Symbolic Integration I: Transcendental Functions, Second Edition (Algorithms and Computation in Mathematics)

Algorithms and Computation in Mathematics ® Volume i Editors Manuel Bronstein Arjeh M. Cohen Henri Cohen David Eisenbud Bernd Sturmfels

Manuel Bronstein

Symbolic Integration I Transcendental Functions

Second Edition

^

Spri rineer

Manuel Bronstein INRIA 2004 route des Lucioles - B.P. 93 06902 Sophia Antipolis Cedex, France e-mail: [email protected]

Mathematics Subject Classification (2000): 12F20,12H05,12Y05,13N15, 28-04, 33B10, 33F10, 68W30

Library of Congress Control Number: 2004110974

ISSN 1431-1550 ISBN 3-540-21493-3

Springer Berlin Heidelberg New York

ISBN 3-540-60521-5 ist edition Springer-Verlag Berlin Heidelberg New York This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Dupfication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9,1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable for prosecution under the German Copyright Law. Springer is a part of Springer Science+Business Media springeronline.com © Springer-Verlag Berlin Heidelberg 2005 Printed in Germany The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typesetting: by the author using a Springer KTgX macro package Production: LE-TEX Jelonek, Schmidt & Vöckler GbR, Leipzig Cover design: design & production GmbH, Heidelberg Printed on acid-free paper

46/3142YL - 5 4 3 2 1 0

Foreword

This book brings together two streams of research in mathematics and computing t h a t were begun in the nineteenth century and made possible t h r o u g h results brought to fruition in t h e twentieth century. Methods for indefinite integration have been important ever since t h e invention of the calculus in the 1700s. In the 1800s Abel and Liouville began t h e earliest mathematical research on algorithmic methods on integration in finite terms leading to what might be considered today as an early mathematical vision of a complete algorithmic solution for integrating elementary functions. In an 1842 publication Lady A d a Augusta, Countess of Lovelace, describing t h e capabilities of Babbage's analytical engine put forth the vision t h a t computational devices could do algebraic as well as numerical calculations when she said t h a t "[Babbage's Analytical Engine] can arrange and combine its numerical quantities exactly as if they were letters or any other general symbols; and in fact it might bring out its results in algebraical notation were provisions made accordingly." Thus these two visions set the stage for a century and a half of research t h a t partially culminates in this book. Progress in the mathematical realm continued through out the nineteenth and twentieth centuries. The Russian mathematician Mordukhai-Boltovskoi wrote t h e first two books on this subject in 1910 and 1913^ W i t h the invention of electronic computers in the late 1930s and early 1940s, a new impetus was given to b o t h the mathematical and computational streams of work. In the meantime in the mathematical world i m p o r t a n t progress had been made on algebraic methods of research. Ritt began t o apply the new algebraic techniques to the problem of integration in finite terms, an approach t h a t has proven crucially important. In 1948 he published t h e results of his research in a little book. Integration in Finite Terms, T h e use of these algebraic ideas were brought to further fruition by Kolchin, Rosenlicht, and, particularly for problems of symbolic integration, by three of Rosenlicht's P h . D . students — Risch, Singer, and Bronstein^. On the Integration in Finite Terms of Linear Differential Equations. Warsaw, 1910 (in Russian) and On the Integration of Transcendental Functions. Warsaw, 1913 (in Russian). Let me hasten to add that there have been important contributions by many others and it is not my intention to give a complete history of the field in this short paragraph, but to indicate some of main streams of work that have led to the current book.

VI

Foreword

On the computational side, matters rested until 1953 when two early programs were written, one by Kahrimanian at Temple University and another by Nolan at Massachusetts Institute of Technology, to do analytic differentiation — the inverse of indefinite integration. There was active research in the late 1950s and early 1960s on list processing packages and languages that laid the implementation foundations for today's computer algebra systems. Slagle's 1961 thesis was an early effort to write a program, in LiSP, to do symbolic integration. With the advent of general computer algebra systems, some kind of symbolic integration facility was implemented in most. These integration capabilities opened the eyes of many early users of symbolic mathematical computation to the amazing potential of this form of computation. But yet none of the systems had a complete implementation of the full algorithm that Risch had announced in barest outline in 1970. There were a number of reasons for this. First and foremost, no one had worked out the many aspects of the problem that Risch's announcement left incomplete. Starting with his Ph.D. dissertation and continuing in a series of beautiful and important papers, Bronstein set out to fill in the missing components of Risch's 1970 announcement. Meanwhile working at the IBM T. J. Watson Research Center, he carried out an almost complete implementation of the integration algorithms for elementary functions. It is the most complete implementation of symbolic integration algorithms to date. In this book, Bronstein brings these mathematical and computational streams of research together in a highly effective manner. He presents the algorithmic details in pseudo-code that is easy to implement in most of the general computer algebra systems. Indeed, my students and I have implemented and tested many of the algorithms in M A P L E and MACSYMA. Bronstein's style and appropriate level of detail makes this a straightforward task, and I expect this book to be the standard starting place for future implementers of symbolic integration algorithms. Along with the algorithms, he presents the mathematics necessary to show that the algorithms work correctly. This is a very interesting story in its own right and Bronstein tells it well. Nonetheless, for those primarily interested in the algorithms, much of the mathematics can be skipped at least in a first study. But the full beauty of the subject is to be most appreciated by studying both aspects. The full treatment of the subject is a long one and it is not finished in this volume. The longer and more difficult part involving the integration of algebraic functions must await a second volume. This volume serves as a good foundation to the topic of symbolic integration and as a nice introduction to the literature for integration of algebraic functions and for other aspects such as integration involving non-elementary functions. Study, learn, implement, and enjoy! B. F. Caviness

Preface to the Second Edition

I have taken the opportunity of this second edition to add a chapter on parallel integration, a method that is used by several computer algebra systems, either before or in place of the complete integration algorithm. I have also added newreferences and exercises that expand on topics such as obtaining continuous integrals or the relations between special polynomials, Darboux polynomials and constants in monomial extensions. I would like to thank all the readers of the first edition who have sent me various corrections and suggestions. While I have tried to incorporate all of them in this edition, I remain responsible for the remaining errors.

Sophia Antipolis, June 2004

M. Bronstein

Preface to the First Edition

The integration problem, which is as old as calculus and differentiation, can be informally stated very concisely: given a formula for a function / ( x ) , determine whether there is a formula for a differentiable function F{x) satisfying

and compute such an F(x), which is called an antiderivative of f{x) and is denoted

F{x) = j f{x)da if it exists. Yet, while symbolic differentiation is a rather simple mechanical process, suitable as an exercise in a first course in analysis or computer programming, the inverse problem has been challenging scientists since the time of Leibniz and Newton, and is still a challenge for mathematicians and computer scientists today. Despite the many great strides made since the 19*^ century in showing that integration is in essence a mechanical process, although quite more complicated than differentiation, most calculus and analysis textbooks give students the impression that integration is at best a mixture of art and science, with flair in choosing the right change of variable or approach being an essential ingredient, as well as a comprehensive table of integrals. The goal of this book is to show that computing symbolic antiderivatives is in fact an algorithmic process, and that the integration procedure for transcendental functions can be carried out by anyone with some familiarity with polynomial arithmetic. The integration procedure we describe is also capable of deciding when antiderivatives are not elementary, and proving it as a byproduct of its calculations. For example the following classical nonelementary integrals

S^'^'^^

/biM'

sm{x)dx

/

X

can be proven nonelementary with minimal calculations.

X

Preface to the First Edition

The algorithmic approach, pioneered by Abel and Liouville in the past century, eventually succeeded in producing a mechanical procedure for deciding whether an elementary function has an elementary antiderivative, and for computing one if so. This procedure, which Risch described in a series of reports [73, 74, 75, 76], unfortunately not all of them published, forms the basis of most of the symbolic integration algorithms of the past 20 years, all of them loosely grouped under the appellation Risch algorithm. The procedure which we describe in this book also has its roots in the original Risch algorithm [75] and its improvements, our main sources besides Risch being [12, 13, 83, 89]. We have tried to keep the presentation as elementary as possible, with the minimal background for understanding the algorithm being an introductory course in algebra, where the topics rings and fields, polynomial greatest common divisors, irreducible polynomials and resultants are covered"^. Some additional background in field theory, essentially algebraic and transcendental extensions, is occasionally used in the proofs associated with the algorithm. The reader willing to accept the algorithm without proof can skip those sections while learning the algorithm. We have also generalized and extended the original Risch algorithm to a wider class of functions, thereby offering the following features, some of them new, to the reader already familiar with symbolic integration: ® The algorithms in this book use only rational operations, avoiding factorization of polynomials into irreducibles. ® Extensions by tangents and arc-tangents are treated directly, thereby real trigonometric functions are integrated without introducing complex exponentials and logarithms in the computations. ® Antiderivatives in elementary extensions can still be computed when arbitrary primitives are allowed in the integrand, e.g. Erf(x), rather than logarithms. ® Several subalgorithms are applicable to a large class of non-Liouvillian extensions, thereby allowing integrals to be computed for such functions. The material in this book has been used in several courses for advanced undergraduates in mathematics or computer science at the Swiss Federal Institute of Technology in Zurich: ® In a one-semester course on symbolic integration, emphasizing the algorithmic and implementation aspects. This course covers Chap. 2 in depth, Chap. 3 and 4 superficially, then concentrates on Chap. 5, 6, 7 and 8. • In the first part of a one-semester course on differential algebra. This course covers Chap. 3, 4 and 5 in depth, turning after Liouville's Theorem to other topics (e.g. differential Galois theory). ® In the last part of a one-semester introductory course in computer algebra, where some algorithms from Chap. 2 and 5 are presented, usually without proofs. ^Those topics are reviewed in Chap. 1.

Preface to the First Edition

XI

In all those courses, the material of Chap. 1 is covered as and when needed, depending on the background of the students. Chap. 9 contains complete proofs of several structure theorems and can be presented independently of the rest of this book. By presenting the algorithm in pseudocode in various "algorithm boxes" throughout the text, we also hope to make this book useful for programmers implementing symbolic integrators: by following the pseudocode, they should be able to write an integrator without studying in detail the associated theory. The reader will notice that several topics in symbolic integration are missing from this book, the main one being the integration of algebraic functions. Including algorithms for integrating algebraic and mixed algebraictranscendental functions would however easily double the size of this book, as well as increase the mathematical prerequisites, since those algorithms require prior famiharity with algebraic curves and functions. We have thus decided to cover algebraic functions in a second volume, which will hopefully appear in the near future. In the meantime, this book is an adequate preparation to the extensive literature on the integration of algebraic functions [8, 9, 11, 14, 29, 73, 74, 76, 91]. Another related topic is integration in nonelementary terms, i.e. with new special functions allowed in the antiderivatives. Here also, the reader should have no difficulty moving on to the research literature [5, 6, 21, 22, 52, 53, 70, 94] after completing this book. Acknowledgements I am thankful to several colleagues and students who have read and corrected many early drafts of this book. I am particularly grateful to Bob Caviness. Thorn Mulders and Paul Zimmermann, who corrected many errors in the final text and suggested several improvements. Sergei Abramov, Cedric Bächler, Johannes Grabmeier, David Stoutemyer, Jacques-Arthur Weil and Clifton Williamson have also helped a great deal with their corrections and suggestions. Of course, I am fully responsible for any error that may remain. Finally, I wish to thank Dr. Martin Peters and his staff at Springer-Verlag for their great patience with this project.

Zurich, November 1996

M. Bronstein

Contents

Foreword Preface to the Second Edition

V VII

Preface to the First Edition

IX

1

Algebraic Preliminaries 1.1 Groups, Rings and Fields 1.2 EucHdean Division and Pseudo-Division 1.3 The EucHdean Algorithm 1.4 Resultants and Subresultants 1.5 Polynomial Remainder Sequences 1.6 Primitive Polynomials 1.7 Squarefree Factorization Exercises

1 1 8 10 18 21 25 28 32

2

Integration of Rational Functions . 2.1 The BernouUi Algorithm 2.2 The Hermite Reduction 2.3 The Horowitz-Ostrogradsky Algorithm 2.4 The Rothstein-Trager Algorithm 2.5 The Lazard-Rioboo-Trager Algorithm 2.6 The Czichowski Algorithm 2.7 Newton-Leibniz-Bernoulli Revisited 2.8 Rioboo's Algorithm for Real Rational Functions 2.9 In-Field Integration Exercises

3

Differential Fields 3.1 Derivations 3.2 Differential Extensions

35 36 39 45 . 47 49 53 54 59 70 72 75 75 79

XIV

Contents 3.3 Constants and Extensions 3.4 Monomial Extensions 3.5 The Canonical Representation Exercises

85 90 99 104

4

The O r d e r Function 4.1 Basic Properties 4.2 Localizations 4.3 The Order at Infinity 4.4 Residues and the Rothstein-Trager Resultant Exercises

107 107 110 115 118 126

5

Integration of Transcendental Functions 5.1 Elementary and Liouvillian Extensions 5.2 Outline and Scope of the Integration Algorithm 5.3 The Hermite Reduction 5.4 The Polynomial Reduction 5.5 Liouville's Theorem 5.6 The Residue Criterion 5.7 Integration of Reduced Functions 5.8 The Primitive Case 5.9 The Hyperexponential Case 5.10 The Hypertangent Case 5.11 The Nonlinear Case with no Specials 5.12 In-Field Integration Exercises

129 129 134 138 140 142 147 154 157 160 163 172 175 178

6

T h e Risch Differential E q u a t i o n 6.1 The Normal Part of the Denominator 6.2 The Special Part of the Denominator 6.3 Degree Bounds 6.4 The SPDE Algorithm 6.5 The Non-Cancellation Cases 6.6 The Cancellation Cases Exercises

181 181 186 193 202 206 211 216

7

Parametric Problems 7.1 The Parametric Risch Differential Equation 7.2 The Limited Integration Problem 7.3 The Parametric Logarithmic Derivative Problem Exercises

217 .....217 245 250 255

Contents 8

The 8.1 8.2 8.3 8.4

Coupled Differential System The Primitive Case The Hyperexponential Case The NonUnear Case The Hypertangent Case

9

Structure Theorems 9.1 The Module of Differentials 9.2 RosenHcht's Theorem 9.3 The Risch Structure Theorems 9.4 The Rothstein-Caviness Structure Theorem Exercises

XV 257 259 261 263 264 269 269 276 282 293 296

10 Parallel I n t e g r a t i o n 10.1 Derivations of Polynomial Rings 10.2 Structure of Elementary Antiderivatives 10.3 The Integration Method 10.4 Simple Differential Fields

297 298 301 305 311

References

317

Index

323

Algebraic Prelimmaries

We review in this chapter the basic algebraic structures and algorithms that will be used throughout this book. This chapter is not intended to be a replacement for an introductory course in abstract algebra, and we expect the reader to have already encountered the definitions and fundamental properties of rings, fields and polynomials. We only recall those definitions here and describe some algorithms on polynomials that are not always covered in introductory algebra courses. Since they are well-known algorithms in computer algebra, we do not reprove their correctness here, but give references instead. For a comprehensive introduction to constructive algebra and algebraic algorithms, including more efficient alternatives for computing greatest common divisors of polynomials, we recommend consulting introductory computer algebra textbooks [2, 28, 39, 64, 97]. Readers with some background in algebra can skip this chapter and come back to it later as needed.

1,1 Groups, Rings and Fields An algebraic structure is usually a set together with one or more operations on it, operations that satisfy some computation rules called axioms. In order not to always list all the satisfied axioms for a given structure, short names have been given to the most common structures. Groups, rings and fields are such structures, and we recall their definitions in this section. Definition 1.1.1. A group (G,o) is a nonempty set G, together with an operation o : G X G —^ G satisfying the following axioms: (i) (Associativity) Va, 6, c G G^, a o (6 o c) = (a o 6) o c. (a) (Identity element) 3e G G such that Va€ G^eo a = ao e = a. (Hi) (Inverses) Va € G^ 3a~-^ G G such that a o a~^ — a~^ o a = e. In addition, o is called commutative (or Abelian^ ifaob = boa for all a^b £ G, and (G, o) is called a commutative group (or Abelian groupj if it is a group and o is commutative.

2

1 Algebraic Preliminaries

Example 1.1.1. Let G = GL{Q,2) be the set of all the 2 by 2 matrices with rational number coefficients and nonzero determinant, and let o denote the usual matrix multiplication. (C, o) is then a group: associativity can easily be checked, the identity element is the identity matrix, and the inverse of a matrix in G is given by a b\ c dj

1 f d ad-bc\-c a

which is in G since the determinant of any element of G is nonzero. Note that (G, o) is not a commutative group since 11\

/01\

/ll

^oij°(,ioJ = ^io and

Example 1.1.2. Let G = A^2,2(Q) be the set of all the 2 by 2 matrices with rational number coefficients, and let o denote the usual matrix addition. It can easily be checked that (G, o) is a commutative group with the zero matrix as identity element. Definition 1.1.2. A ring (fi,+,•) ^-^ ^ ^^^ Ft, together with two operations -i- : R X R ^ R and - : R x R -^ R such that: (i) (i?, +) is a commutative group. (ii) (Associativity) \/a^b,c ^ R^a • {h • c) = {a - b) • c. (Hi) (Multiplicative identity) 3i G R such that \/a E R^i • a = a • i = a. (iv) (Distributivity) Va, b, c G R^ a • {b -\- c) = {a • b) + {a • c) and {a -i- b) - c = (a - c) -{- {b • c). (i?, +, •) is called a commutative ring if it is a ring and • is commutative. In addition, we define the characteristic of R to be 0 if ni ^ e for any positive integer n, the smallest positive integer m such that mi = e otherwise. Let R and S be rings. A map (j) : R -^ S is a ring-homomorphism if (i){eR) = es, 4'{'^R) = T^s, and <j){a^b) — (f){a)-j-(j){b) and (t){ab) = (t){a) • (j){b) for any a^b G R. A ring-isomorphism is a bijective ring-homomorphism. In the rest of this book, whenever (fi,+, •) is a ring, we write 0 for the identity element of R with respect to +, 1 for the identity element of R with respect to -, and for a^b E R^ we write ab instead of a • b. Example 1.1.3. Let R = A42,2(Q) be the set of all the 2 by 2 matrices with rational number coefficients, and let + denote matrix addition and • denote matrix multiplication. {R, +, •) is then a ring, but not a commutative ring (see example L L l ) . Since

1.1 Groups, Rings and Fields 0\

(n

0^

0 \)

1

\\)

n

is nonzero for any positive integer n, R has characteristic 0. Example 1.1.4- Let R — IJQ (the integers modulo 6) with + and • being the addition and multiplication of integers modulo 6. (i?, +, •) is then a commutative ring, and the map (/> : Z ^ Ze defined by (/)(n) = n (mod 6) is a ring-homomorphism. Since 1 + 1 + 1 + 1 + 1 + 1 = 0 i n Z 6 , and nl j^ 0 for 0 < n < 6, ZQ has characteristic 6. Note that 2 • 3 = 0 in Ze, while 2 7^ 0 and 3 7^ 0, so we cannot in general deduce from an equation ab = 0 that either a or b must be 0. Commutative rings where we can make this simplification are very useful and common, so they receive a special name. Definition 1.1,3. An integral domain ( ß , + , •) is a commutative ring where 0 7^ 1 and \/a,be R,a-b = 0=^ a = 0 or 5 = 0. Example 1.1.5. Let R = Z[-\/—b] = {a+6^—5; a, 6 E Z} with + and • denoting complex addition and multipHcation. (i?, +, •) is then an integral domain. We now come to the problem of factoring, i.e. writing elements of an integral domain as a product of other elements. Definition 1.1.4. Let (ß,+,•) be an integral domain, and x,y ^ R. We say that X divides y, and write x \ y, if y — xt for some t £ R. An element X £ R is called a unit if x \ 1. The set of all the units of R is written R*. We say that z E R is a greatest common divisor (gcd) of x i , . . . , x^ and write z = gcd{xi,...,Xn) if: (i) z I Xi for 1 < i < n, (ii) yt G R,t \ Xi for 1 < i < n => t \ z. In addition, we say that x and y are coprime if there exists a unit u E R*, which is a gcd of x and y. Example 1.1.6. Let R = Z [A/—5j as in example 1.1.5, x = 6 and y = 2 + 2 \ / ^ . A norm argument shows that x and y have no gcd in R. Let N : R -^ Z be the map given by N{a + 5\A-5) = a^ + 56^ for a, 6 G Z. It can easily be checked that N{uv) = N{u)N{v) for any u^v £ R^ so u \ v in R implies that N{u) I N{v) in Z. Suppose that z E R is a greatest common divisor of x and y, and let n = N{z) > 0. Then, n \ N{x) = 36 and n \ N{y) = 24, so n | 12 in Z. We have 2 | x and 2 | y in fi, so 4 = iV(2) | n in Z. In addition, 1 + v ^ | y in i?, and 6 = 2 • 3 = (1 + V^){1 - V^) (1.1) so 1 + \ / ^ \ x in R, hence 6 = N{1 + \ / ^ ) | n in Z. Thus, 12 | n in Z, so n == 12. Writing z = a + 6 ^ / ^ for some a, 6 G Z, this implies that N{z) = a? -\- 56^ — 12, hence that a^ = 2 (mod 5). But the squares in Z5 are 0,1 and 4, so this equation has no solution, implying that x and y have no gcd in R.

4

1 Algebraic Preliminaries

Although gcd's do not always exist, whenever they exist, they are unique up to multiplication by units. T h e o r e m 1 . 1 . 1 . Let (jR,+,•) be an integral domain, and x^y E R. If z and t are both gcd^s of x and y, then z = ut and t = vz for some u^v G R*. Proof Suppose t h a t both z and t are gcd's of x and y. Then, t \ z since t \ x, t I y, and z — gcd(x,y). Thus, z = ut for some u e R. Similarly, z | t, so t = vz for some v e R. Hence z = ut = uvz^ so (1 — uv)z = 0. If z / 0, t h e n 1 = uv^ so u^v E R*. If z = 0, then t = vz = 0,soz = lt and t = Iz. D D e f i n i t i o n 1.1.5. Let R be an integral domain. A nonzero element p E R\R* is called prime if for any a^b £ R, p\ ab =^ P \ ct or p \ b. A nonzero element p E R\ R* is called irreducible if for any a^b E R, p = ab = ^ a E R"^ or b E R\ Example 1.1.7. Let R = Z [\/---5] as in example 1.1.5, and check t h a t 2, 3,1 + A / ^ and 1 — ^/^ are all irreducible elements of R. Equation (1.1) t h e n shows t h a t the same element can have several different factorizations into irreducibles. Therefore, integral domains where such a factorization is unique receive a special name. D e f i n i t i o n 1.1.6. A unique factorization domain (UFD) ( ß , + , • ) ^-^ «^^ 'integral domain where for any nonzero x E R\ R*, there are u E R*, coprime irreducibles p i , . . . ,_py^ E R and positive integers e i , . . . , e^ such that X = upl^ • • -p^"'. Furthermore, this factorization is unique up to multiplication of u and the pi 's by units and up to permutation of the indices. Example 1.1.8. Let R = Q[X, Y] be t h e set of all the polynomials in t h e variables X and Y and with rational number coefficients. It is a classical result ([54] Chap. V §6, [92] §5.4) t h a t (Ä, + , •) is a unique factorization domain where + and • denote polynomial addition and multiplication respectively. In any integral domain, a prime is always irreducible. The converse is not always true, but it holds in unique factorization domains. Thus, we can use interchangeably "prime" or "irreducible" whenever we are in a unique factorization domain, so, "the prime factorization of x" and "the irreducible factorization of x" have the same meaning. T h e o r e m 1.1.2 ([54] Chap. II §4). Let ( i ? , + , •) be an integral domain. every prime p E R is irreducible. If R is a unique factorization domain, every irreducible p E R is prime.

Then then

In addition, gcd's always exist in U F D ' s , and can be obtained from t h e irreducible factorizations. T h e o r e m 1.1.3. If R is a UFD, then any x^y E R have a gcd in R.

1.1 Groups, Rings and Fields

5

Proof. Let x^y G R^ and suppose first that x = 0. Then y \ y, y \ 0^ and any t e R that divides x and y must divide y, so y is a gcd of x and y. Similarly, a: is a gcd of x and y if y = 0, so suppose now that x ^ 0 and y 7^ 0, and let X = uflpE^P^^ ^^^ y ~ ^YlpeyP^^ ^^ ^^^ irreducible factorizations of X and y, where /¥ and 3^ are finite sets of irreducibles. We choose the units u and V so that any irreducible dividing both x and y is in X oy. Let then

^^ n

p™^""^'"^^^ e R.

(1.2)

We have pexny

peP(:\y

so 2; I X. A similar formula shows that z \ y. Suppose that t \ x and t \ y for some t E R^ and let t = it; H O G T ^ ^ ^ ^^ ^^^ irreducible factorization where T is a finite set of irreducibles. For p G T, we have x = tb = p^^ab for some a, 6 E ß , so sp e ^ for some 5 G i?*. Replacing t(; by t(;5~^p, we can assume that p G Af, and Cp < Up by the unicity of the irreducible factorization. Similarly, we get p G y and e^ < nip since t | y. Hence, T C ^ n 3^ and e^ < min(np, rup) for any p E T. Thus, Z — tw~^

T T pmm(np,mp)-ep pGT

TT

mm(np,mp)

pG{xny)\T

which means that t | z, hence that z = gcd(x, y).

D

It is a classical result due to Gauss that polynomials can be factored uniquely into irreducibles. T h e o r e m 1.1.4 ([54] Chap. V §6, [92] §5.4). If R is a UFD, then the polynomial ring JR[XI, . . . , X^] is a UFD, Definition 1.1.7. Let (G, o) be a group with identity element e. We say that H C G is a subgroup of (G^o) if: (i) eeH. (a) \/a,b e H\aob £ H. (iii)ya G H,a~'^ G H. In practice, given a subset iJ of a group G, it is equivalent to check the above properties (i), (ii) and (iii), or that H is not empty and that aob"^ G H for any a,b e H. Example 1.1.9. Let G = GL{Q^ 2) as in example 1.1.1 with o denoting matrix multiplication, and let H = SL{Q, 2) be the subset of G consisting of all the matrices whose determinant is equal to 1. The identity matrix is in iJ, so i7 is not empty, and for any a,b E H^ the determinant of a o 6~-^ is the quotient of the determinant of a by the determinant of 6, which is 1, so i^ is a subgroup of G.

6

1 Algebraic Preliminaries

D e f i n i t i o n 1.1.8. Let ( Ä , + , • ) be a commutative ring. A subset I of R is called an ideal if (/, + ) is a subgroup of ( ß , + ) and xa £ I for any x in R and a in I. Let x i , . . . , x^ G R. The ideal generated by { x i , . . . , x ^ } is the smallest ideal of R containing { x i , . . . ^Xn}, and is denoted ( x i , . . . , x ^ ) . An ideal I C R is called principal if I = (x) for some x E R. In fact, the ideal generated by { x i , . . . , x „ } is just the set of all the linear combinations of the x^'s with coefficients in R. T h e o r e m 1.1.5. Let ( A , + , • ) be a commutative Then, (Xi, . . . , Xn) = {aiXi

Proof

H

ring, and x i , . . . , x ^ E R.

h ünXn] tti, . . . , « ^ G

R}.

Let / = {aiXi -|- •. • -|- a^Xn^ a i , . . . , a^ G R}. T h e n Xi £ I for any i.

Let a = ^2^=1 ^i^i ^ -^ ^ ^ ^ ^ = S i L i ^*^* ^ -^- ^ ® have a — h = Y^^=i{^i ~ bi)xi G / , so ( / , + ) is a subgroup of ( i ? , + ) . For any x G i?, we have xa = Yl7=ii^^i)^'i G I , so / is an ideal of R containing { x i , . . . , x ^ } . Let now J be any ideal of R containing { x i , . . . , x ^ } , and let a — YM=I ^i^i ^ ^- ^^^ each i, Xi G J , so üiXi G J since RJ C J , so a G J since (J, + ) is a group. Hence / C J , so / = ( x i , . . . , x ^ ) . D {X,Y). Example LLIO. Let ß = Q [ X , y ] as in example L L 8 , and let / = It can be checked t h a t / is not principal, hence t h a t not every ideal of jR is principal. Naturally, this means t h a t integral domains where every ideal is principal receive a special name. D e f i n i t i o n 1.1.9. A principal ideal domain (PID) ( ß , + , •) ^5 an integral domain where any ideal is principal. Example l.Lll. Let R = Q[X] be the set of all the univariate polynomials in X with rational number coefficients. (A, + , •) is then a principal ideal domain ([54] Chap. V §4, [92] §3.7) where + and • denote polynomial addition and multiplication respectively. The last, and most useful, type of ring t h a t we use in this book, is an integral domain in which Euclidean division can be carried out. D e f i n i t i o n 1.1.10. A Euclidean domain ( A , + , • ) is an integral domain gether with a map v : R\ {0} —> N such that:

to-

(i) Ma.beR \ {0}, u{ab) > u{a). (a) (Euclidean division) For any a^b G R, b j^ 0, there are q^r £ R such that a = bq + r and either r = 0 or ^{r) < y{b). The map u is called the size function of the Euclidean

domain.

Example LL12. T h e ring (Z, + , •) of the integers with the usual addition and multiphcation is a Euclidean domain with z/(a) = |a|, a fact t h a t was known to Euchd, and which is the origin of the name.

1.1 Groups, Rings and Fields

7

Even though the notions of principal ideal domains and EucHdean domains are defined for an arbitrary integral domain, there is in fact a linear hierarchy of integral domains. T h e o r e m 1.1.6 ([92] §3.7). Every Euclidean domain is a PID. T h e o r e m 1.1.7 ([54] Chap. II §4, [92] §3.8). Every PID is a UFD. Since every PID is a UFD, and gcd's always exist in UFD's by Theorem 1.1.3, then gcd's always exist in PID's. We show that in PID's, the gcd of two elements generates the same ideal than them. T h e o r e m 1.1.8. If R is a PID, then (x^y) = (gcd(a:,y)) for any x^y G R. Proof Let x^y e R and z £ R he a generator of the ideal (x,y), i.e. (z) = (x^y). Then, x G (2), so x = zu for some u G R^ which means that z \ x. Similarly, y G {z), so z \ y. In addition, z G {x,y)^ so z = ax -{- by for some a^b G R. Let t G R he such that t \ x and t \ y. Then x = ct and y = dt for some c^d G R. Hence, z = act + bdt = {ac + bd)t so t | z, which implies that z — gcd(ii,'u). D We finally recall some important definitions and results about fields. Definition 1.1.11. A field (F, H-, •) is a commutative ring where {F \ {0},-) is a group, i.e. every nonzero element is a unit (F* = F \ {0}). Example 1.1.13. Let F = Z5 (the integers modulo 5) with + and • being the addition and multiplication of integers modulo 5. (F, +, •) is then a field. Example 1.1.14- Let R he an integral domain and define the relation ~ on R X R\ {0} by (a, b) ~ (c, d) if ad = be. It can easily be checked that ~ is an equivalence relation on R x R\{0} and that the set of equivalence classes is a field with the usual operations a b

c d

ad-^bc bd

a c bd

ac bd

where a/b denotes the equivalence class of (a, 6). This field is called the quotient field of R. For example, the quotient field of Z is Q and the quotient field of the polynomial ring D[x] is the rational function field D{x) when D is an integral domain. Definition 1.1.12. Let F C E be fields. An element a G E is called algebraic over F if p{a) = 0 for some nonzero polynomial p G F[X], transcendental over F otherwise. E is called an algebraic extension of F if all the elements of E are algebraic over F. Definition 1.1.13. A field F is called algebraically closed if for every polynomial p G F[X]\F there exists a G F such thatp{a) = 0. ^ field E is called an algebraic closure of F if E is an algebraically closed algebraic extension of F.

8

1 Algebraic Preliminaries

Note t h a t if JP is algebraically closed, then any p G K[X]\K factors linearly as p = c n r = i ( ^ ~ ^'^T'' ^^^^ -^' P ™ust have one root a in i^ by definition, and p/{X — a) factors linearly over F by induction. The fundamental result about algebraic closures is a result of E. Steinitz which states t h a t they exist and are essentially unique. T h e o r e m 1.1.9 ([54] Chap. VII §2, [92] §10.1). Every field F has an algebraic closure, and any two algebraic closures of F are isomorphic. In view of the above theorem, we can refer to the algebraic closure of a field F , and we denote it F. T h e last result we mention in this section is Hilbert's Nullstellensatz, which is not needed in the algorithm, but is needed in order t o eliminate the possibility of new transcendental constants appearing in antiderivatives. We present it here in b o t h its classical forms. T h e o r e m 1.1.10 ( W e a k N u l l s t e l l e n s a t z , [92] §16.5). Let F be an braically closed field, I an ideal of the polynomial ring F[Xi^..., X^] and V{I) be the subset of F'^ given by V{I) = { ( x i , . . . ,Xn) G F " s.t p{xu . . . , x , ) = 0 / o r allpe Then, V{I) = (D 4=^

1} .

(1.3)

l e L

T h e o r e m 1.1.11 ( N u l l s t e l l e n s a t z , [54] Chap. X §2, [92] §16.5). Let F be an algebraically closed field, I an ideal of the polynomial ring F[Xi^..., Xn] and V{I) be given by (1.3). For any p € F[Xi^... ,X^]^ if p(xi^... ,Xn) = 0 for every ( x i , . . . , Xn) € y{I), then p^ £ I for some integer m > 0.

1.2 Euclidean Division and Pseudo-Division Let K be a field and x be an indeterminate over K. We first describe t h e classical polynomial division algorithm ([92] §3.4), which, given A^B G K[x]^ B j^ 0^ produces unique Q,R G K[x] such t h a t A = BQ + R and either R = 0 or deg{R) < d e g ( ß ) . This shows t h a t t h e polynomial ring K[x] is a Euclidean domain with t h e degree for size function when K is field. Q and R are called the quotient of A by B , and t h e remainder of A modulo B respectively. PolyDivide(A, B)

(* Euclidean Polynomial Division *)

(* Given a field K and A,B e K[x] with B j^ 0, return Q Re K[x] such that A = BQ + R and either R = 0 or deg{R) < deg{B). *) Q i~~0, R<-A while Ä 7^ 0 and (5 ^ deg(i^) - deg{B) > 0 d o

T ^ |^x^ Tetnrn{Q,R)

Q^Qi-T,R^R-BT

1.2 Euclidean Division and Pseudo-Division Example 1.2.1. Here is the Euclidean division oi A — 3x*^ -\- x^ + x - 5 by E = 5 r r 2 - 3 j : + l in Q[x]\

0 ^x 5*^ ^ 25

(5 T

Ä

Q

3x^ + x^ + x + 5 1

fx^ + fx + S

0

52^ 1 i n 25"^ ^ 25

-1

gX 14 25

Thus, /52

--•1-M

111

This algorithm requires t h e coefficients to be from a field because it makes the quotient in K of the two leading coefficients. If K is an integral domain, the leading coefficient of B does not always divide exactly t h e leading coefficient of A, so Euclidean division is not always possible. For example it is not possible in the above example to do a Euclidean division of ^4 by ß in IJ[X\. But it is possible to apply P o l y D i v i d e to 2bA and B in Z[x] since all the divisions in Z will then be exact. In general, given an integral domain D and A,B £ D[x], applying P o l y D i v i d e to h^^^A and B where h = 1C(JB) and Ö — max(—l,deg(yl) — deg(jB)) only generates exact divisions in D , and the Q and R it returns are respectively called the pseudo-quotient of A by B and pseudo-remainder of A modulo B. They satisfy b^'^^A = BQ 4- R and either i? = 0 or deg(fi) < deg(-B). We write pquo{A,B) and p r e m ( ^ , ß ) for the pseudo-quotient and pseudo-remainder of A and B. It is more efficient in practice to multiply A hy b iteratively, as is done in the algorithm below, rather t h a n once by 6^+^.

