that the sequence of classical approximants for a1 1 + a3 a2 a3 (1 + a3 )(1 + a5 ) 1 + a2 + 1 + a4 + a4 a5 (1 + a5 )(1 + a7 ) a6 + 1 + +· · ·

1 + a1 −

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 , . . . . n 28. Pad´ e table. Given the power series ∞ n=0 n!z . Find the 3 × 3 upper left entries in the Pad´e table for the series. 29. Diﬀerential equation. Solve the diﬀerential equation y = 2y + 3y by using the “ method” in Subsection 1.5.1. 30. Moments. Let a be an arbitrary positive number, and let Ψ be the distribution function 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). n −n (b) Find the S-fraction corresponding to the series ∞ . (The series m=0 (−1) cn z can 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 and Ψ(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 and 3 at the points 0, 1 and 2 respectively. 33. ♠ Thiele fractions. Find the shortest Thiele fraction F (z) which has the values 1, 2, 3, b at the points 0, 1, 2 and 3 respectively. What can be said about F (z) for diﬀerent 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 + r stable?

Chapter 2

Basics The transformations sn (w) := an /(bn + w) are linear fractional transformations with sn (∞) = 0. Hence Sn := s1 ◦ s2 ◦ · · · ◦ sn are linear fractional transformations with Sn (∞) = Sn−1 (0), and a continued fraction is essentially a sequence of linear fractional transformations {Sn } with Sn (∞) = Sn−1 (0) for all n. It is therefore natural to deﬁne convergence of continued fractions as convergence of {Sn } in some This deﬁnition sense. Traditionally what is required is that {Sn (0)} converges in C. is easy to grasp, but not quite what we are after. In this chapter we shall deﬁne an alternative concept, general convergence, which essentially requires that {Sn } The only snag is that we have to accept converges to a constant function in C. exceptional sequences to be deﬁned. Two important tools in the convergence theory for continued fractions are tail sequences and value sets which both will be deﬁned in this chapter. The interplay between these two gives a deeper understanding of a continued fraction, but also useful methods for proving convergence and deriving truncation error bounds. Indeed, 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 fraction to another. The idea is that the new one shall have some desired properties lacking in 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

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 bijective onto C; that is, τ (C) =C and τ is univalent (one-tomeromorphic mapping of C 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, if then u, v, w and z are four distinct points in C, u−z v−w τ (u) − τ (z) τ (v) − τ (w) · = · τ (u) − τ (w) τ (v) − τ (z) u−w v−z

(1.1.2)

with the natural limiting forms if one or two of the quantities involved are inﬁnite. This property follows by straight forward checking. 2. From (1.1.2) it follows that whenever w1 , w2 , w3 are three distinct points in then there exists a unique τ and γ1 , γ2 , γ3 are three distinct points in C, C from 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 τ is 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 to some τ ∈ M in all of C. 4. If {τn } converges at n ≥ 3 distinct points, but not to a τ ∈ M, then the limit is the same at at least two of the three points. is not a special point for a linear fractional transformation. The point ∞ ∈ C 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 calls which makes no distinction between ∞ and the complex for a representation of C numbers. The Riemann sphere is a perfect choice.

2.1.1 Linear fractional transformations

55

In the complex plane, |w1 − w2 | is the natural way to deﬁne the distance between one way to do it is to two points (the euclidean metric). On the Riemann sphere C measure the euclidean length of the chord connecting the two points. This gives us the chordal distance (or chordal metric) introduced by Ahlfors ([Ahlf53]): ⎧ 2|w1 − ⎪ w2 | for w1 , w2 ∈ C ⎪ ⎪ 2 ⎨ 1 + |w1 | 1 + |w2 |2 m(w1 , w2 ) := 2 (1.1.3) for w1 ∈ C, w2 = ∞ ⎪ ⎪ 1 + |w1 |2 ⎪ ⎩ 0 for w1 = w2 = ∞ . For our purpose the main advantages of this metric are: (The radius of the (i) it is bounded, since m(w1 , w2 ) ≤ 2 for all w1 , w2 ∈ C. Riemann sphere is taken to be 1.) equipped with this metric is compact; that is, every sequence {wn } on C (ii) C has a convergent subsequence. if and only if m(wn , w) (iii) wn → w ∈C → 0. with respect to the (iv) if τn → τ ∈ M, then the convergence is uniform on C chordal metric. (v) the cross ratio (1.1.2) implies that m(τ (u), τ (z)) m(τ (v), τ (w)) m(u, z) m(v, w) · = · m(τ (u), τ (w)) m(τ (v), τ (z)) m(u, 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 cases w = ∞ and w = ∞ to deﬁne the convergence wn → w in this metric. If w = ∞, if and only if |wn − w| → 0. If w = ∞, this characterization fails. then wn → w We use the chordal distance to deﬁne convergence in M, or rather the metric σ(τ1 , τ2 ) := sup m(τ1 (w), τ2 (w)). w∈C

(1.1.5)

Deﬁnition 2.1. A sequence {τn } from M converges to some τ ∈ M if and only if lim σ(τn , τ ) = 0.

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

56

Chapter 2: Basics

Example 1. We shall see three examples of convergence of a sequence {τn } from M given by τn (w) :=

an w + b n , cn w + dn

Δn := an dn − bn cn = 0

for n = 1, 2, 3, . . . .

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

a ad − bc a aw + b = − = cw + d c c(cw + d) c

d for w = − . c

except possibly at the point w† := −d/c. The So τn (w) → a/c for every w ∈ C \ {−d/c} with respect to the chordal metric, convergence is locally uniform in C and τn (wn ) → a/c whenever lim inf m(wn , −d/c) > 0. For wn := −dn /cn we always have τn (wn ) = ∞ = a/c for all n, so there exist sequences {wn } approaching −d/c for which τn (wn ) → a/c. Example 1C: Let an → a = ∞, cn → c = 0, ∞, and (an dn − bn cn ) → 0. τn (w) still seems to converge to a/c, but we have a problem. This time {dn } may have a large number of limit points d, and w ought to stay away from −d/c for all of them. This can be very restrictive and even impossible since the limit points for On the other hand, for every limit point d there exists {dn } may be dense in C. a subsequence {dnk } converging to d, and for this subsequence we are back to the former case: {τnk } converges (locally uniformly with respect to the chordal metric) \ {−d/c}. So there is some extensive converging to a/c going on. To to a/c in C get hold of this convergence we look at τn (wn ) =

an wn + bn an an dn − bn cn = − 2 . cn wn + dn cn cn (wn + dn /cn )

Let wn∗ := −dn /cn from some n on. Then τn (wn ) → a/c whenever wn stays a uniform distance away from wn∗ ; i.e., whenever lim inf m(wn , wn∗ ) > 0. n→∞

3 '

$

Deﬁnition 2.2. A sequence {τn } from M converges generally to a constant such that if and only if there exists a sequence {w† } from C γ∈C n lim τn (wn ) = γ

n→∞

&

whenever

lim inf m(wn , wn† ) > 0 . n→∞

(1.1.6) %

2.1.1 Linear fractional transformations

57

Remarks. 1. We write τn → γ to denote that {τn } converges generally to the constant and {w† } is called an exceptional sequence for{τn } in this case. γ ∈ C, n 2. The exceptional sequence is not unique. If the sequence {τn } of transformations from M converges generally to γ with exceptional sequence {wn† }, then every sequence {wn∗ } with lim m(wn† , wn∗ ) = 0 is exceptional. It is only its asymptotic behavior that matters. are equivalent if and only 3. We say that two sequences {un } and {vn } from C if m(un , vn ) → 0. Thereby the exceptional sequences for {τn } form an equivalence class, and {wn∗ } or {wn† } are just representatives for this equivalence class. 4. If τn → γ with exceptional sequence {wn† }, it is not always clear what happens to τn (wn† ). It may converge to γ or to some other value, or it may diverge. Its behavior depends on how {wn† } is chosen compared to {τn }. to a constant γ, not 5. A sequence {τn } from M can not converge uniformly in C even in the chordal metric. One always has to accept exceptional sequences. This follows since τn (wn ) = μ = γ for all n whenever wn := τn−1 (μ). Indeed, it is more of a mystery that we can get away with only one equivalence class of exceptional sequences. We shall return to this question in Section 2.1.3 on page 60. Condition (1.1.6) is not always so easy to check since it requires that we know an exceptional sequence {wn† }. The following equivalent deﬁnition makes checking easier: '

$

Deﬁnition 2.3. A sequence {τn } from M converges generally to a constant such if and only if there exist two sequences {vn } and {wn } from C γ∈C that (1.1.7) lim inf m(vn , wn ) > 0 n→∞

and lim τn (vn ) = lim τn (wn ) = γ .

n→∞

&

n→∞

(1.1.8) %

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

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 exceptional sequence {wn† } 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) := τn−1 (γ) and vn if m(vn , γ (n) ) > m(wn , γ (n) ), pn := wn otherwise . Then lim inf m(pn , γ (n) ) > 0 and γ = τn (pn ) → γ. Let wn† := τn−1 (μ) for some with lim inf m(un , w† ) > 0. μ = γ, and let {un } be any sequence from C μ ∈ C, n We shall prove that lim τn (un ) = γ. Since we already know that τn (pn ) → γ and τn (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 := wn† = τn−1 (μ): m(γ (n) , un ) m(wn† , pn ) m(γ, τn (un )) m(μ, τn (pn )) . · = · m(γ, τn (pn )) m(μ, τn (un )) m(γ (n) , pn ) m(wn† , 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 linear fractional transformations, but of course, for τn ∈ M we know more: (i) if {τn (w)} converges at three distinct points to distinct values, then {τn } uniformly with respect to the chordal metric. converges to a τ ∈ M on C, (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 one single point. Here (ii) is fairly trivial, but it may also be very restrictive. General convergence generalizes (ii) by allowing w to vary with n. Thereby we only have to stay away from one single point for each n, and we get convergence to the constant in “all of except at one point for each n”. Therefore there are only two possibilities for a C convergent sequence {τn } from M: either (i) {τn } converges to some (non-singular) τ ∈ M, and we write τn → τ as n → ∞, with exceptional sequence (ii) or {τn } converges generally to some constant γ ∈ C † {wn }. Then τn (wn ) → γ whenever lim inf m(wn , wn† ) > 0, and we write τn → γ as n → ∞. In all other cases we say that {τn } diverges.

2.1.2 Convergence of continued fractions

2.1.2

59

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 . Incidentally, 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 the continued fraction K(an /1); that is, ∞

1 1 2 1 1 2 an := . K n=1 1 1 + 1 − 1 + 1 + 1 − 1 +··· By the recurrence relations (1.2.7) on page 6 and induction, we ﬁnd that A3n−2 B3n−2

= 2n , = 2n+1 − 3 ,

A3n−1 B3n−1 1 2

and so lim S3n−2 (0) = lim S3n−1 (0) = diverges. On the other hand Sn (wn ) =

= =

2n , A3n 2n+1 − 2 , B3n

= 0, = 1,

and lim S3n (0) = 0, and thus {Sn (0)}

An−1 wn + An , Bn−1 wn + Bn

so limn→∞ S3n−2 (w3n−2 ) = limn→∞ S3n−1 (w3n−1 ) = limn→∞ S3n (w3n ) =

1 2 1 2

whenever {w3n−2 } is bounded whenever {w3n−1 } is bounded away from − 1

1 2

whenever {w3n } is bounded away from 0

† † † := ∞, w3n−1 := −1 and w3n := 0 for all n ≥ 1, we have Indeed, with w3n−2

lim Sn (wn ) =

n→∞

1 whenever lim inf m(wn , wn† ) > 0. 2

Therefore, {Sn } converges generally to

1 2

(1.2.2)

with exceptional sequence {wn† }. 3

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

60

Chapter 2: Basics

Deﬁnition 2.4. The continued fraction K(an /bn ) converges generally to with exceptional sequence {w† } if and only if {Sn } converges generally f ∈C n to f with exceptional sequence {wn† }.

Properties of general convergence: 1. The classical convergence to f implies general convergence to f . Hence the new concept includes the classical one, and picks up additional cases, for instance the one in Example 2. 2. The deﬁnitions of general convergence add some insight to the classical concept of convergence. In particular it makes sense to talk about exceptional sequences {wn† } for a (classically) convergent continued fraction. 3. If K(an /bn ) fails to converge generally, we say that K(an /bn ) diverges generally. General convergence has also another important advantage: it is sometimes much easier to prove! (This will be evident in Chapter 4.) We still use the classical deﬁnition of convergence alongside with general convergence, mainly because the Sn (0)–deﬁnition is so well established, but also because it has some merits of its own. For instance: it is simple and uniquely deﬁned, and it is important in many applications. To distinguish between the two, the Sn (0)–convergence will be referred to as just convergence or classical convergence, whereas the convergence in Deﬁnition 2.4 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 all n for the choice wn := Sn−1 (q). That is, {Sn−1 (q)} is an exceptional sequence for K(an /bn ). This is true for every q = f . Hence all these sequences {Sn−1 (q)} belong to the same equivalence class. (See Remark 3 on page 57.) Now, {Sn−1 } is also a sequence of linear fractional transformations. Since m(Sn−1 (q1 ), Sn−1 (q2 )) → 0 whenever q1 = f and q2 = f , no subsequence of {Sn−1 } can converge to a transformation from M. Following Jacobsen and Thron ([JaTh87]) we say that {Sn−1 } is a restrained sequence. Note that {Sn−1 } need not be generally convergent since {Sn−1 (q)} with q = f may have more than one limit point. But every subsequence of {Sn−1 } must have a generally convergent subsequence.

2.1.3 Restrained sequences

61 $

' Deﬁnition 2.5. A sequence {τn } from M is restrained if and only if there such that whenever exists a sequence {wn† } from C lim inf m(vn , wn† ) > 0

and

lim inf m(wn , wn† ) > 0,

lim m(τn (vn ), τn (wn )) = 0.

&

n→∞

(1.3.1) (1.3.2) %

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

Deﬁnition 2.7. A sequence {τn } from M is restrained if and only if no subsequence of {τn } converges to some τ ∈ M.

You are asked to prove that these three deﬁnitions are equivalent in Problem 3 on page 91. Deﬁnition 2.5 shows that {τn } is restrained if and only if the asymptotic behavior of {τn (wn )} is independent of {wn } in the (1.3.2)-sense whenever lim inf m(wn , wn† ) > 0. That is, all sequences {τn (wn )} with lim inf m(wn , wn† ) > 0 are equivalent (in the sense of Remark 3 on page 57). A sequence from this equivalence class is called a generic sequence for {τn }. We evidently have:

Theorem 2.1. Let the sequence {τn } ⊆ M be restrained with generic sequence {zn } and exceptional sequence {wn† }. Then {τn−1 } is restrained with generic sequence {wn† } and exceptional sequence {zn }.

Remarks. 1. If {τn } converges generally to some constant, then {τn } is in particular restrained. 2. If {τn } converges generally to f with exceptional sequence {wn† }, then the constant sequence {f } is a generic sequence for {τn }, and {τn−1 (q)} is an exceptional sequence for every q = f . Therefore {τn−1 (q)} is generic for {τn−1 } when q = f and {f } is exceptional for {τn−1 }.

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−1 wn + An Bn−1 wn + Bn

where A3n−2 = A3n−1 = 2n , A3n = 0, B3n−2 = 2n+1 − 3, B3n−1 = 2n+1 − 2 and B3n = 1 for all n. We showed that {Sn } converges generally to 12 with exceptional sequence {wn† } given by † = ∞, w3n−2

† † w3n−1 = −1 and w3n =0

for all n.

Every sequence {zn } given by zn := Sn (wn ) with lim inf m(wn , wn† ) > 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, {Sn−1 } should therefore be restrained with exceptional sequence { 12 } and generic sequence {wn† }. Let us check this out. Simple computation shows that Sn−1 (w) = − so

Bn w − An , Bn−1 w − An−1

(2n+1 − 2)wn − 2n → −1, (2n+1 − 3)wn − 2n wn − 0 −1 S3n (wn ) = − n+1 → 0, (2 − 2)wn − 2n (2n+2 − 3)wn − 2n+1 −1 S3n+1 →∞ (wn ) = − wn − 0 −1 (wn ) = − S3n−1

whenever lim inf d(wn , 12 ) > 0. 3 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. To determine whether {τn } is restrained or not, the following observation may be of help.

Theorem 2.2. Let the sequence {τn } from M converge to a transformation −1 τ ∈ M. Then σn := τn−1 ◦ τn converges to the identity transformation I(w) ≡ w in C.

2.1.4 Tail sequences Proof :

2.1.4

63

−1 σn = τn−1 ◦ τn where τn → τ and thus τn−1 → τ −1 .

Tail sequences

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

$

the sequence Deﬁnition 2.9. For every t ∈ C, tn := Sn−1 (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 continued 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 a representative 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 sequence for its N th tail. More generally, if {tn } is a tail sequence for {Sn }, then the subsequence {tnk } is a tail sequence for {Snk }.

5. If {tn } and {t˜n } are two tail sequences for K(an /bn ) with tk = t˜k for one index k, then tn = t˜n for all n by (1.4.2) since sn , s−1 n ∈ M are univalent functions. $ ' Theorem 2.3. Let {tn } be a tail sequence for K(an /bn ). Then −1 −1 tn = Sn−1 (t0 ) = s−1 n ◦ sn−1 ◦ · · · ◦ s1 (t0 ) ! " an−1 a2 a1 an = − bn + for n ≥ 1 . bn−1 + bn−2 +· · · + b1 + (−t0 )

&

(1.4.3) %

64

Chapter 2: Basics

Proof :

The result follows since s−1 k (w)

! " ak ak = − bk + = −bk + w (−w)

for k ≥ 1 .

(1.4.4)

$

' Theorem 2.4. Let K(an /bn ) have three distinct tail sequences {tn }, {t˜n } and {t∗n } where lim m(t˜n , t∗n ) = 0

and

lim inf m(tn , t∗n ) > 0.

Then K(an /bn ) converges generally to t0 with exceptional sequence {t˜n }. &

%

Proof : Since lim m(t˜n , t∗n ) = lim m(Sn−1 (t˜0 ), Sn−1 (t∗0 )) = 0 where t˜0 = t∗0 , it follows that {Sn−1 } is restrained with generic sequence {t˜n }. On the other hand we know that lim inf m(tn , t∗n ) = lim inf m(Sn−1 (t0 ), Sn−1 (t∗0 )) > 0. Therefore the constant sequence {t0 } is exceptional for {Sn−1 }. Hence {Sn } is restrained with generic sequence {t0 } and exceptional sequence {t˜n } (Theorem 2.1 on page 61). That is, K(an /bn ) converges generally to t0 . The tail sequence {ζn } deﬁned by ζn := Sn−1 (∞) 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 an an−1 a2 = −bn − Bn−1 bn−1 + bn−2 +· · · + b1

(1.4.6)

for n ≥ 2. In earlier work one has rather used the notation hn := −Sn−1 (∞),

(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 exceptional sequence which often shows up in useful formulas.

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

K a1n = 12 + 11 − 11 + 21 + 11 − 11 + 21 + . . .

n=1

2.1.5 Tail sequences and recurrence relations

65

converges generally to f = 1/2 with exceptional sequence wn† :

0, ∞, −1, 0, ∞, −1, 0, . . .

(starting with w0† = 0). Every tail sequence {tn } with t0 = asymptotic behavior; i.e., lim t3n = 0,

lim t3n+1 = ∞,

1 2

must have the same

lim t3n+2 = −1.

The tail values {f (n) } is the tail sequence starting with t0 = 12 : f (0) =

1 , 2

2 = 3, 1/2 2 1 (1) = s−1 ) = −1 + = − , 2 (f 3 3 1 −1 −1 (2) = , = s3 (f ) = −1 + −2/3 2

(0) ) = −1 + f (1) = s−1 1 (f

f (2) f (3)

and so on. That is, {f (n) } is periodic with period 3, as it just has to be. In Problem 4 on page 92 you are asked to prove that {wn† } as given above and {f (n) } are the only periodic tail sequences for K(an /1). 3

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 = bn Xn−1 + an Xn−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 a solution 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., not and {tn }∞ is a all Xn = 0). Then tn := −Xn /Xn−1 is well deﬁned in C n=0 tail sequence for K(an /bn ).

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

Xn an = −bn + , Xn−1 −Xn−1 /Xn−2

66

Chapter 2: Basics

that is, tn = −bn + an /tn−1 = s−1 n (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 solutions of (1.5.1) in terms of tail sequences. We shall only do so for the particular solutions {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 it is not diﬃcult to extend the results to general solutions {Xn } of (1.5.1), since it turns out that every solution {Xn } is a linear combination of {An } and {Bn }. The formulas 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 history on 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−1 tn =

n

(bk + tk ) ,

(1.5.3)

k=1

An − Bn t0 =

n

(−tk ) ,

(1.5.4)

k=0

An = t0

n k

(bj + tj )

k=1 j=1

Bn =

n k

(bj + tj )

k=0 j=1

n

(−tj ) ,

(1.5.5)

j=k+1 n

(−tj ) ,

(1.5.6)

j=k+1

An = t0 /Σn Bn n k bj + tj Pk and Pk := , where Σn := −tj j=1 t0 −

(1.5.7) (1.5.8)

k=0

t0 − Sn (w) = &

t0 (w − tn ) . wΣn−1 − tn Σn

(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−1 t0 = 1 and B1 + B0 t1 = b1 + t1 . Since an = tn−1 (bn + tn )

2.1.5 Tail sequences and recurrence relations

67

by (1.4.2), and thus, by the recurrence relation, Bn + Bn−1 tn = bn Bn−1 + an Bn−2 + Bn−1 tn = (bn + tn )Bn−1 + tn−1 (bn + tn )Bn−2 = (bn + tn )(Bn−1 + Bn−2 tn−1 ) , equality (1.5.3) follows by induction. To prove (1.5.4) we again use induction. Since A−1 − B−1 t0 = 1, A0 − B0 t0 = −t0 and am = tm−1 (bm + tm ), we have Am − Bm t0 = bm Am−1 + am Am−2 − t0 (bm Bm−1 + am Bm−2 ) = bm (Am−1 − Bm−1 t0 ) + tm−1 (bm + tm )(Am−2 − Bm−2 t0 ) . Hence, if (1.5.4) holds for n := m − 1 and n := m − 2, then

A m − B m t0 = b m = bm

m−1

m−2

k=0

k=0

(−tk ) + tm−1 (bm + tm )

m−1

m−1

k=0

k=0

(−tk ) − (bm + tm )

(−tk )

(−tk ) =

m

(−tk ) .

k=0

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

Bn =

(bk + tk ) − tn Bn−1

k=1

=

n

(bk + tk ) − tn

k=1

=

n k=1

(bk + tk ) − tn

n−1

(bk + tk ) + tn tn−1 Bn−2 = · · · =

k=1 n−1

n−2

k=1

k=1

(bk + tk ) + tn tn−1

(bk + tk ) − · · · +

n

(−tk )

k=1

which is exactly the expression for Bn . The expression (1.5.5) follows since by (1) (1) Lemma 1.4 on page 10, An = a1 Bn−1 where Bk denotes the canonical denominators 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 convergence and estimate the speed of this convergence. For instance, the following corollary follows easily:

68

Chapter 2: Basics $

' Corollary 2.7. (Waadeland’s Tail Theorem.) Let {tn }∞ n=0 be a tail sequence for K(an /bn ) with all tn = ∞. Then K(an /bn ) converges in the classical sense if and only if {Σn } given by (1.5.8) converges in C. then K(an /bn ) converges to f := t0 1− 1 If {Σn } converges to Σ∞ ∈ C, , Σ∞ 1 1 − and f − fn = t0 . Σn Σ∞ &

%

Example 5. For given b, t ∈ C with t = 0, −b and b = 0, the periodic continued fraction t(b + t) t(b + t) t(b + t) b + b + b +· · · has a tail sequence {tn } with all tn := t. Therefore, by (1.5.7) b+t t 1+ An t t = n = t − fn = t − n+1 k b+t Bn b+t 1− −t −t

(1.5.10)

k=0

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. If in particular (b + t)/t is an N th root of unity; i.e., ((b + t)/(−t))N = 1 for some N ∈ N, then {t − fn } is periodic. Otherwise {((b + t)/(−t))n+1 } is a dense sequence of distinct points on the unit circle. 3

Formulas (1.5.3) and (1.5.6) imply that n (bk + tk ) Bn Bn + Bn−1 tn =− + tn = tn − k=1 Bn−1 Bn−1 Bn−1 n (b + t ) P t k k n n = tn − n−1 k k=1 = tn + n−1 Σn−1 (bj + tj ) (−tj )

ζn = −

k=0 j=1

(1.5.11)

j=k+1

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

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 T n = tn

(t0 − T0 )Σn − t0 (t0 − T0 )Σn−1 − t0

where Σn :=

k=0

& Proof :

Since

k n bj + tj . −tj j=1

%

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

Tn = Sn−1 (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 fractions K(an /bn ) for which {tn } is a tail sequence. The only requirement is that tn−1 (bn + tn ) = an . then this continued If {bn } is chosen such that {Σn } given by (1.5.8) converges in C, fraction converges. Its value is t0 if Σ∞ = ∞ and t0 (1−1/Σ∞ ) otherwise (Corollary 2.7). 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 k j+2 1 + t2j−1 1 + t2j k+2 = 2k = 2k−1 (k + 2) · = 2· −t −t j + 1 2 2j−1 2j j=1 j=1

P2k−1 = P2k−2 ·

1 + t2k−1 = −2 P2k−2 = −2k−1 (k + 1) , −t2k−1

it follows that Σ2n = 1 +

n

(P2k−1 + P2k ) = 1 +

k=1

n

2k−1 = 1 +

k=1

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

n

1 − 2n = 2n → ∞, 1−2

70

Chapter 2: Basics

Therefore

∞

K a1n := 21 + 13 + 14 + 14 + 61 + 15 + 18 + 16 + 101 + 71 +· · · = 1 .

n=1

3

2.1.6

Value sets

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

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

an ⊆ Vn−1 bn + Vn

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 for K(an /bn ). If {Vn } is 2-periodic, so that all V2n = V0 and all V2n+1 = V1 , we say that (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 sets is reserved for prevalue sets with the property an /bn ∈ Vn−1 for all n ≥ 1. This is a natural choice when convergence is based on the asymptotic behavior of classical approximants. Our wider view of approximants Sn (wn ) and convergence calls for the 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-empty sets, and thus the nestedness (1.6.2) implies that the limit set K := lim Kn :=

∞ #

Kn

(1.6.4)

n=1

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

2.1.6 Value sets

71

and Sn (wn ) → f whenever wn ∈ Vn • K consists of just one point, say f ∈ C, 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 value sets 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 ∈ Sn−1 (K) ⊆ Vn for every tail sequence {tn } starting with a t0 ∈ K, \ Vk for some k ∈ N ∪ {0} satisﬁes and every tail sequence {tn } with tk ∈ C tn ∈ C \ Vn for all n ≥ k. &

%

Proof : Let ﬁrst t0 ∈ K. Then tn = Sn−1 (t0 ) ∈ Sn−1 (K) ⊆ Sn−1 (Kn ) = Vn . Next, \ Vk . Since tk = sk+1 (tk+1 ) ∈ Vk whereas sk+1 (Vk+1 ) ⊆ Vk , it follows that let tk ∈ C \ Vk+1 . By induction it follows therefore that tn ∈ C \ Vn for all n ≥ k. tk+1 ∈ C '

$

Corollary 2.10. Let {Vn } be a sequence of closed value sets for the generally 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 for some k ∈ N ∪ {0}, then K(an /bn ) has an exceptional over Vk = C \ Vn for all n ≥ k. sequence {wn† } with wn† ∈ C &

%

Proof : We ﬁrst prove that the value f of K(an /bn ) belongs to K. Let {wn† } be an exceptional sequence for K(an /bn ). Under our conditions there exists a subsequence {Vnk } of {Vn } with diamm (Vnk ) ≥ d for some d > 0. Therefore there exists a wnk ∈ Vnk with m(wnk , wn† k ) ≥ d/2 for each k ∈ N. Hence Snk (wnk ) → f ,

72

Chapter 2: Basics

and the result follows since Snk (wnk ) ∈ Knk → K Hence f (n) ∈ Vn for all n by Theorem 2.9. and tk ∈ C \ Vk . Then tk = f (k) , and the corresponding tail sequence Let Vk = C \ Vn for all n ≥ k by Theorem 2.9. {tn } is an exceptional sequence with tn ∈ C Remark. In the particular case where all bn = b = 0 and all Vn = V , then sn (V ) :=

an ⊆V b+V

⇐⇒ s−1 n (−b − V ) = −b +

an ⊆ −b − V, −b − V

so {wn† } 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 we replace the condition on diamm (Vn ) with the condition 0 ∈ Vn for inﬁnitely many n. Another question is: how can we ﬁnd value sets for a given continued fraction?

Example 7. Let the real interval V := [− 12 , 12 ] be a simple value set for K(an /1); that is, an ⊆ V for all n. 1+V 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 [ 2a3n , 2an ] ⊆ [− 12 , 12 ] or an < 0 and [2an , 2a3n ] ⊆ [− 12 , 12 ]; i.e., if and only if an ∈ [− 14 , 14 ]. 3 In this example we started with the value set, and found suﬃcient conditions on a continued fraction for V to be its simple value set. This is the easy way. To ﬁnd value sets for a given continued fraction is harder:

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

K a1n := 21 + 11 − 11 + 12 + 11 − 11 + 21 + . . .

n=1

converges generally and has tail values f (3n) =

1 , 2

f (3n+1) = 3,

2 f (3n+2) = − . 3

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

2.1.7 Element sets

73

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

$ 1

1 %

V3n = w ∈ C; w − < 2

6%

$

V3n+1 = w ∈ C; w − 3 < 1

$ 2

1%

V3n+2 = w ∈ C; w + < . 3 12 We shall return to this idea in Section 5.2 on page 238. 3 Now, K(an /bn ) does not necessarily converge generally, even if it is ε-contractive with respect to V ; i.e., if there exists an ε > 0 such that sn (V ) ⊆ V , but omits a chordal ε-disk m(w, γn ) ≤ ε} (1.6.7) Bm (γn , ε) := {w ∈ C; with some center γn ∈ V for all n ∈ N. (The parabola theorem on page 151 constitutes 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 ) is ε-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 := Sn−1 ◦ Snk+1 = snk +1 ◦ k · · · ◦ snk+1 converges uniformly to I(w) ≡ w (as in Theorem 2.2 on page 62). This is 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 for K(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 from the 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 a sequence {Ωn } (or {En } or {Gn }) of element sets, we say that {Vn } is a sequence

74

Chapter 2: Basics

of value sets for {Ωn } (or {En } or {Gn }). To a given sequence of value sets, there exists 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 ) ) corresponding to {Vn }. &

%

These element sets characterize all continued fractions for which {Vn } is a sequence of 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)

Gn := {b ∈ C; 1/(b + Vn ) ⊆ Vn−1 }

for n = 1, 2, 3, . . .

(1.7.3)

and 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 ). Of course we use the same vocabulary to describe the element sets for K(an /1) and K(1/bn ). For instance, the set E := [− 14 , 14 ] in Example 7 on page 72 is a simple element set corresponding to the simple value set V := [− 12 , 12 ].

Example 9. Let V := D, the closure of the unit disk D := {w ∈ C; |w| < 1}. Then a/(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 every continued fraction K(an /bn ) from Ω. On page 129 we shall prove that every continued fraction from Ω converges. 3

Example 10. Let V0 := B(1, 3)

\ B(0, 2)◦ and V1 := C

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)◦ \ B(− 2 , 3 )◦ , and thus is its interior. Then 1 + V0 = B(2, 3), 1/(1 + V0 ) = C 5 5 3|a| 2a ◦ \ B(− , a/(1 + V0 ) = C ) . Therefore a/(1 + V0 ) ⊆ V1 if and only if 5

5

2a

− − 0 + 2 ≤ 3|a| ;

5 5

i.e., |a| ≥ 10.

Similarly, 1+V1 = C\B(1, 2)◦ , 1/(1+V1 ) = B(− 13 , 23 ), and a/(1+V1 ) = B(− a3 , which is contained in V0 if and only if

2|a|

a

≤ 3;

− − 1 + 3 3

2|a| 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 ). 3 To a given sequence of element sets there also exists an extreme sequence of value sets: '

$

2 Theorem 2.12. For a given sequence {Ωn }∞ n=1 with Ωn ⊆ C , let for be the set of nth tail values for every generally each n ∈ N ∪ {0}, Vn ∈ C convergent continued fraction from {Ωn }. Then {Vn } is a sequence of value sets for {Ωn }. & Proof :

% Let (a, b) ∈ Ωn with a = 0 and f (n) ∈ Vn be arbitrarily chosen. Then f (n) =

an+1 an+2 an+3 bn+1 + bn+2 + bn+3 +· · ·

for some generally convergent continued fraction K(an /bn ) from {Ωn }, and thus also a a an+1 an+2 an+3 f (n−1) := = (n) b + bn+1 + bn+2 + bn+3 +· · · b+f 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 called the sequence of limit sets for {Ωn }, and we use the terms simple limit set and twin limit sets with obvious interpretation.

76

Chapter 2: Basics

More importantly: when does K(an /bn ) from {Ωn } converge? This will be the topic of Chapter 3. The following slight generalization of a result from [Jaco86b] gives a simple answer: '

$

Theorem 2.13. Let (E1 , E2 ) be twin element sets corresponding to the twin value sets (V0 , V1 ), and let E1∗ , E2∗ be closed, bounded subsets of E1◦ and E2◦ respectively. Then every continued fraction K(an /1) from (E1∗ , E2∗ ) converges. &

%

The proof is postponed to Chapter 4. The last example in this section indicates how the knowledge of corresponding element and value sets can be used to estimate continued fraction values when we already know convergence.

Example 11. We shall see later (Worpitzky’s theorem on page 135) that the continued fraction ∞ an −1/4 1/8 a3 a4 := (1.7.4) n=1 1 1 + 1 + 1 + 1 +· · ·

K

with − 14 ≤ an ≤ K(an /1)?

1 4

for all n converges. What can be said about the value f of

Since the interval V := [− 12 , 12 ] is a simple value set for K(an /1) (Example 7), it follows that the value f (2) of the second tail a3 a4 an 1 + 1 +· · · + 1 +· · · lies in this interval. Hence f ∈ S2 (V ) where S2 (w) =

Now, V + thus

9 8

−1/4 1 1+w 1 =− · 9 =− 1/8 4 8 +w 4 1+ 1+w

1−

1/8 w + 98

.

9 8 8 1 9 1 1 = [ 58 , 13 8 ], so 1/(V + 8 ) = [ 13 , 5 ] and − 8 /(V + 8 ) = [− 5 , − 13 ], and

S2 (V ) = −

& ' & ' 1 1 1 3 1 1− , 1− = − ,− . 4 5 13 13 5

That is, −0.23 · · · < − 3

3 1 ≤ f ≤ − = −0.20 . . . . 13 5

2.2.2 Equivalence transformations

2.2

77

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 than the original one.

2.2.2

Equivalence transformations

Deﬁnition 2.12. We say that two continued fractions are equivalent if they have 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 equivalent. 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 continued fractions 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 sequence {rn } of complex numbers with r0 = 1, rn = 0 for all n ∈ N, such that (2.2.1) cn = rn rn−1 an , dn = rn bn for all n ∈ N . &

%

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 the canonical numerators Cn and denominators Dn of K(cn /dn ) can be written C−1 = 1,

D−1 = 0,

Cn = An

n

rk ,

k=0

D n = Bn

n

rk

(2.2.2)

k=0

for all n. Since D0 = B0 = 1 we need r0 = 1. From Theorem 1.2 on page 8 it follows 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 denominators Bn , then K(cn /dn ) has canonical numerators Cn and denominators Dn given by (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 (rn w) for n = 0, 1, 2, . . . ,

(2.2.3)

where Sn (w) are approximants of K(an /bn ), and Tn (w) are approximants of K(cn /dn ). 3. If {tn } is a tail sequence for K(an /bn ), then {tn rn } is a tail sequence for K(cn /dn ).

Example 12. The continued fraction ∞

an z z z/2 z/6 2z/6 2z/10 3z/10 3z/14 4z/14 K := , n=1 1 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +· · · where a1 := 1,

a2k :=

k , 2(2k − 1)

a2k+1 :=

k for k ≥ 1, 2(2k + 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 z z 2z 2z 3z 3z 4z 4z 5z 5z 6z 1 + 2 + 3 + 2 + 5 + 2 + 7 + 2 + 9 + 2 + 11 + 2 +· · · which therefore also converges to Ln(1 + z). 3

2.2.2 Equivalence transformations

79

Example 13. In Section 1.3.2 on page 27 we found that a(c − b) (b + 1)(c − a + 1) z z an z c(c + 1) (c + 1)(c + 2) := 1 − 1+ n=1 1 1 1 − − (a + 1)(c − b + 1) (b + 2)(c − a + 2) z z (c + 2)(c + 3) (c + 3)(c + 4) 1 1 − − −· · · ∞

K

where a, b, c ∈ −N ∪ {0}, converges to the ratio F (a, b; c; z)/F (a, b + 1; c + 1; z) of hypergeometric functions in the cut plane D := {z ∈ C; 0 < arg(1 + z) < 2π}. An equivalence transformation with rn := c + n for n ≥ 1, and multiplication with c, gives the equivalent identity F (a, b; c; z) F (a, b + 1; c + 1; z) a(c − b)z (b + 1)(c − a + 1)z (a + 1)(c − b + 1)z =c− c+1 − c+2 c+3 − −· · ·

c

for z ∈ D. 3

Example 14. A regular C-fraction is a continued fraction of the form ∞

an z a1 z a2 z a3 z K = , n=1 1 1 + 1 + 1 +· · · and if all an > 0, it is called an S-fraction. The equivalence transformation with r0 := 1, r2n−1 := w := 1/z and r2n := 1 for all n ∈ N brings it over to the form a1 a2 a3 a4 1 ; w := . w + 1 + w + 1 +· · · z √ If we instead use r0 := 1, rn := w := 1/ z for all n ∈ N, we ﬁnd that K(an z/1) is equivalent to 1 a1 /w a2 a3 a4 a5 ; w := √ . w + w + w + w + w +· · · z 3 Example 15. The continued fraction in Problem 26 on page 51 has the form a0 +

a0 (a1 − 1) a1 (a2 − 1)/(a1 − 1) a2 (a3 − 1)/(a2 − 1) 1 − (a1 a2 − 1)/(a1 − 1) − (a2 a3 − 1)/(a2 − 1) − an−1 (an − 1)/(an−1 − 1) . · · · − (an−1 an − 1)/(an−1 − 1) −· · ·

80

Chapter 2: Basics

An equivalence transformation with r0 := 1, r1 := 1, rn := an−1 − 1 for n ≥ 2 leads to a0 (a1 − 1) a1 (a2 − 1) a2 (a3 − 1)(a1 − 1) a0 + 1 a2 a3 − 1 − a1 a2 − 1 − − a3 (a4 − 1)(a2 − 1) a4 (a5 − 1)(a3 − 1) a3 a4 − 1 a4 a5 − 1 − −· · · − n which therefore also has classical approximants fn = k=0 ak . 3 Two equivalence transformations are of particular interest, namely the ones we get by using n 1 (−1)n−k+1 rn := ak and rn := for all n ≥ 1 : bn k=1

$

' Corollary 2.15. A. K(an /bn ) ∼ K(1/dn ) where dn := bn

n

(−1)n+1−k

ak

for n = 1, 2, 3, . . . .

(2.2.4)

k=1

B. If bn = 0 for all n ≥ 1, then K(an /bn ) ∼ K(cn /1), where c1 :=

a1 , b1

cn :=

an bn bn−1

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

(2.2.5)

&

%

Remark: The transformation in A can always be performed. That is, every continued fraction K(an /bn ) can be brought to the equivalent form K(1/dn ). The elements dn have the structure dn = bn /an dn−1 , so d1 = b 1 ·

1 , a1

d2 = b2

a1 , a2

d 3 = b3

a2 , a1 a3

d 4 = b4

a1 a3 ,... . a2 a4

The transformation in B can only be applied if all bn = 0, since otherwise cn is not a well deﬁned complex number. The equivalence transformation is extremely useful in the continued fraction theory. Of course, if K(an /bn ) ∼ K(cn /dn ), then K(an /bn ) converges if and only if K(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 forms K(cn /dn ) may still diverge generally.

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−1 rn an and dn := rn bn 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−1 w/r3n + A3n A3n − A3n−1 wB3n /B3n−1 , = B3n−1 w/r3n + B3n B3n (1 − w)

where we know that A3n = 0, A3n−1 = 2n , B3n = 1 and B3n−1 = 2n+1 − 2. Therefore w 1 w 2n T3n (w) = − n+1 · → , 2 −2 1−w 2 w−1 a non-singular transformation from M. Hence {Tn } is not restrained, and K(cn /dn ) diverges generally. 3 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 (tn w) is not a restrained sequence.

Proof : Since {Sn (0)} diverges, there exists a subsequence {nk } of the natural numbers 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 , and fnk −1 and fnk are = 0, ∞ from some k on. So, for n := nk with k suﬃciently large, Sn (tn w) =

An−1 tn w + An −fn−1 Bn w + An −fn−1 w + fn = = Bn−1 tn w + Bn −Bn w + Bn −w + 1

which converges to the non-singular transformation lim Snk (tnk w) =

k→∞

g2 − g 1 w . 1−w

If t0 = ∞ and/or gk ∈ {0, ∞} for k = 1 or 2, there always exist complex numbers c1 , c2 , d1 , d2 such that c1 c2 = 0 and c1 c2 = ∞, d1 + d2 + t 0 Since therefore

c1 c2 = 0, ∞ for k = 1, 2. d1 + d2 + gk c1 c2 d1 + d2 + Sn (tn w)

82

Chapter 2: Basics

is non-restrained, so is {Sn (tn w)}. This means in particular that if K(an /bn ) converges generally with exceptional sequence {wn† }, but not in the classical sense, then one should stay away from equivalence transformations (2.2.1) with lim inf m(rn , wn† ) = 0. Typical in this situation is that {wn† } has limit points at 0 and ∞. Indeed, if {rn } is bounded away from 0 and ∞, the equivalence transformation preserves general convergence:

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, then K(rn−1 rn an /rn bn ) converges generally to r0 f . Proof :

This follows from (2.2.3) since lim inf m(un , vn ) > 0 n→∞

⇐⇒

lim inf m(un /rn , vn /rn ) > 0. n→∞

2.2.3

The Bauer-Muir transformation

For given continued fraction b0 + K(an /bn ) and given sequence {wn } from C, the idea is to construct a new continued fraction d0 + K(cn /dn ) whose sequence of classical approximants {Tn (0)} is exactly {Sn (wn )}. The new continued fraction is only unique up to an equivalence transformation. The Bauer-Muir transform, the way 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 fraction d0 + K(cn /dn ) whose canonical numerators Cn and denominators Dn are given by D−1 = 0 , C−1 = 1 , Cn = An−1 wn + An , Dn = Bn−1 wn + 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:

2.2.3 The Bauer-Muir transformation

83 $

' Theorem 2.18. The Bauer-Muir transform of b0 + K(an /bn ) with respect to {wn } from C exists if and only if λn := an − wn−1 (bn + wn ) = 0

for n = 1, 2, 3, . . . .

(2.3.2)

λ1 c2 c3 where b1 + w1 + d2 + d3 + · · · λn λn and dn := bn + wn − wn−2 . cn := an−1 λn−1 λn−1

(2.3.3)

If it exists, then it is given by b 0 + w0 +

&

%

Proof : Let {Cn } and {Dn } be given by (2.3.1). Then {Cn } and {Dn } are canonical numerators and denominators of a continued fraction d0 + K(cn /dn ) if and only n := Cn−1 Dn − Dn−1 Cn = 0 for all n ≥ 1. if C−1 = D0 = 1, D−1 = 0 and Δ (See Theorem 1.2 on page 8.) The initial conditions for Cn and Dn are satisﬁed. Moreover n Δ

=

Cn−1 Dn − Dn−1 Cn (An−2 wn−1 + An−1 )(Bn−1 wn + Bn ) − (Bn−2 wn−1 + Bn−1 )(An−1 wn + An ) (An−2 wn−1 + An−1 )(Bn−1 wn + bn Bn−1 + an Bn−2 )

=

−(Bn−2 wn−1 + Bn−1 )(An−1 wn + bn An−1 + an An−2 ) −(An−2 Bn−1 − An−1 Bn−2 )(an − wn−1 bn − wn−1 wn )

:= =

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 of Theorem 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 transform converge to the same value. This was proved to be the case for positive continued fractions and positive wn by Perron ([Perr57], p 27), but of course it holds whenever K(an /bn ) converges with exceptional sequence {wn† } and lim inf m(wn , wn† ) > 0, ([Lore94c]).

Example 17. Stable computation. We want to compute the approximants Sn (−6) of the continued fraction 2

K a1n := 30 +1 0.5 + 30 + 1(0.5)

30 + (0.5)3 . + 1 +···

84

Chapter 2: Basics

It will be evident later that K(an /1) converges to a ﬁnite number f > 0 with exceptional sequence {−6}. Computation shows that f = 5.05859 correctly rounded to 6 digits. But what happens to Sn (−6)? As noted in Remark 4 on page 57 we do 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 the eﬀect that {−6}∞ n=0 is not recognized as an exceptional sequence. Our numerically computed 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. It is given by n 1 2 3 12 13 14 15 18 19 20 161 162 163

Sn (−6) −6.09999 −6.03962 −6.07580 −6.06215 −6.06218 −6.06229 −6.06208 −6.06247 −6.06189 −6.06257 5.05858 5.05859 5.05859

Tn (0) −6.03960 −6.07582 −6.05404 −6.06228 −6.06215 −6.06223 −6.06218 −6.06221 −6.06220 −6.06220 −6.06220 −6.06220 −6.06220

0.5 (30 + 0.5) · 0.5 (30 + (0.5)2 ) · 0.5 −5 + −5 − (−6) · 0.5 + −5 − (−6) · 0.5 +· · · 0.5 15 + (0.5)2 15 + (0.5)3 = −6 + , −5 + −2 −2 + +· · ·

−6+

and its classical approximants Tn (0) (which can be computed stably) are exactly Sn (−6). In the table we have computed Sn (−6) and Tn (0). The two columns should have been identical. The computation is done with only 6 digits to illustrate the instability in the computation of Sn (−6). 3

The Bauer-Muir transformation has many applications. Let us look at one more:

Example 18. Functional equation. The continued fraction ∞

K z + n := z +1 1 + z +2 2 + z +3 3 +· · · n=1 n converges to some f (z) with exceptional sequence {wn† } where (wn† /n) → −1 for all z ∈ C. (This is a consequence of Theorem 4.13 on page 188 after an equivalence transformation.) Hence also Sn (1) converges to f (z). Therefore its Bauer-Muir transform 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 +

K cn = 1 + z −2 1 + z +2 1 + z +3 2 + z +4 3 +· · · ,

n=1 dn

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 z z+1 z+2 = f (z − 1) , = (1) 1 1 + g (z) + 2 + 3 +· · · that is, g (1) (z) = −1 + z/f (z − 1). This means that f (z) = 1 +

z−1 z−1 f (z − 1) + 1 =1+ =z· z (1) f (z − 1) + z 2 + g (z) 1+ f (z − 1)

for z = 1. This is a functional equation for f (z). For z = 1 the original continued fraction is 2 3 4 5 f (1) = =1 1 + 2 + 3 + 4 +··· since the constant sequence {1} is a tail sequence for this continued fraction, and ∞ ∞ ∞ n n bk + t k k+1 = = (−1)n (n + 1)! = ∞ −tk −1 n=0 n=0 n=0 k=1

k=1

(corollary 2.7 on page 68). Hence, 4 1+1 = , f (2) = 2 · 1+2 3

4 +1 21 f (3) = 3 · 3 , = 4 13 +3 3

21 +1 136 13 , = f (4) = 4 · 21 73 +4 13

136 +1 1045 f (5) = 5 · 73 , = 136 501 +5 73

and so on. 3

2.2.4

Contractions and extensions

We say that d0 + K(cn /dn ) is a contraction of b0 + K(an /bn ) if its classical approximants {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 in particular 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 of d0 + K(cn /dn ) and b0 + K(an /bn ) respectively. To derive a general expression for a canonical 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 important special cases:

86

Chapter 2: Basics • an even part of b0 + K(an /bn ) is a contraction with classical approximants gk := f2k . (That is, an even part is a continued fraction!) It is the canonical even part if Ck = A2k and Dk = B2k for all k. • an odd part of b0 + K(an /bn ) is a contraction with classical approximants gk := f2k+1 . It is the canonical odd part if Ck = A2k+1 and Dk = B2k+1 for all k. $

' Theorem 2.19. K(an /bn ) has an even part if and only if all b2n = 0. Its canonical even part is then given by a2 a3 b4 /b2 a4 a5 b6 /b4 b2 a1 b2 b1 + a2 − a4 + b3 b4 + a3 b4 /b2 − a6 + b5 b6 + a5 b6 /b4 −· · ·

(2.4.2)

If {tn }∞ n=0 is a tail sequence for K(an /bn ) with all tn = ∞, then t0 , −t1 t2 , −t3 t4 , −t5 t6 , . . . 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 part if and only if f2n = f2n−2 for all n, that is, if and only if all b2n = 0 (Theorem 1.3 on page 9). From Theorem 1.2 on page 8 we ﬁnd that the canonical even part d0 + K(cn /dn ) has elements d0 := 0, d1 := D1 = B2 = b2 b1 + a2 , c1 := C1 − C0 D1 = A2 − A0 B2 = b2 a1 , n /Δ n−1 , (1) /Δ n−1 , cn := −Δ dn := Δ n

where

n := Cn−1 Dn − Dn−1 Cn = A2n−2 B2n − B2n−2 A2n Δ (1) := Cn−2 Dn − Dn−2 Cn = A2n−4 B2n − B2n−4 A2n . Δ n

The recurrence relation for {An } and {Bn } (page 8) and the determinant formula (page 7) lead to n = b2n Δ

2n−1

(−aj ) ,

j=1

) 2n−3 ( (1) = b (b b + a ) + a b (−aj ) . Δ 2n 2n−1 2n−2 2n−1 2n 2n−2 n j=1

This proves (2.4.2). Let {tn } be a tail sequence for K(an /bn ) with all tn = ∞. To see that t0 , −t1 t2 , −t3 t4 , . . . is a tail sequence for (2.4.2) it suﬃces to prove that −t2n−3 t2n−2 = cn /(dn − t2n−1 t2n ) for n ≥ 2; i.e., cn = −t2n−3 t2n−2 (dn − t2n−1 t2n ) and c1 = t0 (d1 − t1 t2 ) .

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 b2 a1 a2 a3 b4 b2 b1 + a2 − b2 (a4 + b3 b4 ) + a3 b4 a4 a5 b6 b2 a6 a7 b8 b4 , − b4 (a6 + b5 b6 ) + a5 b6 − b6 (a8 + b7 b8 ) + a7 b8 −· · ·

b0 +

(2.4.3)

which is more widely used (but which is no longer canonical). This even part was established by Seidel ([Seid55]) although Lagrange had some special cases already in 1774–76 ([Lagr76]). '

$

Theorem 2.20. K(an /bn ) has an odd part if and only if all b2k+1 = 0. Its canonical odd part is then given by a1 a1 a2 b3 /b1 a3 a4 b5 b1 /b3 − b1 b1 (a3 + b2 b3 ) + a2 b3 − a5 + b4 b5 + a4 b5 /b3 a5 a6 b7 /b5 a7 a8 b9 /b7 − a7 + b6 b7 + a6 b7 /b5 − a9 + b8 b9 + a8 b9 /b7 −· · ·

(2.4.4)

If {tn } is a tail sequence for b0 + K(an /bn ) with all tn = ∞, then −t0 t1 /b1 , −b1 t2 t3 , −t4 t5 , −t6 t7 , . . . is a tail sequence for (2.4.4). &

%

The proof follows the same lines as the proof of Theorem 2.19 and is omitted. The odd part (2.4.4) is equivalent to a1 a1 a2 b3 /b1 − b1 b1 (a3 + b2 b3 ) + a2 b3 a3 a4 b5 b1 a5 a6 b7 b3 . b (a + b b ) + a b b (a + b6 b7 ) + a6 b7 −· · · − 3 5 4 5 4 5− 5 7

2.2.5

(2.4.5)

Contractions and convergence

Example 19. For given 0 < q < 1 we study the continued fraction ∞

1 1 1 K (1/q n ) = . n=1 q + q 2 + q 3 +· · · Now, A0 = 0, A1 = 1, B0 = 1 and B1 = q, and the recurrence relations An = q n An−1 + An−2 ,

Bn = q n Bn−1 + Bn−2

(2.5.1)

88

Chapter 2: Basics

show that {A2n }, {A2n+1 }, {B2n } and {B2n+1 } are strictly increasing, positive sequences, and that by induction An <

n

(1 + q k ),

k=1

Bn ≤

n

(1 + q k ) .

k=1

Hence the four sequences are bounded, and thus convergent, and the ﬁnite, positive limits A0 := lim A2n , A1 := lim A2n+1 , B0 := lim B2n , B1 := lim B2n+1 all exist. Moreover, the determinant formula shows that |An−1 Bn − Bn−1 An | = 1;

i.e., |A0 B1 − B0 A1 | = 1 ,

and thus {S2n } and {S2n+1 } converge to non-singular linear fractional transformations A0 w + A1 A1 w + A0 , S ∗ (w) =: S(w) =: B1 w + B0 B0 w + B1 respectively. In particular this means that both {S2n } and {S2n+1 } are totally non-restrained. Still lim S2n (0) = A0 /B0

and

lim S2n+1 (0) = A1 /B1 ,

so the even and odd parts of K(1/q n ) converge to separate values. The canonical even part of K(1/q n ) is q2 q2 q2 1 + q 1+2 − 1 + q 2 + q 3+4 − 1 + q 2 + q 5+6 −· · ·

(2.5.2)

and the canonical odd part is q2 1 q3 q2 − q q(1 + q 2 + q 2+3 ) − 1 + q 2 + q 4+5 − 1 + q2 + q 6+7 − q2 q2 . 1 + q 2 + q 8+9 −· · · − 1 + q 2 + q 4n+1 −· · ·

(2.5.3)

The sequences An−1 w + An A2n−2 w + A2n , Tn (w) := , Bn−1 w + Bn B2n−2 w + B2n A2n−1 w + A2n+1 Un (w) := B2n−1 w + B2n+1 Sn (w) :=

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) converge in the classical sense, so {Tn } and {Un } are restrained, even generally convergent. 3

Remarks

89

There is an important lesson to be learned from this example: contractions are continued 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)} converges for 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 be gained from extending the concept to more general approximants, but it is interesting to study the connection between the linear fractional transformations Sn (w) =

An−1 w + An Bn−1 w + Bn

for K(an /bn ) and Sne (w) =

A2n−2 w + A2n , B2n−2 w + B2n

Sno (w) =

A2n−1 w + A2n+1 B2n−1 w + B2n+1

for its canonical even and odd parts when these continued fractions exist. The recurrence relation (1.2.7) for {An } and {Bn } on page 6 then shows that A2n−2 w + b2n A2n−1 + a2n A2n−2 B2n−2 w + b2n B2n−1 + a2n B2n−2 2n A2n−2 w+a w + a2n b2n + A2n−1 , = = S2n−1 2n b2n B2n−2 w+a b2n + B2n−1

Sne (w) =

and similarly

Sno (w) = S2n

2.3

w + a2n+1 b2n+1

(2.5.4)

.