PolyPseudoDivide(74, B)

(* Euclidean Polynomial Pseudo-Division *)

(* Given an integral domain D and A, B G D[x] with B ^ 0, return pquo(A,jB) and pYem.{A,B). *) b ^ lc(ß), N ^ deg(A) - deg{B) + l,Q^O, R^A while Ä / 0 and (5 ^ deg{R) - deg{B) > 0 do T ^ lc{R)x\ N ^N-l,Q<-bQ + T,R^bR-TB retnrn(b^ Q,b^R)

Example

1.2.2. W i t h A and B as in example 1.2.1, we get 6 = 5, A^ = 2, and

Q 0 3x 15^ + 14

R

Ö T N

3x^ + x^ + X + 5 1 3x 1 14x2 -f 2x + 25 0 14 0 52X + 111

so 25^1 = 5 ( 1 5 x + 14) + (52x + 111).

-1

10

1 Algebraic Preliminaries

1.3 The Euclidean Algorithm Let D be a Euclidean domain and u : D \ {0} —> N its size function. The Euclidean division in D can be used to compute the greatest common divisor of any two elements of D. The basic idea, which goes back to Euclid who used it to compute the gcd of two integers, is that if a = 6g + r, then gcd(a, b) = gcd(6,r). Since gcd(j:, 0) = x for any x G -D, the last nonzero element in the sequence (a^)^>o defined by ao

ai = 6,

and

{qi-,cii) = EuclideanDivision(ai_2 7 G^2-I) for 2 > 2

is then a gcd of a and h. Since for a^ ^ 0 and i > 1, either a^+i = 0 or v[aij^\) < z/(a^), that sequence can only have a finite number of nonzero elements. This yields an algorithm for computing gcd(a, b) by repeated Euclidean divisions. Euclidean(a, b)

(* EucHdean algorithm *)

(* Given a Euclidean domain D and a, 6 G D, return gcd (a ,6). *) while 6 7^ 0 d o (g,r) ^— EuclideanDivision(a, 6) a ^~ b

(*

a =-bq-{- r

*)

b <— r

return a

Example 1.3.1. Applying the Euclidean algorithm to a = x'^ -2x^

~ 6x^ + I2x + 15

and

x^ + x^ ~ 4x - 4

in D = Q[x] gives: a

b

^

r

x^ - 2x^ - 6x2 _^ 22x + 15 x^ -4- x^ - 4x - 4 X — 3 x^ -4- 4x + 3 x^ + x^ - 4x - 4 x^ + 4x + 3 X—3 5x + 5 x^ + 4x + 3 5x-h5 0 k +i 5x-f 5 0 so 5x + 5 is a gcd of a and b in Q[x]. The Euclidean algorithm can be easily extended to return, not only a gcd of a and 6, but also elements s and HVLD such that sa^tb = gcd(a, 6). Such elements always exist since gcd (a, 6) belongs to the ideal generated by a and 6 by Theorem 1.1.8.

1.3 The Euclidean Algorithm ExtendedEuclidean(a,6)

11

(* Extended Euclidean algorithm *)

(* Given a EucHdean domain D and a,b E D, return s,t,g G D such that g — gcd(a, b) and sa -\- tb — g. *) ai 4— 1, a2 ^- 0, 5i ^- 0, 62 ^ 1 while 6 y^ 0 d o (g,r)^— EuclideanDivision(a, 6) a <— 6, 6 ^— r n ^ ai - qbi, r2 ^ a2 - qb2 ai ^— 61, a2 ^— 62, ^1 ^ r i , 62 ^~ r2 r e t u r n ( a i , a 2 , a)

Example

(* a = 5g + r *)

1.3.2. Using the same a and b as in example 1.3.1:

6

a x 4 - - 2x3 __ 5^2 ^ ^2x + 15x^

x^ + x^ - 4x - 4 x^ + 4x + 3 5x + 5

+ x^ -- 4x - 4 X — 3 x^ + 4x + 3 5x + 5 x^ + 4x + 3 X—3 0 5x + 5 i- + i 0

ai

a2

bi

b2

1

0

1

0 1

1

0 1 -x + 3

-x + 3 - x + 3 x2 - 6x + 10

r

Q

1^2 5-^

4 5

-~-x + 3 x^ - 6x + 10 -|x"'^ + | x 2 + | x -- 3

Thus, 5x + 5 is a gcd of a and 6 in Q[x], and ( - X + 3)a + (x^ - 6x + 10)6 = 5x + 5 .

(L4)

If only one of the coefficients 5 or t is needed, a variant of the extended Euclidean algorithm t h a t computes only t h a t coefficient can be used: H a l f E x t e n d e d E u c l i d e a n ( a , b)

(* Half extended Euchdean algorithm *)

(* Given a Euclidean domain D and a b€D, g = gcd(a, 6) and sa = g (mod b). *) ai -^- 1, 61 '^- 0 while 6 7^ 0 d o (g,r) ^ EuclideanDivision(a b) n ^

ai - qbi, ai <— 61, 61 <- ri

return(ai,a)

return 5 g £ D such that

(^^ a = bq -\- r ^)

12

1 Algebraic Preliminaries

This "half" variant of the algorithm is also used as a more efficient alternative to the extended Euclidean algorithm, since the second coefficient can be obtained from the first via a — sa where the division is always exact. ExtendedEuclidean(a, h) (* Extended Euclidean algorithm - "half/full" version *) (* Given a Euclidean domain D and a, 6 G D, return s, t, p G D such that g = gcd(a, h) and sa^th = g. *) {s,g) <— HalfExtendedEucIidean(a,6) (t, r) ^- EuclideanDivision(p — sa, b) return{s, t, g)

(^ sa = g (mod b) *) (* r must be 0 *)

Example 1.3.3. Recomputing the extended gcd of the a and b of example 1.3.1, we get: 1. (5, g) •=• HalfExtendedEuclidean(a, 6) = (—x + 3,5a: + 5) 2. g-sa = x^ - 5x^ + 30^^ - 16x 3. (t, r) = PolyDivide(^ - sa, b) - (x^ - 6x -|-10,0) so we recover (1.4). The extended Euclidean algorithm can also be used to solve the diophantine equation sa-^tb = c (1.5) where a^b^c E D are given and s^t E D are the unknowns. For (1.5) to have a solution, it is necessary and sufficient that c be in the ideal generated by a and 6, i.e. that c be a multiple of gcd(a, 6) in D. The extended Euclidean algorithm first solves the equation sa-i-tb — gcd (a, 6), and there remains only to multiply the solutions by c/gcd(a,6) to get a solution of (1.5). It should be noted that when c is in the ideal generated by a and 5, then (1.5) has as many solutions as the number of elements of D (when a and b are nonzero), since sa + tb = (s -i- bd)a + {t — ad)b for any d E D. Since there can be no confusion with the previous extended Euclidean algorithm, which has only two parameters, we also call this algorithm the "extended Euclidean algorithm". As before, a half-extended version exists when only one of the coefficients is needed. We remark that the versions of the algorithm that we present here, and use extensively in the sequel, all return a solution s or (s^t) such that either 5 = 0 or ^{s) < z/(6). An important consequence of this in polynomial rings (where u{p) = deg(p)) is that if deg(c) < deg(a) + deg(6), then we also

1.3 The Euclidean Algorithm

13

get either t = 0 or deg(t) < deg(a). Indeed, if we had (leg(s) < deg(6) and deg(t) > deg(a), then we would have deg(c) = deg(5a + tb) = deg{tb) = deg(t) + deg(6) > deg(a) + deg(6).

E x t e n d e d E u c l i d e a n ( a , 6, c) (* Extended Euclidean algorithm - diophantine version *) (* Given a Euclidean domain D and a,b,c e D with c e {a ,b), return s,t e D such that sa + tb = c and either s =-- 0 or ^{s) < v{b).*) {s,t,g)^E x t e n d e d E i i c l i d e a n ( a , 6) (^ g = sa + tb *) (* c == gq+ r *) (g,r)<— EuclideanDivision(c, p) if r / 0 t h e n e r r o r "c is not in the ideal g<9nerated by a and 5" s '(^ qs, t ^^ qt if s 7^ 0 and i/(s) > iy(b) t h e n {q,r) ^— EuclideanDivision(5,6) (* s -=• bq + r*) s <— r, t -i^ t -\- qa r e t u r n ( s , t)

Example 1.3.4- Suppose that we want to solve sa -j- tb = x"^ — 1 in Q[x] with the a and b of example 1.3.1. Applying E x t e n d e d E u c l i d e a n we get: 1. (5, t, g) = E x t e n d e d E u c l i d e a n ( a , b) — {~x + 3, :r^ — 6a: + 10, 5x + 5) 2. (g, r) = PolyDivide(x2 - 1, 5x + 5) = {{x ~ l)/5,0) 3. s^qs = {-x^ -h 4x - 3)/5 4.t^qt = {x^ - Ix^ H- 16x - 10)/5 So we get the following solution: a +

—

—

6 = x^ - 1.

H a l f E x t e n d e d E u c l i d e a n ( a , 6, c) (* Half extended Euclidean algorithm - diophantine version *) (* Given a Euclidean domain D and a^b^c G D with cI G (a,^), return s E D such that sa = c (mod 6) and either s — 0 or i/(s) < uib). *) (* sa (s,g)^ H a l f E x t e n d e d E u c l i d e a n ( a , 6) (g,r) ^ EuclideanDivision(c, p) if r 7^ 0 t h e n e r r o r "c is not in the ideal generated by 5 <— gs if s 7^ 0 and z/(s) > 1/(6) t h e n {q,r)
= ^f (mod b) *) (* c = pg + r a and 6"

(* s = 6g + r *)

(1.6)

14

1 Algebraic Preliminaries

As earlier, the "half" variant yields a more efficient alternative to the extended diophantine version, since the second coefficient can be obtained via c — sa where the division is always exact. E x t e n d e d E i i c l i d e a n ( a , 6, c) (* Extended Euclidean algorithm - "half/full" diophantine version *) (* Given a Euclidean domain D and a,b,c G D with c e (a ,b), return s,t e D such that sa + tb = c and either 5 =: 0 or 1/(5) < y{h). *) s ^- H a l f E x t e n d e d E u c l i d e a n ( a , 6, c) (t, r) ^- EuclideanDivision(c — sa, h) r e t u r n ( 5 , t)

(* sa E=

C (mc d5) *) (* r must be 0 *)

Example 1.3.5. Solving sa ^ th = x^ — 1 in Q[x] with the a and h of example 1.3.1, we get 1. s - HalfExtendedEuclidean(a, h, rr^ - 1) = (-x^ -f 4x - 3)/5 2. c - 5a = ^2 - 1 - sa = {x^ - 6x^ + 5x^ + 30x^ - 46^^ - 24x -f 40)/5 3. (t, r) = PolyDivide(c - sa, h) = {{x^ - T:^^ + 16x - 10)/5,0) so we recover (1.6). Since the extended Euclidean algorithm can be used to solve diophantine equations, it is also useful for computing partial fraction decompositions. Let d G D\ {0} and let d = di - •• dn be any factorization of d (not necessarily into irreducibles) where gcd((i^, dj) = I for i ^ j . Then, for any a e D \ {0}, there are ao, a i , . . . , a^ in D such that either a^ = 0 or z/(a^) < iy{di) for i > 1, and a

a

n ir-^ üi

Such a decomposition is called, the partial fraction decomposition of a/d with respect to the factorization d — HlLi ^^' ^^^ computing it reduces to solving equations of the form (1.5), so to the extended Euclidean algorithm. Indeed, write first a = (iao + r by the Euchdean division, where either r = 0 or y(r) < u{d). If n = 1, then a/d = ao + r/d is already in the desired form. Otherwise, since gcd{di^dj) = 1 for i 7^ j , we have gcd{di^d2 • • • d^) = 1, so by the extended Euclidean algorithm, we can find ai and 6 in Ü such that r = ai (^2 • • • dn) + bdi

(1.7)

1.3 The Euclidean Algorithm

15

and either a i = 0 or z^(ai) < u(di). We can recursively find 5o, a 2 , . . . , a^ € D such t h a t either a^ = 0 or i/(a^) < ^{di)^ and

;r-V = ^° + E7d2--'dn

^ d i

Dividing (1.7) by d and adding ao, we get a

r

ai

h

/

T

x

2 = 1 ^i \~^

aoH--:}=ao + "1 J rings, - = («0since + oo) deg(r) + > ^ -7We note t •:7 h a= t in the case + of -Tpolynomial < •deg((i) = a a d i 0(2 • • • ttn ~ ^ ^2 deg((ii) + deg{d2 •' • dn) and deg(ai) < deg{di) in (1.7), then deg(6) < deg(ii2 • --c^n), so 60 = 0. P a r t i a l F r a c t i o n ( a , «ii,... ,dn)

(* Partial fraction decomposition *)

(* Given a Euclidean domain D, a positive integer n and a,di,... ,dn G D \ {0} with gcd{di^dj) = 1 for i 7^ j , return ao, a i , . . . , an G -D such that

di • • -dn

+ES

• ao

. 1 --

2= 1

and either at = 0 or z^(ai) < iy{di) for i > 1. *) (ao, r) <— EucIideanDivisioii(a, di • • - dn) {^ a = (di • • • dn)ao + r *) if n = 1 t h e e r e t u r n ( a o , r ) (ai,t) ^- ExtendedEoclidean(cl2 • • •c^njC^i,'^) (* z^(ai) < i^{di) *) (5o, a 2 , . . . , ttn) ^- P a r t i a l F r a c t i o n ( t , c?2, • • •,
Example

1.3.6. We compute the partial fraction decomposition of „ a

x^ + 3x

with respect t o the factorization d — [x ^ l)(x^ — 2x -{- 1) = did2- Applying P a r t i a l P r a c t i o n to a, di and
1 3x + r 2'

2

16

1 Algebraic Preliminaries

3. {bo, as) = PartialPraction((3x -f l ) / 2 , x^ -2x + l) = (0, (3x + l)/2) so the partial fraction decomposition of / is x^ + 3x

__ - 1 / 2 X+ 1

(3x + l)/2 ^2 - 2x + 1

We can combine this with the Euclidean division to get a refinement of the partial fraction decomposition: let ?7i > 1 and d £ D \ {0}. Then, for any a E D \ {0}, there are a o , a i , . . . , am ^ D such that either aj = 0 or i^{aj) < v{d) for j > 1, and 771

3=1

Such a decomposition is called the d-adic expansion of a/d^. Write a = dq-\~ am by the Euclidean division, where either a^ = 0 or i'(am) < y{d). Then, a _ dq^ um _ ^m ~

ßrn

q

~ ßrn-l

o^, ßrn "

If ?Ti = 1, then the above is in the desired form with ÜQ — q. Otherwise, we recursively find a o , a i , . . . ,am~\ € D such that either a^ = 0 or y(aj) < u{d) for J > 1, and Q

m—l

^0 + E l 3=1

Thus a

q

, am

. sr^ aj

Let now d G D \ {0} and let d = d^^ • - • d^- be any factorization of (i, not necessarily into irreducibles, where gcd{di^dj) = 1 for i 7^ j , and the e^'s are positive integers. Then, for any a E D\ {0}, we can first compute the partial fraction decomposition of a/d with respect to d = bi •• - bn where bi = df'-: a

n IT—^ a^

n Y A ai

and then compute the (i^-adic expansion of each summand to get a

--+Eti^

where ä £ D and either aij = 0 or ^{aij) < ^{di) for each i and j . This decomposition is called the complete partial fraction decomposition of a/d with respect to the factorization d = YTi=i ^T ^ or simply the complete partial

1.3 The Euclidean Algorithm

17

fraction decomposition of a/d when the factorization of d into irreducibles^ is used. PartialPraction(a,c?i,... ,(in, e i , . . . , en) (* Complete partial fraction decomposition *) (* Given a Euclidean domain D, positive integ ers Ti, e i , . . . , e^ and a,
• • •

dn

and either aij = 0 or i^{aij) < u(di). *) ( a o , a i , . . . ,an) <— PartialPractioii(a,
Example 1.3.7. We compute the complete partial fraction fraction decomposition of a x^ + 3a: [x) x"^ — x^ — X -\-l d with respect to the factorization d = {x -\- l){x ~ 1)'^ = did^. Applying Par™ t i a l F r a c t i o n to a, d i , ^2, and the exponents 1 and 2, we get: 1 3x 4-1 ( a o , a i , . . . ,an) = P a r t i a i P r a c t i o n ( x ^ + 3 x , x + l, ( x - 1 ) ^ ) = (0, —-, — - — ) and then:

i J

ßi

di

q

aij

ao

1 1 -1/2 x + 1 0 -1/2 0 2 2 (3x + l)/2 x - 1 3/2 2 0 0 3/2 0 21 3/2 x-l so the complete partial fraction decomposition of / is

x^ + Sx • x^ — X + 1

-1/2

x -\-l +

3/2

[x — 1)

4- x - l

We show in Sect. 2.7 how to compute that decomposition for linear factors without factoring d.

18

1 Algebraic Preliminaries

The algorithm for computing partial fraction decompositions t h a t we presented here dates back to Hermite in the 19*^ century. There are alternative and faster approaches for rational functions t h a t we do not detail here. See [1, 45] for other approaches and their complexities.

1.4 Resultants and Subresultants We describe in this section the fundamental properties of the resultant of two polynomials. Although they originate from IQ^'^-century work on solving systems of nonlinear equations, resultants play a crucial role in integration. Throughout this section, let i? be a commutative ring and x be an indeterminate over R. h aix + ao D e f i n i t i o n 1 . 4 . 1 . Let A,B G R[x] \ {0}. Write A = anx"^ H and B = b^x^ -j- . . . -f. hix + bo where a^ ^ 0, bm j^ 0 and at least one of n or m is nonzero. The Sylvester matrix of A and B is the n + m by n-\-m matrix defined by I On

ai

ao

\ m rows ai

6i

ao

bo

S{A,B) } n rows

bo

\

J)

where the A-rows are repeated m times and the B~rows are repeated n The resultant of A and B is the determinant of S{A^B).

times.

Example l.J^.l. Let i? = Z[t], A = S t x ^ - t ^ - ~ 4 € R[x], and ß = x^~\-t^x-~9 R[x]. The Sylvester matrix of A and B is

S{A,B)--

3t 0 1 0

0 3t t^ 1

-t^ - 4 0 -9 t^

0

\

0 -9

/

G

and the resultant of A and B is det{S{A,

B)) = - 3 t ^ ° - 12t^ + t^ - 54t^ + 8t^ + 729^^ - 216t + 16 .

The first useful property of the resultant of two polynomials is t h a t it can be expressed in terms of their roots.

1.4 Resultants and Subresultants

19

T h e o r e m 1.4.1 ([54] Chap. V §10, [92] §5.9). Let ai,... ,an, ßi,...,ßm, a and b be in R with a j^ 0, b j^ 0, A = a{x — ai) - • • [x — an) and B ~ b{x~ßi)---{x-ßm)Then, n

m

n

resultant(^,ß) = a'"6" [ ] [ ] ( a i - / 3 , ) = a " ] j 5 ( a i ) m

= (-1)"^6^ ]J^(/?,-) = ( - l ) ^ ^ r e s u l t a n t ( ß , ^ ) . j=i

As a consequence, the resultant of two polynomials over an integral domain i^ is 0 if and only if they have a common zero in the algebraic closure of the quotient field of R. Corollary 1.4.1 ([54] Chap. V §10, [92] ^5.8). Suppose that R is an integral domain. Let K be the quotient field of R and K the algebraic closure of K Then, for any A,B £ R[x] \ {0}, resultant(A, B) = 0 4=> 3j eT{ such that A{j) = ^(7) = 0. Proof Let A,B e R[x] \ {0}, and let n

A = a Y[{x - aiY'

m

and

B = b Y[{x - ßjY^

be the prime factorizations of A and B in K[x]. We have a 7^ 0 7^ 6 since A^^^B^ so by Theorem 1.4.1 we get n

m

C = resultant(^,B) = a^b^ J ] JJiai - ßjY'^'^^ where M = X]jLi fj ^^^ ^ — Yl7=i ^*- Since ÜT is a field, if C = 0 then a^Q — ßjQ = 0 for some io and jo- But then A(7) = ^ ( 7 ) = 0 where 7 == a^Q = ßjQ G K. Conversely, if ^(7) = B{j) = 0 for some j E K^ then, since ÜT is a field, 7 =: a^Q = ßjQ for some io and jo, so a^^ — ßj^ = 0, so C = 0. D Another property is that the resultant of two polynomials is in the ideal that they generate. T h e o r e m 1.4.2 ([54] Chap. V §10, [92] §5.8). For any A,B there are 5, T E R[x] such that resultant (A, B) — SA + TB.

e R[x] \ {0},

As a consequence, the resultant of two polynomials over a unique factorization domain is 0 if and only if they have a non-trivial common factor. Corollary 1.4.2 ([92] §5.8). Suppose that R is a unique factorization domain. Then, for any A,B e R[x] \ {0}, resultant(^, E) = 0 ^=^ deg(gcd(A, ß ) ) > 0.

20

1 Algebraic Preliminaries

Subresultants are polynomials obtained from submatrices of the Sylvester matrix. Definition 1.4.2. Let A,B e R[x] \ {0}, n = deg{A),m = deg{B), S be the Sylvester matrix of A and B, and j he an integer such that 0 < j < min(n, m). Let jS be the n + ?n — 2 j by n^m matrix obtained by deleting from S: (i) rows ?7i — j + 1 to m (i.e. the last j rows corresponding to A), (ii) rows ?Ti + n — j + 1 to m -\- n (i.e. the last j rows corresponding to B). Furthermore, for 0 < i < j , let jSi be the square matrix obtained by deleting columns m + n — 2j to m -\- n (i.e. the last 2j + 1 columns) of jS except for column m + n — i — j . The f^ subresultant of A and B is then Sj{A,B)

G R[x].

= Y^det{^Si)x' i=0

It is clear from the definition that deg{Sj{A^ B)) < j for each j . Following the standard terminology [60], we call Sj{A^B) defective if deg{Sj{A^ B)) < j , regular otherwise. In addition, QSQ = 5, so So{A^B) = resultant(^, ß ) . Example 1.4.2. Let A = x^ + 1 and B = x'^ — 1 in Z[a:]. The Sylvester matrix of A and B is 1 0 1 0

S{AB)=\ \ I \ .0

1

^

0

and the submatrices of Definition 1.4.2 are QS = QSQ = S{A^B)^

i^=(l

0 -1 o j '

'^°=[l

-l)

""""^ 1^1 = ( l

0

so the subresultants of A and B are So = det(o5'o) = 4 = resultant (A, B) and Si = detd^So) + det(i5'i)x = —2, which is defective. Another useful property of subresultants is that they commute with ring homomorphisms when the degrees do not decrease, and that they specialize in a predictable way when only one degree decreases: any ring homomorphism a : R ^ S induces the homomorphism of polynomial rings a : R[x] —> S[x] given by

The following theorem describes how Sj{ä{A)^ä{B)) can be computed from Sj{A^B)^ when at least one of the leading coefficients of A or ß is not taken to 0 by a.

1.5 Polynomial Remainder Sequences

21

T h e o r e m 1.4.3 ([64] §7.8). Let a : R -^ S be a ring homomorphism, ä : R[x] -> S[x] be given by (1.8), and A,B £ R[x] \ {0}. If deg{ä{A)) = deg{A) then for 0 < j < m.m{deg{A)^ deg{W{B))). Note in particular that ä{Sj{A^B)) = Sj{'ä{A),'ä{B)) when either A or B is monic, or when deg{cr{A)) = deg(^) and deg{W{B)) = deg(B). Theorem 1.4.3 will be used for specialization homomorphisms, when R is of the form R = D[ti^... ,tn] where the t^'s are independent indeterminates, 5' is a ring containing D, « i , . . . , a^ are given elements of S^ and a : R -^ S is the ring homomorphism that is the identity on D and that takes each ti to ai. In this case, Theorem 1.4.3 states that under certain circumstances, evaluating a subresultant for given values of the parameters ti yields the corresponding subresultant of the two initial polynomials evaluated with the same values. Example 14.3. Let A = Mx^ - t^ - 4 and B = rc^ + t^x - 9 in Z[t][x]. The Sylvester matrix of A and B is / 3t 0 1

S{A,B)

0 3t t^

-t'^ 0 -9

V o l

t^

and the submatrices of Definition 1.4.2 are QS = o'S'o = 5(A,B), 3t

0

-t2-4

1

t^

-9

0\

^

/3t

O J ' ^^°= Vl

-^3-4\

-^9

/3t

0

j ^ndi^i^^^

, ^

^3

so the subresultants of A and ß are 5 o ( ^ , ß ) = resultant;^ ( A ß ) = det(o6'o) = = -3t^^ - 12t'^ + 1 ^ - 54t^ + 8t^ + 729^2 - 216t + 16 , Si{A,B) = det{iSi)x + det(i5o) - 3t^x + t^ - 27t + 4. Consider now the evaluation map t -^ 1, i.e. the homomorphism cr : Z[t] —> Z given by a{t) = 1 and a{n) = n for n G Z. We have ^{A) = 3x^ — 5, and ä{B) = x^ + X -- 9, so Theorem 1.4.3 implies that So{ä{A),ä{B)) Si{-ä{A),ä{B))

=resultant:,(3x2-5,x2+x-9) =ä(5o(A,ß)) =469, = ä(3t^x + t^ - 27t + 4) = 3x - 22 .

1.5 Polynomial Remainder Sequences We now introduce polynomial remainder sequences, which are generalizations of the Euclidean algorithm for computing gcd's and resultants. Let D be an integral domain and x be an indeterminate over D throughout this section.

22

1 Algebraic Preliminaries

Definition 1.5.1. Let A,B e D[x] with B j^ 0 and deg{A) > deg{B). A Polynomial Remainder Sequence (PRS) for A and B is a sequence {Ri)i>o in D[x] satisfying (i) Ro =A, Ri= (a) For i > 1;

B,

_rO P^^^+l

j pTem{Ri^uRi)

zfR,=0 ifRi^O

where {ßi)i>i is a sequence of nonzero elements of D. It is clear from the definition that either Ri-j-i = 0 or deg{Ri^i) for z > 1, hence,

< deg{Ri)

(i) A PRS has finitely many non-zero elements. (ii) If Ri / 0, Rj 7^ 0, deg{Ri) = deg(Äj) and i, j > 1, then i = j {i.e. only Ro and Ri may have the same degree). Definition 1.5.2. Let A,Be such that a A = bB.

D[x]. A is similar to B if there are a^b e D\{0}

From the definition of a PRS, we see that various choices for the ßi^s yield different types of PRS. For example, the PRS obtained with ßi = 1 is just the sequence of the successive pseudo-remainders of A and B, and is called the Euclidean PRS of A and B. The PRS obtained with ßi set to the gcd in D of the coefiicients of prem(i?i_i, i?i) is called the primitive PRS of A and B. An important fact is that if D is a unique factorization domain, then the last nonzero element of a PRS is similar to a gcd of A and B. T h e o r e m 1.5.1. Suppose A,B e D[x] with B ^ 0 be any PRS of A and B gcd{Rj^ Rj^i) for 0 < i^j to gcd{A,B).

that D is a unique factorization domain, and let and deg{A) > deg{B). Let (Äo,i?i, •. • ,fi/c,0,...) with Rk j^ 0. Then gcd{Ri, Ri^i) is similar to < k. In particular (i = 0, j = k), Rk is similar

Proof Let i be such that 0 < i < k^ G = gcd{Ri^Ri^i) and H = gcd{Ri^i,Ri_^2)' Since i < k^ Ri-}-i 7^ 0, so, from the definitions of a PRS and of a pseudo-remainder, there are a^ß e D\ {0} and Q e D[x] such that aRi = Ri-\-iQ + ßRii-2 • Hence H \ aRi^ but H \ aRi^i so H \ aG since aG is a gcd of aRi and aRi^i. From the above equation we also get G \ ßRi^2- But G \ ßRi-^i so G I ßH. So there are Qi,Q2 € D[x] such that aG = HQi and ßH = GQ2. This implies that aßG = GQ1Q2, hence that Qi,Q2 G D^ so G is similar to H. Thus, the theorem holds for j = i + l. Since similarity is transitive, it holds for 0 < i < J < Ä:. Since similarity is symmetric, it holds for 0 < i ^^^ j < A;. It is trivial for i = j , so it holds for 0 < i^j < k. D

1.5 Polynomial Remainder Sequences

23

Thus, any PRS of ^ and B contains gcd{A^ B). In addition, all the nonzero subresultants of A and B are similar to some element in the PRS. The following fundamental theorem of PRS gives explicit formulas for the similarity coefficients. T h e o r e m 1.5.2 (Fundamental P R S Theorem,[39] Chap. 7,[60]). Let A and B j^ 0 be in D[x] with deg(^) > deg(jB)^ and let (i?o, fii,..., Rk^ 0^ • • •) be any PRS of A and B with Rk -=/=- 0. For i = 1^... ,k, let Ui — deg{Ri) and ri be the leading coejficient of Ri. Then, for any j m { 0 , . . . , deg{B) — 1}^ Sj{A,B)

= { TiRi if j ^Ui 0 otherwise

where X-^-nj—rii-^x

n

nr = (-i)^vLT-+-

-l)^'rf'-'-"'-i

n

nj_i-nj +i .l+nj-i-rij

A

nj„i-nj4-i

.l+^i-i-^j

i-l

(1.9)

i-l

^i == ^ ( n ^ - - n i _ - i + l ) ( n j _ i - n i _ _ i + l ) ,

ui = ^{nj-i~ni){nj-ni).

3=1

(1.10)

j=l

The Subresultant PRS of A and B is a particular PRS, introduced by Collins [23] and Brown [16], for which r/^ = 1 in Theorem 1.5.2. It is obtained with the following recursion for ßf. Ro^A,

Ri = B,

71 = - 1 ,

A = ( - l \<5i +

l

and

for z > 1 where 6i = deg(i?i_i) — deg{Ri). Its key property is given by the following theorem. T h e o r e m 1.5.3 ([16] §7, [23, 60]). Let A and B be in D[x] with deg{A) > deg{B), {RQ, i?i, i?27 • • •, ß/c, 0, • • •) be the subresultant PRS of A and B with Rk 7^ 0; and ui = deg{Ri) for i = 1 , . . . , /c. Then, Vj G {0,... ,deg(B) - 1}, where TI is given by formula (1.9).

Sj{A,B)

Ri ifj = ni-i-l = { nR^ if j = m 0 otherwise

24

1 Algebraic Preliminaries

This theorem yields the so-called suhresultant algorithm for computing t h e resultant of A and B: if deg(A) > d e g ( 5 ) , then resultant(A, B) = So{A, B) by definition, so we compute t h e subresultant P R S of A and B. If deg(fifc) > 0, then A and B have a common factor, so resultant(yl, ß ) = 0. Otherwise is equal to either Rk if deg[Rk~i) = 1, Theorem 1.5.3 implies t h a t SQ{A^B) or TkRk if 1. In t h a t last case, the computation of Tk can be simplified since n^ = 0: (1.10) becomes Gk = YljZi'^j-i'^j-) ^^ (—1)"^^ = YljZii-^T'"^'''• A factor of - 1 appears in this product whenever b o t h n j „ i and rij are odd. Furthermore, since deg{Rk) = 0, r^ = Rk and (1.9) becomes k-l

T, = {-ly-RT-

-'n

Ä

-Uj + i

^1+^'i

If deg(74) < d e g ( 5 ) , we compute t h e subresultant P R S of B and A, and resultant(yl,5) = (-l)^^g^^)^^s(ß)resultant(ß, A) by Theorem 1.4.1.

SubResuItaiit(A,ß)

(* Subresultant algorithm *)

(* Given an integral domain D and A^ B E D[x] with B ^ 0 and deg(74) > deg(B), return result ant (A, 5 ) and the subresultant PRS (ßo,-Ri,-..,i?fc,0) of A and B. *) Ro ^ A.Ri^B i ^ 1, 71 ^ 1 ^1 <- deg(A) - deg{B) while Ri ^Q do n <~ lc{Ri) {Q,R)^ PolyPseudoDivide(Ä-i,ÄO Ri-^i <— R/ßi (* this division is always exact *) i <-i+ 1 5, ^ d e g ( i ? , _ i ) - d e g ( ß O /C ^

2 -~ 1

if deg(Efc) > 0 t h e n r e t u r n ( 0 , (ßo,-ßi, • • •, Äfc,0)) if deg(.ßfc-i) = 1 t h e n Teturn{Rk, {Ro,Ri,... ,Rk,0)) s ^- 1, c ^— 1 (* s will be (-1)°'S s ß f * ' ^ ' ' - ' ' ~ ' c will be n for j «— 1 t o /c — 1 d o (* compute TkRk *) if deg(i?j_i) is odd and deg(i?j) is odd t h e n s < s return(scfif''*'^'-'\(Äo,ßi,

1.6 Primitive Polynomials

25

Example 1.5.1. Here is the subresultant algorithm for ^ = x^ + 1 and B = x^ — 1 in Zbl: ,1+.. i Ri 7i 6^ A ^i 0 x^ + 1 1 1 x^-1 - 1 0 - 1 1 2 -2 -1 2 -1 -2 3 0

1 -8

We get k = 2j deg(i?2) = 0 and deg(i?i) = 2, so we compute s and c: j deg(Ä^_i) deg(Ä^-) 5 C 1 2 2 11 2 2 0 11 SO R = scRl = 4 = resultant(a::^ + l,a:^ — 1). Example 1.5.2. Here is the subresultant algorithm for A = 3tx^ — t^ — A and 5 = ^2 + t^x - 9 in D[x] where D = Z[t]: i

Ri

7i

Si

A

r^

,1+..

0

3t A 1 B -1 0 -1 1 1 2 3t^x -h t^ - 27t + 4 - 1 1 1 3^4 9t8 3 R -St"^ 1 9t« Ä Ä 2 4 0

where i? = -3t^O-12t^+t^--54t4 + 8t3 + 729t2-216t + 16 G D. We get A: = 3, deg(i?3) = 0 and deg(fi2) = 1, so Ä = resultanta^(A, E), as in example 1.4.3.

1.6 Primitive Polynomials Let D be a unique factorization domain, and x be an indeterminate over D. Then, gcd's always exist in D by Theorem 1.1.3, and we study in this section the properties of the gcd of the coefficients of an element of D[x]. Definition 1.6.1. Let A = E L o ^ ^ ^ ' ^ ^ N \ i^}- ^^^ content of A is content(^4) = gcd(ao,..., a^) € D . We also say that A is primitive if content (A) £ D*. Finally, the primitive part of A is given by pp(A)

A e D{x]. content (A)

By convention, content(O) = pp(0) = 0 and 0 is not primitive.