(2.5.5)

Remarks

1. The convergence concept. The natural way to deﬁne convergence of a continued fraction K(an /bn ) is to truncate it after n terms to get either fn :=

a1 a2 an b1 + b2 +· · · + bn

or

a1 a 2 an fn := b1 + b2 +· · · + bn + an+1

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 of the type Sn (w) were in use. In 1918 Hamel ([Hamel18]) suggested that convergence should be deﬁned in terms of Sn (w) instead of Sn (0). His rather vague ideas was made into a proper deﬁnition of strong convergence by Thron and Waadeland in 1982 ([ThWa82] p 42). But even this was not quite right. The deﬁnition of general convergence dates back to 1986 ([Jaco86]). It is really a convergence concept for sequences of functions from M (or rather functions from quasi-normal families [Lore03a]). Now, there exist continued fractions for which {fn } converges and {fn } diverges and vice versa. More generally, for any sequence {ϕn } from M with ϕ0 (w) ≡ w s˜n := ϕ−1 n−1 ◦ sn ◦ ϕn

and

S˜n := s˜1 ◦ · · · ◦ s˜n = Sn ◦ ϕn

could in principle have been used to deﬁne K(an /bn ), and the convergence properties of {S˜n } may diﬀer from the ones of {Sn }. 2. Tail sequences. In early works on continued fractions the critical tail sequence hn := −Sn−1 (∞) pops up in formulas of diﬀerent types, mainly because hn = Bn /Bn−1 . Otherwise tail sequences have not been recognized as objects of its own right until Waadeland in a series of papers starting in 1966 ([Waad66]) used periodic tail sequences of periodic continued fractions. His aim was to continue analytic functions deﬁned by limit periodic continued fractions to larger domains. Later on, Thron and Waadeland ([ThWa80a]) applied sequences of tail values for periodic continued fractions to accelerate the convergence of limit periodic ones. These applications were the inspiration for the names of these tail sequences: right tail sequence for the tail values, and wrong tail sequence for the other tail sequences of a convergent continued fraction. In this book the tail sequences are used for a number of purposes, such as proving convergence, estimating the value and ﬁnding truncation error bounds for a given continued fraction. The vital point is the asymptotic behavior of the sequence of tail values on the one hand and the remaining tail sequences which all work as exceptional sequences for a restrained continued fraction on the other hand. This use 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 chain sequences {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 sequence for {an }. The reason why this works so well is that {gn − 1} actually is a tail sequence for K(−an /1). The ﬁrst formulas in Theorem 2.6 on page 66 was published in [JaWa82], and the last one by Waadeland ([Waad84]) for continued fractions K(an /1) and extended by Jacobsen ([Jaco86a]). 3. Value sets. The early convergence results were obtained by manipulating the recurrence relation Xn = bn Xn−1 + an Xn−2 . The idea of value sets possibly started with Weyl ([Weyl10]) and Hamel ([Hamel18]). The concept was applied with great success by a number of authors, such as Jones, Leighton, Paydon, Thron, Scott,

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]) and Thron and Waadeland ([ThWa82], p 43). Indeed, the idea of general convergence came 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} (n)

for n = 0, 1, 2, . . . . That is, Wn is the set of all classical approximants fm for the nth tail of every continued fraction from {Ωk }. It was best in the sense that Wn ⊆ 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 value sets are too large in most cases; we can do better. We want our value sets to be “ small” 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 ⊆ V˜n for all n for every sequence {V˜n } of closed value sets for {Ωn }. 5. Transformations of continued fractions. One can construct a number of different types of transformations for continued fractions, for instance based on approximants, classical or non-classical. But one must always keep in mind what the transformation actually does, because it may change certain convergence properties. Equivalence transformations are totally based on classical approximants. Hence K(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 have Sn (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 transformation τ with τ (0) = −1, τ (1) = 0 and τ (∞) = 12 . if and only if lim m(an , γ) = 0. 2. Chordal metric. Prove that lim an = γ ∈ C 3. Restrained sequences. Prove that the three deﬁnitions 2.5, 2.6 and 2.7 of a restrained sequence beginning on page 61 are equivalent.

92

Chapter 2: Basics 4. Periodic tail sequences. Prove that {wn† } in Example 4 on page 64 is a tail sequence for K(an /1). Prove that {wn† } and {f (n) } are the only periodic tail sequences for K(an /1). 5. Computation of tail sequences. Given the periodic continued fraction ∞

K 21 = 21 + 12 + 21 +· · · .

n=1

(a) Prove that Sn (w) =

(1 + 2 · (− 12 )n )w + 2(1 − (− 12 )n ) . (1 − (− 12 )n )w + 2 + (− 12 )n

(b) Prove that {Sn } converges generally to 1. ﬁnd the tail sequence {tn } of K(2/1). Distinguish (c) For t0 := 1 + h ∈ C, \ {0, −3}. Which ones of between the three cases h = 0, h = −3 and h ∈ C these tail sequences are exceptional sequences for K(2/1)? What is the critical tail sequence for K(2/1)? (d) Find an explicit expression for Sn−1 (w), and prove that {Sn−1 } converges generally 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 ∞

K

∞

K

an n(n + 2) 1·3 2·4 3·5 := = . n=1 1 n=1 1 1 + 1 + 1 +· · · (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 tail sequence {tn } for K(an /1) starting with t0 := 1 + h = 1. 7. ♠ Critical tail sequence. Let hn := −Sn−1 (∞) 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 continued fraction ∞

Ka

n

n=1 bn

=

2 −4 2 −4 2 −4 2 4 2 4 2 4 = . 4 + 1 + 4 + 1 + 4 + 1 +· · · 4 − 1 + 4 − 1 + 4 − 1 +· · ·

(a) Find the periodic tail sequences of K(an /bn ). (b) Find the ﬁrst few terms of its critical tail sequence {hn }.

Problems

93

(c) Show that 1 ≤ h2n ≤ 1 3

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 (t˜n ) for the continued fraction 2 + 0.5 4 + (0.5)2 2 + (0.5)3 4 + (0.5)4 2 + (0.5)5 4 − 1 4 1 4 + − + −· · · where {tn } and {t˜n } are the periodic tail sequences of K(an /bn ) from (a). Compare these sequences of approximants. (g) Construct the Bauer-Muir transform of the continued fraction in (f) with respect to {tn } and with respect to {t˜n }. 9. Tail sequences. In each of the following cases, ﬁnd a tail sequence for the given continued fraction, and use this to ﬁnd its value. ∞

(a)

K (x + n)(x1 + n + 2) .

n=1 ∞

(b)

+ n) K (x −1

.

n=1 ∞

(c)

2

K n2x−−x1 2

2

n=1

.

∞

(d)

K b1

n=1 n

where bn =

(n + a)2 + n + a − 1 . n+a+1

10. Tail sequences. Prove that if the continued fraction 1+

a1 (a1 − z)(a1 + 1) a1 a2 (a2 − z)(a2 + 1) a2 a3 1+ z z + 1 + + 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−

(b + 1)(b + z + 1) (a + 1)(a + z + 1) a(a + z) a + b + z + 1− a + b + z + 2 − a + b + z + 3 − (b + 2)(b + z + 2) (a + 2)(a + z + 2) − a + b + z + 4 − a + b + z + 5 −· · ·

converges, where a, b, z ∈ C are chosen such that the terms are = 0, then it converges to either 0 or −z. (Perron [Perr57], p 25.)

94

Chapter 2: Basics

12. Tail sequence. Let t2n := 1 and t2n+1 := 22n+1 + 1 for n = 0, 1, 2, . . . , and let ∞

K a1

n

n=1

:=

4 6 10 18 2n + 2 . 1 + 1 + 1 + 1 +· · · + 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 1 and the odd part to 2. (Part (c) was proved by Angell and Hirschhorn ([AnHi05]) in a completely diﬀerent way.) 13. Continued fraction identity. Determine the continued fraction K(an /1) which has 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} a value 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 + n1 )w 2 + (d − n1 )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 exceptional sequence {wn† } and a generic sequence {zn } for {τn }. (c) Give an illustration of Theorem 2.1 on page 61 by explicit computation of τn−1 . (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 if −1 ◦ τn → I. For τn → τ in M, then σn := τn−1 τn :=

an w + bn c n w + 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-periodic continued fraction ∞ p 1 1 p 1 1 p an = n=1 1 1 + 1 − 1 + 1 + 1 − 1 + 1 +· · ·

K

Problems

95

(a) Find its 3-periodic tail sequences. (b) Find an expression for Sn . (Compute the coeﬃcients for n = 3m, n = 3m + 1 and 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 converges generally but not classically. 19. Periodic continued fraction. We return to the 3-periodic continued fraction ∞

K a1

n

n=1

=

2 1 1 2 1 1 2 1 + 1 − 1 + 1 + 1 − 1 + 1 +· · ·

from Example 2 on page 59. (a) Write h9 := −S9−1 (∞) as a terminating continued fraction by means of formula (1.4.6) on page 64. n of the terminating continued fraction (b) Show that the canonical denominators B 3n−1 = B3n−1 for n = 0, 1, 2. in (b) satisfy B (c) Compute Sn (wn ) for the following values of wn . In which cases does {Sn (wn )} converge? (i) (iii) (v)

wn = 1, wn = 2

(n+2)/3

,

(ii)

wn = n,

(iv)

wn = 1/n,

wn = 2−n/3 .

20. The reversed periodic continued fraction. Let a1 a2 a3 a1 a2 a3 a1 b1 + b2 + b3 + b1 + b2 + b3 + b1 +· · · be a 3-periodic continued fraction. The reversed (or dual) continued fraction is then the 3-periodic continued fraction b3 +

a3 a2 a1 a3 a2 a1 a3 . b2 + b1 + b3 + b2 + b1 + b3 + b2 +· · ·

Prove that the canonical denominators B3n−1 are the same for each n ∈ N for the two 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 its reversed continued fraction ap ap−1 a2 a 1 bp + . bp−1 + bp−2 +· · · + b1 + bp +· · ·

96

Chapter 2: Basics

22. ♠ General convergence of periodic continued fraction. Give an example other 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 of length ≥ 3 ? 23. ♠ From series to continued fraction. (Euler [Euler48].) Prove that the continued fraction ρ1 ρ2 ρ3 ρ0 + where all ρk = 0 1 − 1 + ρ2 − 1 + ρ3 −· · · has canonical numerators and denominators given by k n An = ρ0 + ρj , Bn = 1 for n = 0, 1, 2, . . . . k=1

j=1

24. Equivalence transformation. Prove that K(an /(−bn )) ∼ −K(an /bn ). 25. Equivalence transformation. (Stern [Stern32].) Prove that b1 b2 b3 b0 b0 b1 ∼ when all bn = 0. b1 + b2 + b3 +· · · b0 + b1 + b2 +· · · 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 a1 a2 /b1 b 0 b 1 + a1 a3 a4 − . b1 (b1 b2 + a2 )/b1 + b3 + b4 +· · ·

(∗)

Show further that if {tn }∞ n=0 is a tail sequence for b0 + K(an /bn ) with t0 , t1 = ∞, then −t0 t1 , t2 , t3 . . . . is a tail sequence for (∗). 27. ♠ Shift transformation. Let N > 1 be a ﬁxed integer. Show that the canonical contraction of b0 + K(an /bn ) with Cn = An ,

D n = Bn

for n = 0, 1, 2, . . . , N − 1

and Cn = 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 aN −1 bN +1 aN b1 +· · · + bN −1 + bN +1 bN + aN +1 −aN +1 aN +2 /bN +1 aN +3 . + (bN +2 bN +1 + aN +2 )/bN +1 + bN +3 +· · ·

(Perron [Perr57], p 16.) Show further that if {tn }∞ n=0 is a tail sequence for b0 + K(an /bn ) with tN = ∞ and tN +1 = ∞, then t0 , t1 , . . . , tN −1 , −tN tN +1 , tN +2 , tN +3 , . . . is a tail sequence for this contraction.

Problems

97

28. ♠ The Wall transformations. Let {fn }∞ n=0 be the sequence of classical approximants 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 approximants for the continued fraction a1 −

a3 a4 a5 a6 a 7 a8 a9 a1 a2 1 + a2 + 1 + a4 − 1 + a5 + 1 + a7 − 1 + a8 + 1 + a10 −· · ·

(b) f0 , f2 , f3 , f5 , f6 , f8 , . . . , f3n , f3n+2 , . . . is the sequence of classical approximants for the continued fraction a1 a2 a3 a4 a 5 a6 a7 a 8 a9 1 + a2 − 1 + a3 + 1 + a5 − 1 + a6 + 1 + a8 − 1 + a9 +· · · (c) f0 , f1 , f3 , f4 , f6 , f7 , . . . , f3n , f3n+1 , . . . is the sequence of classical approximants for the continued fraction a 3 a4 a5 a 6 a7 a8 a1 a2 1 + 1 + a3 − 1 + a4 + 1 + a6 − 1 + a7 + 1 + a9 −· · · (d) If two of the continued fractions in (a)–(c) converge, then also the third one converges. (e) K(an /1) converges if and only if the three continued fractions in (a)–(c) converge. (Wall ([Wall57]), where also a number of similar results are presented.) 29. ♠ Extensions. Show that 1 1 a1 a2 a3 a4 b1 − a2 + 1 − 1 + b2 − a3 + 1 − 1 + b3 − a4 + 1 −· · · and

1 a2 a3 a1 1 1 b1 − 1 − 1 + 1 + b2 − a2 − 1 + 1 + b3 − a3 − 1 +· · · are extensions of K(an /bn ), (McLaughlin and Wyshinski [McWy07]). 30. ♠ Continued fractions with canonical denominators equal to 1. Let b0 + ∞ for K(an /bn ) have critical tail sequence {hn } (i.e., hn = −Sn−1 (∞)) with hn = n ≥ 1. Prove that the continued fraction b0 +

a1 /h1 a2 /h1 h2 a3 /h2 h3 1 + b2 /h2 + 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 ) be have classical approximants Sn (0) = fn , let N ∈ N with N ≥ 2 and let g ∈ C chosen such that −1 (g) = ρ := −SN

A N − BN g = 0, ∞. AN −1 − BN −1 g

98

Chapter 2: Basics Prove that b0 +

a1 aN +1 /ρ ρ aN +2 aN −1 aN b1 +· · · + bN −1 + bN − ρ + 1 − bN +1 + aN +1 /ρ + bN +2 +· · ·

is an extension of b0 + K(an /bn ) with classical approximants ⎧ for n = 0, 1, . . . , N − 1 , ⎨ fn ∗ g for n = N , fn = ⎩ for n = N + 1, N + 2, . . . . fn−1 32. ♠ The Khovanskii transform. The Khovanskii transform ([Khov63], p 23) of K(an /1) is given by a3 a5 a2 a4 a1 1 + 2a2 − 1 − 1 + 2a3 + 2a4 − 1 − 1 + 2a5 + 2a6 a2n+1 a2n . 1 1 + 2a − −· · · − 2n+1 + 2a2n+2 −· · · Prove that if both K(an /1) and its Khovanskii transform converge (in the classical sense), then they converge to the same value. 33. The Bauer-Muir transform. (a) Find the Bauer-Muir transform of z2 − 1 +

22 22 42 42 62 62 1 + z 2 − 1 + 1 + z 2 − 1 + 1 + z 2 − 1 +· · ·

with respect to {wn } given by n+1 − 1 2 wn := z + 1 1 (3 + 2z − z 2 ) n(z + 1) + 2

if n is odd, if n is even.

(b) Assume that the continued fraction in (a) and its Bauer-Muir transform converge 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 a1 a2 + h a2 1 + b + 1 + b +· · · a1 a1 + h a2 a2 + h h+ 1 + b + 1 + b +· · · converge. Prove that then they converge to the same value. (Ramanujan [Bern89], p 122), ([Jaco90]).

Chapter 3

Convergence criteria The convergence theory for continued fractions relies on some basic concepts and ideas. 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 in two meanings of the word. For one thing they are old, well-proven results. But they are also classical in the sense that they aim for classical convergence. The proofs we oﬀer are not always the classical ones, though. We have chosen to see the theorems in a more modern setting whenever convenient. In particular we use value sets to prove uniform convergence of {Sn (w)} both with respect to w and with respect to a family of continued fractions. This allows us to produce some newer results as well. 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 continued fractions 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 convergence criterion. They are useful because of their generality and their simplicity. For bounds 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

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 |bn | < ∞. Later on, von Koch tion K(1/bn ) diverges in the classical sense if ([Koch95a], [Koch95b]) proved that {A2n+m }n and {B2n+m }n converge to ﬁnite values for which (1.1.1) holds in this case. From their proofs we can even extract alittle more information. We say that a sequence {cn } converges absolutely if |cn − cn−1 | < ∞. Absolute convergence implies convergence to a ﬁnite value n since k=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 &

A1 B0 − A0 B1 = 1.

(1.1.1)

%

Proof : We shall use the classical proof of this theorem. It is related to Example 19 on page 87. Let |bn | < ∞. Now, {An } and {Bn } are solutions of the recurrence relation (1.1.2) Xn = bn Xn−1 + Xn−2 for n = 1, 2, 3, . . . . 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 > 0 for m := n − 1 and m := n − 2, then |Xn | ≤ (|bn | + 1)λ

n−1

n

k=1

k=1

(|bk | + 1) = λ

(|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 } are bounded under our conditions. Therefore the two series bn An−1 and {B n bn Bn−1 converge absolutely. Since solutions {X } of (1.1.2) satisfy X n n −Xn−2 = |An − An−2 | < ∞ and |Bn − Bn−2 | < ∞. In other bn Xn−1 , this means that 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−1 B2n − A2n B2n−1 = (−1)2n = 1

for all n .

3.1.1 The Stern-Stolz Divergence Theorem

101

Finally, since lim S2n (w) =

n→∞

A1 w + A0 , B1 w + B0

lim S2n+1 (w) =

n→∞

A0 w + A1 , B0 w + 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. but their limits Indeed, lim S2n (w) and lim S2n+1 (w) exist for every w ∈ C, depend totally on w. In particular, lim S2n (0) =

n→∞

A0 and B0

lim S2n+1 (0) = lim S2n (∞) =

n→∞

n→∞

A1 B1

since Bm = 0 where A0 /B0 = A1 /B1 by (1.1.1). (Am /Bm is well deﬁned in C implies 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

A B

n n−2 − An−2 Bn bn

|Sn (0) − Sn−2 (0)| =

=

. Bn Bn−2 Bn Bn−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 of a continued fraction K(an /bn ). The equivalence transformation in Corollary 2.15 on page 77 brings K(an /bn ) to the form K(1/dn ). Hence, K(an /bn ) diverges if ∞ ∞

n

(−1)n−k+1

(1.1.4) |dn | = ak S :=

b n

< ∞, n=1

n=1

k=1

and its sequence of even and sequence of odd classical approximants both converge to distinct values. The series S in (1.1.4) is called the Stern-Stolz Series of K(an /bn ). It is invariant under equivalence transformations of K(an /bn ), and it can also be written ∞

∞

a2 a4 · · · a2n

a1 a3 · · · a2n−1

b + (1.1.5) S=

2n

b2n+1

. a2 a4 · · · a2n a1 a3 · · · a2n+1 n=1 n=1 4. In principle it may occur that K(an /bn ) converges generally even if its SternStolz 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, this problem can not occur if (−1)n−k+1

inf |rn | > 0 and sup |rn | < ∞ for rn := Πnk=1 ak

.

(1.1.6)

102

Chapter 3: Convergence criteria

5. A continued fraction of the form K(an /1) diverges generally if its Stern-Stolz Series 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 that |bn | < ∞, then the K(an /bn ) converges separately. Hence, if all bn = 0 and even and odd parts of K(1/bn ) converge separately. (The condition bn = 0 is needed for the existence of the even and odd parts.) Indeed, Theorem 3.1 is actually 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 least one of the following three conditions hold: (i)

∞

|bn bn−1 /an | = ∞ ,

(1.1.7)

n=2

(ii) (iii) &

an

< ∞, lim inf

n→∞ bn bn−1

∞ |bn bn−1 | = ∞. n|an | n=2

(1.1.8) (1.1.9) %

Proof : (i): The series in (1.1.7) and the Stern-Stolz Series are invariant under equivalence transformations, so it suﬃces to consider continued fractions K(1/bn ) and to prove that ∞ |bn bn−1 | = ∞

=⇒

n=2

∞

|bn | = ∞ .

n=1

√ This implication follows easily since pq ≤ 12 (p + q) for positive numbers p and q, and thus ∞ ∞ ∞ 1 (|bn | + |bn−1 |) ≤ |bn bn−1 | ≤ |bn |. 2 n=2 n=2 n=1 (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 ∞ qn =∞ n n=2

=⇒

∞ √ n=2

qn = ∞

3.1.2 The Lane-Wall Characterization

103

√ for an arbitrary sequence {qn } with qn ≥ 0. If lim sup qn > 0, then clearly q = √ √n ∞. 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 of classical approximants. But how about convergence ? When is divergence of the Stern-Stolz Series suﬃcient for convergence of K(an /bn )? The following theorem by Lane and Wall ([LaWa49]), gives a very useful answer: '

$

Theorem 3.3. (The Lane-Wall Characterization.) Let K(an /bn ) with classical approximants fn = Sn (0) = ∞ satisfy ∞

|fn+1 − fn−1 | < ∞.

(1.2.1)

n=1

Then K(an /bn ) converges if and only if

∞

n

(−1)n−k+1

ak

b n

= ∞.

n=1

(1.2.2)

k=1

&

%

Proof : Since the approximants fn and the Stern-Stolz Series (1.2.2) are invariant under equivalence transformations, we may assume that K(an /bn ) has the form K(1/b Theorem we know that K(1/bn ) diverges n ). By the Stern-Stolz Divergence |bn | = ∞. Since {f2n } and {f2n+1 } converge if |bn | < ∞. Assume that absolutely, 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. The determinant formula and the recurrence relation (1.1.2) on page 100 for {An } and

104

Chapter 3: Convergence criteria

{Bn } show that An−1 An+1 − An+1 Bn−1 − An−1 Bn+1 Bn B Bn−1 δn = − n+1 =− · An An+1 An+1 Bn − An Bn+1 Bn−1 − Bn+1 Bn Bn An Bn−1 − An−1 Bn Bn = −bn+1 · = bn+1 An+1 Bn − An Bn+1 Bn−1 Bn−1

(1.2.3)

where all Bn = 0 since all fn = ∞. That is, for all n ∈ N. (1.2.4) |bn | < ∞, a contradiction, and thus We shall prove that |δn | < ∞ implies L0 = L1 . We ﬁrst observe that δn Bn−1 = bn+1 Bn

1 Bn−1 Bn−1 Bn−1 1 · = = = . (1.2.5) Bn bn Bn−1 + Bn−2 δn−1 Bn−2 + Bn−2 δn−1 + 1 Bn−2 /Bn−1 Therefore it follows from (1.2.4) that 1 δn δn (δn−2 + 1) Bn−3 Bn−1 · · = = Bn δn−1 + 1 Bn−2 /Bn−1 δn−1 + 1 Bn−2 1 δn (δn−2 + 1) δn (δn−2 + 1)(δn−4 + 1) Bn−5 · · = = (δn−1 + 1)(δn−3 + 1) Bn−4 /Bn−3 (δn−1 + 1)(δn−3 + 1) Bn−4

bn+1 = δn

and so on. That is, b2n+1 =

δ2n (δ2n−2 + 1)(δ2n−4 + 1) · · · (δ2 + 1) B1 · (δ2n−1 + 1)(δ2n−3 + 1) · · · (δ3 + 1) B2

(1.2.6)

and

δ2n+1 (δ2n−1 + 1)(δ2n−3 + 1) · · · (δ1 + 1) B0 . (1.2.7) · (δ2n + 1)(δ2n−2 + 1) · · · (δ2 + 1) B1 −1, ∞ and all Bn = 0, ∞, this shows that if |δn | < ∞, then Since all δn = |bn | < ∞. b2n+2 =

∞ Remark. If fn = ∞ for n ≥ m ∈ N only, and n=m+1 |fn+1 − fn | < ∞, then the conclusion of Theorem 3.3 still holds. The proof only needs the minor modiﬁcation that the process of using (1.2.5) to produce (1.2.6)-(1.2.7) stops at Bm /Bm+1 and Bm+1 /Bm+2 . The absolute convergence in (1.2.1) is vital. As Wall proved in [Wall56], there exists a divergent continued fraction for which the even and odd parts converge to distinct values and (1.2.2) holds:

3.1.2 The Lane-Wall Characterization

105

Example 1. Let K(an /1) be given by an := (−1)n−1 n2

for n = 1, 2, 3, . . . .

Then K(an /1) ∼ K(1/bn ) where 1/b1 := a1 = 1 and 1/bn−1 bn := 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−1 bn | = n1 ≤ 12 (|bn−1 | + |bn |), and thus

∞ n=1

∞

|bn | =

∞

|b1 | 1 1 1 + (|bn−1 | + |bn |) ≥ + = ∞; 2 2 2 n=2 n n=2

i.e., (1.2.2) holds for K(an /1). The even part of K(an /1) is the continued fraction ∞

K

a1 a2 a3 a4 a5 a2n−2 a2n−1 cn ∼ 1 + a2 − 1 + a3 + a4 − 1 + a5 + a6 −· · · − 1 + a2n−1 + a2n −· · · n=1 1 where a1 1 −a2 a3 = − , c2 := = 2, 1 + a2 3 (1 + a2 )(1 + a3 + a4 ) (n − 1)2 (2n − 1) −a2n−2 a2n−1 = cn := (1 + a2n−3 + a2n−2 )(1 + a2n−1 + a2n ) 2n − 3 c1 :=

for n ≥ 3. Similarly, the odd part of K(an /1) is ∞

a1 − where

K

a1 a 2 a3 a4 a5 a6 dn ∼ a1 + n=1 1 1 + a2 + a3 − 1 + a4 + a5 − 1 + a6 + a7 −· · ·

−a1 a2 2 = , 1 + a2 + a3 3 −a2n−1 a2n n2 (2n − 1) dn := = . (1 + a2n−2 + a2n−1 )(1 + a2n + a2n+1 ) 2n + 1

d1 :=

Both K(cn /1) and K(dn /1) converge, as you are asked to prove in Problem 3 on page 166. Indeed, since all dn > 0, it follows that K(dn /1) converges to a positive number. 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. 3

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) converge. If lim inf |an | < ∞, then K(an /1) converges.

106

Chapter 3: Convergence criteria

Proof : Let L0 := lim S2n (0) and L1 := lim S2n+1 (0), and assume that L0 = L1 . Without loss of generality we assume that L1 = ∞ and L2 = ∞. (Otherwise ˜/(1 + a/(1 + K(an /1))) if we consider a continued fraction a/(1 + K(an /1)) (or a necessary) for appropriately chosen a, a ˜ = 0.) Since Sn (−1) = Sn−2 (0), the two sequences {S2n+m }n for m = 0, 1 converge generally to Lm , respectively, with exceptional sequences {ζ2n+m }n . Let δn be as deﬁned in the previous proof. Then δn → 0, and by (1.2.3) δn := −

fn+1 − fn−1 1 Bn ζn = · =− → 0. fn+1 − fn an+1 Bn−1 an+1

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 either even or odd numbers, say nk are even. Then lim Snk (∞) = 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 approximants Sn (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 bounds for 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

1 1 − Σn Σ∞

(1.3.4)

3.1.3 Truncation error bounds

107

when K(an /bn ) converges to f , where Σn :=

n

Pk → Σ∞ ∈ C

for Pk :=

k=0

k bj + tj −tj j=1

(1.3.5)

for an arbitrary tail sequence {tn } for K(an /bn ) with all tn = ∞. Of course, −1 fn+m − fn = t0 (Σ−1 n − Σn+m ) holds even if K(an /bn ) diverges. 3. From the invariance of the cross ratio u−z v−w Sn (u) − Sn (z) Sn (v) − Sn (w) · = · Sn (u) − Sn (w) Sn (v) − Sn (z) u−w v−z

(1.3.6)

(n)

with u := 0, z := fk (the kth classical approximant of the nth tail), w := ∞ and v := ζn := Sn−1 (∞), we obtain (n)

−fk fn − fn+k ·1= (n) fn − fn−1 ζn − f k (n)

when fk

= 0, ∞ and ζn = 0, ∞, and thus fn , fn−1 = ∞. I.e., (n)

fn+k − fn =

−fk

(n)

fk

− 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) (fn − fn−1 ). f (n) − ζn

(1.3.8)

If K(an /bn ) converges to f = ∞, it is natural to use 1 1 1 =− − f fn fn

or

1 1 1 − =− f Sn (wn ) Sn (wn )

as a measure for the truncation error. Bounds for this error can be found from the truncation error of a continued fraction a/(b + K(an /bn )) which then converges to 0, regardless of the choice of a = 0 and b. Now, a lot of the truncation error bounds derived up through the ages concern classical approximants. On the other hand we have seen that a clever choice of approximants Sn (wn ) can give faster convergence, so we really want bounds for |f −Sn (wn )|. Fortunately we have the following method to adjust existing truncation error 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 (f − fn−1 ). ζ n − wn

(1.3.9) %

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 is a 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 critical tail ζn . We distinguish between a priori bounds which can be computed in advance, before we compute the approximants, and a posteriori bounds which are expressed in terms of 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 normally tighter than a priori bounds. A priori bounds on the other hand make it possible to estimate in advance the order of the approximant and thus the number of digits needed 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)]. Let K(an /1) be a convergent continued fraction from E. By (1.4.6) on page 64 we know that an−1 a2 an . −ζn = hn = 1 + 1 + 1 +···+ 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) | ≤ g and |ζn − f (n) | ≥ 1 − 2g, and thus by (1.3.8) |f − fn | ≤

g |fn − fn−1 | 1 − 2g

for n = 1, 2, 3, . . . .

3

3.1.4

Mapping with linear fractional transformations

is also a circle in the complex plane C if ∞ ∈ C, A circle C on the Riemann sphere C but if ∞ ∈ C, then C \ {∞} is a straight line in C. Therefore circles and lines on C are also called generalized circles as a common name. Linear fractional transformations τ ∈ M have beautiful mapping properties. Indeed, they

3.1.4 Mapping with linear fractional transformations

109

onto circles on C, 1. map circles on C

onto points 2. map points symmetric with respect to a generalized circle C in C symmetric with respect to τ (C). (Two points u and v are symmetric with respect 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 the perpendicular bisector of the line segment connecting u and v.) where ∂V (the boundary of We shall focus on the family V of closed sets V on C V ) 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 + eiα H where ξ ∈ C, α ∈ R (the set of real numbers), H := {w ∈ C; Re w > 0}, and H is its closure in C.

∞ ∈ V ◦ (the interior of V ): We say that V is the complement of (or the exterior |w − Γ| ≥ ρ} with of) an open disk, and we write V = B(Γ, −ρ) := {w ∈ C; ρ > 0, when V is the exterior of B(Γ, ρ)◦ . Let τ ∈ M be given by

τ (w) :=

aw + b cw + d

with

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. Then τ can be written τ (w) = ad w + db , and therefore c=0

=⇒

Otherwise we have:

τ (B(Γ, μ)) = B( ad Γ + db , | ad |μ) , τ (ξ + eiα H) = τ (ξ) + ei(α+β) H where β := arg

a d

.

(1.4.2)

110

Chapter 3: Convergence criteria $

' Lemma 3.6. Let τ be given by (1.4.1) with c = 0, and let V ∈ V. If − dc ∈ ∂V , then τ (V ) = B(Γ1 , μ1 ) where Γ1 :=

Δ(cΓ + d)/c a μ|Δ| − and μ1 := 2 2 c |cΓ + d| − |cμ| |cΓ + d|2 − |cμ|2 if V = B(Γ, μ) with Γ ∈ C and μ ∈ R \ {0}

(1.4.3)

and Γ1 :=

1 1 2 Δ e−iα /c2 a

2 Δ/c − 2 | := and |μ

1 d −iα d −iα

c Re[(ξ + c )e ] Re[(ξ + c )e ]

(1.4.4)

if V = ξ + e H with ξ ∈ C and α ∈ R. iα

& Proof: This time τ can be written τ (w) =

%

Δ/c2 a − c w+β

where 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. Im 6

v+ = β + Γ +

Let ﬁrst V := B(Γ, μ). If β + Γ = 0, then the boundary of (β + V ) is a circle with center at the origin and radius |μ|. Therefore 1/∂(β + V ) is a circle with center at the origin and radius 1/|μ|. Therefore cΔ2 /∂(β +V ) is a circle with center at the origin and radius | cΔ2 /μ|. Therefore V is mapped onto B( ac , −| cΔ2 |/μ), as claimed in (1.4.3).

β+Γ |β+Γ| |μ|

? |μ| β+Γ

Let β + Γ = 0. Since v± := (β + Γ)(1 ± |μ|/|β + Γ|)

6

Re

v− = β + Γ −

β+Γ |β+Γ| |μ|

are the two points on ∂(β + V ) furthest away from 0 and closest to 0 respectively (see the picture), it follows that the circle 1/(β + ∂V ) has and radius r given by center Γ

3.1.4 Mapping with linear fractional transformations

Im 6

β+Γ := 1 1 + 1 = Γ 2 v− v+ |β + Γ|2 − μ2

1 1 1

μ

r :=

−

=

. 2 v− v+ |β + Γ|2 − μ2

ξ+β

β+V

α

111

Since τ (V ) is bounded if and only if 0 ∈ β + V , this proves (1.4.3).

v α

Re

Next, let V := ξ+eiα H. Then V +β = ξ + β + eiα H. The point on ∂(V + β) closest to the origin is v := eiα Re[(ξ + β)e−iα ]. Therefore 1/∂(V + β) is the circle through the origin with center 1 1 2v and radius 2|v| . Hence the expression for τ (V ) follows from (1.4.5).

∂(β + V ) We conclude this section with some useful observations ([Lore94a]), strongly inspired by 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 ∈ M map the unit disk D into itself with lim sup rad τn (D) < 1.

(1.4.6)

n→∞

\ D. Then {Tn } is restrained with exceptional {Tn−1 (∞)} from C sequence −1 m(|Tn (∞)|, 1) < ∞, and {Tn } has If moreover lim rad Tn (D) > 0, then an exceptional sequence from ∂D. &

%

Proof : Let Γn and Rn be the center and radius of the disk Tn (D). The nestedness Tn+1 (D) = Tn (τn+1 (D)) ⊆ Tn (D) shows that {Rn } is a non-increasing sequence of positive numbers. Hence Rn → R ≥ 0. If R = 0, the limit point case, then the sequence {Tn } is clearly restrained. Indeed, it converges generally to the limit point γ ∈ lim Tn (D) ⊆ D. Assume that R > 0, the limit circle case. A linear fractional transformation mapping D onto D can always be written on the form eiα (w−q)/(1−qw) for some real constant α and complex constant q ∈ D. Therefore Tn can be written Tn (w) = Γn + Pn

(Γn − Pn Qn ) + (Pn − Qn Γn )w w − Qn = 1 − Qn w 1 − Qn w

(1.4.7)

112

Chapter 3: Convergence criteria

where |Pn | = Rn and |Qn | < 1. Similarly, τn can be written

τn (w) = γn + pn

w − qn 1 − qn w

for 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 − Pn Qn )(−Qn ) − (Pn − Qn Γn ) · 1| rn+1 2 |1 − Qn γn+1 |2 − |Qn |2 rn+1 Rn (1 − |Qn |2 )rn+1 . 2 |1 − Qn γn+1 |2 − |Qn |2 rn+1

Therefore (1 − |Qn |2 )rn+1 Rn+1 ≤ 2 Rn (1 − |Qn |(1 − rn+1 ))2 − |Qn |2 rn+1 (1 + |Qn |)rn+1 (1 − rn+1 )(1 − |Qn |) =1− 1 − |Qn | + 2|Qn |rn+1 1 − |Qn | + 2|Qn |rn+1 (1 − r)(1 − |Qn |) ≤1− =: 1 − δn 1 − |Qn | + 2r|Qn | =

∞ where /Rn ) = R/R1 > 0, it follows that δn < ∞; δn > 0. Since n=1 (Rn+1 −1 i.e., (1 − |Qn |) < ∞. Since Tn (∞) = 1/Qn where|Qn | → 1, this proves that −1 Tn−1 (∞) = ∞ for n ≥ n0 for some n0 ∈ N and that ∞ n=n0 (|Tn (∞)| − 1) < ∞, ∞ or, equivalently, n=1 m(|Tn−1 (∞)|, 1) < ∞. From the expression (1.4.7) for Tn it follows that its derivative is Tn (w) = Pn

1 − |Qn |2 (1 − Qn w)2

(1.4.8)

where |Pn | = Rn → R < 1 and |Qn | → 1. Hence Tn (w) → 0 for every w ∈ D, and {Tn } is restrained. It remains to prove that {Tn−1 (∞)} is an exceptional sequence. But this follows trivially from (1.4.8) since w := 1/Qn = Tn−1 (∞) gives Tn (w) = ∞. Since |Tn−1 (∞)| → 1, there also exist exceptional sequences {wn† } with |wn† | = 1 for all n; i.e., wn† ∈ ∂D.

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 exists such that a sequence {wn } ⊆ C

(1.4.9) lim inf |wn | − 1 > 0 and lim inf |τn (wn )| − 1 > 0. Then {Tn } converges generally to some constant γ ∈ D with an exceptional \ D. If the limit point case fails to occur for sequence {wn† } with wn† ∈ C {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 point case 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) ⎧ (1 − |Qn |2 )|u − v| ⎪ ⎪ R if u, v = ∞, n ⎪ ⎪ ⎨ |1 − Qn u| · |1 − Qn v| 1 − |Qn |2 (1.4.10) |Tn (u) − Tn (v)| = if u = ∞, Rn ⎪ ⎪ ⎪ |Q | · |1 − Q v| n n ⎪ ⎩ 0 if u = v = ∞. 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 |) (1 − |Qn |) < ∞ under our conditions, from some n on,say n ≥ n0 . Since ∞ this means that |T (w ) − T (wn )| < ∞; i.e., {Tn (wn )}∞ n+1 n+1 n n0 converges n=n0 absolutely to some γ ∈ D. Let w ∈ D be arbitrarily chosen. Then (1 − Qn w) is bounded away from 0. Hence also |Tn (wn ) − Tn (w)| ≤ A(1 − |Qn |) for some A > 0 from some n on. Therefore {Tn (w)} converges absolutely to γ for every w ∈ D, and thus {Tn } converges generally to γ. Remark: The conditions in Lemma 3.8 are in particular satisﬁed if all τn map D into itself and |τn (w0 )| ≤ r < 1 for all n for some given point w0 ∈ D, since then rad(τn (D)) ≤ 1+r 2 for all n. This generalizes a result by Hillam and Thron \ D and z ∈ D. ([HiTh65]) who required that τn (w) = z for all n for some w ∈ C 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−1 n−1 ◦ sn ◦ ϕn

114

Chapter 3: Convergence criteria

maps D into D. Moreover, sn = ϕn−1 ◦ τn ◦ ϕ−1 , so −1 Sn := s1 ◦ s2 ◦ · · · ◦ sn = ϕ−1 0 ◦ τ1 ◦ τ2 ◦ · · · ◦ τn ◦ ϕn = ϕ0 ◦ Tn ◦ ϕn .

Therefore, if {ϕn } is totally non-restrained, then {Sn } converges generally if and only if {Tn } converges generally. Hence the following result is for instance a consequence of Lemma 3.8:

$

' Corollary 3.9. Let {ϕn }∞ n=0 be a totally non-restrained sequence of linear fractional transformations, and let {Vn }∞ n=0 given by Vn := ϕn (D) be a sequence of value sets for the continued fraction K(an /bn ). If lim sup rad ϕ−1 n−1 ◦ sn ◦ ϕn (D) < 1

(1.4.11)

n→∞

such that and there exists a sequence {wn } from C lim inf distm (wn , ∂Vn ) > 0 and lim inf distm (sn (wn ), ∂Vn−1 ) > 0, n→∞

n→∞

(1.4.12) then K(an /bn ) converges generally to some value f ∈ V0 with exceptional \ Vn for all n. sequence {wn† } with wn† ∈ C &

%

Here distm (w, V ) := inf{m(w, v); v ∈ V } denotes the chordal distance between a and a set V ⊆ C. point w ∈ 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 the idea earlier in his important 1894-paper ([Stie94]). Therefore we follow Jones and Thron ([JoTh80], p 89) in naming the theorem:

3.1.6 A simple estimate

115 $

' Theorem 3.10. (The Stieltjes-Vitali Theorem.) Let {Fn } be a sequence of holomorphic functions in a region D, such that (i) there exist two distinct values p, q ∈ C for which Fn (z) = p and Fn (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 which has 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 complex variable, 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 b + k − a b + 1 a < b+k b+n+1

for n > 0.

(1.6.1)

k=1

& Proof :

% For u := (b + k + 1)/(b + k) we consider the integral

t−a−1 t−a u u− −a − 1 −a 1 1 −a −a 1 1 = − a+1 (u − u) + a (u − 1) b + k a b + k − a 1 . − = a(a + 1) b + k + 1 b+k

I :=

u

t−a−2 (u − t) dt =

It is clear from the deﬁnition of I that I > 0. Hence b + k − a b + k a , < b+k b+k+1 and (1.6.1) follows by taking products on both sides.

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 positive continued 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 to ﬁnite, non-negative values. If all bn > 0, then (2.1.1) holds with strict inequalities. &

%

Proof : Let ﬁrst b1 > 0. Then Bn > 0 for n ≥ 1 by the recurrence relation. Therefore the Euler-Minding series n ∞ n+1 k=1 ak (fn − fn−1 ) where fn − fn−1 = (−1) (2.1.2) B n Bn−1 n=1 is an alternating series. Since

fn − fn−1 an Bn−2 Bn − bn Bn−1

= ≤ 1,

fn−1 − fn−2 = Bn Bn

(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 n0 th tail K∞ n=n0 +1 (an /bn ) of K(an /bn ). Since K(an /bn ) = Sn0 (K∞ n=n0 +1 (an /bn )) where ⎧ a1 a3 · · · an0 ⎪ ⎨ a2 a4 · · · an0 −1 w if n0 is odd

Sn0 (w) =

a2 an0 a1 = ⎪ 0 + 0 +···+ w ⎩ a1 a3 · · · an0 −1 a2 a4 · · · an0 w if n0 is even

(2.1.1) holds also now.

(2.1.4)

3.2.1 Positive continued fractions

117

The following classical result due to Seidel ([Seid46]) and Stern ([Stern48]) is a direct consequence of Theorem 3.12 combined with the Stern-Stolz Divergence Theorem on page 100 and the Lane-Wall Characterization on page 103:

Theorem continued fraction K(1/bn ) converges if and 3.13. A positive only 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 absolutely (Theorem 3.12), and thus K(1/bn ) converges (the Lane-Wall Characterization). Next let bn < ∞. The general divergence follows then from the Stern-Stolz Divergence Theorem. Seidel and Stern required that all bn > 0. That bn ≥ 0 suﬃces was proved by Broman ([Brom77]). An equivalence transformation gives the formulation: '

$

Theorem 3.14. (The Seidel-Stern Theorem.) A positive continued fraction 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

(−1)n+k+1

ak

= ∞.

(2.1.5)

k=1

&

%

Positive continued fractions and tail sequences. It is clear that if the positive continued fraction K(an /bn ) with all bn > 0 converges generally, then its sequence {f (n) } of tail values is a positive sequence. What is more interesting is that a kind of converse result is true. The following was published by Khrushchev ([Khru06b]) under the name of Markov’s Theorem:

Theorem 3.15. Let {tn } be a positive tail sequence for the generally convergent positive continued fraction K(an /bn ). Then K(an /bn ) converges to t0 and {tn } is its sequence of tail values.

118

Chapter 3: Convergence criteria

Proof : Since K(an /bn ) converges and all tn = 0, −bn , it follows from Corollary 2.7 on page 68 that {Σn } given by Σn :=

n

Pk where Pk :=

k=0

k bm + tm −tm m=1

as n → ∞. Since |Pk | ≥ 1 for all n, the possibility that converges to some Σ∞ ∈ C Σ∞ is ﬁnite is ruled out. Hence, by Corollary 2.7 t0 is the value of K(an /bn ). Truncation error bounds. Both R+ and R+ Let R+ be the set of positive numbers, and R+ be its closure in C. are simple value sets for the family of positive continued fractions. Indeed, R+ is the 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 diam Sn (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 ≤ 0 if n odd, (2.1.9) (f − fn−1 ) is ≥ 0 if n even. We can even exploit this oscillatory character further to get |f − fn∗ | ≤

1 1 (f2n+1 − f2n ) for fn∗ := (f2n+1 + f2n ). 2 2

(2.1.10)

These simple a posteriori bounds secure reliable computation of positive continued fractions. We shall see later (Theorem 3.24 on page 128 and Corollary 3.41 on page 148) that the following a priori bounds hold:

3.2.1 Positive continued fractions

119 $

' Theorem 3.17. If K(an /bn ) is a positive continued fraction, then diam Sn

(R+ )

n √ 1 + 4ak − 1 √ ≤ 2a1 1 + 4ak + 1 k=2

and

1/b1 (1 + bk bk−1 ) k=2

diam Sn (R+ ) ≤ n

if all bn = 1,

(2.1.11)

if all an = 1.

(2.1.12)

&

%

Example 3. In Example 12 on page 26 we used the continued fraction Ln 2 =

1 1/2 1/(2 · 3) 2/(2 · 3) 2/(2 · 5) 3/(2 · 5) 3/(2 · 7) 1+ 1 + 1 1 1 1 1 + + + + +· · ·

to estimate the value of Ln 2. The ﬁrst seven approximants fn = Sn (0) were given in 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) * √ 1+ 1+2−1 |Ln 2 − f6 | ≤ 2 · √ ·* 1+2+1 1+

2 3

−1

2 3

+1

* ·*

1+

4 3

1+

4 3

−1

*

1+ ·* +1 1+

6 5

−1

6 5

+1

(2.1.13)

which is ≈ 2.8 · 10−3 . Now, a2k =

1 1/4 k/2 = + , 2k − 1 4 2k − 1

for k ≥ 1. Hence an ≤

1 4

a2k+1 =

1 1/4 k/2 = − 2k + 1 4 2k + 1

1/4 5

= 0.3 for n ≥ 6, and so √ √ √ √ √ 3−1 5− 3 7− 3 √ ·√ √ × |Ln 2 − fn | ≤ 2 · √ ·√ 3+1 5+ 3 7+ 3 n−5 √ √ √ 9− 5 2.2 − 1 √ · √ ×√ 9+ 5 2.2 + 1 +

(2.1.14)

for n ≥ 4, which is an easier bound to compute. In the table below we compare the actual value of |Ln 2 − fn | to the three bounds given above. n 10 11 12 13

fn .693147157853 . . . .693147184962 . . . .693147179886 . . . .693147180688 . . .

|Ln 2 − fn | 2 · 10−8 4 · 10−9 7 · 10−10 1 · 10−10

(2.1.8) 2 · 10−7 3 · 10−8 5 · 10−9 8 · 10−10

(2.1.11) 5 · 10−8 1 · 10−8 2 · 10−9 3 · 10−10

(2.1.14) 1 · 10−7 2 · 10−8 4 · 10−9 8 · 10−10

120

Chapter 3: Convergence criteria

The improved approximant f3∗ := 12 (f7 + f6 ) gives for instance Ln 2 ≈ 0.693136 with |Ln 2 − f3∗ | ≤ 1.5 · 10−5 . The correct value is of course Ln 2 = 0.69314718055 . . . . 3

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+1

and

wn <

an+1 an+2 bn+1 + 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 recurrence relation for {Bn } that Bn > 0 for n ≥ 0. Hence ζn := Sn−1 (∞) = −Bn /Bn−1 < 0. For w > ζn , the derivative of Sn satisﬁes n (−a ) < 0 for n odd A B − B A k n−1 n n−1 n k=1 Sn (w) = = (Bn−1 w + Bn )2 (Bn−1 w + Bn )2 > 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 an+1 Sn+1 (wn+1 ) = Sn , 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 , an+2 bn+1 + 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 Sn (w) that

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 with all “ <” 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 − a an+1 −x= >0 1+x 1+x

and

an+1 an+1 − a + x(an+1 − an+2 ) > 0. an+2 − x = 1 + x + an+2 1+ 1+x

Remark: As in the previous remark, the inequalities in (2.1.17) are for instance reversed if (an − a) < 0 and (an − a) < x(an+1 − an ) for all n.

Example 4. The positive continued fraction ∞

1

1

1

1

2+ n 2+ 1 2+ 2 2+ 3 an := K n=1 1 1 + 1 + 1 +· · · + 1 +· · · has elements 0 < an → 2, and x = 1 is the positive root of the equation x = 1 2/(1 + x). Here (an − 2) = n1 > 0 and (an − 2) = n1 > x(an+1 − an ) = 1 · ( n+1 − n1 ) . Hence (2.1.17) holds. Since K(an /1) converges by the Seidel-Stern Theorem (see Remark 3 on page 101 and Theorem 3.2(ii) on page 102), this means that |f − Sn (x)| < |Sn+1 (x) − Sn (x)| and |f − fn∗ | ≤ 12 (S2n+1 (x) − S2n (x))

for fn∗ := 12 (S2n+1 (x) + S2n (x)).

(2.1.18)

122

Chapter 3: Convergence criteria

In the table below we have listed the approximants fn := Sn (0) and the truncation errors (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 3.19 and the idea of convergence acceleration from Example 3 on page 11. The value of K(an /1) is 1.37587055 correctly rounded to 8 decimals. n 5 6 7 8 9 10

fn 1.4599... 1.3342... 1.3974... 1.3649... 1.3814... 1.3730...

f − fn −0.084... 0.041... −0.021... 0.010... −0.005... 0.002...

fn+1 − fn −0.125... 0.063... −0.032... 0.016... −0.008... 0.004...

f − Sn (x) −0.0030... 0.0013... −0.00059... 0.00026... −0.00012... 0.000056...

Sn+1 (x) − Sn (x) −0.0043... 0.0019... −0.00085... 0.00039... −0.00017... 0.00008...

3

3.2.2

Alternating continued fractions

A continued fraction b0 + K(an /bn ) with real elements an and bn is called a real continued 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 part of the alternating continued fraction K(an /1) with a2 > −1 be a positive continued fraction. Then K(an /1) converges if and only if its even part converges. "

!

The canonical even part of K(an /1) is given by ∞

K cn =: 1 +a1a2 + 1 +−aa32a+3 a4 + 1 +−aa54 a+5 a6 +· · · .

n=1 dn

(2.2.1)

Since {an } is alternating in sign, this is a positive continued fraction if and only if a2n−1 > 0 and a2n < 0 for all n and a2n−1 + a2n ≥ −1 for n ≥ 2.

(2.2.2)

3.2.2 Alternating continued fractions

123

Proof of Theorem 3.20: Clearly, K(an /1) diverges if its even part diverges, so let its even part converge. That is, f2n = (A2n /B2n ) → f where 0 ≤ f < ∞ by the Seidel-Stern Theorem. Now,

A2n+1 B2n − A2n B2n+1 B2n+2

f2n+1 − f2n

= ·

A2n+2 B2n − A2n B2n+2 B2n+1

f2n+2 − f2n

B2n+2

B2n+2

=

=

B2n+1 B2n+2 − a2n+2 B2n

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 1 2 3 4 5 6 2n 2n + 1 1 1 − 1 + 1 − 1 + 1 − 1 + 1 −···+ 1 − 1 +··· has even part

(2.2.3)

∞

K cn := 01 + 1 0· 2 + 3 0· 4 + 5 0· 6 + 7 0· 8 +· · · n=1 dn which diverges by the Seidel-Stern Theorem. Hence the alternating continued fraction (2.2.3) diverges. The alternating continued fraction n n + 1/2 1 1/2 1 3/2 2 5/2 3 7/2 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 +· · · + 1 − 1 +· · · has even part

(2.2.4)

1 · 1 32 · 2 52 · 3 72 · 4 1 2 1/2 + 1/2 + 1/2 + 1/2 + 1/2 +· · ·

which is equivalent to ∞

K c1n := 12 + 1 1· 2 + 3 1· 4 + 5 1· 6 + 7 1· 8 +· · · .

n=1

Now, K(cn /1) converges by the Seidel-Stern Theorem and Theorem 3.2(i) on page 102 since ∞ ∞ 1 1 > = ∞. √ cn+1 2n n=1 n=1 Hence the convergence of (2.2.4) follows by Theorem 3.20. 3

124

Chapter 3: Convergence criteria

3.2.3

Stieltjes continued fractions

A continued fraction of the form ∞

b0 +

K an1z = b0 + a11 z + a12z + a13 z +· · · ;

n=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 number of continued fraction expansions of useful functions have this form. S-fractions are also essential in the Stieltjes moment theory (Section 1.5.2 on page 39). Luckily they have nice convergence properties ([Stie94]): '

$

Theorem 3.21. (Stieltjes continued fractions.) The even and odd parts of an S-fraction converge locally uniformly with respect to z in the cut plane D := {z ∈ C; | arg(z)| < π} to holomorphic functions. The S-fraction itself converges for z ∈ D if and only if n ∞

(−1)n−k+1

ak

= ∞.

(2.3.2)

n=1 k=1

&

%

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(an z/1).) We shall ﬁrst see that for ﬁxed z, the closed half plane V (z) := eiα H where

H := {w ∈ C; Re w > 0}

(2.3.3)

is a simple value set for K(an z/1). Let w ∈ V (z). Then − π2 + α ≤ arg w ≤ so − π2 + α < arg(1 + w) < π2 + α, and thus 2α − ( π2 + α) < arg

π 2

+ α,

an z < 2α + ( π2 − α), 1+w (1)

i.e., an z/(1 + V (z)) ⊆ V (z). This means that the classical approximants fn (z) of its ﬁrst tail satisfy fn(1) (z) := (1)

a2 z a3 z an+1 z ∈ V (z). 1 + 1 +· · · + 1 (1)

In particular fn (z) = −1. Therefore fn (z) := Sn (0) = a1 z/(1 + fn−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 StieltjesVitali Theorem on page 115, and {fn } converges locally uniformly in D if and only if (2.3.2) holds.

3.2.3 Stieltjes continued fractions

125

Remarks. 1. A very nice consequence of Theorem 3.21 is that an S-fraction converges to a holomorphic 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), the angular opening (or ray if α = 0) (−i e2iα H) ∩ (iH) if α ≥ 0, Vα := (2.3.4) (i e2iα H) ∩ (−iH) if α ≤ 0 between the rays R+ and e2iα R+ is also a value set for every S-fraction K(an z/1). If α = 0, then K(an z/1) is a positive continued fraction which was studied in Section 3.2.1. Let α = 0. Our next theorem shows that then every point in the interior of Vα , 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| < π2 , Vα◦ is the limit set for the family of continued fractions K(an /1) from the element set E := e2iα R+ .

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 (Corollary 2.10 on page 71). Since also the ﬁrst tail value f (1) for K(an /1) is the value of a continued 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 if arg(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 proves that f ∈ Vα◦ . Next, let f := r eiθ ∈ Vα◦ 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 of a 2-periodic continued fraction r1 e2iα r2 e2iα r1 e2iα r2 e2iα r1 e2iα 1 + 1 + 1 + 1 + 1 +· · ·

(2.3.5)

for some r1 > 0 and r2 > 0. Such a continued fraction clearly converges to a value in Vα◦ by the arguments above. A necessary condition for f to be its value is that f = r eiθ =

r1 e2iα 1 + r eiθ r2 e2iα 2iα = r e . 1 1 + 1 + r eiθ 1 + r2 e2iα + r eiθ

(2.3.6)

126

Chapter 3: Convergence criteria

Without loss of generality (complex conjugation) we assume that α > 0. Then 0 < θ < 2α, and 1 + r eiθ = ρ eiψ for some ρ > 0 and 0 < ψ < θ. Hence we need r eiθ =

r1 ρ ei(2α+ψ) . ρ eiψ + r2 e2iα

(2.3.7)

Let r2 > 0 be chosen such that arg(ρ eiψ + r2 e2iα ) = 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 ([HePf66]) have proved a posteriori truncation error bounds for clas sical approximants fn (z) := Sn (0) of Stieltjes fractions. These bounds are based on the value set Vα given by (2.3.4). What they actually proved though, is in fact a bound for the diameter of Sn (Vα ):

'

$

Theorem 3.23. (The Henrici-Pﬂuger Bounds.) For given z = 0 with α := 12 arg z ∈ (− π2 , π2 ), let Vα be given by (2.3.4). Then

diam Sn (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(an z/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 2α

Sn (R)

2α fn−1

Sn (Re2iα )

α = 0. Without loss of generality (complex conjugation) we assume that α > 0. The set Kn := Sn (Vα ) is bounded by the circular arcs Sn (R+ ) and Sn (e2iα R+ ). These two arcs meet at Sn (0) = fn = fn (z) and Sn (∞) = fn−1 = fn−1 (z) at the angle 2α since Sn is a conformal map. Indeed, 2α is the 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 region, for instance as illustrated in the present ﬁgure. Now, if Kn is not “ too fat”, then diam(Kn ) = |fn − fn−1 |. Otherwise one measures the diameter along the perpendicular bisector between fn−1 and

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 diam(Kn ) is less than or equal to the diameter 2Rn of two equally large circles which intersect at fn and fn−1 under an angle 2α. The ﬁgure illustrates the situation. Here C is the center of the left circle, and B is the midpoint between the two centers.

fn θ/2 Rn

θ/2 A

π−θ

Let θ2 denote the angle ∠CAfn . Then also ∠Afn C = θ2 and the angle ∠Afn B is equal to π2 − θ2 which again is less than α; i.e., π − θ < 2α. Therefore, since π < π − θ < 2α < π, the common radius 2 Rn of these two circles is given by

C B

α fn−1

|fn − fn−1 | |fn B| = sin θ 2 sin θ |fn − fn−1 | |fn − fn−1 | = < . 2 sin(π − θ) 2| sin 2α|

Rn = |fn C| =

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(an z/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 and Warner ([GrWa83]), and use the half plane V := V (z) = eiα H which also is a value set for K(an z/1) for given z, even though it is normally quite a lot larger.

128

Chapter 3: Convergence criteria $

' Theorem 3.24. (The Thron-Gragg-Warner Bounds.) For given z := rei2α with − π2 < α < π2 and r > 0, n a1 r 1 + 4ak r/ cos2 α − 1 diam(Sn (V )) ≤ 2 cos α 1 + 4ak r/ cos2 α + 1 k=2

(2.3.12)

for the S-fraction K(an z/1) and V := eiα H. &

%

Since ∞ ∈ V , it follows in particular that

n a1 r 1 + 4ak r/ cos2 α − 1 |fn+m − fn−1 | ≤ 2 cos α 1 + 4ak r/ cos2 α + 1 k=2

(2.3.13)

for m, n ∈ N, which essentially was the original result by Thron and Gragg-Warner. If K(an z/1) converges to f (z), then |f (z) − fn−1 (z)| and |f (z) − Sn (w)| for w ∈ V are 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 z/6 2z/6 2z/10 3z/10 K an1z = 1z + z/2 1 + 1 + 1 + 1 + 1 +· · · 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). Let √ now z := 1 + i = 2 eiπ/4 . Then still |Ln(1 + i) − fn (1 + i)| ≤ |fn (1 + i) − fn+1 (1 + i)| by (2.3.8), and by (2.3.13) we get the a priori bounds * √ √ n+1 1 + 4ak 2/ cos2 π8 − 1 2 2 * |Ln(1 + i) − fn (1 + i)| ≤ √ cos π8 1 + 4ak 2/ cos2 π8 + 1 k=2 * * √ √ 2 1 + w − 1 1 + 43 w − 1 2 2 1 + 2w − 1 3 ≤ ·* ·* × ·√ cos π8 1 + 2w + 1 1 + 23 w + 1 1 + 43 w + 1 * * 1 + 45 w − 1 1 + 65 w − 1 n−5 ·* ×* 1 + 45 w + 1 1 + 65 w + 1

(2.3.14)

(2.3.15)

´ 3.2.4 The Sleszy´ nski-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 order of the true truncation errors compared to the one in (2.3.14) and the two in (2.3.15): n 10 11 12 13

|Ln(1 + i) − fn | 5.7 · 10−7 1.5 · 10−7 3.1 · 10−8 7.8 · 10−9

|fn+1 − fn | 7.1 · 10−7 1.8 · 10−7 3.7 · 10−8 9.3 · 10−9

(2.3.15)1 1.1 · 10−5 2.4 · 10−6 6.0 · 10−7 1.4 · 10−7

(2.3.15)2 1.7 · 10−5 4.5 · 10−6 1.2 · 10−6 3.2 · 10−7

3

´ The Sleszy´ nski-Pringsheim Theorem

3.2.4

A continued fraction K(an /bn ) with |bn | ≥ |an | + 1

for n ≥ 1

(2.4.1)

´ is called a Sleszy´ nski-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, ´ ´ nski ([Sles89]) proved that 0 ∈ D, so fn = Sn (0) ∈ D for all n. In 1889 Sleszy´ (2.4.1) was suﬃcient for the (classical) convergence of K(an /bn ) (although he never used 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 with ϕn (w) ≡ w and wn := ∞. But the classical proof gives some additional information on the convergence: '

$

´ Theorem 3.25. (The Sleszy´ nski-Pringsheim Theorem.) A ´ Sleszy´ nski-Pringsheim continued fraction converges absolutely to some value f with 0 < |f | ≤ 1. Moreover, {|An |} and {|Bn |} are strictly increasing sequences, and 0 < |Sn (w)| ≤ 1 for all w ∈ D. &

%

Proof :

Solutions {Xn } of the recurrence relation Xn = bn Xn−1 + an Xn−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 |) n (|bk | − 1) > 0 ≥ · · · ≥ (|Xm | − |Xm−1 |) k=m+1

(2.4.3)

130

Chapter 3: Convergence criteria

for n ≥ m. Now, {An } and {Bn } are such solutions with |A1 | − |A0 | = |a1 | > 0 ∞ and |B0 | − |B−1 | = 1 > 0. Therefore {|An |}∞ 0 and {|Bn |}0 are strictly increasing sequences. Moreover, |Bn | − |Bn−1 | ≥

n

(|bk | − 1) ≥

k=1

n

|ak |

(2.4.4)

k=1

by (2.4.3). The determinant formula on page 7 therefore gives that

n

An |Bn | − |Bn−1 | 1 1 An−1

k=1 |ak |

(2.4.5)

Bn − Bn−1 = |Bn Bn−1 | ≤ |Bn Bn−1 | = |Bn−1 | − |Bn | where ∞ n=1 (1/|Bn−1 |−1/|Bn |) = 1−1/ lim |Bn |. Hence the absolute convergence of fn = Sn (0) = An /Bn is established. That |Sn (w)| ≤ 1 for w ∈ D and thus |f | ≤ 1 follows from the nestedness of {Sn (D)}. This also holds true for the ﬁrst tail of (1) K(an /bn ); i.e., |Sn−1 (w)| ≤ 1 and |f (1) | ≤ 1. Since |a1 /(b1 +q)| ≥ |a1 |/(|b1 |+1) > 0 for |q| ≤ 1, it therefore follows that |Sn (w)| > 0 and |f | > 0. Example 7. We consider the continued fraction ∞

b0 +

2

2

2

z z . K an := 1 + 2 2z− z + z6 + 10 n=1 bn +· · · + 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 n0 th tail of b0 + K(an /bn ) converges, and thus the continued fraction itself converges. Hence (2.4.6) converges for every z ∈ C. Indeed, its value is ez ((2.2.1) on page 268). 3

´ nski-Pringsheim continued fraction, Remark: If K(an /bn ) is equivalent to a Sleszy´ then also K(an /bn ) converges absolutely, and its classical approximants are still contained in the open unit disk D. Therefore K(an /bn ) converges if (2.4.7) |b1 | ≥ |a1 | + |a2 |, |bn | ≥ |an | + |an+1 | for n ≥ 2, since then

a2 / |a2 a3 | a3 / |a3 a4 | a1 / |a2 | an ∼ n=1 bn b1 / |a2 | + b2 / |a3 | + b3 / |a4 | + · · · ∞

K

´ which is a Sleszy´ nski-Pringsheim continued fraction. Similarly, if |b1 | ≥ |a1 | + |a2 |,

|bn | ≥ |an+1 | + 1

for n ≥ 2,

then K(an /bn ) converges since ∞

/a2 K an ∼ ab11/a n=1 bn 2

1/a3 1/a4 + b2 /a3 + b3 /a4 + · · ·

(2.4.8)

´ 3.2.4 The Sleszy´ nski-Pringsheim Theorem

131

´ is a Sleszy´ nski-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 Sleszy´ nski-Pringsheim continued fraction:

´ Theorem 3.26. D \ {0} is the limit set for the family of Sleszy´ nskiPringsheim continued fractions.

Proof : It follows from Theorem 3.25 that the value of a continued fraction is contained in D \ {0}. We need to prove ´ is the value of a Sleszy´ nski-Pringsheim continued fraction. a := 1 − b. Then the periodic continued fraction

´ Sleszy´ nski-Pringsheim that every f ∈ D \ {0} For a ﬁxed b > 2 let

∞

K a = ab + ab + ab +· · ·

n=1 b

´ is a Sleszy´ nski-Pringsheim continued fraction which converges to a root of the equation a/(b + x) = x. Since this value must be ∈ D, it follows that K(a/b) converges to −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 a a a b1 + b + b + b +· · · converges to a1 /(b1 − 1) = f . Truncation error bounds. ´ In [BeLo01] we studied the speed of convergence of Sleszy´ nski-Pringsheim continued fractions. From (2.4.4) it follows immediately that |Bn | ≥ Σn :=

n

Pk

where Pk :=

k

(|bm | − 1).

(2.4.9)

m=1

k=0

Therefore, by (2.4.5) |f − fn | ≤

∞

|fk − fk−1 | ≤

∞

k j=1

|aj |

≤

∞

k j=1

|aj |

,

(2.4.10)

∞ ∞ 1 1 1 1 Σk − Σk−1 = = − − . Σk Σk−1 Σk−1 Σk Σn Σ∞

(2.4.11)

k=n+1

k=n+1

|Bk Bk−1 |

k=n+1

Σk Σk−1

and since |aj | ≤ |bj | − 1 and Pk = Σk − Σk−1 , |f − fn | ≤

k=n+1

k=n+1

132

Chapter 3: Convergence criteria

Alternatively, (2.4.4) and (2.4.9) lead to |f − fn | ≤

1 1 − ∗ Σ∗n Σ∞

where Σ∗m :=

m

Pk∗ and Pk∗ :=

k

|aj | .

(2.4.12)

j=1

k=0

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 have full 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 1 − . Σ∞ 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 n (−ak ) fn − fn−1 = − k=1 < 0, Bn Bn−1

(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. an /bn ) be the two continued fractions given by Let K(˜ an /˜bn ) and K( a ˜n := an := −|an |,

˜bn := |bn |

and bn := |an | + 1.

(2.4.16)

an /˜bn ) as described in case 2. We Then K( an /bn ) is as described in case 1 and K(˜ shall prove (with obvious notation): $

' ´ Theorem 3.27. Let K(an /bn ) be a Sleszy´ nski-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),

&

1 1 − . |f − fn | ≤ f˜n − f˜ ≤ fn − f = Σn Σ∞

(2.4.17) %

´ 3.2.4 The Sleszy´ nski-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), f˜n − f˜ = −

n k k=1

+ ˜k B ˜k−1 |am | B

m=1

with obvious notation. The ﬁrst inequality in (2.4.17) follows therefore from (2.4.10) ˜n , and thus if we can prove that |Bn | ≥ B ˜n ≥ B n = Σn |Bn | ≥ B

for n ≥ 1.

(2.4.18)

˜n . Then X−1 = X0 = 0 and Let Xn := |Bn | − B ˜n = |bn |Xn−1 − |an |Xn−2 , Xn ≥ |bn Bn−1 | − |an Bn−2 | − B 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 Sleszy´ nskiPringsheim continued fractions which converge arbitrarily slowly. (Just choose tn := n k −1 and bn > 1 so that Σn = k=0 j=1 (bj − 1) converges as slowly as you want, and set an := −(bn − 1) for all n.) Hence the convergence is not uniform with ´ respect to the family of Sleszy´ nski-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 Sleszy´ nski-Pringsheim continued fraction. Then n k=1 |ak | rad Sn (D) = (|Bn | − |Bn−1 |)(|Bn | + |Bn−1 |) n |ak |/(|bk | − 1) 1 1 . ≤ ≤ ≤ k=1 |Bn | + |Bn−1 | |Bn | + |Bn−1 | Σn + 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 S˜n (D) ≤ rad Sn (D) = &

1 , Σn + Σn−1

(2.4.20) %

134

Chapter 3: Convergence criteria

Proof : The equality in (2.4.19) follows from (1.4.3) on page 110. The ﬁrst inequality follows from (2.4.4). The second inequality is a consequence of (2.4.1), and the 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 ˜n | − |B ˜n−1 | ≥ |B n | − |B n−1 | = Σn − Σn−1 . |Bn | − |Bn−1 | ≥ |B n = Σn , so the equality is clear. Moreover, n | = B Of course, |B ˜ n | − |B ˜n−1 | = B ˜n − B ˜n−1 = (|bn | − 1)B ˜n−1 − |an |B ˜n−2 |B n ˜n−1 − B ˜n−2 ) ≥ n − B n−1 , ≥ (|bn | − 1)(B (|bk | − 1) = B k=1

and ﬁnally ˜n − B ˜n−1 ) ≥ 0 Qn := (|Bn | − |Bn−1 |) − (B since

˜n−1 ) − |an |(|Bn−2 | − B ˜n−2 ) Qn ≥ (|bn | − 1)(|Bn−1 | − B n ≥ (|bn | − 1)Qn−1 ≥ · · · ≥ Q1 (|bk | − 1) ≥ 0. k=2

´ Remark. There exist Sleszy´ nski-Pringsheim continued fractions for which the limit n an /bn ) the radius R point case for Sn (D) fails to occur. For instance, for K( converges to a positive value if Σ∞ < ∞. Example 8. For given q ∈ C with |q| = 1, the continued fraction ∞

K z = 2qz n=1 2q n

z z + 2q 2 + 2q 3 + · · ·

´ is a Sleszy´ nski-Pringsheim continued fraction for |z| ≤ 1. So let |z| ≤ 1. Then K(z/2q n ) 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 ∞ k=n+1

∞ |z|k |z|k ≤ Σk Σk−1 (k + 1)k

1 = (k + 1)k

k=n+1 ∞

k=n+1

1 k

−

1 1 . = k+1 n+1

For |z| < 1 we also have the bound |f (z) − fn (z)| <

∞ k=n+1

|z|n+1 /(1 − |z|) |z|k = . (n + 2)(n + 1) (n + 2)(n + 1)

3.2.5 Worpitzky’s Theorem

135

We can also majorize the expression by a telescoping series to get

|f (z) − fn (z)| ≤ <

∞ |z|k |z|k − k k+1

k=n+1 ∞ k=n+1

|z|k |z|k+1 |z|n+1 = − . k k+1 n+1

3

3.2.5

Worpitzky’s Theorem

Worpitzky was a high school teacher who published deep results in the yearly report from his school. His convergence result from 1865 ([Worp65]) can be stated as follows: $

' Theorem 3.29. (Worpitzky’s Theorem.) Let 0 < |an | ≤

1 4

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 directly ´ from the Sleszy´ nski-Pringsheim Theorem on page 129 since 2K(an /1) is equivalent ´ to the Sleszy´ nski-Pringsheim continued fraction 4an 4a1 4a2 4a3 . 2 + 2 + 2 +· · · + 2 +· · ·

(2.5.2)

This equivalence also means that V := 12 D is a simple value set for the continued ´ nskifractions K(an /1) from E := 14 D, and the truncation error bounds for Sleszy´ Pringsheim continued fractions can easily be adjusted to K(an /1) from E. We still chose to include Worpitzky’s Theorem since it is so easy to apply and so widely known and used. The following version of Worpitzky’s Theorem is more general. It can be derived from the work of Pringsheim ([Prin05]) and Wall ([Wall48], p 45-50):

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) converges to some value f with 0 < |f | ≤ g0 , and {gn D}∞ n=0 is a sequence of value sets for K(an /1). &

%

´ nski-Pringsheim continued This follows since g0−1 K(an /1) is equivalent to the Sleszy´ fraction an /gn−1 gn /g0 g1 a2 /g1 g2 a3 /g2 g3 K dcnn := a11/g . 1/g 1/g 1/gn +· · · + + +· · · + 1 2 3 The choice gn :=

1 2

(2.5.4)

for all n gives back the original Worpitzy Theorem.

The sequence {−gn }. an /1) given The sequence {−gn }∞ n=0 is a tail sequence for the continued fraction K( by (2.5.5) an := −gn−1 (1 − gn ) for all n. This continued fraction is special since • it deﬁnes En as {a ∈ C; |a| ≤ | an |} = | an |D, cn /dn ) where cn := −(1 − gn )/gn and dn := 1/gn , so • it is equivalent to g0 K( ´ nski-Pringsheim continued fraction as described in case K( cn /dn ) is a Sleszy´ 1 on page 132. It follows from Theorem 2.6 on page 66 (and also K( an /1) has approximants fn and denominators n = Σ n , fn = g0 − g0 , f = g0 − g0 , B n ∞ Σ Σ where

n := nΣn Σ

k=1 gk

=

n k=0

Pk

from Property 1 on page 78) that n given by B g0 g0 f − fn = − Σ∞ Σn k (2.5.6) 1 − gj with Pk := . gj j=1

∞ = ∞, then {−gn } is the sequence of tail values for K( an /1) (Corollary 2.7 If Σ on page 68). Otherwise K( an /1) still converges, but it has another sequence {f(n) } (n) ∈ Vn according to Theorem 3.30, and f(n) = −gn . That is, of tail values. Also f (n) −gn < f , and thus also f(n−1) =

an gn−1 (1 − gn ) =− < −gn−1 (1 − gn ) < 0. 1 + f(n) 1 + f(n)

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

∞

where Pk :=

Pk < ∞,

k=0

k 1 − gj . gj j=1

Then there exists a sequence {gn∗ } with gn (1 − gn+1 ) < gn∗ < gn for all n ≥ 0 ∗ (1 − gn∗ ) = gn−1 (1 − gn ) and such that gn−1 ∗ = ∞ Σ ∞

∗ := where Σ n

n

Pk∗ and Pk∗ :=

k 1 − gj∗ . gj∗

j=1

k=0

&

%

Remark: Corollary 2.8 on page 69 shows that n − g0 (g0 − g0∗ )Σ gn∗ = −f(n) = gn ∗ (g0 − g0 )Σn−1 − g0

for n ≥ 1

∞ + g0 (Corollary 2.7 on page 68). That is, where g0∗ = −f = −g0 /Σ gn∗ = gn

n ∞ − Σ Σ n−1 ∞ − Σ Σ

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 given an | = | a|, and by (2.5.5). If all gn = g, then g := 12 gives the maximal value for | we are back to the original Worpitzky Theorem. If {gn } is allowed to vary, we can actually obtain | an | > 14 for all n. For instance, if gn := then

| an | =

1 1 + 2 4n + 2

1 1 + 2 4n − 2

for all n,

1 1 − 2 4n + 2

=

1/4 1 + 2 . 4 4n − 1

(2.5.8)

(2.5.9)

This is not an optimal choice, though. Indeed, there exists no optimal choice of this type:

138

Chapter 3: Convergence criteria

Lemma 3.32. Let 12 < gn < 1 be chosen such that | an | := gn−1 (1−gn ) > 14 for 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 ).

n and Pn be given by (2.5.6), and set Let Σ

Proof :

gn := gn (1 + εn ) where

εn :=

Pn Pn−1 . n−1 2Σ

n−1 Then gn < gn . Now, 0 < (1 − gn )/gn < 1, so 0 < Pn < Pn−1 < 1 and Pn−1 < Σ for all n ≥ 2. Therefore gn = gn + gn Moreover

P 2 Pn Pn−1 = gn + (1 − gn ) n−1 < gn + (1 − gn ) = 1. n−1 n−1 2Σ 2Σ

gn εn an | = gn−1 (1 − gn ) (1 + εn−1 ) 1 − | an | − | −1 1 − gn gn εn gn εn εn−1 − = | an | εn−1 − 1 − gn 1 − gn

where P2 P3 Pn−2 Pn−1 Pn−2 gn εn gn εn εn−1 − = − n−1 − n−1 n−2 n−2 n−1 n−1 Σ 1 − gn 1 − gn 2Σ 2Σ 4Σ Pn−1 2 n−1 − 2Pn−1 Σ n−2 − Pn−1 = Pn−2 2Pn−2 Σ n−1 Σ n−2 4Σ Pn−1 n−1 − Σ n−2 ) − P2 Pn−2 2Pn−1 (Σ > n−1 n−1 Σ n−2 4Σ Pn−1 2 = 2Pn−1 Pn−2 − Pn−1 Pn−2 n−1 Σ n−2 4Σ P 2 Pn−2 = n−1 2 − Pn−1 > 0. n−1 Σ n−2 4Σ

εn−1 −

(k)

Indeed, it was proved in [JaMa90] that there is a hierarchy of sequences {gn }∞ n=0 for k = −1, 0, 1, 2, . . . given by (k) gnk +n

k 1 1 := + 2 m=0 4Lm (n)

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 :=

k 1 1 + 4 m=0 (4Lm (n))2

(2.5.10)

from some n on, and the continued fraction K(an /1) diverges when k−1 1 1+ε 1 − an := − − 4 m=0 (4Lm (n))2 (4Lk (n))2

(2.5.11)

for all n suﬃciently large, where k ∈ N and ε > 0 are arbitrarily chosen. In 1+ε particular K(an /1) with an := − 14 − 16n 2 diverges for ε > 0. Limit sets. ∞ = ∞, then every f ∈ V0 \{0} is the value of a continued fraction K(an /1) from If Σ an /1) with an given by (2.5.5) {En }. This follows by the following arguments: K( converges to −g0 , and thus its ﬁrst tail converges to −g1 . Let f ∈ V0 \ {0, −g1 } be arbitrarily chosen, and set a := f · (1 − g1 ). Then |a| ≤ g0 (1 − g1 ) and a a2 a3 a4 1 + 1 + 1 + 1 +· · · ∞ < ∞, this is no longer true. Then V0 is “ too large”. Indeed, converges to f . If Σ no value f ∈ V0 \ V0∗ with V0∗ 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 {gn D \ {0}}n=0 is the sequence of limit sets for the family of continued fractions 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 ). an ≤ a ˜n < 0 is equivalent to g0 K(˜ cn /d˜n ) where c˜n := a ˜n /gn gn−1 and K(˜ an /1) with ´ d˜n := 1/gn , and thus K(˜ cn /d˜n ) is a Sleszy´ nski-Pringsheim continued fraction as ˜n and approximants described in case 2 on page 132. Therefore the denominators B f˜n for K(˜ an /1) satisfy ˜n ≥ Σ n ˜n+1 ≥ B B

and

− gn < f˜n < f˜n−1 < 0

(2.5.12)

140

Chapter 3: Convergence criteria

n still is given by (2.5.6). For the general case we get from Theorem 3.27 where Σ (again with obvious notation): '

$

Theorem 3.34. Let {gn } and K(an /1) be as in Theorem 3.30, and let K(˜ an /1) and K( an /1) be given by a ˜n := −|an | and an := −gn−1 (1 − gn ) for all n. Then g0 g0 |f − fn | ≤ f˜n − f˜ ≤ fn − f = − . Σn Σ∞

&

(2.5.13) % $

' Corollary 3.35. For given positive r ≤ 1, let κ := |an | ≤

r 4

for K(an /1), then

|f − fn | ≤

&

⎧ 1/2 ⎪ ⎪ ⎪ ⎪ ⎨n + 1

√ 1+ 1−r √ . If all 1− 1−r

if r = 1,

√ ⎪ ⎪ ⎪ 1−r ⎪ ⎩ n+1 κ −1

(2.5.14) if r < 1.

√ Proof : Let gn := g := (1 − 1 − r)/2 for all n. Then |an | ≤ g(1 − g) = n = n + 1 if κ = 1 and κ = (1 − g)/g. Therefore Pk = κk and Σ n = Σ

n k=0

κn =

%

r 4

and

κn+1 − 1 κ−1

n → Σ ∞ = ∞. If r = 1, then κ = 1 and g = 1 . If r < 1, otherwise. In both cases Σ 2 1 then κ > 1 and g < 2 and √ (κ − 1)g g0 1−r 1 − 2g = n+1 = n+1 = n+1 . κ − 1 κ − 1 κ −1 Σn The result follows therefore from Theorem 3.34.

3.2.5 Worpitzky’s Theorem

141

The radius of Sn (gn D). '

$

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 n k=1 |ak | rad Sn (gn D) = 2 |Bn | /gn − gn |Bn−1 |2 (2.5.15) n g0 |ak | g0 ≤ . ≤ n + Σ n + Σ n−1 n−1 gk−1 (1 − gk ) Σ Σ k=1 an /1) be given by a ˜n := −|an | and an := Let further K(˜ an /1) and K( −gn−1 (1 − gn ). Then rad Sn (gn D) ≤ rad S˜n (gn D) ≤ rad Sn (gn D) =

g0 . n−1 n + Σ Σ

(2.5.16)

&

%

Proof : The equivalence (2.5.4) shows that the radius Rn of Sn (gn D) is equal to g0 · (radius of Tn (D)) where Tn (w) =

Cn−1 w + Cn c1 c2 c3 cn =: d1 + d2 + d3 +· · · + dn + w Dn−1 w + Dn

an 1 ´ and dn := . Here K(cn /dn ) is a Sleszy´ nski-Pringsheim gn gn−1 gn continued fraction, and the approximants of K(an /1) satisfy with cn :=

Sn (w) :=

An−1 w + An ; Bn−1 w + Bn

An := Cn

n k=0

gk ,

Bn := Dn

n

gk

k=1

for n ≥ 2, so by Theorem 3.28 on page 133 n g0 nk=1 |ck | g0 |ck | g0 ≤ Rn = ≤ 2 2 |Dn | − |Dn−1 | |Dn | + |Dn−1 | |dk | − 1 |Dn | + |Dn−1 | k=1

where by (2.4.9) |Dn | ≥

n k

(|dm | − 1) =

k=0 m=1

n k 1 n . −1 =Σ gm m=1

k=0

Therefore (2.5.15) follows since |ak | ≤ gk−1 (1 − gk ). Similarly we have g0 K(˜ an /1) ∼ an /1) ∼ K( cn /dn ) where c˜n := −|an |/gn gn−1 , d˜n := 1/gn , K(˜ cn /d˜n ) and g0 K( cn |. Hence also (2.5.16) follows from cn := −(1 − gn )/gn and dn := 1/gn = 1 + | Theorem 3.28.

142

Chapter 3: Convergence criteria

Remark: The limit point case Sn (gn D) → {f } occurs for every continued fraction ∞ = ∞. Indeed, if Σ ∞ < ∞, then Vn := gn D K(an /1) from {En } only if Σ ∗ contains both −gn and −gn (as given in Theorem 3.31), and thus Sn (Vn ) contains both −g0 and −g0∗ for all n for the continued fraction K( an /1) from {En } with an := −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| ≤

π 2

− ε} ∪ {0}

(2.6.1)

for some arbitrarily given 0 < ε < π/2. Let Vε := Gε ∪ {∞} Then sn (Vε ) =

1 ⊆ Vε bn + Vε

for ε ≥ 0.

for

(2.6.2)

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 Theorem 3.13 on page 117 that a positive continued fraction K(1/bn ) converges if and only if bn = ∞, where the case bn < ∞ is described by the Stern-Stolz Theorem on page 100. Van Vleck ([VanV01]) showed that this is true, not only for positive continued 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 if and only if |bn | = ∞. If K(1/bn ) diverges, then it diverges generally, and its even and odd classical approximants 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 page 101).

3.2.6 Van Vleck’s Theorem

143

Let |bn | = ∞. Following Jones and Thron ([JoTh80], p 89) we shall use the Stieltjes-Vitali Theorem to prove that then K(1/bn ) converges. Let βn := arg(bn ) and

dn (z) := |bn |eiβn z

for all n .

(2.6.5)

(If bn = 0, we just set βn := 0.) Then |βn | ≤ π2 − and dn (z) ∈ G /2 if | arg(dn (z))| ≤ π − 2 ; i.e., 2 $ π− % dn (z) ∈ G /2 if z ∈ D := z ∈ C; |Re(z)| ≤ . (2.6.6) π − 2 Now, dn (z) ≥ 0

if z ∈ D∗ := {z ∈ D; Re(z) = 0} .

(2.6.7)

∗

Since for every ﬁxed z ∈ D |dn (z)| = |bn | = ∞ , |bn |e−βn Im(z) > e−|Im(z)|π/2 it follows from the Seidel-Stern Theorem on page 117 that K(1/dn (z)) converges to a 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 D which converges for z ∈ D∗ . By the Stieltjes-Vitali Theorem on page 115 it follows therefore that K(1/dn (z)) converges for all z ∈ D◦ . In particular it converges for z = 1. Hence K(1/bn ) converges. Remark. Equivalence transformations show for instance that the conclusions of Van 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 −1 −1 −1 −1 −1 + i + 1 + i + −1 + i + 1 + i + −1 + i + 1 + i + · · · is equivalent to the continued fraction 1 1 1 1 1 i . 1 + i + 1 − i + 1 + i + 1 − i + 1 + i + 1 − i +··· (Use rn := −i in the equivalence transformation.) This new continued fraction has |bn | = ∞, the original the form i · K(1/bn ) where all bn ∈ Gε for ε := π4 . Since continued fraction converges. It is rather easy to ﬁnd its value f . It must be f = ix where x satisﬁes the equation 1

. 1 1+i+ 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). 3 x=

144

Chapter 3: Convergence criteria

Limit set.

Theorem 3.38. For given 0 ≤ ε ≤ π/2, Vε \ {0, ∞} is the limit set for the family 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 tail value f (1) are ﬁnite and belong to Vε . Hence also f = 1/(b1 + f (1) ) = 0, and thus f ∈ Vε \ {0, ∞}. We need to prove that every f ∈ Vε \ {0, ∞} is the value of a continued fraction K(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 1 1 −i iε = e =: f ∈ ∂Vε = b2 + f i(r2 + r)e−iε r2 + r and

1 b1 + f

=

1 = i r e−iε = f . −i(r1 + r21+r )eiε

That is, the 2-periodic sequence f, f, f, f, f, . . . is a tail sequence for the 2-periodic continued fraction ∞

K 1 := b11 + b12 + b11 + b12 + b11 + b12 +· · · . n=1 bn Now, K(1/bn ) converges (the Van Vleck Theorem), and its value must belong to Vε . Moreover, its value must be a solution of the quadratic equation * 1 = x; i.e., x = 12 (−r1 r2 ± r12 r22 + 4ir2 e−iε ). 1 b1 + b2 + x

3.2.6 Van Vleck’s Theorem

145 Only one of the two solutions belongs to Vε . Hence K(1/bn ) converges to f . This proves that ∂Vε \ {0, ∞} belongs to the limit set. If ε = π/2 we are ﬁnished, since then Vε◦ = ∅. Let ε < π/2, and let f := r eiα ∈ Vε◦ with −( π2 −ε) < α < π2 −ε and 0 < r < ∞ be arbitrarily chosen. Then 1 1 −iα ∈ Vε◦ , and it suﬃces to ﬁnd f = re a b ∈ Gε and an f (1) ∈ ∂Vε such that f = r eiα = b+f1 (1) ; i.e., b + f (1) = f1 . This can for instance be done by drawing lines L1 and L2 through f1 parallel to the two boundary lines (see the picture). The points where L1 and L2 intersects the boundary of Vε can then be our points f (1) and b.

Im 6 f (1)

ε L1

1 f

ε

= b + f (1)

Re

L2 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 1 1 1 1 1 0 + 1 + 0 + 1 + 0 +· · ·

and

1 1 1 1 1 1 + 0 + 1 + 0 + 1 +· · ·

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 if ε ≥ π/4, |fn − fn−1 | diam Sn (Vε ) ≤ |fn − fn−1 |/ sin 2ε if ε < π/4. & Proof :

% As always, H denotes the right half plane, so Vε can be written Vε = (eiε H) ∩ (e−iε H).

Now, Sn (eiε H) and Sn (e−iε H) are two circular disks since −b1 ∈ e±iε H, and thus Sn (e±iε H) is bounded. The boundaries of these two disks intersect at Sn (0) = fn

146

Chapter 3: Convergence criteria

and Sn (∞) = fn−1 under the inner angle 2α := π − 2ε. That is, Sn (Vε ) is a lensshaped set as illustrated in the ﬁgure on page 126. Therefore the result follows just 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 which gives no information on the speed of convergence. We therefore give an alternative proof which also opens up for an extension of Van Vleck’s classical result and for truncation error bounds. It is due to Jensen ([Jens09]), and was later found, independently, by Lange ([Lange99a]).

Proof 2 of Van Vleck’s Theorem: value set for K(1/bn ). Therefore

Also the half plane V0 := H is a simple

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 all Kn circular disks. Since 0 ∈ V0 , we have fn ∈ Kn , so all Bn = 0, and thus ζn := Sn−1 (∞) = −Bn /Bn−1 = ∞ for all n ∈ N. ζn is symmetric to (−ζ n ) with respect to ∂V0 = i R+ . The center Cn of Kn is symmetric to ∞ with respect to ∂Kn = ∂Sn (V0 ). Therefore, by Property 2 on page 109, Cn = Sn (−ζ n ). Hence, the radius of Kn is

An An − An−1 ζ n

Rn = |Sn (0) − Cn | =

− Bn Bn − Bn−1 ζ n

=

|(An Bn−1 − An−1 Bn )ζ n | 1 · |B n /B n−1 | = n |Bn (Bn − Bn−1 ζ n )| |Bn Bn−1 ( BBn−1 + BB n )| n−1

1 |B n /Bn | = . = |Bn B n−1 + B n Bn−1 | 2|Re(Bn B n−1 )| Now, set Qn := Re(Bn B n−1 ). Then the recurrence relation for {Bn } gives Qn = Re(bn |Bn−1 |2 + Bn−2 B n−1 ) = Qn−1 + |Bn−1 |2 Re(bn ).

(2.6.9)

Since Q1 = Re(b1 ) > 0 and Re(bn ) ≥ 0, this means that Qn > 0 and non-decreasing as n increases. Let Q := lim Qn . If Q = ∞, then Rn → 0, and the convergence is clear. Therefore also |bn | = ∞ (the Stern-Stolz Theorem).

3.2.6 Van Vleck’s Theorem

147

Let Q < ∞. Then

∞

∞

1 1

1 1 1 1

− − − = = < ∞.

Qn−1

Q Q Q Q Q n n−1 n 1 n=2 n=2

(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. Re(bn ) Then λ := sin ε > 0 and |bn | ≤ cos(π/2−ε) = Re(bn )/λ, so

An Bn−2 − Bn An−2 bn

Re(bn )

=

|fn − fn−2 | =

Bn Bn−2 ≤ λ|Bn Bn−2 | Bn Bn−2 Qn − Qn−1 |Bn−1 |2 Re(bn ) = λ · |Bn Bn−1 | · |Bn−1 Bn−2 | λ|Bn Bn−1 | · |Bn−1 Bn−2 | 1 Qn − Qn−1 1 1 1 ≤ · = · − λ Qn Qn−1 λ Qn−1 Qn

=

(2.6.11)

where we have used the equality (2.6.9). Hence |fn − fn−2 | < ∞, and the even and odd approximants of K(1/bn ) converge absolutely. Van Vleck’s Theorem follows therefore 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. diam Sn (H) → 0 if and only if |bn | = ∞.

Then

Proof : If |b n | < ∞, then K(1/bn ) diverges, and diam Sn (H) can not vanish as 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−1 Bn−2 | n ≥ Qn−1 (1 + Re(bn )|ζn−1 |) ≥ Q1 (1 + Re(bk )|ζk−1 |)

(2.6.12)

k=2

where Q1 = Re(b1 ) > 0 and Re(bk ) ≥ λ|bk |. Therefore it suﬃces to prove that |bk ζk−1 | = ∞. For convenience we set δk−1 := −b k ζk−1 = bk Bk−1 /B k−2 . Then |δn | < ∞, then |bn | < ∞ {δk } is exactly as in (1.2.3) on page 104, and thus, if |bk ζk−1 | = as in the proof the Lane-Wall Characterization. Therefore |δn | = ∞.

148

Chapter 3: Convergence criteria $

' Corollary 3.41. For given K(1/bn ) with Re(b1 ) > 0 and Re(bn ) ≥ 0 for n > 2, 1/Re(b1 ) 1 ≤ n diam Sn (H) = Re(Bn B n−1 ) k=2 (1 + |ζk−1 |Re(bk )) 1/Re(b1 ) (1 + Re(bk−1 )Re(bk )) k=2

≤ n &

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(b n ) ≥ 0 for n ≥ 2, then K(1/bn ) |ζn−1 | · Re(bn ) = ∞. Although converges if Re(Bn B n−1 ) → ∞, which happens if this extends Van Vleck’s Theorem, the criterion is more diﬃcult to check. Beardon and Short ([BeSh07]) have proved that the limit point case may fail to occur for a convergent 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 ) for which {bn } stays away from a circular disk B(−2γ, 2r)◦ symmetric about the real axis. The unifying factor for this family is that the disk B(γ, r) is a simple value set for all its continued fractions. '

$

Theorem 3.42. (The Thron-Lange Theorem). For given γ ∈ R, let r := 1 + γ 2 , 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 , and every continued fraction K(1/bn ) from G converges with n 4 + σ σ |γ| 2k − σ ≤ 2r . (2.7.1) ; σ := 1 − diam Sn (V ) ≤ 2r 2k + σ 2n + 2 + σ r k=2 &

%

The convergence of K(1/bn ) was proved by Thron. Actually, it is a corollary of a much more general result from [Thron49]. Lange ([Lange99b]) has proved a number of corollaries of Thron’s general theorem, but our particular version can be found in 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 Corollary 3.9 on page 114 with ϕn (w) := ϕ(w) := γ + rw and wn := ∞. However, it is

3.2.7 The Thron-Lange Theorem

149

also a consequence of Lange’s truncation error bound (2.7.1) since this vanishes as n → ∞, 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 if b ∈ G. Evidently, 1/(b + V ) ⊆ V if and only if b + V = B(b + γ, r) ⊆

γ r 1 , = B(−γ, −r) =B 2 V γ − r2 γ 2 − r 2

(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 generality since K(1/(−bn )) is equivalent to −K(1/bn ). Let K(1/bn ) be a continued fraction from 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 given by r , Rn = 2 |Bn−1 | (|γ − ζn |2 − r2 ) and thus

Rn+1 1 |γ − ζn |2 − r2 = · Rn |ζn |2 |γ − ζn+1 |2 − r 2

where ζn+1 = −bn+1 +

1 . ζn

(2.7.2)

(2.7.3)

Now, bn+1 ∈ B(−2γ, −2r), and we shall ﬁrst prove that n + 1 n+1 γ, − r for n ≥ 1 ζn ∈ B n n

(2.7.4)

which by Lemma 3.6 is equivalent to nγ 1 nr ∈B − , ζn n+1 n+1

for n ≥ 1.

(2.7.5)

(2.7.4) holds trivially for n = 1 since ζ1 = −b1 ∈ (−G) = B(2γ, −2r). Assume it holds for a given n ∈ N. Then we can write nr 1 nγ + (1 − μ)eiθ , bn+1 = −2γ + 2(1 + λ)r eiϕ =− ζn n+1 n+1

(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 nr iθ γ + ρ eiψ + e = 2− n+1 n+1 n+1

150

Chapter 3: Convergence criteria

n + 2 r(n + 2) n+2 nr = . That is, ζn+1 ∈ B γ, − r . n+1 n+1 n+1 n+1 Hence (2.7.4) holds for all n by induction. where ρ ≥ 2(1 + λ)r −

With the notation (2.7.6), the ratio (2.7.2) takes the form 2

| ζγn − 1|2 − |ζrn |2 Rn+1 1 |γ − ζn |2 − r2 = · = Rn |ζn |2 |γ + bn+1 − ζ1n |2 − r 2 |γ + bn+1 − ζ1n |2 − r2

nγ 2 nγr nγ nr iθ 2

iθ 2

2

− n+1 + n+1 (1 − μ)e − 1 − r − n+1 + n+1 (1 − μ)e = .

− γ + 2(1 + λ)r eiϕ + nγ − nr (1 − μ)eiθ 2 − r 2 n+1 n+1 Let un := γ −

(2.7.7)

nr nr nγ γ + (1 − μ)eiθ = + (1 − μ)eiθ . n+1 n+1 n+1 n+1

Then |un | ≤

nr r + nr γ + < = r, n+1 n+1 n+1

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

2

nγ γ γ

2 2 nγ

− 1 − − + un − + un −

r

γ − n+1 n+1 n+1 n+1 = |1 + γ 2 − γun |2 − |un − γ|2 r 2 = |r2 − γ un |2 − |un − γ|2 r 2 = r 4 + γ 2 |un |2 − 2r 2 γ Re(un ) − (γ 2 + |un |2 − 2γ Re(un ))r2 = r 4 + γ 2 |un |2 − r 2 γ 2 − r 2 |un |2 = (r 2 − |un |2 )(r2 − γ 2 ) where r 2 − γ 2 = 1. Therefore Rn+1 r 2 − |un |2 r + |un | r + max |un | ≤ = ≤ Rn (3r − |un |)(r − |un ) 3r − |un | 3r − max |un | where |un | ≤ γ/(n + 1) + nr/(n + 1). That is, (n + 1)r + γ + nr Rn+1 (2n + 1)r + γ 2n + 2 − σ ≤ = = . Rn 3(n + 1)r − γ − nr (2n + 3)r − γ 2n + 2 + σ Therefore Rn+1

n n n Rk+1 2k + 2 − σ (1 + σ/2) + k − σ = R1 ≤r =r Rk 2k + 2 + σ (1 + σ/2) + k k=1

k=1

k=1

which proves the ﬁrst inequality in (2.7.1). The second one follows from Lemma 3.11 on page 115.

3.2.8 The parabola theorems

3.2.8

151

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 Theorem. 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

∞

∞

a1 a3 · · · a2n−1

a2 a4 · · · a2n

S :=

a2 a4 · · · a2n +

a1 a3 · · · a2n+1

n=1

(2.8.2)

n=1

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 ) ≤

1 2

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| < π2 , 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 absolutely to distinct ﬁnite values, and {S2n } and {S2n+1 } converge generally to these values. &

% Im 6

α − 12

0

Vα

Bn = 0 and ζn = ∞ for n ≥ 1.

Re

Both Vα◦ and Eα◦ contain the origin. The boundary of the half plane Vα is a line passing through − 12 . The boundary of Eα is a parabola with axis along the ray arg z = 2α, focus at the origin and vertex at the point − 14 ei2α cos2 α. It intersects the real axis at z = − 14 . The half plane Vα is illustrated to the left and the parabolic region Eα on the next page. Since 0 is an interior point in Vα , it follows that the approximants Sn (0) = An /Bn of a continued fraction K(an /1) from Eα are interior points in Vα . In particular Sn (0) = ∞, and thus

152

Chapter 3: Convergence criteria

3

2

y

1

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 considerably. It also generalizes the Seidel-Stern Theorem on page 117 which says that a positive 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 ≥ − 14 . Proof of the Parabola Theorem:

We shall ﬁrst prove that s(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 onto the closed circular disk with center at γa := (ae−iα )/ cos α and radius ρa := |a|/ cos α (Theorem 3.6 on page 110). This disk is contained in Vα if and only if γa ∈ Vα and γa has a distance δa to ∂Vα such that δa ≥ ρa . Now

Im 6

γa

α − 12

Re

0

δa =

1 cos α + Re(γa e−iα ) , 2

where δa > 0 if and only if γa ∈ Vα . (See the ﬁgure.)

∂Vα

So s(Vα ) ⊆ Vα if and only if −iα 1 ae |a| cos α+Re e−iα ≥ , 2 cos α cos α

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 sets Kn := Sn (Vα ) are bounded disks. If diam(Kn ) → 0, the limit point case, then the 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, and τn (∞) = ϕ−1 ◦ s2n−1 ◦ s2n (−1) = ϕ−1 ◦ s2n−1 (∞) = ϕ−1 (0) = −1 + eiα cos α =: k where |k e−iα | = |i sin α| < 1. It follows therefore by Lemma 3.8 on page 113 with wn := ∞ that the sequence {T n } given by Tn := τ1 ◦ τ2 ◦ · · · ◦ τn converges generally to 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 that K(an /1) converges if and only if S = ∞. That {S2n } and {S2n+1 } converge generally, 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 theorems for continued fractions K(an /1). It is best in several respects. For instance, one can not enlarge the set Eα , not even by adding just one point, without destroying the property that every continued fraction K(an /1) from Eα with divergent Stern-Stolz Series converges, ([Lore92]). The Parabola Theorem has an extension of type similar to the extension of Worpitzky’s Theorem on page 136. It is due to Thron, ([Thron58]). This time we have a sequence {Vα,n }∞ n=0 of half planes given by Vα,n := −gn + eiα H = {w ∈ C; Re(we−iα ) ≥ −gn cos α} ∪ {∞}

(2.8.6)

where − π2 < α < are given constants.

π 2,

g0 > 0 and 0 < ε ≤ gn ≤ 1 − ε for n ≥ 1

(2.8.7)

154

Chapter 3: Convergence criteria $

' Theorem 3.44. (The Parabola Sequence Theorem.) For given constants (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 } given by (2.8.6). Let K(an /1) be a continued fraction from {Eα,n }, and let S be the sum of its Stern-Stolz Series (2.8.2). If S = ∞, then K(an /1) converges to a ﬁnite value. If S < ∞, then {f2n } and {f2n+1 } converge absolutely to distinct ﬁnite 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 the disk 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 and γα,n has a distance δα,n ≥ ρα,n to ∂Vα,n−1 ; i.e., δα,n = gn−1 cos α + Re

an e−iα e−iα 2(1 − gn ) cos α

≥

|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 ), then K(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−1 2n ◦ s2n+1 ◦ iα s2n+2 ◦ ϕ2n+2 maps D into D with τn+1 (∞) = ϕ−1 (0) = −1 + 2(1 − g )e cos α =: 2n 2n kn . Now, |kn | is bounded away from 1 since kn e−iα = (1−2g2n ) cos α+i sin α where |1− −iα 2 | ≤ (1−2ε)2 cos2 α+sin2 α = 1−4ε(1−ε) cos2 α < 1. 2g2n | ≤ 1−2ε, and thus |kn e Therefore we also now get |Tn (∞) − Tn−1 (∞)| < ∞. Since Tn = ϕ−1 0 ◦ S2n ◦ ϕ2n −1 where ϕ0 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 n := Σ

n m=0

Pm

where

Pm :=

m 1 − gk gk

k=1

(2.8.9)

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 Sequence n be given by (2.8.9). Then Theorem, and let Pm and Σ diam Sn (Vα,n ) ≤ &

|a1 |/((1 − g1 ) cos α) . n Pj−1 gj−1 (1 − gj ) cos2 α 1+ j−2 |aj |Σ j=2

(2.8.10)

%

Proof : Since a1 /(1 + Vα,1 ) is bounded, we know that Sn (Vα,n ) is a sequence of nested circular disks. Therefore fn−1 = An−1 /Bn−1 = ∞ which means that Bn = 0 and ζn = −Bn /Bn−1 = ∞ for n ≥ 1. We can therefore write Sn (w) = fn−1 +

An Bn−1 − Bn An−1 , 2 Bn−1 (w − ζn )

and thus the radius Rn of Sn (Vα,n ) is given by

An Bn−1 − Bn An−1

1 1

sup |Sn (w) − Sn (∞)| = sup Rn = 2 2 w∈∂Vα,n 2 w∈∂Vα,n Bn−1 (w − ζn )

where ζn ∈ Vα,n since ζ1 = −1 ∈ Vα,1 . Now, points on ∂Vα,n can be written gn eiα (−1 + i vn ) cos α with vn ∈ R. Therefore ζn can be written ζn = gn eiα (−1 − un + ivn ) cos α where un > 0, vn ∈ R, and so

(2.8.11)

1

An Bn−1 − Bn An−1

. Rn =

2 2 Bn−1 gn un cos α

By means of the determinant formula on page 7 we therefore get that |an | gn−1 un−1 Rn = . Rn−1 |ζn−1 |2 gn un Since an ∈ Eα,n , we may also write an = gn−1 (1 − gn )e2iα (xn + iyn ) cos2 α where yn2 ≤ 4xn + 4. Then gn (1 + un ) cos α = −Re(ζn e−iα ) where an −iα −1+ −Re(ζn e−iα ) = −Re e ζn−1 an e−2iα = −Re −e−iα + ζn−1 e−iα gn−1 (1 − gn )(xn + iyn ) cos2 α = cos α − Re . gn−1 (−1 − un−1 + i vn−1 ) cos α

(2.8.12)

156

Chapter 3: Convergence criteria

This means that xn + iyn 1 + un−1 − ivn−1 xn (1 + un−1 ) − yn vn−1 = (1 − gn ) 1 + . 2 (1 + un−1 )2 + vn−1

gn un = gn (1 + un ) − gn = (1 − gn ) 1 + Re

(2.8.13)

Therefore

gn−1 (1 − gn ) x2n + yn2 Rn = 2 · 2 Rn−1 gn−1 {(1 + un−1 )2 + vn−1 }

=

x2n

gn−1 un−1 /(1 − gn ) xn (1 + un−1 ) − yn vn−1 1+ 2 (1 + un−1 )2 + vn−1

(2.8.14)

yn2

+ un−1 . 2 (1 + un−1 )2 + vn−1 + xn (1 + un−1 ) − yn vn−1

Since yn2 ≤ 4xn + 4, the last denominator can be written (1 + un−1 )2 + (vn−1 − yn /2)2 − yn2 /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 + yn2 ≤ x2n + 4xn + 4 = (xn + 2)2 , x2n + yn2 x2n + yn2 Rn 1 ≤ ≤ = un−1 2 2 √ Rn−1 xn + 2 + un−1 1 + xn + yn + un−1 2

.