26

1 Algebraic Preliminaries

Note that the content and primitive part are defined, like the gcd, up to multiplication by a unit. We make the convention throughout this book however that the choice of unit is made consistently in order that A = content(A)pp(yl) for any A G D[x]. In addition, primitivity depends on the ring D, and nonprimitive polynomials can become primitive when D is embedded into a larger UFD: 4x + 6 is not primitive as an element of Z[x], but becomes primitive as an element of Q[x]. In fact, if D is a field, then every nonzero polynomial is primitive. Let P E D[x] \ D he irreducible. Since P = content(P)pp(P) and pp(P) is not a unit, it follows that content(P) must be a unit, hence that P is primitive. The main property of contents is that they are multiplicative. This classical result is due to Gauss and is known as Gauss' Lemma: L e m m a 1.6.1 ([54], Chap. V, §6, [92], §5.4). content(AP) = content(^) content(P)

for any A^ B £ D[x] .

As a result, a product of primitive polynomials is itself primitive. This has an effect on the leading coefficients of prime factorizations in D[x]: let A £ D[x] be nonzero, and A = ufÜLiP/ lYi=i^t' ^e its prime factorization where u e D*^ each pj is an irreducible of P , and each Pi is an irreducible of D[X]\D. Each Pi is primitive as noted above, hence HlLi ^t'' ^^ primitive, so content (A) = uv YiT=i P/ ^^'^ some t» G P* by Lemma L6.L If yl is primitive, the unicity of the prime factorization over D implies that m = 0, so we can choose an appropriate unit for the content so that the prime factorization of pp(A) is of the form pp(A) = IIILi ^t' where the Pi are coprime and deg(P^) > 0. We use this fact in the following definition, as well as whenever we use primitive parts in the integration algorithm. Definition 1.6.2. Let A e. D[x] and pp(^) = Yl7=i ^t' ^^ ^^^ prime factorization of its primitive part where e^ > 1 for each i. We define the squarefree part of A to be n i=l

and for k G Z^ k >0, the A:-deflation of A to be j^-k

_ TTpmax(0,e,:~fc) ^

TT

pe,

i\ei>k

Note that A ^ = PP(^)- For convenience we call A ^ simply the deflation of A^ and denote if by A~~^ i.e.

A-=A-^ = l[pr

1.6 Primitive Polynomials

27

As consequences of the definition we have the following useful relations: A*A-=MA), A"^' = A~'' ^ where i,j>0

(1.11) and i -^ j = k .

A special case of the above relation is

which together with (1.11) implies that A-'^+i = - 3 ^

for Ä; > 0 .

(1.13)

Although the squarefree part and deflations are defined in terms of the prime factorization, they can be computed by gcd computations in D[x]. The basic idea is that a prime factor of A divides dA/dx once less than A. T h e o r e m 1.6.1. Let A^P E D[x]\D

and n > 0 be an integer. Then,

(i) P^+i I A = ^ P ^ I gcd{A,dA/dx), (ii) if P is prime and char(D) = 0, then P ^ | gcd{A, dA/dx) = > P"+^ | A, Proof, (i) Suppose that P"'+^ | A, then there exists B G D[x] such that A = P ^ + i p . Hence,

^dx= P - - ^ ^dx^ + (n + l ) P - Pdx? so P"^ I dA/dx, which implies that P"^ | gcd{A, dA/dx). (ii) Suppose that D has characteristic 0, P is prime, and P^ | gcd(^, dA/dx). Let m > 0 be the unique integer such that P ^ | A and P ^ + i ]/A. Then, there exists B G D[x] such that A = P ^ P and P)(B. As in part (i), we have dx dx dx We have m > n since P^ | A. Suppose that m = n. Then, dx dx dx We have P ^ | dyl/dx, so P^ | nP''-^B{dP/dx), hence P | nB{dP/dx). But P is prime and P J/P, so P | n{dP/dx). In characteristic 0, n{dP/dx) is nonzero and has a smaller degree than P, so P)( n{dP/dx). Hence m j^ n, so m > n, which imphes that P""^^ \ A. D An immediate consequence of Theorem 1.6.1 is that when D has characteristic 0, then A-=gcci(A,^)

(1.14)

for any primitive A, and A* can then be computed by (1.11). The further deflations of A can be computed recursively with (1.12). Squarefree parts and deflations are thus easier to compute than prime factorizations. We use this in the next section where we introduce the notion of squarefree factorization.

28

1 Algebraic Preliminaries

1»7 Squarefree Factorization Let D be a unique factorization domain, and x be an indeterminate over D. D[x] is then a unique factorization domain, so every A E D[x] has a factorization into irreducibles. Such a factorization is usually difficult to compute in general, but there are other factorizations that are easier to compute and that can be used instead for many purposes. We introduce in this section the squarefree factorization^ which is the one primarily used by integration algorithms. Definition 1.7.1. A G D[x] is squarefree if there exists no B G D[x] \D such that B'^ \A in D[x]. Equivalently, A is squarefree if e^ = 1 for i = 1 , . . . , n in any prime factorization of A over D. Definition 1.7.2. Let A G D[x]. A squarefree factorization of A is a factorization of the form A = YllLi ^l 'where each Ai is squarefree andgcd{Ai^ Aj) G D fori / j . Note that there is no need to require a separate leading coefficient in D* and prime factors in D as in the prime factorization, since the elements of D are automatically squarefree by our definition. In addition, if we have a squarefree factorization of the primitive part of A of the form pp{A) = Hl^i ^h then A = (content(^) Ai) J | A] is a squarefree factorization of A, so it is sufficient to compute squarefree factorizations of primitive parts. In addition, we assume that D has characteristic 0 (see [39, 40, 97] for squarefree factorizations in positive characteristic). In characteristic 0, a squarefree factorization of A separates the zeroes of A by equal multiplicities, since a zero of A must be a zero of exactly one Ai, and its multiplicity in A is then i. We use this fact in order to express the Ai^s in terms of the deflations of A and vice-versa. L e m m a 1.7.1. Let A G D[x] \D, pp{A) = HlLi -^t ^^ ^ prime factorization ofpp{A), m = m a x ( e i , . . . , e^) and Ai = Hjic^i Pj f^^ 1 < i <m. Then, ß)

A-'- = UT=k^i ^ r ^ = ^fc+i^i+2 • • • ^ " " ^ for any integer Ai=

^_^,

forl
(Hi) pp{A) = JXiLi ^\ ^^ ^ squarefree factorization o/pp(^).

k>0. (1.15)

1.7 Squarefree Factorization

29

Proof, (i) We have m

m

n ^'= n n^r= n pr'=A-^-

iz=k-\-l

i=k-\-l j\ej—i

j\ej>k

ill) From (i) we have Ai • ^m.

-^

(iii) Each Ai is squarefree, since it is a product of coprime irreducibles. In addition, gcd(A^,^j) G D for i ^ j since each prime factor of pp(^) appears in its factorization with a unique exponent. Finally, using A; = 0 in (i), we get pp(^) = A~^ = YYiLi -A*, which is a squarefree factorization of pp(A). D Since deflations and squarefree parts can be computed by gcd's as explained in the previous section, we get the following squarefree factorization algorithm for a primitive A: by (1.14), we have A~^ = A~~ = gcd{A^dA/dx), which gives us A~^* = A* = pp{A)/A~. Once we have A~^* and A~^+^ for Ä: > 0, the sequence can be continued by gcd

(A--^*,

A-'=+>) = gcd (Afc+i •••Am, Ak+2Al+, • • • A--'=-i) = 4--=+^* ,

and Ak-^i and A~^+^ are obtained by (1.15) and (1.13) respectively. We continue this sequence until A~~^'+'^ G .D, which implies that A~^' is squarefree, in which case /c = m —1, and Am = A~^'. This squarefree factorization algorithm uses only rational operations plus gcd computations in D[x]. Squarefree (A)

(* Musser' s squarefree factorization *)

(* Given a unique factorization domain D of characteristic 0 and A G D[x], return A i , . . . , Am G D[x] such that A = YYk'-i A^ is & squarefree factorization of A. *) c ^- content(yi), S <— A/c S^gcd{S,dS/dx) S'* <- S/Ski-l while deg(5'"') > 0 d o Y ^gcd{S\S~) Ak ^ S^/Y S"" ^Y S-
(* 5 = pp(A) *)

(* S- =A-\S*

=A-'--^" *) (* Y = A~>^'* *) (* Ak = A"''-^*/A"''-* *) (* 5^* =A-^'* *) (* S~ = A"^+i *)

30

1 Algebraic Preliminaries

Example 1.7.1. Applying Squarefree to A = x^ + ^x^ + 12x^ + 8^^ G Q[x], we get c = 1, 5 = ^ , dS/dx = 8x^ + 36x^ + 48x^ + 16:c, 5 - = gcd

5,

dS

c^ + 4x^ + 4x

dx

and 5* = 5 / 5 - = x^ + 2x. Then,

5-

5*

k

y

^. 1

1 x^ + 2a: x^ + 4a:^ + 4x x^ + 2x 2 :c^ + 2x x2 + 2 x2-|-2 X 3 ^2 + 2 1 x2 + 2 Hence, ^ - x^ + 6x^ + 12x'* + 8x^ = x\x^

4- 2)

Yun [95] showed t h a t it is possible to be more efficient t h a n the above algorithm by reducing the degree of the polynomials appearing in the gcd inside the loop. His idea is to consider t h e following sequence of polynomials:

Yk = Y,{i-k

+ 1)

i=k

Y,{i

dAi A - ^ - i ^ Ai

dx

dAi ~k + l)Ak • • • A , _ i — ^ A , + i •••Am

ioik>l

(1.16)

i=k

whose properties are summarized in the following lemma. L e m m a 1.7.2. With the above

notation,

gcd{A-^-^*,Yi)

G D,

dA'-''^/dx

A~'Yi,

(1.17)

AiVi+i

(1.18)

and with Ai as defined in Lemma Yi

=

1.7.1,

dA->

-

dx

for 1 < i < m. Proof. Let 1 < i < j < m. Then, gcd{Aj,Ai---Aj-i^Aj+i

dAi

•-•Am)£D

since Aj is squarefree and the Ai are pairwise relatively prime. Since Aj \Ai---

dAk Ak-i-j-Ak+i

•••Am

for j ^ fc,

1.7 Squarefree Factorization this implies t h a t gcd{Aj,Yi) G D , hence t h a t gcd{A~'''-^*,Yi) Let 1 < i < m. Using Lemma 1.7.1 and (1.11) we get

31

G D.

^ - - " f : ü - + idx) ^Aj^ = ^ - - ^ From d^-.-i*

d / -

^ \

-

d^.^-.-i'

dx

dx \ ^^

I

^-^

\j=k

J

j=k

dx

Aj -^

we get dA-'-^*

_ ^

.

dAj A-'-^''

= 2^ 0 - 0 dx

^dAjA"'-'

Aj

D Since A ^-^ that

= AiA

'-* and gcd(A '•*^Yi^i) gcd Ur-^

,r,

dA-"-'' ^—j

E D , we conclude from (1.18) = Ai

(1.19)

which yields Yun's squarefree factorization algorithm: assuming as before t h a t A is primitive, we have A~ = gcd{A^dA/dx), which gives us A-"°* =A* = p p ( ^ ) M "

and

Yi = ^ ^ ^

by (1.17).

Once we have A~^-^* and Yk^ Ak is computed by (1.19), and Y^+i and ^4"^* are obtained by (1.18) and (1.15) respectively. We continue this sequence until Yk = dA"^-^ / d x , which implies t h a t A''^-^ is squarefree, in which case k = m^ and A^ = A~^'~^ = A~^-^*. T h e difference between this squarefree factorization algorithm and t h e previous one, is t h a t Yk —dA~^--^*/dx appears in the main gcd computation instead of A~^.

1 Algebraic Preliminaries

32

Squarefree(A)

(* Yun's squarefree factorization *)

(* Given a unique factorization domain D of characteristic 0 and A e D[x], return Ai ...,Am e D[x] such that A = 0 ^ = ^ A^ is a squarefree factorization of A. *) c ^- content (/I), S S' ^ dS/dx S- ^ gcd{S, S')

s* 4-

s/s-

Y ^ k ^ l

S'/S-

^A/c

while (z <- y -- dsydx) Ak ^ gcd(5* ,Z) s* ^ syAk

(* 5 = p p ( ^ ) *)

^ 0 do

(* 5* = A-^-^*,Y^Yk {* (1.19) (* 5* =A->'* (* y = Yu+i

Y ^ Z/Ak ki~k^l Ak^S* r e t u r n ( c A l , . . . ,Ak)

*) *) *) *)

Example 1.7.2. Here is a step-by-step execution of Yun's algorithm on the A of example l.7A.Wegetc=l,S = A,S' = dS/dx = 82:'^ + 36x^ + 48x^ + 16x, and S- = gcd(5', 5^) = x^ + 4^^ + 4x. Then, A:

5*

Y

Z

1 x^ + 2x 8 x ^ + 4 5x^ + 2 2 x'^ + 2x 5x^ + 2 3 x2 + 2

^3

1

2x2

X

0

x2 + 2

2x

Hence, A = x^ + 6x^ + 12x^ + Sx^ = x^{x^ + 2)^ The second arguments to the repeated gcd computations inside the loop are in the Z-column, and their degrees are smaller than in the corresponding AS"-column of example 1.7.1.

Exercises Exercise 1.1. Use the Euclidean Algorithm to compute the gcd of 217 and 413 in Z. Exercise 1.2. Find integers x,y such, that (a) 12x + 1 9 ^ = 1. (b) 3x + 2y = 5.

1.7 Squarefree Factorization

33

Exercise 1.3. Find the inverse of 14 in Z37. Exercise 1.4. Compute the gcd of 2x^ — ^x'^ — x + | and a;^ + | x — ^ in Exercise 1.5. Compute the pseudo-quotient and pseudo-remainder of x^ — 72: 4-7 by 3 x 2 - 7 in ^[^]_ Exercise 1.6. Compute the quotient and remainder (or pseudo-quotient and pseudo-remainder) of 7x^ + 4x"^ -\-2x-\-l by 2x^ + 3 in Zs[x], Zii[x], Z[x] and Q[x]. In each case determine over which kind of algebraic structure you are computing. Exercise 1.7. Compute the primitive PRS and the subresultant PRS of x"^ + x^ - t and x^ + 2^^ + 3tx - t + 1 in Z[t][x]. Exercise 1.8. Compute the gcd of 4x^ + 13x^ + löx^ + 7x + 1 and 2x^ + x^ — 4x - 3 in a) q[x] and b) Z[x]. Exercise 1.9. Compute a squarefree factorization of x^ — 5x^ + ßx"^ + ix"^ — 8. Exercise 1.10. Prove that 2 is irreducible but not prime in Z [\/—5]. Exercise 1.11. Prove that similarity as defined in Definition 1.5.2 is an equivalence relation. Exercise 1.12. Prove that if a, 6 are in a Euclidean domain D, and a = qh-\-r for some q^r G D^ then gcd(a,6) = gcd(6,r). Exercise 1.13. Use the Extended Euclidean algorithm and Theorem 1.4.1 to prove Theorem 1.4.2. Exercise 1.14. Use a loop invariant to prove that the Extended Euchdean algorithm is correct. Exercise 1.15. Let D be a unique factorization domain, F its quotient field (see example 1.1.14) and x be an indeterminate over D. Show the following consequence of Lemma 1.6.1: for any A^ B £ D[x] with A primitive, A divides B in D[x] if and only if A divides B in F[x]. Exercise 1.16. Let R be an integral domain and x i , . . . , x^ G ß . We say that 2; G fi is a least common multiple (1cm) of xi,... ^Xn if Xi | z for 1 < i < n and \/t E R^ Xi \ t foi 1 < i < n ==4> z

\t.

Show that if i? is a UFD, then any x, y have an 1cm in R.

Integration of Rational Functions

We describe in this chapter algorithms for the integration of rational functions. This case, which is the simplest since rational functions always have elementary integrals, is important because the algorithms for integrating more complicated functions are essentially generalizations of the techniques used for rational functions. Since the algorithms and theorems of this chapter are special cases of the Risch algorithm, we postpone the proof of their correctness until the later chapters on integrating transcendental functions. Throughout this chapter, let K he SLfield-^of characteristic 0, x an indeterminate over K, and ' denote the derivation d/dx on K{x)^ so x is the integration variable. By a rational function w.r.t. x, we mean a quotient of two polynomials in X, allowing other expressions provided they do not involve x. For example, log{y)/{x — e — TT) is a rational function w.r.t. x, where K = Q(log(y), e, TT). We see from this example that computing in the algebraic closure K of K^ while possible in theory, is in general ineffective or impractical. Thus, modern algorithms try to avoid computing in extensions of K as long as possible.

Introduction The problem of integrating rational functions seems to be as old as differentiation. According to Ostrogradsky [68], both Newton and Leibniz attempted to compute antiderivatives of rational functions, neither of them obtaining a complete algorithm. Leibniz' approach was to compute an irreducible factorization of the denominator over the reals, then a partial fraction decomposition where the denominators have degree 1 or 2 in x, and then to integrate each summand separately. However, he could not completely handle the case of a quadratic denominator. In the early 18*^ century, Johan Bernoulli perfected the partial fraction decomposition method and completed Leibniz' method The reader unfamiliar with algebra can think of K throughout this chapter as either the set of rational, real or complex numbers.

36

2 Integration of Rational Functions

{Acta Eruditorum^ 1703), giving what seems to be the oldest integration algorithm on record ([61] Chap. IX p. 353). Amazingly, it remains the method found in today's calculus textbooks and taught to high-school and university students in introductory analysis courses. The major computational problem with this method is of course computing the complete factorization of a polynomial over the reals. This problem was already an active research area in the 19*^ century, and as early as 1845, the Russian mathematician M. W. Ostrogradsky [68] presented a new algorithm that computes the rational part of the integral without factoring whatsoever. Although his method was taught to Russian students, and appears in older Russian analysis textbooks ([36] Chap. VIII §2), it was not widely taught in the rest of the world, where competing or similar methods were independently discovered^. Thus, Hermite [43] published in 1872 a different algorithm that achieved the same goal, namely computing the rational part of the integral without factoring. And more recently, E. Horowitz independently discovered essentially Ostrogradsky's method and presented it with a detailed complexity analysis [45]. The problem of computing the transcendental part of the integral without factoring remained open for over a century, and was finally solved in recent papers [56, 83, 89].

2.1 The Bernoulli Algorithm This approach, both the oldest and simplest, is not often used in practice because of the cost of factoring in R[a:], but it is important since it provides the theoretical foundations for all the subsequent algorithms. Let / G R(a:) be our integrand, and write / = P + A/D where P^A^D G M[x], gcd(A, D) = 1, and deg(^) < deg{D). Let n

m

D = c Y[{x - aiY'^' [ ] ( x 2 + bjX + CjY^ i=i

j=i

be the irreducible factorization of D over E, where c, the a^'s, ö^'s and c^'s are in M and the e^'s and /^'s are positive integers. Computing the partial fraction decomposition of / , we get

/ = ^+EE7-£^+EE ' ^-^ (x — aA^ ^-^ ^^ [x^ + hnX + Co-) k

where the A^/c's, Bjk^s and Cjk^s are in M. Hence,

J^ J

ht^J (^~^^f ht^J ^^'^h^^^j)k' i=\k=\'' ^ "^ j=ik=i

I would like to thank Prof. S.A. Abramov, Moscow State University, for bringing this point to my attention.

2.1 The Bernoulli Algorithm

37

Computing J P poses no problem (it will for any other class of functions), and for the other terms we have f Ajk _ f Ak{x - a,)^-V(l - ^) if A: > 1 J (x-üi)^ \Aiilog{x-ai) if/c = l

(2.1)

and, noting that 6| — Acj < 0 since x'^ + bjX + Cj is irreducible in R[x] BjlX

+ Cnl

Bjl

T

/ 9

,

X

/

4c,-&2

VV^'^.-^'l

and for /c > 1, ßj/^x + Cjfc _ (2Cjfc — bjBjk)x + bjCjk — 2cjBjk (x2 + 6^x 4- c^)^ (A; - l)(4q - b^.){x^ + 6j:r + c^)^-^ (2A:-3)(2C,fc-6,ß,-fc) (Ä: - l)(4c^- - b]){x'^ + 6jx + Cj)^-^ ' This last formula can be used recursively until /c = 1, thus producing the complete integral. Example 2.LI. Consider / = l/{x'^-\-x) G Q(x). The denominator of/ factors over E as x"^ + X = x{x^ + 1), and the partial fraction decomposition of / is 1

l_

x^ -\- x

X

X

x^ + 1

So from the above formulas we get

Example 2A.2. Consider / = l/{x'^ + 1)^ G Q(x). The denominator of / factors over M as (x^ + 1)"^, and the partial fraction decomposition of / is l/{x^ + l)'^, so from the above formulas with j = 1^ k = 2^ bi = B12 = 0 and ci = C12 = 1, we get f dx 2x f 2dx X 1 J p n F " 4(x2 + 1) ^ i 4(x2 + 1) = 2(x2 + 1) "^ 2 ^"^^^^(^^ • A variant of Bernoulli's algorithm that works over an arbitrary field K of characteristic 0, is to factor D linearly over the algebraic closure of K^ D = ri?=i(^ — ^iY''^ ^^d then use (2.1) on each term of the following partial fraction decomposition of / :

38

2 Integration of Rational Functions

i=l

j=l ^

^'

We note that this approach is then equivalent to expanding / into its Laurent series at all its finite poles, since at a: = a^, the Laurent series is -A79

yi,>.

A.ä-\

f =

^ \ \ 1 \— {x — ai)'^ {x — ai) (x — aiY' where the TI^J'S are the same as those in (2.3). Thus, this approach can be seen as expanding the integrand into series around all its poles (including 00), then integrating the series termwise, and then interpolating for the answer, by summing all the polar terms, obtaining the integral of (2.3). Example 2.1.3. Consider / = l/{x^ ^x) G Q(x). The denominator of/ factors over Q ( A / ^ ) as X^ + X = x{x + \/'~\){x — A/—T), and the partial fraction decomposition of / is 1

___ _1

X^ + X

X

1/2 X + \ / ^

1/2 X -

So an integral of / is

/ x^ + x

log(x) - - log {x + v ^ ) - - log {x • ' ' 2 ^ ^ 2

Note that there exists an integral of / expressible without \/--T, namely (2.2). Since this algorithm can be based, on power series expansions, we call it a local approach. Its major computational inconvenience is the requirement of computing with algebraic numbers over K that might not appear in the integral, namely the coefficients of the Laurent series. This is the case in the previous example, in which the algorithm computes in Q ( ^ / ^ ) , but there exists an integral that is expressible over Q(x) only. On the other hand, there are integrals that cannot be expressed without the introduction of new algebraic constants, like j{dx/{x^ ~ 2)), which cannot be expressed without using \/2 ([75] Prop. 1.1), so in general we may need to introduce an algebraic extension of K at some point. In order to have a complete and efficient algorithm, we have to answer the following questions: Ql. How much of the integral can be computed with all calculations being done in K{x)l Q2. Can we compute the minimal algebraic extension of K necessary to express the integral? An algorithm that never makes an unnecessary algebraic extension and does not compute irreducible factorizations over K will be called "rational".

2.2 The Hermite Reduction

39

2.2 The Hermite Reduction We can see from the variant of Bernoulli's algorithm discussed above, that any / G K{x) has an integral of the form m

/ = ^ + E^^^^g(^^)

/

(2.4)

i=l

where v^ui^... ^ u^ G K{x) and C i , . . . , c^ G ÜT. i» is called the rational part of the integral, and the sum of logarithms is called the transcendental part of the integral. Hermite [43] gave the following rational algorithm for computing v: write the integrand as / = A/D where ^4, D G K[x] and gcd(74, D) = 1. Let D = D1D2 • •' D^ be a squarefree factorization of D. Using a partial fraction decomposition of / with respect to Di, I^^, • • •, -Dn? write

k=l

k

where P and the A^'s are in K[x] and either Ak — 0 oi deg{Ak) < deg(D^) for each k. Then,

/-/-s/^

so the problem is now reduced to integrating a fraction of the form Q/V^ where deg(Q) < deg(y^) and V is squarefree, which implies that gcd(y, V) = 1. Thus, if A; > 1 we can use the extended Euclidean algorithm to find B^C G K[x] such that Q BV + CV l~k and deg(ß) < deg(F). This implies that deg(BF') < deg(l/2) < deg(y'=), hence that deg(C) < Aeg{y'^~^). Multiplying both sides by (1 — k)/V'' we get

Q_ \/k

{k - l)BV'

(l-fc)C

yk

1/fc-i

^ "

•

Adding and subtracting B'/V''~^ to the right hand side we get Q _ ( B' yk yyk-i

{k-l)BV'\ yk

{l-k)C~B' yk-i

j

And integrating both sides yields

1^ J yk

= -^+ yk-i ^ J

{l-k)C-B'

[•

yk-i

Since deg((l — k)C — B') < deg(y^ ^), the integrand is thus reduced to a similar one with a smaller power of V in the denominator, so, repeating this

40

2 Integration of Rational Functions

until Ä: = 1, we obtain y G K{x) and E e K[x] such t h a t deg(£^) < deg(V) and Q/V^ = y^ + E/V. Doing this to each term Ai/D\, we get g,h e K{x) such t h a t f = g' -\- P ^ h and h has a squarefree denominator and no polynomial part, so J /i is a linear combination of logarithms with constant coefficients. The V of (2.4) is t h e n merely g -i- J P. Hermite did not provide any new technique for integrating /i, so question Q2 remained open at t h a t point.

HermiteReduce(^,D)

(* Hermite Reduction - original version *)

(* Given a field K and A,D G K[x] with D nonzero and coprime with A, return g^h G K(x) such that ^ = -^ + ^ and h has a squarefree denominator. *) ( D i , . . . , Dn) ^ SquarePree(i:^) (P, v4i, ^ 2 , • • •, ^ n ) ^ PartialFraction(A, Di, D i , . . . ;jjn h<~P + Ai/Di for k ^- 2 to n such that deg(Dfc) > 0 do

V^Dk for j ^— /c — 1 t o 1 s t e p —1 do (B,C) ^ E x t e n d e d E u c l i d e a n ( ^ , l / , - A / e / i ) g^g + B/V^ h^h + Ak/V r e t u r n ( p , h)

Example 2.2.1. Here is H e r m i t e R e d u c e on x^ - 24x^ - 4x^ + 8x - 8 ^ ~~ x^ + 6x6 + 12x4 + 8^2

^ ^V^) •

A squarefree factorization of the denominator of / is D =: x^ + 6x^ + 12x^ + 8x^ = x2(x2 + 2)^ =

DIDI

and the partial fraction decomposition of / is:

/ = X-

1

-4X

- 6x3 __ 18^2 __ i 2 x + 8 (x2 + 2)3

Here is the rest of the Hermite reduction for / :

i

V

3

A.

B

C

2

X

1

x-1

1

-1

3 x2 + 2 2 x4 - 6x3 _ 28^2 _ i2x + 8 6x 3 x2 + 2 1 -x + 3 x^ - 6x ~ 2

-4+3X-2 1

2.2 The Hermite Reduction

41

Thus, 24x^ - 4^2 + 8x - 8 , 1 6x dx = -\, ^ ^ (x2 +, ^,n 2)2 x» + 6J:6 + 12^4 + 8x2

x-3 f dx 77——: + x^ + 2

We also mention t h e following variant of Hermite's algorithm t h a t does not require a partial fraction decomposition of / : let D = D1D2 • • • D'^ be a squarefree factorization of D and suppose t h a t m > 2 (otherwise D is already squarefree). Let then V = Dm smd U = D/V^. Since gcd{UV\V) = 1, we can use t h e extended Euclidean algorithm to find B^C G K[x] such t h a t

1 and deg{B)

^

= BUV

+ CV

< d e g ( y ) . Multiplying both sides by (1 — m)/{UV'^) A l/ym

__{l-m)BV' ym

so, adding and subtracting B'/V^"^ A l/ym

( B' \ym-i

'

gives

(l-m)C [/F^-i

to t h e right h a n d side, we get

{m~l)BV'\ ym

(l-m)C~-UB' f/l/m-i

J

and integrating b o t h sides yields A

/

UV^

B 1/m-i

f

{l~m)C-UB'

/

ijy

so the integrand is reduced to one with a smaller power of V in the denominator. This process is repeated until the denominator is squarefree. Since the exponent of one of the squarefree factors is reduced by 1 at every pass, the number of reduction steps in the worst case is 1 + 2 + • • • + (m — 1), which is 0{m'^) so we call this variant the quadratic Hermite reduction. HermiteReduce(/l,D)

(* Hermite Reduction - quadratic version *)

(* Given a field K and A,D E K[x] with D nonzero and coprime with A, return g^h E K{x) such that ^ = ^ + ^ ^^d h has a squarefree denominator. *) g ^0, (Di,..., Dm) ^ SquareFree(i:>) for 2 <— 2 t o m such that deg{Di) > 0 d o V ^ Di, U <~ D/V' for j ^ z — 1 t o 1 s t e p —1 do {B,C) ^ ExtendedEucIidean(C/f^,F,-A/i) g^g + B/V^,A^-jC~-U^ D ^UV r e t u r n ( ^ , A/D)

2 Integration of Rational Functions

42

Example 2.2.2. Given the same integrand as in example 2.2.1, the quadratic Hermite reduction makes the following steps, where D3 = x'^ -\- 2:

i V

u3

B

2

Dl 1

1

X

c -x'-

3 D3

X

2

3 03

X

1 -x + S

%x

a;4 2

A

- x^ + 18x^ - 8x - 8 x^ + :r^ - 18x^ + 8:r 4- 8 - ^ + x^ - 2a; - 2 x^ + x^ - 2^2 -2x + A

-a;^ + X - 2

x"^ + 2

Thus,

I

x^ - 2^x^ - 4x2 •Sxx8-h 6x6 + 12x4 + 8x2

6a: dx — —h rj, (a:2 + 2)2

+ x2 + 2 "^ ./

dx x

as in example 2.2.1, but no partial fraction decomposition was required. Suppose that the denominator D of the integrand has a squarefree factorization of the form D = DiD^- • - D^ where each Di has positive degree (this is the worst case for the Hermite reduction). In both of the above versions, the number of reduction steps needed is quadratic in m. There is however another variant, due to Mack [62], which requires only m — 1 reduction steps, so we call this variant the linear Hermite reduction. In addition, Mack's variant does not require either a partial fraction decomposition of / or a squarefree factorization of its denominator (which is computed during the reduction). Let D = D1D2 •' • D'^ be a squarefree factorization of the denominator of / (we do not need to actually compute it), and recall the notations defined in Definition 1.6.2, namely

= \[p.

and

P-

max(0,e-i—fc)

for any P G K[x]\K^ where pp(P) = IlILi ^t^' ^^ ^ prime factorization of pp(P). Since we are working over a field K., we can assume that D = pp(i^). As in the squarefree factorization algorithms, we first compute D~ = gcd(D, D^) and D* = D/D~. If deg{D~) = 0, then D is squarefree, otherwise, since D- = D-^D-^ by (1.11), D-' = D-^Y2 by Lemma 1.7.2 where Y2 is given by (1.16), and Di = D^/D-'' by Lemma 1.7.1, we get D*D-^Y2 D-

D^D-^Yo D-^'D-^

.Y2 = D1Y2 e K[x].

(2.5)

Furthermore, gcd(Di,D ) = 1 as a consequence of Lemma 1.7.1, and gcd.{y2-,D~ ) = 1 by Lemma 1.7.2, which implies that

.M^-^.u-

gcA{DiY2,D-*)

= l.

2.2 The Hermite Reduction

43

Therefore, we can use the extended Euclidean algorithm to find B^C G K[x] such that

As previously, dividing both sides hy D = D*D A D

BD-' D-^

so, adding and subtracting B'/D~

C DiD-

to the right hand side, we get BD-'\

C - DiB'

and integrating both sides yields A D

B D-

J

fC-DiB' DiD'

Since DiD" = {DiD2)Dl • • - 1 ) ^ " ^ the integrand is reduced to one whose denominator has a squarefree factorization with at most m — 1 different exponents, as opposed to m for the initial integrand. Thus, repeating this process at most m~l times yields an integrand with a squarefree denominator. A further optimization is that the parameters of the next iteration can be computed from the current ones: the new integrand is C-~DiB' DiD-

A D

where A = C-~DiB'

=

C~^B'

and D=

D^D~-=D^D2Dl...D^-\

We then have D =DiD2...Dm

= D''

which means that D* remains unchanged throughout the reduction. In addition, D~ =DsDl...Dl^-^ = D-' which means that D~ is replaced by its deflation at each step throughout the reduction. We remark that it is possible to perform all the variants of Hermite's reduction over a UFD rather than a field, the result being expressed over its quotient field. In that case, it is necessary for Mack's variant that the denominator D be primitive (this is not necessary for the previous variants).

44

2 Integration of Rational Functions

HeriiiiteReduce(yl, D)

(* Hermite Reduction - Mack's linear version *)

(* Given a field K and A,D e K[x] with D nonzero and coprime with A^ return g^h ^ K{x) such that ^ = -^ + /i and h has a squarefree denominator. *) g^Q

D-^Scd{D,^) D* ^ D/Dwhile deg(D") > 0 do D - ^ g c d ( D - , ^

--) D-* ^D-/D-^ {B, C) ^- E x t e n d e d E u c l i d e a n ( --D^'^/D-

A^C-^D-'/D-"

-,D- -\A) (* new numerator *)

9^g + ßfD{^W

D- ^ D-^ returii(p, A/D*)

= D"2 *)

Example 2.2.3. Consider the same integrand as in example 2.2.1. Mack's algorithm proceeds as follows: 1.^-0 2. D- = gcd(D, dD/dx) - x^ + 4.x^ + 4x 3. D* = D/D- = x^ + 2x 4. First reduction step: D-^ = gcd(2:^ + Ax^ + Ax, 5x^ + 12^2 + 4) = x^ + 2 5. D-* = D-/D-^ =x^ i-2x 6. ( ß , C) = ExtendedEuclidean(-5x^ - 2, x^ + 2x, A) = (Sx^ + 4, x^ - 2x^ + 16x + 4) 7.A = x^8.

2x2 ^_ 16^ ^ 4 ^ iß^ = x^ - 2x2 + 4 B

8x^+4

9. D- =D-^ =x^ -h2 10. Second reduction step: D-2 = gcd(x2 + 2,2x) = 1 ll.i:)-* = D-/5'3=x2 + 2 12. {B, C) = ExtendedEuclidean(-2x2, x2 + 2, x^ - 2x2 + 4) = (3, x^ + 2) 13. A = x2 + 2 14. B 8x2+4 3 ^ ~ ^ "^ ü ^ ~ x5 + 4x3 + 4x "^ x2 + 2

2.3 The Horowitz-Ostrogradsky Algorithm 15. D- =D-^

45

=1

Thus, x^ - 24^4 - 4x2 + 8a: - 8 , x^ + 6x6 ^ 12x4 ^ g^2

Sx^ + 4 ^5 _j_ 42^3 ^ 4^

3

f dx X

which is equivalent to the result obtained from the Hermite reduction, but only 2 reduction steps were needed instead of 3.