(2.8.15)

2 xn +yn

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 ) − 2 xn + 1 · |vn−1 | xn (1 + un−1 ) − yn vn−1 ≥ , 2 2 (1 + un−1 )2 + vn−1 (1 + un−1 )2 + vn−1 where a standard minimalization procedure shows that the right hand side attains √ its minimum for vn−1 := (1 + un−1 ) xn + 1; that is, √ xn (1 + un−1 ) − 2 xn + 1 · |vn−1 | 2 (1 + un−1 )2 + vn−1 −1 xn (1 + un−1 ) − 2(1 + un−1 )(xn + 1) = ≥ . 2 2 (1 + un−1 ) + (1 + un−1 ) (xn + 1) 1 + un−1 With this inequality we get from (2.8.13) that (1 − gn )un−1 1 = gn un ≥ (1 − gn ) 1 − , 1 + un−1 1 + un−1 i.e., gn 1 ≤ un 1 − gn

1+

1 un−1

.

(2.8.16)

3.2.8 The parabola theorems

157

Now, ζ1 = −1, and thus, by (2.8.11), using (2.8.16), that

1 g1 = . Hence it follows by induction, u1 1 − g1

n 1 gn−1 gk gn gn ≤ + + ··· + un 1 − gn 1 − gn 1 − gn−1 1 − gk k=1

n−1 /Pn . = (Pn−1 + Pn−2 + · · · + 1)/Pn = Σ Therefore the bound (2.8.15) is again bounded by Rn ≤ Rn−1

1 1 = . n−2 Pn−1 /Σ Pn−1 gn−1 (1 − gn ) cos2 α 1+ 1+ n−2 x2n + yn2 |an |Σ

n Now, diam Sn (Vα,n ) = 2Rn = 2R1 k=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 the Parabola 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 diam Sn (Vα ) ≤

j=2

& Proof :

n

For all gn :=

1 2

2|a1 |/ cos α 1+

cos2 α 4(j − 1)|aj |

.

(2.8.17)

%

n = n + 1. we get Pk = 1 and Σ

Remarks. 1. The bound (2.8.17) depends on {an }. If also |an | ≤ M for all n, then this bound is again bounded by n−1

j=1 (1

2M/ cos α + (cos2 α)/(4M j))

→ 0 as n → ∞.

(2.8.18)

That is, every continued fraction K(an /1) from Eα ∩ B(0, M ) converges, and the convergence is uniform with respect to K(an /1) from Eα ∩B(0, M ). Moreover, the limit point case for Sn (Vα ) occurs for all these continued fractions.

158

Chapter 3: Convergence criteria

2. Also the truncation error bound (2.8.10) depends on K(an /1). If we add the condition that |an | ≤ Mn for all n, then diam Sn (Vα,n ) ≤

n−1 j=1

M1 /((1 − g1 ) cos α) . Pj gj (1 − gj+1 ) cos2 α 1+ j−1 Mj+1 Σ

(2.8.19)

j−1 Mj+1 ) = ∞, then every continued fraction K(an /1) from {Eα,n ∩ If Pj /(Σ B(0, Mn )} converges to some ﬁnite value f ∈ Vα,0 , uniformly with respect to K(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 j−1 = ∞, which happens if (2.8.19) vanishes as n → ∞ if and only if P j /Σ and only if Σ∞ = ∞ (standard property of positive series). ∞ < ∞, then the limit point case fails to occur for the continued n → Σ 4. If Σ fraction K( an /1) with an := −gn−1 (1 − gn ) from {Eα,n }, since the limit set must contain both −g0 and f = −g0 . As in Theorem 3.31 on page 137, we can replace {gn } by some smaller constants {gn∗ } to get smaller value sets ∗ Vα,n := −gn∗ + eiα H ∗∞ := ∞ n (1 + g∗ )/g ∗ = ∞. If 0 < ε ≤ gn ≤ 1 − ε for {Eα,n } where Σ k k n=0 k=1 for all n, then 0 < ε2 ≤ gn∗ ≤ 1 − ε < 1 − ε2 for all n.

Proof of Theorem 3.24 on page 128: For ﬁxed z := r e2iα with |α| < π/2, the S-fraction K(an z/1) is a continued fraction from Eα given by (2.8.3), and the half plane Vα in (2.8.1) is a simple value set for K(an z/1). By (2.8.14) (with all gn = 12 ) it follows therefore that the radius Rn of Sn (Vα ) satisﬁes x2n + yn2 un−1 Rn = . 2 Rn−1 (1 + un−1 )2 + vn−1 + xn (1 + un−1 ) − yn vn−1 Now, with this notation, an z = an r e2iα = 14 e2iα xn cos2 α and ζn = 12 eiα (−1 − un + i vn ) with un > 0 by (2.8.11). That is, xn = 4an r/ cos2 α > 0, yn = 0 and xn un−1 Rn ≤ Rn−1 (1 + un−1 )2 + xn (1 + un−1 ) which obtains its maximum with respect to un−1 for un−1 := √ xn 1 + xn Rn √ √ ≤ Rn−1 (1 + 1 + xn )2 + xn (1 + 1 + xn ) √ xn 1 + xn √ = 2(1 + xn ) + (2 + xn ) 1 + xn xn x √n = √ = = 2 1 + xn + 2 + xn (1 + 1 + xn )2 √1

xn

√

1 + xn . That is,

1 * + 1+

1 xn

2

3.2.8 The parabola theorems

159

where xn = 4an r/ cos2 α. Since R1 = a1 r/( 12 cos α), this proves that , −2 n n 4a1 r 1 Rk 1 ≤ diam Sn (Vα ) = 2Rn = 2R1 √ + 1+ Rk−1 cos α xk xk k=2

k=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 √ = = √ 2 . * 2 1 + xn + 1 ( 1 + xn + 1) 1 √1 + 1 + xn xn Limit sets. Also for the Parabola Theorem it turns out that every point in Vα \ {0, ∞} is the value of a continued fraction K(an /1) from Eα :

Theorem 3.47. For ﬁxed α ∈ R with |α| < π2 let Eα and Vα be given by (2.8.3) and (2.8.1). Then Vα \ {0, ∞} is the limit set for the family of continued fractions K(an /1) from Eα .

Proof : Every convergent continued fraction K(an /1) from Eα has its value in Vα \ {0, ∞} (the Parabola Theorem). Let w0 ∈ Vα \ {0, ∞} be arbitrarily chosen, and let w ∈ ∂Vα and μ ≥ 0 be the (uniquely determined) numbers which make w0 = w + μeiα . Let further w1 := −1 − w + μ eiα ,

a1 := −w 2 + μ2 e2iα ,

a2 := −(1 + w) 2 + μ2 e2iα .

Then also w1 ∈ Vα \ {0, ∞}. Moreover, −w 2 ∈ ∂Eα since w can be written w = 1 iα 2 ∈ ∂Eα . Therefore − 2 + it e (see Problem 19 on page 169). Similarly, −(1 + w) a1 , a2 ∈ Eα and the 2-periodic continued fraction ∞

a1 a2 a1 a2 a1 an := K n=1 1 1 + 1 + 1 + 1 + 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, −w 2 + μ2 e2iα w 2 − μ2 e2iα a1 = = =w + μ eiα = w0 , iα 1 + w1 1−1−w + μe w − μ eiα −(1 + w) 2 + μ2 e2iα a2 = = −(1 + w) + μ eiα = w1 . 1 + w0 1+w + μ eiα That is, S2n (w0 ) = S2n+1 (w1 ) = w0 for all n, and therefore K(an /1) converges to w0 .

160

Chapter 3: Convergence criteria

3.3

Additional convergence theorems

3.3.1

Simple bounded circular value sets

´ The Sleszy´ nski - Pringsheim Theorem required that V = D. This easily generalizes to 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 value f ∈ 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 (b + Γ) d

and ρ1 :=

|a| ρ d

(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. In other words, Ω is the element set corresponding to the value set V . This was ﬁrst proved by Lane ([Lane45]). Since moreover s(∞) = 0 ∈ V ◦ for all (a, b) ∈ Ω with a = 0, the convergence follows from Corollary 3.9 on page 114. (See the remark on page 113.) ´ If we set Γ := 0 and ρ := 1, we get back the Sleszy´ nski-Pringsheim criterion on page 135. 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 \V. tional sequence {wn† } from C &

%

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)d makes 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 general convergence follows from Corollary 3.9 on page 114 with wn := ∞. Since V contains more than two points, the value f of K(an /1) belongs to V . In particular f = ∞, \ V since so {Sn−1 (∞)} is an exceptional sequence. Finally, wn† := Sn−1 (∞) ∈ C † Sn (wn ) = ∞ ∈ 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 we recognize from the generalized version of the Worpitzky Theorem. If Γ = 0, then E = a∗ E ∗

ρ2 Γ d where a∗ := Γ(1 + Γ) 1 − = 2 |1 + Γ| 1+Γ $ ρ % ρ . and E ∗ := ξ ∈ C; |ξ − 1| + |ξ| ≤ |1 + Γ| |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 = } with 0 < k < 1 and k < is a closed convex curve, symmetric about the real axis with O ∩ R = {μ, ν} given by μ :=

1− +1 1− if ≤ 1, μ := if ≥ 1 and ν := . 1−k 1+k k+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) = ,

162

Chapter 3: Convergence criteria

and thus O is convex. The symmetry follows since ξ ∈ O implies that its complex conjugate ξ also is an element in O. For ξ ∈ O ∩ R, +1 , k+1 1− =⇒ < 1, 0 < ξ < 1 =⇒ 1 − ξ + ξk = =⇒ ξ = 1−k 1− =⇒ ≥ 1. ξ ≤ 0 =⇒ 1 − ξ − ξk = =⇒ ξ = 1+k ξ ≥ 1 =⇒ ξ − 1 + ξk = =⇒ ξ =

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 on the 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) from E 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 have assumed that |Γ| = |1 + Γ|. Since E is a compact set, there therefore exists an ε > 0 such that |a| ≤ (1 − ε)d for all a ∈ E. Hence it follows from Theorem 3.49 that K(an /1) converges generally to some value f ∈ V := B(Γ, ρ). It has an exceptional sequence {Sn−1 (−1 − w)} for some w ∈ V with −1 − w = f , where Sn−1 (−1 − w) = −1 −

an an−1 a1 ∈ −1 − V 1 + 1 +·+ 1+w

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 following truncation error bound:

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 n−1 |Γ| + ρ Mk (3.1.7) |f − Sn (Γ)| ≤ ρ |bn + Γ| − ρ k=1

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 M n−1 |1 + Γ| − ρ

(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 Γ(1 + Γ) − ρ2 = , 2 2 |1 + Γ| − ρ |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 Parabola Theorem since Lange ([Lange95]) proved that E is a subset of the parabolic region Eα with α := arg(Γ + 12 ). The importance of the Oval Theorem lies therefore with its better truncation error bound for this smaller family of continued 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. In particular ∞ ∈ V , so if V is a simple value set for K(an /bn ), then an /(bn + ∞) =

164

Chapter 3: Convergence criteria

0 ∈ V . Hence we are back to the classical type of value sets which contain the classical approximants. If we stick to continued fractions of the form K(an /1) or K(1/bn ), we have the following interesting observation: $

' Theorem 3.54. \ (−1 − V ) and A. Let V be a simple value set for K(an /1). Then W := C W ∩ V are also simple value sets for K(an /1). \ (−1/V ) and B. Let V be a simple value set for K(1/bn ). Then W := C W ∩ V are also simple value sets for K(1/bn ). &

%

Proof : A. Since sn (V ) := an /(1 + V ) ⊆ V , we have V ⊆ s−1 n (V ) = −1 + an /V ; i.e., (−1 − V ) ⊆ an /(−V ) = sn (−1 − V ), and thus W ⊇ sn (W ). Clearly, sn (V ) ⊆ V and 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 −

1 1 ⊆ = sn (−1/V ) V bn − 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 the disk B(−1 − Γ, |μ|) is a simple value set for K(an /1), and K(an /1) converges by 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 ◦ . Therefore \ (−1/V ) is a circular disk, and we can apply Theorem 3.49. W := C 3. If V := w0 + eiα H 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 are back in the Parabola Theorem. 4. If V := w0 + eiα H 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 the \ (−1/V ) is a disk, and we can again exterior of a disk, and thus W := C apply Theorem 3.49.

Remarks

3.4

165

Remarks

1. Lemma 3.7. This crucial lemma is inspired by the work of W.J.Thron. In 1965 he and his student K.L.Hillam ([HiTh65]) published the result that {Tn (0)} converges \ D, such that τn (w2 ) = w1 for all n. if there exist two points, w1 ∈ D and w2 ∈ C The present version dates from 1994 ([Lore94a]) where it also was proved that the conclusions still hold if we replace (1.4.6) by lim inf rad(τn−1 (D)) > 1. (If τn−1 (D) is unbounded, we set rad τn−1 (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 became more 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 deriving such estimates. This started a series of useful publications in this area, such as [HePf66], [JoTh76], [CrJT94], [BaJo85], [FiJo72] and [CJPVW7]. For further references we refer to ([JoTh76], p 298). 3. The parabola theorems. The ﬁrst parabola theorem was published by Scott and Wall in 1940, ([ScWa40]). It was valid for the special case α = 0. It was proved by exploiting what they called the fundamental inequalities. The result was generalized almost immediately ([PaWa42] and [LeTh42]). The most general parabola theorem is due to Jones and Thron, ([JoTh68]) who also proved convergence for cases of continued fractions K(an /bn ) with half planes as value sets. 4. Classical convergence theorems. The Worpitzky Theorem was proved already in 1865, but remained unknown to workers in the ﬁeld until Pringsheim rediscovered it more than 30 years later. It was not until 1905, through Van Vleck, that Worpitzky got 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]), but there 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. Sleszy´ nski is the right one to credit, since ´ he already proved the theorem in 1888, see [Sles89]. 5. Boundary versions of convergence theorems. The limit regions in Worpitzky’s Theorem and the Parabola Theorem are created by using the whole element set in both cases. A natural question to ask is: what happens to the limit regions if we restrict the elements to the boundary of the element sets? The answer is given in [Waad89]: the Worpitzky limit set is reduced to the annulus 16 ≤ |f | ≤ 12 whereas the half plane minus {0, ∞} remains unchanged in the parabola case. In [Waad92] an investigation of the Oval Theorem is carried out. The process and the result are somewhat more complicated, but the limit region is a circular disk with a hole (not centered). Proper limits of parameters make the oval region approach the parabolic region of the Parabola Theorem and the corresponding limit sets approach the half plane limit set in the parabola case. A probabilistic aspect also turned up in the Worpitzky case. When we computed values (or rather high order approximants) of continued fractions from the boundary of the Worpitzky disk, they all seemed to lie in a more narrow annulus that expected from the theoretic results. The reason for this turned out to be that the probability density for the values is very small close to the boundary of the annulus, ([Waad98]).

166

Chapter 3: Convergence criteria

6. More convergence theorems. It is clear that with the help of simple value sets one can produce large numbers of convergence theorems. This has been done by a number of authors, in particular by Jones and Thron ([Thron48], [Thron74], [JoTh68]). Let us also mention Lange’s strip convergence regions ([Lange94]) and Overholts polygons ([Over82]) which are based on non-circular value sets, and the general work by C´ ordova Yevenes ([Cord92]). 7. Twin convergence theorems. The situation where {Vn } is a periodic sequence of value sets has also been studied in detail. In particular the 2-periodic case is very interesting. Here Thron has given a large number of convergence results based on circular value sets, in later years in collaboration with Jones and Lange, ([Thron43], [Thron49], [Thron59], [LaTh60], [Lange66], [JoTh68], [JoTh70]). In [Lore08a] the cases where 0 ∈ Vn are also included.

3.5

Problems

1. Convergence to ∞. Prove that if the continued fraction K(an /bn ) with b2n−1 = 0 for all n converges, then it converges to ∞. (Broman [Brom77].) 2. Convergence of continued fractions. Determine whether K(an /bn ) converges or diverges and whether its even and odd parts converge or diverge in each of the following 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 positive continued fractions K(cn /1) and K(dn /1) given by 1 , 3 2 d1 := , 3 c1 :=

(n − 1)2 (2n − 1) for n ≥ 3, 2n − 3 2 n (2n − 1) dn := for n ≥ 2 2n + 1

c2 := 2,

cn :=

converge to ﬁnite values. (Hint: Theorem 3.2(i) on page 102 and the Seidel-Stern Theorem on page 117 may be of help.) 4. Convergence of continued fractions. Determine whether K(1/bn ) converges or diverges when (b) bn := 1 + (−1)n . (c) bn := in . (a) bn := (−1)n n. (d) bn := 1/n. (e) bn := (2i)n . (f) bn = (sin n)/n2 . 5. Convergence of continued fractions. Determine whether K(an /1) converges or diverges when (a) an := (−1)n n. (b) an := 2 + (−1)n . (c) an := (−2)n . 2 (e) an := 1/n . (f) an = (sin n)/n2 . (d) an := 1/n.

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+

K 2ba = b + 2ba + 2ba + 2ba +· · ·

n=1

√ a + b2 . Use this to Prove that if a > 0 and b > 0 then √ b + K(a/2b)converges to −4 ﬁnd a rational approximation to 13 with an error less than 10 . 8. Positive continued fractions. Let b > 0 and p > 0. Prove that K(np /b) converges if and only if p ≤ 2. (Perron [Perr57], p 48.) 9. Oscillations of approximants. The continued fraction ∞

K a1

n

n=1

=

1 1/2 1/(2 · 3) 2/(2 · 3) 2/(2 · 5) 3/(2 · 5) 1+ 1 + 1 1 1 1 + + + +· · · +

n/(2(2n − 1)) n/(2(2n + 1)) 1 1 + +· · ·

converges to Ln(2). Suggest three sequences {Sn (wn )} of approximants converging to Ln(2). Does any of these sequences have an oscillating character which can be used 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 if n = m? (d) Use the value set V found in (a) to derive upper bounds for the truncation error |f − Sn (w)| for suitably chosen w ∈ C. 11. Truncation error bounds. Use Theorem 3.46 on page 157 to derive a priori truncation error bounds for ∞

Ln(1 + i) =

K a1 i = 1i + i/21 + i/(21· 3) + 2i/(21 · 3) + 2i/(21 · 5) +· · · n

n=1

where a2n = n/(2(2n − 1)) and a2n+1 = n/(2(2n + 1)) for all n ≥ 1. Compare these with the Gragg-Warner truncation error bounds in Theorem 3.24 on page 128.

168

Chapter 3: Convergence criteria

12. ♠ Alternating continued fraction. Let K(an /1) have real elements an such that (−1)n an > 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 sequence Sn (wn ) of approximants for ∞

K a1

n

n=1

=

12 5 · 22 32 5 · 42 52 5 · 62 72 1 + 1 + 1 + 1 + 1 + 1 + 1 +· · ·

(hopefully) converges faster to the value of K(an /1) than Sn (0). Compute the ﬁrst 6 approximants of Sn (0) and Sn (wn ). Use the oscillating character to determine an error bound for S6 (0), and if possible for S6 (w6 ). 14. Convergence of continued fractions. Let α be a positive number. For which −α ) converge, and for which values values of α does the continued fraction K∞ n=1 (1/n does 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 + 1 |) converges. 4

(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], p 108.) 16. From Van Vleck to Stieltjes. Prove the following implications: (a) Van Vleck’s Theorem on page 142 (b) Theorem 3.39 on page 145

=⇒

=⇒

Theorem 3.21 on page 124.

the Henrici-Pﬂuger Bounds on page 126.

17. Convergence of even and odd parts. Prove that if the even and odd parts of K(an /1) with an ∈ C \ {0} converge to distinct values, then an → ∞. (Bowman and McLaughlin [BoML04].) 18. The Cardioid Theorem. Use the Parabola Sequence Theorem to prove that if |an | − Re(an ) ≤ k gn−1 (1 − gn ) and

∞ n

for n = 1, 2, . . .

|ak |n−k+1 = ∞,

n=1 k=1

then K(an z/1) converges locally uniformly to a holomorphic function in the cardioid domain D := {z = r e2iθ ∈ C; r < k1 cos θ}. (Jones and Thron [JoTh80], p 134.)

Problems

169

19. ♠ The Parabola Theorem. Let Vα and Eα be as in the Parabola Theorem. Prove that (a) a ∈ Eα if and only if a = c2 for some c ∈ C with |Im(c e−iα )| ≤ 2

−iα

(b) a ∈ Eα if and only if a = −c for some c ∈ C with |Re(c e

1 2

)| ≤

cos α. 1 2

cos α.

20. ♠ A convergence theorem. Let {ρn } be a sequence of positive numbers. Prove that (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 ) corresponding to 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

,

,

b+n−1 b+n for a ﬁxed constant 0 ≤ b ≤ 1 converges to − 1 + 1/b. bn :=

b+n+1 + b+n

(e) Prove that diamSn (Vn ) ≤ 2ρ0

n

ρj ρj−1 (1 + 3ρj−1 ρj−2 ) 2 2ρ ρ ρ j j−1 j−2 + 2ρj ρj−1 − ρj−1 ρj−2 + 1 j=2

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 ﬁnite value 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 with τn (∞) = −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).

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 value sets, 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 value sets, 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 to prove the following: (a) The value of any continued fraction 1/4 −1/4 a3 a4 , 1 + 1 + 1 + 1 +· · · with |an | ≤ 1/4 for all n must lie in the disk

w − 2 ≤ 1 .

5

10 (b) The value of any continued fraction i/4 a2 a3 , 1 + 1 + 1 +· · · with |an | ≤ 1/4 for all n must lie in the disk

w − i ≤ 1 .

3

6 (c) Show that the disk in (b) is a limit set for the family of continued fractions from (a). 26. ♠ The Parabola Theorem. Prove that Eα in the Parabola Theorem on page 151 also is given by 1 cos2 α i(θ+2α) 2 . Eα := r e ; 0≤r≤ 1 − cos θ

Chapter 4

Periodic and limit periodic continued fractions Periodic continued fractions are well understood. We can determine when they converge, when they diverge, their values if they converge and the asymptotic behavior of their tail sequences. Indeed, it is all a matter of iterations of linear fractional transformations, and we have closed expressions for their approximants Sn (w). Only few continued fraction expansions of interesting functions are periodic, but large numbers are limit periodic, where the tail of K(an /bn ) looks more and more like a periodic continued fraction the further out it starts. Then K(an /bn ) essentially inherits the asymptotic properties from the periodic one in most cases. This means that also limit periodic continued fractions are reasonably well understood. Equivalence transformations are sometimes useful to bring a continued fraction to a wanted form. Naturally, there is no point in transforming a given continued fraction 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 are analyzed 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

172

4.1 4.1.1

Chapter 4: Periodic and limit periodic continued fractions

Periodic continued fractions 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 are p-periodic; i.e. if (1.1.1) an+p = an , bn+p = bn for all n ∈ N , and p is the smallest positive integer for which this holds. For such continued fractions the approximants can be written Snp+m (w) = Sp[n] ◦ Sm (w) = Sm ◦ (Sp(m) )[n] (w)

(1.1.2)

where F [n] := F ◦ F ◦ · · · ◦ F is the nth iterate of a function F , and −1 ◦ Sp ◦ Sm (w) = Sp(m) (w) := Sm

am+1 am+2 am+p . bm+1 + bm+2 +· · · + bm+p + w

(1.1.3)

The convergence behavior of K(an /bn ) therefore depends on how sequences of it(m) erates of linear fractional transformations Sp behave asymptotically. For convenience we also use the notation S0 (w) := w

4.1.2

(0) and Sm := Sm (w).

(1.1.4)

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)

then x must be a ﬁxed point for τ ; i.e. If {τ [n] } converges generally to some x ∈ C, 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 see that ∞ is a ﬁxed point for τ if and only if c = 0. For c = 0 and a + d = 0, the ﬁxed points are a − d ± (a − d)2 + 4bc a − d ± (a + d)2 − 4Δ x, y = = 2c 2c (1.2.2) a − d ± (a + d)u where u := 1 − 4Δ/(a + d)2 = 2c √ (with Re . . . ≥ 0 as always), and for c = 0 and a = −d a (1 ± −Δ/a2 ) if a = 0, c (1.2.3) x, y = if a = 0. ± b/c

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 1 (cx + d)(cw + d) 1 = = τ (w) − x τ (w) − τ (x) (ad − bc)(w − x) cx + d cx + d 2c 1 = c+ = + . ad − bc w−x a+d w−x That is, ϕ ◦ τ (w) = q + ϕ(w) where q := 2c/(a + d) and ϕ(w) :=

1 , w−x

(1.2.5)

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 form q + w, this time with q := b/d. The advantage of this form is that iterations of τ˜ just gives τ˜[n] (w) = w + nq, and thus we know how iterations of τ = ϕ−1 ◦ τ˜ ◦ ϕ behave. It is also clear that τ [n] = ϕ−1 ◦ τ˜[n] ◦ ϕ, so the following result can be stated: '

$

Theorem 4.1. Let τ given by (1.2.1) have only one ﬁxed point x. Then w−x 2c where x = a−d , q := a+d if c = 0 x + nq(w−x)+1 [n] 2c τ (w) = 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) − τ (x) cy + d w − x = = · τ (w) − y τ (w) − τ (y) cx + d 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 w−x cy+d and := cx+d ϕ(w) := , so w−y τ˜(w) := ϕ ◦ τ ◦ ϕ−1 (w) = w.

(1.2.8)

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 ﬁxed points ∞ and b/(d − a), where a = d. Since τ (w) = (a/d)w + (b/d), we have b b a b b a w− = w+ − = d−a d d d−a d d−a b a and ϕ(w) := w − , ϕ ◦ τ = · ϕ(w) for := d d−a τ (w) −

(1.2.9)

so τ˜(w) := ϕ ◦ τ ◦ ϕ−1 (w) = w, and thus τ˜[n] (w) = n w, which tells the whole story 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 ⎧ n n ⎨ (x − y)w − xy(1 − ) for := cy + d if c = 0, [n] n n (1 − )w + x − y cx + d τ (w) = ⎩ a n b if c = 0. ( d ) w + (1 − ( ad )n ) d−a

&

4.1.3

%

Classiﬁcation of linear fractional transformations

We classify τ ∈ M according to the behavior of iterations τ [n] as n → ∞. A linear fractional transformation τ is conjugate (or similar) to τ if there exists a linear fractional transformation ϕ such that τ = ϕ ◦ τ ◦ ϕ−1 . As seen in the previous section, 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 (coinciding if necessary), the ratio := (τ ) of τ is a complex number with 0 < || ≤ 1 given by ⎧ ⎪ if c = 0, ⎨(cy + d)/(cx + d) := a/d (1.3.1) if c = 0 and |a| ≤ |d|, ⎪ ⎩ d/a if c = 0 and |a| > |d|. &

%

(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 of uniqueness will not be a problem for our applications. With u as given by (1.2.2),

4.1.3 Classiﬁcation of linear fractional transformations can be written

⎧ ⎨1 − u = 1+u ⎩−1

if c = 0, a + d = 0, if c = 0, a + d = 0.

175

(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 two and τ is conjugate to τ˜(w) := w which has distinct ﬁxed points x, y ∈ C, ﬁxed points at 0 and ∞. Since τ˜[n] (w) = n w → 0 for w = ∞, the sequence {τ [n] } converges generally to x with exceptional sequence {y}∞ n=1 . We say that 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 has and τ is conjugate to τ˜(w) = w, exactly two distinct ﬁxed points x, y ∈ C, [n] but {˜ τ } 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. We say that x and y are indiﬀerent ﬁxed points for τ . • τ ∈ M is the identity transformation if τ (w) = I(w) ≡ w. Then every point is a ﬁxed point for τ , and = 1. τ = I is only conjugate to itself, and in C τ [n] (w) = w. • τ ∈ M is a parabolic transformation if τ = I and = 1. Then τ has only one and τ is conjugate to τ˜(w) := w + q with q = 0. Since ﬁxed point x = y ∈ C, [n] it follows that τ [n] (w) → x for τ˜ (w) = w + nq → ∞ for every ﬁxed w ∈ C, every 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 if and only if w is a ﬁxed point for τ −1 , and straight forward computation shows that 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 since for τ with ratio and ﬁxed points x, y, τ [n] has ratio n and ﬁxed points x, y. The only exception occurs when τ is elliptic with ratio , where N = 1 for some N ∈ N. Then τ [N ] is the identity transformation. Indeed, τ [nN ] = I for 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 point x.

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 the attracting 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 + 3 2w + 5

has ﬁxed points 1 and −3/2. Since |2 · 1 + 5| = 7 and |2 · (−3/2) + 5| = 2, we set x := 1, y := − 32 , 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 + 32 ( 27 )n )w + 32 (1 − ( 27 )n ) 3 → 1 for w = − = y. 2 (1 − ( 27 )n )w + ( 27 )n + 32

3 Even though τ [n] has ratio n when τ has ratio , the ratio is not multiplicative in general:

Example 2. The transformation s1 (w) := (−1/4)/(1 + w) is parabolic with ﬁxed point x = − 12 and ratio 1, and the transformation s2 (w) := 2/(1 + w) is loxodromic with ﬁxed points 1 and −2 and ratio −1/2. Their composition is −1/4 1 w+1 =− · 1 + 2/(1 + w) 4 w+3 √ which is loxodromic with ﬁxed points x, y = 18 (−13 ± 153) and ratio = (3 + √ √ 1 (−13 + 153))/(3 + 18 (−13 − 153)). 3 8 s1 ◦ s2 (w) =

4.1.4

General convergence of periodic continued fractions

We say that a p-periodic continued fraction K(an /bn ) is of parabolic or loxodromic or 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 . The convergence properties for iterations of Sp are described in the previous section. By (m) (1.1.3) we see that Sp is conjugate to Sp , and that x (or y) is a ﬁxed point for Sp if and only if −1 −1 x(m) := Sm (x) (or y (m) := Sm (y)) (1.4.1) (m)

is a ﬁxed point for Sp

.

4.1.4 Convergence of periodic continued fractions

177

If Sp is loxodromic, then x and x(m) shall always denote the attracting ﬁxed point (m) of Sp and Sp 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 to the attracting ﬁxed point of Sp . B. If Sp is elliptic or the identity transformation, then {Sn } is totally non-restrained, and K(an /bn ) diverges generally. &

%

Example 3. Straight forward checking shows that the 1-periodic continued fraction K(a/b) with a, b ∈ C, a = 0 is of • elliptic type if b = 0 or • parabolic type if

a b2

a b2

< − 14 ,

= − 14 ,

• loxodromic type otherwise. Hence K(a/b) converges generally if and only if b = 0 and (a/b2 ) lies in the cut plane C \ (−∞, − 14 ). Its value is then x = 2b (−1 + u)

where u :=

1 + 4a/b2 (with Re u ≥ 0).

If b = 0, then = −1, and {Sn } is the 2-periodic sequence fractional transformations. 3

a , w

w,

a , w

(1.4.2) w, . . . of linear

The picture does not change much if we consider tail sequences {Sn−1 (t0 )} instead of approximants {Sn (w)}. Our classiﬁcation is independent of inversion, and −1

−1

−1 −1 (t0 ) = (Sp(m) )[n] (Sm (t0 )) = (Sp(m) )[n] (tm ). tnp+m = Snp+m (m)−1

Hence {tnp+m }n is the sequence of iterates of Sp

evaluated at tm .

(1.4.3)

178

Chapter 4: Periodic and limit periodic continued fractions $

' Theorem 4.4. Let K(an /bn ) be a p-periodic continued fraction, and let be a tail sequence for K(an /bn ). {tn } ⊆ C A. If Sp is parabolic with ﬁxed point x, then −1 (x) =: x(m) for 0 ≤ m ≤ p − 1. lim tpn+m = Sm

n→∞

(1.4.4)

B. If Sp is loxodromic with attracting ﬁxed point x and repelling ﬁxed point y, and t0 = x, then −1 (y) =: y (m) for 0 ≤ m ≤ p − 1. lim tpn+m = Sm

n→∞

&

(1.4.5) %

Of course, if t0 = x, then the p-periodic sequence {x(m) }∞ m=0 is the sequence of tail values for K(an /bn ).

Example 4. We consider the 1-periodic continued fraction K(a/b) from Example 3. In the parabolic case, K(a/b) has one and only one periodic tail sequence {x}∞ n=0 , where x = − 2b is the value of K(a/b). Indeed, {− 2b }∞ is its sequence of tail values, n=0 and every tail sequence for K(a/b) converges to − 2b . ∞ 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 other tail sequence {tn } for K(a/b). In particular {y}∞ n=0 is an exceptional sequence. 3

Example 5. The 4-periodic continued fraction ∞

an 1 3 2 1 1 3 2 1 1 K := n=1 1 1 + 1 − 1 − 1 + 1 + 1 − 1 − 1 + 1 +···

(1.4.6)

converges generally to the attracting ﬁxed point x = 1 of the loxodromic transformation 1 4 + 2w 1 3 2 S4 (w) = = . 1 + 1 − 1 − 1+w 5+w The repelling ﬁxed point of S4 is y = −4, so y (0) = −4, 5 9 y (2) = s−1 2 (− 4 ) = − 5 ,

5 y (1) = s−1 1 (−4) = − 4 , 9 y (3) = s−1 3 (− 5 ) =

2 3

,

4.1.5 Convergence in the classical sense

179

(3) and y (4) = s−1 ) = −4 = y (0) , and so on. Therefore, every tail sequence {tn } 4 (y with t0 = 1 satisﬁes ⎧ ⎪ ⎪ −4 if m = 0 , ⎪ ⎪ ⎪ ⎪ ⎨ −5 if m = 1 , 4 lim t4n+m = n→∞ ⎪ ⎪ if m = 2 , − 95 ⎪ ⎪ ⎪ ⎪ ⎩ 2 if m = 3 . 3

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 parabolic transformation τ converge to the ﬁxed point for τ for every w ∈ C. Therefore (m) and m ∈ {0, 1, . . . , p − 1}, and thus limn→∞ Snp (w) = x(m) for every w ∈ C lim Sn (w) = x for every w ∈ C. In particular Sn (0) → x. Therefore:

Theorem 4.5. A periodic continued fraction of parabolic type converges to x in the classical sense and Sn (w) → x for every w ∈ C.

not even with respect to Still, {Sn (w)} does not converge uniformly to x in C, the chordal metric. We always have exceptional sequences when {Sn } converges generally. For example, every tail sequence {tn } with t0 = x is an exceptional sequence. The loxodromic case. This is a case where K(an /1) converges generally, but may diverge in the classical sense:

Example 6. We have seen in Example 2 on page 59 that the three-periodic continued fraction ∞

K a1n := 21 + 11 − 11 + 12 + 11 − 11 + 21 + · · ·

n=1

180

Chapter 4: Periodic and limit periodic continued fractions 1 , 2

converges generally to transformation

but diverges in the classical sense. The linear fractional

S3 (w) :=

1 1 2w 2 = 1 + 1 − 1+w 1 + 2w

is loxodromic with attracting ﬁxed point 1/2. But S3 (0) = 0, and thus S3n (0) = 0 for all n whereas limn→∞ S3n+m (0) = 1/2 for m := 1, 2. 3

# Theorem 4.6. Let K(an /bn ) be a p-periodic continued fraction of loxodromic 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 the classical sense. "

!

Proof : K(an /bn ) converges generally to x (Theorem 4.3). Hence K(an /bn ) converges to x in the classical sense if its exceptional sequences have no limit point at −1 (y) = 0; i.e., 0. Now, {Sn−1 (y)} is an exceptional sequence. It is p-periodic, so if Sm 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 for all 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 for which Sk (0) = y. Hence K(an /bn ) diverges in this case. Remark: This type of divergence of periodic continued fractions was ﬁrst described by Thiele ([Thie79]), and is therefore called Thiele oscillation. If Thiele oscillation occurs for a p-periodic continued fraction of loxodromic type, then lim Snp+m (0) =

n→∞

x whenever Sm (0) = y, y whenever Sm (0) = y.

Since all y (m) = 0 in Example 5, the continued fraction in this example converges in the classical sense.

Corollary 4.7. A 1- or 2-periodic continued fraction K(an /bn ) of loxodromic type with b1 b2 = 0, converges to x in the classical sense.

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 \ (−∞, − 14 ], * b1 b1 x = (−1 + u1 ), y = (−1 − u1 ) where u1 := 1 + 4a1 /b21 . 2 2 Hence y = 0.

(1.5.1)

Next, let K(an /bn ) be 2-periodic with b1 b2 = 0. The ﬁxed points of S2 (w) are 2a1 − λ − λu2 2b1 where λ := a1 + a2 + b1 b2 and u2 := 1 − 4a1 a2 /λ2 . x=

2a1 − λ + λu2 , 2b1

y=

(1.5.2)

Since x = y, this means that u2 = 0 and λ = 0. Therefore y = 0 only if (2a1 − λ)2 = λ2 u22 ; i.e., if b1 b2 = 0 which we have excluded. Moreover, also y (1) := a2 /(b2 +y) = 0, and the result follows. Example 7. The continued fraction z z z z 2 + 2 + 2 + 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, since then S1 is parabolic. But for z on the cut z < −1, S1 is elliptic and the continued fraction diverges generally (Theorem 4.3B). 3

4.1.6

Approximants on closed form

The expressions for τ [n] (w) in Section 4.1.2 can be used to ﬁnd closed expressions for Sn (w) = (An−1 w + An )/(Bn−1 w + Bn ) when K(an /bn ) is p-periodic. But a periodic tail sequence also simpliﬁes the formulas in Theorem 2.6 on page 66. For instance: '

$

Corollary 4.8. Let K(a/b) be 1-periodic, and let x and y be the two possibly coinciding ﬁxed points for s(w) := a/(b + w), where x is attracting or indiﬀerent. Then x = 0, ∞, −b, = (b + y)/(b + x) = x/y, y = −b − x and Bn + Bn−1 x = (b + x)n , An − Bn x = (−x)n+1 , n n n + An −k , Bn = (−x)n −k , x − =x −k . An = x(−x)n Bn k=1 k=0 k=0 &

%

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 − x where 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 in Corollary 4.8 imply that An = −n( 2b )n+1 , x−

An b/2 =− , Bn n+1

Bn = (n + 1)( 2b )n , nb/2 An =− . Bn n+1

If |b| < 2, then {An } and {Bn } therefore converge to 0, and if |b| ≥ 2, they converge to ∞. 2. For = 1 we can sum the geometric sums in the formulas in Corollary 4.8 to get −n − 1 (−y)n − (−x)n An = x(−x)n = xy , 1− y−x −n − (−x)n+1 − (−y)n+1 = , Bn = (−x)n 1− y−x An An x(1 − ) x(−n − 1) , . x− = −n = Bn − Bn −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 transformation Sp : $

' Theorem 4.9. Let K(an /bn ) be p-periodic with a p-periodic tail sequence t := {tn } for which all tn = ∞, and let

p−1

X := X(t) :=

(−tm )

and

m=0

Y := Y (t) :=

p

(bm + tm ) .

m=1

tn } is a second p-periodic Then either X/Y = or X/Y = −1 . If t := { tail sequence for K(an /bn ) with all tn = ∞, then X(t) = Y ( t ) &

and

X( t ) = Y (t).

(1.6.1) %

4.1.7 Connection to the Parabola Theorem

183

Remark. We recognize the expression p Y bm + tm Pp := = X −tm m=1

(since t0 = tp )

from Theorem 2.6 on page 66. (m)

Proof of Theorem 4.9: Evidently tnp+m is a ﬁxed point for Sp x(m) . The derivative of Sp at x(p) is (using the chain rule)

, say tnp+m =

Sp (x(0) ) = Sp (x(p) ) = s1 (x(1) ) · s2 (x(2) ) · · · sp (x(p) )

(1.6.2)

where sm (x(m) ) =

−am −x(m−1) = . (m) 2 (bm + x ) bm + x(m)

(1.6.3)

This proves that Sp (x(0) ) = X/Y . That X/Y = follows now from Remark 5 on page 176. Next, assume that { tn } is a second p-periodic tail sequence with all (m) tnp+m must be a second ﬁxed point for Sp ; i.e., tnp+m = y (m) . tn = ∞. Then := X( t ) and Y := Y ( t ) we similarly ﬁnd that X/ Y = 1/. That is, Writing X = X/Y = Y /X. Moreover, Bp + Bp−1 tp =

p

(bm + tm )

(1.6.4)

m=1

by Theorem 2.6 on page 66. Hence by Deﬁnition 4.1 of on page 174 =

p Y bm + y (m) Bp + Bp−1 y (p) = = . (p) (m) Y Bp + Bp−1 x b +x m=1 m

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 + eiα H is a simple value set for every continued fraction K(an /1) from Eα := {a ∈ C; |a| − Re(a e−2iα ) ≤

1 2

cos2 α}

(1.7.1)

for given α ∈ R with |α| < π/2. It follows from this Parabola Theorem that every periodic 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.

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 a2 , s2 (w) := 1+w 1+w

(1.7.2)

is a parabolic transformation. & Proof :

% That S2 is parabolic follows since S2 (w) =

−x2 (1 + x)2 −x2 (1 + w) = 1 − 1+w w − 2x − x2

has only one ﬁxed point, namely x. Since ∞ = x ∈ ∂Vα , it can be written x = − 12 + it eiα for some t ∈ R, and thus |a1 | − Re(a1 e−2iα ) = |x|2 − Re[(− 14 + it eiα + t2 e2iα )e−2iα ] =

1 4

+ t2 + t sin α +

1 4

cos 2α − t sin α − t2 =

1 2

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) (1.7.2) is parabolic with ﬁxed point x. Similarly, S2 := 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 .

4.1.7 Connection to the Parabola Theorem

185 $

' Theorem 4.11. K(an /1) is a p-periodic continued fraction of parabolic type from Eα if and only if p = 1 or p = 2 and ∞

an −x = K n=1 1 1

2

&

(1 + x)2 x2 1 − − 1 −· · ·

where x = Sp (x) ∈ ∂Vα .

(1.7.4) %

Proof : That (1.7.4) is a parabolic continued fraction from Eα follows from Theorem 4.10. Let K(an /1) be a p-periodic continued fraction of parabolic type from Eα . It is clear that am ∈ ∂Eα for all m since there are no p-periodic continued fractions of elliptic type from Eα . Since K(an /1) is from Eα , it converges to a ﬁnite value, and so do also all its tails (the Parabola Theorem). Hence, the attracting (m) ﬁxed point x(m) of Sp is = ∞ for all m. Moreover, all x(m) ∈ ∂Vα since all (m−1) (m) (1 + x ) ∈ ∂Eα . That is, am = x x(m) = − 12 + iμm eiα am =

−(− 12

with μm ∈ R for m = 1, 2, . . . , p iα 2

+ iλm e )

with λm ∈ R,

and since am = x(m−1) (1 + x(m) ), −( 14 − iλm eiα − λ2m e2iα ) = (− 12 + iμm−1 eiα )( 12 + iμm eiα )

iλm eiα + λ2m (cos α + i sin α)eiα = 2i (μm−1 − μm )eiα − μm−1 μm e2iα , which gives the two equations λm + λ2m sin α = 12 (μm−1 − μm ) − μm−1 μm sin α ,

λ2m cos α = −μm−1 μm cos α

where cos α = 0. That is, λm = 12 (μm−1 − μm ) and λ2m = 14 (μm−1 − μm )2 = −μm μm−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 and x(2n+1) = − 12 − iμ0 eiα = −(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 the line a + xieiα R.

186

Chapter 4: Periodic and limit periodic continued fractions

Proof :

The boundary of Vα is the line w = − 12 + ti eiα

∂Vα :

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 = −2 − 12 + tieiα · ieiα = −2xieiα for x := − 12 + tieiα . ∂t

4.2 4.2.1

Limit periodic continued fractions Deﬁnition

A continued fraction K(an /bn ) is limit periodic with period length p, or just limit p-periodic or limit periodic for short, if the limits ˜m , lim anp+m =: a

n→∞

lim bnp+m =: ˜bm

n→∞

for m = 1, 2, . . . , p

(2.1.1)

and p is the smallest positive integer for which (2.1.1) holds (although we exist in C, sometimes ﬁnd it convenient to relax this last condition). An arbitrary continued fraction K(an /bn ) can always be made limit periodic by an equivalence transformation. 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 continued for m = 1, 2, . . . , p. fractions for which limn→∞ anp+m /(bnp+m−1 bnp+m ) exists in C Let us assume that the limits in (2.1.1) are ﬁnite with all a ˜m = 0. (Other cases will be treated later on.) Then the (np)th tail of K(an /bn ) looks more and more like the corresponding p-periodic continued fraction ∞

K a˜bn = a˜˜b1 + a˜˜b2 +· · · + a˜˜bp + a˜˜b1 + a˜˜b2 +· · · + a˜˜bp + a˜˜b1 + a˜˜b2 +· · · n=1 ˜ n

1

2

p

1

2

p

1

(2.1.2)

2

as n increases. The convergence of K(˜ an /˜bn ) is determined by the classiﬁcation of the linear fractional transformation a ˜1 a ˜2 a ˜p S˜p (w) := . ˜b1 + ˜b2 +· · · + ˜bp + w

(2.1.3)

So we expect the convergence properties of K(an /bn ) to depend on this classiﬁcation too. We say that K(an /bn ) is a limit periodic continued fraction of parabolic, loxodromic, elliptic or identity type depending on the classiﬁcation of S˜p . We further

4.2.2 Finite limits, loxodromic case

187

say that is the ratio for K(an /bn ) if is the ratio for S˜p . The important point in our investigations is actually that lim Sp(np+m) = S˜p(m)

n→∞

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 a ˜m+1 a ˜m+2 a ˜m+p S˜p(m) (w) = ˜bm+1 + ˜bm+2 +· · · + ˜bm+p + w

(2.1.5)

(m) for all m ≥ 0, and y (m) be a second ﬁxed point of S˜p if it exists, and

x(m−1) = s˜m (x(m) ),

y (m−1) = s˜m (y (m) ) (m)

(m)

(2.1.6) (m)

for all m with obvious notation. We shall also let xn , yn and n denote (np+m) (m) two ﬁxed points (coinciding if necessary) and the ratio for Sp , where xn is attracting or indiﬀerent. Here, as always, (n)

Sk (w) := Sn−1 ◦ Sn+k (w) =

4.2.2

an+1 an+k . bn+1 +· · · + bn+k + w

(2.1.7)

Finite limits, loxodromic case

Let K(an /bn ) with approximants Sn (w) satisfy (2.1.1) with ﬁnite limits a ˜m , ˜bm , ˜ where Sp is a loxodromic transformation from M. Then lim x(m) = x(m) , n

n→∞

lim y (m) n→∞ n

= y (m) and

lim n(m) =

n→∞

(2.2.1)

(m) where is the ratio for S˜p and thus for S˜p . In particular

x(m) = y (m) for all m and || < 1.

(2.2.2)

We shall also allow that one or more a ˜m = 0, as long as the limits (2.1.1) still exist and (2.2.2) holds. Then S˜p(m) is a singular transformation, but we still say that in C K(an /bn ) is a limit p-periodic continued fraction of loxodromic type.

188

Chapter 4: Periodic and limit periodic continued fractions

'

$

Theorem 4.13. Let K(an /bn ) be a limit p-periodic continued fraction of loxodromic type with ﬁnite limits (2.1.1). with the A. Then K(an /bn ) converges generally to some value f ∈ C (m) ∞ p-periodic exceptional sequence {y }m=0 . B. K(an /bn ) converges in the classical sense if all y (m) = 0. ! (m) if t0 = f x −1 C. lim Snp+m (t0 ) = for m = 1, . . . , p . n→∞ y(m) if t0 = f &

%

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 ﬁxed points xn → x and yn → y, where || < 1, x = y and yn is the repelling and ﬁxed point of τn from some n on. Then Tn → f for some f ∈ C, Tn → x. "

!

Proof : Without loss of generality we assume that x = 0 and y = ∞. (Otherwise we choose a ψ ∈ M with ψ(x) = 0 and ψ(y) = ∞, and consider transformations ψ ◦ τn ◦ ψ −1 .) Then xn = ∞ from some n on, and thus τn (xn ) = n → . Since xn → 0 and || < 1, there exist constants n0 ∈ N and μ, ρ > 0 such that |τn (w)| ≤ μ < 1 for all n ≥ n0 and |w| ≤ ρ. That is, τn (ρD) ⊆ ρD for n ≥ n0 . Without loss of (k) generality we assume that n0 = 1. (Otherwise we look at Tn := τk+1 ◦· · ·◦τk+n and (k) (k) Tn := τk+n ◦ · · · ◦ τk+1 for suﬃciently large k ∈ N, and use that Tk+n = Tk ◦ Tn (k) and Tk+n = Tn ◦ Tk .) The chain rule leads to Tn (wn ) = τ1 (w1 ) · τ2 (w2 ) · · · τn (wn ), Tn (w 0 ) = τn (w n−1 ) · τn−1 (w n−2 ) · · · τ1 (w 0 ) k := τk (w k−1 ) for all k. where wk−1 := τk (wk ) and w 0 ∈ ρD implies that all Now, wn ∈ ρD implies that wk ∈ ρD for all k ≤ n, and w w k ∈ ρD. Hence |Tn (w)| ≤ μn → 0 and |Tn (w)| ≤ μn → 0 for |w| ≤ ρ. Therefore rad Tn (ρD) → 0 and rad Tn (ρD) → 0.

4.2.2 Finite limits, loxodromic case

189

The nestedness Tn+1 (ρD) ⊆ Tn (ρD) implies that {Tn } converges generally to some f ∈ 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 x such 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 generally with exceptional sequence {y}.

Proof : That {Tn } converges generally follows from Lemma 4.14. Now, also τk−1 ∈ M with ratio n → and ﬁxed points xn → x and yn → y, only this time xn is the −1 repelling one for large n. Since Tn−1 = τn−1 ◦ τn−1 ◦ · · · ◦ τ1−1 , it follows from Lemma −1 4.14 that {Tn } 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, (m) (m) with constant exceptional sequence {Snp }∞ ∈C n=0 converges generally to some f (m) (m) {y }. We need to prove that Sm (f ) = f (0) for all m ∈ N. This clearly holds if (m) {f (np+m) }∞ for m = 1, 2, . . . , p. Since by Lemma 4.14 n=0 has no limit point at y (np+m) (m) (m) =x = y , the conclusion in A follows with f := f (0) . limn→∞ f B. This is a straight forward consequence of the result in A. (m)

(m) C. Since {Snp }∞ with constant exceptional sequence n=1 converges generally to f (m) (m) −1 ∞ {y }, the sequence {(Snp ) }n=1 converges generally to y (m) with constant exceptional 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 a ˜m = ∞ and ˜bm = 0, converges in the classical sense.

˜ = ∞ and bn → ˜b = 0. Proof : Let ﬁrst K(an /bn ) be limit 1-periodic with an → a For suﬃciently large n, bn = 0, and the ﬁxed points of sn (w) are xn :=

bn (−1 + un ), 2

yn :=

bn (−1 − un ) 2

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 a ˜1 , a ˜2 = ∞ and ˜b1 , ˜b2 = 0. By (1.5.2) (2n+m)

on page 181 and the subsequent arguments, the ﬁxed points of S2 (m)

(m) (m)

xn(m) =

2a2n+m+1 − λn + λn un 2b2n+m+1

yn(m) =

2a2n+m+1 − λn − λn un 2b2n+m+1

(m)

(m)

for suﬃciently large n, where λn m = 0, 1.

are

,

(m) (m)

(m) →a ˜1 + a ˜2 + ˜b1˜b2 = 0 and yn → y (m) = 0 for

Example 8. The continued fraction K(an /1) where an → 0 is limit 1-periodic of loxodromic type, since sn (w) := an /(1 + w) has ﬁxed points xn , yn and ratio n given by −1 + un −1 − un 1 − un , yn = , n = xn = 2 2 1 + un √ Its where un := 1 + 4an → 1. Hence K(an /1) converges to some value f ∈ C. tail values f (n) → 0, and every tail sequence {tn } with t0 = f converges to −1. 3

Example 9. The continued fraction ∞

K an := 3 + 11/1 n=1 1

2

4 + 3/22 3 + 1/32 4 + 3/42 3 + 1/52 1 1 1 1 + + + + +· · ·

in Example 4 on page 13 is limit 2-periodic with corresponding 2-periodic continued fraction ∞ 3 4 3 4 3 a ˜n := . n=1 1 1 + 1 + 1 + 1 + 1 +· · · Since 3 3(1 + w) S˜2 (w) = = 4 5+w 1+ 1+w

K

has ﬁxed points x(0) = 1 and y (0) = −3 where |5 + y (0) | < |5 + x(0) |, it follows that K(an /1) is of loxodromic type, and that x(0) is the attracting ﬁxed point for S˜2 . Since Therefore: K(an /1) converges to some value f ∈ C. x(1) :=

4 =2 1 + x(0)

and

y (1) :=

4 = −2, 1 + y (0)

4.2.2 Finite limits, loxodromic case

191

every tail sequence {tn } for K(an /1) satisﬁes 1 if t0 = f, 2 lim t2n = lim t2n−1 = n→∞ n→∞ −3 if t0 = f, −2

if t0 = f, f. if t0 =

3 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 ∞

K a˜1n = 12 + 11 − 11 + 12 + 11 − 11 +· · ·

n=1

(m) from Example 6 on page 179 and Example 2 on page 59. S˜3 is loxodromic with (m) (m) attracting ﬁxed point x and repelling ﬁxed point y given by

2 x(0) = 12 , x(1) = 3, x(2) = − , y (0) = 0, y (1) = ∞, y (2) = −1. 3 and its tail sequences satisfy Hence K(an /1) converges generally to some f ∈ C, x(m) if t0 = f, lim t3n+m = n→∞ if t0 = f, y(m) just as the tail sequences for K(˜ an /1). Now, K(˜ an /1) diverges by Thiele oscillation. This does not necessarily imply that K(an /1) diverges. It depends on how {a3n+m }∞ ˜m for n=1 approaches its limit a m = 1, 2, 3. It may actually be very complicated to determine whether K(an /1) converges or diverges in the classical sense. 3

Example 11. Let K(an /bn ) satisfy an → 0, bn → 0, but an /bn bn−1 → 2. Then bn = 0 from some n on, so K(an /bn ) is equivalent to K(cn /dn ) where cn := an /bn bn−1 and dn := 1 from some n on. That is, K(cn /dn ) is limit 1-periodic Thereof loxodromic type, and thus converges in the classical sense to some f ∈ C. fore also K(an /bn ) converges to f . Since s(w) := 2/(1 + w) has attracting ﬁxed point 1 and repelling ﬁxed point −2, the tail sequences {t˜n } for K(cn /dn ) satisfy 1 if t˜0 = f lim t˜n = n→∞ −2 if t˜0 = f. The tail sequences {tn } for K(an /bn ) satisfy tn = bn t˜n from some n on. Hence tn → 0, or more precisely, bn if t0 = f tn ∼ −2bn if t0 = f. 3

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 ) where the limits a ˜m and ˜bm in (2.1.1) are ﬁnite, but now we assume that S˜p given by (2.1.3) is a parabolic transformation from M. We shall not allow a ˜m = 0 in this section. The situation is signiﬁcantly diﬀerent from the loxodromic case. Periodic continued fractions of parabolic type converge (Theorem 4.5 on page 179), but among their closest neighbors are the elliptic ones which diverge generally. This is reﬂected in the 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 = − 14 . 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−1 bn → − 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 on page 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 − π2 < 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 1 − gk . gk

(2.3.2)

k=1

(Otherwise we replace {gn } by a sequence {gn∗ } as explained in the remarks on page 136.) Since an → − 14 , the limit point case then occurs for Sn (Vα,n ) where Vα,n := −gn + eiα H. Hence K(an /1) converges. We can say even more (the proof is postponed to page 209): '

$

Theorem 4.17. Let K(an /1) with an → − 14 be a continued fraction from 1 {Eα,n }∞ n=1 given by (2.3.1). If gn → 2 , then {Sn (w)} converges to some iα with w = − 1 , and every tail ﬁnite value f ∈ −1 + ε + e H for all w ∈ C 2 1 sequence for K(an /1) converges to − 2 . If gn−1 (1 − gn ) ≥ 14 from some n on, then also Sn (− 12 ) → f . &

%

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 = − 12 . 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 tail sequence {tn } of K(cn /dn ) behaves like tn ∼ −dn /2 and Tn (dn w) → f for every 3 w ∈ 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 = −(− 12 )2 is the line − 14 + i eiα R. Hence K(an /1) converges in particular if an → − 14 through a compact subset of the open half plane − 14 + eiα H, (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 | ≤ ρ˜n } with ρ˜n := (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 divergent continued fractions K(an /1) from {En }. &

%

To prove this, we ﬁrst observe that gn → 12 : '

$

Lemma 4.19. Let ρn := gn−1 (1 − gn ) − where 0 < gn < 1 for all n. Then &

1 2

1 4

> 0 for n = 1, 2, 3, . . . .

< gn →

1 2

(2.3.5)

strictly monotonely. %

194

Chapter 4: Periodic and limit periodic continued fractions

◦ Proof : With such a choice of {gn }, we have − 14 ∈ E0,n for all n, where E0,n is 1 ◦ given by (2.3.1). Since K((− 4 )/1) converges, its tail values − 12 ∈ V0,n = −gn + H 1 for all n. Hence gn > 2 for n ≥ 0.

It also follows from (2.3.5) that (1 − gn ) > ( 14 )/gn−1 , and thus gn < 1 −

1/4 1/4 1/4 <1− < · · · < 1 + Sn (g0 ) gn−1 1 − gn−2

where Sn (g0 ) → − 12 (Theorem 4.5). Hence lim sup gn ≤ 12 . This shows that gn → The monotonicity follows since gn < 1 −

1/4 < gn−1 gn−1

⇔

gn−1 −

1 2 < gn−1 4

1 2

0 < (gn−1 − 12 )2 .

⇔

Remark. It follows similarly that if ρn ≥ 0 for all n, then

1 2

≤ gn →

1 2

monotonely.

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 := 1 −gn−1 (1 − gn ) for all n. Since gn > 2 , the sequence {Pn } is a monotonely de Pn = ∞. Let z = 1 be some N th creasing sequence of positive numbers with n root of unity; i.e., z = 1 with z N = 1 for some N ∈ N. Then Pn z diverges by oscillation. Let further 1 + tn −gn 1 − gn , so that tn := = z, gn + z − gn z −tn gn ∞ n k diverges by oscillation. Since {tn } is a tail sequence for and thus n=0 k=1 1+t −tk the continued fraction K(an /1) given by an := tn−1 (1 + tn ) =

−gn−1 (1 − gn )z · , gn−1 + z − gn−1 z gn + z − gn z

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 close enough 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 brings the continued fraction to the form K(cn /1). We therefore study continued fractions K(an /1). Our analysis thereby covers continued fractions K(an /bn ) where a2n+1 a2n =a ˜1 and lim =a ˜2 , n→∞ b2n−1 b2n b2n b2n+1 ˜m /(1 + w) . with S˜2 := s˜1 ◦ s˜2 parabolic when s˜m (w) := a lim

n→∞

(2.3.6)

4.2.3 Finite limits, parabolic case

195

K(an /1) is limit 2-periodic of parabolic type with ﬁnite limits a ˜1 , a ˜2 if a2n−1 → a ˜1 = −x2

and a2n → a ˜2 = −(1 + x)2

(2.3.7)

a2 )2 = 4˜ a1 a ˜2 = 0. The transformations S˜2 for some x = −1, 0, ∞; that is, if (1+˜ a1 +˜ (1) ˜ and S2 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 parts of K(an /1) converge, then K(an /1) itself converges. And the even and odd parts of 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 Remark 1 on page 184 the boundary of Eα is made up of parabolic pairs. Indeed, to every parabolic pair (−x2 , −(1 + x)2 ) with x ∈ C \ (−∞, − 12 ], there exists a parabolic region 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 ) > 14 and thus Eα ⊆ Eα,n for all n. '

$

Theorem 4.20. Let K(an /1) be a limit 2-periodic continued fraction of parabolic type (with ﬁnite limits) from {Eα,n } given by (2.3.1). If gn → 12 , then K(an /1) converges to some value f ∈ −1 + ε + eiα H, and every tail sequence {tn } for K(an /1) satisﬁes lim t2n = x, lim t2n+1 = −1 − x where x is given by (2.3.7). If gn−1 (1 − gn ) ≥ 14 from 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 be omitted.

More general cases. For longer periods than p = 2, p-contractions can often give some answers, for example in combination with the following lemma:

Lemma 4.21. Let K(an /1) be limit p-periodic with ﬁnite limits a ˜m := (m) ∞ limn→∞ anp+m = 0 for m = 1, 2, . . . , p. If {Snp }n=0 converges generally for m = 0, 1, . . . , p − 1, then K(an /1) converges generally to f (0) .

196

Chapter 4: Periodic and limit periodic continued fractions

Proof : Since {am } is bounded and bounded away from 0, there always exist such that sequences {un } and {vn } from C f = lim Snp+m (wnp+m ) n→∞ a np+m+1 = lim Snp+m+1 (wnp+m+1 ) = lim Snp+m n→∞ n→∞ 1 + wnp+m+1 for both {wn } := {un } and {wn } := {vn }, where a an+1 n+1 > 0. lim inf m(un , vn ) > 0 and lim inf m , 1 + un+1 1 + vn+1

4.2.4

Finite limits, elliptic case

Since periodic continued fractions of elliptic type diverge, it is easy to believe that also limit periodic ones diverge. However, as pointed out by Gill ([Gill73]) there also exist convergent ones. Here is a rather simple example (compared to other examples in the literature) of this phenomenon:

Example 13. A limit 1-periodic continued fraction K(an /1) is of elliptic type if ˜ < − 14 . an → a

(2.4.1)

˜/(1 + w) has the two indiﬀerent ﬁxed points Then S˜1 (w) = a * x = − 12 + iq, y = − 12 − 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) for all n ≥ 0. Then {tn } is a tail sequence for the limit 1-periodic continued fraction K(an /1) of elliptic type given by an := tn−1 (1 + tn ). Now, t0 − fn =

t0 Σn

where

Σn :=

n k 1 + tj . −tj j=1

k=0

(See formula (1.5.7) on page 66.) We have 1 + x − 2|x|2 /j 1 + tj−1 = −tj−1 −x + 2|x|2 /j 1 + 2x/j x j − 1 + 2iq x −x − 2|x|2 /j = · = · = −x + 2|x|2 /j 1 − 2x/j x j + 1 + 2iq x

(2.4.3)

4.2.4 Finite limits, elliptic case and thus

197

k k (1 + 2iq)(2 + 2iq) 1 + tj x · = −tj (k + 1 + 2iq)(k + 2 + 2iq) x

j=1

which means that

k

∞ ∞ k+1 ∞ 1 + tj−1

1 + tj 1

=

= 1 + |1 + 2iq| · |2 + 2iq| .

−tj

−tj−1

2 k + O(k) j=1 j=2 k=0

k=0

k=1

That is, Σn converges to a ﬁnite number, and K(an /1) converges to some f = t0 . 3 (np)

Still, K(an /bn ) diverges if {Sp tion.

} converges fast enough to an elliptic transforma$

' Lemma 4.22. Let τn ∈ M have ratio n and ﬁxed points xn , yn for all n ∈ N, where 1/|yn | < ∞ and |n − | < ∞ (2.4.4) |xn | < ∞, for some = 1 with || = 1, and let Tn := τ1 ◦ · · · ◦ τn for all n ∈ N. Then {Tn } is totally non-restrained, and {Tn (−n w)} converges to T (w) for all for a T ∈ M. w∈C &

%

To prove this we shall use the following auxiliary result: $

' Lemma 4.23. For a given sequence {rn } of positive numbers with r := ∞ r < ∞, let {(Xn , Yn )}∞ n n=0 satisfy n=1 0 ≤ Xn ≤ (1 + rn )Xn−1 + rn Yn−1 , 0 ≤ Yn ≤ rn Xn−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 inequalities in (2.4.5); i.e., when Xn−1 Xn 1 + rn rn = Mn where Mn := ; rn 1 + rn Yn Yn−1

198

Chapter 4: Periodic and limit periodic continued fractions

i.e., when

Xn Yn

= Mn,1

X0 Y0

where Mn,1 := Mn Mn−1 · · · M1 .

Now, the matrix product Mk+1 Mk is 1 + rk+1,k Mk+1 Mk = rk+1,k

rk+1,k 1 + rk+1,k

where rk+1,k := rk + rk+1 + 2rk rk+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,1 rk+1 = rk,1 (1 + 2rk+1 ) + rk+1 . We shall prove by induction that rn,1 =

n 1 1 (1 + 2rj ) − . 2 j=1 2

(2.4.6)

Evidently r1,1 = r1 = 12 (1 + 2r1 ) − 12 , and if (2.4.6) holds for n = k, then rk+1,1 = rk,1 (1 + 2rk+1 ) + rk+1 =

k+1 k+1 1 1 1 1 (1 + 2rj ) − (1 + 2rk+1 ) + rk+1 = (1 + 2rj ) − 2 j=1 2 2 j=1 2

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 points xn , 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 where ϕ(x) = 0 and ϕ(y) = ∞, and look at Tn := τm+1 ◦ · · · ◦ τm+n = Tm−1 ◦ Tm+n for suﬃciently large m.) Since all n = 1, we have all xn = yn , and by Theorem 4.2 on page 174, τn can be written τn (w) =

(n − xynn )w + (1 − n )xn (xn − n yn )w − (1 − n )xn yn = (1 − n )w + n xn − yn − y1n (1 − n )w − n xynn + 1

with the natural limit form if yn = ∞. That is, (1)

τn (w) =

(2)

(1 + δn )w + δn (3)

(4)

δn w + 1 + δ n

for n = 1, 2, 3, . . .

(2.4.7)

4.2.4 Finite limits, elliptic case

199

(k) where |δn | < ∞ for k = 1, 2, 3, 4 under our conditions. Let {rn } be a sequence of positive numbers such that |δn(k) | ≤ rn

for all n and k = 1, 2, 3, 4

and r :=

rn < ∞.

We want to prove that Tn (w) =

An w + Bn lim An −n = C(1) , lim Bn = C(2) where lim Cn −n = C(3) , lim Dn = C(4) Cn w + Dn

(2.4.8)

with C(1) C(4) − C(2) C(3) = 0 for r suﬃciently small. Since Tn = Tn−1 ◦ τn , we ﬁnd that An = (1 + δn(1) )An−1 + δn(3) Bn−1 , Bn = δn(2) An−1 + (1 + δn(4) )Bn−1 , Cn = (1 + δn(1) )Cn−1 + δn(3) Dn−1 ,

Dn = δn(2) Cn−1 + (1 + δn(4) )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) | = |δn(1) −(n−1) An−1 + δn(3) −(n−1) Bn−1 | ≤ rn |An−1 | + rn |Bn−1 | since || = 1, where {An } and {Bn } are bounded. Hence {An −n } converges absolutely 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) := lim Bn and C(4) := lim Dn also exist. Finally, by (2.4.9) Δn := An −n Dn − Bn Cn −n = [(1 + δn(1) )An−1 + δn(3) Bn−1 ]−(n−1) · [δn(2) Cn−1 + (1 + δn(4) )Dn−1 ] − [(1 + δn(1) )Cn−1 + δn(3) Dn−1 ]−(n−1) · [δn(2) An−1 + (1 + δn(4) )Bn−1 ] = [(1 + δn(1) )(1 + δn(4) ) − δn(2) δn(3) ]Δn−1 n (1) (4) (2) (3) = · · · = Δ1 [(1 + δk )(1 + δk ) − δk δk ] k=2

which stays bounded away from 0 when rn2 = 1 − 2rn > 0; i.e., all rn < 12 . Our main theorem follows immediately:

rn < ∞ and all rn < 1 with (1 − rn )2 −

200

Chapter 4: Periodic and limit periodic continued fractions $

' Theorem 4.24. Let τn ∈ M have ratio n and ﬁxed points xn , yn , and let Tn := τ1 ◦ τ2 ◦ · · · ◦ τn for all n ∈ N. If τn → τ ∈ M where τ is elliptic with ratio and ﬁxed points x, y, and m(xn , x) < ∞ and m(yn , y) < ∞, |n − | < ∞, then Tn (−n w) → T (w) for a T ∈ M. In particular {Tn } is totally nonrestrained. & (np)

Therefore, a limit periodic continued fraction of elliptic type diverges if {Sp 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 a an sn (w) := and s(w) := bn + w b+w

have ﬁnite ﬁxed points xn , yn = − 12 (bn ± b2n + 4an ), x, y = − 12 (b ± b2 + 4a). Therefore |xn − x| < ∞ and |yn − y| < ∞. Moreover, the ratios of sn and s are given by bn + y n xn x = , = , n = bn + x n yn y |n − | < ∞. where yn , y = 0 since an , a = 0, so also |an − a| < ∞. Then K(an /1) Example 14. Let 0 = an → a < − 14 so fast that is a generally divergent limit periodic continued fraction of elliptic type. 3

4.2.5

Inﬁnite limits

Let K(an /bn ) be limit p-periodic where one or more of the limits (2.1.1) are equal to ∞. Then it may be a good idea to use an equivalence transformation to bring it

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 → 0 can 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

|ak |(−1)

n+k+1

(2.5.2)

n=0 k=1

converges to a ﬁnite value (the Stern-Stolz Theorem on page 100). Hence we concentrate on the case S = ∞. We follow an idea from [JaWa86], and analyze the even and odd parts of (2.5.1). They are given by a2 a3 a4 a5 a1 1 + a2 − 1 + a3 + a4 − 1 + a5 + a6 −· · · a1 a 2 a3 a4 a1 − 1 + a2 + a3 − 1 + a4 + a5 −· · ·

(even part) (odd part).

If they both converge to the same value, then K(an /1) converges. This is in particular the case in the following situation:

Theorem 4.26. K(an /1) converges if an → ∞ with lim

n→∞

a2n−1 =: α ∈ C, a2n

lim

n→∞

a2n where |α|, |β| = 1. =: β ∈ C a2n+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 of K(an /1) is equivalent to a1 /a2 a3 /a4 a5 /a6 . 1/a2 + 1 − 1/a4 + a3 /a4 + 1 − 1/a6 + a5 /a6 + 1 −· · · The transformation s(w) :=

(2.5.3)

−α has ﬁxed points x = −α, y = −1 and α+1+w

α+1+y = α. Hence s is loxodromic with repelling ﬁxed point −1. α+1+x with constant exceptional Therefore (2.5.3) converges generally to some f (e) ∈ C (e) sequence {−1}. Hence it also converges to f in the classical sense. ratio =

Next, let |α| > 1. The even part of K(an /1) is also equivalent to a2 /a1 a4 /a3 1 . 1/a1 + a2 /a1 − 1/a3 + 1 + a4 /a3 − 1/a5 + 1 + a6 /a5 −· · ·

(2.5.4)

202

Chapter 4: Periodic and limit periodic continued fractions

−1/α is also loxodromic with repelling ﬁxed 1/α + 1 + w point y = −1. Hence {S2n } still converges generally and in the classical sense to with exceptional sequence {−1}. some f (e) ∈ C

The transformation s˜(w) :=

with exceptional sequence Similarly, {S2n+1 } converges generally to some f (o) ∈ C (o) (e) {−1} if |β| = 1. It remains to prove that f = f . By (2.5.4) on page 89 Sn(e) (−a2n ) = S2n−1 (0). Since (−a2n ) → ∞, and thus stays asymptotically away from the exceptional se(e) quence {−1}, it follows that Sn (−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) or elliptic (α = 1) type, and the question of convergence is more subtle. Example 15. The continued fraction Rη (a, b) :=

a 12 b2 22 a2 32 b2 42 a2 52 b2 η + η + η + η + η + η +· · ·

where a, b, η are complex numbers = 0, is called Ramanujan’s AGM-fraction. The reason for this is the surprising identity √ 1 Rη ( a+b 2 , ab) = 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 if a2 = b2 ∈ C \ [−∞, 0]. Let a2 = b2 . Then R1 (a, b) has the form K(an /1) where an → ∞ and b2 a2 a2n a2n−1 = 2 and lim = 2. lim n→∞ a2n−1 n→∞ a2n a b Therefore Theorem 4.26 shows that R1 (a, b) also converges whenever |a| = |b|. This was 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 continued fractions of elliptic type. Since a2n+2 )2 1 b2

b

2 ( 2n + 1

−1 = − 2 = < ∞, a2n+1 a a 2n 4n2 it follows from Corollary 4.25 that (2.5.4) diverges generally. Hence R1 (a, b) diverges generally in this case. This was also ﬁrst proved by J. Borwein et al ([BBCM07]) and later on by Lorentzen ([Lore08b]). 3

4.3.1 Periodic continued fractions with multiple limits

4.3

203

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 asymptotic behavior of {Sn } is quite interesting. Example 16. The 1-periodic continued fraction K(−1/1) is of √ elliptic type since S1 (w) = s1 (w) = −1/(1 + w) has the two ﬁxed points (−1 ± i 3)/2, and thus the ratio √ 1−i 3 √ = e−2iπ/3 . = 1+i 3 Since 3 = 1, the sequence {Sn } of linear fractional transformations is 3-periodic. Indeed, −1 S1 (w) = s1 (w) = , 1+w 1+w , S2 (w) = s1 ◦ s1 (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-periodic sequence {fn } : −1, ∞, 0, −1, ∞, 0, −1, . . . . We follow Bowman and Mc Laughlin ([BoML06]) and say that K(−1/1) has 3 limits. Also {Sn−1 } 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, − 12 , 1, −2, − 12 , . . . .

Indeed, the continued fraction K(an /1) can be regarded as N p-periodic of identity type. 3 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, this means 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 (N p)-periodic sequence of linear fractional transformations. pIf moreover K(an /bn ) has a tail sequence {tn } with all tn = ∞ and n=1 (−tn )N = 1, then also {An } and {Bn } are (N p)-periodic. &

%

204

Chapter 4: Periodic and limit periodic continued fractions (m)

Proof : Since {Snp }n is N -periodic for m = 0, 1, . . . , p − 1, it follows that {Sn } is (N p)-periodic. (In principle it may also have a shorter period length, of course.) Indeed, SN p (w) = I(w) ≡ w is the identity transformation, and thus SnN p = I for all n ∈ N, so there exists a γ = 0, ∞ such that AN p = 0,

AN p−1 = γ,

, BN p = γ

and BN p−1 = 0.

On the other hand it follows from Theorem 2.6 on page 66 that AN p − BN p t0 =

Np

(−tn );

i.e.,

− γ t0 = −t0

n=0

Np

(−tn ).

n=1

p Hence, if N n=1 (−tn ) = 1, then γ = 1, and thus also {An } and {Bn } are (N p)periodic. This was proved by Bowman and Mc Laughlin who studied such continued fractions in a series of papers ([BoML06], [BoML07], [BoML08]). √ Example 17. The constant sequence {x} with x := (−1 + i 3)/2 = e2iπ/3 is a tail sequence for K(−1/1) from Example 16. Since 3 = 1 and p = 1, this means that {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 an An Bn An /Bn

−1

0

1 0

0 1

1 −1 −1 1 −1

2 −1 −1 0 ∞

3 −1 0 −1 0

4 −1 1 −1 −1

5 −1 1 0 ∞

6 −1 0 1 0

7 −1 −1 1 −1

8 −1 −1 0 ∞

9 −1 0 −1 0

··· ··· ··· ··· ···

3

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 (totally non-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 T˜m ∈ M for m = 1, 2, . . . , N .

4.4.2 Fixed circles for τ ∈ M

205

Proof : From Lemma 4.22 we know that Tn (−n w) → T (w) for a T ∈ M. Since N = −N = 1, this means that limn→∞ TnN +m (w) = T (m w) =: Tm (w) for each ﬁxed 1 ≤ m ≤ N . Remark: This means that {TnN +m (w)}n converges for each m ∈ N, but to values depending on w.

Example 18. The limit 1-periodic continued fraction K(1/bn ) where bn → ˜bi is of elliptic type if ˜b ∈ R with |˜b| < 2. In particular this holds for ˜b = 0 when 2 S˜1 (w) := 1/(˜b + w) = 1/w, and thus has the ratio = −1. That is, = 1. |bn | < ∞ is totally non-restrained and {S2n−1 } and Therefore K(1/bn ) with {S2n } converge to functions T1 ∈ M and T2 ∈ M respectively in this case. This was exactly what we proved in Van Vleck’s Theorem on page 142. 3

4.4

Fixed circles

4.4.1

Introduction

In order to get a more geometric idea of the asymptotic behavior of {Sn (w)} and {Sn−1 (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. is called a ﬁxed circle for τ ∈ M if it is invariant under the mapping A circle C in C τ ; i.e., if τ (C) = C. If ∞ ∈ C, it may also be called a ﬁxed line for τ . The crucial point is that if w is a point on such a circle C, then iterations of τ will move this point 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 each not a ﬁxed point for τ , there is a unique ﬁxed circle for τ passing through w∈C w. 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 ) = C˜ϕ(w)

(4.2.1)

where C˜v denotes the ﬁxed circle for τ˜ through v. We shall use this fact to describe the ﬁxed circles for τ . is a ﬁxed circle If τ = I; i.e., τ (w) ≡ w, then the case is clear: every circle C on C for τ . But this fact is of minor interest. Nothing happens to w under iterations of

206

Chapter 4: Periodic and limit periodic continued fractions

I. Not much happens if = −1 either. Then τ is conjugate to τ˜(w) = −w which moves a point from w to −w, and back again to w in the sense that τ˜[2n−1] (w) = −w and 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 τ is conjugate to τ˜(w) := w which has ﬁxed points at 0 and ∞. τ˜ moves a point w = 0, ∞ the angle θ around on the circle C˜w through w centered at the origin. Hence every circle C˜ with center at the origin is a ﬁxed circle for τ˜, and no other circle has this property. A circle with center at 0 can also be seen as a circle with center at ∞. Let ϕ ∈ M be the transformation which makes τ = ϕ−1 ◦ τ˜ ◦ ϕ. Then the circles ˜ on the Riemann sphere are ﬁxed circles for τ . They are no longer C := ϕ−1 (C) necessarily concentric, but they are disjoint. For τ (w) :=

aw + b cw + d

with Δ := ad − bc = 0, c = 0,

ϕ(w) := (w − x)/(w − y), and thus ϕ−1 (w) = (x − yw)/(1 − w). The ﬁxed circle for τ˜ through ϕ(∞) = 1 is the unit circle ∂D. Therefore the ﬁxed line L for τ is given by L = ϕ−1 (∂D). That is, L is given by ϕ−1 (eit ) =

x − y 1 + eit x − y eit x+y + = for 0 ≤ t < 2π, it 1−e 2 2 1 − eit

where 1 + eit e−it/2 + eit/2 cos t/2 = it1 = −it/2 = it it/2 1−e −i sin t/2 e −e

for t1 ∈ R ∪ {∞} =: R.

That is, L is the perpendicular bisector of the line segment connecting

x

y

C

L

˜ x and y. The other ﬁxed circles C = ϕ−1 (C) for τ have their centers on the line through x and y, and each one of them winds around x or y, and none of them intersects or touches L or each other. Moreover, since 0 and ∞ are symmetric points with respect to every circle C˜ centered at the origin, it follows that the ﬁxed points x and y are symmetric with respect to each ﬁxed circle (remarks on page 109). We distinguish between the following two 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 n w from M.

4.4.3 Fixed circles and periodic continued fractions

207

Hence also {τ [n] } is a sequence of distinct elliptic transformations from M. Iterations of τ move a point w = x, y around and around on Cw , indeﬁnitely, without ever 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 for all n. Hence {˜ τ [n] } is an N -periodic sequence from M with τ˜nN = I for all n. All τ [n] (ϕ(w))} is an N -periodic other transformations τ˜[n] are elliptic. For w = x, y, {˜ sequence of equidistant points on the circle C˜ϕ(w) with center at the origin, and so {τ [n] (w)} is an

L

N -periodic sequence of (not necessarily equidistant) points on the ﬁxed circle ϕ−1 (C˜ϕ(w) ) = Cw for τ .

τ [3] (w) τ [2] (w) τ (w)

The parabolic case: Let τ be parabolic with ﬁxed point x. If x = ∞, then τ (w) = w+q for some q = 0. Since τ (qt + λ) = q(t + 1) + λ, it follows that every with λ ∈ C is invariant under τ . No other line λ+q R has this property. If x = ∞, then τ is w circle on C conjugate to τ˜(w) := w + q, and the line L through x and the point ζ := τ −1 (∞) = −d/c is a ﬁxed line Cw for 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 = x monotonely along Cw towards x without ever passing x. 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 point y˜ = ∞. Let ﬁrst ∈ R. Then −1 < < 1, and we say that τ is hyperbolic. In this case R iα is a ﬁxed line for τ˜. Indeed, every line e R through the origin is invariant under τ˜, and these are the only ﬁxed circles for τ˜. If ∈ R, then there are no ﬁxed circles for τ˜. Roughly speaking, τ˜[n] (w) spirals in towards the attracting ﬁxed point at the origin. This means that a loxodromic transformation τ has ﬁxed circles if and only if it is hyperbolic. In the hyperbolic case, its ﬁxed circles are exactly all the circles passing through both its ﬁxed points x, y. If > 0, then iterations of τ move a point w monotonely along the ﬁxed circle Cw towards x. If < 0, then iterations of τ also move 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) = Sp[n] ◦ Sm (w) = Sm ◦ (Sp(m) )[n] (w).

(4.3.1)

208

Chapter 4: Periodic and limit periodic continued fractions (m)

(m)

−1 If Sp has ﬁxed circles C, then Sp has the ﬁxed circles Sm (C). We let Cw denote (m) (m) (m) (m) be the ﬁxed points of Sp and the ﬁxed circle of Sp through w, and x , y (0) x := x .

The parabolic case. for λ ∈ C. Iterations The ﬁxed circles for τ˜(w) := w + q are the lines λ + q R τ˜[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 the (m) opposite direction since (˜ τ −1 )[n] (w0 ) = w0 − nq. Therefore, iterations Snp (w) of (m)

CS2 (w)

the parabolic transformation Sp moves a point w = x(m) monotonely towards x(m) (m) (m) along the ﬁxed circle Cw of Sp . Similarly, (m)−1

(m)

moves w monotonely along Cw Snp wards x(m) from the other direction.

to-

(m)

x

Cw

CS1 (w)

L

p=3

Now, Snp+m = Sm ◦ Snp , and CSm (w) = (m) Sm (Cw ) is the corresponding ﬁxed circle for Sp for each m ∈ {1, 2, . . . , p}. Hence, Sn (w) jumps periodically between the p ﬁxed circles CSm (w) for Sp , gradually moving towards x, and a tail sequence {tn } with t0 = x jumps (m) periodically between the p circles Ctm , since (m)−1

(m)−1 Snp (tm )

∈

(m) Ctm .

−1 tnp+m := Snp+m (t0 ) = Snp

−1 ◦ Sm (t0 ) =

We have proved:

'

$

Theorem 4.29. Let K(an /bn ) be p-periodic of parabolic type. For each m ∈ {1, 2, . . . , p} and w ∈ C\{x}, {Snp+m (w)}n is a monotone sequence on (m) (m) −1 CSm (w) and {Snp+m (w)}n is a monotone sequence on CS −1 (w) . {Snp (w)}n (m)−1

and {Snp sides. &

m

(m)

(w)}n approach x(m) monotonely along Cw

from opposite %

Example 19. The 1-periodic continued fraction K((− 14 )/1) is of parabolic type, and every and S1 has the ﬁxed point x = − 12 . The ﬁxed circles for S1 are the line R 1 circle tangent to the real line at x = − 2 . Iterations of S1 move every point w = − 12 monotonely along Cw towards − 12 , and K((− 14 )/1) converges to − 12 . {− 12 } is the sequence of tail values for K((− 14 )/1). Hence it follows from Remark 2 on page 182

4.4.3 Fixed circles and periodic continued fractions

209

that An = −n( 12 )n+1 and Bn = (n + 1)( 12 )n , so that Sn (w) =

−(n − 1)( 12 )n w − n( 12 )n+1 1 1/2 1 − . =− + 2 2n 2n2 w + n2 + n n( 12 )n−1 w + (n + 1)( 12 )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. 3 This observation, combined with the following lemma, makes it easy to prove Theorem 4.17 on page 192.

Lemma 4.30. Let K(an /bn ) be a p-periodic continued fraction of parabolic be a closed set = C, containing at least type with value x, and let V ⊆ C two points. If Sp (V ) ⊆ V , then x ∈ ∂V .

Proof : x ∈ V since K(an /bn ) converges to x. Assume that x ∈ V ◦ . Then every ﬁxed circle C of Sp contains an arc A ⊆ V with x an inner point in A. For every tail sequence {tn } for K(an /bn ), the points tnp lies on such an arc from some n which is on. Since Sp (V ) ⊆ V , this means that t0 = Snp (tnp ) ∈ V for all t0 ∈ C impossible. Hence x ∈ ∂V .

Proof of Theorem 4.17: Since an → − 14 , the convergence of Sn (wn ) with wn ∈ Vα,n := −gn + eiα H

for n = 0, 1, 2, . . .

(4.3.2)

follows from the Parabola Sequence Theorem on page 154. Therefore {Sn } and thus also {Sn−1 } are restrained. Let gn → 12 . Then Vα,n → Vα := − 12 + eiα H. Assume that K(an /1) has a tail sequence {tn } with a limit point x = − 12 . The ﬁxed circle 1 is a circle tangent to the real line at − 12 or Cx of s˜(w) := − 4 /(1 + w) through x the line itself. And every term in the two sequences {˜ s[n] ( x)} and {(˜ s−1 )[n] ( x)} on Cx is also a limit point for {tn } since =⇒ tnk −1 = snk (tnk ) → s˜( x), tnk → x =⇒ tnk +1 = s−1 ˜−1 ( x), tnk → x nk (tnk ) → s

tnk −2 → s˜[2] ( x), · · · tnk +2 → (˜ s−1 )[2] ( x), · · ·

Since s˜[n] ( x) → − 12 from one side and (˜ s−1 )[n] ( x) → − 12 from the other side of − 12 1 along Cx , a subarc of Cx with − 2 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 .

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 ∞ n 1 − gk Σ∞ := = ∞ (remarks on page 136). (4.3.3) gk n=0 k=1

Then the limit point case occurs for Sn (Vα,n ) (the Parabola Sequence Theorem). We know from the remark on page 194 that 12 ≤ gn → 12 . Therefore − 12 ∈ 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 (m) with t0 = x, y approaches y (m) along Ctm . If > 0, then this convergence is (m) monotone on CSm (0) and Ctm respectively. If < 0, they converge in an alternating manner. If Sp is loxodromic with ∈ R, then {Snp+m (0)}n and {tnp+m }n with t0 = x spiral in towards their limits x and y (m) respectively. The elliptic case. A p-periodic continued fraction of elliptic type diverges, of course, and {Sn } is totally non-restrained (Theorem 4.3). But elliptic transformations have ﬁxed circles, so: (m)

−1 • the points Snp+m (w) ∈ CSm (w) and Snp+m (w) ∈ CS −1 (w) for all n, and they m move around and around indeﬁnitely as n increases. −1 ∞ • If the ratio of Sp is an N th root of unity, then {Snp+m }∞ n=1 and {Snp+m }n=1 are N -periodic for each m ∈ {1, 2, . . . , p}. Otherwise the points {Snp+m (w)}∞ n=1 (m) −1 and {Snp+m (w)}∞ are dense on C and C respectively. −1 S (w) n=1 m S (w) m

Example 20. The continued fraction K(−2/1) is 1-periodic of elliptic√type since S1 (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 N th root of unity. If this is as it seems, then the sequence {Sn (w)} of approximants and the tail sequence {Sn−1 (w)} with w = x, y consist of distinct points dense on the ﬁxed circle Cw for S1 . For instance, {fn } and {tn } are dense on R. Below we have recorded the ﬁrst few terms of {fn } with t0 ∈ R and {tn } with t0 := 1: {fn } : {tn } : 3

2 2 −2, 2, − , −6, , 3 5 1 7 1, −3, − , 5, − , 3 5

10 14 6 34 22 , ,− ,− , ,... 7 3 17 11 23 3 17 11 23 45 ,− ,− , , − ,... . 7 3 17 11 23 −

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 {Sn−1 (w)} are p-periodic. This can not happen with p = 1 since S1 (w) = a1 /(b1 +w). Indeed, since S2 (w) = a1 (b2 + w)/(b1 b2 + a2 + b1 w) ≡ w only if b1 = b2 = 0 and a1 = a2 , and thus K(an /bn ) is 1-periodic of elliptic type, we need p ≥ 3. Example 21. For the 3-periodic continued fraction ∞

K an = − 22 − 12 − 11 − 22 − 21 − 11 − 22 − · · · ,

n=1 bn

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 by 1, −4, − 12 , 1, −4, . . . . 3

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, terminating periodic continued fractions for approximating square roots. But the ﬁrst study of inﬁnite periodic continued fractions seems to be due to Daniel Bernoulli ([Berno75]) who studied the 1-periodic continued fraction K(1/b). Periodic continued fractions was also the topic of E. Galois’ ﬁrst published paper ([Galo28]). Here he studied reversed regular periodic continued fractions. The ﬁrst comprehensive study of periodic continued fractions in general seems to be due to Stolz ([Stolz86]). Later on, 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 the transformations τ from M. An equivalent criterion is based on the trace tr(τ ) := a + d of aw + b where Δ := ad − bc = 1. τ (w) := cw + d The connection between and tr(τ ) when Δ = 1 is . Δ 1−u = 1 − tr(τ )−2 . = where u := 1 − 1+u tr(τ )2 3. Limit periodic continued fractions. The systematic work on limit periodic continued fractions started with Van Vleck ([VanV04]). He studied S-fractions K(an z/1) where 0 < an → a. Later on Pringsheim ([Prin10]) and Perron ([Perr13]) made substantial contributions to the theory. This was later simpliﬁed in ([Perr29]).

212

Chapter 4: Periodic and limit periodic continued fractions

4. Tail sequences and the linear recurrence relation. A sequence {tn } is a tail sequence for K(an /bn ) if and only if tn = −Xn /Xn−1 for all n where {Xn } is a non-trivial solution of the recurrence relation Xn = bn Xn−1 + an Xn−2

for n = 1, 2, 3, . . . .

Theorem 4.13 on page 188 can therefore also be seen as a consequence of Poincar´e’s 0. work on recurrence relations ([Poin85]) if all a ˜m = 5. Periodic sequences of element sets. The remark on page 189 shows that if K(˜ an /˜bn ) is a p-periodic continued fraction of loxodromic type, then there exists a p-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 Thron and 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 indicated that our deﬁnition of continued fractions, as a sequence {Sn } of linear fractional transformations with Sn (∞) = Sn−1 (0), could be generalized to give deﬁnitions invariant under conjugation.

4.6

Problems

1. ♠ Diagonalization of matrices. In Problem 12 on page 49 we saw that if an w + bn an bn , τn := was identiﬁed by Tn := cn d n cn w + d n then τ1 ◦ τ2 was identiﬁed by the matrix product T1 T2 . q +w in (a) Prove that if τ1 is parabolic, then the composition ϕ ◦ τ ◦ ϕ−1 (w) = a b (1.2.6) on page 173 correspond to diagonalization of the matrix T1 := . c 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 ) converges if an = 2 and bn = 4eiθ for all n. 4. Convergence of periodic continued fractions. Given the continued fraction ∞

b+

Ka

n=1 2b

=b+

a a a 2b + 2b + 2b +· · ·

Problems

213

√ (a) Prove that if a > 0 and b > 0 then b√ + K(a/2b)converges to a + b2 . Use this to 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 every continued 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-periodic continued fraction ∞

K ab

n

n=1 n

=

a1 a2 ap a1 a2 ap a 1 b1 + b2 +· · · + bp + b1 + b2 +· · · + bp + b1 +· · ·

is the p-periodic continued fraction bp +

ap ap−1 a1 ap ap−1 a1 ap . bp−1 + bp−2 +· · · + bp + bp−1 + bp−2 +· · · + bp + bp−1 +· · ·

Let {Sn (w)} and {Sn (w)} be the approximants of K(an /bn ) and its reverse, respectively. Prove that K(an /bn ) converges generally if and only if its reversed continued fraction 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) = −

w + 1/2 1 + 2 2nw + n + 1

for n = 1, 2, 3, . . . .

Draw a picture of the ﬁxed line for Sn and the ﬁxed circle for Sn through w0 := − 12 + 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 the left when t0 := −1. 8. ♠ Approximants for 2-periodic continued fractions K(an /1). For given x = −1, 0, ∞, ﬁnd general formulas for the classical approximants of the 2-periodic continued fraction −x2 (1 + x)2 x2 (1 + x)2 x2 . 1 − 1 1 − 1− − 1 −· · ·

214

Chapter 4: Periodic and limit periodic continued fractions

9. ♠ Divergence of limit periodic continued fractions of parabolic type. Let K(an /1) have a tail sequence {tn } given by tn := −

1 1/4 + i μ − 2 n+1

for n = 0, 1, 2, . . .

for some μ > 0. Prove that then K(an /1) is a divergent limit 1-periodic continued 2 fraction of parabolic type with an = − 14 − 1/16+μ . n(n+1) 10. General and classical convergence. (a) Show that if K(a/b) is a 1-periodic continued fraction which converges generally 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 generally 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 general sense but not in the classical sense. 11. Thiele oscillation. For which values of z does the 4-periodic continued fraction ∞

K a1

n

n=1

=

1 1 2 2z 1 1 2 2z 1 + 1 − 1 − 1 + 1 + 1 − 1 − 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 limiting behavior of its tail sequences in this case. 12. 2-periodic continued fractions and the Parabola Theorem. Prove that K(an /1) with all a2n−1 := a and a2n := a converges if and only if |a| − Re a ≤ 12 . (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 fraction K( 5z /1) (i) converge? (ii) diverge? 4

/1). (b) Find the periodic tail sequences for K( 5z 4

(c) For which values of z ∈ C with |z| = 1 does the continued fraction K(an z/1) converge when an := (5n2 + 1)/(4(n + 1)2 ) for all n? 14. ♠ Convergence of limit periodic continued fractions. For given non-negative integers p, q and r, let a2n−1 :=

p

αk nk ,

a2n :=

k=0

b2n−1 :=

q k=0

p

γk nk ,

k=0

βk nk ,

b2n :=

r k=0

δk n k

Problems

215

be polynomials in n for n = 1, 2, 3, . . . , where all αk , γk , βk and δk are complex numbers with αp γp βq δr = 0. Prove that K(an /bn ) converges if either (a), (b), (c) or (d) holds. (a) q + r > p. (b) q + r = p and

/ 0 αp γp ∈ 14 , ∞ . 2 (αp + γp + βq δr )

(c) q + r = p − 1, αp = γp and βq δr /αp ∈ [−∞, 0] . β + (d) q + r = p − 2, αp = γp and arg αp = arg αp−1 − αp q−1 βq βq δr /αp ∈ [−∞, 0].

δr−1 δr

and

Hint: You may use the Parabola Theorem on page 151 and Theorem 3.4 on page 105. 15. Convergence of a limit periodic continued fraction. Prove that K(an /1) with a2n := 1 − 4n2 ,

a2n+1 := −4n2

converges. (Hint: tn := (−1)n n is a tail sequence for K(an /1).) What is the value of K(an /1)? Does K(−an /1) converge? 16. Convergence of a limit periodic continued fraction. Prove that K(an z/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 + is the ﬁxed line for τ . d)/2c)R 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. Prove that the circle with center Γ ∈ C and radius ρ > 0 is a ﬁxed circle for τ if and only if 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 for K((− 14 )/1). − 12 , then all tn (a) Prove that if t0 = − 12 , then tn = − 12 for all n, and that if t0 = are distinct points on the ﬁxed circle Ct0 , approaching − 12 monotonely from the left. (b) Let x ∈ C \ {0, −1} be arbitrarily chosen. Prove that the 2-periodic continued fraction −x2 −(1 + x)2 −x2 −(1 + x)2 an = 1 1 + 1 1 + 1 + +· · ·

K

is of parabolic type with tail values f (2n) = x and f (2n+1) = −(1 + x).

216

Chapter 4: Periodic and limit periodic continued fractions (1)

(c) Let Sn (w) be the approximants of K(an /1) in (b), and let Cw and Cw de(1) note the ﬁxed circles through w for S2 and S2 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) −(1 + x), and t2n ∈ Ct0 , t2n+1 ∈ Ct1 for t0 = x. (e) Prove that the ﬁxed line for S2 in (b) is the line through x and S2−1 (∞) = 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 at γ 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 := − 12 + 23 7 i, ρ := 76 . 21. Fixed circle for elliptic transformation. Let C be the circle with center at − 12 + 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 q 2 − t2 = r 2 .

Chapter 5

Numerical computation of continued fractions If K(an /bn ) converges generally to f , then its approximants {Sn (wn )} converge to f as long as {wn } stays asymptotically away from its exceptional sequence. Therefore we can use Sn (wn ) as an approximation to f . However, the choice of {wn } makes quite 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 to f . 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 of K(an /bn ) to be polynomials or rational functions, and use a rational approximation w n to wn to get the rational approximants Sn (w n ). Even more important from a practical point of view is the control of the truncation error |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 this chapter 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 backward algorithm is stable in most cases, and good choices for {wn } even stabilize it further. 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

218

Chapter 5: Computation of continued fractions

5.1 5.1.1

Choice of approximants 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) 1

1 1

f (1) − Sn−1 (w)

= −

=

Sn (w) f Sn (w) a1

(1.1.1)

(k)

is a measure for the truncation error. Here, as always, Sn (w) denotes the nth approximant of the kth tail of K(an /bn ). For f = ∞ we want bounds for the truncation error |f − Sn (wn )|, and the sequence {ζn } with ζn := Sn−1 (∞) = −Bn /Bn−1 is an exceptional sequence for K(an /bn ). We know that Sn (wn ) → f whenever {wn } stays asymptotically away from an exceptional sequence {wn† }. But some choices of {wn } give faster convergence than others. Since f = Sn (f (n) ), it seems reasonable to expect that Sn (wn ) converges faster the better wn approximates f (n) . Since by Lemma 1.1 on page 6 Sn (f (n) ) − Sn (w) f − Sn (w) = f − Sn (v) Sn (f (n) ) − Sn (v) =

Bn−1 v + Bn w − f (n) v − ζn w − f (n) · = · Bn−1 w + Bn v − f (n) w − ζn v − f (n)

(1.1.2)

(with the natural limit forms if f (n) = ∞ or ζn = ∞), we have f − Sn (wn ) →0 f − Sn (vn )

⇐⇒

κn (wn ) w − f (n) → 0 where κn (w) := . κn (vn ) w − ζn

(1.1.3)

We say that {Sn (wn )} accelerates the convergence of K(an /bn ) when lim

n→∞

f − Sn (wn ) = 0. f − Sn (0)

(1.1.4)

The quantity κn (wn ) is a measure for how well Sn (wn ) approximates f for a given continued fraction. It is essentially the ratio of wn ’s distances to the “good tail sequence” {f (n) } and the “ bad tail sequence” {ζn }. Hence even a rather lousy approximation to f (n) may work well. In particular we do not need high precision in the computation of wn to obtain a reasonably good approximation Sn (wn ) to f . But {f (n) } is in general unknown. Still, if we have some information on its asymptotic behavior, as we for example often have for limit periodic continued fractions, then this can be used to ﬁnd favorable wn . To determine such asymptotic behavior one may need to transform the continued fraction. The good thing about transformations like equivalence transformations or contractions, is that we know exactly what happens to tail sequences. It is also clear that if {wn } accelerates the

5.1.2 The ﬁxed point method

219

convergence of the kth tail of K(an /bn ), then it also accelerates the convergence of K(an /bn ).

5.1.2

The ﬁxed point method

This method gives the “ obvious” approximants for limit periodic continued fractions. 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 from Theorem 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 † = y (m) = x(m) for m = 1, 2, . . . , p for the exceptional se• limn→∞ wnp+m † quences {wn } for K(an /bn ).

Therefore it is a very good idea to use the approximants Snp+m (x(m) ). Indeed, by (1.1.3) f − Snp+m (x(m) ) lim = 0 for f = ∞ (1.2.1) n→∞ f − Snp+m (wnp+m ) for every other sequence {wn } with lim inf n→∞ m(wnp+m , x(m) ) > 0. In particular, f − Snp+m (x(m) ) = 0 for x(m) = 0, f = ∞. n→∞ f − Snp+m (0) lim

(1.2.2)

(If x(m) = 0, then Snp+m (0) can already be seen as a result of the ﬁxed point method.) The eﬀect of this choice of approximants has been demonstrated in a number 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 is no extra work to speak of to compute Snp+m (x(m) ) instead of Snp+m (0), we strongly recommend the use of Snp+m (x(m) ) (or even faster converging approximants to be suggested 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 )

220

Chapter 5: Computation of continued fractions

converges. To get an impression of what to expect, we shall investigate some special cases of the situation K(an /1) with

an → − 14 and wn := − 12 , −2iα

an ∈ Eα,n := {a ∈ C; |a| − Re(a e

and

) ≤ 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 ) ≥ 1 for all n. It follows then from the Parabola Sequence Theorem on page 154 that 4 K(an /1) converges to a ﬁnite value f ∈ −g0 + eiα H. Moreover Sn (− 12 ) → f (Theorem 4.17 on page 192). The question is now whether f − Sn (− 12 ) → 0. n→∞ f − Sn (0) lim

By (1.1.2), f − Sn (− 12 ) κn (− 12 ) f (n) + = = f − Sn (0) κn (0) f (n)

1 2

·

ζn ζn +

1 2

∼

f (n) + 12 . 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(− 12 , 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): let w := − 12 + R eiψ ∈ Vn ; that is, R ≤ gn − 12 . Since an = − 14 + r eiθ for some 0 ≤ r ≤ ρn , we get

an 1

r eiθ + 12 R eiϕ

1

− 14 + r eiθ

1 + w + 2 =

1 − 1 + R eiϕ + 2

=

1 + R eiϕ

2 2 ≤

gn−1 (1 − gn ) − 14 + 12 gn − ρn + 12 (gn − 12 ) = 1 1 − gn − (gn − 12 ) 2

1 4

= gn−1 − 12 .

Hence an /(1 + w) ∈ Vn−1 . Without loss of generality we assume that ∞ n=0

Pn = ∞

for Pn :=

n 1 − gk gk

k=1

(1.2.6)

5.1.2 The ﬁxed point method

221

(Theorem 3.31 on page 137). Then {Sn (V0,n )} with V0,n := −gn + H converges to a limit point f (Theorem 3.45 on page 155). In particular Sn (wn ) → f for wn ∈ Vn ⊆ V0,n . Since Sn (wn ) ∈ V0 for all n, also f ∈ V0 . Similarly f (n) = (n) limk→∞ Sk (wn+k ) ∈ Vn for all n. Therefore, if an ∈ B(− 14 , ρn ) for all n, then |f (n) + 12 | ≤ gn − 12 and |ζn + 12 | > gn − 12 (ζn ∈ Vn since Sn (ζn ) = ∞ ∈ V0 ), and we can guarantee that

f − S (− 1 )

n

2

lim sup

≤ 1. f − Sn (0) n→∞

(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 ) −

1 4

≤ ρn

and

gn − gn −

1 2 1 2

→ 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

f − S (− 1 )

n

2

lim

= 0. n→∞ f − Sn (0)

(1.2.9)

&

%

Proof : This time |f (n) + 12 | ≤ gn − 12 since {B(− 12 , gn − 12 )} is a sequence of value sets for K(an /1). But also {B(− 12 , gn − 12 )} is a sequence of value sets for K(an /1) since |an + 14 | ≤ ρn ≤ ρn . Therefore |ζn + 12 | > gn − 12 , and the result follows since

f − S (− 1 ) f (n) + 1 g − n n

2

2

≤

∼

f − Sn (0) ζn + 12 gn −

1 2 1 2

→ 0.

The conditions in this theorem are really quite restrictive. Indeed, since gn → 1 2 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:

222

Chapter 5: Computation of continued fractions

Corollary 5.3. Let K(an /1) with |an + 14 | = o(n−2 ) have a ﬁnite value f . Then (1.2.9) holds.

Proof :

Let gn :=

1 2

+ 1/(4n + 2) for n ≥ 0. Then

ρn := gn−1 (1 − gn ) − and gn−1 − gn =

1 4

=

1 16n2

4 16n2 − 4

−4

= (gn−1 − 12 )(gn − 12 )

(1.2.10)

= 4(gn−1 − 12 )(gn − 12 ) = 4ρn .