2.3 The Horowitz-Ostrogradsky Algorithm Ostrogradsky's algorithm also computes rationally the rational part of the integral, but it reduces to solving systems of linear algebraic equations over K instead of solving polynomial diophantine equations of the form (1.5). Let the integrand / be of the form A/D and suppose in addition that deg(A) < deg(D). As previously, let D — D1D2 • • • D ^ be a squarefree factorization of the denominator of / (the algorithm does not actually compute it). Using the notations P* and P ~ of Definition 1.6.2, we have D~ = gcd(D,D') and D* = D/D~'. From looking at the steps in the Hermite reduction, it is clear that if f = g^ -\-h where h has a squarefree denominator, then the denominator of g divides D~ and the denominator of h divides D*, so we can write g = B/D~ and h = C/D* where B^C G K[x] are unknown. Furthermore, since deg{A) < deg(i}), we can assume that deg(ß) < deg(D~) and deg((7) < deg(D*). Writing / = ^^ + ft,, we get

A__R_

BD-'

C

and multiplying by D = D*D~ ^ A = B'D* ^ B

+ CD- .

(2.6)

Since D~ \ D'^D~ by (2.5), the above is a linear equation for B and C with polynomial coefficients. Furthermore, it must always have a solution in K[x], since the Hermite reduction returns such a solution. Since we have bounds on the degrees of B and C, we can replace B and C in this equation by deg(D~)-l

y ^

deg(jD*)-l

bix'^

and j=0

where the 6^'s and Cj's are undetermined constants in K. Equating both sides of (2.6) yields a system of linear algebraic equations for the 6^'s and Cj's, and any solution of this system gives B and C, hence g and h.

46

2 Integration of Rational Functions

H o r o w i t z O s t r o g r a d s k y ( ^ , D)

(* Horowitz-Ostrogradsky algorithm *)

(* Given a field K and A, D e K[x] with deg(A) < deg(D), D nonzero and coprime with A, return g, h G K{x) such that ^ = -j^ -{- h and h has a squarefree denominator. *) D--gcd(A^) D* ^ D / I } " n ^— deg(D~) — 1 m ^- deg{D*) — 1 d <- deg{D)

B^Etobi^'^ ^ ^ Y^JLo^j^^ (* exact division *) H ^ A - B'D* + BD'^D-'/D- - C D = 0,0
Example 2.3.1. Given the same integrand as in example 2.2.1, the HorowitzOstrogradsky algorithm proceeds as follow: 1. D- = gcd{D, dD/dx) = x^ + 4x^ + 4x 2. D* = D/D- = :r^ + 2x 3. n = deg(D-) - 1 = 4,m = deg(D*) - 1 = 2 4.

H = A-~D^ (f^^iA +(E^^^M ^^-D-

Y.c,x^

= (1 - C2) x^ + (64 - ci) x^ + (263 - Co " 4c2) x^ +(362 - 664 - 4ci - 24) x^ + 4(61 -bs-co-

C2) x^

+(56o - 262 - 4ci - 4) x^ + 4(2 - co) x + 2(6o - 4) 5. The system obtained from equating iJ to 0 has the unique solution {bo, 61, 62,63,64, Co, ci, C2) = (4,6, 8, 3,0,2,0,1). Thus, x^ - 24x'^ - 4:x^-h 8x - 8 , 3x^ + Sx^ + 6x + 4 f x'^ + 2 , dx = —^ h / - dx x^ + 6^6 + 12x4 _^ 3^2 ^5 _^ 4^3 _^ 4^ J -^x^ +^ 2x 3x^ + 8x2 + 6x + 4 ^ ^^ x^ + 4x^ + 4x J X which corresponds to the result returned by the Hermite reduction. While the complexity of this algorithm is very good for rational functions [45], it does not generalize as easily as the Hermite reduction to larger classes of functions, so we use the linear Hermite reduction in the general algorithm later.

2.4 The Rothstein-Trager Algorithm

47

2.4 The Rothstein—Trager Algorithm Following the Hermite reduction, we consider now the integration of fractions of the form / = A/D with deg(^) < deg(D) and D squarefree. If a i , . . . , a^ G K are the zeros of D in K^ the partial fraction decomposition of / must be of the form

/= E

X

•

1

where ai^... ^a^ £ K. By analogy with complex-valued functions, a^ is called the residue of f at x = ai. From the naive algorithm, we know that n

/

f=

^ailog{x-ai).

The problem is thus to compute the residues of / without factoring D. Trager [89] and Rothstein [83] independently proved the following theorem. T h e o r e m 2.4.1 ([83, 89]). Lett be an indeterminate over K{x) and A^D be in K[x] with deg{D) > 0, D squarefree and gcd(^, D) = 1. Let R = resultant^(L), A - tD')

e K[t].

(2.7)

Then, (i)

the zeros of R in K are exactly the residues of A/D at all the zeros of D in K, (a) let a e. K be a zero of R, and Ga — gcdi{D^A — aD') E K{a)[x]. Then, deg{Ga) > 0, and the zeros of Ga in K are exactly the zeros of D at which the residue of A/D is equal to a. (Hi) Any field containing an integral of A/D in the form (2.4) also contains all the zeros of R in K.

Since this theorem is a special case of results that will be proven in Chaps. 4 and 5, we delay its proof until then. A direct consequence of this theorem is that / ^ = "^

E

alog{gcd{D, A-aD'))

(2.8)

a\R{a)=0

where the sum is taken over the distinct roots of R. The Rothstein-Trager algorithm for integrating rational functions with a squarefree denominator and no polynomial part is given by formulas (2.7) and (2.8). With appropriate modifications, the Rothstein-Trager algorithm can, like the Hermite reduction, be applied to rational functions over a UFD rather than a field. Part (in) of Theorem 2.4.1 shows that the splitting field of R is the minimal algebraic extension of K necessary to express the integral of A/D using logarithms, thereby essentially answering question Q2. Of course it may be

48

2 Integration of Rational Functions

possible to express an integral over a smaller constant field using other functions t h a n logarithms, for example jdx/{x^ -f 1) = arctan(x), but since an antiderivative of a function can be formally adjoined t o a field (Chap. 3), question Q2 is meaningful only when related to specific forms of the integral. If inverse trigonometric functions are allowed in the integral, then Rioboo's algorithm (Sect. 2.8) shows t h a t the integral can be expressed in a field containing the real and imaginary parts of all the roots of R. This result, together with part (iii) of Theorem 2.4.1, provides a complete answer to question Q2 for elementary integrals of rational functions (elementary integrals will be defined formally in Chap. 5). Note t h a t this algorithm requires a gcd computation in K{a)[x] where a, a zero of Ä, is an algebraic constant over K. A prime factorization R = u Rl^ • • • R^' over K is thus required, and we must compute the corresponding gcd for a zero of each Ri. Since the answer can be presented as a formal sum over t h e zeros of each Ri, there is no need to actually compute the splitting field of R.

IntRationaILogPart(A,D)

(* Rothstein-Trager algorithm *)

(* Given a field K of characteristic 0 and A,D E K[x] with deg{A) < deg(D), D nonzero, squarefree and coprime with A, return J A/Ddx. *) t
(* factorization into irreducibles *)

(* algebraic gcd computation *)

Example 2.4-1' Consider

x^ -3x^ + 6

The denominator of / , D = x^ — 5x^ + ^x"^ + 4 is squarefree (and in fact irreducible over Q), and the Rothstein-Trager resultant is resultanta,(x^ - 5x^ + 5^^ + 4, x^ - 3x^ + 6 - t {6x^ - 20x^ + lOx)) = 45796(4^2 + 1 ) ^ Let a be an algebraic number such t h a t Ao? + 1 = 0, we find Ga = gcd{x^ - 5x^ + hx^ + 4, x^ - 3x2 + 6 - ^^^^5 __ 20:^3 _^ ^^^^^ = x^ -\- 2ax'^ - 3x - 4 a .

2.5 The Lazard-Rioboo-Trager Algorithm

49

Thus, y ^

x^ - 5^4 + 5^2 + 4

a log(a:"^ + 2ax^ - 3x - 4a)

a\4a^-\-l=0

^

2

log {x^ + x^y^

- 3x -

log (x-^ - x ^ v ^ -

3J:

2A/^)

+

2A/^)

2.5 The Lazard-Rioboo-Trager Algorithm While t h e Rothstein-Trager algorithm computes in the smallest algebraic extension required to express the integral, Träger^ and Rioboo [56] independently discovered t h a t the prime factorization of the Rothstein-Trager result a n t , and t h e gcd computations over algebraic extensions of K can be avoided, if one uses t h e subresultant P R S algorithm to compute the resultant of (2.7). Their algorithm is justified by the following theorem. T h e o r e m 2 . 5 . 1 ([56, 66]). Let K he the algebraic closure of K, t be an indeterminate over K{x), and A^B^C £ K[x] \ {0} be such that gcd(A, C) = g c d ( B , C ) = 1, deg(yl) < deg(C) and C is squarefree. Let R = resultant3,(C, A - tB) £ K[t] and {Ro^ Ri^..., Rk 7^ 0 , 0 , . . . ) be the subresultant PRS with respect to x of C and A~ tB if deg{B) < deg{C), or of A - tB and C if deg{B) > d e g ( C ) . Let a E K be a zero of multiplicity n > 0 of R. Then, either (i) n = deg{C),

in which case gcd(C, A-aB)=C

e

(a) n < deg{C), in which case there exists deg^{Rm) = n, and gcd{C,A~aB) where pp-j.(i?m) is the primitive

K{a)[x]. a unique m

= pp^{Rm){a,x)

> 1 such

that

G K{a)[x]

part of Rm with respect to x.

Proof. Let R = resultanta^((7, A — tB) and (i?o, i ? i , . . . , i?A; 7^ 0 , 0 , . . . ) be the subresultant P R S with respect to x {i.e. in iir[t][x]) of C and A — tB ii deg{B) < deg((7), or of yl - tB and C if d e g ( ß ) > deg(C). Let q = deg(C), c G K* be t h e leading coefficient of C, and C = c Ylt=ii^ ~ A ) be the linear ^Although he did not publish it, Trager programmed this algorithm in his axiom implementation of rational function integration.

50

2 Integration of Rational Functions

factorization of C over K^ where the ßis By Theorem 1.4.1 we have

R=

are distinct since C is squarefree.

d'\{{A{ßi)-tB{ßi))

where p = dieg^{A — tB). Hence, the leading coefficient of R is ± c ^ 11?=! ^{ßi)-> which is nonzero since g c d ( ß , C ) = 1. Thus Ä ^ 0, so let a G A" be a zero of multiplicity n > 0 of i?. We note t h a t the trailing monomial of R is ^^ riir=i-^(ft)) which is nonzero since gcd{A^C) = 1, so a 7^ 0. Since a has multiplicity n, then there is a subset I^ C { 1 , . . . ,g} of cardinality n such t h a t A{ßi) - aB{ßi) = 0 if and only if i e la. Hence G^ = Hi^j^i^ - ßi) divides A — aB in K[x]. But x — ßi ^ A — aB for i ^ /Q,, so Ga is a gcd of C and A — aB in K{a)[x]. Hence, deg^{gcd{C^A — aB)) = n, which implies t h a t n < deg(C). (i) If n = deg(C), then gcd(C, A — aB) is a divisor of C of degree deg(C), hence gcd(C, A — aB) = C. (ii) Suppose t h a t n < deg(C). Then, A — aB / 0, otherwise we would have gcd(C, A — aB) = gcd(C, 0) = C which has degree greater t h a n n. Let Sn G Ä'[t][x] be the n*^ subresultant of C and A — tB with respect t o x^ (J : K[t] —> K{a) be the ring-homomorphism t h a t is the identity on K and maps t to a, and ä : Ä'[t][x] —> K{a)[x] be given by ä (J2 CLJX^) — Yl cr{cij)x^Since A^B^C do not involve t, ö'(C) = C and 'ä{A — tB) = A — aB^ hence d e g ( ä ( C ) ) = g, and Theorem 1.4.3 implies t h a t 'ä{Sn) = c^Sn where r is a nonnegative integer and Sn is the n*^ subresultant of C and A — aB. Let {QoiQi^- • • ,Qi 7^ 0 , 0 , . . . ) be the subresultant P R S in if(a)[x] of C and A - aB if d e g ( ß ) < d e g ( a ) , or of yi - aB and C if deg{B) > d e g ( C ) . By Theorem 1.5.1, Qi is a gcd of C and A — aB, so deg^(Q^) = n. Hence, 'ä{Sn) is similar to Qi by Theorem 1.5.2, which implies t h a t ö'{Sn) ^ 0 and it is a gcd of C and A — aB, and in particular deg{'ä(Sn)) = n, Since deg^(5^) < n by definition, and deg(ä(6'^)) < deg3,(5^), we have deg^{Sn) = n. By Theorem 1.5.2, Sn is similar to some Rjn for m > 0, which implies t h a t deg^{Rm) = TI. Since deg(jRo) ^ deg(C) > n, we have m > 1, which implies t h a t m is unique, since deg(i?^) > deg(i?^^i) for i > 1 in any P R S . Write piSn = p2PPxi^rn) with pi,/>2 ^ -^[^] satisfying gcd{pi,p2) = 1. Then, cr(/)i)ä(S'^) = a{p2)ä{pp^{Rm)). Note t h a t ä ( 5 ^ ) / 0 and ö'(pp2,(i?^)) / 0 since pp^{Rm) is primitive. In addition we cannot have (j{pi) = (j{p2) = 0 since gcd{pi,p2) = 1. Hence, (j{pi) 7^ 0 and cr(p2) 7^ 0, so '^(pPx(^m)) = PPx{^m){o^-> x) IS a gcd of C and A — aB. D Now let A, 1} G K[x]\{0} with gcd(A, D) = 1, I ) squarefree and deg(A) < d e g ( D ) . Applying Theorem 2.5.1 with A = A, B = D' and C = D , we get i? = resultantx(D, A — tD') and for any root a of ß of multiplicity z > 0, we have i < deg(.D) and: (i) if i = deg(D), then gcd{D, A - aD') = D,

2.5 The Lazard-Rioboo-Trager Algorithm

51

(ii) if i < deg{D)^ then gcd(D, A — aD') = p p ^ ( f i ^ ) ( a , x ) where TTI > 1 is the unique strictly positive integer such t h a t deg^{Rm) = iThus, it is not necessary to compute the gcd's appearing in t h e logarithms in (2.8), we can use the various remainders appearing in t h e subresultant P R S instead. Since deg(D^) < deg(i9), we are in the case where deg(E) < deg((7), so we use the subresultant P R S of D and A — tD'. As long as the result is returned as a formal sum over the roots of some polynomials, all the calculations are done over K , no algebraic extension is required, and the formal algebraic numbers introduced by the sum are in the smallest possible algebraic extension needed to express the integral. In practice, we perform a squarefree factorization R = H l L i Q\^ ^^ ^^^ roots of multiplicity i of R are exactly the roots of Qi. Evaluating ppx{Rm) foit = a where ce is a root of Qi is equivalent to reducing each coefficient with respect to x of pp^(jR^) modulo Qi. We do not really need to compute pp^(Rm), it is in fact sufficient to ensure t h a t Qi and t h e leading coefficient with respect to x of Rm do not have a nontrivial gcd, which implies then t h a t the remainder by Qi is nonzero"^. Since multiplying the argument of any logarithm in (2.8) by an arbitrary constant does not change the derivative, we can make pp^{Rm){o^, x) monic in order to simpHfy the answer, although this requires computing an inverse in i f [a], but not computing a gcd in iir[a][x]. Since the Qi's are not necessarily irreducible over K, K[a] can have zero-divisors, but the leading coefficients of the pp^{Rm){c^^xys are always invertible in K[a\ (Exercise 2.7). This normalization step is optional.

I n t R a t i o n a l L o g P a r t ( A , D)

(* Lazard-Rioboo-Trager algorithm *)

(* Given a field K of characteristic 0 and A,D ^ K[x] with deg(A) < deg(D), D nonzero, squarefree and coprime with A, return J A/Ddx. *) t ^r- a. new indeterminate over K {R, (ßo, ß i , . . . , ß/c, 0)) ^ SubResultanta; (D, A - t ^) ( Q i , . . . , Qn) <- SquarePree(i?) for i ^-- 1 t o n such that deg^(Qi) > 0 do if i = deg{D) t h e n Si ^ D else Si <— Rm where deg^(Arn) = i, 1 < m < k {Ai,...,Aq) ^ SquareFree(lcx(5'i)) for j ^ 1 to q do Si ^— Si/gcd(Aj,Qiy (* exact quotient *) return(X:r=iX]a|Q,(a)=o«log(*5'i(a,rr)))

^This requirement, which was overlooked in the original publication of Theorem 2.5.1, has been pointed out by Mulders [66] (see Exercise 2.5).

2 Integration of Rational Functions

52

Example 2.5.1. Consider t h e same integrand as in example 2.4.1. D is squarefree, and t h e subresultant P R S of D and A — tD' is i

Ri

0

x^ - Sx^ + 5 x ^ + 4

1

-6tx^ + x^ + 20tx^ - 3^2 - lOtx + 6

2

(-60^2 + 1) x^ 4- 2tx^ + (120t2 - 3) x^ + 26tx + 144^^ + 6

3 (800t^ - 14t) x^ - (400^2 - 7) ^2 - (2440t2 - 32t) x + 792^^ - 16 4

(-11200^4 - 2604^2 + 49) x^ + 25600t4 + 5952^2 - 112

5

(-119840t^ - 59920^3 - 7490t) x - 23968t^ - 11984t2 - 1498

6

2930944t^ + 2198208^4 + 549552^2 + 45796

The Rothstein-Trager resultant is R = Rß^ and its squarefree factorization is R = 2930944t^ + 2198208t^ + M9552t^ + 45796 = 45796 (4^^ + 1 ) ^ = 45796 Q | and t h e remainder of degree 3 in x in the P R S is Rs = (800t^ - 14t)x^ - (400^2 - 7)x^ - (2440t^ - 32t)x + 792^^ - 16. Since gcd(lc,(Ä3), Q3) = gcd(800t2 - 14t, 4t2 + 1) = 1, Ss = R3. Evaluating for t at a root a of Qsia) = 4a^ + 1 = 0 we get 53(a, x) = - 2 1 4 a x ^ + 107x=^ + 642ax - 214 so an integral is x"^ - 3 x 2 + 6 c^ - 5x4 _|_ 5^2 _|_ 4

=

^

a log(-214ax^ + 107x2 + 642ax - 214).

a|4a2 + l=0

Making 6*3(a, x) monic we get Ss{a^ x) = —214 a (x^ H- 2ax2 — 3x — 4a) so t h e integral is t h e n as t h a t in example 2.4.1.

IntegrateRationalFunction(/)

(* Rational function integration *)

(* Given a field K of characteristic 0 and / G K{x), return J f dx. *) (g^h) ^— HermiteReduce(numerator(/),denominator(/)) (Q^R) <~~- PolyDivide(numerator(/i),denominator(/i)) if R = 0 t h e n r e t u r n ( ^ + / Q ^^) r e t u r n ( ^ + J Q dx + I n t R a t i o n a l L o g P a r t ( ß , denominator(/i)))

2.6 The Czichowski Algorithm

53

Example 2.5.2. Let us compute the integral of 36

r_A_

•^ ~ D " x5 - 2x4 - 2x3 + 4:r2 + X - 2 ^ ^V^)' 1. HermiteReduce(A, D) returns g = (12x + 6)/{x'^ ~ 1) and h = 12/E where E = x'^ — x — 2, 2. PolyDivide(12,E) returns Q = 0 and Ä = 12, 3. IntRationalLogPart(12, JB) returns Ylaia'^-i6=o ^^^^(^ ~ V^ — 3a/8), so we have • dx — x^ - 2x4 - 2x3 + 4x2 + X - 2 12x + 6 .^^ ,

,~^

E

/

1

3a

«logf:^-^-^). 2~ ^

(2.9)

A simpler form for the logarithmic part will be obtained by the algorithm of Sect. 2.7.

2.6 T h e Czichowski Algorithm Czichowski has pointed out that the resultant and subresultant computations of the Rothstein-Trager and Lazard-Rioboo-Trager algorithms can be replaced by a Gröbner basis computation in i^[x,2;]. Since Gröbner bases are beyond the scope of this book, we do not present a proof of his theorem here, but give it without proof together with the corresponding algorithm. Interested readers can consult [7, 25] for an introduction to Gröbner bases, and [26] for a proof of the following theorem. T h e o r e m 2.6.1 ([26]). Lett be an indeterminate over K{x), A^ D be in K[x] with deg(D) > 0, D squarefree and gcd(^, D) = 1, and B be the reduced Gröbner basis with respect to pure lexicographic ordering with x > t of the ideal generated by D and A — tD' in K[t^ x]. Write B = { P i , . . . , Pm} where each Pi is in K\t^x] and for each i the highest term of Pi^i is greater than the highest term of Pi in the pure lexicographic ordering with x > t. Then, (i) \ca:{Wx{Pi)) = 1 fori
54

2 Integration of Rational Functions

IntRationalLogPart(A,D)

(* Czichowski algorithm *)

(* Given a field K of characteristic 0 and A, D G K[x] with deg{A) < deg(D), D nonzero, squarefree and coprime with A, return J A/Ddx. *) (* Compute the reduced Gröbner basis *) ( P i , . . . , Pm) ^ R e d u c e d G r ö b n e r [D, A - t ^ , pure lex, x > t) (* ( P i , . . . , P m ) must be sorted by increasing highest term *) for 2 ^- 1 t o m — 1 d o Qi ^- contents; (Pi)/content;c(Pi+i) (* exact quotient *) Si ^ p p ^ ( P i + i ) retura(^^~^5;]^IQ^(^)^oalog(5i(a,x)))

Example 2.6.1. Consider the same integrand as in example 2.4.1. T h e reduced Gröbner basis of (D, A - tdD/dx)

= {x^ - 5x^ + hx'^ + 4, - 6 t x ^ + x^ + 20tx^ - 3x^ - lOtx + 6)

w.r.t. pure lexicographic ordering with x > t is B = { P i , P2} = {4^2 + 1, x^ + 2tx^ - 3x - 4t} . Thus, Qi = 4t2 + 1/1 = 4t^ + 1 and Si = P2, which yields the integral

/

x^ - 3 x ^ + 6 x^ - 5x4 + 5:^2 _^ 4

y^

a log(x^ + 2ax^ - 3x -- 4a)

a|4a2 + l=0

which is the same integral t h a t was obtained in example 2.4.1.

2.7 Newton—Leibniz—Bernoulli Revisited We have seen t h a t t h e difficulty with using formula (2.1), which dates back t o Newton and Leibniz, was the computation of the Laurent series expansions at the poles of the integrand. However, Bronstein and Salvy [15] gave a rational algorithm for computing those series. T h a t algorithm can be then used t o make the full partial fraction approach rational, so we describe it here. The basic result is t h a t the ^^^'s of

(x — aiY'-

(x — aiY

(x — ai)

can be computed as functions of the a^'s without factoring. T h e o r e m 2 . 7 . 1 . Let A, P) G K[x] with D monic and nonzero, gcd(yl, D) = 1, and let D = D1D2 • • • D^^ be a squarefree factorization of D. Then, using only

2.7 Newton-Leibniz-Bernoulli Revisited

55

rational operations over K, we can compute Hij € K[x] for 1 < j < i < n such that the partial fraction decomposition of A/D is

D

z^

z^

i = l a\Diia)=0

\(x-aY

X- a

^^

'

where P is the quotient of A by D. Proof We first describe the construction of the iJ^^'s: let i G { 1 , . . . , n}, Ei =

D/Dl and where li is a differential variable over K{x) {i.e. u and all its derivatives u'^u"^... are independent indeterminates over K{x)). Each Di is squarefree and coprime with the other D/c's by the definition of squarefree factorization, so gcd{Eij Di) = gcd(D^, Di) = 1. Thus, use the extended Euclidean algorithm to compute Bi^Ci E K[x] such that BiEi = 1 (mod Di)

and

dD^ = 1 (mod A ) •

(210)

For j = 1 , . . . , z, compute hi'~'^^/{i — j)\ and write it as

where Py- is a polynomial with coefScients in K. Let then

and finally let Hij = Qij S r ' ^ ' Ci'~'

(mod Di)

(212)

where Bi and Ci are given by (210). We prove now that the iJ^j's given by (212) satisfy the theorem. Let K be the algebraic closure of K^ a E K he di root of D^, Di^a = Di/{x — a), and

Since hi^^ is just hi evaluated at t^ = Di^oc^ we have

f^-i) _ p,, ( x , A , « , i ) U ' ^ " « ' • • •'A^:«'0 (^-i)! D^'-^Et'^' 2,a where P^^ is the same polynomial as in (2.11). We have Di = {x — a)Di^a^ so for k > 0,

56

2 Integration of Rational Functions

since (x — a)^-^'' = 0 for j > 1. Hence,

for Ä: > 0, which impUes that Qij{a) =

Pijla,D',{a),-

2

'

3

'

'

i-j+1

I

= Pi^ (^a,D,,^{a),D[Ja),D'l^ia),...,Dii-'\a)) In addition, (2.10) implies that Bi{a) = l/Ei{a) Hij{a) =

.

and Ci{a) = 1/Dl{a), so

Qij{a)Bi{ay-^+'Ci{af'-^

2,a

V

/

i^-Jy•

Since x — a does not divide the denominator of /i^^«, hi^a has a Taylor series at X = a, and by Taylor's formula,

fc>0

so the Laurent series of AjD at x = a is

D

(x - ay

(i - j)! (a: - a)^

^

—"^ {^ - ^^V

which proves the theorem.

D

This theorem yields an algorithm for computing Laurent series expansions of rational functions. We can make an additional improvement: it is possible that for given i and j ' s , Gij = gcd{Hij, Di) is nontrivial. This means that for a root a of C^j, the coefficient of l/{x — a)^ in the expansion of A/D is 0. When this happens, we replace Di by Dij = Di/Gij^ and return the partial fraction decomposition of A/D in the form

p+EE

E

where all the summands are guaranteed to be nonzero.

2.7 Newton-Leibniz-Bernoulli Revisited

57

L a u r e n t Series (A, D,F,n) (* Contribution of F to the full partial fraction decomposition of A/D *) (* Given a field K of characteristic 0 and A,D,F G K[x] with D monic, nonzero, coprime with A, and F the factor of multiplicity n in the squarefree factorization of D, return the principal parts of the Laurent series of A/D at all the zeros of F. *) if deg(F) = 0 t h e n r e t u r n 0 14 <— a differential indeterminate, a <— 0

E^D/F'',

h^A/iu^'E)

{B, G) <- E x t e n d e d E u c l i d e a n ( E , F, 1) {^ BE + GF = l *) (C, G) ^ E x t e n d e d E u c l i d e a n ( F ^ F, 1) (* CF' + C F = 1 *) for j ^- 0 t o n — 1 do P^t^^+^'E^+^'/i (* P eK[x,u,u',u",...,u^^^] *) vj ^ F^^^'^/{j + 1) Q i~ eval(P, u ^ vo,..., u^^^ —> Vj) h ^ h'/ij +1) F'^F/gcd{F,Q) if cleg(7^*) > 0 t h e n H ^ QB^+iC"-+^ mod F* return a

Example

2.7.1. Consider

Applying L a u r e n t S e r i e s to A = 36, D = (x - 2)(x^ - 1)^, F = x^ - 1 and n = 2 we get: 1. a = 0, E = X - 2, /i = 36/(^2(^x - 2)), 2. (^2 - l ) / 3 - ( x / 3 + 2/3)(x - 2) = 1, so E = - ( x + 2 ) / 3 , 3. (x/2)2x - (x2 - 1) - 1, so C = x / 2 , 5. 6. 7. 8.

7;o = F ' = 2:r, Q = eval(36, u-^ VQ) = 36, so gcd(x2 - 1, Q) = 1, F * = a:2 - 1, i J = - 3 6 {x + 2 ) / 3 (x/2)2 mod x^ - 1 = - 3 x - 6,

9.h = h' = ( ( - 7 2 x + 144)^' 10. P = ^^(x - 2)2/i = ( - 7 2 x l l . t ; i =F"/2 = 1, 12. Q = eval(P, u -^ VQ, U' -^ 13. P * = ( x 2 - l ) / ( x - l ) = x

- 36ii)/(w^(x - 2)^), + 144)i^' - 36ii, vi) = - 1 4 4 x + 144, so gcd(x2 - 1, Q) = x - 1, + l,

58

2 Integration of Rational Functions

14. H = (-144x + 144) {{x + 2)/3)2 {x/2f

mod (x + 1) = - 4 , so a = er +

Hence, the sum of the Laurent series of / at the roots of x^ — 1 = 0 is 36 x^-2^4-2x3+4x2+x-2

^ V I Z_^

~3^~^I (x-a)2|

4 x+1

'

\Q;^ —1=0

FullPartialFraction(/) (* Full partial fraction decomposition of / *) (* Given a field K of characteristic 0 and / G K{x), return the full partial fraction decomposition of / . *) D
Example 2.7.2. Applying FuliPartiaiPraction to ^ ^ x5 - 2x4 ^ 2x3 + 4x2 + X - 2 ^ Q(^) we get: 1. 2. 3. 4. 5.

D = x^ - 2x4 __ 2x^ + 4x2 + X __ 2^ {Q,R) = PolyDivide(36, D) == (0,36), DiD^ = SquarePree(D) = (x - 2)(x2 - 1)2, LaurentSeries(36, D, x — 2,1) returns 4/(x — 2), LaurentSeries(36, D, x2 — 1, 2) returns Y^

—3a — 6 (x — a;)2 I

x+1

as seen in example 2.7.1. Hence, the full partial fraction decomposition of / is 36 c^ - 2x4 - 2x3 + 4x2 + X - 2

/ II

Y^ Z—/ Z^

—3a — 6 I fry. (x _- rv\2 a)2 II

^ O 4- 11 x- +

x- 2 (2.13)

2.8 Rioboo's Algorithm for Real Rational Functions IntegrateRationalFiiiiction(/)

59

(* Pull partial fraction integration *)

(* Given a field K of characteristic 0 and / G K{x), return the full partial fraction decomposition of J f dx. *)

^ + X] Z l

J2

(^'!}%

i = l j = l a\Di.j(cx)=0

return

^

P + \^

^

FulIPartialPractioii(/)

^

\^

+EE

Hii{a)log(x

E jT. (1 -

i=2 j=2 cx\Dij(cx)=0 ^

— a)

Hij{a) i)(x ~

ay-^

-^^^

'

Example 2.7.3. For the fraction / of example 2.7.2, F u I l P a r t i a l F r a c t i o n returns (2.13), so the integral of / is

x^ - 2x4 _ 2x3 + 4x2 + :z: - 2

dx =

41og(x-2)-41og(x + l ) +

>

^-—^

« 2 - 1= 0

X—a

.

Compare with the algorithm of the previous sections, which returns (2.9) for the same integrand. Since the resulting integral is returned in the form (2.4) with the fraction v also expanded into partial fractions with linear denominators, this algorithm is not a better alternative t h a n the other rational algorithms in this chapter, but it makes the partial fraction algorithm factor-free nonetheless. Thus, all the approaches to rational fraction integration can be implemented using only rational operations.

2.8 Rioboo's Algorithm for Real Rational Functions The algorithms of this chapter give the integral of a rational function in the form (2.4), i.e. using logarithms whose arguments may involve algebraic quantities over the ground field. In the case where the ground field i^ is a subfield of the reals, those algebraic numbers can be complex, so complex arithmetic is necessary for computing a definite integral. This may cause branch problems in the numerical computation, since the arguments t o the logarithms may have complex zeros, while the initial integrand has no pole in the path of integration. As a result, a direct apphcation of the fundamental theorem of

60

2 Integration of Rational Functions

calculus can yield an incorrect value, since the antiderivative is not necessarilycontinuous on the interval of integration. For example, consider the definite integral

x^-Sx^ + e

,

,

•dx.

,

(2.14)

It is easily checked that the integrand is continuous and positive on the real line, hence that the above integral must be a positive real number. The indefinite integral as computed by the algorithms of this chapter is -7 7r—dx:= x ^ - 5 x 4 + 5x2+ 4

> Z-^

a logfx^ + 2ax^ - 3x - 4a). sv -r J

(2.15) \ J

a|4a2 + l=0

The zeros of 40^ + 1 are a = ±i/2 where i^ = — i^ so, applying (incorrectly) the fundamental theorem of calculus to the above integral with x = 2 and X = 1, we would get for the definite integral ^ log(2 + 2t)~'-

log(2 - 2i)\ - (^log(-2

- 0 - ^ log(-2 + i)

OTT

—

/1\

^ arctan

- R^ —3.46 . 4 \2j As explained above, this result cannot be the correct area. Thus, it is preferable to return a real function given a real integrand. We describe in this section an algorithm of Rioboo [72] that expands a result of the form (2.4) into a real function without introducing new real poles, provided that the initial integrand is real. We use the following properties of fields which do not contain \ / ^ : if x^ + 1 is irreducible over if, then, for any P, Q £ if [x], p2 + Q2 _ 0 = ^ P = Q = 0 .

(2.16)

Indeed, if P^ + Q2 ^ Q and Q 7^ 0, then (P/Q)^ - ~ 1 , so Q | P, which implies that P/Q G K is a, square root of —1, in contradiction with x^ + 1 irreducible over if. We first present the classical algorithm for rewriting complex logarithms as real arc-tangents. L e m m a 2.8.1. Let u e K{x) be such that v? ^ —\. Then,

Proof. Writing i = y—T7 ^^ immediate calculation yields

d . I - j ~ log

dx

fu-^i\ \u — i J

fu — i\

d fu + i

\u -^ i J dx \u — i^ u — i\ du {u ~ i) — (u i- i) u-\-iJ dx (lA — 2)2 du/dx d .: —7^ 7 = 2 -—- arctan(ii) . u^ -\-l dx

2.8 Rioboo's Algorithm for Real Rational Functions

61 D

Directly using (2.17) for rewriting complex logarithms with real arc-tangents is possible, b u t does not eliminate the problem of obtaining discontinuous antiderivatives, since the resulting integral always has singularities at the poles of u^ while its derivative does not. For example, applying it to the integral (2.15) gives (we write f ^ g for df/dx = dg/dx): y^

alog(a:^ + 2ax^ - 3x - 4a)

a|4a2 + lzzr0

- log {x^ - 3x + i{x^ - 2)) - 2 log (^ log

x^ -~3x + i{x'^ - 2) x^ — 3x — i{x'^ — 2)

arctan

- 3x — i{x x^

-3x"

•2)) (2.18)

Using this to compute the definite integral (2.14) via the fundamental theorem of calculus we get 7r/4 —arctan(2) ^ —0.32, which is also incorrect. The reason is t h a t (2.17) introduced discontinuities at ± \ / 2 , as can be seen from the graph of arctan((x^ - 3x)/{x'^ - 2)) (Fig. 2.1).

-n/2f x^Sx^+G Fig. 2 . 1 . A discontinuous formal integral of x^-5x'^-\-5x^-\-4 —^

To avoid this problem, Rioboo gave an improvement to Lemma 2.8.1 where the argument of the arc-tangent is a polynomial in x instead of a fraction.