Then B(− 14 , ρn ) ⊆ E0,n given by (1.2.3) for all n. Since ρn := |an + 14 | = 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 + 14 |. Then also an ∈ B(− 14 , ρn ) for all gn , belongs n, so K( an /1) converges, and its nth tail value, which we denote by − to B(− 12 , gn − 12 ) for all n (Lemma 5.1). This means that 1 − gn ≤ gn ≤ gn . Indeed, gn−1 (1 − gn ) ≤ − 14 , it follows from Lemma 4.19 on page 193 and since all an = − the subsequent remark that 12 ≤ gn → 12 monotonely, so 12 ≤ gn ≤ gn . ρn /ρn ) → 0. We want to prove that Now, an ∈ B(− 14 , ρn ) where ( δn :=

gn − gn −

1 2 1 2

→ 0.

Straight forward computation, using (1.2.10), shows that gn−1 (1 − gn ) − ρn = ρn ρn =

=

1 4

=

1 gn−1 2 (

− gn ) − ( gn−1 − 12 )( gn − 12 ) (gn−1 − 12 )(gn − 12 )

1 gn−1 − gn ) 2 ( − δn−1 δn = (gn−1 − 12 )(gn − 12 ) gn−1 − 12 δn−1 − δn gn − 12 1 − δn δn−1 2 gn−1 − 12

gn−1 − 12 ) − ( gn − 12 ) 1 ( − δn−1 δn 1 2 (gn−1 − 2 )(gn − 12 )

where 1 < (gn−1 − 12 )/(gn − 12 ) → 1. Assume that a subsequence δnk −1 → δ > 0. Since (gnk −1 − 12 ) → 0, this means that also δnk → δ. Therefore δn → δ. However, this implies that gn−1 − gn ) − ( gn−1 − 12 )( gn − 12 ) ρn = 12 (

∼ δ · 12 (gn−1 − gn ) − δ 2 (gn−1 − 12 )(gn − 12 ) = δ(2ρn − δρn ) = δ(2 − δ)ρn

which is impossible since ρn = o(ρn ) and 0 < δ ≤ 1. Hence δn → 0.

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 :=

1 1 + 2 4n + 2

and gn :=

1 1 + 2 nδ+1

or gn :=

1 + qn 2

from some n on in Theorem 5.2, where δ > d − 2 and r < q < 1 are ﬁxed constants. Then |ζn + 12 | > gn and |f (n) + 12 | < gn from some n on, and thus it follows from Corollary 5.3 and (1.2.8) that

f − S (− 1 )

o(n−δ ) when |an + 14 | ≤ C n−d , n

2

(1.2.11)

=

f − Sn (0) when |an + 14 | ≤ C rn o(ρn ) 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 with an /bn−1 bn → − 14 , 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 approximate the value of K(an /bn ). This contraction is limit 1-periodic, and can possibly be 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 limit p-periodic with ﬁnite limits ˜m = 0, lim anp+m = a

n→∞

lim bnp+m = ˜bm

n→∞

for m = 1, 2, . . . , p.

Then its N th tail looks more and more like the N th tail of the corresponding periodic continued fraction K(˜ an /˜bn ) when N increases. We have proved (Theorem 4.13 on page 188) that if K(an /bn ) is of loxodromic type, then the sequence of tail values for K(an /bn ) has the same asymptotic behavior as the sequence {f˜(n) } of tail 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 periodic an /˜bn ) be two generally continued fractions. More generally, let K(an /bn ) and K(˜ convergent continued fractions where m(an , a ˜n ) → 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 conditions. 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 of K(an /bn ).

224

Chapter 5: Computation of continued fractions

Example 1. The continued fraction K(n2 /1) converges to some f = ∞ (the Parabola 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) which converges f˜(n) := n. (This follows since f˜(n−1) (1 + f˜(n) ) = n2 − ∞with ntail values ˜(k) )/(−f˜(k) ) = ∞.) It seems reasonable to believe that 1 and (1 + f n=0 k=1 approximants Sn (wn ) with wn := n is a good choice for K(an /1). The table below indicates that Sn (n) approximates the value f = 0.4426950409 (correctly rounded to 10 decimals) reasonably well, and far better than Sn (0). For instance, in order to get an absolute error less than 0.002, we need n to be ≈ 500 for Sn (0), but only ≈ 10 for Sn (n). n 1 2 3 23 24 25 501 502 503

Sn (0) 1.00000000 0.20000000 0.71428571 0.48644179 0.40302141 0.48305030 0.44476903 0.44063101 0.44476080

Sn (n) 0.50000000 0.42857143 0.44827586 0.44271504 0.44267739 0.44271070 0.44269504 0.44269504 0.44269504

|f − Sn (0)| −5.6 · 10−1 2.4 · 10−1 −2.7 · 10−1 −4.4 · 10−2 4.0 · 10−2 −4.0 · 10−2 −2.1 · 10−3 2.1 · 10−3 −2.1 · 10−3

|f − Sn (n)| −5.7 · 10−2 1.4 · 10−2 −5.6 · 10−3 −2.0 · 10−5 1.8 · 10−5 −1.6 · 10−5 −2.1 · 10−9 2.0 · 10−9 −2.0 · 10−9

3

˜n := 110+210/n Example 2. The limit periodic continued fraction K(˜ an /1) with a for 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 of K((110 + 215/n)/1) we use approximants Sn (f˜(n) ). The ﬁxed point method gives approximants 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 errors. The true value of K((110 + 215/n)/1) is f = 20.18548937, correctly rounded to 8 decimals. n 1 2 3 4 97 98 99 100

Sn (0) 325.00000000 1.48741419 148.35485214 3.11185304 20.19073608 20.18071637 20.18983340 20.18153746

Sn (10) 29.54545455 15.64551422 24.22152021 17.57667608 20.18551791 20.18546367 20.18551253 20.18546851

Sn (10 n+2 ) n+1 20.31250000 20.09345794 20.25574154 20.13065841 20.18549003 20.18548878 20.18548991 20.18548889

5.1.3 Auxiliary continued fractions n 1 2 3 4 97 98 99 100

f − Sn (0) −304.8 18.7 −128.2 17.1 −5.2 · 10−3 4.8 · 10−3 −4.3 · 10−3 4.9 · 10−3

225 f − Sn (10) −9.4 4.5 −4.0 2.6 −2.9 · 10−5 2.6 · 10−5 −2.3 · 10−5 2.1 · 10−5

f − Sn (10 n+2 ) n+1 −0.13 0.092 −0.070 0.055 −6.6 · 10−7 5.9 · 10−7 −5.4 · 10−7 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 get an absolute error ≤ 0.005, we need n ≈ 100 for Sn (0), n ≈ 50 for Sn (10) and n ≈ 10 for Sn (10 n+2 n+1 ). 3 But when does (f (n) − f˜(n) ) → 0 hold more generally? Let {Ωn }∞ n=1 be a sequence of element sets for continued fractions K(an /bn ). Let further {Vn }∞ n=0 be a sequence of value sets for {Ωn } and let dist(x, V ) denote the euclidean distance between a We then follow ([Jaco83], [Jaco87]) and deﬁne: point x ∈ C and a set V ⊆ C. $

' Deﬁnition 5.1. The sequence {Ωn }∞ n=1 of element sets is a tusc with reof value sets if there exists a sequence {λn }; spect to its sequence {Vn }∞ n=0 0 < λn → 0, such that diam Sn(m) (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 convergence sets”. 2. Let {Ωn } be a tusc with respect to {Vn } where lim inf diamm (Vn ) > 0. (As always, diamm (Vn ) is the chordal diameter of Vn .) Then every continued fraction K(an /bn ) from {Ωn } converges generally (the limit point case occurs). The convergence of {Sn (wn )} is uniform with respect to wn ∈ Vn and K(an /bn ) from {Ωn }. 3. If {En } is a tusc with respect to {Vn } for continued fractions K(an /1), where lim inf diamm Vn > 0, then {En } is bounded; i.e., M := sup{|an |; an ∈ En , n ∈ N} < ∞, since the euclidean diameter diam

am+1 ≤ λ1 1 + Vm+1

for all m ∈ N ∪ {0}.

226

Chapter 5: Computation of continued fractions (m)

If moreover dist(−1, Vn ) ≥ ε > 0 for all n, then |Sn (w)| ≤ M/ε for all w ∈ Vm+n and |f (m) | ≤ M/ε for every continued fraction from {En }. We can therefore 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 Parabola Sequence Theorem, where we assume that n(m) := Σ

n

(m) Pk → ∞ as n → ∞, uniformly with respect to m

k=0 (m) for Pk :=

m+n

1 − gj . gj j=m+1

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 diam Sn(m) (Vα,m+n ) ≤

n−1 j=1

where

n−1 j=1

1+

M/(ε cos α (m) Pj ε2 cos2 α 1 + (m) · M Σ j−1

(m) (m) Σ Pj n−1 j (m) → ∞ =Σ = j (m) (m) Σ Σ j−1

j=1

j−1

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 fractions from {Ωn }, where {Ωn } is a tusc with respect to {Vn }, dist(dn , Vn ) ≥ ε > 0 for all n and all K(cn /dn ) from {Ωn }, and lim inf diamm (Vn ) > 0. If {an } ˜n ) → 0 and (an˜bn − a ˜n bn ) → 0, then (f (n) − f˜(n) ) → 0. is bounded, (an − a &

%

Proof : Let {an } be bounded, (an − a ˜n ) → 0 and (an˜bn − a ˜n bn ) → 0. We shall ﬁrst prove that for each given n ∈ N, Dn(m) := |Sn(m) (wm+n ) − S˜n(m) (wm+n )| → 0 as m → ∞

(1.3.3)

for wm+n ∈ Vm+n . Without loss of generality, {Vn } is bounded since

am+1

|Sn(m) (w)| =

≤ M/ε for w ∈ Vm+n (m+1) bm+1 + Sn−1 (w)

where M := sup |am |.

5.1.4 The improvement machine

227

For n := 1 and wm+1 , w ˜m+1 ∈ Vm+1 , (m)

D1

am+1 a ˜m+1

−

bm+1 + wm+1 ˜bm+1 + w ˜m+1 (1.3.4)

a ˜m+1 bm+1 + am+1 w ˜m+1 − a ˜m+1 wm+1

m+1˜bm+1 − a =

, (bm+1 + wm+1 )(˜bm+1 + w ˜m+1 ) (m)

= |S1

(m) (wm+1 ) − S˜1 (w ˜m+1 )| =

where the denominator is ≥ ε2 > 0 and am+1 w ˜m+1 − a ˜m+1 wm+1 = am+1 (w ˜m+1 − wm+1 ) + wm+1 (am+1 − a ˜m+1 ). (m)

(m)

˜m+1 ) → 0. In particular, D1 → 0 as m → ∞ Hence D1 → 0 if also (wm+1 − w for w ˜m+1 := wm+1 . We shall prove (1.3.3) by induction on n, so assume that it holds for all 1 ≤ n ≤ ν − 1. For n := ν we then get

Dn(m) =

am+1 (m+1)

bm+1 + Sn−1 (wn+m )

−

a ˜m+1

˜bm+1 + S˜(m+1) (wn+m ) n−1

(m+1) (m+1) where vm+1 := Sn−1 (wm+n ) ∈ Vm and v˜m+1 := S˜n−1 (wm+n ) ∈ Vm with (m) (vm+1 − v˜m+1 ) → 0 as m → ∞ by our induction hypothesis. Hence, also Dn → 0 as m → ∞ by (1.3.4). This proves (1.3.3). (m) (m) (m) Now, |f (m) − f˜(m) | ≤ |f (m) − Sn (wm+n )| + |f˜(m) − S˜n (wm+n )| + Dn ≤ 2λn + (m) Dn 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 ﬁnite limits, and assume that we have accelerated its convergence by using Sn (wn ) instead of Sn (0), where 0 = (wn − f (n) ) → 0, for instance by using the ﬁxed point method. If the ﬁnite limits λnp+m+1 =: rm n→∞ λnp+m lim

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]). We demonstrate 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 S˜1 (w) := a ˜/(˜b + w) if a ˜ = 0 and x := 0 if a ˜ = 0. (The case p > 1 can be found in ([JaWa88], [JaWa90]).)

228

Chapter 5: Computation of continued fractions $

' Theorem 5.5. Let K(an /bn ) be a limit 1-periodic continued fraction of loxodromic type, with ﬁnite limits ˜b = 0 and a ˜, attracting ﬁxed point x, and ﬁnite value f . Let further {wn } be a sequence from C with wn = 0 and 0 = (wn − f (n) ) → 0. If limn→∞ λn+1 /λn = r for {λn } given by (1.4.1), then (1)

f − Sn (wn ) =0 n→∞ f − Sn (wn ) lim

wn(1) := wn +

for

λn+1 . ˜ (b + x + rx)

&

%

a )/2 (Theorem 4.13 Proof : Let λn+1 /λn → r. Since f (n) → x = (−˜b + ˜b2 + 4˜ on page 188), we may without loss of generality assume that all f (n) = ∞, and bn + f (n) = 0. Since 0 = εn := (wn − f (n) ) → 0, it follows that λn → 0, and thus that |r| ≤ 1. Without loss of generality we may therefore assume that bn + wn = 0 and 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 it follows that

(n)

)−f

(n−1)

εn − εn εn−1 ,

εn bn+1 + f (n+1) + wn εn+1 /εn λn+1 = · , λn εn−1 bn + f (n) + wn−1 εn /εn−1

so εn εn−1

=

λn+1 λn (bn

(1.4.2)

+ f (n) ) +

λn+1 εn λn wn−1 εn−1

bn+1 + f (n+1) + wn εn+1 /εn

(1.4.3)

.

We multiply this identity by wn−1 and solve for (wn−1 εn /εn−1 ): wn−1

εn εn−1

=

λn+1 λn (bn

+ f (n) )wn−1

bn+1 + f (n+1) − wn−1 λλn+1 + wn εn+1 εn n

.

This shows that {wn εn+1 /εn } is a tail sequence for the continued fraction K(cn /dn ) given by cn :=

λn+1 (bn + f (n) )wn−1 λn

dn := bn+1 + f (n+1) − wn−1

→ r(˜b + x)x = r˜ a =: c˜, λn+1 λn

˜ → ˜b + x − xr =: d.

5.1.4 The improvement machine

229

Since K(cn /dn ) is limit 1-periodic of loxodromic type, where S˜1 (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−1 εn /εn−1 → rx or wn−1 εn /εn−1 → −(˜b + x). However, if wn−1 εn /εn−1 → −(˜b + x), then εn /εn−1 → −(˜b + x)/x which is impossible since 0 < εn → 0 and |˜b + x|/|x| > 1. Hence wn−1 εn /εn−1 → rx. This proves that εn /εn−1 → r when x = 0. If x = 0, then it follows from (1.4.3) that εn λn+1 ∼ λn εn−1

˜b εn = ˜b εn−1

as 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 /n−1 = −(˜b + x + rx) = 0. Since by (1.1.2) on page 218 (1)

(1)

f (n) − wn f − Sn (wn ) ζn − wn = (n) · , f − Sn (wn ) f − wn ζn − wn(1) (1)

where lim(ζn − wn ) = lim(ζn − wn ) = (y − x) = 0, ∞ ( y := −b − x is the repelling ﬁxed point for K(an /bn )) and (1)

−εn − λn+1 /(˜b + x + rx) f (n) − wn = → 0, (n) −εn f − wn the result follows. Remarks. 1. Indeed, lim λn+1 /λn = r

n→∞

⇐⇒

lim n+1 /n = r

n→∞

(1.4.4)

under the conditions of Theorem 5.5 since ⇐= follows from (1.4.3). (1)

2. If wn = f (n) for inﬁnitely many indices, then {wn } does not lead to convergence acceleration, since f − Sn (wn ) = 0 for these indices. This is a drawback since {f (n) } is not known exactly. On the other hand, this will not disturb (1) the computations too much, since {Sn (wn )} still converges reasonably fast to the correct value. 3. The expression for λn is also found in the description of the Bauer-Muir transformation. In fact, our improvement machine is closely related to the following idea: ◦ Find {wn } such that (f (n) − wn ) → 0. (1)

(1)

(1)

◦ Construct the Bauer-Muir transform b0 +K(an /bn ) of K(an /bn ) with respect to {wn }.

230

Chapter 5: Computation of continued fractions (1)

◦ Find numbers {wn } which accelerate the convergence of this new con(1) (1) (1) tinued fraction b0 + K(an /bn ), and repeat the process, etc. The main diﬀerence is that by using the improvement machine, we do not have to (1) (1) compute the Bauer-Muir transform K(an /bn ). 4. A simple, but important special case is the case where all bn = 1 and an → a with | arg(a + 14 )| < π, and all wn = x, the attractive ﬁxed point for K(an /1). Then λn = an − a, and an+1 − a = r =⇒ n→∞ an − a lim

(1)

f − Sn (wn ) =0 n→∞ f − Sn (x) an+1 − a for wn(1) := x + 1 + x + rx lim

when the value f of K(an /1) is ﬁnite. Example 3. The Stieltjes fraction 12 z 22 z 32 z 42 z 52 z an z := 1 · 3 3 · 5 5 · 7 7 · 9 9 · 11 n=1 1 1 + 1 + 1 + 1 + 1 +· · · ∞

K

converges to a holomorphic function f (z) in D := {z ∈ C; | arg(z + 1)| < π} since ˜. Clearly f (0) = 0, so let 0 = z ∈ D. Then K(an z/1) is limit 1an → 1/4 =: a periodic of loxodromic type. Indeed, s˜(w) := a ˜z/(1 + w) has attracting √ ﬁxed point x := (u − 1)/2 and repelling ﬁxed point y := (−u − 1)/2 where u := 1 + z. Hence the approximants √ (1.4.5) Sn (wn(0) ) where wn(0) := x = (u − 1)/2; u := 1 + z can be a starting point for the improvement machine. Since λ(0) ˜z = n = an z − a (0)

z/4 z/4 = 2 4n − 1 (2n − 1)(2n + 1)

(0)

(1)

in this case, we ﬁnd that λn+1 /λn → 1, and thus the approximants Sn (wn ) with wn(1)

:=

wn(0)

(0)

λ an+1 z − a z/(4u) ˜z =x+ + n+1 = x + 1 + 2x u (2n + 1)(2n + 3) (1)

converge faster to f (z) than Sn (x). Now, also (wn − f (n) ) → 0, and (1)

(1) λ(1) n := an z − wn−1 (1 + wn ) z/(4u) z/(4u) 1+x+ = an z − x + 2 4n − 1 (2n + 1)(2n + 3) (1 + x)z/(4u) xz/(4u) − = an z − x(1 + x) − (2n + 1)(2n + 3) 4n2 − 1 z 2 /(16u2 ) − 2 (4n − 1)(2n + 1)(2n + 3)

5.1.4 The improvement machine

231

(0)

where an z − x(1 + x) = λn , so an z − x(1 + x) − Hence λ(1) n = (1)

(4n2

z/(4u) 1 z/4 xz/(2u) = 1 − = . 2 2 4n − 1 u 4n − 1 4n2 − 1

xz/u z 2 /(16u2 ) − , 2 − 1)(2n + 3) (4n − 1)(2n + 1)(2n + 3)

(1)

(2)

so also λn+1 /λn → 1. Hence {Sn (wn )} where wn(2) := wn(1) +

(1)

(1)

λn+1 λ = wn(1) + n+1 1 + 2x u

converges even faster. We can continue the process, but for this example we prefer to stop here. From (2.6.1) in the appendix we know that the principal value of arctan z has the continued fraction expansion Arctan z = z/(1 + K(an z 2 /1)) for z 2 ∈ D. √ √ That is, K(an z/1) has the value f (z) = −1 + z/Arctan z. For z := 1, this is f (1) = −1 + 4/π = 0.273239544735, correctly rounded to 12 decimals. For (1) comparison, the table below shows the ﬁrst approximants Sn (0), Sn (x), Sn (wn ) (2) and Sn (wn ) for K(an z/1) for z := 1; that is, √ √ 1/ 2 2−1 (1) , wn := x + , x := 2 4(2n + 1)(2n + 3) √ 1/32 2− 2 wn(2) := wn(1) + − . 4(2n + 1)(2n + 3)(2n + 5) (2n + 1)(2n + 3)2 (2n + 5) n 5 6 7 8 9 10 11 12 13 14

Sn (0) .2732919254 .2732305259 .2732410964 .2732392779 .2732395906 .2732395368 .2732395461 .2732395448 .2732395448 .2732395447

Sn (x) .2732398207 .2732395100 .2732395493 .2732395441 .2732395448 .2732395447

(1)

Sn (wn ) .2732395555 .2732395435 .2732395449 .2732395447

(2)

Sn (wn ) .2732395411 .2732395451 .2732395447

Each column stops when we have reached the correctly rounded value with 10 decimals, and this value is repeated for all larger indices. We already see the improvement. Now, the continued fraction converges quite fast for z := 1. For values closer to the boundary of D the convergence is slower, and thus the improvement more valuable. For instance, for z := −2 + i/100 we get:

232

Chapter 5: Computation of continued fractions n 100 1000 5000

|f − Sn (0)| ca 1 9.2 · 10−3 1.9 · 10−10

|f − Sn (x)| 1.0 · 10−5 1.2 · 10−8 9.5 · 10−19

(1)

|f − Sn (wn )| 1.5 · 10−6 1.6 · 10−11 2.7 · 10−22

(2)

|f − Sn (wn )| 2.9 · 10−8 3.5 · 10−14 1.1 · 10−26

This indicates that to obtain a bound for |f − Sn (wn )| of the order 10−8 , we need (1) maybe n ≈ 4000 for wn := 0, n ≈ 1000 for wn := x, n ≈ 500 for wn := wn and (2) n ≈ 100 for wn := wn . 3

5.1.5

Asymptotic expansion of tail values

If we can use the improvement machine repeatedly, as we did in the previous example, we actually develop the ﬁrst terms in some asymptotic expansion of f (n) , where the method chooses the terms in the expansion. But we can also take more control over the expansion, and choose its basis functions uj (n), such that (hopefully) f (n) ∼

∞

kj uj (n)

as n → ∞

where

j=σ

lim

n→∞

uj+1 (n) = 0 for all j. uj (n)

(1.5.1)

We demonstrate the idea in the following situation: let K(an /bn ) be a convergent continued fraction where an = a(n) and bn = b(n) are polynomials in n. That is, K(an /bn ) is limit periodic. Assume that its tail values f (n) have asymptotic expansions ∞ kj n−j as n → ∞ (1.5.2) f (n) = f (n) ∼ j=σ

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 to (N ) use the approximants Sn (wn ) for K(an /bn ) where wn(N ) :=

σ+N

kj n−j .

(1.5.4)

j=σ

This idea was suggested by Wynn ([Wynn59]). That f (n) really has an asymptotic expansion of the form (1.5.1) is not always the case. But by the Birkhoﬀ-Trjzinzki theory it follows that if the process ∞ of ﬁnding σ and kj works, then K(an /bn ) has a tail sequence {tn } with tn ∼ σ kj uj (n) as n → ∞ where uj (n) is as speciﬁed for σ ≤ j ≤ σ + N . We just have to make sure that it is the sequence of tail values. (More information on this can be found in the remark section on page 262.)

5.1.5 Asymptotic expansion of f (n)

233

Example 4. The continued fraction ∞

2 2

an 2z 1 z K = n=1 bn 1− 3

22 z 2 32 z 2 − 5 − 7 −· · ·

in the cut plane D where 0 < arg(1−z 2 ) < 2π (formula converges to f (z) := Ln 1+z 1−z (2.4.9) in the appendix on page 271). We assume that (1.5.2) holds. Then (1.5.3) can be written ∞

kj (n − 1)−j

2n − 1 +

j=σ

∞

kj n−j = −(n2 − 2n + 1)z 2

(1.5.5)

j=σ

for n ≥ 2 where −j

(n − 1)

=n

−j

(1 − 1/n)

−j

=n

−j

m ∞ −j 1 . − m n m=0

(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 =

−z 2 ,... 8u

where u2 = 1 − z 2 .

(2)

That is, we get two possibilities for wn := k−1 n + k0 + k1 /n, depending on our choice for u. To make the right choice we observe that K(an /1) is equivalent to K(cn /1) where cn+1 =

an+1 −n2 z 2 z2 = → − , bn bn+1 4n2 − 1 4

so the sequence {f(n) } of tail values for K(cn /1) converges to the attracting ﬁxed 2 point x of the loxodromic transformation s(w) := − z4 /(1+w); i.e., to x = (u−1)/2 with Re(u) > 0. Since f (n) = bn f(n) , we let this be our choice for u. Since K(an /bn ) (N ) is limit periodic of loxodromic type for z ∈ D, it follows that Sn (wn ) converges than Sn (0). faster to Ln 1+z 1−z 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 order √ (N ) approximants Sn (0) and Sn (wn ) for N = 0, 1, 2, where u := 1 − z 2 and wn(0) := (u − 1)n,

wn(1) := wn(0) +

1−u , 2

wn(2) := wn(1) −

z2 . 8un

Each column stops when the value, correctly rounded to 8 decimals, is equal to the rounded value of Ln 1+z 1−z for all indices ≥ n.

234

Chapter 5: Computation of continued fractions n 5 6 7 8 9 10 11 12 13

Sn (0) 1.94498141 1.94571873 1.94587082 1.94590209 1.94590850 1.94590981 1.94591008 1.94591013 1.94591015

(0)

Sn (wn ) 1.94602817 1.94593035 1.94591370 1.94591078 1.94591026 1.94591017 1.94591015

(1)

Sn (wn ) 1.94589653 1.94590818 1.94590985 1.94591010 1.94591014 1.94591015

(2)

Sn (wn ) 1.94591264 1.94591045 1.94591019 1.94591015

The truncation errors are then for instance n 8 9 10 11 12

f − Sn (0) 8.1 · 10−6 1.7 · 10−6 3.4 · 10−7 6.9 · 10−8 1.4 · 10−8

(0)

f − Sn (wn ) −6.4 · 10−7 −1.2 · 10−7 −2.1 · 10−8 −4.0 · 10−9 −7.4 · 10−10

(1)

f − Sn (wn ) 4.7 · 10−8 7.7 · 10−9 1.3 · 10−9 2.2 · 10−10 3.7 · 10−11

(2)

f − Sn (wn ) −5.6 · 10−9 −8.1 · 10−10 −1.2 · 10−10 −1.9 · 10−11 −3.0 · 10−12

It is more interesting to see what happens for z-values where K(an /bn ) converges more 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 10 20 50 100 200 300 400 500 1000

|f − Sn (0)| 4.3 11.3 9.1 3.1 2.6 9.6 · 10−1 6.2 · 10−1 3.7 · 10−1 1.9 · 10−2

(0)

|f − Sn (wn )| 1.7 · 10−1 8.0 · 10−2 2.7 · 10−2 1.0 · 10−2 2.9 · 10−3 1.1 · 10−3 4.5 · 10−4 2.0 · 10−4 5.6 · 10−6

(1)

|f − Sn (wn )| 5.4 · 10−3 1.2 · 10−3 1.6 · 10−4 3.0 · 10−5 4.2 · 10−6 1.0 · 10−6 3.3 · 10−7 1.2 · 10−7 1.6 · 10−8

(2)

|f − Sn (wn )| 3.8 · 10−4 4.5 · 10−5 2.4 · 10−6 2.3 · 10−7 1.6 · 10−8 2.6 · 10−9 6.2 · 10−10 1.8 · 10−10 1.2 · 10−11

To get an error less than ca 5 · 10−5 , one actually needs n to be more than 2000 for (0) (1) (2) Sn (0), ca 750 for Sn (wn ), ca 100 for Sn (wn ) and ca n = 20 for Sn (wn ). 3 The method also works in situations where anp+m = am (n),

bnp+m = bm (n) for m = 1, 2, . . . , p.

(m) −j We then try to ﬁnd the ﬁrst few terms of expansions f (np+m) ∼ kj n for each m ∈ {1, 2, . . . , p}. For further examples we refer to ([Wynn59]).

5.1.6 The square root modiﬁcation

5.1.6

235

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 replace its nth tail by a periodic continued fraction: f (n) =

an+1 an+2 an+3 an+1 an+1 an+1 ≈ bn+1 + bn+2 + bn+3 +· · · bn+1 + bn+1 + bn+1 +· · ·

(1.6.1)

(or a p-periodic continued fraction). If this last continued fraction converges to some value wn , then hopefully f (n) is close to this value. Hence we use the approximants Sn (wn ). This idea was suggested by Gill ([Gill80]). He used it to accelerate continued fractions K(an /1) where an → − 14 . In [JaJW87] it was used to accelerate the convergence of continued fractions K(an /1) with an → ∞. In both these cases the classical convergence may be rather slow, so a method for convergence acceleration is useful.

Example 5. The continued fraction K(an /1) with z 1 for n = 1, 2, 3, . . . an = − + 4 8n 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 Theorem on page 151, since all an ∈ Eα for α := 12 arg 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 → − 14 monotonely, and its nth tail looks similar to the periodic continued fraction an+1 an+1 an+1 1 + 1 + 1 +· · · which converges to √ , 1 + 4an+1 − 1 z 1 1 =− + . wn := 2 2 2 2(n + 1)

(1.6.2)

We therefore expect Sn (wn ) to converge faster, something which is conﬁrmed by the table below. For z := 1, the correct value of K(an /1) is −0.172160228791, rounded to 12 decimals.

236

Chapter 5: Computation of continued fractions n 1 2 3 4 5 6 7 8 9 10 11 12

Sn (0) −0.12500000 −0.15384615 −0.16379310 −0.16793893 −0.16987898 −0.17086247 −0.17139164 −0.17168993 −0.17186450 −0.17196993 −0.17203530 −0.17207677

Sn (− 12 ) −0.25000000 −0.20000000 −0.18421053 −0.17801047 −0.17523168 −0.17386837 −0.17315351 −0.17275888 −0.17253187 −0.17239677 −0.17231404 −0.17226212

Sn (wn ) −0.16666667 −0.17036664 −0.17144140 −0.17183361 −0.17199842 −0.17207476 −0.17211279 −0.17213282 −0.17214386 −0.17215017 −0.17215390 −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 converging continued fractions are for instance: n 100 500 1000 5000 10000 100000

|f − Sn (0)| 4.03 · 10−1 1.28 · 10−1 5.47 · 10−2 1.15 · 10−3 6.15 · 10−5 2.52 · 10−14

|f − Sn (− 12 )| 5.90 · 10−1 1.64 · 10−1 5.95 · 10−2 1.15 · 10−3 6.16 · 10−5 2.52 · 10−14

|f − Sn (wn )| 6.15 · 10−3 8.03 · 10−4 2.25 · 10−4 2.03 · 10−6 7.69 · 10−8 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 for n = 5000, whereas we need considerably more than n = 10 000 for |f − Sn (0)| to be so small. 3

Example 6. Let an := n for all n ∈ N. Then K(an /1) converges to a ﬁnite value 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 corresponding truncation errors. Here f = 0.525135276161 is the value of K(an /1), correctly rounded to 12 decimals.

5.1.6 The square root modiﬁcation n 1 2 3 4 5 11 12 51 52 101 102

Sn (0) 1.00000000 0.33333333 0.66666667 0.44444444 0.58333333 0.53312330 0.51912183 0.52514011 0.52513106 0.52513529 0.52513526

237

Sn (wn ) 0.50000000 0.53518376 0.52051760 0.52752523 0.52380952 0.52504318 0.52519962 0.52513526 0.52513529 0.52513528 0.52513528

f − Sn (0) −4.75 · 10−1 1.92 · 10−1 −1.42 · 10−1 8.07 · 10−2 −5.82 · 10−2 −7.99 · 10−3 6.01 · 10−3 −4.83 · 10−6 4.21 · 10−6 −1.53 · 10−8 1.39 · 10−8

f − Sn (wn ) 2.51 · 10−2 −1.00 · 10−2 4.61 · 10−3 −2.39 · 10−3 1.33 · 10−3 9.21 · 10−5 −6.43 · 10−5 1.24 · 10−8 −1.06 · 10−8 1.97 · 10−11 −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 1 2·3 4·5 1 + 2 − 1 + 3 + 4 − 1 + 5 + 6 −· · · which is equivalent to K(cn /1) with c1 := 1/3, c2 := −1/4 and −(2n − 2)(2n − 1) (1 + 2n − 3 + 2n − 2)(1 + 2n − 1 + 2n) n − 1/2 1 1 1 =− =− + →− as n → ∞ . 4n 4 8n 4 We recognize this continued fraction from the previous example where we also used the square root modiﬁcation. 3 cn :=

Of course, the square root modiﬁcation also works for limit periodic continued fractions of loxodromic type. In our next example we show a method to compute π: Example 7. The beautiful continued fraction ∞

K an n=1 1

where a1 := 4, an+1 :=

n2 for n ≥ 1 4n2 − 1

(1.6.4)

converges to π (appendix, page 267). K(an /1) converges rather fast, but it is easy to accelerate its convergence. The ﬁxed point method gives for instance the approximants * √ Sn (x) where x := 12 ( 1 + 4 · 14 − 1) = 12 ( 2 − 1), and the square root modiﬁcation Sn (wn )

where wn :=

1 1 + 4an+1 − 1) = 2

1 2(

,

8n2 − 1 − 1 4n2 − 1

ought to approximate π even better. The table below shows that this is indeed the case:

238

Chapter 5: Computation of continued fractions n 1 2 3 4 5 6 7 8 9 10

Sn (x) 3.3137084990 3.1344464996 3.1421356273 3.1415404008 3.1415983790 3.1415919726 3.1415927393 3.1415926423 3.1415926551 3.1415926534

Sn (wn ) 3.1651513899 3.1409646632 3.1416290537 3.1415898074 3.1415929165 3.1415926266 3.1415926566 3.1415926532 3.1415926536 3.1415926536

π − Sn (x) −1.7 · 10−1 7.1 · 10−3 −5.4 · 10−4 5.2 · 10−5 −5.7 · 10−6 6.8 · 10−7 −8.6 · 10−8 1.1 · 10−8 −1.5 · 10−9 2.1 · 10−10

π − Sn (wn ) −2.4 · 10−2 6.3 · 10−4 −3.6 · 10−5 2.8 · 10−6 −2.6 · 10−7 2.7 · 10−8 −3.0 · 10−9 3.5 · 10−10 −4.3 · 10−11 5.4 · 10−12

Both {Sn (x)} and {Sn (wn )} are alternating sequences – a fact that gives simple a posteriori truncation error bounds. 3

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 the truncation 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 to reﬂect the fast convergence due to the choice of {wn }. We shall therefore look for (n) ∈ Vn and wn ∈ Vn a sequence {Vn }∞ n=0 of closed value sets for K(an /bn ) with f (n) for all n. Then f = Sn (f ) ∈ 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 )| ≤ diam Sn (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 following lemma: '

$

Lemma 5.6. Let ∞ = fn∗ := Sn (wn ) → f = ∞ for the generally convergent continued fraction K(an /bn ) with value f , tail values {f (n) } and critical tail sequence {ζn } with ζn = ∞. Then f − fn∗ =

an (f (n) − wn ) 1 − wn−1 /ζn−1 ∗ ∗ · (fn − fn−1 ) −λn f (n) − ζn

when λn := an − wn−1 (bn + wn ) = 0 and wn = 0, ∞ for all n. &

%

5.2.1 The ideas Proof :

239

∗ Let wn∗ := s−1 n (wn−1 ) = −bn + an /wn−1 . Then wn = ∞ and

∗ = Sn (wn ) − Sn−1 (wn−1 ) = Sn (wn ) − Sn (wn∗ ) fn∗ − fn−1

=

An−1 wn + An An−1 wn∗ + An (An−1 Bn − An Bn−1 )(wn − wn∗ ) − = , Bn−1 wn + Bn Bn−1 wn∗ + Bn (Bn−1 wn + Bn )(Bn−1 wn∗ + Bn )

and thus, since f = Sn (f (n) ), f (n) − wn Bn−1 wn∗ + Bn Sn (f (n) ) − Sn (wn ) f − fn∗ = = · . ∗ fn∗ − fn−1 Sn (wn ) − Sn (wn∗ ) wn − wn∗ Bn−1 f (n) + Bn Now, ζn = −Bn /Bn−1 = −bn + an /ζn−1 , and both wn = ζn and f (n) = ζn since fn∗ = ∞ and f = ∞. Since also Bn−1 = 0 (because ζn = ∞), we get −1 −1 wn−1 − ζn−1 Bn−1 wn∗ + Bn −bn + an /wn−1 − ζn = = a n Bn−1 f (n) + Bn f (n) − ζn 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 bound for |f − Sn (wn )|. Idea 3: We can combine an already known truncation error bound |f − Sn (w n )| ≤ λn for K(an /bn ) with the identity

w

n − ζn wn − f (n)

· n )| |f − Sn (wn )| =

|f − Sn (w wn − ζn w n − f (n)

(2.1.2)

( (1.1.2) on page 218). If wn ∈ Vn and f (n) ∈ Vn , then |wn − f (n) | ≤ diam Vn . If we can establish a bound

w n − ζn

≤ kn , (wn − ζn )(w n − f (n) ) then |f − Sn (wn )| ≤ kn λn diam Vn . In Chapter 3 we started with the value sets, and the purpose was to prove convergence for a large family of continued fractions. The truncation error bounds we obtained, were then valid for every continued fraction from this family and for every wn ∈ Vn . What we now want, is to ﬁnd “ small” value sets {Vn } for a given continued fraction K(an /bn ) to get “ small” error bounds. It takes considerably more eﬀort to establish such bounds, so the truncation error bounds in Chapter 3 are 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 whole point! If f (n) was known, then f = Sn (f (n) ) makes the need for {wn } and truncation

240

Chapter 5: Computation of continued fractions

error bounds quite obsolete.) So, how can we make sure that also f (n) ∈ Vn for all n? Well, for one thing, if f (n) ∈ Vn from some n on, then f (n) ∈ Vn for all n. But more importantly, the following holds:

Lemma 5.7. Let {Vn } be a sequence of closed value sets for K(an /bn ). If ˜n ∈ Vn for all n such that Sn (w ˜n ) → f , there exists a sequence {w ˜n } with w then f (n) := Sn−1 (f ) ∈ Vn for all n.

Proof :

˜n ) ∈ V0 for all n, we have f ∈ V0 , and similarly Since Sn (w (n)

˜n+k ) := Sk (w

an+1 an+k ∈ Vn for all n and k bn+1 +· · · + bn+k + w ˜n+k

(2.1.3)

implies that (n)

f (n) = Sn−1 (f ) = lim Sn−1 ◦ Sn+k (w ˜n+k ) = lim Sk (w ˜n+k ) ∈ Vn . k→∞

k→∞

This means that if K(an /bn ) actually converges generally to this value f , then its tail values f (n) ∈ Vn . This gives us the tool we need to develop a priori truncation error bounds.

5.2.2

Truncation error bounds

Let K(an /bn ) be a generally convergent continued fraction, and let {wn } be a sequence of complex numbers which approximate its tail values. We want to ﬁnd a sequence {Vn }∞ n=0 of “ small” value sets for K(an /bn ) where wn ∈ Vn , at least from some n on. Since circular disks behave so nicely under linear fractional transformations, we shall 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 determine ρn > 0 such that an /(bn + B(Γn , ρn )) ⊆ B(Γn−1 , ρn−1 ) for all n. We ﬁrst note the following result due to Lane ([Lane45]):

5.2.2 Truncation error bounds

241 $

' Lemma 5.8. Let Vn := B(Γn , ρn ) for Γn ∈ C and ρn > 0 for n ≥ 0, and let Ωn be the set of all pairs (a, b) ∈ C2 for which |b + Γn | > ρn and

a(b + Γn ) |a|ρn

− Γ ≤ ρn−1

+ n−1 |b + Γn |2 − ρ2n |b + Γn |2 − ρ2n

(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(Γ∗n , ρ∗n ) where b + B(Γn , ρn ) Γ∗n :=

a(b + Γn ) , |b + Γn |2 − ρ2n

ρ∗n :=

ρn |a| . |b + Γn |2 − ρ2n

(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 given in Lemma 5.8. Then n−1 |Γ0 | + ρ0 Mk |Sn+m (w) − Sn (Γn )| ≤ ρn |bn + Γn | − ρn

for w ∈ Vn+m , (2.2.4)

k=1

$

where Mk := max

% w

; w ∈ Vk and Vn := B(Γn , ρn ) for all n. bk + w

&

%

Proof : Since Sn = s1 ◦ s2 ◦ · · · ◦ sn where sk (w) := ak /(1 + w), the chain rule shows that its derivative at w ∈ Vn is given by Sn (w) = s1 (wn,1 ) · s2 (wn,2 ) · · · sn (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 sk (wn,k ) =

−ak −wn,k−1 = . (bk + wn,k )2 bk + wn,k

(2.2.6)

242

Chapter 5: Computation of continued fractions

Therefore |Sn (w)|

n n−1 wn,k

|wn,k−1 | |wn,0 |

= =

bk + wn,k

|bk + wn,k | |bn + wn,n | k=1

≤

k=1

n−1

|Γ0 | + ρ0 |bn + Γn | − ρn

Mk ,

k=1

and thus, for w∗ ∈ Vn , |Sn (w∗ ) − Sn (Γn )| ≤ |w∗ − Γn | · sup |Sn (w)| ≤ ρn · w∈Vn

n−1 |Γ0 | + ρ0 Mk . |bn + Γn | − ρn k=1

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

n−1 |Γ0 | + ρ0 Mk . |bn + Γn | − ρn k=1

2. It follows from the proof of (2.2.4) that diam Sn (B(Γn , ρn )) ≤ 2ρn

n−1 |Γ0 | + ρ0 Mk . |bn + Γn | − ρn k=1

3. For given {ρ∗n } with ρn ≤ ρ∗n < |bn + Γn | for all n, n−1 n−1 |Γ0 | + ρ0 |Γ0 | + ρ∗0 ∗ ρn Mk ≤ ρn Mk∗ |bn + Γn | − ρn |bn + Γn | − ρ∗n k=1

where

$

Mk∗ := max

k=1

% w

; |w − Γk | ≤ ρ∗k . bk + w

4. Just as in Remark 1 on page 163, we ﬁnd that Mk =

|bk + Γk |2 − (bk + Γk ) − ρ2k | + ρk . |bk + Γk |2 − ρ2k

5.2.3 The Oval Sequence Theorem

5.2.3

243

The Oval Sequence Theorem

The idea is now to let Γn := wn and choose ρn such that (an , bn ) ∈ Ωn given in Lemma 5.8. Now, Lemma 5.8 oﬀers no guarantee for Ωn = ∅. Indeed, it would be helpful 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 stay with this case. Then we need ρn < |1 + Γn |, and we replace Ωn by En × {1} where

$

En := a ∈ C;

% |a|ρn a(1 + Γn )

. + − Γ ≤ ρ

n−1 n−1 |1 + Γn |2 − ρ2n |1 + Γn |2 − ρ2n

(2.3.1)

If Γn−1 = 0, then a ∈ En if 0 ≤ |a|

|1 + Γn | + ρn |a| = ≤ ρn−1 . 2 2 |1 + Γn | − ρn |1 + Γn | − ρn

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 the curve ∂En = a∗n On where a∗n :=

Γn−1 dn 1 + Γn

with dn := |1 + Γn |2 − ρ2n

(2.3.3)

and On := {ξ ∈ C; |ξ − 1| + kn |ξ| = n }

where kn :=

ρn ρn−1 , n := . (2.3.4) |1 + Γn | |Γn−1 |

Let On be the closed set bounded by On . Since kn < 1, we ﬁnd that On = ∅ if 1 ∈ On : i.e., if kn > n . If kn = n , then On consists of the point ξ = 1 only. Hence we assume that (2.3.5) kn < 1 and kn < n . Then 1 ∈ O◦ , and we recognize On = ∂On as a cartesian oval. As a corollary to Theorem 5.9 we now get: $

' Corollary 5.10. (The Oval Sequence Theorem.) Let K(an /1) be a continued fraction from {En } given by (2.3.1). Then |Sn+m (w) − Sn (Γn )| ≤ ρn

n−1 |Γ0 | + ρ0 Mk for w ∈ Vn+m |1 + Γn | − ρn k=1

% $ w

where Mk := max

; w ∈ Vk and V : n := B(Γn , ρn ) for all n. 1+w &

%

244

Chapter 5: Computation of continued fractions

Remark. From Remark 4 on page 242, we ﬁnd that Mk =

|Γk (1 + Γk ) − ρ2k | + ρk . |1 + Γk |2 − ρ2k

(2.3.6)

If Γk > 0 and ρ2k ≤ Γk (1 + Γk ), this reduces to Mk =

5.2.4

Γk + ρk . 1 + Γk + ρk

(2.3.7)

An algorithm to ﬁnd value sets for a given continued fraction 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} approximating {f (n) }. We are looking for circular disks in C centered at Γn := wn (or at a 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 looking for radii ρn > 0 such that an ∈ En given by (2.3.1). The shape of the ovals. If Γn−1 = 0, then En = a∗n O given by (2.3.3)–(2.3.4) has the following properties: 1. En is convex and symmetric about the line e2iαn R where αn :=

1 2

arg a∗n =

1 2

arg(Γn−1 (1 + Γn )).

(2.4.2)

(Lemma 3.50 on page 161.) 2. The intersection En ∩(e2iαn R) is called the axis of En . Lemma 3.50 shows that En ∩ (e2iαn R) = e2iαn In where In := [un , vn ] with un := a∗n μ and vn := a∗n ν; that is, (|Γn−1 | − ρn−1 )(|1 + Γn | − ρn ) if |Γn−1 | ≤ ρn−1 , un := (|Γn−1 | − ρn−1 )(|1 + Γn | + ρn ) if |Γn−1 | ≥ ρn−1 , (2.4.3) vn := (|Γn−1 | + ρn−1 )(|1 + Γn | − ρn ), as essentially pointed out in [JaTh86, p100]. 3. Straight forward checking shows that B(a∗n , rn∗ ) ⊆ En where

when Γn−1 = 0,

rn∗ := (ρn−1 |1 + Γn | − ρn |Γn−1 |)(1 − ρn /|1 + Γn |).

The boundary of this disk is tangent to ∂En at vn e2iα .

(2.4.4) (2.4.5)

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 vn e2iα if it exists. This happens if and only if rn := vn − |Γn−1 (1 + Γn )| (2.4.6) = ρn−1 |1 + Γn | − ρn |Γn−1 | − ρn ρn−1 > 0. Then B(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 based on the disk B(a∗n , rn∗ ) in (2.4.4), and strategies based on the disk an := Γn−1 (1 + Γn ) B( an , rn ) where and rn := ρn−1 |1 + Γn | − ρn |Γn−1 | − ρn ρn−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

Γ Ψ

n−1

1+

|1 + Γn | 1 + Γn−1 n−1

Γ

Γj

n−1

Γn−2

+

+ · · · +

1 + Γn−1 1 + Γn−2 1 + Γj j=0

σn :=

(2.4.10)

for n = 0, 1, 2, . . . gives Ψn = Ψ for all n. 2. Let 1 ρn := 2σn

* 2 Ψn+1 − Ψn+1 − 4δn+1

where

δn+1 := sup {σm−1 σm |am − am |; am := Γm−1 (1 + Γm )} m≥n+1

(2.4.11)

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 {ρn σn }∞ n=n0 is non-increasing since 2ρn σn = Ψn+1 −

* Ψ2n+1 − 4δn+1 =

4δ * n+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 rn σn σn−1 = (ρn−1 σn−1 )σn |1 + Γn | − (ρn σn )σn−1 |Γn−1 | − (ρn σn )(ρn−1 σn−1 ) ≥ (ρn−1 σn−1 )(Ψn − ρn−1 σn−1 ) = δn . an |σn σn−1 ≤ δn ≤ rn σn σn−1 ; i.e., |an − an | ≤ rn for n ≥ n0 + 1. That is, |an − $

' Theorem 5.11. For a given continued fraction K(an /1) and a given sequence {Γn } of complex numbers = −1, let {σn } be a sequence of positive numbers 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 ) for n ≥ n0 . &

%

Remarks. 1. The expression (2.4.11) for ρn−1 satisﬁes ρn−1 =

2δn /σn−1 2δn ≤ =: ρ∗n−1 . Ψn σn−1 Ψn + Ψ2n − 4δn

(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 of bounded sets, then the Oval Sequence Theorem still applies for n ≥ n0 > 0 if we replace |Γ0 | + ρ0 by sup{|w|; w ∈ V0 } in the error bound.

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 (n0 )

Sn0 +k (w) = Sn0 ◦ Sk

(n0 )

(w) =

An0 −1 Sk

(w) + An0

(n ) Bn0 −1 Sk 0 (w)

(n0 )

=

fn0 −1 Sk

(w) − fn0 ζn0

+ Bn 0

(n ) Sk 0 (w)

(n )

(n )

− ζn0

(Lemma 1.1 on page 6). Then Sn+m (w) − Sn (v) =

0 0 ζn0 (fn0 − fn0 −1 )(Sn+m−n (w) − Sn−n (v)) 0 0

(n )

(n )

0 0 (Sn+m−n (w) − ζn0 )(Sn−n (v) − ζn0 ) 0 0

,

(n )

0 where |Sn+m−n (w) − ζn0 | > |Γn0 − ζn0 | − ρn0 if ζn0 ∈ Vn0 , and 0

fm =

a1 a 2 am 1 + 1 +· · · + 1

and ζm = −1 −

am am−1 a2 1 + 1 +· · · + 1

are easy to compute for low values of m. Hence, for w ∈ Vn+m , it follows from the Oval Sequence Theorem that n−1 |ζn0 (fn0 − fn0 −1 )| · ρn (|Γn0 | + ρn0 ) Mj |Sn+m (w) − Sn (Γn )| < (|Γn0 − ζn0 | − ρn0 )2 (|1 + Γn | − ρn ) j=n +1

(2.4.13)

0

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 )| <

n−1 |a1 |ρn (|Γ1 | + ρ1 ) Mj (|1 + Γ1 | − ρ1 )2 (|1 + Γn | − ρn ) j=2

(2.4.14)

for w ∈ Vn+m and n ≥ 2. 2. For n0 = 1 we can also replace a1 by some a ˜1 = 0 which allows n0 = 0. Then n−1

a

|Γ0 | + ρ0

1

Mk |Sn+m (w) − Sn (Γn )| ≤ ρn a ˜1 |1 + Γn | − ρn

(2.4.15)

k=1

where V0 = B(Γ0 , ρ0 ) := a ˜1 /(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 )| < n−1 |a1 a2 |ρn (|Γ2 | + ρ2 ) Mj 2 |1 + a2 + Γ2 | − ρ2 ) (|1 + Γn | − ρn ) j=3

for w ∈ Vn+m for n ≥ 3. 4. These formulas easily generalize to continued fractions K(an /bn ).

(2.4.16)

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 that an → a with | arg(a + 14 )| < π. To ﬁnd bounds for the truncation error |f − Sn (x)| where x is the attracting ﬁxed point √ u−1 where u := 1 + 4a, (2.5.1) x := 2 we shall ﬁnd a sequence {B(x, ρn )} of value sets for K(an /1) by means of our algorithm. 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 |am − a|. (2.5.2) m≥n+1

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 with tail values f (n) and attracting ﬁxed point x ∈ C. 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 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 since the constant sequence {y} with y := −1 − x = x is an exceptional sequence for (n) K(an /1), and thus f (n) = limm→∞ Sm (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 := 12 (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 immediately that K(an /1) in Corollary 5.13 is of loxodromic type, and that x is its attracting ﬁxed point.

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 region En reduces to the disk En = B(0, ρn−1 (1 − ρn )) (see (2.3.2)), a situation we recognize from the extension of Worpitzky’s Theorem on page 136. 3. The radius ρn in (2.5.2) can be written ρn =

2d 2dn+1 n+1 =: ρ∗n . < 2 Ψ Ψ + Ψ − 4dn+1

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 ∞

K an n=1 1

where a1 := 4, an+1 :=

n2 for n ≥ 1 −1

4n2

which converges to π (appendix, page 267). K(an /1) is limit 1-periodic of loxo√ dromic type with a := lim an = 14 and attracting ﬁxed point x := 12 ( 1 + 4a − 1) = √ 1 1 15 2 ( 2 − 1). Now, d1 = a1 − 4 = 4 and dn+1 := sup |am − a| = an+1 − a = m≥n+1

n2 1 1/4 − = 2 for n ≥ 1, −1 4 4n − 1

4n2

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) 4ρ1 , ρ0 := . |1 + x|2 − ρ21 |1 + x|2 − ρ21

In particular . √

n2 − 12 1 1/2 2 − 1

(n) < 2 dn+1 = for n ≥ 1. −

≤ ρn = −

f 2 2 2 4n − 1 4n2 − 1 From the Oval Sequence Theorem it follows for instance that n−1 |Γ0 | + ρ0 |x(1 + x) − ρ2k | + ρk 1/2 |π − Sn (x)| ≤ · . 4n2 − 1 |1 + x| − ρn |1 + x|2 − ρ2k k=1

250

Chapter 5: Computation of continued fractions

That is, ρ1 < 16 , Γ0 ≈ 3.333, ρ0 < 0.47 and Mk =

|x(1 + x) − ρ2k | + ρk < 0.2 for k ≥ 2, |1 + x|2 − ρ2k

so |π − Sn (x)| ≤ 0.44 · 3

(0.2)n−2 4n2 − 1

for n ≥ 2.

The positive case. Of course, Theorem 5.12 holds with Ψ = 1 when all an > 0. But it is a restriction to require that an ∈ B(a, rn ), as we have done in Corollary 5.12. (See Property 4 on 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 ﬁxed point x. Then {B(x, ρn )} is a sequence of value sets for K(an /1) if for each n ∈ N either ρn−1 ≤ x and |an − a + ρn−1 ρn | ≤ ρn−1 (1 + x) − ρn x or ρn−1 ≥ x and an − a + ρn−1 ρn ≤ ρn−1 (1 + x) − ρn x.

(2.5.3)

&

%

√ Proof : Since an > 0 and x = 12 ( 1 + 4a − 1) ≥ 0, it follows from Property 2 on 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, if either 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 the following simpler result: $

' Corollary 5.15. Let K(an /1) with 0 < an → a = ∞ have attracting ﬁxed point x. Then {B(x, ρn )} is a sequence of value sets for K(an /1) if |an − a + ρn−1 ρn | ≤ ρn−1 (1 + x) − ρn x &

for n ≥ 1.

(2.5.5) %

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 of Remark 3 on page 249, on could for instance try ρn := 2dn+1 .

Example 9. Let K(an /1) be given by a1 := 1, Then an →

1 4

a2k :=

k , 4k − 2

a2k+1 :=

k 4k + 2

for k ≥ 1.

√ =: a, x = 12 ( 2 − 1) and

a1 − a =

3 , 4

a2k − a =

1/4 −1/4 and a2k+1 − a = 2k − 1 2k + 1

for k ≥ 1. The choice ρn := 2 dn+1 gives ρ0 :=

3 1 1/2 , ρ1 := , ρ2k := ρ2k+1 := 2 2 2k + 1

for k ≥ 1.

Condition (2.5.5) therefore looks like √ √ | 34 + 32 · 12 | < 32 · 12 ( 2 + 1) − 12 · 12 ( 2 − 1) for n := 1, √ √

1/4 1/4

( 2 + 1)/4 ( 2 − 1)/4

+ − for n := 2k,

≤ 2k − 1 4k 2 − 1 2k − 1 2k + 1 √ √

−1/4 1/4

( 2 + 1)/4 ( 2 − 1)/4

≤ + − for n := 2k + 1

2k + 1 (2k + 1)2 2k + 1 2k + 1 for k ≥ 1, which actually is satisﬁed. Hence K(an /1) has a sequence {Vn } of value sets given by V0 := B(x, 32 ), V1 := B(x, 12 ), V2k := V2k+1 := B x,

1/2 ; k ≥ 1. 2k + 1

Hence it follows for instance from the Oval Sequence Theorem that |f − Sn (x)| ≤

√ n−1 1 ( 2 − 1) + 32 2 ρn 1 √ Mk ( 2 + 1) − ρn k=1 2

(2.5.6)

where by (2.3.7) √

M2m = M2m+1 =

1 2( 2 √ 1 ( 2 2

√ ( 2 − 1)(2m + 1) + 1 = √ . + 1) + 1/(4m + 2) ( 2 + 1)(2m + 1) + 1

− 1) + 1/(4m + 2)

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 to simplify the computation. This is done in the last column with k0 := 4. This bound is easier to handle if we want to determine the order n of Sn (x) needed to reach a

252

Chapter 5: Computation of continued fractions

given accuracy. A slight improvement is obtained if we do not insist on having the √ center 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 √ n−1 2 Mk . (2.5.7) |f − Sn (x)| ≤ ρn 1 √ 2 ( 2 + 1) − ρn k=1 n 5 6 7 8 9 10 11 12 13 14 15

Sn (x) .6931772536 .6931515697 .6931478287 .6931472790 .6931471956 .6931471829 .6931471809 .6931471806 .6931471806 .6931471806 .6931471806

|f − Sn (x)| 3.0 · 10−5 4.4 · 10−6 6.5 · 10−7 9.8 · 10−7 1.5 · 10−8 2.3 · 10−9 3.7 · 10−10 5.8 · 10−11 9.2 · 10−12 1.5 · 10−12 2.4 · 10−13

(2.5.6) 1.1 · 10−3 1.8 · 10−4 4.0 · 10−5 6.6 · 10−6 1.4 · 10−6 2.3 · 10−7 4.7 · 10−8 7.9 · 10−9 1.6 · 10−9 2.7 · 10−10 5.2 · 10−11

(2.5.6)modiﬁed 1.1 · 10−3 1.8 · 10−4 4.3 · 10−5 7.1 · 10−6 1.6 · 10−6 2.6 · 10−7 5.5 · 10−8 9.6 · 10−9 1.9 · 10−9 3.2 · 10−10 6.3 · 10−11

(2.5.7) 9.2 · 10−4 1.5 · 10−4 3.3 · 10−5 5.5 · 10−6 1.1 · 10−6 1.9 · 10−7 3.9 · 10−8 6.6 · 10−9 1.3 · 10−9 2.2 · 10−10 4.3 · 10−11

3

Limit 1-periodic S-fractions. Let z = 0 with | arg z| < π be kept ﬁxed in the S-fraction K(an z/1) where 0 < an → a ≥ 0. Then K(an z/1) is limit 1-periodic of loxodromic type with attracting ﬁxed 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 cartesian ovals 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 un z/|z| and vn z/|z| where un < 0 if |x| < ρn−1 and un = a|z| + |x|ρn − |1 + x|ρn−1 − ρn ρn−1 if |x| ≥ ρn−1 , vn = a|z| − |x|ρn + |1 + x|ρn−1 − ρn ρn−1 . Now, an z ∈ En if un ≤ an |z| ≤ vn . Hence, just as for the positive case we get: $

' Theorem 5.16. Let K(an z/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 a sequence of value sets for K(an z/1) if for each n ∈ N either ρn−1 ≤ |x| and |(an − a)|z| + ρn−1 ρn | ≤ ρn−1 |1 + x| − ρn |x| or ρn−1 > |x| and (an − a)|z| + ρn−1 ρn ≤ ρn−1 |1 + x| − ρn |x|. &

(2.5.8) %

5.2.5 Value sets and the ﬁxed point method

253

Example 10. The S-fraction K(an z/1) given by a1 := 1,

a2k :=

k , 4k − 2

a2k+1 :=

k 4k + 2

for k ≥ 1,

converges to Ln(1 + z) for | arg(z + 1)| < π (see page 270). For z = 0 its value is clearly 0. In the following we assume that z = 0 and | arg z| < π. Since an → 14 =: a and dn := supm≥n |am z − az| = |an z − az| in this case, we ﬁnd that x := x(z) = √ 1 2 ( 1 + z − 1), d1 =

3|z| |z|/4 |z|/4 , d2k = and d2k+1 = for k ≥ 1. 4 2k − 1 2k + 1

As in the previous example we choose ρn := 2 dn+1 ; i.e., ρ0 :=

3|z| |z| |z|/2 , ρ1 := , ρ2k := ρ2k+1 := for k ≥ 1. 2 2 2k + 1

Then condition (2.5.8) looks like 3|z| 3|z|2 3|z| |z| + ≤ |1 + x| − |x| for n := 1, 4 4 2 2 |z|/4 |z|/2 |z|/2 |z|/2 |z|/2 + · ≤ |1 + x| − |x| for n := 2k, 2k − 1 2k − 1 2k + 1 2k − 1 2k + 1 and for n := 2k + 1 ≥ 3 it only requires that

−|z|/4 |z|/2 2

|z|/2 |z|/2

+ (|1 + x| − |x|) if < |x|.

≤

2k + 1 2k + 1 2k + 1 2k + 1 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, |z| ≤ 2Ψ, 2k + 1

|z| |z|

for n = 2k + 1 ≥ 3 if 2k + 1 ≤ or − 1 +

≤ 2Ψ. 2|x| 2k + 1 for n = 2k ≥ 2 if 1 +

Hence there exists a sequence {Vn } of value sets for K(an z/1) with Vn := B(x, ρn ) for n ≥ n0 where 0 if 1 + |z| ≤ 2|1 + x| − 23 |x|, n0 = 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 exactly what we obtained in Example 9. 3

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 particular case where we choose Γ2n := x(0) and Γ2n+1 := x(1) for all n, (2.4.9) holds with equality 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, 1 am | (2.5.10) ρn−1 := Ψ − Ψ2 − 4σ1 σ2 dn where dn := sup |am − 2σn−1 m≥n 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 with tail values f (n) and ﬁnite attracting ﬁxed points (x(0) , x(1) ). Let further n0 ∈ N be such that 4σ1 σ2 dn0 +1 ≤ Ψ2 . Then there exists a sequence {Vn } of value 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 ∞

K an := 3 + 11/1 n=1 1

2

4 + 3/22 3 + 1/32 4 + 3/42 3 + 1/52 1 1 1 1 + + + + +· · ·

from Example 4 on page 13. K(an /1) is limit 2-periodic of loxodromic type with attracting ﬁxed points (1, 2). Therefore σn given by (2.5.9) is σ2n := 3 + 2 = 5 and σ2n+1 := 2 + 1 = 3, and Ψn = Ψ = 2 · 3 − 1 · 2 = 4. Hence σ1 σ2 = 15 and am | gives d1 = 1, d2 = 34 and d2n−1 = d2n = 3/(4n2 ) for n ≥ 2. dn := supm≥n |am − Therefore 4σ1 σ2 dn < Ψ2 for n ≥ 3, and thus K(an /1) has a sequence {Vn } of value sets with V2n := B(2, ρ2n ), V2n+1 := B(1, ρ2n+1 ) for n ≥ 1 where ρ2n

2 2 1 Ψ − Ψ2 − 4σ1 σ2 d2n+1 = − := 2σ2 5 5

ρ2n+1

2 2 1 Ψ − Ψ2 − 4σ1 σ2 d2n+2 = − := 2σ1 3 3

. 1−

45/16 (n + 1)2

1−

45/16 . (n + 1)2

.

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. 3

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 we can choose p−2 p−2 n |1 + Γj | |Γj | (2.5.12) σnp+p−1 := σp−1 := n=−1 j=0

j=n+1

and the other σn s are determined from σp−1 by cyclic shifts. For instance, for k = 3 the three σn given by σ0 = |Γ1 Γ2 | + |1 + Γ1 ||Γ2 | + |1 + Γ1 ||1 + Γ2 | , σ1 = |Γ2 Γ0 | + |1 + Γ2 ||Γ0 | + |1 + Γ2 ||1 + Γ0 | , σ2 = |Γ0 Γ1 | + |1 + Γ0 ||Γ1 | + |1 + Γ0 ||1 + Γ1 | . k−1 k−1 With this choice we get Ψn = Ψ := j=0 |1 + Γj | − j=0 |Γj | > 0. Also now Theorem 5.11 implies that f (np+m) ∈ Vnp+m := B(x(m) , ρnp+m ) for n ≥ n0 suﬃciently large. We refer to [Jaco87] for value sets for more general continued fractions.

5.2.6

Value sets B(wn ρn ) for limit 1-periodic continued fractions of loxodromic or parabolic type

Let K(an /1) be limit 1-periodic of loxodromic type, and let wn ≈ f (n) be chosen to make Sn (wn ) → f = ∞ fast, where f is the value of K(an /1). To take full advantage 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 ≥ 1 from Example 7 on page 237. The square root modiﬁcation gave approximants Sn (wn ) with , 1 √ , 1 8n2 − 1 1/2 wn := 2 · 1 + − 1 = − 1 2 4n2 − 1 2 4n2 − 1 √ √ √ 2−1 2/8 1/4 1 + 2 . −1 = 2 1+ 2 ≈ 2 4n − 1 2 4n − 1 We therefore choose √ Γn := wn :=

√ 2−1 2/8 + 2 2 4n − 1

for n ≥ 1.

256

Chapter 5: Computation of continued fractions

We ﬁrst want to ﬁnd positive σn s such that Ψn := σn (1 + wn ) − σn−1 wn−1 > 0 is non-decreasing. Since √ 1 1 2 − 1 + wn+1 − wn = 1 + 8 (2n + 1)(2n + 3) (2n − 1)(2n + 1) √ √ (2n + 3) − (2n − 1) 2 2/2 =1− =1− 8 (2n − 1)(2n + 1)(2n + 3) (2n − 1)(2n + 1)(2n + 3) is monotonely increasing, we can take all σn := 1 to get √ 2/2 Ψn+1 = 1 + wn+1 − wn = 1 − 2 (4n − 1)(2n + 3)

for n ≥ 1.

Now, an+1 := Γn (1 + Γn+1 ) √ √ √ √2 − 1 2/8 2 + 1 2/8 = + 2 + 2 4n − 1 2 (2n + 1)(2n + 3) √ 2/4 1 1/4 = + + 4 (2n − 1)(2n + 3) (2n − 1)(2n + 1)(2n + 3) 1/32 , + (2n − 1)(2n + 1)2 (2n + 3) so an+1 − an+1

√ 1/32 (2 − 2)/4 − = (2n − 1)(2n + 1)(2n + 3) (2n − 1)(2n + 1)2 (2n + 3) √ (2 − 2)/4 < , (2n − 1)(2n + 1)(2n + 3)

and we choose ρn := 12 (Ψn+1 −

* Ψ2n+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 2/2 2− 2 ≤ 1− (2n − 1)(2n + 1)(2n + 3) (2n − 1)(2n + 1)(2n + 3) which holds for n ≥ 1. Hence K(an /1) has a sequence {Vn } of value sets with √ √2 − 1 2/8 Vn := B + 2 , ρn for n ≥ 1, 2 4n − 1 and |f

(n)

− wn | ≤ ρn <

ρ∗n

√ δn+1 (2 − 2)/2 √ for n ≥ 1. := 2 = Ψn+1 (4n2 − 1)(2n + 3) − 2/2

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) √

√

4( 2+1 + 8·32 ) 4(1 + Γ1 ) 2 √ ≈ 3.160305, Γ0 := √ ∗2 ≤ √2+1 2 (2− 2)/2 2 |1 + Γ1 | − ρ1 √ ( 2 + 8·32 )2 − ( 3·5− ) 2/2 ρ∗0 :=

4ρ∗1 ≈ 0.051153. |1 + Γ1 |2 − ρ∗2 1

The Oval Sequence Theorem and Remark 3 on page 242 thereby imply that |Γ0 | + ρ∗0 ∗ Mk ≤ |1 + Γn | − ρ∗n k=1 √ n−1 (2 − 2) · (3.160305 + 0.051153) k=1 Mk∗ n−1

|f − Sn (Γn )| ≤ ρ∗n

√ ( 2 + 1)[(4n2 − 1)(2n + 3) − where

√ 2 ] 2

√

+

√

Γk + ρ∗k Mk∗ := < 1 + Γk + ρ∗k

2−1 2 √ 2+1 2

+ +

2 (2n 4 √ 2/8 4k2 −1 √ 2/8 4k2 −1

+ 3) −

+ +

1/4 4n2 −1

(2.6.1) −2+

√ 2

√ (2− 2)/2

√ 2/2 √ (2− 2)/2 √ (4k2 −1)(2k+3)− 2/2 (4k2 −1)(2k+3)−

.

The table below compares this error bound to the true truncation errors. In the last column we have replaced Mk by M5 for k > 5. This bound is well suited to determine n in Sn (Γn ) to achieve a wanted accuracy. The true value of K(an /1) is π, and the approximants Sn (Γn ) converge very fast to this value. n 5 6 7 8 9 10 11 12 13 14 15

|f − Sn (Γn )| 2.6 · 10−7 2.7 · 10−8 3.0 · 10−9 3.5 · 10−10 4.3 · 10−11 5.4 · 10−12 7.1 · 10−13 9.5 · 10−14 1.3 · 10−14 1.8 · 10−15 2.5 · 10−16

(2.6.1) 7.4 · 10−7 7.6 · 10−8 8.5 · 10−9 1.0 · 10−9 1.2 · 10−10 1.6 · 10−11 2.0 · 10−12 2.7 · 10−13 3.7 · 10−14 5.2 · 10−15 7.2 · 10−16

(2.6.1)modiﬁed 7.4 · 10−7 7.6 · 10−8 8.5 · 10−9 1.0 · 10−9 1.2 · 10−10 1.6 · 10−11 2.1 · 10−12 2.8 · 10−13 3.9 · 10−14 5.4 · 10−15 7.6 · 10−16

3

Our next example demonstrates how diﬃcult it can be to ﬁnd useful value sets for limit periodic continued fractions of parabolic type.

258

Chapter 5: Computation of continued fractions

Example 13. The continued fraction K(an /1) with 1 (−1/2)n an := − + 4 4n + 2

for n = 1, 2, 3, . . .

is limit 1-periodic of parabolic type. Let all Γn := − 12 . Then |1 + Γn | = |Γn−1 | = 12 , 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 $ 1/2 (1/2)m % 1 − 1 − 4δn where δn := sup (4m2 − 1) ρn−1 = 2n − 1 4m + 2 m≥n i.e., δn = sup {( 12 )m+1 (2m − 1)} = ( 12 )n+1 (2n − 1) for n ≥ 3 m≥n

and δ1 = δ2 = 38 . In particular 4δn < 1 for n ≥ 4, so K(an /1) has a sequence 1 {Vn }∞ n=0 of value sets with Vn := B(− 2 , 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. 3

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 truncation error bounds |f − Sn (w n )| ≤ λn to get |f − Sn (wn )| ≤

λn ρn |w n − ζn | . |wn − ζn | |w n − f (n) |

(2.7.1)

For instance, if {Vn } with Vn := (−gn +eiα H) for 0 < gn < 1 and ﬁxed − π2 < α < π2 is a sequence of value sets for K(an /1), then we know from Theorem 3.45 on page 155 that |a1 |/((1 − g1 ) cos α) diam Sn (Vn ) ≤ n (2.7.2) Pj−1 gj−1 (1 − gj ) cos2 α 1+ j−2 |aj |Σ j=2

where n := Σ

n

Pm

and Pm :=

m=0

m 1 − gk . gk

(2.7.3)

k=1

Another typical example is valid for S-fractions K(an z/1) with an > 0 and z := r e2iα for some ﬁxed r > 0 and − π2 < α < π2 . Then the Thron-Gragg-Warner bound n 2a1 r 1 + 4ak r/ cos2 α − 1 =: λn ≤ (2.7.4) |f − Sn (w)| cos α 1 + 4ak r/ cos2 α + 1 k=2

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 idea is 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 √ √ 2−1 2/8 + 2 |f − Sn (wn )| for wn := Γn := 2 4n − 1 was established by means of the Oval Sequence Theorem in Example 12. We shall now 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 := 12 , then |f − Sn (∞)| ≤ 8

n

1+

j=2

n−1 −1 1/4 4j 2 − 1 −1 =8 . 1+ (j − 1)|aj | 4j 3 j=1

In Example 12 we found that |f

(n)

− wn | ≤

ρ∗n

√ (2 − 2)/2 √ = (4n2 − 1)(2n + 3) − 2/2

for n ≥ 1.

Since all an > 0, we deﬁnitely have a2 an an−1 < −1. ζn = −1 − 1 + 1 +· · · + 1 n := ∞ that Hence |wn − ζn | > |1 + wn | − ρ∗n , and thus it follows from (2.7.1) with w ρ∗n 4j 2 − 1 −1 8 . 1 + |1 + wn | − ρ∗n j=1 4j 3 n−1

|f − Sn (wn )| <

(2.7.5)

Now, K(an /1) can also be considered as an S-fractions with z := 1. The ThronGragg-Warner bound (2.7.4) can therefore also be used; i.e., n−1 1 + 4k 2 /(4k 2 − 1) − 1 ρ∗n |f − Sn (wn )| < . (2.7.6) 8 |1 + wn | − ρ∗n 1 + 4k 2 /(4k 2 − 1) + 1 k=1

The table below shows the true truncation error |f − Sn (wn )| and the three truncation error bounds (2.6.1), (2.7.5) and (2.7.6).

3

n 5 6 7 8 9 10 11 12 13 14 15

|f − Sn (Γn )| 2.6 · 10−7 2.7 · 10−8 3.0 · 10−9 3.5 · 10−10 4.3 · 10−11 5.4 · 10−12 7.1 · 10−13 9.5 · 10−14 1.3 · 10−14 1.8 · 10−15 2.5 · 10−16

(2.6.1) 7.4 · 10−7 7.6 · 10−8 8.5 · 10−9 1.0 · 10−9 1.2 · 10−10 1.6 · 10−11 2.0 · 10−12 2.7 · 10−13 3.7 · 10−14 5.2 · 10−15 7.2 · 10−16

(2.7.5) 3.6 · 10−4 1.8 · 10−4 9.9 · 10−5 5.9 · 10−5 3.8 · 10−5 2.5 · 10−5 1.7 · 10−5 1.2 · 10−5 9.0 · 10−6 6.8 · 10−6 5.2 · 10−6

(2.7.6) 1.7 · 10−6 1.8 · 10−7 2.0 · 10−8 2.3 · 10−9 2.9 · 10−10 3.7 · 10−11 4.8 · 10−12 6.4 · 10−13 8.7 · 10−14 1.2 · 10−14 1.7 · 10−15

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 approximants a1 a2 An−1 wn + An an = . (3.1.1) Sn (wn ) = b1 + b2 +· · · + bn + wn Bn−1 wn + Bn 1. The forward recurrence algorithm where An and Bn are computed by means of the recurrence relation Ak = bk Ak−1 + ak Ak−2 ,

Bk = bk Bk−1 + ak Bk−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 + qk

for 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 n (−ak ) a1 a1 a2 a1 a2 a3 a1 a2 a3 a4 − + − + · · · − k=1 (3.1.4) Sn (0) = B0 B1 B1 B2 B2 B3 B3 B4 Bn−1 Bn which normally only is suited to ﬁnd classical approximants for continued fractions K(1/bn ), even though it can be modiﬁed to give general approximants Sn (wn ). The emphasis in section 1.1.3 was on the complexity of the algorithm; i.e., then number of operations needed to reach Sn (wn ). Even more important is the stability of the computation, in particular if we want to compute high order approximants Sn (wn ). The forward algorithm is clearly unstable if for instance bk Ak−1 ≈ −ak Ak−2 or bk Bk−1 ≈ −ak Bk−2 for some indices. One can also run into very large or very small values for |An | or |Bn | fairly soon, which complicates the 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 than the other two. So, what are we actually doing in this algorithm? Well, let us assume that 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) of Sn (w) is closer to f than it should be. What we do, is: we start with a given value and compute w∈C wn,n := w,

wn,k−1 := sk (wn,k )

for k = n, n − 1, . . . , 1.

(3.1.5)

Remarks

261

That is, we compute the ﬁrst (n + 1) terms of the tail sequence {tn,k }k where (k)

tn,k = wn,k = sk+1 ◦ sk+2 ◦ · · · ◦ sn (w) = Sn−k (w).

(3.1.6)

We know a lot about tail sequences for a generally convergent continued fraction K(an /bn ): • there are two kinds of tail sequences for K(an /bn ); the sequence {tn } of tail values where t0 = f , and all the other tail sequences which in the literature often are called the wrong tail sequences. • all the wrong tail sequences have the same asymptotic behavior in the sense that lim m(tn , t˜n ) = 0 when t0 = f and t˜0 = f . • if we start with a given value for t0 and compute the tail sequence by the recurrence (3.1.7) tn+1 := s−1 n (tn ) for n = 0, 1, 2, . . . , then the computed sequence { tn } will always behave like a wrong tail sequence asymptotically, even if we started with t0 = f . The reason is that even minor inaccuracies makes tn = f (n) , at least if lim inf m(f (n) , tn ) > 0 n→∞

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 start with tn and compute tn−1 = sn (tn ). Therefore, the computation of the sequence of tail values by this algorithm is highly stable, whereas computation of wrong tail sequences 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 bounded closed value set for K(an /1) with V ◦ = ∅ and V ∩ (−1 − V ) = ∅. Let w ∈ V0 . Then the part {tn,k }nk=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 tail values 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 the convergence may become more exact.” The correction he had in mind was exactly the idea of using Sn (w) instead of Sn (0). He was comparing the continued fraction to the series (fn − fn−1 ) where fn := Sn (0).

262

Chapter 5: Computation of continued fractions This idea has been applied by a number of authors, but the ﬁrst systematic investigation 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 work by Gill ([Gill75]). Later on, Gill and the authors of this book have extended the theory in a number of papers.

2. Birkhoﬀ-Trjzinski theory. In order to accelerate the convergence of a continued fraction K(an /bn ), we use approximants Sn (wn ) where wn approximates its tail value 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 = bn Pn−1 + an Pn−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

∞

αm n−m/ω , b(n) ∼ nλ2 /ω

m=0

∞

βm n−m/ω as n → ∞,

m=0

where λi are integers, ω ∈ N and α0 β0 = 0. They proved that every solution of the recurrence relation can be written as a linear combination of two solutions Pn := P (n) which have asymptotic expansions of the form eQ(ρ,n) s(ρ, n)

where

Q(ρ, n) := δ0 n ln(n) +

ρ

δj n(ρ+1−j)/ρ ,

j=1 t (ln n)j nrt−1 /ρ qj (ρ, n) s(ρ, n) := nθ j=0

qj (ρ, n) :=

∞

cj,m n−m/ρ

m=0

where the integers ρ ≥ 1, rj and δ0 ρ and the complex numbers δj , θ and cj,m with cj,0 = 0, r0 := 0 and −π ≤ Im δ1 < π can be determined by substituting this into the recurrence relation and comparing coeﬃcients, ([Birk30], [BiTr32]). For more information we refer to [Wimp84].

5.5

Problems

1. Limit periodic continued fraction. Let K(1/bn ) be the continued fraction where bn = 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 if n = m? (d) Use the value set V found in (a) to derive upper bounds for the truncation error |f − Sn (w)| for suitably chosen w ∈ C.

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 cartesian oval in the Oval Sequence Theorem on page 243 with all Γn := Γ and ρn := ρ. Use this to prove that K(an /1) converges to a ﬁnite value and to ﬁnd truncation error bounds for suitably chosen approximants of K(an /1). 3. The logarithm of (1 + i). Use the Oval Sequence Theorem on page 243 with all Γn := Γ and ρn := ρ to ﬁnd bounds for |f − Sn (w)| for the continued fraction 2i/10 3i/10 K a1 i := 1i + i/21 + i/61 + 2i/6 1 + 1 + 1 +· · · n

√ when w = ( 1 + i − 1)/2.

4. Limit periodic continued fraction. We want to improve the speed of convergence of the continued fraction

K a1

n

=

6 + (0.9) 6 + (0.9)2 6 + (0.9)3 6 + (0.9)4 . 1 1 1 1 + + + +· · ·

(a) Prove that K(an /1) converges to a ﬁnite value f . (b) Suggest a value wn = w for all n such that lim

n→∞

f − Sn (w) = 0. f − Sn (0)

Does the sequence {Sn (w)} have an oscillating character which can be used to obtain upper bounds for the truncation error |f − Sn (w)|? (c) Suggest values for wn such that lim

n→∞

f − Sn (wn ) =0 f − Sn (w)

where w is the value from a). Does {Sn (wn )} have such an oscillating character? 5. The case an → ∞. Suggest expressions for wn such that the approximants Sn (wn ) of 12 5 · 22 32 5 · 42 52 5 · 62 72 an = 1 1 + 1 + 1 + 1 + 1 + 1 + 1 +· · · (hopefully) converge faster to the value f of K(an /1) than Sn (0). Compute the ﬁrst 6 approximants of Sn (0) and Sn (wn ) and use the oscillating character to determine an error bound for S6 (0). Find an upper bound for |f − Sn (wn )|.

K

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 than 0.05.

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 . (m)

Use the improvement machine on page 227 to derive approximants Sn (wn ) for K(an /1) such that (m)

f − Sn (wn ) =0 lim n → ∞ f − S (w (m−1) ) n n

for m = 1, 2, 3, . . . .

8. ♠ The Oval Theorem. Set all Γn := Γ and ρn := ρ in the oval sequence theorem on page 243, where Re Γ > − 12 and ρ < |1 + Γ|. (a) Prove that

$

E := a ∈ C;

% |a|ρ a(1 + Γ)

− Γ + ≤ρ 2 2 2 2 |1 + Γ| − ρ |1 + Γ| − ρ

is the simple element set for continued fractions K(an /1) corresponding to the simple value set V := B(Γ, ρ). (b) Prove that every continued fraction K(an /1) from E converges to some value in V . (Hint: the Parabola Theorem on page 151 or the lemma on page 113 may be of help.) (c) Prove that |f − Sn (wn )| ≤ 2ρ

|Γ| + ρ M n−1 |1 + Γ| − ρ

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. Let E be as in Problem 8 with ρ := Re Γ + 12 > 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 Re Γ > 0. (d) Prove that E is contained in the parabolic region Eα from the Parabola Theorem 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 is kept ﬁxed and Γ → ∞.

Appendix A

Some continued fraction expansions This is a catalogue of some of the known continued fraction expansions. The list is in no way complete. Still it can be useful, both to ﬁnd a continued fraction expansion of some given function and to “sum” a given continued fraction. We have not attempted to ﬁnd the origin of each result. The references we give are therefore just pointing to books or papers where the expansion also can be found.

A.1 A.1.1

Introduction 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 parameters (· · · ) in the set D. In the literature the set D is often far too restrictive, if it is given at all. We have determined a (possibly larger) set Dc where the continued fraction converges. This is done by methods presented in this book. However, it may well happen that the equality (1.1.1) fails in a subset of Dc , even if f (· · · ) is interpreted as an analytic continuation of the expression in question. In some cases we therefore give expressions for sets Dc and Df such that K(an (· · · )/bn (· · · )) converges in Dc and the equality holds in Df ⊆ Dc . What happens outside these sets is not checked. The identities normally holds also at points where an (· · · ) = 0 unless otherwise stated. The elements an (· · · ) and bn (· · · ) of almost all continued fractions in this appendix are polynomials in the parameters. The classical approximants are then rational functions of the parameters, and can therefore not converge to multivalued functions. If the left hand side 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

266

Appendix A: Some continued fraction expansions

stated. The principal part is often written with a capital ﬁrst letter, such as Ln z, Arctan z etc.

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 be derived from one another by simple transformations. We have for instance f = b0 + K(an /bn ) ⇐⇒ c ·

c a1 a2 1 = ; f b0 + b1 + 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 2 a1 a2 =1− . f +1 1 + b0 + b1 + b2 +· · ·

(1.2.2)

Another simple transformation is maybe most easily described for S-fractions. Assume that f (z) = b0 + K(an z/1). Then f (z −1 ) = b0 + K(an z −1 /1). Equivalence transformations lead to ( ) a1 a2 a3 a4 a5 f z1 = b0 + z + 1 + z + 1 + z +· · · a1 /z a2 a3 a4 a5 = b0 + 1 + z + 1 + z + 1 +· · · a1 /ξ a2 a3 a4 a5 , = b0 + ξ + ξ + ξ + ξ + ξ +· · · where ξ 2 = 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 +

a1 z 2 a 2 z 2 a 3 z 2 . 1 + 1 + 1 +· · ·

(1.2.3a)

f (iz) = b0 −

a1 z 2 a2 z 2 a3 z 2 . 1 − 1 − 1 −· · ·

(1.2.4b)

Then

Of course, every time we have a continued fraction expansion f = b0 + K(an /bn ) with all an , bn = 0, we can take its even or odd part and obtain a “new” continued fraction converging to the same value f . Some of these variations will be listed, in particular if they turn out to be nice and simple.

A.2

Elementary functions

A.2.1

Mathematical constants π =3+

1 1 1 1 1 1 1 1 1 1 1 1 , 7 + 15 + 1 + 292 + 1 + 1 + 1 + 2 + 1 + 3 + 1 + 14 +· · ·

(2.1.1)

A.2.1 Mathematical constants

267

([JoTh80], p 23). This is the regular continued fraction expansion of π. 4 12 22 32 42 , 1 + 3 + 5 + 7 + 9 +· · · ([JoTh80], p 25), (see also (3.6.1)). π=

1 1·2 2·3 3·4 π =1+ , 2 1 + 1 + 1 + 1 +· · ·

([Khru06a]).

(2.1.2)

(2.1.3)

For the Riemann zeta function we have 1 ζ(2) 2

=

π2 1 14 24 34 = , 12 1 + 3 + 5 + 7 +· · ·

(2.1.4)

([Bern89], p 150). ζ(2) =

1 12 1 · 2 22 2 · 3 32 3 · 4 42 π2 =1+ , 6 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +· · ·

(2.1.5)

([Bern89], p 153). Apery’s constant: ζ(3) = 1 +

1 13 13 23 23 33 33 43 43 , 4 + 1 + 12 + 1 + 20 + 1 + 28 + 1 + 36 +· · ·

(2.1.6)

([Bern89], p 155), (see also (4.7.37)). Euler’s number:

1 1 1 1 1 1 1 1 , 1 − 1 + 2 − 3 + 2 − 5 + 2 − 7 +· · · ([JoTh80], p 25), (see also (3.2.1)). e=

e=2+

1 1 1 1 1 1 1 1 , 1 + 2 + 1 + 1 + 4 + 1 + 1 + 6 +· · ·

(2.1.7)

(2.1.8)

([JoTh80], p 23). 2 1 1 1 1 , 1 + 6 + 10 + 14 + 18 +· · · ([Khov63], p 114). (See also (3.2.2)). e=1+

e=2+

2 3 4 5 , 2 + 3 + 4 + 5 +· · ·

(2.1.9)

(2.1.10)

([Perr57], p 57). 1 2 1 1 1 1 , 1 − 3 + 6 + 10 + 14 + 18 +· · · ([Khov63], p 114). (See also (3.2.2) for z = −1). e=

√

e=1+

1 1 1 1 1 1 1 1 1 , 1 + 1 + 5 + 1 + 1 + 9 + 1 + 1 + 13 +· · ·

(2.1.11)

(2.1.12)

268

Appendix A: Some continued fraction expansions

([Euler37]). √ 3

1 1 1 1 1 1 1 1 1 1 , 2 + 1 + 1 + 8 + 1 + 1 + 14 + 1 + 1 + 20 +· · · ([Euler37]). (See also (2.2.4).) e=1+

coth

1 2

=

(2.1.13)

e+1 1 1 1 1 =2+ , e−1 6 + 10 + 14 + 18 +· · ·

(2.1.14)

√ 5−1 1 1 1 = , 2 1 + 1 + 1 +· · ·

(2.1.15)

([Khru06b]). The golden ratio:

([JoTh80], p 23). Catalan’s constant: G :=

∞

k=0 (−1)

k

/(2k + 1)2 :

12 22 22 42 42 62 62 , 3 + 1 + 3 + 1 + 3 + 1 + 3 +· · · ([Bern89], p 151). (See also (4.7.30) with z := 2.) 2G = 2 −

2G = 1 +

1 12 1 · 2 22 2 · 3 32 3 · 4 , 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 +· · ·

(2.1.16)

(2.1.17)

([Bern89], p 153). (See also (4.7.32) with z := 12 .)

A.2.2

The exponential function

1 z z z z z z z 1 − 1 + 2 − 3 + 2 − 5 + 2 − 7 +· · · 1 z 1z 1z 2z 2z 3z 3z 4z 4z = ; z ∈ C, 1 − 1 + 2 − 3 + 4 − 5 + 6 − 7 + 8 − 9 +· · · ([JoTh80], p 207). (See also (4.1.4).) The odd part of this continued fraction is ez = 1 F1 (1; 1; z) =

ez = 1 +

2z z2 z 2 z2 z2 ; 2 − z + 6 + 10 + 14 + 18 +· · ·

z ∈ C,

(2.2.1)

([Khov63], p 114). ez = 1 +

z 1z 2z 3z ; z ∈ C, 1 − z + 2 − z + 3 − z + 4 − z +· · ·

(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 z 1z 2z 3z ; z ∈ C, 1 − 1 + z − 2 + z − 3 + z − 4 + z +· · ·

(2.2.3)

A.2.3 The general binomial function

269

([Khov63], p 113). e1/z = 1 +

1 1 1 1 1 1 1 1 z ∈ C, z − 1 + 1 + 1 + 3z − 1 + 1 +· · · 1 + 5z − 1 + 1 +· · ·

(2.2.4)

([Khru06b]). Lambert’s continued fraction z z2 z2 z 2 ez − e−z = ; z ∈ C, z −z e +e 1 + 3 + 5 + 7 +· · ·

(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 mentioned in 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 = 2 F1 (−α, 1; 1; −z) =

1 αz (1 + α)z (1 − α)z (2 + α)z (2 − α)z (3 + α)z (3 − α)z 1− 1 + 2 3 2 5 2 7 + + + + + +· · ·

(2.3.1)

for α ∈ C and | arg(z + 1)| < π, ([JoTh80], p 202). (See also (3.1.6).) The odd part of this continued fraction is (1 + z)α = (2.3.2) 2αz (12 − α2 )z 2 (22 − α2 )z 2 (32 − α2 )z 2 2 + (1 − α)z − 3(z + 2) − 5(z + 2) − 7(z + 2) −· · · for α ∈ C and | arg(z + 1)| < π, ([Khov63], p 105). (2.3.2) is also the odd part of 1+

(1 + z)α = (1 − α)z (1 + α)z (2 − α)z (2 + α)z (3 − α)z 1 αz 1 − 1(1 + z) − 2 2 2 − 3(1 + z) − − 5(1 + z) − −· · · for α ∈ C and | arg(z + 1)| < π, ([Khov63], p 101). (1 + z)α =

1 1(1 + α)z(1 + z) 2(2 + α)z(1 + z) αz 1 − 1 + (1 + α)z − 2 + (3 + α)z − 3 + (5 + α)z −· · ·

([Khov63], p 101). C2 ; Re(z) > − 12 }. (1 + z)α =

(2.3.3)

(2.3.4)

Dc := {(α, z) ∈ C2 ; Re(z) = − 12 , z = −1}, Df := {(α, z) ∈

1(1 − α)z 2(2 − α)z 3(3 − α)z 1 αz , 1 − 1 + αz + 2 − (1 − α)z + 3 − (2 − α)z + 4 − (3 − α)z +· · ·

(2.3.5)

270

Appendix A: Some continued fraction expansions

([Khov63], p 102). Dc := {(α, z) ∈ C2 ; |z| = 1}, Dc := {(α, z) ∈ C2 ; |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 expansion for α α 2 z+1 = 1+ (2.3.6) z−1 z−1 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

α =1+

2α α2 − 12 α2 − 22 α2 − 32 , z − α + 3z + 5z + 7z +· · ·

(2.3.7)

for α ∈ C and z ∈ C \ [−1, 1], ([Perr57], p 153). z − 1 (1z − 1)(z − 1) (1z + 1)(z − 1) 1z + 2 3z + + (2z − 1)(z − 1) (2z + 1)(z − 1) (3z − 1)(z − 1) (3z + 1)(z − 1) 2 5z 2 7z + + + +· · ·

z 1/z = 1 +

(2.3.8)

for | arg z| < π, ([Khov63], p 109). The even part of (2.3.8) is z 1/z = 1+

2(z − 1) (12 z 2 − 1)(z − 1)2 z2 + 1 − 3z(z + 1) −

(2.3.9)

(22 z 2 − 1)(z − 1)2 (32 z 2 − 1)(z − 1)2 5z(z + 1) 7z(z + 1) − −· · · for | arg z| < π, ([Khov63], p 110).

1 + az 1 + bz

α =1+

2α(a − b)z (a − b)2 (12 − α2 )z 2 2 + (a + b − α(a − b))z − 3(2 + (a + b)z) −

(2.3.10)

(a − b)2 (22 − α2 )z 2 (a − b)2 (32 − α2 )z 2 5(2 + (a + b)z) − 7(2 + (a + b)z) −· · · ∈ C \ [−1, 1], ([Perr57], p 264). (Note that for α ∈ C and 2+(a+b)z (a−b)z x := 1 + az, y := 1 + bz.)

A.2.4

2+(a+b)z (a−b)z

=

x+y x−y

for

The natural logarithm

1

dt 1 + zt z 1z 1z 2z 2z 3z 3z 4z 4z = 1 + 2 + 3 + 2 + 5 + 2 + 7 + 2 + 9 +· · · z 12 z 12 z 22 z 22 z 32 z 32 z 42 z 42 z = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 +· · ·

Ln(1 + z) = z 2 F1 (1, 1; 2; −z) = z

0

for | arg(1 + z)| < π, ([JoTh80], p 203). (See also (3.1.6).)

(2.4.1)

A.2.4 The natural logarithm

271

z 1 z 1/2 z 1 z 2/3 z (2.4.2) 1 + z − 1 + 1 + z − 1 + 1 + z − 1 + 1 + z − 1 + 1 + z −· · · for | arg(1 + z)| < π, ([JoTh80], p 319). Here the continued fraction has the form K(an (z)/bn (z)) where all a2n (z) = −z, a4n−1 (z) = 1 and a4n+1 (z) = n/(n + 1). (2.4.1) is the even part of (2.4.2). The odd part of (2.4.2) can be written Ln(1 + z) =

" ! z z 2z z 3z 2z 4z 3z 5z 4z Ln(1 + z) = 1+ 1+z 2 + 3 + 2 + 5 + 2 + 7 + 2 + 9 + 2 +· · · " ! z 1 · 2z 1 · 2z 2 · 3z 2 · 3z 3 · 4z 3 · 4z z 1+ = 1+z 2 + 3 + 4 + 5 + 6 + 7 + 8 +· · ·

(2.4.3)

for | arg(1 + z)| < π. The even part of (2.4.1) is 12 z 2 22 z 2 32 z 2 2z 1(2 + z) − 3(2 + z) − 5(2 + z) − 7(2 + z) −· · · for | arg(1 + z)| < π, ([Khov63], p 111). Ln(1 + z) =

12 z 22 z 32 z 42 z 52 z z , 1 + 2 − z + 3 − 2z + 4 − 3z + 5 − 4z + 6 − 5z +· · · ([Khov63], p 111). Dc := {z ∈ C; |z| = 1}, Df := {z ∈ C; |z| < 1}. Ln(1 + z) =

The connection Ln(1 + z) = −Ln

1 1+z

= −Ln 1 −

z 1+z

(2.4.4)

(2.4.5)

(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 z 1z 1z 2z 2z 3z 3z 1 + z − 2 − 3(1 + z) − 2 − 5(1 + z) − 2 − 7(1 + z) −· · ·

Ln(1 + z) =

(2.4.7)

for | arg(1 + z)| < π, ([Khov63], p 110), and from (2.4.5) z 12 z(1 + z) 22 z(1 + z) 32 z(1 + z) , 1 + z − 2 + 3z − 3 + 5z − 4 + 7z −· · · ([Khov63], p 111). Dc := {z ∈ C; Re(z) = − 12 }, Df := {z ∈ C; Re(z) > − 12 }. Ln(1 + z) =

Ln

1 + z 1−z

= 2z

2 1 3 2 F1 ( 2 , 1; 2 ; z )

2z = Ln 1 + =z 1−z

1 −1

dt 1 + tz

2z 12 z 2 22 z 2 32 z 2 42 z 2 = 1 − 3 − 5 − 7 − 9 −· · · for | arg(1 − z 2 )| < π, ([JoTh80], p 203). (See also (3.1.6).) From this we also get z + 1

Ln

z−1

= Ln

1 + 1/z 1 − 1/z

=

2 12 22 32 42 z − 3z − 5z − 7z − 9z −· · ·

(2.4.8)

(2.4.9)

(2.4.10)

for z ∈ C \ [−1, 1], ([Perr57], p 155). Of course, also other continued fraction expansions for Ln(1 + z) can be used to derive expressions for Ln((1 + z)/(1 − z)).

272

Appendix A: Some continued fraction expansions

A.2.5

Trigonometric and hyperbolic functions

2 sin z z z2 z2 z2 z2 0 F1 (3/2; −z /4) =z = ; z∈C 2 cos z 1 − 3 − 5 − 7 − 9 −· · · 0 F1 (1/2; −z /4) ([JoTh80], p 211), (See also (3.1.1).) The odd part of (2.5.1) is

tan z :=

1 · 9z 4 5z 3 2 1 · 3 · 5 − 6z − 5 · 7 · 9 − 14z 2 − 5 · 13z 4 9 · 17z 4 ; z ∈ C, 9 · 11 · 13 − 22z 2 − 13 · 15 · 17 − 30z 2 −· · ·

(2.5.1)

tan z = z +

(2.5.2)

Another type of expansion is tan

z 1 2 − z 2 3 2 − z 2 52 − z 2 72 − z 2 zπ = ; z ∈ C, 4 1 + 2 + 2 + 2 + 2 +· · ·

(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

y−1 (1 + i tan z)α − (1 − i tan z)α = −i , (1 + i tan z)α + (1 − i tan z)α y+1

(2.5.4)

where y := ((1 + i tan z)/(1 − i tan z))α can be expanded according to (2.3.7). Combined with (1.2.2) we get

tan αz =

α tan z (α2 − 12 ) tan2 z (α2 − 22 ) tan2 z (α2 − 32 ) tan2 z , 1 − 3 5 7 − − −· · ·

(2.5.5)

([Khov63], p 108). Dc := {(α, z) ∈ C2 ; Re(cos z) = 0}, Df := {(α, z) ∈ C2 ; |Re(z)| < π/2}. coth

1 z

=1+

1 1 1 1 ; z ∈ C, 3z + 5z + 7z + 9z +· · ·

(2.5.6)

([Khru06b]). πz z 2 12 (z 2 + 12 ) 22 (z 2 + 22 ) 32 (z 2 + 32 ) πz coth =1+ 2 2 1+ 3 5 7 + + +· · · for all z ∈ C, ([ABJL92], entry 44).

(2.5.7)

ab (a2 + 12 )(b2 + 12 ) (a2 + 22 )(b2 + 22 ) a tanh(πb/2) − b tanh(πa/2) = (2.5.8) a tanh(πa/2) − b tanh(πb/2) 1+ 3 5 + +· · · for all a, b ∈ C, ([ABJL92], entry 47). sinh(πz) − sin(πz) 2z 2 4z 4 + 14 4z 4 + 24 4z 4 + 34 = sinh(πz) + sin(πz) 1 + 3 5 7 + + +· · · for all z ∈ C, ([ABJL92], entry 49).

(2.5.9)

A.2.6 Inverse trigonometric and hyperbolic functions

A.2.6

273

Inverse trigonometric and hyperbolic functions i 1 + iz Arctan z = z 2 F1 ( 12 , 1; 32 ; −z 2 ) = − Ln 2 1 − iz 2 2 2 2 2 2 2 2 z 1 z 2 z 3 z 4 z = 1 + 3 + 5 + 7 + 9 +· · ·

(2.6.1)

for | arg(1 + z 2 )| < π; i.e. z ∈ C \ i((−∞, −1] ∪ [1, ∞)), ([JoTh80], p 202). (See also (3.1.6).) This continued fraction can also be written z 3 32 z 2 22 z 2 52 z 2 42 z 2 72 z 2 62 z 2 3 + 5 + 7 + 9 + 11 + 13 + 15 +· · · for z ∈ C \ i((−∞, −1] ∪ [1, ∞)), ([Khov63], p 117). Arctan z = z −

Arctan z =

(2.6.2)

1 · 2z 2 1 · 2z 2 3 · 4z 2 3 · 4z 2 5 · 6z 2 z 1(1 + z 2 ) − 3 − 5(1 + z 2 ) − 7 − 9(1 + z 2 ) − 11 −· · ·

(2.6.3)

for z ∈ C \ i((−∞, −1] ∪ [1, ∞)), ([Khov63], p 121). (This follows from (2.6.6) with z replaced by z(1 + z 2 )−1/2 .) Since Artanh z = iArctan(−iz), we also get continued fraction expansions for Artanh z from (2.6.1)– (2.6.3). Also expressions for z Arcsin z = Arctan √ , 1 − z2

Arccos z = Arctan

√1 − z 2 z

can be obtained. For instance, from (2.6.1) we get z Arcsin z 12 z 2 22 z 2 32 z 2 42 z 2 √ = 1(1 − z 2 ) + 3 + 5(1 − z 2 ) + 7 + 9(1 − z 2 ) +· · · 1 − z2

(2.6.4)

for | arg(1 − z 2 )| < π, ([Khov63], p 118) and Arccos z 1 12 (1 − z 2 ) 22 (1 − z 2 ) 32 (1 − z 2 ) √ , (2.6.5) = z+ 3z 5z 7z + + +· · · 1 − z2 ([Khov63], p 119). Here Dc := {z ∈ C; Re z = 0} and Df := {z ∈ C; Re z > 0}. We also have 2 1 1 3 Arcsin z 2 F1 ( 2 , 2 ; 2 ; z ) √ =z 1 1 1 2 2 1−z 2 F1 ( 2 , − 2 ; 2 ; z )

z 1 · 2z 2 1 · 2z 2 3 · 4z 2 3 · 4z 2 5 · 6z 2 5 · 6z 2 = 1 − 3 − 5 − 7 − 9 − 11 − 13 −· · · for | arg(1 − z 2 )| < π, ([JoTh80], p 203), and thus, since Arccos z = Arcsin 0 ≤ z ≤ π2

(2.6.6) √

1 − z 2 for

z 1 · 2(1 − z 2 ) 1 · 2(1 − z 2 ) 3 · 4(1 − z 2 ) 3 · 4(1 − z 2 ) Arccos z √ , = 1− 3 5 7 9 − − − −· · · 1 − z2

(2.6.7)

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)/ z 2 − 1 = (Arccos z)/ 1 − z 2 for 0 ≤ z ≤ π2 . A neat formula can be obtained from (3.2.6) in the following way

iz + 1 iz − 1

iα

iz + 1 = exp(2αArctan(1/z)) = exp iαLn iz − 1 2α α2 + 12 α2 + 22 α2 + 32 =1+ z − α + 3z + 5z + 7z +· · ·

(2.6.8)

for | arg(1 + 1/z 2 )| < π, i.e. z ∈ i[−1, 1], ([Wall57], p 346). z 2z 2 2(1 + z 2 ) 4z 2 4(1 + z 2 ) Arsinh z 2 3 = z F (1, 1; ; −z ) = 2 1 2 1+ 1 + 1 1 (1 + z 2 )1/2 + 1 + +· · ·

(2.6.9)

for z ∈ R, ([ABJL92], entry 37).

Arctan z = z 2 F1 ( 12 , 1; 32 ; −z 2 ) =

z 1z 2 2(1 + z 2 ) 3z 2 4(1 + z 2 ) 1+ 1 + 1 1 + 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]). The ﬁrst one is taken from ([Perr57], p 279). It converges for all pairs (a, z) ∈ C2 , but the equality 0 = −a − z +

z 2z 3z 1 − a − z + 2 − a − z + 3 − a − z +· · ·

(2.7.1)

given by Perron is only claimed for z = 0 and a ∈ N ∪ {0}. In our next continued fraction the constant sequence {1} is always a tail sequence, so 1=

z+1 z+2 z+3 z+4 ; z + z + 1 + z + 2 + 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 sequence of 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 (z + a)2 − a2 (z + 2a)2 − a2 (z + 3a)2 − a2 a + a a 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.)

A.3.1 Hypergeometric functions

a=

275

ab (a + d)(b + d) (a + 2d)(b + 2d) , a + b + d − a + b + 3d − 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) ∈ C3 ; Re((a − b)/d) < 0 or a = b}. (See also (3.1.6) with z = 1, a replaced by (a + d)/2d, b replaced by a/2d and c replaced by (a + b + d)/2d.) For b = a replaced by a + 1 and d := 1, (3.7.4) can be transformed into a = 2a + 1 −

(a + 1)2 (a + 2)2 (a + 3)2 ; 2a + 3 − 2a + 5 − 2a + 7 −· · ·

a ∈ C,

(2.7.5)

([Perr57], p 105). az =

abz (a + 1)(b + 1)z (a + 2)(b + 2)z , b − (a + 1)z + b + 1 − (a + 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 = c replaced by b − 1.) z+a z+a+1 z + 2a z + 3a = , z+1 z − 1 + z + a − 1 + z + 2a − 1 +· · ·

(2.7.7)

([Bern89], p 115). Dc := {(a, z) ∈ C2 ; a = 0 and z/a = 0, −1, −2, . . .} ∪ {(a, z) ∈ C2 ; a = 0 and |z| = 1}, Df := {(a, z) ∈ Dc ; if a = 0, then |z| > 1}. (See also (3.1.5) with z replaced by 1/a, a replaced by z/a + 1 and c replaced by z/a.) If we instead let z = 1 and replace a by z − 1, c by z − 3 in (3.1.5) we get z2 + z + 1 z z+1 z+2 z+3 z+4 = , z2 − z + 1 z − 3 + z − 2 + z − 1 + z + z + 1 +· · ·

(2.7.8)

([Bern89], p 118). Dc := C, Df := {z ∈ C; z = 0, −1, −2, . . .}. For z := 1, a := z − 1 and c := z − 4 in (3.1.5) we get z 3 + 2z + 1 z z+1 z+2 z+3 z+4 = , (z − 1)3 + 2(z − 1) + 1 z − 4 + z − 3 + z − 2 + z − 1 + z +· · ·

(2.7.9)

([Bern89], p 118). Dc := C, Df := {z ∈ C; z = 0, −1, −2, . . .}.

A.3 A.3.1

Hypergeometric functions General expressions c

0 F1 (c; z) 0 F1 (c

([JoTh80], p 210).