62

2 Integration of Rational Functions

T h e o r e m 2.8.1 ([72]). LetA.B

e K[x]\{0}

d ^ fA^iB\ "1 ^ dx log^\A-iBj

be such thatA^-^B^

^ 0. Then,

d ^ f-B + iA = 1dx- log ""y-B-iA

and, for any C^D E K[x] such that BD — AC = gcd(A^B), .d^ fA + iB\ ^d fAD^BC\ i —— —- = 2 dx -—- arctan \gcd{A,B)J — dx log ""yA-iBj

C ^ 0 and

d^ fD + iC + ^dxT" ^og "^ \D - iC

where i^ — —1. Proof. We have A + iB _ {-{) {-B + iA) _ -B-^iA A-iB ~ i {--B - iA) ~ ~~BiA so, taking logarithmic derivatives on both sides, d ^ fA-i-iB\ d , f-B l^g ~A ^ h = 1 - log

dx

\A~iBj

dx

+ iA

\-B-iA

Let now G = gcd{A, B) and C,D G K[x] be such that C ^ 0, C^ + I}^ / 0 and BD-AC = G. Write P = {AD + BC)/G. We note that P e K[x] since G I ^ and G I 5 . We have A + iB A~iB

.d^ dx

fD--iCA + iB\D + iC \D + iCA-iBj D-iC 'AD-hBC-i-i{BD-AC)\ D + iC AD + BC-i {BD ~AC)JD - iC P + i\ [D + iC^ D-iC

fA + iB\ \A — iB J

.d^ dx

fP + i\ \P —iJ

^d

,^.

d^ dx

fD + iC \D — iC

Hence, by Lemma 2o8.1, d^ I-

fA-^iB\

dx ^°^ X I l S

.d,

fD + iC

= ' d^ "'^^^^'^^^ -^'TJ''^\D-^C D

Note that (2.16) implies that Theorem 2.8.1 is always applicable in fields not containing \f-i. Furthermore, it provides an algorithm for rewriting

2.8 Rioboo's Algorithm for Real Rational Functions

63

as a sum of arc-tangents with polynomial arguments: since G = gcd(A, 5 ) , we have deg(G) < d e g ( 5 ) . If d e g ( ß ) = deg(G'), then B\A,soG = B, which implies t h a t D = 1 and (7 = 0, hence t h a t P = {AD + BC)IG = A/B e K[x] and t h a t -— = 2 —- a r c t a n ( P ) ax ax by Lemma 2.8.1. If deg{A) < deg{B),

then

f-B-ViA df d ^ -r- log \ —B — lA dx — i dx by Theorem 2.8.1, so we can assume t h a t deg(yl) > deg{B) > deg(G). By the extended Euclidean algorithm, we can find C^ D G K[x] such t h a t BD —AC = G and deg{D) < d e g ( ^ ) . In addition, D / 0 since d e g ( ^ ) > deg(G'). This implies t h a t C / 0, since deg{B) > deg(G), hence t h a t C^ + D^ / 0 as we have seen earlier. Hence, by Theorem 2.8.1, the derivatives of / and of fAD + BC\ ., fD + iC 2 a r c t a n ^ ^ ^ j + . l o g ^ ^ — ^ are equal. We can apply the algorithm recursively to the remaining logarithm, and m a x ( d e g ( C ) , d e g ( D ) ) < m a x ( d e g ( A ) , d e g ( ß ) ) guarantees t h a t this process terminates. LogToAtan(A, B) (* Rioboo's conversion of complex logarithms to real arc-tangents *) (* Given a field K of characteristic 0 such that -s/--T ^ K, and A, B G ^l^] with B j^ 0, return a sum / of arctangents of polynomials in K[x] such that df d ., fA-hiB —- = -— ^ log ' dx dx \A — iB *) if B\ A t h e n r e t u r n ( 2 arctan(A/5)) if deg{A) < deg{B) t h e n r e t u r n L o g T o A t a n ( - B , A) (D, C, G) ^ E x t e n d e d E u c I i d e a n ( ß , -A) {^ BD - AC = G *) r e t u r n ( 2 arctan((ylD + BC)/G) + LogToAtan(i:), C))

Example 2.8.1. Plugging in a = ± i / 2 in (2.15), we get \r^

1 / ^

^

2

o

. N

^1

/ ( x ^ - 3 x ) + 2(x2-2)

Applying L o g T o A t a n to A = x^ — Zx and ß = x^ — 2, we get

64

2 Integration of Rational Functions

A

B

C

x^ — 'ix x^-2

xß

x2/2 - 1/2 x/2 2 2x

2

D

G

(yl£) + ß C ) / G

^2/2 - 1/2 1 a;V2 - 3x3/2 + a:/2 x3 2x 1

SO the integral is x'^ - 3 x ^ + 6

(ix = arctan

x^ - 3x^ + X

arctan(x^) + arctan(x) (2.20)

which differ from (2.18) only by a step function (Fig. 2.2).

Fig. 2.2. A continuous formal integral of

x6~5a;4-}-5x2+4 "

Using (2.20) to compute the integral (2.14), we get the correct answer:

1 x^ ~ 5x4 4 - 5 ^ 2 + 4

dx = arctan(5) + arctan(8) + arctan(2) arctan ( — - j -- a r c t a n ( l ) — arctan(l) arctan(5) + arctan(8) ,

TT

TT

arctan(5) + arctan(8) Pä 2 . 8 1 .

TT

2.8 Rioboo's Algorithm for Real Rational Functions

65

T h e above algorithm returns a real primitive given an expression of t h e form (2.19). B u t the integration algorithms return a sum of terms of t h e form Y^

alog{S{a,x))

(2.21)

a\R{a)=0

where R G K[t] is squarefree, and S G K[t^x]. In order to complete Rioboo's algorithm, we need to convert such a sum t o one where all t h e complex logarithms are in the form (2.19). This conversion can be done whenever K is a real field, which is an algebraic generalization^ of the subfields of t h e real numbers. D e f i n i t i o n 2 . 8 . 1 ([3]). Let K be a field. K is a real field if —1 cannot be written as a sum of squares of elements of K. K is a real closed field if any real algebraic extension of K is isomorphic to K. E is a real closure of K if E is a real closed algebraic extension of K. Example 2.8.2. R, Q, Q{^/p) for any prime number p > 2, and Q ( a ) where a is an indeterminate over Q, are all real fields. Q ( - > / ^ ) is not a real field 2

since —1 = 1^ + y^—2 is a sum of squares. If K has characteristic p > 0, t h e n — 1 = J2^Zi 1^, so any real field must have characteristic 0. T h e o r e m 2 . 8 . 2 ([54], Chap. XI, §2). Any real field has a real closure. This theorem is also proven in [92], §11.6 b u t for countable real fields only. Note t h a t t h e real closure of K is not unique, even up to isomorphism, unless K is already ordered. T h e o r e m 2 . 8 . 3 ([54], Chap. XI, §2, [92], §11.5). Let L be a real closed field. Then, (i) L has a unique ordering, given by: x > 0 4 = ^ x = y^ for some y G L. (a) L ( \ / ^ ) is the algebraic closure of L. Let K be a real field for the rest of this section, and let IK be a real closure of K, and K = K{i) where i^ = —1. W i t h a slight abuse of language, we say t h a t a £ K is "real" if a G K. Let / be a sum of t h e form (2.21) where R = Y^jTjX^ G K[x], S = Ylj^k^jk^'^^^ ^ i f [ t , x ] , and let u^v be indeterminates over K{x). We first separate t h e sum (2.21) into one over the real roots of R and one over t h e other roots: f = g+

Y.

c^log{S{a,x)).

(2.22)

where ^The reader wishing to avoid this extra algebraic machinery can skip this definition and the following theorems, and think of K in the rest of this section as a given subfield of the real numbers, with real closure K = M.

66

2 Integration of Rational Functions 9=

Yl

a\og{S{a,x))

oceK,R{(x)=0

is a real function. We then compute P, Q ^ K[u^v\ such that = ^rj{u-\-ivy

= P{u,v) ^iQ{u,v),

(2.23)

and A^B G K[u^v^x] such that S{u + iv,x) = ^Sjkiu^ivYx^

= A{u,v,x)

-\~iB{u,v,x).

(2.24)

Since K = IK(z), it is a vector space of dimension 2 over K with basis (l,i), so for a £ K, R{a) = 0 if and only if P(a, b) = Q{a, b) = 0 where a = a + ib. Furthermore, a ^ IK if and only if 6 7^ 0. Hence, we can rewrite (2.22) as f = 9+

Yl

{a + ib)log{S{a-i-ib,x))

(2.25)

a,beK,bj^O P{a,b)=Q{a,b)=0

Let a be the field-automorphism of K such that a(i) = —i and a{z) = z for any z e K, and define W : K[x] —^ K[x] by ä{J2ajX^) = J^^i^j)^'^- Let a^b eK. Applying W to (2.24) we get A{a^ b^x) — i B{a, 6, x) = cr{A{a, 6, x) + i J5(a, 6, x)) = W{S{a + i 6, x)) = S{a(a + 25),a:) = 5(a — ib^x). Applying a to (2.23) we get P{a, b) - i Q{a, b) = a{P{a, b) + i Q{a, b)) = a{R{a ^ib)) = R{a{a + i5)) = R{a - ib) which implies that R{a + i 6) = 0 if and only if R{a — ib) = 0. Hence, for any pair (a, b) appearing in the sum (2.25) with 6 / 0 , the pair (a, —b) must appear also, and is a different pair, so we can rewrite (2.25) as f = g-h

Y

{(« + ^ ^) log(*5'(ß -^ib,x))

+ {a-~i b) log{S{a ~ i 6, x))}

a,beK,b>0 P{a,b)=Q{a,b)=0

= 9+ 22 {^ (log(A(a, 6,x) + 2ß(a, 6,x)) + log(A(a, b^x) — iB{a,6,

x)))

a,6GK,6>0 P(a,6)=Q(a,6)=:0

^ ib (log(A(a, 6, x) + i ß ( a , 6, x)) ~ log(74(a, 6, x) — iB{a^ 5,x)))} .

2.8 Rioboo's Algorithm for Real Rational Functions

E

/ == g + h +

-^l»g(1^^'1^'o!"'''1'l

67

(2.26)

a,beK,b>0 P{a i,b)=Q{a,b) ==0

where /z=

^

a log(A(a,6,x)^ + E(a,5,x)^)

a,6GK,6>0 P(a,6)=Q(a,6)=0

is a real function. Since the remaining nonreal summands in (2.26) are all of the form (2.19), we can use Theorem 2.8.1 and its associated algorithm to convert them to real functions. Note that, since converting (2.19) to real functions requires computing the gcd of A and B^ we have, in theory, to use algorithm LogToAtan over an algebraic extension K{a,h) of K where P{a, b) = Q{a^b) = 0, which means that we have to solve this nonlinear algebraic system. However, the following theorem of Rioboo shows that, when the complex logarithms to expand arise from the integration of a real rational function, it is not necessary to solve this system. T h e o r e m 2.8.4 ([72]). Let K be a real field, K be a real closure of K, C,D e K[x] with deg(J9) > 0^ deg(D) > deg(C); D squarefree and gcd(C, D) = 1. Suppose that the R and S of (2.21) are produced by the Rothstein-Trag er or Lazard-Rioboo-Trager algorithm applied to C/D, and let P^Q be given by (2.23) and A,B by (2.24). If a,b £ K satisfy P{a,b) = Q{a,b) = 0 and b^O, then gcd{A{a^b,x),B{a^b^x)) = 1 in K{a^b)[x]. Proof Let a, 5 € IK be such that P{a^ b) = Q{a^ b) = 0 and 6 7^ 0 where P and Q are given by (2.23). Then, R{a + ib) = 0 where i^ ^ - 1 and i? = P + iQ is a squarefree factor of the Rothstein-Trager resultant of C — tD' and D. Furthermore, since A and B are given by (2.24), S{a + ib,x) = A{a^b^x) + iP(a, 6,x) is a gcd in K{a + ib)[x] of C — (a + ib)D' and J9, so there exist E and F in K{a + ib)[x] such that C{x) - (a + ib)D'{x) = E{a + z6, x)S{a + ib, x) and D{x) = F{a + i6, x)S{a + ib, x). Writing E = Ei-^ iE2 and F = Fi + iF2 where ^ 1 , ^ 2 , ^ 1 , ^ 2 e K{a, b)[x], we get C{x) - {a + ib)D\x) and

= {Ei{a,b,x) + iE2{a,b,x)){A{a,b,x)

-i-iB{a,b,x)) (2.27)

68

2 Integration of Rational Functions D{x) = (Fi(a, b, x) + iF2(a, 6, x)){A{a, 6, x) + 25(a, 6, x)).

(2.28)

Taking the imaginary part of (2.27) and the real part of (2.28) we get -~hD\x) = Elia, 6, x)B{a, 6, x) + ^2(0,6, :2:)^(a, 6, x)

(2.29)

and D{x) = Fl (a, 6, x)A(a, 6, x) - F2(a, 6, a:)B(a, 6, x).

(2.30)

Since D is squarefree, gcd(D,D^) = 1, so there exist Gi, G2 € K[x] such that G1D + G2D' = 1. Multiplying (2.30) by bGi, (2.29) by -G'2 and adding both yields b = {GiD-hG2D')b = bGiFi{a,b,x)A{a,b,x)

—

bGiF2{a,b,x)B{a,b,x)

—G2Ei{a, 6, x)B{a, 6, x) — (^2-E'2(a, 6, x)yl(a, 6,2:) = (6GiFi(a,6,a:) - G'2^2(ß,^,^))^(a, &,^) - ( ^ 2 ^ 1 (a, 5, x) + 6^1^2(0,6, x))ß(a, 6, x)

which is a linear combination of A{a,b,x) and B{a,b,x) with coefficients in isr(a,6)[x]. Since 6 ^ 0 , this implies that gcd(A(a, 6,x), ß(a, 5,x)) = 1 in ir(a,6)[x]. D As a consequence, we can perform Rioboo's conversion to arc-tangents generically, i. e. expand once . ' A(u,v,x) + i B(u,v,x) % log ^ A{u, v^x) — i B{u, V, x) where u and v are independent indeterminates, obtaining a real function (j){u,v,x). We can then rewrite (2.26) as f = gi-h^

Y^

b(j){a,b,x)

a,5GK,6>0 P(a,6)=g(a,6)=0

where Theorem 2.8.4 guarantees that (j){u,v,x) specializes well, i.e. that no division by 0 occurs when we replace u and v by the various solutions a and 6 > 0 in ]K of P{u,v) = Q{u,v) — 0. By presenting the answer in terms of formal sums, we do not need to actually solve this system, or to introduce any algebraic number. In practice, whenever the real roots of P{u, v) = Q{u,v) = 0 can be computed efficiently (for example if they are all rational numbers), then it can be more efficient to first compute the roots, and then call LogToAtan, rather than perform the reduction with generic parameters.

2.8 Rioboo's Algorithm for Real Rational Functions

69

LogToReal(Ä, S) (* Rioboo's conversion of sums of complex logarithms to real functions *) (* Given a real field K^ R G K[t] and S G K[t,x], return a real function / such that

£ = ^ E «M5(a,a.)). cx\R{cx)=0

*) write R{u -i-iv) as P(u, v) -\-i Q{u, v) write S{u -\- iv^x) as A{u^ v^x) -\- iB(u, v, x) return y^

a log {A{a, b, x)^ -f B{a, b, xf)

+ b L o g T o A t a n ( ^ , B){a, 6, x)

a,beK,b>0 P(a,b)=Q(a,b)=0

+

^

alog(S'(a,x)).

aeK,R(a)=0

Example

2.8.3. Applying L o g T o R e a l to t h e integral (2.15), we have

R{t) = 4^2 + 1 G Q[t],

5 ( t , x)=x^

+ 2tx^ -3x-4te

Q[t, x],

and 1. R{u-i-iv) ==4(iA + iv)2 + l =Au'^-4v'^ + l-i-8iuv, so P = Au'^ - Av'^ ^ 1 and Q = 8uv^ 2. S{u + iv,x) = x^ + 2{u + ^ 'y)^:^ — 3x — 4{u + iv) = x^ i- 2ux'^ — 3x — 4w + i(2'ux^ ~ 4v), so A = x^ + 2ux'^ — 3x - 4u and ß = 2vx'^ - 4t;, 3. H = resultant^(p,g) = 2b6u'^ + 64w^ whose only real root is 0. P ( 0 , f ) = 1 — 4t'^, whose only real positive root is 1/2, 4. yl(0, l / 2 , x ) = x^ - 3x, B ( 0 , l / 2 , x ) = x^ - 2, and L o g T o A t a n ( j : ^ 3x,x^ — 2) returns fx^ 2 arctan I

-3x^

+ x\

^ / sx ^ 1 + 2 arctan(x ) + 2 arctan(x)

as seen in example 2.8.1, so multiplying by 6 = 1/2 we get the same integral as in example 2.8.1. Instead of solving the system P{u^v) = Q{u^v) = 0 in step 3, we can call L o g T o A t a n ( x ^ + 2ux'^ — 3x — 4tx, 2vx'^ — Av)^ which returns /

N

r^

f X

U\

iu.v.x) = 2 arctan \— ^ ' ' ^ \2v vJ , x'^ 2u o 4u^ + 4v^ - 1 + 2 arctan ( 1 x H ^— x 2v V 2v

u V

70

2 Integration of Rational Functions X

5

., U

+ 2 arctan

o,2 I , , 2 4 | " ' - T - t v - ± 3

\— x A \4:V

V? -i-v^ - l . ^ x

V

V

3ii 2^ x

OL6

V

81/2 _,_ 3^2 __ 3 x + Av

^ V

and the integral would be returned formally as —7

z

T;

dx =

>^

b(h(a,b,x)

a,beR,b>0 4a^-4b'^-\-l=8ab=0

which is a real function. Plugging in a = 0 and 6 = 1/2 in this result, we get the same integral as previously.

IntegrateReaIRationalFuiiction(/)

(* Real rational function integration *)

(* Given a real field K and / G K{x)^ return a real function g such that dg/dx = / . *) m

Y^ i=l

a log{Si{a,x))


a Ri{a)=0

return('ü + X ] ^ i L o g T o R e a l ( ß i , 5'i))

Although there is no equivalent algorithm that guarantees the obtention of an integral with no new poles for more general functions, Rioboo's algorithm has been used together with rectifying transformations in order to compute integrals of rational functions in sin(a:) and cos(x). Such integrands can always be transformed into rational functions via the change of variable u = tan(x/2), but the transcendental terms produced by Rioboo's algorithm can have new singularities when that change of variable is reversed after the integral has been computed. Rectifying transformations [47, 49] are then applied to those transcendental terms in order to remove the spurious singularities. See Exercise 2.8 for an application of this technique.

2.9 In-Field Integration We outline in this section minor variants of the integration algorithms that are used for deciding whether a rational function is either a ® derivative of a rational function, ® logarithmic derivative of a rational function.

2.9 In-Field Integration

71

Those problems are important because they arise from the integration of more general functions. Furthermore, deciding whether a rational function is a logarithmic derivative is useful when solving linear ordinary differential equations with rational function coefficients^. Recognizing Derivatives The first problem is, given / G K{x)^ to determine whether there exists u E K{x) such that du/dx = f. To compute such an u^ we simply apply either the Horowitz-Ostrogradsky algorithm, or any variant of the Hermite reduction, to / , obtaining g G K{x) and A, D G K[x] such that D is squarefree and / = dg/dx + A/D. At that point, / = du/dx for u G K{x) if and only if D \ A/in which case u = g -\- J{A/D)dx. There are also a couple of criterions that can determine whether / is the derivative of a rational function without computing an integral of / : ® Compute the squarefree factorization D1D2 ... D^ of the denominator of / , and for each i the polynomial Hn G K[x] of Theorem 2.7.1, using the LaurentSeries algorithm. Write Di = GiEi where Gi — gcd{Hii^ Di) and gcd(jE^, Hii) = 1. Since the residues of / at the roots of Gi are all 0, and the residue of / at a root a of Ei is Hii{a) ^ 0, / is the derivative of a rational function if and only if E^^ = 1 for each i, which is equivalent to Di I Hii for each i. ® Compute the squarefree factorization D1D2 .. • D^ of the denominator of / , and write / as a sum J\ A2=1

*

If / is the derivative of a rational function, then Di | Ai, since the residues of / at the roots of Di would be nonzero otherwise. If Di | Ai, then / is the derivative of a rational function if and only if each Ai/D] is the derivative of a rational function for i > 1, and we can use Mafik's criterion [63], which states that A/D'^ is the derivative of a rational function for m > 1 if and only if D I Wronskian ( -

\dx ^ dx ^

'

dx

While those criterions are not practical alternatives to either the Hermite reduction or the Horowitz-Ostrogradsky algorithm, they are of theoretical interest. No generalization of those criterions is known for more general functions, which makes the problem of recognizing derivatives more difficult in general (see Sect. 5.12). ^The differential Galois group of y*^^^ + an-i{x)y^^ -^-^ + ... is unimodular if and only if ön-i is the logarithmic derivative of a rational function.

72

2 Integration of Rational Functions

Recognizing Logarithmic Derivatives The second problem is, given / G K{x), to determine whether there exists u G K{xY such that du/dx = uf. It will be proven later (see Exercise 4.2) that / is the logarithmic derivative of a rational function if and only if / can be written as / = A/D where D is squarefree, deg(A) < deg(Z}), gcd(^, D) = 1, and all the roots of the Rothstein-Trager resultant are integers. In that case, any of the Rothstein-Trager, Lazard-Rioboo-Trager or Czichowski algorithm produces u G K{x) such that du/dx — uf.

Exercises Exercise 2 . 1 . Compute c^ - x"^ + 4x^ + x^ - a: + 5 , dx using the Hermite reduction and the Rothstein-Trager algorithm. Exercise 2.2. Compute

/

8x^ + x« - 12x^ - 4x^ - 26x^ - Gx^ + SOx^ + 23^2 - 2x - 7 dx ^10 _ 2^8 - 2x^ - Ax^ + 7x4 + 10x3 + 3x2 - 4x - 2

using the Lazard-Rioboo-Trager algorithm. Exercise 2.3. a) Compute 72x'^ + 256x^ - 192x^ - 1280x4 _ ^i2x^ ^ 1440x2 _^ 575^ _ 9g dx 9x8 j^ 36^7 _ 32:^6 - 252x5 „ 73^4 ^ 458x3 _^ 288x2 „_ IQSX + 9 using the Rothstein-Trager or the Lazard-Rioboo-Trager algorithm. With that integral compute the symbolic definite integral for —2 < x < —2/3 and compare it with the result obtained by direct numerical integration. b) Apply the Rioboo algorithm to the above result and compute again the definite integral for — 2 < x < — 2/3. Exercise 2.4. a) Compute dx l + x4 ' b) Find a closed form for J dx/{l + x^) for n G N.

2.9 In-Field Integration

73

E x e r c i s e 2.5 ([66]). Compute x'^ H- x^ + x^ + X + 1 + x4 + 2x3 4- 2^2 - 2 + 4 \ / - l + \ / 3 ax using the Lazard-Rioboo-Trager algorithm. W h a t happens if the subresultants are not made primitive before evaluating them? E x e r c i s e 2 . 6 . Write procedures for the Hermite Reduction, the L a z a r d Rioboo-Trager algorithm and the Rioboo algorithm using your favourite programming language or computer algebra system. E x e r c i s e 2 . 7 . Modify the Lazard-Rioboo-Trager algorithm so t h a t in t h e result, the polynomials inside the logarithms are monic in x. Note t h a t t h e polynomials Qi{t) indexing the sums of logarithms are not necessarily irreducible (show first t h a t the leading coefficients of the polynomials inside t h e logarithms must be units in K\t]/{Qi{t))). E x e r c i s e 2.8 ([47]). a) Compute ?tdx 5 — 4cos(x) using the change of variable u = t a n ( x / 2 ) followed by rational function integration and the Rioboo algorithm. b) Find the real singularities of your integral (note t h a t the integrand is continuous in R). c) Apply to your integral the transformation t h a t sends arctan(a t a n ( x / 2 ) + 6) to X ( 2ab cos(x) — (1 + 6^ — a^) sin(x) - 4- arctan f \ J \ J \ J V (1 + «)^ + ^^ + (1 + ^^ - «^) cos(x) + 2absin(x) 2 and verify t h a t the resulting function is an integral of 3/(5 — 4cos(x)) t h a t is continuous in M.

Differential Fields

We develop in this chapter the algebraic machinery in which the integration algorithms can be presented and proved correct. The main idea, which originates from J. F. Ritt [78], is to define the notion of derivation in a pure algebraic setting {i.e. without using the notions of "function", "limit", and "tangent line" from analysis) and to study the properties of such formal derivations on arbitrary objects. This way, we can later translate an integration problem to solving an equation in some algebraic structure, which can be done using algebraic algorithms. Since an arbitrary transcendental function can be seen as a univariate rational function over a field with an arbitrary derivation, we first need to study the general properties of derivations over rings and fields. This will allow us to generalize the rational function integration algorithms to large classes of transcendental functions (Chap. 5).

3.1 Derivations Although the integration algorithm we present in later chapters works only over diff"erential fields of characteristic 0, the rings and fields in the first two sections of this chapter are of arbitrary characteristic. Given a map in any ring, we call it a derivation if it satisfies the usual rules for differentiating sums and products. Definition 3,1.1. Let R be a ring (resp. field). A derivation on R is a map D : R —^ R such that for any a^b G R: (i) D{a + b) = Da^ Db, (ii) D{ab) = aDb + bDa. The pair (fi, D) is called a differential ring (resp. fieldj. The set Consti)(i?) = {a E R such that Da = 0}

76

3 Differential Fields

is called the constant subring (resp. subfieid) of R with respect to D. A subset S C R is called a differential subring (resp. subfieid) of R if S is a subring (resp. subfieid) of R and DS C S. When there is no ambiguity about the derivation in use, we often say that i? is a differential ring (field) rather than the pair {R, D). We first show that the usual algebraic properties of the derivations of analysis are consequences of the above definition. T h e o r e m 3.1.1. Let (R^D) be a differential ring (resp. field). Then, (i) D{ca) — cDa for any a E R and c E Consto^R). (a) If R is a field, then a bDa — aDb b 62 for any a^b E R, b ^ 0. (Hi) Const/) (i?) is a differential subring (resp. subfieid) of R. (iv) Da^ = nd^~^Da for any a E R\ {0} and any integer n > 0 (resp. any integer n). (v) Logarithmic derivative identity: if R is an integral domain, then D{u\K..ul-) L^...

for any ui,...

__

Du^

Un

j_

^

Ui

,Un E R* and any integers ei,

(vi) ^

dP Ui

for any i^i,..., i^^ in R and polynomial P with coefficients in Constu{R). Proof, (i) Let a E R and c G Consti)(i?). Then, D{ca) = cDa + aDc = cDa since Do = 0. (ii) Suppose that i? is a field, and let a^b E R with 6 7^ 0, and c = a/b. Then, a = be, so by property (ii) of Definition 3.1.1, Da = D{bc) = bDc + cDb = bD^ + ^ Db. Hence, a _,\

D-b = -b VDa--Db] b

J

=

bDa — aDb 62

•

(iii) Let C = Constj9(i?). From property (i) of Definition 3.1.1, D(0) = D{0-h 0) - D{0) + D{0), soO E C. From property (ii) of Definition 3.1.1, D{1) = D{1 X1) = D ( 1 ) + I } ( 1 ) , so 1 € C. Since DC = {0}, this implies t h a t DC c C.

Let ÜER. Then, Da + D{-a) = D{a + {--a)) = D{0) = 0, so D{-~a) = ~Da. Let c,d E C. Then, D{c ~ d) = Dc + D{-d) = Dc - Dd = 0 - 0 = 0, so c- d E C. Also, D{cd) = cDd + dDc = 0 ^ 0 = 0, so cd E C, hence C is

3.1 Derivations

77

a differential subring of R. Suppose that jR is a field and that d ^ 0. Then, D{l/d) = —Dd/d^ = 0, so l/(i G C, which implies that C is a differential subfield of R. (iv) Let a £ R\ {0}. For n = 1, Da^ = Da = la^Da. Suppose that Da"" = na^~^Da for some n > 1. Then, Da^+i = D{a''a) = a'^Da + aDa"" = a'^Da + aina'^-^Da) = {n + l)a'^Da so (iv) holds for any integer n > 1. Suppose that i? is a field. Then, DoP = D{1) = 0 = Oa^^Da^ so (iv) holds for n = 0. For n < 0 we have r^ n

n 1

^«~"

-na-^~'Da

,_.

(v) is left as Exercise 3.1 at the end of this chapter. (vi) Let X i , . . . , Xn be indeterminates, P £ Const£>(i?)[Xi,..., Xn] and write n

(e) = (ei,...,e„,)

i=l

where a^e) € C = Const/:)(i?). Using property (ii) of Definition 3.1.1 and the fact that D is C-hnear, we get

\ ( e ) = (ei,...,e,.)

i-1 n

IZ (e)=:(ei,...,e,,,)

-(e) 1 ] eiuf-^Dui i=l

n

J l li^'^' .^^

^^Tr-(^i,--.,'^n)i^'^i.

ax.

D

In general, a ring can have more than one derivation defined on it. For example, Q[X, Y] has at least the derivations 0, d/dX and d/dY. But it has a lot more derivations, for instance D = d/dX + d/dY. In fact, any linear combination of derivations with coefficients in R is again a derivation on R. L e m m a 3.1.1. The set Q{R) of all the derivations on R is a left-module over R. Proof Let D^ D2 ^ f^{R) and c e R. Let D = cDi + D2, i.e. D : R-^ R is defined by Da ~ cDia + D2a for any a £ R. Let a^b £ R. Then, D{a + b) = cDi{a + b) + D2{a + 6) = cDia + cDib + D2« + D2b = Da + Db,

78

3 Differential Fields

and D{ab) •=• cDi{ab) + D2{ab) = caDib + cbDia + aD2b + 6D20 = a{cDib + D2Ö) + b{cDia + D2a) = aDb + 61}a so D E [2{R). Since the zero-map on R (which maps every element of R to 0) is a derivation on J?, this implies that Ü{R) is a left-module over R. D Definition 3.1.2. Let {R,D) be a differential ring. An ideal I of R is a differential ideal if DI C I. L e m m a 3.1.2. Let (i?, D) be a differential ring, I be a differential ideal of R, and TT : R -^ R/I be the canonical projection. Then, D induces a derivation D* on R/I such that D* OTT = IT o D. Proof. Define D* as follows: for x G R/I, let a G R be such that 7r(a) = x, and set D*x = 7r(Da). Suppose that 7r(a) = 7r(6) = x for a, 6 G R. Then, a — be.1^ so D{a — b) E I since J is a differential ideal. This implies that Da — Db E / , hence that 7r{Da) = 7r(D6), so D* is well-defined. We have D* o TT = TT o L) by the definition of D*. Let x^y E R/I and let a^b E R he such that 7r(a) = x and 7r(6) = y. Then, 7r(a + 6) = x + y and 7r(a6) = xy, so I}*(x + y) = 7r(i:>(a + b)) = 7r{Da + Db) = 7T{Da) + 7r(i:)6) = D'^a + 1:^*5 and D*{xy) = Ti{D{ab)) = Ti{aDb + bDa) = 7r(a)7r(D6) + 7r(6)7r(Da) == xß*y + yD*x so D* is a derivation on R/I.

D

Example 3.1.L Let Ä be any ring and D be the zero-map on R. Then any ideal of jR is a differential ideal, and the induced derivation J9* is the zero-map on R/L Example 3.1.2. Let X be an indeterminate and D be d/dX on R = Q[X]. The only differential ideals of R are (0) and (1), and the induced derivations are D and the zero-map respectively. Example 3.1.3, Let {Rj D) be a diflFerential ring, X be an indeterminate and Ä : Il[X] -> i?[X] be the map defined by n

n

It can be checked that Z\ is a derivation on R[X] and that for any integer m > 0, the ideal J ^ = {X^) is a differential ideal. For m = 1, the map TT : i?[X] -^ R\X]/{X) ::± jR is the substitution X -^ 0, and the induced derivation A* on R satisfies Zl*7r(p) = n{Ap) = D{p{{))) for any p G R[X] so A* = D on i^.

3.2 Differential Extensions

79

3.2 Differential Extensions We study in this section the problem of extending a given derivation to a larger ring or field. As in the previous section, the rings and fields in this section can have arbitrary characteristic. In classical algebra, roots of equations or new indeterminates are added to a given ring in order to create a larger ring. An obvious question is then, if the initial ring admits a derivation D, can it be extended to a new derivation on the larger ring? If this is the case, and the new derivation is compatible with D, we say that the larger differential ring is a differential extension of the initial one. The following definition formalizes the notion of "compatibility with D". Definition 3.2.1. Let (R^D) and {S^A) be differential rings. We say that (5, A) is a differential extension of (ß, D) if R is a subring of S and Aa = Da for any a £ R. We first show that any derivation on an integral domain has a unique extension to its quotient field, and this extension is given by the usual rule for differentiating quotients. T h e o r e m 3.2.1. Let R be an integral domain, F the quotient field of R and D a derivation on R. Then there exists a unique derivation A on F such that (F, A) is a differential extension of (ß, D). Proof Define A : F ^ F SiS follows: for any x E F, write x = a/b where a^b e R, b y^ 0^ and let Ax — {bDa — aDb)/b'^. Suppose that x = a/b = c/d for a, b^c^d G R. Then, ad = bc^ therefore: bPa - aPb ^^^

dDc - cDd _ d%Da - d^aDb - b'^dDc + b^cDd ^2 "~~ b^d^ ____{bDd + dDb){bc - ad) + abdDd _ _ - bcdDb + bd{dDa - bDc) __ D{bd){bc - ad) + bd{dDa + aDd - bDc - cDb) b^d^ D{bd){bc - ad) + bdD{ad - be) 0 _ _

which impHes that A is well defined. Let now x,y e F and write x ~ a/b^y c/d where a, b^c^d G R. By a calculation similar to the one above, we get: ^ad + bc

bdDiad + be) - (ad + bc)D(bd)

__ bcfPa + abdPd + bcdPb + b^dPc - abdPd - ad?Dh - bcdPb - b'^cDd bPa~aPb and

dPc-cPd

3 Differential Fields ^ac

hdDiac) — acD(hd) abdDc + bcdDa — abcDd — acdDb b^d^ ajdDc - cDd) cjbDa - aPb) xAy + yAx bd? "*" d^ '

so Zl is a derivation on F. Take a E R^ and write a = a / 1 . This implies that Aa — {IDa — aDl)/l^ — Da^ so (F, A) is a differential extension of (i?, D). Suppose that there are two derivations Zli and A2 on F such that (F, Ai) and (F, A2) are both differential extensions of (R^D), and let x £ F. Write X = a/b where a,b £ R and 6 7^ 0. From part (ii) of Theorem 3.1.1 we have a bAia — aZ\i6 bDa — aDb ^1^ = ^11 = p = p =

bA2a — aA2b a p =A2-^=A,x

SO Al = ZI2, which shows that A as defined above is the only derivation on F such that (F, A) is a differential extension of ( ß , D). D Definition 3.2.2. Let R be a ring and X an indeterminate over R. For any derivation D on R, we define the coefficient lifting of D to be the map KJJ : R[X] -> R[X] given by

KniY^^iX') = Y.^Da,)X\ i=0

i=0

The map KD simply applies the derivation D to every coefficient of a polynomial over R. Note that the degree is not necessarily preserved under KD. L e m m a 3.2.1. KO is a derivation on R[X]. Proof. Let p, g E R[X] and write p = n

YA=O

^i^'^ ^^^ Q = SlLo ^i^^- Then,

n

n

and 2n

2n

KD{pq) = J2^( E a^&i)^' = E E ^ M i ) ^ ' k=ö

. .^^ i-\-j=k

2n

i-j-j=k 2n

= E E «'^^^•^' + E E ^^-^«^^

so

tZD

is a derivat;ion on Ä[X].