+ 1; z)

=c+

z z z ; c + 1 + c + 2 + c + 3 +· · ·

(c, z) ∈ C2 ,

(3.1.1)

276

Appendix A: Some continued fraction expansions

az (b + 1)z (a + 1)z (b + 2)z (a + 2)z 2 F0 (a, b; z) =1− F (a, b + 1; z) 1 − 1 1 1 1 − − − −· · · 2 0

(3.1.2)

for (a, b, z) ∈ C3 with | arg(−z)| < π, ([JoTh80], p 213). The even part of this one is

2 F0 (a, b; z) 2 F0 (a, b

+ 1; z)

=1+

(a + 1)(b + 1)z 2 (a + 2)(b + 2)z 2 az (b + 1)z − 1 − (a + b + 3)z − 1 − (a + b + 5)z − 1 −· · ·

(3.1.3)

for (a, b, z) ∈ C3 with | arg(−z)| < π. (c − a)z (a + 1)z =c− + 1; c + 1; z) c+1 + c+2 (c − a + 1)z (a + 2)z (c − a + 2)z ; (a, c, z) ∈ C3 , c+3 c+5 + c+4 − +· · · − c·

1 F1 (a; c; z)

1 F1 (a

(3.1.4)

([JoTh80], p 206). 1 F1 (a

+ 1; c + 1; z)

1 F1 (a; c; z)

=

(a + 1)z (a + 2)z (a + 3)z c c − z + c + 1 − z + c + 2 − z + c + 3 − z +· · ·

(3.1.5)

for (a, c, z) ∈ C3 , ([JoTh80], p 278). a(c − b)z (b + 1)(c − a + 1)z =c− + 1; c + 1; z) c+1 − c+2 (a + 1)(c − b + 1)z (b + 2)(c − a + 2)z (a + 2)(c − b + 2)z c+3 c+4 c+5 − − −··· − c·

2 F1 (a, b; c; z)

2 F1 (a, b

(3.1.6)

for (a, b, c, z) ∈ C4 with | arg(1 − z)| < π, ([JoTh80], p 199). The N¨ orlund fraction has the form c·

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

+ 1, b + 1; c + 1; z)

= c − (a + b + 1)z +

(a + 1)(b + 1)(z − z 2 ) c + 1 − (a + b + 3)z +

(a + 2)(b + 2)(z − z 2 ) (a + 3)(b + 3)(z − z 2 ) c + 2 − (a + b + 5)z + c + 3 − (a + b + 7)z +· · ·

(3.1.7)

([LoWa92], p 304). Dc := {(a, b, c, z) ∈ C4 ; Re(z) = 12 }, Df := {(a, b, c, z) ∈ C4 ; Re(z) < 1 }. The Euler fraction has the form 2 c·

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

2 F1 (a, b

(3.1.8)

([LoWa92], p 308). Dc := {(a, b, c, z) ∈ C4 ; |z| = 1}, Dc := {(a, b, c, z) ∈ C4 ; |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 continued fraction expansions for 2 F0 (a, 1; z) and 2 F1 (a, 1; c + 1; z). Similarly, a := 0 in (3.1.3) or (3.1.4) gives continued fraction expansions for 1 F1 (1; c + 1; z). A diﬀerent expansion is

A.3.3 Special examples with 2 F0

277

Γ(1 − a)Γ(c + 1) (1 − z)c−a − Γ(c − a + 1) (−z)c 1(1 − c)(z − 1) 2(2 − c)(z − 1) c , 1 − c + (a − 1)z + 3 − c + (a − 2)z + 5 − c + (a − 3)z +· · ·

2 F1 (a, 1; c

+ 1; z) =

(3.1.9)

([Bern89], p 164). Dc := {(a, c, z) ∈ C3 ; |z − 1| = 1}, Df := {(a, c, z) ∈ C3 ; |z − 1| < 1 and c = 0, 1, 2, . . . }. From this follows after some computation, ([Bern89], p 165) that ez Γ(c + 1) c 1−c 1 2−c 2 3−c − zc z + 1 + z + 1 + z + 1 +· · · ez Γ(c + 1) c 1(1 − c) 2(2 − c) 3(3 − c) = − zc z + 1 − c − z + 3 − c − z + 5 − c − z + 7 − c −· · ·

1 F1 (1; c

+ 1; z) =

(3.1.10)

for | arg z| < π, ([Bern89], p 165) (the second continued fraction is the even part of the ﬁrst one).

A.3.2

Special examples with 0 F1

The Bessel function of the ﬁrst kind and order ν is Jν (z) :=

∞ z ν (−1)k (z/2)2k (z/2)ν z2 = 0 F1 (ν + 1; − 4 ), 2 k! Γ(ν + k + 1) Γ(ν + 1)

(3.2.1)

k=0

so that by (3.1.1) 2 Jν+1 (z) z 0 F1 (ν + 2; −z /4) = · Jν (z) 2(ν + 1) 0 F1 (ν + 1; −z 2 /4)

=

z z2 z2 z2 2(ν + 1) − 2(ν + 2) − 2(ν + 3) − 2(ν + 4) −· · ·

(3.2.2)

for z ∈ C, ν = −1, −2, −3, . . . , ([JoTh80], p 211).

A.3.3

Special examples with 2 F0

The connection (see for instance ([Wall48], p 352, p 355)) 2 F0 (a, b; −z)

∼

1 Γ(a)

∞ 0

1 e−t ta−1 dt = (1 + tz)b Γ(b)

∞ 0

e−t tb−1 dt (1 + tz)a

(3.3.1)

implies that (3.1.2) – (3.1.3) lead to continued fraction expansions for ratios of such integrals. In particular, the incomplete gamma function Γ(a, z) satisﬁes

∞ e−t ta−1 dt ∼ e−z z a−1 2 F0 (1 − a, 1; −1/z) , (3.3.2) Γ(a, z) := z

([EMOT53], p 266). Hence, by (3.1.2)

278

Appendix A: Some continued fraction expansions

e−z z a 1 − a 1 2 − a 2 3 − a 3 z + 1 + z + 1 + z + 1 + z +· · · e−z z a 1(1 − a) 2(2 − a) 3(3 − a) = 1 + z − a − 3 + z − a − 5 + z − a − 7 + z − a −· · ·

Γ(a, z) =

(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 integral in (3.3.2) is taken for real z. Then Γ(a, z) is the analytic continuation of this function to the given domain. The complementary error function erfc z satisﬁes

∞ 2 2 2 1 1 erfc z := √ e−t dt = √ Γ( 12 , z 2 ) ∼ √ e−z z −1 2 F0 ( 12 , 1; −1/z 2 ) π z π π

(3.3.4)

([EMOT53], p 266), which means that by (3.1.2) % $ 1 2 2 4 6 8 2 erfcz = √ e−z 2z + 2z + 2z + 2z + 2z +· · · π % 1·2 3·4 5·6 7·8 z 2 −z2 $ , = √ e 2 2 2 2 2 1 + 2z − 5 + 2z − 9 + 2z − 13 + 2z − 17 + 2z −· · · π

(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 even part of the ﬁrst one. If we integrate this complementary error function we get similar expressions:

∞ 2 2 in−1 erfc t dt (3.3.6) i−1 erfc z = √ e−z , i0 erfc z = erfc z , in erfc z = π z for n = 1, 2, 3, . . . . Therefore 2 n+1 n in−1 erfc z 2 F0 ( 2 , 2 ; −1/z ) = 2z n+1 n 2 in erfc z 2 F0 ( 2 , 2 + 1; −1/z )

= 2z +

(3.3.7)

2(n + 1) 2(n + 2) 2(n + 3) 2z + 2z + 2z +· · ·

([JoTh80], p 219). Dc := {(n, z) ∈ C2 ; Re z = 0}, Df := {(n, z) ∈ C2 ; Re z > 0}. For the exponential integral

−Ei(−z) :=

∞ z

e−t e−z 1 dt ∼ 2 F0 (1, 1; − z ) , t z

(3.3.8)

([EMOT53], p 267), we get by (3.1.2) and its even part e−z 1 1 2 2 3 3 4 z + 1 + z + 1 + z + 1 + z + 1 +· · · e−z 12 22 32 42 =− 1 + z − 3 + z − 5 + z − 7 + z − 9 + z −· · · for | arg z| < π, ([Khov63], p 145). Ei(−z) = −

(3.3.9)

A.3.2 Special examples with 1 F1

279

Similarly, for the logarithmic integral

z z 1 1 2 2 dt = Ei(Ln z) = li z := Ln t Ln z 1 Ln z 1 Ln z −· · · − − − − 0 =−

z 12 22 32 42 1 − Ln z − 3 − Ln z − 5 − Ln z − 7 − Ln z − 9 − Ln z −· · ·

(3.3.10)

for | arg(−Ln z)| < π. The plasma dispersion function is

∞ −t2 √ 2 1 e dt = i π e−z erfc(−iz) P (z) := √ π −∞ t − z 1·2 3·4 5·6 7·8 2z , = 1 − 2z 2 − 5 − 2z 2 − 9 − 2z 2 − 13 − 2z 2 − 17 − 2z 2 −· · ·

(3.3.11)

([JoTh80], p 219). Dc := {z ∈ C; Im(z) = 0}, Df := {z ∈ C; Im(z) > 0}. 1∞

2 ta e−bt−t /2 dt 1 ∞0 ta−1 e−bt−t2 /2 dt 0

( a a+1 ) 1 a a+1 a+2 a+3 a 2 F0 2 , 2 ; − b2 ( ) = = · , (3.3.12) 1 b 2 F0 a2 , a−1 b + b + b + b +· · · ; − 2 b2

([Perr57], p 297). Dc := {(a, b) ∈ C2 ; Re b = 0}, Df := {(a, b) ∈ C2 ; Re b > 0}.

A.3.4

Special examples with 1 F1

From ([EMOT53], p 255) it follows that 1 F1 (a; c; z)

=

Γ(c) Γ(a)Γ(c − a)

1 0

etz ta−1 (1 − t)c−a−1 dt

(3.4.1)

for Re(c) > 0, Re(a) > 0. Hence (3.1.4), (3.1.5) and (3.1.10) lead to continued fraction expansions of ratios of such integrals. The error function is given by

z 2 2 2 e−t dt = √ z 1 F1 ( 12 ; 32 ; −z 2 ), ([EMOT53], p 266) erf (z) := √ π 0 π 2 2 = √ ze−z 1 F1 (1; 32 ; z 2 ), ([JoTh80], p 282). π Hence,

(3.4.2)

2

erf (z) =

2 e−z z 2z 2 4z 2 6z 2 8z 2 √ π 1 − 3 + 5 − 7 + 9 −· · · 2

z 4z 2 8z 2 12z 2 2 e−z = √ 2 2 2 π 1 − 2z + 3 − 2z + 5 − 2z + 7 − 2z 2 +· · · for z ∈ C, ([JoTh80], p 208 and 282). The error function is related to Dawson’s integral √

z 2 i π erf(−iz), ([JoTh80], p 208) et dt = 2 0

(3.4.3)

(3.4.4)

280

Appendix A: Some continued fraction expansions

and to the Fresnel integrals

z π C(z) := cos t2 dt, 2 0 by

C(z) + iS(z) =

z

e

it2 π/2

z

sin

S(z) := 0

,

−2 iπ

π 2

√−iπ/2 · z

dt = 0 √ 0 π 1+i erf (1 − i)z . = 2 2

t2 dt

(3.4.5)

2

e−u du (3.4.6)

The incomplete gamma function

z z a −z e−t ta−1 dt = e 1 F1 (1; a + 1; z) γ(a, z) := a 0 z a e−z (a + 1)z (a + 2)z az 1z 2z a − a + 1 + a + 2 − a + 3 + a + 4 − a + 5 +· · · (1 + a)z (2 + a)z (3 + a)z az z a e−z = a − 1 + a + z − 2 + 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+1 e−iρ CL (η)1 F1 (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

L(L + 2)((L + 1)2 + η 2 ) (L + 1)(L2 + η 2 )1/2 FL (η, ρ) = FL−1 (η, ρ) (2L + 1)(η + L(L + 1)/ρ) − (2L + 3)(η + (L + 1)(L + 2)/ρ) − (L + 1)(L + 3)((L + 2)2 + η 2 ) , (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})×C2 ; ρ = 0}. It is well known that ∞ k=0

zk e−z (−z)k = = e−z 1 F1 (1; a + 1; z) , k! (a + k) (a)k+1 a ∞

(3.4.10)

k=0

([Bern89], p 166). This means for instance that ∞ k=0

e−z (−z)k 1(1 − a) 2(2 − a) Γ(a) = a − k! (a + k) z z + 1 − a − z + 3 − a − z + 5 − a −· · ·

(3.4.11)

for (a, z) ∈ C2 with | arg z| < π, and ∞ k=0

zk = 1 · 3 · · · (2k + 1)

,

1 1·2 3·4 5·6 π z/2 e − 2z z + 1 − z + 5 − z + 9 − z + 13 −· · ·

(3.4.12)

A.3.5 Special examples with 2 F1

281

for | arg z| < π, ([Bern89], p 166), and ze−z

2

∞ k=0

(2z 2 )k = 1 · 3 · · · (2k + 1)

√

z

2

e−t dt

0

(3.4.13)

2

π e−z 1·2 3·4 5·6 − 2 , = 2 2z + 1 − 2z 2 + 5 − 2z 2 + 9 − 2z 2 + 13 −· · · ([Bern89], p 166). Dc := {z ∈ C; Re z = 0}, Df := {z ∈ C; Re z > 0}.

A.3.5

Special examples with 2 F1

z 0

tp dt z p+1 = 2 F1 q 1+t q

p+1 p+1 , 1; 1 + ; −z q q q

z p+1 (0q + p + 1)2 z q (1q)2 z q 0q + p + 1 + 1q + p + 1 + 2q + p + 1 + (1q + p + 1)2 z q (2q)2 z q 3q + p + 1 + 4q + p + 1 +· · · for p, q > 0 with | arg(1 + z q )| < π, ([Khov63], p 126). =

Incomplete beta functions are given by

x xp Bx (p, q) := tp−1 (1 − t)q−1 dt = 2 F1 (p, 1 − q; p + 1; x) p 0

(3.5.1)

(3.5.2)

for p > 0, q > 0 and 0 ≤ x ≤ 1, ([EMOT53], p 87). Hence, by (3.1.6) and (3.1.8) px 2 F1 (p + 1, 1 − q; p + 2; x) Bx (p + 1, q) = Bx (p, q) p + 1 2 F1 (p, 1 − q; p + 1; x) 1(1 − q)x (p + 1)(p + q + 1)x px = p + 1− p + 2 − p+3 − 2(2 − q)x (p + 2)(p + q + 2)x ; p+4 − p+5 −· · · for p > 0, q > 0, ([JoTh80], p 217), and

(3.5.3) | arg(1 − x)| < π

(p + q + 1)(p + 1)x px Bx (p + 1, q) = Bx (p, q) p + 1 + (p + q)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 given by

1 Γ(α + m + 1) π Pαm (z) := (z + (z 2 − 1)1/2 cos t)α cos mt dt π Γ(α + 1) 0 (3.5.5) m/2 1 z+1 1−z . = 2 F1 −α, α + 1; 1 − m; Γ(1 − m) z − 1 2

282

Appendix A: Some continued fraction expansions

From ([Gaut67], formula (6.1) on p 55) it follows that Pαm (z) (m + 1 + α)(m − α) (m + α)(m − α − 1) = 2mz 2(m + 1)z − − Pαm−1 (z) − 2 − 2 (z − 1)1/2 (z − 1)1/2 (m + 2 + α)(m + 1 − α) 2(m + 2)z − ··· − 2 (z − 1)1/2 √ (m + α)(m − α − 1) z 2 − 1 (m + 1 + α)(m − α)(z 2 − 1) ∼ − 2mz 2(m + 1)z − − (m + 2 + α)(m + 1 − α)(z 2 − 1) (m + 3 + α)(m + 2 − α)(z 2 − 1) 2(m + 2)z 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 given by

∞ Γ(α + 1) cosh mt m Qm (z) := (−1) dt α Γ(α − m + 1) 0 (z + (z 2 − 1)1/2 cosh t)α+1 √ imπ (3.5.7) ( α+m+2 α+m+1 πe 1 m/2 Γ(α + m + 1) 2) 3 · F , ; α + ; 1/z 1 − = . 2 1 2 2 2 (2z)α+1 z2 Γ(α + m + 32 ) In ([JoTh80], p 205) it is proved that $ (α + m + 2)2 (α + m + 3)2 1 Qm α (z) (2α + 3)z − = m Qα+1 (z) α+m+1 (2α + 5)z − (2α + 7)z − % (α + m + 4)2 (α + m + 5)2 (α + m + 6)2 (2α + 9)z − (2α + 11)z − (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 used to 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 the hypergeometric functions themselves.

1

xs e1−x dx =

0

∞ 1 1 1 1 1 ; s ∈ C, = s + s + 1 + s + 1 + s + 1 +· · · (s + 1)n n=1

(3.6.1)

([Khru06b]).

1 0

2 · (−1)k 1 1 2 22 32 xs 4 = , = dx = 1 + x2 s + s + s + s +· · · s + 2k + 1 (s + 2k)2 − 1 k=0 k=1 ∞

∞

(3.6.2)

A.3.6 Some simple integrals

283

([Khru06b]). Dc := {s ∈ C; Re s = 0}, Df := {s ∈ C; Re s > 0}.

∞

e−t dt 1 12 22 32 = ; t+z z + 1 − z + 3 − z + 5 − z + 7 −· · ·

0

| arg z| < π,

(3.6.3)

([BoSh89], p 20).

∞

e−t/z dt (1 + t)n 0 z nz 1z (n + 1)z 2z (n + 2)z 3z = 1+ 1 + 1 + 1 1 +1+ + 1 +· · · z nz 2 2(n + 1)z 2 3(n + 2)z 2 = 1 + nz − 1 + (n + 2)z − 1 + (n + 4)z − 1 + (n + 6)z −· · ·

(3.6.4)

for (n, z) ∈ R × C with | arg z| < π, ([BoSh89], p 157). The second continued fraction is the even part of the ﬁrst one.

∞ (−1)k e−tz 1 1·2 2·3 3·4 = 2z , dt = 2 z + z + z + z +· · · (z + 2k)(z + 2k + 2) cosh t k=0

∞ 0

(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

∞

e−tz sn t dt =

0

1 · 22 · 3k 2 3 · 42 · 5k 2 1 , 12 (1 + k2 ) + z 2 − 32 (1 + k2 ) + z 2 − 52 (1 + k2 ) + z 2 −· · ·

(3.6.6)

([Wall48], p 374). Dc := {(k, z) ∈ C2 ; |k| = 1}, Df := {(k, z) ∈ C2 ; |k| < 1},

∞

e−tz sn2 t dt =

0

22 (1

2 2 · 32 · 4k 2 4 · 52 · 6k 2 , 2 2 2 2 2 2 + k ) + z − 4 (1 + k ) + z − 6 (1 + k2 ) + z 2 −· · ·

(3.6.7)

([Wall48], p 375). Dc := {(k, z) ∈ C2 ; |k| = 1}, Df := {(k, z) ∈ C2 ; |k| < 1},

∞

e−tz cn t dt =

0

1 12 22 k2 32 42 k 2 52 , z + z + z + z + z + z +· · ·

(3.6.8)

([Perr57], p 220). Dc := {(k, z) ∈ R × C; Re z = 0}, Df := {(k, z) ∈ R × C; Re z > 0},

∞

e−tz dn t dt =

0

1 12 k 2 22 32 k2 42 52 k 2 , z+ z + z + z + z + 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 −tz e dt = dn t

1 · 22 · 3k 4 3 · 42 · 5k 4 1 2 · 12 (2 − k2 ) + z 2 − 2 · 32 (2 − k 2 ) + z 2 − 2 · 52 (2 − k2 ) + z 2 −· · ·

(3.6.10)

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−tz dt =

ra ar rcb (a + 1)r 2rcb (a + 2)r z +1+ z + 1 1 + z + +· · ·

(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).

∞

te−tz 1 14 24 34 dt = , sinh t z + 3z + 5z + 7z +· · · 0 ([Wall48], p 371). Dc := {z ∈ C; Re z = 0}, Df := {z ∈ C; Re z > 0}.

∞ 0

2te−tz dt = et + e−t

∞ 0

(3.6.12)

∞ te−tz (−1)n dt = 2 cosh t (z + 1 + 2n)2 n=0

(3.6.13) 1 4·1 4 · 12 4 · 22 4 · 22 4 · 32 = 2 , z − 1 + 1 + z 2 − 1 + 1 + z 2 − 1 + 1 +· · · ([Perr57], p 30). Dc := {z ∈√C; | arg(z 2 − 1)| < π}, Df := {z ∈ C; Re z > 0 and z ∈ (0, 1]}. For instance, for z =: 5 we get 2

∞

√

0

1 12 12 22 22 32 32 4te− 5t dt = , ([Perr57], p 30). cosh t 1 + 1 + 1 + 1 + 1 + 1 + 1 +· · ·

A.3.7

Gamma function expressions by Ramanujan

(3.6.14)

Ramanujan produced quite a number of continued fraction expansions of ratios of gamma functions. These ratios have all proved to be connected to hypergeometric functions, ([Rama57], [Bern89]). We use Ramanujan’s notation 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)×

(3.7.1)

ε

× Γ(a + b − c + d)Γ(a − b − c + d) and so on. That is, ε = ±1, and the product is taken over all diﬀerent combinations of the εs. p 12 − q2 22 − p2 32 − q 2 42 − p2 1−R = 1+R z+ z + z + z + z +· · · z + p + εq + 1 z − p + εq + 3 Γ Γ 4 4 · , where R = z + p + εq + 3 z − p + εq + 1 ε Γ ε Γ 4 4

(3.7.2)

([Bern89], p 156). Dc := {(p, q, z) ∈ C3 ; Re z = 0}, Df := {(p, q, z) ∈ C3 ; Re z > 0}.

A.3.7 Gamma function expressions by Ramanujan

285

From this it follows that ∞ ! k=1

∞

= 0

(−1)k+1 (−1)k+1 + z + q + 2k − 1 z − q + 2k − 1

"

cosh(qt)e−tz 1 12 − q 2 22 32 − q 2 42 dt = , cosh t z+ z +z + z + z +· · ·

(3.7.3)

([Bern89], p 148), Dc := {(q, z) ∈ C2 ; Re z = 0}, Df := {(q, z) ∈ C2 ; Re z > 0}, and !

∞

tanh 0

sinh(at)e−tz dt t cosh t

" =

a 12 22 − a2 32 42 − a2 , z+ z + z +z + z +· · ·

(3.7.4)

([Wall48], p 372), Dc := {(a, z) ∈ C2 ; Re z = 0}, Df := {(a, z) ∈ C2 ; Re z > 0}, and tanh

$1 2

∞ 0

sinh(2at)e−tz % dt = t cosh t a 12 − a2 22 − a2 32 − a2 42 − a2 , z+ z + z + z + z +· · ·

(3.7.5)

([Wall48], p 371), Dc := {(a, z) ∈ C2 ; Re z = 0}, Df := {(a, z) ∈ C2 ; Re z > 0}. Solving (3.7.2) for 1/R gives 1 2p 12 − q 2 22 − p2 32 − q2 42 − p2 =1+ , R z − p+ z + z + z + z +· · ·

(3.7.6)

([Perr57], p 34). Dc := {(p, z) ∈ C2 ; Re z = 0}, Df := {(p, z) ∈ C2 ; Re z > 0}. The values p := q := 1/2 lead to 2 z Γ z 1·3 3·5 5·7 2 4 =1+ 4 2 z+2 2z − 1 + 2z + 2z + 2z +· · · Γ 4

for Re(z) > 0

(3.7.7)

and thus, for z := 4n or z := 4n − 2 where n ∈ N, we have 1 nπ

2 · 4 · · · · (2n) 1 · 3 · · · · (2n − 1)

2 =1+

2 1·3 3·5 5·7 , 8n − 1 + 8n + 8n + 8n +· · ·

(3.7.8)

([Perr57], p 34), 2n2 π 2n − 1

1 · 3 · · · · (2n − 1) 2 · 4 · · · · (2n)

2

2 1·3 3·5 5·7 8n − 5 + 8n − 4 + 8n − 4 + 8n − 4 +· · ·

(3.7.9)

b 1−t dt 1+t a + 1 (a + 1)(a + 2) (a + 2)(a + 3) a+1 0 = , b

1 a 2b + 2b 2b + +· · · 1 − t a−1 t dt 1+t 0

(3.7.10)

=1+

for n ∈ N, ([Perr57], p 34).

1

ta

286

Appendix A: Some continued fraction expansions

([Perr57], p 299). Dc := {(a, b) ∈ C2 ; Re(b) = 0}, Df := {(a, b) ∈ C2 ; Re(b) > 0}. From this follows directly that also b dt 1−t ta 1+t 1−t a + 1 (a + 1)(a + 2) (a + 2)(a + 3) 0 =1+ , b

1 2b + 2b 2b + +· · · dt 1−t a t 1+t 1 − t2 0

1

(3.7.11)

([Perr57], p 300). Dc := {(a, b) ∈ C2 ; Re(b) = 0}, Df := {(a, b) ∈ C2 ; Re(b) > 0}. A formula of the same character as (3.7.2) is z + εp + εq + 1 + z + εp + εq + 3 Γ = Γ 4 4 ε 8 12 − q2 12 − p2 32 − q 2 32 − p2 , 1 (z 2 − p2 + q 2 − 1) + 1 + z 2 − 1 + 1 + z 2 − 1 +· · · 2

(3.7.12)

([Bern89], p 159). Dc := {(p, q, z) ∈ C3 ; | arg(z 2 − 1)| < π}, Df := {(p, q, z) ∈ C3 ; Re z > 0} \ {(p, q, z) ∈ C3 ; 0 < z ≤ 1}. z + εq + 1 4 4 12 − q2 32 − q 2 52 − q 2 = , z + 2z + 2z + 2z +· · · z + εq + 3 Γ 4

Γ ε

(3.7.13)

([Bern89], p 140). Dc := {(q, z) ∈ C2 ; Re z = 0}, Df := {(q, z) ∈ C2 ; Re z > 0}. For q := 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 32 52 , 8n − 2 + 8n − 2 + 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 32 52 , (3.7.15) 8n + 2 + 8n + 2 + 8n + 2 +· · ·

([Perr57], p 36). A formula closely related to (3.7.13) is $

∞

1−

exp 0

cosh 2at −tz dt % e = cosh 2t t 1+

2(12 − a2 ) 32 − a2 52 − a2 , z2 + 1 + z 2 +· · ·

([Wall48], p 371). Dc := {(a, z) ∈ C2 ; Re z = 0}, Df := {(a, z) ∈ C2 ; Re z > 0}. The most involved of Ramanujan’s formulas of this type is

(3.7.16)

A.3.7 Gamma function expressions by Ramanujan

287

8abcdh R−Q = R+Q 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 } +

(3.7.17)

64(a2 − 22 )(b2 − 22 )(c2 − 22 )(d2 − 22 )(h2 − 22 ) , 5{2S4 − (S2 − 2 · 2 · 3)2 − 4(22 + 2 + 1)2 } +· · · where S4 := a4 + b4 + c4 + d4 + h4 + 1, S2 := a2 + b2 + c2 + d2 + h2 − 1, and a + ε(b + c) + ε(d + h) + 1 a + ε(b + d) + ε(c + h) + 1 Γ Γ · , R := 2 2 ε ε a + ε(b − c) + ε(d + h) + 1 a + ε(b + c) + ε(d − h) + 1 Q := Γ Γ · , 2 2 ε ε (3.7.18) ([Bern89], p 163). The expansion (3.7.17) only holds if the continued fraction terminates. 4(a2 − 12 )(b2 − 12 )(c2 − 12 ) 2abc 1−R = 2 2 2 2 1+R z − a − b − c + 1 + 3(z 2 − a2 − b2 − c2 + 5) + 4(a2 − 22 )(b2 − 22 )(c2 − 22 ) for (a, b, c, z) ∈ C4 where 5(z 2 − a2 − b2 − c2 + 13) +· · · (3.7.19) z + a + ε(b + c) + 1 z − a + ε(b − c) + 1 Γ Γ 2 2 · , R := z − a + ε(b + c) + 1 z + a + ε(b − c) + 1 ε Γ ε Γ 2 2 ([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 (a2 − 12 )(b2 − 12 ) (a2 − 22 )(b2 − 22 ) (a2 − 32 )(b2 − 32 ) z + 3z 5z 7z + + +· · · z + ε(a − b) + 1 + z + ε(a + b) + 1 where R = Γ Γ , 2 2 ε ε

(3.7.20)

([Bern89], p 155). Dc := {(a, b, z) ∈ C3 ; Re z = 0}, Df := {(a, b, z) ∈ C3 ; Re z > 0}. In particular ∞ ! k=0

1 1 − z − a + 2k + 1 z + a + 2k + 1

" = lim

b→0

11−R b1+R

a 12 (12 − a2 ) 22 (22 − a2 ) 32 (32 − a2 ) = z+ 3z 5z 7z + + +· · · for Re(z) > 0, ([Bern89], p 149). ∞ k=1

(−1)k+1 (a + 1)2 (b + 1)2 (a + 2)2 (b + 2)2 1 = (a + k)(b + k) (a + 1)(b + 1) + a+b+3 a+b+5 + +· · ·

(3.7.21)

(3.7.22)

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 ab 2 2 − b 2 2 2 − a2 4 2 − b 2 4 2 − a 2 = 2 ; 2 1+R z − 1 − a + 1 + z 2 − 1 + 1 + z 2 − 1 +· · · z + ε(a + b) + 3 2 z + ε(a − b) + 3 (3.7.23) Γ Γ 4 4 R := , z + ε(a + b) + 1 z + ε(a − b) + 1 ε Γ ε Γ 4 4 ([Bern89], p 158). Dc := {(a, b, z) ∈ C3 ; | arg(z 2 − 1)| < π}, Df := {(a, b, z) ∈ C3 ; Re z > 0} \ {(a, b, z) ∈ C3 ; 0 < z ≤ 1}. Dividing (3.7.23) by a and letting a → 0 in this equality gives ∞ ! k=0

(−1)k (−1)k − z − b + 2k + 1 z + b + 2k + 1

"

∞

= 0

e−tz

sinh(bt) dt cosh t

b 22 − b 2 22 42 − b 2 = 2 , z − 1 + 1 + z 2 − 1 + 1 +· · ·

(3.7.24)

([Bern89], p 150). Dc := {(b, z) ∈ C2 ; | arg(z 2 − 1)| < π}, Df := {(b, z) ∈ C2 ; Re z > 0} \ {(b, z) ∈ C2 ; 0 < z ≤ 1}. Of course, dividing this again by b and letting b → 0 gives

2

∞ k=0

1 (−1)k 22 22 42 42 = , (z + 2k + 1)2 z 2 − 1 + 1 + z 2 − 1 + 1 + z 2 − 1 +· · ·

(3.7.25)

([Bern89], p 151). Dc := {z ∈ C; | arg(z 2 − 1)| < π}, Df := {z ∈ C; Re z > 0} \ {z ∈ C; 0 < z ≤ 1}.

1 + 2z

∞ (−1)k 1 1·2 2·3 3·4 = , z + 2k z + z + z + z +· · ·

(3.7.26)

k=1

([Bern89], p 151). Dc := {z ∈ C; Re z = 0}, Df := {z ∈ C; Re z > 0}. 1 + 2z 2

∞ (−1)k 1 12 1 · 2 22 2 · 3 32 , = (z + k)2 z + z + z + z + z + z +· · ·

(3.7.27)

k=1

([Bern89], p 152). Dc := {z ∈ C; Re z = 0}, Df := {z ∈ C; Re z > 0}.

∞

c 0

sinh at sinh bt −tz e dt = sinh ct

1(z 2

+

ab 4 · 12 (12 c2 − a2 )(12 c2 − b2 ) 2 2 − a − b ) − 3(z 2 + 5c2 − a2 − b2 ) −

c2

(3.7.28)

4 · 22 (22 c2 − a2 )(22 c2 − b2 ) , 5(z 2 + 13c2 − a2 − b2 ) −· · · where the coeﬃcients for c2 are 2k 2 + 2k + 1 in the denominators ([Wall48], p 370). Dc := {(a, b, c, z) ∈ C4 ; Re zc = 0}, Df := {(a, b, c, z) ∈ R4 ; Re zc > 0 and Re(z + c − a − b) > 0}.

∞

c 0

a 12 (12 c2 − a2 ) 22 (22 c2 − a2 ) sinh at −tz e dt = , sinh ct z+ 3z 5z + +· · ·

(3.7.29)

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) ∈ C3 ; Re(z/c) > 0}.

∞ 0

e−tz dt = (cosh t + a sinh t)b

1 · b(1 − a2 ) 2(b + 1)(1 − a2 ) 3(b + 2)(1 − a2 ) 1 , z + ab + z + a(b + 2) + z + a(b + 4) + z + a(b + 6) +· · ·

(3.7.30)

([Wall48], p 369). Dc := {(a, b, z) ∈ C3 ; Re a = 0}, Df := {(a, b, z) ∈ C3 ; Re a > 0 and Re(b + z) > 0}.

1 4 · 1ab a+b+1 ; − sinh2 t e−tz dt = 2 z + (a + b + 1)z + 0 4 · 2(a + 1)(b + 1)(a + b) 4 · 3(a + 2)(b + 2)(a + b + 1) (a + b + 3)z (a + b + 5)z + +· · · +

∞

2 F1

a, b;

(3.7.31)

([Wall48], p 370). Dc := {(a, b, z) ∈ C3 ; Re z = 0}, Df := {(a, b, z) ∈ C3 ; Re z > 0}. ζ(3, z + 1) :=

∞ k=1

1 (z + k)3

1 13 13 23 23 = , 2(z 2 + z) + 1 + 6(z 2 + z) + 1 + 10(z 2 + z) +· · ·

(3.7.32)

([Bern89], p. 153). Dc := {z ∈ C; | arg(z 2 + z)| < π} = {z ∈ C; Re z = − 12 } \ [−1, 0], Df := {z ∈ C; Re z > − 12 } \ [− 12 , 0]. The even part of this continued fraction is ζ(3, z + 1) =

1(2z 2

1 16 2 + 2z + 1) − 3(2z + 2z + 3) −

26 36 . 2 2 5(2z + 2z + 7) − 7(2z + 2z + 13) −· · ·

(3.7.33)

Dc := {z ∈ C; Re(z 2 + 12 ) = 0}, Df := {z ∈ C; Re(z 2 + 12 ) > 0}. (The numbers 1, 3, 7, 13, . . . in the denominators are n2 + n + 1 for n = 0, 1, 2, . . . .) ∞ !

1 1 + − z + a + b + 2k + 1 z − a − b + 2k + 1 k=0 " 1 1 − z + a − b + 2k + 1 z − a + b + 2k + 1 ∞ 8ab(z + 2k + 1) = {(z + 2k + 1)2 − a2 − b2 }2 − 4a2 b2

(3.7.34)

k=0

=

1(z 2

2ab 2(12 − a2 ) 2(12 − b2 ) 2 2 2 − 1) + b − a + 1 + 3(z − 1) + b2 − a2 +

4(22 − b2 ) 4(22 − a2 ) , 2 1 + 5(z − 1) + b2 − a2 +· · · ([Bern89], p 158). Dc := {(a, b, z) ∈ C3 ; | arg(z 2 − 1)| < π} = {(a, b, z) ∈ C3 ; Re z = (0, 1)}. Dividing by 2a and 0 and z ∈ (−1, 1)}, Df := {(a, b, z) ∈ C3 ; Re z > 0 and z ∈ letting a → 0 in (3.7.34) leads to

290

Appendix A: Some continued fraction expansions

∞ ! k=0

1 1 − (z − b + 2k + 1)2 (z + b + 2k + 1)2

" =

∞ k=0

4b(z + 2k + 1) {(z + 2k + 1)2 − b2 }2

b 2 · 12 4 · 22 2(12 − b2 ) 4(22 − b2 ) = , 1(z 2 − 1) + b2 + 1 1 + 3(z 2 − 1) + b2 + + 5(z 2 − 1) + b2 +· · · (3.7.35) ([Bern89], p 158). Dc := {(b, z) ∈ C2 ; Re z = 0 and z ∈ (−1, 1)}, Df := {(b, z) ∈ C2 ; Re z > 0 and z ∈ (0, 1)}. The even part of this continued fraction is ∞ ! k=0

1 1 − (z − b + 2k + 1)2 (z + b + 2k + 1)2

" =

∞

4b(z + 2k + 1) {(z + 2k + 1)2 − b2 }2

k=0

b 4(12 − b2 )14 4(22 − b2 )24 4(32 − b2 )34 = . 2 2 2 2 2 2 1(z − b + 1) − 3(z − b + 5) − 5(z − b + 13) − 7(z 2 − b2 + 25) −· · ·

(3.7.36)

Dc := {(b, z) ∈ C2 ; Re z = 0}, Df := {(b, z) ∈ C2 ; Re z > 0}. (The numbers 1, 5, 13, 25, . . . in the denominators have the form 2n2 + 2n + 1 for n = 0, 1, 2, 3, . . . .) 2a2 4a4 + 14 4a4 + 24 4a4 + 34 u−v = u+v 1z + 3z 7z + 5z + +· · · Γ2

where z + 1

(3.7.37) 2 z + 2a + 1 z − 2a + 1 k=0 Γ Γ 2 2 ([ABJL92], entry 48). Dc := {(a, z) ∈ C; Re z = 0}, Df := {(a, z) ∈ C; Re z > 0}. u :=

∞ $

1+

2 % 2a , v := z + 2k + 1

u−v a3 a6 − 16 a6 − 26 = 2 2 u+v 1(2z + 2z + 1) + 3(2z + 2z + 3) + 5(2z 2 + 2z + 7) +· · · where u :=

∞

1+

k=1

a z+k

3 ,

v :=

∞ k=1

1−

a z+k

(3.7.38)

3 ,

([ABJL92], entry 50). Dc := {(a, z) ∈ C2 ; Re z = − 12 }, Df := {(a, z) ∈ C2 ; Re z > − 12 }. (The numbers 1, 3, 7, . . . in the denominators are the numbers n2 +n+1 for n = 0, 1, 2, . . . .) ∞ (−1)k y2k+1 z 12 z 2 22 z 2 32 z 2 = r + 2k + 1 1 + a + 3 + a + 5 + a + 7 + a +· · · k=0 where y := ( 1 + z 2 − 1)/z and r := a/ 1 + z 2

2

(3.7.39)

for (a, z) ∈ C2 with | arg(z 2 + 1)| < π; i.e., z ∈ C \ i((−∞, −1] ∪ [1, ∞)), ([ABJL92], entry 14). With the same notation and same region for (a, z), also ∞ 1 (−1)k y 2k z 1 · 2z 2 2 · 3z 2 3 · 4z 2 y+r y+ = , y r + 2k 2 + a + 4 + a + 6 + a + 8 + a +· · · k=1

([ABJL92], entry 15) and

(3.7.40)

A.4.1 Basic hypergeometric functions 1+

1 z2

(b−1)/2

(2y)b

291

∞ (−1)k (b)k y2k = k!(r + b + 2k) k=0

z 2(b + 1)z 2 3(b + 2)z 2 1 · bz 2 , a + b + a + b + 2 + a + b + 4 + a + b + 6 +· · · ([ABJL92], entry 17), where b ∈ C.

A.4

(3.7.41)

Basic hypergeometric functions

In this chapter we use the standard notation 2 ϕ1 (a, b; c; q; z)

:=

∞ (a; q)n (b; q)n n z (c; q)n (q; q)n n=0

where (d; q)0 := 1, (d; q)n := (1 − d)(1 − dq) · · · (1 − dq n−1 ) for n ∈ N. For convenience we always assume that q ∈ C with |q| < 1, although the continued fraction may well converge, even to the right value, for other values of q ∈ C.

A.4.1

General expressions (1 − c)

2 ϕ1 (a, b; c; q; z) 2 ϕ1 (a, bq; cq; q; z)

=

(1 − a)(c − b)z (1 − bq)(cq − a)z (1 − aq)(cq − b)qz 1 − cq 1 − cq 2 1 − cq 3 + + + 2 2 2 2 2 (1 − bq )(cq − a)qz (1 − aq )(cq − b)q z 1 − cq 4 1 − cq 5 + +· · ·

1−c+

(4.1.1)

for (a, b, c, z) ∈ C4 , ([ABBW85], p 14). (1 − c)

2 ϕ1 (a, b; c; q; z) 2 ϕ1 (aq, bq; cq; q; z)

= b0 + K(an /bn )

where an := (1 − aq n )(1 − bq n )cq n−1 (1 − zabq n /c)z

(4.1.2)

bn := 1 − cq n − (a + b − abq n − abq n+1 )q n z for (a, b, c, z) ∈ C4 .

q(1 − c)

(a − cq)(1 − bq)qz 2 ϕ1 (a, b; c; q; z) = (1 − c)q + (a − bq)z − ϕ (a, bq; cq; q; z) (1 − cq)q + (a − bq 2 )z − 2 1

(a − cq 2 )(1 − bq 2 )qz (a − cq 3 )(1 − bq 3 )qz 2 3 (1 − cq )q + (a − bq )z − (1 − cq 3 )q + (a − bq 4 )z − · · · for (a, b, c, z) ∈ C4 , ([ABBW85], p 18).

(4.1.3)

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 2 ϕ1 (a, q; cq; q; z) or 2 ϕ1 (aq, q; cq; q; z).

A.4.2

Two general results by Andrews G(a, b, c; q) aq + cq bq + cq 2 aq 2 + cq 3 bq 2 + cq 4 =1+ G(aq, b, cq; q) 1 + 1 1 1 + + +· · · ∞ k(k+1)/2 k c (− a ; q)k q a where G(a, b, c; q) := (q; q)k (−bq; q)k

(4.2.1)

k=0

for (a, b, c) ∈ C3 , ([ABJL89], p 80). H(a1 , a2 ; z; q) (1 + aq 2 z)qz (1 + aq 3 z)q 2 z = 1 + bqz + H(a1 , a2 ; qz; q) 1 + bq 2 z + 1 + bq3 z +· · · where a := −1/a1 a2 and b := −1/a1 − 1/a2 and qz qz ;q ;q a1 a2 ∞ ∞ × H(a1 , a2 ; z; q) := (qz; q)∞ (1 − z) ∞ (1 − zq2k )(z; q)k (a1 ; q)k (a2 ; q)k q k(3k+1)/2 (az 2 )k × qz qz k=0 (q; q)k ;q ;q a1 a2 k 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). We refer in particular to ([ABJL92]) for more details. From (4.1.1) we ﬁnd that

(−a; q)∞ (b; q)∞ − (a; q)∞ (−b; q)∞ a−b = · (−a; q)∞ (b; q)∞ + (a; q)∞ (−b; q)∞ 1−q =

bq bq 2 3 2 2 , ;q ;q ;a a a bq b , ; q; q 2 ; a2 2 ϕ1 a a

2 ϕ1

(4.3.1)

a − b (a − bq)(aq − b) (a − bq 2 )(aq2 − b)q (a − bq 3 )(aq3 − b)q 2 1 − q+ 1 − q3 1 − q5 1 − q7 + + +· · ·

for (a, b) ∈ C2 , ([ABBW85], p 14). (a − bq)(b − aq) (a − bq 3 )(b − aq 3 ) 1 (a2 q 3 ; q 4 )∞ (b2 q 3 ; q 4 )∞ = (a2 q; q 4 )∞ (b2 q; q 4 )∞ 1 − ab + (1 − ab)(q 2 + 1) + (1 − ab)(q 4 + 1) +· · ·

(4.3.2)

A.4.3 q-expressions by Ramanujan

293

for (a, b) ∈ C2 , ([ABBW85], entry 12). F (b; a) aq aq 2 aq 3 =1+ 2 F (b; aq) 1 + bq + 1 + bq + 1 + bq3 +· · · 2 ∞ ak q k where F (b; a) := (−bq; q)k (q; q)k

(4.3.3)

k=0

for (a, b) ∈ C2 , ([ABBW85], entry 15). If we set a := 0 in (4.2.1) we get ϕ(c) cq bq + cq2 cq 3 bq 2 + cq 4 cq 5 =1+ ϕ(cq) 1+ 1 1 + 1 + + 1 + ··· ∞ k2 k q c , where ϕ(c) := (q; q)k (−bq; q)k )

(4.3.4)

k=0

([ABJL92], entry 56). If we moreover set b := −c, this reduces to ∞

(−c)k q k(k+1)/2 =

k=0

1 cq c(q 2 − q) cq 3 c(q4 − q 2 ) , 1+ 1 + 1 1 + 1 + +· · ·

(4.3.5)

([ABBW85], p 22). G(z) qz q3 z q2 z q6 z q3 z q9 z =1− G(qz) 1 + q + 1 + q 2 − 1 + q 3 + 1 + q 4 − 1 + q 5 + 1 + q 6 −· · · ∞ (−z)k q k(k+1)/2 where G(z) := (q 2 ; q 2 )k

(4.3.6)

k=0

for z ∈ C, ([ABJL92], formula 9.1). 1 q q3 (q2 ; q 3 )∞ q5 q7 = , 3 2 3 (q; q )∞ 1 − 1 + q − 1 + q − 1 + q − 1 + q 4 −· · ·

(4.3.7)

([ABJL92], entry 10). q3 q5 (q3 ; q 4 )∞ 1 q , = (q; q 4 )∞ 1 − 1 + q 2 − 1 + q 4 − 1 + q 6 −· · ·

(4.3.8)

(−q 2 ; q 2 )∞ 1 q q2 + q q3 q4 + q2 q5 = , 2 (−q; q )∞ 1 + 1 + 1 + 1 + 1 + 1 +· · ·

(4.3.9)

([ABJL92], entry 11).

([ABJL92], entry 12). (q; q 2 )∞ 1 q + q2 q2 + q4 q3 + q6 , = 3 6 3 {(q ; q )∞ } 1 + 1 + 1 + 1 +· · · ([ABJL89], thm 7).

(4.3.10)

294

Appendix A: Some continued fraction expansions

1 q q2 q3 q4 (q; q 5 )∞ (q 4 ; q 5 )∞ = , (q 2 ; q 5 )∞ (q 3 ; q 5 )∞ 1 + 1 + 1 + 1 + 1 +· · ·

(4.3.11)

([ABJL89], (5)). 1 q + q 2 q 4 q 3 + q 6 q 8 q 5 + q 10 (q; q 8 )∞ (q 7 ; q 8 )∞ = , 3 8 5 8 (q ; q )∞ (q ; q )∞ 1+ 1 + 1 + 1 + 1 + 1 +· · ·

(4.3.12)

([ABJL89], thm 6). ∞ k=1

a (1 − a)qz (1 − q)aqz (1 − aq)q 2 z (a; q)∞ ak = k (q; q)k (1 + q z) 1+ 1 1 1 + + + (1 − q 2 )aq 2 z (1 − aq 2 )q 3 z 1 1 + +· · ·

for (a, z) ∈ C2 , ([Wall48], p 376).

(4.3.13)

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Index (a)n Pochhammer symbol, 27 (n) Ak , 9 An canonical numerator, 6 (n) Bk , 9 Bn canonical denominator, 6 F [n] , 172 I(w), 172 N th root of unity, 194 Pk , 66 Sn , 5 (m) Sp , 172 Δn , 7 , 174 Σ∞ , 68 Σn , 66 3 C, u, 15 C, 3 D unit disk, 26 H, 109 N, 9 R real numbers, 109 B(a, r), 75 Bm (γn , ε), 73 V, 109 Ln z, 26 diam(V ), 71 dist(x, V ), 225 distm (w, V ) chordal distance, 114 rad(D), radius of D, 111 R+ , 118 ∼ equivalence between continued fractions, 77 √ . . ., 25 {ζn } critical tail sequence, 64 {hn } critical tail sequence, 64 {tn } tail sequence, 63 {wn† } exceptional sequence, 56 f (n) tail value, 6 fn classical approximant, 6

o(kn ), 221 o(rn+1 ), 12 p-periodic continued fraction, 177 sn , 5 Tn , 111 M family of M¨ obius transformations, 5 S Stern-Stolz Series, 101 R+ , 118 B(C, −r), 109 B(C, r), 109 ε-contractive, 73 2 F1 (a, b; c; z) hypergeometric function, 28 ´ Sleszy´ nski-Pringsheim Theorem, 129 ´ Sleszy´ nski-Pringsheim continued fraction, 129 a posteriori bounds, 108 a priori bounds, 108 absolute convergence of continued fraction, 100 absolute convergence of sequence, 100 absolutely continuous measure, 41 alternating continued fraction, 122 analytic continuation, 35 approximant, 3, 5 arithmetic complexity, 11 asymptotic expansion, 40 attracting ﬁxed point, 175 auxiliary continued fraction, 223 axis of cartesian oval, 244 backward recurrence algorithm, 11 Bauer-Muir transform, 82 best rational approximation, 17 binomial series, 50 Birkhoﬀ-Trjzinski theory, 262 canonical contraction, 85 canonical denominator, 7 canonical numerator, 7 cartestian oval, 161

306

307

Index chain sequence, 90 Chebyshev polynomials, 42 chordal diameter, 71 chordal distance, 55 chordal metric, 55 classical approximant, 6 classical convergence, 60 conjugate transformation, 174 continued fraction, 3 continued fraction expansion, 25 continued fraction of elliptic type, 176 continued fraction of identity type, 176 continued fraction of loxodromic type, 176 continued fraction of parabolic type, 176 continued fraction, deﬁnition, 5 contraction, 85 convergence acceleration, 218 convergence neighborhood, 213 convergence of continued fraction, 6 convergents, 5 correspondence, 33 corresponding periodic continued fraction, 186 critical tail sequence, 64 cross ratio, 54 determinant formula, 7 diagonalization of matrix, 212 diamm , 71 diﬀerential equation, 38 diophantine equation, 21 distribution function, 40 divergent continued fraction, 6 dual continued fraction, 95 element sets, 73 elements of a continued fraction, 5 ellipse, arc length of, 30 elliptic transformation, 175 empty product, 2 empty sum, 2 equivalence transformation, 77 equivalent continued fractions, 77 equivalent sequences, 57 euclidean algorithm, 16 Euler-Minding formula, 7 Euler-Minding summation, 11 even part, 86 exceptional sequence, 56, 57 extension, 85

Favard’s Theorem, 43 Fibonacci numbers, 50 ﬁxed point, 172 ﬁxed circle for τ , 205 ﬁxed line for τ , 205 ﬁxed point method, 219 forward recurrence algorithm, 11 fraction term, 5 functional equation, 85 fundamental inequalities, 165 general convergence, 56, 60 general divergence, 60 generalized circle, 108 generic sequence, 61 golden ratio, 50 greatest common divisor, 16 Henrici-Pﬂuger Bounds, 126 history of continued fractions, 46 hyperbolic transformation, 207 hypergeometric functions, 27 identity transformation, 62 improvement machine, 227 indiﬀerent ﬁxed points, 175 interpolation, 44 inverse diﬀerencies, 44 iterate, 172 J-fractions, 43 Jacobi continued fraction, 43 Khovanskii transform, 98 Kronecker delta, 42 Legendre polynomials, 42 limit p-periodic continued fraction of loxodromic type, 187 limit periodic continued fraction, 186 limit point case, 70 limit sets, 75 linear fractional transformation, 5 loxodromic transformation, 175 M¨ obius transformation, 5 measure, 41 modiﬁed approximant, 6 modifying factor, 6 moment, 41

308 moment problem, 40, 41 odd part, 86 orthogonal polynomials, 42 oval, 161 Oval Sequence Theorem, 243 Pad´e approximant, 35 Pad´e approximants, diagonal, 37 Pad´e table, 35 Parabola Sequence Theorem, 154 Parabola Theorem, 151 parabolic pair, 184 parabolic transformation, 175 period length, 186 periodic continued fraction, 172 Pochhammer symbol, 27 positive continued fraction, 116 prevalue set, 70 probability measure, 41 Ramanujan’s AGM-fraction, 202 ratio for continued fraction, 176 ratio of τ , 174 rational approximation, 17 real continued fraction, 122 recurrence relations for An , Bn , 6 regular C-fraction, 30, 79 regular continued fraction, 4, 14 regular continued fraction expansion, 15 repelling ﬁxed point, 175 restrained continued fraction, 62 restrained sequence, 61 reversed continued fraction, 213 reversed periodic continued fraction, 95 reversed terminating continued fraction, 48 right tail sequence, 90 root of unity, 194 S-fraction, 41, 124 Seidel-Stern Theorem, 117 separate convergence, 102 sequence of tail values, 6 similar transformation, 174 simple element set, 74 simple value set, 70 singular transformation, 5 square root modiﬁcation, 235 stable polynomials, 45 Stern-Stolz Divergence Theorem, 100

Index Stern-Stolz Series, 101 Stieltjes continued fraction, 124 Stieltjes moment problem, 41 Stieltjes-Vitali Theorem, 115 strong convergence, 90 successive substitutions, 30 sum of divergent series, 39 symmetric points with respect to circle, 109 tail, 6 tail sequence, 63 tail values, 6 terminating continued fraction, 10 Thiele continued fraction, 43 Thiele oscillation, 180 Thron-Gragg-Warner Bounds, 128 Thron-Lange Theorem, 148 totally non-restrained, 62 transformations of continued fractions, 77 truncation error, 106 truncation error estimate, 165 truncation error bounds, 106 tusc, 225 twin element sets, 74 twin value sets, 70 unit disk, 26 value set, 70 Van Vleck’s Theorem, 142 Vitali’s Theorem, 114 Worpitzky’s Theorem, 135 wrong tail sequence, 90