D

3.2 Differential Extensions

81

If R is an integral domain, then R[X] is an integral domain, so, by Theorem 3.2.1, KD can be extended uniquely to a derivation on its quotient field R{X) (example 1.1.14), and we also write KJJ for this extension to R{X). L e m m a 3.2.2. Let (R^D) be a differential ring, {S^A) a differential extension of (Ä, D), and X an indeterminate over R. Then, dP A{P{a)) = KD{P){a) + {Aa) — {a) for any a G S and any P G i?[X]. Proof This follows directly from the sum and product derivation rules: write P = J2^=o ^iX^ where the a^'s are in R. Then Aai = Doi for each z, so / n

\

n

n

\i=0

= no{P){a)^{Aa)

dP — {a) D

We can now prove the main result about differential extensions: given a simple extension F{t) of a differential field (F, D), if t is algebraic over F , then D can be extended in a unique way to F(t), otherwise D can be extended in several ways to F{t) but choosing a value for Dt makes the extension unique. We prove this in two theorems, one for the transcendental and one for the algebraic case. Theoremi 3.2.2. Let (F^D) be a differential field, and t be transcendental over F. Then, for any w G F{t), there exists a unique derivation A on F{t) such that At = w and {F(t)^ A) is a differential extension of (F, D). Proof By Lemma 3.2.1, KD is a derivation on F[t], and by Theorem 3.2.1, it has a unique extension to a derivation on F(t). Since d/dt is also a derivation on F(t), the map A = njj -\- w d/dt is a derivation on F{t) by Lemma 3.1.1. We have, At = K^t + w dt/dt = D{l)t -^ w - 1 — w^ and for a G F^ we get Aa = Koct + u) da/dt = Da + w-0 = Da, so (F(t), A) is a differential extension

oiiF,D). Suppose that there are two derivations Ai and A2 on F{t) such that (F(t),Z\i) and {F{t),A2) are both differential extensions of {F^D), and that Alt = A2t = w. Let x G F(t) and write x = a/b where a^b E F[t] and b ^ 0. Using part (ii) of Theorem 3.1.1 and Lemma 3.2.2 applied to both a and b with a == t, we get a

bAia — aAib

^ 1 ^ = ^' 6 = — p —

b{KDCi + w da/dt) — a{KDb + w db/dt)

= =

p bA2a — aA2b .a . p = Ä,-=Ä,x

82

3 Differential Fields

so Al = Zi2, which shows that A as defined above is the only derivation on F{t) such that At = w and {F{t)^ A) is a differential extension of (F, D). D Example 3.2.1. Let F be any field, Oj? be the map that sends every element of F to 0, and x be transcendental over F. Let D be an extension of Op to F{x) satisfying Dx = 1. Since {F{x)^ d/dx) is a differential extension of (F, Oj?) and dx/dx = 1, Theorem 3.2.2 implies that D = d/dx^ i.e. the only derivation on F{x) that is 0 on F and maps x to 1 is d/dx. Example 3.2.2. Let {F^D) be a differential field and t be transcendental over F . Let Zl be an extension of D to F(t) satisfying Zlt = 0. Since (F(t),/^£)) is a differential extension of (F, D) and Kjjt — 0, Theorem 3.2.2 implies that A — H.-D.^ i.e. the only extension of D to F(t) for which t is constant is KJ). We now turn to algebraic extensions of differential fields. The assumption that F is separable over F in the next theorem is needed for the case where F has nonzero characteristic, and F separable over F means that the minimal irreducible polynomial over F for any element of F has no multiple roots. In characteristic 0, algebraic extensions are always separable, so the reader interested in this case only can ignore the separability hypothesis. In addition, we use Zorn's Lemma in the proof to allow for non-finitely generated extensions. That part of the proof can be skipped if one considers only finitely generated algebraic extensions. T h e o r e m 3.2,3. Let (F^D) be a differential field, and E a separable algebraic extension of F. Then, there exists a unique derivation A on E such that (F, A) is a differential extension of (F^D). Proof. Suppose first that F = F{a) for some a £ E. Let X be an indeterminate over F , and P G F[X] be the minimal irreducible polynomial for a over F . Then, since F is separable over F , dP/dX{a) ^ 0, so let nn{P){a) dP/dX{a)

•

Since F c:^ F[a]^ there exists Q G F[X] such that w = Q{a). By Lemma 3.2.1, K]j is a derivation on F[X]. Since d/dX is also a derivation on F[X], the map A = K£) +Q • d/dX is a derivation on F[X] by Lemma 3.LL Let n : F[X] --> F[X]/{P) c:^ E he the canonical projection. We have dP 7r{AP) = 7r{KDP + Qjx)-

dP ^D{P){a) + w —{a) = KD{P){a)-KD{P){a) =0

hence AP G ker(7r) = (P), so ker(7r) is a differential ideal, which implies by Lemma 3.L2 that A induces a derivation A* : F -^ F such that iroA = zl*o7r„ Finally, for a G F , we get

3.2 Differential Extensions A*a = A^ixa = irAa =

83

+ Q -prp) = 7r{Da) = Da dX. so (£", Z\*) is a differential extension of {F^D). Let now E be any algebraic extension of F , and let 5 be the set of all pairs {K^ A) such that (i^, A) is a differential extension of (F, D) and ÜT C E. Define a partial ordering on 5 by {Ki^Ai) < {K2^A2) if (^^2,^2) is a differential extension of (i^i,Z\i). Since {F^D) G 5*, 5 is not empty, so let C •= {(i^i, Zii)} be a totally ordered subset of S, and let K = \^^Ki and define Zi on ÜT by Ax — Aix \i x ^ Ki. Since C is totally ordered, {K^ Zl) is a welldefined differential extension of (F^D). (K,A) is also a differential extension of [Ki^ Ai) for each i, so (ÜT, Zl) € 5 is an upper bound for C with respect to <. Hence every totally ordered subset of S has an upper bound in 5, so there exists a maximal element (Kmax, A^^x) ^ Shy Zorn's Lemma ([54] App. 2, §2, [92] §9.2). By the definition of S^ K^g,x ^ E and (i^max, Zlmax) is a differential extension of {F,D). Let x £ E. By what we have just proven, there exists a derivation A on ii^max(^) such that {Kmg,x{x)^A) is a differential extension of (i<^max,zlmax), SO (i^max,zl^ax) < {Kms.x{x),A) in 5, which impHcs that i^max = Kmax{x) siucc (Kmax, ^max) is a maximal element. Hence x £ Kmax, so E = Kmax5 hence (F, Zlmax) is a differential extension of (F, D). Suppose now that there are two derivations Ai and A2 on E such that (F, Al) and (F, ZI2) are both differential extensions of (F, D). Let x e E and P € F[X] be its minimal irreducible polynomial over F . Since P{x) = 0, we have by Lemma 3.2.2: 0 := A,{P{x))

TT{KDCL

dP = nD{P){x) + (Z\,x) — ( x ) .

Since F is separable over F , dP/dX{x)

7^ 0, so

Since this holds for any x € F , zli == Zi2 5 so there exists a unique derivation Zl on F such that (F, Z\) is a differential extension of (F, D\ D Example 3.2.3. Let {F, D) be a differential field of characteristic 0 and C = Consti:>(F). Let a be algebraic over C and P G C[X] be its minimal irreducible polynomial over C Then D has a unique extension to F{a) and we must have dP 0 = D{P{a)) = Kn{P){a) + {Da)^{a)

dP = {Do.) —{a)

SO Da = 0, which means that any algebraic element over the constants is itself a constant w.r.t D. Example 3.24. Let F = Q(x) and a be a root oi V^ - x G F\Y], i.e. a represents the function iby^. Then, d/dx has a unique extension to Q(x,a) and we must have

84

3 Differential Fields d , r,

^

^^^r,

^^ ^

da d(Y'^ — x) , ,

da

SO

da 1 dx 2a which is the usual derivative w.r.t. rr for a =

±y^.

As a consequence of Theorem 3.2.3, we can always replace any field in a tower of differential extensions by a separable algebraic extension, and we still have a valid tower of differential extensions. Corollary 3.2.1. Let {K,D) he a differential field, {F,A) he a differential extension of {K^ D), F he the algehraic closure of F and E C F he a separahle algehraic extension of K. Then, D can he extended uniquely to E, A can he extended uniquely to EF, and (EF^A) is a differential extension of (E^D). Proof The picture is as follows: EF (F^A)

D can be extended uniquely to E by Theorem 3.2.3. Similarly, EF is a separable algebraic extension of F ([54] Chap. VII, §4), so A can be extended uniquely to EF. Let X be an indeterminate over F , and Q E ir[X]. Considering Q as an element of F[X]^ we have K.A{Q) = ^ D ( Q ) since A extends D. Let X e E and P G K[X] be its minimal irreducible polynomial over K. Since E is separable over /T, dP/dX{x) j^ 0 and we have 0 = D{P{x)) =

KD{P){X)

dP + j^{x)

Dx

so {dP/dX){x) • We get similarly Ax = —K,/^{P){x)/{dP/dX){x) by considering P as an element of F[X]. This impUes that Dx = Ax since n£f{P) = K ^ ( P ) . Therefore, {EF, A) is a differential extension of (E, D). D It turns out that in an algebraic extension of a differential field, derivation commutes with conjugation. This technical point implies that the trace map commutes with the derivation, and gives a formula for the trace of a logarithmic derivative.

3.3 Constants and Extensions

85

T h e o r e m 3.2.4. Let [K^D) he a differential field. (i) Let F he a separahle algehraic extension of K. Then, any field automorphism of F over K commutes with D. (ii) Let E he a finitely generated separahle algehraic extension of K, and Tr : E —^ K and N : E -^ K he the trace and norm maps from E to K. Then, Tr commutes with D and rr[

(Da\

\ aJ

DNja)

=

N{a)

for any a^E

.

Proof, (i) Let i^ be a separable algebraic extension of K. By Theorem 3.2.3, D extends uniquely to a derivation of F. Let (j be a field automorphism of F over K and Da = a~^ o D o a. Since a is an automorphism, it follows that Da is a derivation of F. In addition, a is the identity on i^, so Dx = D^x for any x E K. By the unicity clause of Theorem 3.2.3, D = D^^ which implies that a o D = a o Da = D o a. (ii) Let JE be a finitely generated separable algebraic extension of K^ and Tr : E —^ K and N : E ^ K he the trace and norm maps from E to K. Let K be an algebraic closure of K containing £^, d i , . . . , cr„ be the distinct embeddings of E in K over K^ and F = (aiE) • • • (cfnE) be the normal closure of E in K. F is also separable over K ([54] Chap. VII, §4), and for each i, ai can be extended to a field automorphism of F over K. By Theorem 3.2.3, D extends uniquely to a derivation on F such that (F, D) is a differential extension of {K, D). Let a e E.By part (i) applied to F we have D^a""') = {DaY', so n

n

n

D{Tr{a)) = i ? ( ^ a - ' ) = J ^ D C a " ' ) = ^^(Dfl)-' = Tr{Da) 2=1

i=l

1=1

so D oTr = Tr o D. Furthermore, ai

n

-(^)^E(^r^E^ i=i

\

/

i=i

D

3.3 Constants and Extensions We study in this section the effect of extending differential fields on their constant subfields. We first mention the obvious fact that constants remain constants in an extension. L e m m a 3.3.1. Let {F^D) he a differential field and (E^A) a differential extension of (F^D). Then, Const£)(F) C Const^(E).

86

3 Differential Fields

Proof. Let c G Constp(F). Then c £ E and Ac = Dc = 0 since A extends D, so c G Consta (^). D In the next few lemmas, we consider the new algebraic constants that can appear in a differential extension. We first show that an algebraic constant must in fact be algebraic over the initial constant field, and conversely that any separable algebraic element over the initial constant field must also be a constant. L e m m a 3.3.2. Let {F^D) he a differential field and {E.,A) a differential extension of (F, D). Then, (i) c G Consta(F) is algebraic over F = ^ c is algebraic over ConstJD{F). (a) c E E is algebraic and separable over Const/) (F) = ^ c G Const^(F). Proof, (i) Suppose that c G Consta (£") is algebraic over F , and let

be the minimal polynomial for c over F. We have P{c) — 0, so dP 0 = A{P{c)) = nD{P){c) + {Ac) — {c) =

KD{P){C)

= {Dbn-i)c^~^ +

...^Dbo

so Dbi = 0 for i = 0 . . . n — 1 by the minimality of P . Hence P is in Const£)(F)[X], so c is algebraic over Consti:)(F). (ii) Let c G F be algebraic and separable over Constj3(F), and let P G Consti:) (F) [X] be its minimal polynomial over Consti:)(F). Then, dP = {Dc) — {c).

dP 0 = DiP{c)) = Kn{P){c) + {Dc)~{c)

Since c is separable over ConstD(F), dP/dX{c) ^ 0, so Dc — 0, which means that c G Consta (F). D As a consequence, when making a separable algebraic extension of a differential field, the new constants are exactly the elements of the extension that are algebraic over the initial constant subfield. In particular, the constants of the algebraic closure of a differential field of characterisric 0 form exactly an algebraic closure of the initial constant subfield. Corollary 3.3.1. Let (F^D) be a differential field and E a separable algebraic IT"

extension of F. Let also C = Constx)(F) and C be the algebraic closure of C in E, i.e. the subfield of all the elements of E that are algebraic over C. Then D can be extended uniquely to E and ConstD{E) = C . In addition, if E is algebraically closed, then Const£)(F) is an algebraic closure of C.

3.3 Constants and Extensions

87

Proof. The picture is as follows: • E

c^ {F,D)

c D can be extended uniquely to E by Theorem 3.2.3. Let c E ConstD(^). Since E is algebraic over F , c is algebraic over F , so c € C by Lemma 3.3.2. Hence, ConstD(E) C C n E. Conversely, let c e C 0 E. Then, since E is separable over F^ c E Consti:)(F) by Lemma 3.3.2, so Const£)(F) = C f) E. Suppose that E is algebraically closed. Then, C C E since C C F C. E. Hence, Consta {E) = CnE = C. D As a consequence of Corollary 3.2.1, we can always replace any field in a tower of differential extensions by its algebraic closure if the latter is separable. We show now that if the constant fields were equal in the initial tower, then they remain equal after such a replacement. The hypothesis that F is perfect in the next lemma just ensures that its algebraic closure is separable over it. All fields of characteristic 0 are perfect, so the reader interested in the characteristic 0 case only can ignore that hypothesis. L e m m a 3.3.3. Let (F^D) he a perfect differential field, (E^A) be a differential extension of (F^D), E be an algebraic closure of E, and F C E be an algebraic closure of F. Then, (EF^A) is a differential extension of {F^D), and ConstD{F) = ConstA{E) => ConstD(F) = C o n s t a ( E F ) . Proof Since F is perfect, F is separable over F , so {EF^ A) is a differential extension of {F,D) by Corollary 3.2.L Suppose that Constp(F) = Const^(F) = C, and let C be the algebraic closure of C. Since F is algebraically closed. Consti3(F) = C by Corollary 3.3.L Since EF is algebraic overj;, Const^(FF) =_Ü fl F F by Corollary 3.3.L But C C F , so Const zi (FT) =C = Const i^ (F). D As expected, adjoining a constant to a differential field extends the constant field by that constant only. L e m m a 3.3.4. Let (F, D) be a differential field and (F, A) be a differential extension of (F^D). Then, Const^(F(t)) = ConstD(F)(t) for any te

ConstA{E).

Proof Let C = Const/:) (F). Since Const^(F(t)) is a field containing C and t, it contains C{t). Since At = 0, we have A = KJJ on F[t] by Lemma 3.2.2.

88

3 Differential Fields

Suppose first that t is algebraic over F and let m = [F{t) : F]. Then, any u G F{t) can be written as ti = Yl^i^ ^^^* with a^ E F , so Au = KJJU = X^i^^ {Dai)t'^. Since l , t , . . . ,t"^~-^ are linearly independent over F , Zlii = 0 if and only if a^ G C for each z, so ConstA{F{t)) = C{t). Suppose now that t is transcendental over F , and let u = J2^=o^^^^ ^ -^WThen, Au = tvjju = ^^^^{Daijf'^ so Ziii = 0 if and only ii ai E C for each i. This implies that Const^(F[t]) = C[t]. Let now c E Const^(F(t)) and write c = u/v where u^v E F[t], gcd(u, v) = 1 and f 7^ 0. Dividing ix and v by the leading coefficient of v if necessary, we can assume that the leading coefficient of ^' is 1, which implies that either Av = 0 or deg(Ziv) < deg('ü). Suppose that Av 7^ 0. Since u/v is a constant, we have ^u vAu — uAv 0 = A- = V

V^

so 2; = vAu = uAv is a common multiple of u and v in F[t]. Since deg(2:) = deg(?i) + deg{Av) < deg(?i) + deg(^') = deg{uv) and lcm(i^, v) \ z, we have deg(lcm(ii, v)) < deg{uv). But lcm('u, v) gcd{u^ v) = uv, so deg(gcd('u, f)) > 0 in contradiction with gcd{u^v) = 1, Hence Av = 0, which implies that vAu = 0, hence that Au = 0. Therefore u^v G Const^(F[t]). But Constz^(F[t]) = C[t], so c = u/v G C{t). U Finally, we need a few results from differential algebra about properties involving constants which are preserved under differential extensions. Definition 3.3.1. Let (F, D) he a differential field and y i , . •. ^yn ^ F . The Wronskian ofyi,. ...y^ is W{yi, ...,yn)= det{M{yi, • • • ,yn)) where I

Dyi

Dy2

•••

Dyn

\D^-'yi

D^-'y2

•••

D^~'yn/

M{yi,...,yn)

\

(3.1)

The vanishing of the Wronskian is a well-known test in analysis for hnear dependence of functions over the constants. It turns out to have the same property in arbitrary differential fields. L e m m a 3.3.5 ([51]). Let (F, D) be a differential field. Then, y i , . . . , y n ^ F are linearly dependent over Const/:)(F) if and only if W ( y i , . . . , yn) = 0Proof We have W{yi,... ^y^) = 0 if and only if ker(A^) 7^ {0} where M is the matrix given by (3.1). Write C = Consta(F) and suppose first that y i , . . . , y^ are hnearly dependent over C. Then there are c i , . . . , c^i G C, not all 0, such that X^ILi ^iV^ " ^- Differentiating this an arbitrary number of times, we get that

3.3 Constants and Extensions

89

n

for any m > 0. This implies that ( c i , . . . , c„) G ker(A^), hence that ker(A^) ^ {0}andW^(yi,...,2/„) = 0. We proceed by induction on n for the converse. For n = 1, we have W{yi) = yi so if W{yi) = 0, then yi = 0 is Hnearly dependent over C. Suppose now that n > 1, that the lemma holds for any n — 1 elements in F , and that W ( y i , . . . ,yn) = 0. Then, ker(A^) / {0}, so let ( x i , . . . ,x^) be in kei{J\4) where Xi j^ 0 for some i. Renumbering the y^'s if necessary, we can assume that xi 7^ 0, hence that xi = 1 since ker(A^) is a vector space over F. Since ( x i , . . . ,Xn) G ker(A^), we have Yl^=i ^i^^Vi = 0 for 0 < j < n. Differentiating those equations for 0 < j < n — 1 and using them together with Dxi = Dl = 0, we get

(

n

\

n

n

n

i=l

J

i=l

i=l

i=2

so {Dx2,. ..,Dxn) e ker(A1(y2,---,yn))- If Dx2 = ... = Dx^ = 0, then x i , . . . , Xn £ C, so y i , . . . , y^ are linearly dependent over C. If Dxi ^ 0 for some i > 1, then ker(A^(y2, • • •, Vn)) i=- {0}, so ^2, • • •, Vn are linearly dependent over C by induction, which implies that ?/i,..., ?/^ are linearly dependent over C. D As a consequence, linear independence over the constants is independent of the constant field, hence preserved under differential extensions. Corollary 3.3.2. Let (F, D) he a differential field and (£", A) he a differential extension of (F^D). If S C F is linearly independent over Consti:)(F)^ then S is linearly independent over Const/^(£'). Proof. Let S C F he linearly independent over ConstD(F) and { s i , . . . , 5^} be any finite subset of S. Then, W ( s i , . . . , 5 ^ ) 7^ 0 by Lemma 3.3.5. But s i , . . . , s^ G F , so by Lemma 3.3.5 applied to (F, Zl), { s i , . . . ,s^} is linearly independent over Const/i(£'). Since this holds for any finite subset of 5, S is linearly independent over Const^(E). D The following lemma states that if an algebraic system of equations and inequations is satisfied by constants, then it is also satisfied by algebraic constants. L e m m a 3.3.6 ([51]). Let (F^D) he a differential field with algebraically closed constant field, (E^A) he a differential extension of (F^D), X i , . . . , X m he independent indeterminates over F, g G F [ X i , . . . , Xm] cind S he any suhset of F[Xi,..., Xjn]- If there are c i , . . . , c ^ G Const^(£') such that g{ci,... , c ^ ) 7^ 0 and / ( c i , . . . , Cm) = 0 for any f in S, then there are also such c i , . . . , c^ in Const D{F).

90

3 Differential Fields

Proof. Let C = ConstD{F)

and

V{S) = { ( a i , . . . , a ^ ) G C ^ suchthat / ( a i , . . . , a ^ ) = 0 for all f e S} . Since F is a field containing C, it is a vector space over C, so let ß be a vector space basis for F over C. Then, B generates F[Xi,..., Xm] as a free module over C [ X i , . . . , X ^ ] so write each / in 5 as / = J^beB ^f,b^ where hf^b G C[Xi,..., Xm] and all but finitely many of the hf^b are identically 0. Let / C C [ X i , . . . , Xm] be the ideal generated by all the hf^b and V{I) = { ( a i , . . . ,am) G C^ such that h{ai,...

, a ^ ) = 0 for all h e 1} .

By construction, we have V{I) C F(S'). Let c i , . . . , c^^ G Const^(-E) be such that g{ci,..., c^) 7^ 0, which implies that ö' 7^ 0, and / ( c i , . . . , c^) = 0 for all f E S. Then, for each / G 5, y]/i/,b(ci,...,c^)6 = 0 which implies that /i/^5(ci,... ,Cm) = 0 for each b E B, since >B is linearly independent over Const^(E) by Corollary 3.3.2. Suppose that 1 G / . Then, there are polynomials a/^5 G C [ X i , . . . ^Xm]^ all but finitely many of which identically 0, such that 6Gi3

Evaluating that equality at ( X i , . . . , X m ) == (ci, • •., c^) yields 1 = 0. Therefore 1 ^ / , so by Hubert's Nullstellensatz (Theorem LLIO) V{I) ^ 0. Write now g — Ylbeßdbb where ^5 G C [ X i , . . . ,X^] and all but finitely many of the gb are identically 0. Suppose that ^ ( a i , . . . , a ^ ) = 0 for every ( a i , . . . , a ^ ) G V{I). As previously, this implies that ^5(01,... ,am) = 0 for every b e B and every ( a i , . . . , a ^ ) G V{I)^ hence, by Hilbert's Nullstellensatz (Theorem L L l l ) , that there exist positive integers rib such that g^^ G / for each b E B. Since /i(ci,... ,c^) = 0 for every /i G / , we get 5f^'(ci,...,c^) = 0, hence gbi^ii • • • 5^m) — 0 for every 6 G i5, in contradiction with g{ci^..., Cm) 7^ 0. Hence there exist ( a i , . . . , am) G ^(1) such that ^ ( a i , . . . , a^) 7^ 0. Since V{I) C V(5), this proves the lemma. D

3.4 Monomial Extensions We want to study simple transcendental differential extensions of the form k(t) where there is some amount of similarity between the derivations D and d/dt, which will allow us to apply the algorithms for integrating rational functions to such extensions. Recall that if /c is a differential field, K a differential extension of k, and t an element of K, then k{t) is a differential field itself if it is closed under the derivation D of Ü^. A condition for some similarity with

3.4 Monomial Extensions

91

d/dt is that D transforms polynomials in t into polynomials in t, i.e. that k\t] is closed under D^. Therefore, we study here differential extensions where the derivatives of polynomials are polynomials. In addition, we now restrict our study to fields of characteristic 0, so for the rest of this chapter, fc is a differential field of characteristic 0, K is a differential extension of A:, and D denotes the derivation on K. We first show that the requirement that Dt G k\t] is equivalent to k\t] being a differential subring of k{t). L e m m a 3.4.1. Let t E K. Then, Dt G k[t] 4=^ k[t] is closed under D. Proof. Suppose that Dt G k\t], and let p G k[t]. By Lemma 3.2.2, Dp = nD{p) +

{Dt)^^ek[t]

so k[t] is closed under D. Conversely, if k[t] is closed under D, then Dt G k\t] since t E k\t\. D Note that we did not require that t be transcendental over k in the above lemma. We can now define the class of differential extensions for which the integration algorithm will be presented later. This class is general enough to model the usual elementary transcendental functions of calculus. It consists of simple transcendental extensions for which k\t] is closed under D. Definition 3.4.1. We say that t £ K is a monomial over k (w.r.t. D), if (i) t is transcendental over k, (ii) Dte k[t]. In addition, we define then the D-degree oft to be S{t) = deg^(Dt)^ and the D-leading coefficient oft to be X{t) = lct{Dt). We call t linear if 6{t) < 1, nonlinear otherwise. Furthermore we let Tit G k[X] be the polynomial such that Dt = nt{t). Since the derivative of polynomials are polynomials in monomial extensions, we often need to know the degree and leading coefficient of a derivative. L e m m a 3.4.2. Let t be a monomial over k, and p G k\t]. (i) deg(Dp) < deg(p) -f max(0, 6{t) - 1). (ii) Ift is nonlinear and deg(p) > 0, then equality holds in (i), and the leading coefficient of Dp is deg{p) lc(p) A(t). Proof. If p = 0, then Dp = 0 and (i) is satisfied under the convention that deg(O) = —(X), so suppose that p ^ 0 and let n = deg(p). (i) We have Dp = noip) + {Dt){dp/dt) by Lemma 3.2.2. If n = 0, then dp/dt = 0, so deg{Dp) = deg{tvD{p)) < n < n -{- max(0, 5{t) — 1). Otherwise n > 0, so deg{dp/dt) = n — I and deg{{Dt)dp/dt) = n + ö{t) — 1. Hence, ^This condition is probably not even necessary (Exercises 3.7 to 3.11) but the integration algorithms have not been generalized to the extensions of [71].

92

3 Differential Fields deg{Dp) < max I deg{tvjj{p)),deg{{Dt) —) | < max(n,n + ö{t) — 1) = n + max(0, ö{t) - 1).

(ii) Suppose that t is nonlinear and n > 0. Then, (5(t) > 1 and deg((Dt) dp/^t) = n + 6{t) - 1 > n > deg(KD(p)) so deg(I^p) = n + (5(t) — 1. Since the leading coefficient of dp/dt is na^ where a is the leading coefficient of p, the leading coefficient of Dp is naX{t). D Let t E K he a, monomial over k for the rest of this section. It is well-known that for D — d/dt, every squarefree polynomial has no common factor with its derivative, and this fact forms the basis of the various squarefree factorization algorithms. This fact is not always true for more arbitrary derivations, so we introduce a name for the polynomials for which it still holds. Definition 3.4.2. We say that p G k[t] is normal with respect to D if gcd(p, Dp) = 1. We say that p is special with respect to D i/gcd(p. Dp) = p i.e. p \ Dp. In addition, we introduce the following notations for the sets of special and special monic irreducible polynomials: "^klty.k == {p ^ ^M such that p is special}, ^khhk ~ {P ^ "^klty.k such that p is monic and irreducible} . When the monomial extension is clear from the context, we omit the subscripts and simply write S and S^^^. A polynomial is not necessarily normal or special, but an irreducible polynomial p £ k[t] must be either normal or special, since gcd(p, Dp) is a factor of p. Note that k C S^ and that p G k[t] is both normal and special if and only if (p) = (1), which is equivalent to say that p £ k"". Special polynomials generate differential ideals, so there is an induced derivation on the quotient rings (Lemma 3.L2). More importantly, this induced derivation turns out to be an extension of D. L e m m a 3.4.3. Let p £ S \ k. Then, (p) is a differential ideal of k[t] and (Ä:[t]/(p), D*) is a differential extension of (k^D) where D* is the induced derivation. Proof. Let p £ k\t]\k he special. Then, p | Dp by definition, so (p) is a differential ideal of k\t]. By Lemma 3.L2, D* o TT = TT o D, where D* is the induced derivation on k\t]/{p) and TT :fc[t]-^ ^M/(p) is the canonical projection. Hence, D^'a = D*7r(a) = 7r{Da) = Da for any a £ k, which implies that {k[t]/{p)^D*) is a differential extension of (Ä:,D). D

3.4 Monomial Extensions

93

L e m m a 3 . 4 . 4 . Let p i , . . . ^Pm ^ k[t] be such that gcd{pi^pj) = 1 for i j ^ j , and let p = H ^ ^ I P ^ * where the e^ 's are positive integers. Then,

(

m

\

i=l

m

J i=l

Proof. Let a, 5 E k[t] and suppose t h a t gcd(a, 6) = L Then, gcd(a5, D{ab)) = gcd(a, D{ab)) gcd(6, D{ab)) = gcd(a, aDb + bDa) gcd(6, aDb + bDa) — gcd(a, bDa) gcd(6, aDb) = gcd(a, Da) gcd(6, Db). So by induction, gcd(p, Dp) = H I ^ i gcd(p^'% Z)(p|')). In addition, gcd(pf ,D(pf)) = g c d ( p r , e , p r " ' ^ P ^ ) = Pf"^ gcd(pi, e^Dpi) = p^^""^ gcd(pi, Dpi) which proves the lemma.

D

As a consequence, any normal polynomial must be squarefree. In addition, we get t h e multiplicative properties of special and normal polynomials, in particular t h a t tS is a multiplicative monoid generated by k and S^^^. T h e o r e m 3.4.1. (i)

Any finite product of normal and two by two relatively prime is normal. Any factor of a normal polynomial is normal. (ii) p i , . . . ,pn G cS = ^ Yri=iPi ^ ^• (Hi) p G S \{0} =4> q G S for any q G k[t] which divides p.

polynomials

Proof, (i) Let p i , . . . ,Pm G k[t] be normal and such t h a t gcd(pi,pj) = 1 for i ^ j , and let p = Oliil^*- ^Y L e m m a 3.4.4 we have

i=\

)

i=\

since each pi is normal. Hence, p is normal. Let p G k\t\ be normal and write p = qh where q,h G k[t]. Since p is squarefree, we have gcd(g,/i) = 1, hence by Lemma 3.4.4, 1 = gcd(p. D p ) = gcd(g, Dq) gcd(/i, Dh)^ so gcd(g, Dq) = 1, which implies t h a t q is normal. (ii) Let a^b £ S^ then Da = ap and Db = bq for some p , g G k[t]. Hence, D{ab) — aDb + bDa = abq + bap = ab{p + q) so ab e S. P a r t (ii) follows by induction. (iii) Let p G 5 \ {0}, r G k[t] be an irreducible factor of p, and n b e t h e maximal exponent such t h a t r^ | p . Then, n > 1, since r | p, and p = r""/! for some h e k[t] with gcd(r, /i) = 1, so by Lemma 3.4.4,

94

3 Differential Fields r^/i = p = gcd(p, Dp) = gcd(r^/i, Dir^'h))

= r ^ - ^ gcd(r, Dr) gcd(/i, Dh).

Hence rh = gcd(r, D r ) gcd(/i,-D/i), which implies t h a t gcd{h^ Dh) = h and gcd(r, D r ) = r, hence t h a t r G S. Therefore, every irreducible factor of p must be special. Let now q E k[t] be any factor of p. If g E fc, then q G S hy definition. Otherwise, g is a nonempty finite product of irreducible factors of p, so it is special by part (ii). D As mentioned above, every normal polynomial must be squarefree. T h e converse is not always true however, and there is an important connection between t h e normality of a squarefree polynomial and the differential properties of its roots. This relationship is described by the following two theorems. T h e o r e m 3.4.2. Letk be the algebraic closure ofk, andp G k[t] be squarefree. Then, p normal 4=4> Da y^ ldt{a) for all roots a E k of p. Proof. Let p G k[t] be squarefree, and let a i , . . . , a^ € fc be t h e distinct roots of p, where n = deg(p) > 0. T h e factorization of p over k is then n

P = c J J ( t - ai) i=i

where c G k* is the leading coefficient of p. By Lemma 3.4.4 we have n

gcd{p,Dp)

n

= c J | g c d ( t - ai,D{t-ai))

= c J\gcd{t

i=l

- ai,Ht{t)

-

Dai).

i=l

Hence p is normal if and only if gcd(t — ai,Ht{t) — Dai) = 1 for each i. This is equivalent tot — ai does not divide Ht{t) — Dai in k[t]^ hence t o Dai ^ Ht{oii) for all i. D T h e o r e m 3 . 4 . 3 . Let k be the algebraic closure ofk, andp G k\t\ \ { 0 } . Then, p G S <=^ Da = T~it{(^) for all roots a G k of p. Proof

Let p G k[t] \ {0}, and let n

p=

cYl{t~ai)''' i=l

be the irreducible factorization of p over fc, where c G k* is t h e leading coefficient of p and e^ > 0 for each i. By Lemma 3.4.4 we have n

gcd(p, Dp) =cY[{t-

n

aiY^'^

n

= c]J{t-

Y[ gcd(i - ai, D{t - a^)) n

aiY'-^

IIgcd{t

- ai,nt{t)

- Dai) •

3.4 Monomial Extensions

95

Hence p is special if and only if gcd(t — ai^Ht{t) — Dai) = t — ai for each i. This is equivalent to t ~ ai | li.t{i) — Dai in fc[t], hence t o Dai — T~it{oii) for all i. D We make frequent use in t h e future of the trivial remark t h a t special and normal polynomials remain such when we make an algebraic extension of k. C o r o l l a r y 3 . 4 . 1 . Let E be an algebraic extension ofk. Then, t is a monomial over E. Furthermore, S^^yj. C SE[t]:E ^'^^ '^f P ^ k[t] is normal, then it remains normal when viewed as an element of E[t]. Proof, t is transcendental over k and E is algebraic over /c, so t is transcendental over E. Also, Dt £ k[t] C E[t] so t is a monomial over E. Let p E k[t], k be t h e algebraic closure of k containing E^ and a i , . . . , a^ be t h e distinct roots of p in k. If p is normal (resp. special), then Dai ^ Ht{o^i) (resp. Dai = Ht{ai)) for each i by Theorem 3.4.2 (resp. Theorem 3.4.3), so p is normal (resp. special) when viewed as an element of E[t] again by Theorem 3.4.2 (resp. Theorem 3.4.3). D As a consequence, in t h e case where all the elements of k are constants, the special irreducible polynomials are exactly the factors of H t , and t h e normal polynomials are exactly t h e squarefree polynomials t h a t are coprime with Wf C o r o l l a r y 3 . 4 . 2 . Suppose that Da = 0 for any a e k. (i) Let p £k\t] be monic and irreducible. Then, p G S^^^ <=^ p \T-it(ii) Let p G k[t] be squarefree. Then, p normal 4==> gcd(p, Ti^) = 1Proof, (i) Let p E k[t] be monic and irreducible, and suppose first t h a t p G S^^^. Let a G khe any root of _p. Then, Da = Ht{a) by Theorem 3.4.3. But a is algebraic over k^ so Da = 0 by Lemma 3.3.2, hence Ht{a) — 0. Since this holds for any root of p and p is irreducible, p \lrii. Conversely, let p G k\t\ be a monic irreducible factor of H t , and \<^t a G k be any root of p. Then, yitipL) = 0. B u t a is algebraic over k so Da = 0 as before, hence Da — Ht{a) so pG 5^"" by Theorem 3.4.3. (ii) Let p G k[t] be squarefree. Suppose first t h a t p is normal and let a G k be any root of p. Then, Da ^ T~it{oi) by Theorem 3.4.2. B u t Da = 0 as before since a is algebraic over k^ so 7it{a) ^ 0. Since this holds for any root of p, gcd(p, TYt) = L Conversely, suppose t h a t gcd(p, Ti^) = 1 and \et a Gk be any root of p. Then 1-Lt(a) ^ 0. But Da = 0, so p is normal by Theorem 3.4.2. D In particular, applying the above corollary t o t h e case D = d/dt, we have Dt = 1 = Hti so every squarefree polynomial in k[t] is normal with respect to d/dt. We have made no assumptions on t h e possible extensions of t h e constant field in a monomial extension, so we now look at t h e possible new constants of k{t). It t u r n s out t h a t new constants and special polynomials are closely related. Recall t h a t a monomial t is called nonlinear when deg^(Dt) > 2.

96

3 Differential Fields

L e m m a 3.4.5. Ifc£ ConstD{k{t)) then both the numerator and denominator of c are in S. Furthermore, ifcj^O andt is nonlinear, then both the numerator and denominator of c have the same degree. Proof. Let c £ Const£)(Ä:(t)) and write c = a/b where a^b E k[t], b ^ 0 and gcd(a, b) = 1. Then, bDa - aDb SO bDa = aD6, which implies that a \ Da and b \ Db, hence that a^b £ S. Suppose now that c / 0, t is nonlinear, and that deg(a) ^ deg(6). Since 1/c G ConstD(Ä;(t)), we can assume that deg(a) > deg(6). Write then c = p + e/h where p,e E k\t], deg(p) = deg(a) — deg(6) > 0, and e = 0 or deg(e) < deg(6). Then, ^ ^ bDe — eDb ^ r . ^, ^ 0 = Dc = Dp+ =Dp^q+(3.2) where g, r G k[t]^ {bDe — eDb) = qb'^ + r and either r == 0 or deg(r) < 2deg(6). Since t is nonlinear, we have 8{t) > 1 and deg(jDp) = deg(p) + 5{t) — 1 by Lemma 3.4.2, so deg(Dp) > ö{t) — 1, which means in particular that Dp ^ 0. Hence, e ^^^ 0, so deg(6) > 0, which implies that deg(eD6) = deg(e) + deg(6) + b(t) - 1 by Lemma 3.4.2, so deg(eD6) < 2deg(6) + b(t) - L Either e G A:, in which case deg(6De) < deg(6), or e ^ A:, in which case deg(6jDe) = deg(6) + deg(e) + 8(t) - 1 by Lemma 3.4.2, so deg(6De) < 2deg(6) + 8(t) - 1 in both cases. Hence deg(6De — eDb) < 2 deg(6) + 6(t) — 1, which implies that deg(g) = deg{bDe - eDb) - 2deg6 < 6{t) - 1 < deg(Dp) in contradiction with (3.2), so deg(a) = deg(6).

D

Thus, the existence of new constants in k{t) \ k implies that S^^^ is nonempty. The converse, whether the existence of nontrivial special polynomials imply the existence of a new constant, is a more difficult problem: a theorem of Darboux essentially states that if A; is a purely transcendental extension of its constant field, then the existence of sufficiently many elements of S^^^ is equivalent to the existence of a new constant in k{t) \k (see Exercise 3.6 and [27, 85, 93]). Fortunately, the situation is easier for the key monomial extensions appearing in the integration problem, where any element of S^^^ produces a new constant, as the next lemmas show. L e m m a 3.4.6. Suppose that Dt G k, and let p G k\t] be nonzero. Then, p^S

^=> D

[ P

\HP)

Proof. Let p G k[t] be nonzero, and write q = p/lc{p). If Dq = 0, then q \ Dq^ so g G vS, which implies that p e Shy Theorem 3.4. L Conversely, suppose that p E S. Then, q £ S hj Theorem 3.4.1, and write q = YYi=ii^ ~^^T'' where the

3.4 Monomial Extensions

97

a-i's are in the algebraic closure of k and the e^'s are positive integers. Then, Dai = T~it{o^i) = Dt for each i by Theorem 3.4.3, so n

D L e m m a 3.4.7. Suppose that Dt/t E k, and let p G k[t] he nonzero. Then,

pes ^=^ D fry-T^r-rrl = 0• Proof. Let p G k[t] be nonzero, and write q = p/lc{p). We have deg(g) = deg(p) and Dq - nqDt/t D (^) in for any integer n. Suppose that Diq/t'^^^^'f^) = 0. Then, Dq = deg{q)qDt/t, so q I Dg, which implies that g G ^S, hence that p G «S by Theorem 3.4.1. Conversely, suppose that p G 5, and write q = YYi=ii^ ~~ ^iY'' where the a^'s are in the algebraic closure of k and the e^'s are positive integers. Then, Dai = {Dt/t)ai for each i by Theorem 3.4.3, so n

Dq = J2 ^ii^^ - Dai)(t - aiY"~~^ jQ(t - ajf'^ ^Y^e,

—

which implies that D{q/t^^^^'i'^) = 0 .

(t-a,r^(t^a,r^

D

We need for later use to define one particularly interesting class of special polynomials. We first define some useful terminology. Definition 3.4.3. We say that u G k is a logarithmic derivative of a /c-radical if there exist v E k* and an integer n j^ 0 such that nu = Dv/v. Note that if n < 0, then we can write {~n)u = Dw/w where w = v~-^^ so we can always assume that the coefficient n is positive in the above definition. Example 3.4-i- Let k = Q{x) with derivation D = d/dx^ and u = l/(2x) G k. Since 2u = Dx/x^ li is a logarithmic derivative of a Q(x)-radical. In fact, u is the logarithmic derivative of y ^ , which is a radical over Q{x). On the other hand, Dv/v ^ Z for any -?; G /c*, so 1 is not a logarithmic derivative of a Q(x)-radical.

98

3 Differential Fields

It is clear from the definition that if we extend k to some extension field E^ then the logarithmic derivatives of /c-radicals become logarithmic derivatives of £^-radicals. However, when E is algebraic over k^ then an element of k that is not a logarithmic derivative of a /c-radical cannot become a logarithmic derivative of an E-radical. L e m m a 3.4.8. Let E be algebraic over k, and a G k. If a is not a logarithmic derivative of a k-radical, then it is not a logarithmic derivative of an E-radical. Proof. Suppose that a E k is not a logarithmic derivative of a Ä^-radical, and that there exist a E E* and an integer n / 0 such that na = Da/a. Since E is algebraic over k, let p G k[X] be the minimal polynomial of a over k^ and write p = X^ + Y^^=o ctiX'^ where the a^'s are in k and m > 1. Then, m—l

0 = D{p{a)) = mnaa^

+ V^ {Dai -\- inaa^)a* = q{a) i=0

where a = mna X ^ + EI^Ö ( ^ « ^ ^ + » " « « 4 ) ^ ' e fc[X]. Since p is the minimal polynomial for a over k^ p \ g, so q = mnap, which implies that Dai + inaoi = m n a a ^ for 2 = 0 , . . . , m — 1. Since p is irreducible and a / 0, a^ / 0 for some j in { 0 , . . . , m — 1}. We then have, Doj n[m — jja in contradiction with a not a logarithmic derivative of a Ä:-radical since n 7^ 0 and m. j^ j . D Definition 3.4.4. We say that q G k[t] is special of the first kind (with respect to D) if q E S and for any root a of q in the algebraic closure of k, Pa{o^) is not a logarithmic derivative of a k{a)-radical, where Dt-Da Pa = ——

,, ,,, € k{a)[t\.

In addition, we introduce the following notations: *^i,/c[t]:/c = {p ^ ^k[t]:k such that p is Special of the first kind} , ^^i\\t]'k ~ {P ^ *^i,/c[t]:/c such that p is monic and irreducible} . When the monomial extension is clear from the context, we omit the extension subscripts and simply write Si and Sf^. Note that since a is a root of the polynomial Dt — Da^ t — a \ Dt — Da in k{a)[t], so Pa{o:) is always defined. In addition, we remark that k* C Si by definition, and that we could have replaced "/c(a;)-radical" by "/c-radical" in the above definition in view of Lemma 3.4.8. Theorem 3.4.1 and Corollary 3.4.1 are easily generalized to special polynomials of the first kind, showing that *Si is a multiplicative semigroup generated by k* and Sf^.

3.5 The Canonical Representation

99

T h e o r e m 3.4.4.

(i)

Pi,.^-,Pn^'Si=>JXI;^^PieSi.

(a) p E Si ==4> q G Si for any q £ k[t] which divides p. (Hi) If E is algebraic over k, then Si^j^^tyj^ C Si^ß^ty^ Proof Let k be the algebraic closure of k. (i) Let pi^... ,pn G Si. Then, q = Pi- • -Pn ^ S hj Theorem 3.4.L Let a e k be a root of q. Then, a is a root of pi for some i and pi G Si, so Pa{(^) is not a logarithmic derivative of a Ä:(a)-radical, which implies that q E Si. (ii) Let p e Si and q € k[t] be any factor of p. Then, q E S hy Theorem 3.4.L If g G /c, then g ^ 0 (since p ^ 0), so q E Si. Otherwise, q ^ k, solet a E k be a root of q. Then, a is a root of p, so Pa(a) is not a logarithmic derivative of a A:(a)-radical, which implies that q E Si. (iii) Let p E Si and E be an algebraic extension of k. Then, p is special in E[t] by Corollary 3.4.1. Let a E k he a. root of p. Then, Pa(«) is not a logarithmic derivative of a A:(a)-radical, so it is not a logarithmic derivative of an E{a)radical by Lemma 3.4.8. Hence, p is special of the first kind when viewed as an element of E[t]. D

3.5 T h e Canonical Representation Given p E k[t], we want to separate the special and normal components of p. The following definition formalizes that separation. Definition 3.5.1. Let p E k[t]. We say that p — PsPn ^-^ CL splitting factorization of p if PniPs E k\t], Ps E S, and every squarefree factor of pn is normal A consequence of Theorems 3.4.2 and 3.4.3 is that a splitting factorization of p over k is also a splitting factorization of p over any algebraic extension of k, since Da = Ht{a) for all the roots of ps and Da ^ Ht{a) for all the roots of pn in k. For the same reason, we always have gcd(pn,Ps) = 1 in a splitting factorization of p, and such a factorization is unique up to multiplication by units in fc, like a prime factorization. It is clear that a prime factorization of p yields a splitting factorization of p, but it turns out that a splitting factorization can always be computed by gcd's only, like a squarefree factorization. T h e o r e m 3.5.1. Let p E k[t]. Then,

ß) gcd(p. Dp) gcd(p, dp/dt) is the product of all the coprime special irreducible factors of p. (ii) If p is squarefree, then p = PsPn ^^ CL splitting factorization of p, where Ps = gcd(p. Dp) and pn = p/ps-

100

3 Differential Fields

Proof, (i) Let p G k[t]^ A^i,... ,7V^ be all its coprime normal irreducible factors, and ^ i , . . . , 5^ be all its coprime special irreducible factors in k[t]. The prime factorization of p has then the form p = '^n?=i ^j'^ YllLi ^i\ ^^ by Lemma 3.4.4 applied to both D and d/dt^ we have ^ _

gcd(p, Dp) ;cd(p, dp/dt)

nu sf nT=i Nr' uu g^d(^j, J^,) nr=i gcd(iv,, pm uu <'•"' uti ^r' uu gcd{s„ds,/dt) uZi gcd(iv„ dm/dt) n;=i gcd{s,,dSj/dt) iiT=i gcd{Ni,dNi/dt) • Each Ni and Sj is irreducible, so gcd(A^i, dNi/dt) = gcd{Sj^dSj/dt) = 1. Each A^^ is normal with respect to D, so gcd{Ni^DNi) = L Each 6'j is special, so gcd{Sj^DSj) = Sj. Therefore, S = n ? = i '^j^ which is the product of all the coprime special irreducible factors of p. (ii) Suppose that p £ k[t] is squarefree. Then gcd{p^dp/dt) = 1, so, by (i), Ps = gcd(p, Dp) is the product of all the coprime special irreducible factors of p. But p is squarefree, so pn = p/Ps has no special irreducible factor, which impHes by Theorem 3.4.1 that Pn is normal. D This theorem gives us two algorithms for computing splitting factorizations: the first is to compute S = gcd{p, Dp)/ gcd{p^ dp/dt) and q = p/S. If S £ k^ then p has no special irreducible factor, so return pn = p^Ps = 1Otherwise deg{q) < deg(p), so recursively compute a splitting factorization q = qnqs of q and return p^ = q^^p^ = Sqs. Split Fact or (p, D)

(* Splitting Factorization *)

(* Given a derivation D on k[t] and P ^ k[t], return (Pn ,Ps) e k[t]^ such that p = p-,^Ps, Ps is special, and each squarefree factor of pn is normal. *) S 4- gcd(p, Dp)/ gcd(p, dp/dt) if deg{S) --= 0 t h e n return(p, 1) iQn,qs) ^ SplitFactor(p/5', D) return{qn ,Sqs)

(* exact division *) (* exact division *)

Example 3.5.1. Let k = Q(x) with D = d/dx^ and let t be a monomial over k satisfying Dt = - t ^ - — t + —

(3.3)

i.e. t represents a transcendental function solution of the above differential equation. Applying SplitFactor to

3.5 The Canonical Representation

101

we get: 1. Dp = -20xH^ + 2x^(8x + l)t^ + 3x^{4x + 7)t^ 7 15 -(12a:^ + 25^2 - -x)t^ + {7x^ " "T^ " ^)^^ +(2^2 4- 5a: + 4 - —)t - 2x + — zx zx 2. gcd(p, L^p) = t^ - (2x + 3)t/(4x2) + (2x - l)/(4x2) 3. dp/dt = 20xH'^ - 16x2(x +1)^^ 4-Bx^(2x - 3)t2 + 2x(2x2 + 7x + 2)t - 4^^ 4x + l 4. gcd(p,dp/dt) =t — 1/x

5. 5 = t 2 + t / x - ( 2 x - l ) / ( 4 x 2 ) 6. pi = p/S' = 4.xH^ - 4.x^{x + 2)t^ + 4x2(2x _^ ^^^ _„ 4^2 7. recursive call, SplitFactor(pi, D): a) Dpi = -12xH^ + 2x3(4x + 7)t^ - 2x^{3x + 2)t2 - 2x(2x2 - 2x - l)t + 4x2 _ Q^ b) gcd(pi,Dpi) = t - 1 / x c) dpi/dt = 12xH^ - Sx^{x + 2)t f 8x^ + 4x2 d) gcd(_pi, dpi/dt) = t — 1/x e)Si = l S. qn=pi, Qs = S -1 = S So a splitting factorization of p is P = PnPs = {AxH^ - 4x^(x + 2)t2 + 4x2(2x + l)t - 4x2) A2 _^ 1^ __ 2 x _ l ^ In addition, the roots of Ps are 2x

2V X

which are indeed algebraic functions solutions of (3.3), as expected from Theorem 3.4.3. The second algorithm is to compute first a squarefree factorization p ~ P1P2 • • 'Prn of P? ^^d then compute Si = gcd(pi, Dpi) and A^^ = Pi/Si. By Theorem 3.5.1, _p = PsPn is a splitting factorization of p, where p^ = S1S2 - • - S^ and pn = N1N2 • • • N^. This approach has the advantage of also giving us squarefree factorizations for ps and. Pn- Furthermore, Yun's algorithm can be used for the initial squarefree factorization.

102

3 Differential Fields

SplitSquarefreeFactor(p, D)

(* Splitting Squarefree Factorization *)

(* Given a derivation D on k[t] and p G k[t], return (iVi,... ,Nm) and (5'i, ...,Sm)m /c[t]^ such that p = {NiN^ • • • Ar;;^)(5iS'| • • • S^^) is a splitting factorization of p and the Ni and Si are squarefree and coprime. *) (pi, •. • ,Pm) <- Squarefree(p) for 2 <— 1 t o m d o Si 4 - gcd(p^,Dpi)

Ni ^ Pi/Si r e t u m ( ( i V i , . . . , iV^), ( 5 ' i , . . . , Sm))

(* exact division *)

Example 3.5.2. Applying S p l i t S q u a r e f r e e F a c t o r to the polynomial p of the previous example with t h e same monomial extension, we get: 1. p = PIPI

= {AXH^

- ^x{x

- l)t^

- {^x - l)t + 2x - 1) {xt -

1)2

2. Dpi = -UxH^ 3. 4. 5. 6. 7.

+ 2xiAx - 9)t^ + (16x - 9)t2 _ ( 4x - 7 + ^ ^ ) t - 1 + ; ^ \ 2a: y 2x

5 i = gcd(pi, i^pi) =e^ t/x - {2x - l ) / ( 4 x 2 ) ATi = p i / 5 i = 4 x 2 ^ - 4 ^ 2 Dp2 = -xt^ - t/2 + 1/2 6'2 = gcA{p2,Dp2) = 1 iV2-p2/'^2 = ^ t - l

So we get the splitting factorization p = PnPs with squarefree factorizations of pn and Ps: Pn = iViiVl = 4j:2(t - l ) ( x t - 1)2 and

o 1 2a:-1 p , = 5 i - t^ + - t - — - y - . X 4x^

We can now define a decomposition of the elements of k{t) t h a t generalizes the canonical representations f = p + a/d of rational functions. Let / G k{t) \ {0} and write / = a/d where a^d E k[t]^ gcd(a, d) = 1 and d is monic (such a representation is unique). Let d = dgdn he a, splitting factorization of d with respect to D with dg and dn monic, which makes this factorization unique. Then, there are unique p, 6, c G k[t] such t h a t deg(6) < deg(
ds

dn'

We call this decomposition, which is unique, the canonical representation of f with respect to D. We also introduce the notations fp — p (the polynomial part of / ) , fs = b/dg (the special part of / ) , and / „ = c/d^ (the normal part

off).

3.5 The Canonical Representation C a n o i i i c a l R e p r e s e n t a t i o n ( / , D)

103

(* Canonical Representation *)

(* Given a derivation D on k[t] and / G k(t), return {fp,fs,fn) G k[t] x /c(t)^ such that f = fp + fs + fn ^s, the canonical representation of / . *) {a,d) ^- (numerator(/), denominator(/)) (q,r) ^- PolyDivide(a,(i) {dn,ds)^ SplitFactor(c^, D) (6,c) <— ExtendedEucIidean(
(* d is monic *)

(* deg(5) < deg{ds) *)

We need to define a few more terms t h a t are often used later. A rational function over C is called simple if it has only simple affine poles, i.e. poles of order one only. This is equivalent to having a squarefree denominator. Since normal polynomials are t h e analogue of squarefree polynomials in monomial extensions, it is n a t u r a l to call an element of k{t) simple if it has a normal denominator. Similarly, a usual polynomial can be seen as a rational function with no affine poles, or a rational function with no denominator. The useful analogue in monomial extensions is a function with no normal affine poles, i.e. with poles at most at infinity and at special polynomials. This means a function whose denominator is special. D e f i n i t i o n 3 . 5 . 2 . Let f G k{t). We say that f is simple with respect to D if the denominator of f is normal w.r.t. D. We say that f is reduced with respect to D if the denominator of f is special w.r.t. D. In addition we write k{t) for the set of all the reduced elements of k{t). Obviously, k[t] C k{t). It will be shown in the next chapter t h a t , like k[t], k{t) is a differential subring of k{t). There is an application of splitting factorization t h a t will be useful in the sequel: its use in separating the constant from the nonconstant roots of a polynomial over a differential field. Let K he 8. differential field of characteristic 0, X an indeterminate over K^ p £ K[X] and suppose t h a t we want to separate the constant roots ofp from the others. It turns out t h a t this is just a splitting factorization with respect to the coefficient lifting KD of D on i f [X]. T h e o r e m 3 . 5 . 2 . Let (K^D) be a differential field of characteristic 0, K the algebraic closure of K and X an indeterminate over K. For any p G K[X] \ {0}^ letp — PsPn be a splitting factorization ofp w.r.t. KJJ. Then, for any root a of p in K, ( Da = 0 4=> ps{a) = 0 , [Da^O 4=4> pn{a)=0. Proof

Let a e K he SL root of p. Then, by Theorems 3.4.2 and 3.4.3,

(Da = nt{a) {Da^Htia)

^=^ p.(a) = 0, <=^ Pn{a)=0.

But KDX = 0, so Ht = 0, which proves the theorem.

D

104

3 Differential Fields

Exercises Exercise 3.1 (Logarithmic derivative i d e n t i t y ) . Let R be an integral domain, D a derivation on Ä, t i i , . . . , u^ G i?* and e i , . . . , e^ € Z be integers. Show that ~ ^1

— = ei • • • Un

h en

1 Ui

. Un

Exercise 3.2. Let Q be the field of rational numbers, x be an indeterminate over Q, and D be the derivation d/dx on Q(x). Let P = F^ - 2^^ G Q(x)[y], and ?/ be a root of P , which is irreducible over Q(a:). Show that y

Const£)(Q(a:,y)) = Q(a)

where a= — .

Exercise 3.3. Let (F^D) be a differential field and (E^A) a differential extension of (F^D). Show that if S* C Const^(jB) is algebraically independent over Consti:)(P), then S is algebraically independent over F. Exercise 3.4. Let (fc, D) be a differential field of characteristic 0, and u e khe such that u^ ^ —1. Let £ be a differential extension of k such that ^f—i G E^ and let ti, t2 G -E be solutions of the following differential equations: ^ Dv , u + \J—\ Dt\ = where t» = -== ,

^ jüt2 =

-Du

i.e. t\ is a logarithm of v and ^2 is an arc-tangent of u. Show that ti\f-i is a constant with respect to D.

— 2t2

Exercise 3,5. Let (fc, D) be a differential field of characteristic 0, t be a monomial over k, and E be an algebraic extension of k. Show that

(Compare with Corollary 3.4.1). Exercise 3.6. Let (ÜT, D) be a differential field, {E^ A) a differential extension of (K", D), and t i , . . . ,tn G £ be algebraically independent over K. a) Show that the map KJ} : K f t i , . . . , t^] —> i ^ [ t i , . . . , t^] given by

'«D f E Mr • • • ^:" j = E(^««)*l' • • • *n" is a derivation on i^[ti,..., t^]. b) Show that if K{ti,... ,tn) is closed under Zi, then n

4=1

on jFf(ti,...,tn).

7

3.5 The Canonical Representation

105

c) Suppose that K C Const^(£^), that k = i ^ ( t i , . . . ,tn-i) is closed under Ä and that tn is a monomial over k. (i) Show that I (A) = {h e K[ti,...

,tn-i] s.t. hA is a derivation on K[ti,...

,tn]}

is an ideal of K[ti,... ,t^_i] and that I (A) / {0}. (ii) Let h G I{A)^ p G k[tn] and g € i^[ti,... ,tn_i] be such that P = qp is in K[ti^ • • • ,^n] (for example g can be the common denominator of the coefficients of p). Show that if p is special with respect to A, then P divides hAP in K[ti,... ,t„] (Hint: use Exercise 1.15). Such a P is called a Darhoux polynomial for hA^ and the existence of sufficiently many two-by-two coprime Darboux polynomials is equivalent to the existence of new constants in k{t)\k (see [27, 85, 93]). Although we have defined normal and special polynomials in monomial extensions only, Rao [71] has defined them in any simple transcendental differential extension as follows: let (A:,D) be a differential field of characteristic 0 and {k{t),A) be a differential extension of {k,D) where t is transcendental over k. Then, At G fc(t), so let a^b £ k[t] be such that 6 7^ 0, gcd(a, 6) = 1 and At = a/b. Define then p G k[t] to be normal with respect to A if gcd{p^bAp) = 1, and special with respect to A if p \ bAp. The following exercices all relate to this definition. Exercise 3.7. Prove that if At = a/b for a^b £ k[t]^ then bAp £ k[t] for any p£ k[t]. Exercise 3.8. Prove that all the parts of Theorem 3.4.1 remain true with the above definition. Exercise 3.9. Prove the following analogue of Theorem 3.4.2: let k be the algebraic closure of fc, and p £ k[t] be squarefree. Then, p normal <==^ b{a)Aa 7^ a{a) for all roots a £ k of p. Exercise 3.10. Prove the following analogue of Theorem 3.4.3: let k be the algebraic closure of A:, and p £ k[t]\ {0}. Then, p special <^=^ b{a)Aa = a{a) for all roots a £ k of p. Exercise 3.11. Prove that if p G k[t] is special, then gcd(p, 6) = 1.

The Order Function

We introduce in this chapter the order function at an element, which will be our main tool later when we prove the correctness of the integration algorithm. The usefulness of this function is that it maps elements of arbitrary unique factorization domains into integers, so applying it on both sides of an equation produces equations and inequalities involving integers, making it possible to either prove that the original equation cannot have a solution, or to compute estimates for the orders of its solutions. Therefore it is used in many contexts besides integration, for example in algorithms for solving differential equations. While we use only the order function at a polynomial in the integration algorithm, we introduce it here in unique factorization domains of arbitrary characteristic, and study its properties in the general case when the order is taken at an element that is not necessarily irreducible.

4.1 Basic Properties Throughout this section and the next one, let Z) be a unique factorization domain of arbitrary characteristic, D* be its group of units (Definition 1.1.4), F be its quotient field (see example 1.1.14), and a G D be such that a ^ 0 and a ^ D*. Definition 4.1.1, The order at a is the map z/^ : D --> Z U {+00} given by: (i) Ua{0) = + 0 0 ,

(a) for X £ D\ {0}, v'aix) = max{n G N such that d^ | x}. Even though the map Va takes only nonnegative values, we define it as a map into Z U {+00} in order to extend it eventually to the quotient field of D. We first show that the set Sa{x) = {n G N such that a'^ | x} is finite and nonempty for x G i^ \ {0}. Since a ^ 0 and a is not a unit in D, let p G D be an irreducible factor of a. Then there is an irreducible factorization of x in which p appears with some exponent e > 0. Let n > e and suppose that

108

4 The Order Function

p^ I X. T h e n x = p^y for some y E D. Let y = uf^iLiPl' be the irreducible factorization of y, where the p^'s are coprime and li is a unit. We have t h e n a factorization x = up^ YYiLiVT ^^ ^ where p appears with exponent at least n > e, in contradiction with D being a unique factorization domain. Thus any q in Sa{x) satisfies g < e, so Sa{x) is finite. In addition 0 G Sa{x), so Ua is well-defined on D. L e m m a 4 . 1 . 1 . Let x^y E D. (i) (ii) (Hi) (iv)

Then,

Ua{xy) > z^a(^) + ^a{y) OL'^d equaUty holds if a is irreducible. Ua{x-^y) > mm{iya{x),iya{y)) and equality holds if Ua{x) ^ i^a{y)Ifx\y, then Va{x) < Pa{y)Ua{gcd{x,y)) = min(z/a(a:), z^a(y))-

Proof. All the statements are trivial if either x or y is 0, so suppose t h a t X 7^ ^ ¥" y- Let n = Uaix) and m — i^aiv)- Then x = coP' and y — daT" for some c^d G D , and a divides neither c nor d. (i) we have xy — cddP'^^ so Vaixy) > n -{- m. Suppose t h a t a is irreducible. Then a j/cd since it does not divide c or «i, so a ^ + ^ + i j(xy^ which implies t h a t Va{xy) = n + m. (ii) we can assume without loss of generality t h a t n < m. We have t h e n X -j- y = 0^(0-^- dd^~^) so Va{x + y) > n = min(n, m ) . Suppose t h a t n j^ m^ then 771 — n > 0, so a I dd^"^^ which implies t h a t a)({c-\- da^~^) since a )(c. Hence, //^(x ^ y) = n. (iii) Suppose t h a t x \y. Then y = xz for some z £ D. Hence i^aiv) — ^a(^^) ^ //^(a:) + Ua{z) by part (i), so //«(?/) > ^aG^). ( i v ) L e t ö ' = gcd(x,y).Thenöf | x and g \ y, so Ua{g) < i^aix) a,nd i^aig) < ^a{y) by part (iii). Hence Va{g) < min(zy^(a:), !/«(?/))• Let z = a"^i^(^-(^)'^"(?^)) e D. Then, z | x and z \y, so z \ g. Hence, z/a(ö') > ^a{z) = mm{ua{x),Ua{y)) by part (iii), so Va{g) = min(z/a(:z^), i^a(l/)). • Example 4.1.1. In Z we have z/6(12) = ^IS) = 1 and i/6(12 x 18) = UQ{216) = 3, which shows t h a t the equality in (i) above does not always hold if a is not irreducible. On the other hand, z/3(12) = 1,1^3(18) = 2 and 1/3(216) = 3 == 1 + 2 , as well as z/2(12) = 2,1/3(18) = 1 and z/2(216) = 3 = 1 + 2. The following lemma shows t h a t multiplying a or the argument of Ua by a unit does not change the order function. This property is necessary in order to extend the definition of z/^ to F. L e m m a 4 . 1 . 2 . Let u e D"" and x e D. (i)

Va{ux)

= Z/a(x) =

Then:

Vua{x).

(ii) Va{u) = 0 . Proof, (i) If X = 0, then Uaiux) = i/a(x) = i^uai^) = +00, so suppose t h a t X j^ 0. T h e n a^"'^^^ | x, so a^"-^^^ | ux so Va{x) ^ i^ai^x). Since this inequality holds for any unit, and u'"^ is also a unit in D , we have z/a('w^) < Uaiw^ux) =

4.1 Basic Properties

109

z/a(x), SO Vaix) = Ua{ux). Similarly, a^''-^^^ \ x implies that (uaY"'^^^ | x since ^^aXx) Ig ^ unit, so Ua{x) < lyuai^)- As previously, this inequality applied to ua and u~^ implies that I'uai^) — ^u-'^ua{^) = ^a{x)j so Uai^) = ^ua{x)(ii) By (i), Vaiu) = Va{u^)- But Uai'u'^) > ^^^aiu) by Lemma 4.1.1, so Uaiu) G {0, +00}. Since ?i / 0, we must have i^a('^) = 0D In the following definition and the rest of this chapter, we say that any two elements y and z in D have no common factor when gcd(y, z) is a unit in D. Definition 4.1.2. Let x G F* and write x = y/z where y^z E D have no common factor and z ^ 0. We then define Va{x) = i^aiv) — i^a{z)Let X E F* and y^z^t^w G D be such that y and z have no common factor, t and w have no common factor, and x = y/z = t/w. Then yjt = z/w = u £ D*, so i^aiy) = i^a{ut) = Uait) ^.ud z^a(^) = ^ai^w) = Vai"^) by Lemma 4.1.2, so ^a{y) — ^a{z) — ^^a(0 — ^a('^)? which shows that Va Is wcll-defined on F . In addition, VaiX) — 0 by Lemma 4.1.2, so choosing y = x and z = 1 when X G D, we see that the above definition is compatible with the definition of i/a on D. We note that parts (i) and (ii) of Lemma 4.1.1 do not remain valid over F: ^5/3) = lyßilß) = 0, but i/6(5/3 x 1/2) = z/6(5/3 + 1/2) = - 1 < 0. Those statements remain however true when a is irreducible. T h e o r e m 4.1.1. Let x^y G F and suppose that a is irreducible in D. Then^ (i)

Ua (xy)

= Va {x) + l^a (v) •

(ii) if X ^ 0 then I'aix'^) = mUa{x) for any m GZ. (Hi) Ua{x -i-y) > min(z/a(x), J^a{y)) and equality holds if Va{x) /

faiv)-

Proof. Let x^y E F and write x = b/c^y = d/e where b^c^d^e G D, b and c have no common factor, d and e have no common factor and c 7^ 0 ^ e. Since a is irreducible, we have I'aifg) = ^a{f) + ^a{g) for any / , ^ G D by Lemma 4.1.1. (i) Let h — gcd(6(i, ce), / = bd/h and g = ce/h. We have f^g^heD^ f and g have no common factors, and xy = bd/ce = f/g^ so ^a(a:t/) = Va{f)

~ J^a{g) = ^a{f)

= l^aifh)

- Va{gh)

+ ^a{h)

= Va{bd) -

- {Va{g) +

^aih))

Va{ce)

= {^a{b) - I/a(c)) + {Va{d) - Va[^)) = Z^a(^) + ^a{v) •

(ii) a:^ = 1 is a unit in D, so z^a(l) = 0 by Lemma 4.1.2. Suppose that the statement holds for ?7i > 0. Then, Uaix"^^^)

= Uaix'^x)

= Vai^"^)

+ ^a{x)

= mUa{x)

+ M^)

= (m +

l)Ua{x)

SO it holds for m + 1. Thus (ii) holds for all m > 0. For m < 0 we have 0 = Ua{l) = Z / a ( x ^ X ~ ^ ) = Uaix"^)

holds for all m G Z.

- mUa{x),

SO Z/a(x^) = miya{x).

T h u S (ii)

110

4 The Order Function

(iii) X -^ y = {be -\- cd){ce)'~^. Although be + cd and ce may have common factors, we have Ua{x + y) = i^aibe + cd) + Ua ((ce)~^) = Uaibe + cd) - Ua{ce) by parts (i) and (ii). We can suppose without loss of generality that Ua{x) < ^a{y)i which implies that z^a(^) — ^a(c) < ya{d) — ^^a(e), hence that Vaib) + z^a(e) < ^a(d) + z^a(c). Thus, z/a(&e) < i/«( Va(be) by Lemma 4.1.1, so z/a(^ + y) ^ ^a(^e) — z/a(ce) = z^a(&) — i^a(c) = ^aip^)- Suppose that Vaip^) < ^a{y)j then iJa{be) < Va{dc) as above, so Va{be + cd) = i^a{be) by Lemma 4.1.1, so Ua{x + y) — i^aibe) — z/a(ce) = Uai^). • Parts (i) and (ii) of the above theorem show that the restriction that y and z have no common factor in Definition 4.1.2 can be removed if a is irreducible: for any y^z G F such that x = y/z, we have h'a{x) = h'a{yz~^) = ^a(y) — ^a(^)In the case of polynomial rings, we need to study the effect of enlarging the constant field on the order function. It turns out that when an irreducible polynomial splits in an algebraic extension, then the order at the new irreducible factors remains the same as before for arguments that are defined over the ground field. T h e o r e m 4.1.2. Let F be a field, E be a separable algebraic extension of F and X be an indeterminate over E. If p G F[x\ is irreducible over F, then ^p{f) = ^qif) /^^ ^'^y irreducible factor q G E[x] of p in E[x], and any

fGFix). Proof Let q G E[x] be any irreducible factor ofp in E[x] and write p = qr with r G E[x]. Let h G F[x] and n = Up{h) > 0. Then p"" \ h, so h = p'^s = q^'r^'s with s G F[x]^ which implies that q^ \ h. In addition, p^~^^ / / i , so p does not divide 5, which implies that gcd(p, s) = 1 since p is irreducible in F[x]. Thus, I = ap -]- bs = arq + bs for some a^b e F[x], so gcd(g, s) = 1. Suppose now that q^ I h for some m > n. Then, h = p^s = q^r'^s = q^t for some t G E[x]^ so r^s = q^~~^t^ which implies that q \ r^s in E[x]. Since q is irreducible in E[x] and gcd(g, s) = 1, g | r"', which implies that n > 0 (otherwise q would be a unit) and that q \ r, hence that q^ | p, in contradiction with p squarefree in E[x] (since E separable over F). Hence g"^ / / i for m > n, so i'q{h) — n. Let now / G F{x) and write / = a/b for a^b £ F[x] and 6 7^ 0. By Theorem 4.1.1 and the above proof, ^p(/) = ^'p(a) — iyp{b) = z/g(a) — z^g(6) =

^.(/).

•

4.2 Localizations Definition 4.2.1. We define the localization at a to be Oa = f^ {x e F such that Up{x) > 0} p\a

where the intersection is taken over all the irreducible factors of a in D.

4.2 Localizations

111

Intuitively, the localization at a is the set of all the fractions in F t h a t can be written with a denominator having no common factor with a. If a is irreducible, the localization, which is then a local ring, can also be seen as the set of all t h e fractions in F with nonnegative order at a. L e m m a 4.2.1. (i) (ii) (Hi) (iv) (v) (vi)

Oa is a subring of F containing D. xeOa=^ iya{x) > 0. X € aOa <=^ ^a(^) ^ 1; where aOa is the ideal generated by a in OaIf a is irreducible, then x ^ Oa <==r- z^a(^) ^ 0. If a is irreducible, then xa~^"'^^^ G Oa for any x E F*. If A is any derivation on D, then AOa ^ ^a-

Proof, (i) Let p G D be any irreducible factor of a, and x , y G Oa- Then ^p{x) > 0 and iyp{y) > 0. By Lemma 4.1.2, iyp{—y) = i^p{y)^ so iyp{x — y) > 0 and Up{xy) > 0 by Theorem 4.1.1. Since this holds for any irreducible factor p of a, X — y £ Oa and xy E Oa- Let c E D. Then, iyp{c) > 0 for any irreducible factor p of a, hence D C O^, so in particular 0,1 G Oa, and Oa is a subring of F containing D. (ii) Let X G Oa and write x = b/c where b^c £ D have no common factor. Let p E D he any irreducible factor of a. Then h'p{x) > 0, so i^p(6) — i^p{c) > 0. Since z/p(6) and i'p{c) cannot be both nonzero (otherwise p would divide both b and c) and since they are both nonnegative, this implies t h a t I'pic) = 0, hence p / c , so a / c , so Va{c) = 0, which implies t h a t Ua{x) = i^aib) > 0. (iii) Let x G aOa^ then x = ay for some y £ Oa- Write y = b/c where b^c E D have no common factor. From the proof of part (ii) we have p )( c for any irreducible factor p of a, so c and ab have no common factor. Hence, ^a{x) = I'aicdj/c) = ^a{cib) — ^a(c). But i^a(c) = 0 from the proof of part (ii), and Va{cib) > Ua{ci)-{-iya{b) > 1 by Lemma 4.1.1. Hence, Uaix) > 1. Conversely, let X E F* he such t h a t Uai^) > 1, and write x = b/c where b,c E D have no common factor. At most one of Uaib) and z^a(c) can be nonzero, and Ua{x) =• i'a{b) — Va{c) > 1, SO Va{c) = 0 and Va{b) > 1, which implies t h a t a I 6, so 6 = ad for (i G D. Let p G D be any irreducible factor of a. Then, p I 6, so p / c , hence Up{d/c) = Vp{d) — i'p{c) = Vp{d) > 0. Since this holds for any irreducible factor of a, we get d/c E Oa^ hence x = b/c = ad/c E aOa(iv) We have x E Oa = ^ J^a{x) > 0 by part (ii). Conversely, suppose t h a t a is irreducible and let x G F be such t h a t Uaix) > 0. Let p be any irreducible factor of a in D. Then p = ua foi u E D*, so Up{x) = i^uai^) = i^a{x) by Lemma 4.1.2. Thus z/p(x) > 0, so x G O^. (v) Suppose t h a t a is irreducible, and let x E F*. Then, i/a(^ö~^''^^^) = i^aix) — ^a{x) = 0, SO xa~^°'^^^ G Oa by part (iv). (vi) Let A he any derivation on D. Then, A can be extended uniquely to a derivation on F by Theorem 3.2.1. Let x E Oa and write x = b/c where b,c E D have no common factor and c ^ 0. Let p G D be any irreducible

112

4 The Order Function

factor of a. Then, z/p(x) = i^p{b) — Jyp{c) > 0 since x G Oa^ so i^p{c) = 0, which imphes that z/p(zlx) = Up (

2

~ ) = ^pi^^b - bAc) - 2z/p(c) = iyp{cAb - bAc) > 0 .

Since this holds for any irreducible factor of a, we get Ax G Oa, hence AOa C

a.

•

Example 4.2.1. In D = Z, Oe = ^2 n C^s = {a: G Q such that x = b/c where 6, c G Z, 2 / c and 3 / c } so 1/3 ^ Oß although 1/^(1/3) = 0, which shows that parts (iv) and (v) of the above lemma do not always hold if a is not irreducible. This makes it worth noticing that both directions of part (iii) of the lemma hold for non-irreducible a's. When D is a principal ideal domain, for any proper nonzero ideal / of D, the canonical projection TTJ : D —^ D/I can be extended naturally to the localization Oa for any generator a of / . The next definition constructs this extension. Definition 4.2.2. Let D be a principal ideal domain, and I be a proper nonzero ideal of D, i.e. I ^ D and I :^ [0), and a G D be a generator of I, i.e. I = (a). We define the value at a to be the map Tia '• ^a ~^ D/I given by: let x £ Oa and write x = b/c where b^c G D have no common factor. We define 7ra{x) to be 7rj{bd) where d^e G D are such that cd i- ae = 1 and TTJ is the canonical projection from D onto D/I. In order to show that TT^ is well-defined, we need to show that such d and e always exist, and that the value of na{x) is independent of the choice of 6, c, d and e. First, a 7^ 0 since I ^ (0), and a ^ D* since I y^ D^ so Oa is defined. Let X G Oa and write x = b/c where b^c G D have no common factor. Let p be any irreducible factor of a. Since x G (9a we have Vp{x) — i^p{b)—iyp{c) > 0. But at least one of Up{b) and ^'p(c) must be 0 since b and c have no common factor, so z/p(c) = 0, which implies that p / c . Since this holds for any irreducible factor p of a, we have gcd(a,c) = 1, so there are d^e G D such that cd -\- ae = 1. Suppose now that cd ^ ae = cf + ag = 1 for some ^, e, / , ^r G D. Then, a{g — e) = c{d ~ / ) . Let p be any irreducible factor of a. We then have z/p(c) -h Up{d - f) = i^pic {d- f)) = Up{a {g - e)) = Up{a) + yp{g - e) > Vp{a). But i^pic) = 0 as previously, so Vp{d~ f) > Vp{a)., which implies that any irreducible p G D that appears in the factorization of a with a positive exponent n must appear with an exponent ?n > n in the factorization of (i — / , hence that a I d — / , i.e. d — f G I, so 7ri{d — f) = 0. Since TT/ is a ring-homomorphism,

4.2 Localizations

113

we get 7rj{bd) = nj{bf) so the value of 7Ta{x) does not depend on the choice of d and e. Suppose finally t h a t x = b/c = yjd where b^c^b'^d G D and gcd(6, c) — gcd(6', d) = 1. As previously, this implies t h a t V = ub and d = uc for some u £ D"". Let (i, e G D be such t h a t C(i + ae = 1. Then, c'(i^ -\- ae = 1 for d' = ti~-^(i E D , and we have b'd' = ubu~^d = bd, so the value of 7ra(x) does not depend on the choice of b and c, so rca is well-defined on OaWe next show t h a t TTa is an extension of TTJ to Oa which induces an isomorphism between Oa/aOa and D / I , z.e. t h a t we have the following diagram: Oa/aOa

T h e o r e m 4 . 2 . 1 . Let D be a principal ideal domain, I be a proper ideal of D and a £ D be a generator of I. Then,

nonzero

(i) 7Ta{b) = 7Tj{b) for any b e D (i.e. TTa extends TTJ). (a) ker(7ra) = aOafrom Oa onto D/I, hence a ring(Hi) TTa 'is a surjective ring-homomorphism isomorphism between Oa/aOa and D/I (a field-isomorphism if I is maximal). (iv) If A is a derivation of D and AI C I, then A* o TT^ = iVa o A where A* is the induced derivation on D/I (Lemma 3.1.2). Proof, (i) Let b E D and write b = b/c with c = 1. Then C(i + ae = 1 for d = 1 and e = 0, so 7Ta{b) = TTi{bd) = TTj{b). (ii) Let X G Oa and write x = b/c where b^c e D have no common factor. Let d,e E D he such t h a t cd ^ ae = 1. Suppose first t h a t x G aOa- Then, ^a{x) = z^a(^) — ^a(c) > 0 by L e m m a 4.2.1, so z^a(^) > ^a(c) > 0, so a | 6, hence b e i , which implies t h a t bd G / , therefore t h a t 7ra{x) = ixi{bd) = 0. Conversely, suppose t h a t TTa{x) = 0. Then iTi{bd) = 0, so bd G / , which implies t h a t a \ bd. But gcd(a, c?) = 1 since cd + ae = 1, hence a | 6, so Uai^) > 0. Also, gcd(a,c) = 1 since cd -{- ae = 1, so a / c , so Ua{c) — 0, hence Va{x) ~ Va{b) — ^a[d) > 0, SO X G aOa by Lemma 4.2.1. (iii) Since TT/ is surjective and TTQ is an extension of TTJ by (i), it follows t h a t TT^ is surjective. Another consequence of (i) is 7ra(l) = '7^/(1) = 1- Let x , x ' G Oa and write x = b/c,x' = V/d where b^c^b'^d G D, b and c have no common factor, and V and d have no common factor. Write also xx' = b"/d' where V,d' G D have no common factor. Then, bb' = gb" and cd = gd' for some g e D. Let d,e,d\e^ £ D he such t h a t cd + ae = 1 and dd' + ad = 1. Multiplying those two equalities together, we get cddd' -\- ah = 1 where h =

114

4 The Order Function

c'd'e + cde' + aee' G D. Hence, c"{gdd') + a/i = 1, so, using the fact that TTJ is a ring-homomorphism: lTa{xx') =lTa(!^)=

lTl{h"gdd') = llj{hh'dd')

= IT I {hd)n I {h'd') = lla{x)lTa{x') .

Write now x + x^ = h"jd' where h'^^d' G D have no common factor. Then, hd + h'c — ghl' and cd = gd' for some g £ D. Let (i, e^d\d G D be such that C(i + ae = c^d' + ad = 1. As above, this impHes that d'{gdd') + a/i = 1 for some /i G D, so 7r,(x + xO = iTai^) = iTi{y'gdd') = nj{{bd + b^c)dd') =

TTl{hd)lTl{dd')^lTl{h'd')7Tl[cd)

= TXa{x)Tll{dd') + lTa{x')TTi{cd) .

From 1 = cd-\- ae, we get 1 = 7r/(l) = 7ri{cd) + 7rj(ae) = 'Kj{cd) since a G L Similarly, Tii{dd') = 1, hence 7ra(a: + x') = 7ra(x) + 7ra(x'), so TTQ is a ringhomomorphism. Since ker(7ra) = aOa by part (ii), this implies that iXa is a ring-isomorphism between Oa/aOa and D/I. If / is maximal, then D/I is a field, so TTa must be a field-isomorphism. (iv) Let Zl be a derivation on D and suppose that AI C / . Then, the induced derivation Zl* on D/I satisfies Zi* o TT/ = TT/ o Zl by Lemma 3.1.2. Since AOa ^ ^a by Lemma 4.2.1, TTa o zi is defined on Oa- Let x £ Oa and write X = b/c where b^c £ D have no common factor. Then, gcd(a, c) = 1 as explained earlier, so 1 = ad + ce for some d,e G D, which implies that 1 = ^a(l) = 7ra{a)7Ta{d) + 7ra(c)7ra(e) = 7ra{d)7Ta{e), hcucc that 7ra(c) is a unit in D/I. In addition, b = ex, so 7ra{b) = TTa (ex) = 7ra(e)7ra(x), so applying Zl*, we get Zl*^,(5) = Zl*(7r,(e)7ra(x)) = 7ra(c)zl*7r,(x) + 7r,(x)zl*7r,(e).

(4.1)

From 6 = ex also follows Zl6 = eZlx + xZic, and applying TTa, we get 7Ta{Ab) = 7raic)7Ta{Ax) + na{x)lTa{Ac)

.

(4.2)

But 7Ta{Ab) = TTj{Ab) = Zl*7r/(6) = Zl*7ra(6) and 7ra(Zlc) = Zl*7ra(c) in a similar way, so (4.2) becomes zl*7ra(&) = iTa{c)TTa{Ax) + 7ra(x)zl*7ra(e).

(4.3)

Equating (4.1) and (4.3) yields 7ra(e)Zl*7ra(x) = 7ra(c)7ra(zlx). Since T^aid) is invertible in D / / , we get Zl* o TTQ = TTa o Zl. D In the case when D is a Euclidean domain, we call Ha (x) the remainder of X at a. It can be computed by the following algorithm, which has the same complexity as the extended Euclidean algorithm in D.

4.3 The Order at Infinity R e m a i n d e r (a:, a)

115

(* Local remainder at a point *)

(* Given a Euclidean domain D, a G Z}\{0} with a return n a(x) as an element of D. *)

i

D* and X e O a ,

( 6 , c ) ^ E x t e n d e d E u c l i d e a n ( a , denominator (x), 1) (g,r) ^ PolyDivide(numerator(x) c, a) r e t o r n ir

4.3 T h e Order at Infinity In the case of polynomial rings, we introduce an extra order function, called the order at infinity, which has properties similar to the order functions of the previous sections. While the usual degree function for polynomials can be used instead, the properties of the order at infinity can be generalized later to points at infinity on algebraic curves for which the degree is not defined. Let D be an integral domain of arbitrary characteristic and x an indeterminate over D throughout this section. For a G D[x], we use lc(a) to denote the leading coefficient of a, i.e. if a = ao + aix + • —h anX^ with a^ 7^ 0, then lc(a) = a^. Definition 4.3.1. The order at 00 is the map Z/QO • D{x) —> Z U {+CXD} given hy z^oo(O) = +00, and Uoo{h/c) = deg(c) - deg(6) for h,ce D[x] \ {0}. Suppose that / = b/c = d/e for 6, c, d, e G D[x]. Then be = cd^ so deg(5) -f deg(e) = cieg(c) + deg((i), so deg(c) ~ deg(6) = deg(e) — deg( min(z/oo(/),z^oo(5')) (^nd equality holds if Voo{f) 7^ ^oo{9)(Hi) if f j^ 0 then i^ocif^) = '^'^oo{f) for any m G Z. Proof. Let f^g E D{x) and write / = b/c^g = d/e where b^c^d^e G D and Cy^O^e.

(i) fg = bd/ce so Uooifg) = deg(ce) - deg(5 deg(ce) — deg(6e) = deg(c) - deg(6) = z/co(/). Suppose that z^oo(/) < ^oo{g), then deg{cd) < deg(6e) as above, so deg(öe + cd) = deg(6e), so z^oo(/ + g) = deg(ce) - deg(5e) = Poo{f)(iii) This is trivial for m = 1. Suppose that it holds for m > 0. Then, i^oo(/'"+') = vUn) = VooiD + v^{f) = (m + l)i/oo(/) so it holds

116

4 The Order Function

for m + 1. Thus (iii) holds for m > 1. For m = 0, / ° = 1 = 1/1, so z/oo(l) = d e g ( l ) - d e g ( l ) = 0. For m < 0 we have: 0 = z/oo(l) = ^ o o ( / " " / ~ " ' ) = ^oo{f'^)-mUoo{f), so i^ooif^) = 'muooif)- Thus (iii) holds for any m e Z. D Since Z/QO satisfies properties similar to Ua, it is natural to define the notions of the localization and value m a p at infinity in a manner similar to what was done in the previous section at a point. D e f i n i t i o n 4 . 3 . 2 . We define the localization at oo to be 0^

= {f e D{x) such that z^oo(/) > 0}.

Intuitively, Ooo, which is a local ring, is the set of all the rational functions in D{x) for which the degree of the denominator is at least t h a t of the numerator, i.e. which have no pole at infinity. As expected, Ooo satisfies properties similar to Oa for an irreducible a G D[x]. L e m m a 4.3.1. (i) (n)

Ooo ^5 a subring of

D{x).

fex-^Ooo ^=^ z^oo(/)>l where X~^OOQ is the ideal generated (iii) fx""^^^^ e Ooo for any f G D{xy.

by x~^ in Ooo-

Proof (i) Let f^g G Ooo, and write g = b/c for 6,c G D[x]. Then i^ooif) ^ 0 and TJooig) > 0, so deg(6) < deg(c). But d e g ( - 6 ) = deg(6), so Uoo{-~g) = ^oo[--b/c) > 0, so z^cx)(/ — Ö') > 0 and i^ooUd) ^ 0 by Theorem 4.3.1. Hence, f — g G Ooo and fg G Ooo- In addition, 0 G Ooo since z/oo(0) = + o o , and 1 G Ooo since z/oo(l) = 0, so Ooo is a subring of D{x). (ii) Let / G x"^Ooo, then / = g/x for some g G Ooo, so z^oo(/) = ^oo{g) i^oo(^) = ^oo{g) 4- 1 > 1. Conversely, let / G D{x) be such t h a t i^oo(/) ^ 1, and let g = fx. If / = 0, then x / = 0 G Ooo, so / G x~-^Ooo. Otherwise, / / 0 so ö' ^ 0 and we have Vooi^g) = i^oo{f) + ^^00(2:) = i^ooU) - 1 ^ 0^ which implies t h a t g G Ooo, hence t h a t / = g/x G x^-^Ooo(iii) Let / G D{xY. Then, v^o {fx^-^f^) = z/oo(/) - ^oo(/) = 0, so fx^-^f^ G Ooo. • D e f i n i t i o n 4 . 3 . 3 . Let F be the quotient field of D. We define the value at 00 to be the map TTOO ' Ooo —^ F given

TTooif)

by:

ric(6)/lc(c),

i/z/oo(/)-0,

1

i/z/oo(/)>0.

0,

where b^c E D[x] and f == b/c. Suppose t h a t / = b/c = d/e where b^c^d^e G D[x] and t h a t z^oo(/) — 0Then, be = cd^ so lc(5)lc(e) = lc{c)lc{d)^ so lc(6)/lc(c) = lc((i)/lc(e), which implies t h a t TTOO is well-defined on Ooo-

4.3 The Order at Infinity

117

T h e o r e m 4.3.2. (i) ker(7roo) = x'^ö^o(ii) TToo is a surjective ring-homomorphism from Ooo onto the quotient F of D, hence a field-isomorphism between OOO/X~^OQO and F.

field

Proof (i) Let / G Ooo- If / € x"^(9oo, then Uooif) > 1 by L e m m a 4.3.1, so ^oo{f) = 0 from the definition of TTOO- Conversely, suppose t h a t / ^ x'^Ooo, which impHes t h a t i^oo(/) = 0, and write / = b/c where b,c ^ D[x]. T h e leading coefficients of b and c are never 0 by definition, so iToo{f) 7^ 0. Hence ker(7roo) = x~^Ooo(ii) Let F be the quotient field of D and a; G F . If cj = 0, then tu = 7roo(0). Otherwise, write a; = b/c with b^c E D and 6 7^ 0 7^ c. Then, deg(5) = deg(c) = 0, so z^oo(^/c) = 0, so TToo{b/c) = b/c = UJ. Hence, TTOO is surjective. Taking LÜ = 1 yields 7roo(l) = 1- Let f^g G Ooo. Then, I'ooif) ^ 0 and i^ooig) > 0. Suppose t h a t Uooif) > 0. Then, i>oo{fg) = ^ooU) + ^00(p) > 0, so TTooifg) = 0 = 7Too{f)7Too{g) sluce TTooU) = 0. Similarly, iToo{fg) = 0 = 7Too{f)7^oo{g) if J^oo{g) > 0 so suppose t h a t i^ooif) = '^oo{g) = 0. Write / = b/c and g = d/e where 6, c, d, e G D[x]. Then, I'ooifg) = 0 by Theorem 4.3.1, so

,. ,

Icibd)

lcib)lc{d)

"-(^^) = H ^ ) = W ) W ) ^

-oo(/)7roo(5) .

Suppose t h a t Voo{f) > 0 and I'ooig) > 0- Then, i^ooif + fi') > 0 by Theorem 4.3.1, so 7roo(/ + fi^) = 0 == TTooif) + ^00(0^) since 7roo(/) = TToo{g) = 0. We can now assume without loss of generality t h a t Uooif) — 0, i.e. t h a t deg(6) = deg(c). Suppose first t h a t Voo{g) > 0- Then, deg{d) < deg(e), so deg{cd) < deg(6e), so deg(6e + cd) = deg(6e) = deg(ce) and lc(6e + cd) = lc{be) = lc(6)lc(e). We also have i^ooif -^ g) = z/oo((^e + cd)/ce) = 0, so ._

,

IcjbMe)

lc{b)

since TTooig) = 0- Suppose finally t h a t i^ooig) = 0- Then, deg{d) = deg(e), so deg(cd) = deg(6e), so deg(5e + cd) < deg{be) — deg(ce). If deg(öe + cd) = deg(6e), then i^cx)(/ -\- g) = 0 and lc(6e + cd) = lc(6)lc(e) + lc(c)lc(d), so lc(6e + cd)

''-^f-^'^ =

lc{b)

lc{d)

-H^er^^

If deg(6e + cd) < deg(6e), then z/oo(/ + ^f) > 0 and lc(6)lc(e) = lc(c)lc((i), so ^oo(/) = ^oo{g)^ so TTooif) -^ TTooig) = 0 = ^ o o ( / + g)• Hence, TToo is a ringhomomorphism. Since ker(7roo) = x~^^Ooo by part (i), this implies t h a t TTOO is a field-isomorphism between OQO/X'^^OOO and the quotient field of D . D

118

4 The Order Function

ValoeAtInfinity(/)

(* Value at infinity *)

(* Given a Euclidean domain D, and / G Ooo, return 7roo(a:). *) if / = 0 then return 0 a ^— nunierator(/), b ^— denoniinator(/) if deg(6) > deg(a) then return 0 return(lc(a)/lc(6))

4.4 Residues and the Rothstein—Trager Resultant We present in this section the properties of the order function that are used for integration, namely the relation between the orders of a function and its derivative at a point, and the basic theory of residues in monomial extensions, up to the fundamental property of the Rothstein-Trager resultant. This relation and the various residue formulas let us connect the poles of a function to the poles of the functions that appear in its integral. Throughout this section, let Ü^ be a differential field of characteristic 0 with derivation D, and t be a monomial over K. We first define the notion of a residue at a normal polynomial. Definition 4.4.1. Let p G K[t] \K

be normal, and 7Zp be the set

np = {f e K{t) such that pf e Op} . We define the residue at p to be the map residue^ : IZp —> K[t]/{p) given by p residuep(/) =^'Kp{f — ) . Let q £ K[t] be any irreducible factor of p. Then q jf Dp since p is normal, so I/Dp e Oq. Since this holds for any irreducible factor of p, we have 1/Dp £ Op. For / E T^p, pf £ Op^ so fp/Dp G Op, which means that residucp is welldefined. Since 7rp(a) = a for any a G if, we identify K and 7rp{K) C K[t]/{p) when dealing with residues. Thus, when we say in the rest of this section that / has a residue a G ÜT, we mean the residue of / is the image of an element of K by TTp. T h e o r e m 4.4.1. Let p G K[t] \K be normal Then, IZp is a vector space over K, ker(residuep) = Op, and residucp is a K-vector space isomorphism between IZp/Op and K\t]/{p). Proof. We have 0,1 G T^p since 0,p G Op. Let f.gG^lp and c G K C Op. Then, pf^pg G Op, so cpf -\-pg £ Op since Op is a ring. Hence, cf + g E IZp, so Tip is a vector space over K. Let f e Op. We have 1/Dp G Op as earlier, so f/Dp G Op, so pf/Dp G pOp, which implies that residuep(/) = 7Tp{pf/Dp) = 0 by Theorem 4.2.L Hence, Op C ker(residuep). Conversely, let / G ker(residuep).

4.4 Residues and the Rothstein-Trager Resultant

119

Then, 7rp{fp/Dp) = 0, so pf/Dp G pOp by Theorem 4.2.1, which impUes t h a t f/Dp G O^. But Dp e Op, so / = Dp{f/Dp) e Op. Hence ker(residuep) = Op. Let f^g E Tip and c G K. Since iTp is a ring-homomorphism by Theorem 4.2.1 we have iesidnep{cf

p ^ g) = 7Tp{{cf + 9)-^)

p = 'rCp{c)7Tp{f — ) +

p 7Tp{g—)

= TTp{c)iesidvLep{f) + residuep(^) . But c £ K, so 7rp{c) = c, hence residuep(c/ + ^) = c residuep(/) +residuep(5f) in K[t]/(p), so residucp is a K-vector space homomorphism. Let cu E K[t]/{p). Since iip is surjective by Theorem 4.2.1, there exists g E Op such t h a t TTp{g) = (jj7Tp{Dp). Let / = g/p. Then pf e Op so f e 1Zp, and residuep(/)=7r,(/-^) =

^ ^ = c .

hence residue^ is surjective. Since ker(residuep) = Op, this impUes t h a t D residucp is a K-vector space isomorphism between IZp/Op and K\t\/{p). Example 4-4-^- Let ii" = Q, t be a monomial over K with Dt = 1 {i.e. D = d/dt), and p = t £ K[t] is normal and irreducible. We have / = 1/t G Tip b u t P = 1/t^ ^ Tip, so Tip is not a ring, even when p is normal and irreducible. The following formula gives a useful relation between the residue at a normal polynomial and at any of its nontrivial factors. L e m m a 4 . 4 . 1 . Let p G K[t] \K be normal, and q G K[t] \ K be a factor p. Then, Tip C Tlq and residueg(/) = 7rg(residuep(/)) for any f GTlp.

of

Proof Since q \ p, we have Op C Oq and TTq{7rp{g)) = T^q{g) for any g £ Op. Write p = qr with r G K[t]. Since p is normal, p is squarefree, so gcd(g, r) = 1, which means t h a t 1/r G Oq. Let / G Tip. Then, pf G Oq, so g / — pf /r G Og, which implies t h a t / G T^g. Since p is normal, gcd(p. Dp) = 1, so let a, 6 G iir[t] be such t h a t aDp -^bp = 1. We have arDq + {aDr + 6r)g = a{rDq + g D r ) + 6rg = aDp + 6p = 1, so gcd{rDq,q)

= 1 and 7rq(a) = TTq{l/{rDq)).

7rg(residuep(/)) = 7Tq (iVp f/—

Then,

j j = 7Tq{7rp{fap)) = Tiq{faqr)

120

4 The Order Function

Example 4-4-^- Let K = Q, t be a monomial over K with Dt = 1 {i.e. D = d/dt), and / = (t - 2)/(t2 - 1) € Ä"[t]. Then, residuet2_i(/) = 7rt2__i

2t

/

l-2t 2

while residue,_i(/) = 7r,_i ( ^

j = - 1 = 7r,_, ( 1 ^

and residuet+i(/) = TTt+i f ^ — j ] = 2 "" ^*+i T h e o r e m 4.4.2. Let f G K{t) \ {0} anc^ p e K[t] be irreducible. (i) If p is normal^ then Vp{Df) = i^p(/) — I if ^p{f) 7^ 0^ Up{Df) > 0 if ^pif) = 0- Furthermore, n,{p'-^^^f^Df) (ii) (Hi) p£Si

=

Mf)n,{p-^'^^f^f)ix,{Dp).

peS^iyp{Df)>iyp{f). and Up{f) / 0 = ^ T^p{Df) = Vpif).

Proof. Let p E K\t] be irreducible, / G K{t) \ {0} and n = ^p{f)- Let 5' = fp""^. By Lemma 4.2.1, ^ e Op. Also, D / = ngp^'-^Dp ^p^'Dg.

(4.4)

Write ^ = b/c where b.,c £ K\t] and gcd(6,c) = 1. We have i'p{g) = ^pif) + i^p{p~^) = n — n = 0, so i/p(6) — z/p(c) = 0. But at most one of iyp{b) and iyp{c) can be nonzero since gcd(6, c) = 1, so iyp{b) = z^p(c) = 0. We have Dg =

cDb - bDc 2

so yp{Dg) = Up{bDc — cDb) — 2z/p(c) = i/p{bDc — cDb) > 0 since bDc — cDb £ K[t]. By Lemma 4.2.1, this implies that Dg G Op. Suppose that n = 0. Then f = g^ so Df = Dg e Op, so VpiDf) = yp{Dg) > 0. This implies that Vp{pDf) > 0, hence that TTp{pDf) = 0 by Theorem 4.2.1. This is valid regardless of whether p is normal or special, so (i) and (ii) hold when n = 0. Suppose now that n ^ 0. (i) p is normal, so gcd(p, Dp) = 1, so Up{Dp) — 0. This implies that Dp G Op and iyp{ngp^~^ Dp) = yp{g) + n - l = n ~ l < n < Up{p'^Dg) so from (4.4) and Theorem 4.1.1 we get i/p{Df) = n ~ 1. We then have pi-njjf ^ Q^ |3y Lemma 4.2.1, and from (4.4) we get iXp{p^~'^Df) = 7Tp{ngDp -\- pDg). Since g^Dp^p and Dg are all in Op and Tip is a ringhomomorphism, we have

4.4 Residues and the Rothstein-Trager Resultant TTpip^-'^Df)

= n7Tp{g)7rp{Dp) + TTp{p)TTp{Dg) =

121

n7Tp{p~''f)iTp{Dp)

since TTp{p) = 0. (ii) p ^ S so p\ Dp^ which means t h a t Up{Dp) > 1. Hence, Up{ngp^~^Dp) > n. Since Vp{p'^Dg) = n + Vp{Dg) > n, from (4.4) and Theorem 4.1.1 we get Up{Df) > n. (iii) Let p E Si, and suppose t h a t n ^ 0. Assume first t h a t p = t — a for a £ K. Then, Pa{(^) is not a logarithmic derivative of a i<"-radical, where Pa = {Dt — Da)/{t — a ) = Dp/p. Let h = Dg + npo,g. Since p\Dp, Pa G K[t], hence pa G Op. In addition, g G Op and D ^ G Op as seen above, so /i G Op. Since TTp is a ring-homomorphism, we have 7Tp{h) = 7rp{Dg + npo,g) = 7Tp{Dg) + n'Kp{g)iTp{p^). We have i/p{g) = 0, so g ^ pOp by L e m m a 4.2.1, which implies t h a t T[p{g) ^ 0 by Theorem 4.2.1. Suppose t h a t 7rp{h) = 0. Then, using t h e facts t h a t (p) is a differential ideal of K[t] (Lemma 3.4.3) and t h a t D* o Hp = iTp o D (Theorem 4.2.1) where D* is the induced derivation on K[t]/{p) (Lemma 3.1.2), we get: iTp{Dg) _ D'^TTpig) __ D'^u -npa{a) = -~mTp{pa) where u = T^p{g) G K[t]/{t — a). B u t K\t\/{t — a) c^ K , and D* is an extension of D by Lemma 3.4.3, so u E. K and D*u = Du, which implies t h a t Pa{oL) is a logarithmic derivative of a if-radical, in contradiction with p being of the first kind. Hence TTp{h) / 0, so Vp{h) = 0 since h £ Op. From (4.4) we have Df = {Dg + npag)p'^ = hp"", so Pp{Df) = Up{h) -\-jyp{p'') = n. Let now p have arbitrary degree m > 0, let K be the algebraic closure of K, and p = (t ~ o^i) • • • (t — a ^ ) be the factorization of p in K[t]. t is a monomial over K and p and the t — a^'s are in ^ I ; ^ W . K ' by Theorem 3.4.4. Then, z/t-^,.(/) = n for each i by Theorem 4.L2, so Ut-ai{Df) = n by the previous proof. Hence, Up{Df) = n by Theorem 4.1.2. D Example 4-4-^- Let i f = Q, t be a monomial over K with Dt = 1 {i.e. D d/dt), p = t £ K[t] is normal and irreducible, and / = t"^ + 1 G i f for an arbitrary integer m > 0. We have Vt{f) = 0, but Df = mV^'~^, Vt{Df) = m — 1. This shows t h a t one cannot give a general upper bound iyp{Df) when i/p(/) = 0.

= (t) so on

Theorem 4.4.2 has several useful consequences: K{t) must be a differential subring of i f (t), and we get formulas for the orders and residues of logarithmic derivatives, and for the residue at a given p. C o r o l l a r y 4 . 4 . 1 . Let f G K{t). (i) f simple w.r.t D =4> i^p{f) > —1 for any normal irreducible p G i f [t]. (ii) f G K{t) <==^ ^p{f) ^ 0 foT' any normal irreducible p G i f [t]. (iii) Kit) is a differential subring of K{t).

122

4 The Order Function

Proof. Let / G K{t) and write / = a/b with a^b ^ K[t]^ gcd(a,6) = 1, and b ^ 0. Let p ^ K[t] be normal irreducible. (i) If / is simple, then b is normal hence squarefree. If p | 6, then p / a, so ^p(/) =^ ~^p{^) — ~~1 since b is squarefree. (ii) If / G K{t)^ then 6 G tS, so p | 6, which implies that i^pif) = J^p{a) > 0. Conversely, suppose that i^qif) > 0 for any normal irreducible q G K[t]^ and let p G ür[t] be a normal irreducible factor of b. Then, p Ya^ so i^pif) = —Vp{b) < 0 in contradiction with our hypothesis. Hence, all the irreducible factors of b are special, so ö G tS by Theorem 3.4.1, which implies that / G K{t). (iii) K{t) is not empty since K[t] C K{t). Let f,g£ K{t) and p G iir[t] be normal irreducible. Then, i/p(/) > 0 and iyp{g) > 0 by part (ii). We have ^p{-9) = J^pid) by Lemma 4.1.2, so Up{f - g) > mm{up{f),Up{-g)) > 0. Hence, f — g E K{t) by part (ii). In addition, i'p{fg) — ^p{f) + ^p(ö') ^ 0, so fg G K{t), hence K(t) is a subring of K(t). If z/p(/) = 0, then jyp{Df) > 0 by Theorem 4.4.2. Otherwise, z/p(/) > 0, so i^piDf) = i/p(/) - 1 > 0 by Theorem 4.4.2. Thus Df G K{t) in any case, so K{t) is a differential subring ofK{t). D Corollary 4.4.2. Let / G K(t) \ {0} and p G Ä'[t] 5e irreducible. Then, (i) yp{Df/f)>-l. (ii) Up{Df/f) = —1 4=4> z^p(/) 7^ 0 and p is normal. (iii) If p is normal, then yp{Df) ^ —1 and residuep(D///) = i'p{f)Proof. Letp G K\t] be irreducible, / G iC(t)\{0}, n = Pp{f) and m = Pp{Df). (i) By Theorem 4.4.2, either m > n or m = n — 1, so Up{Df/f) = m — n> —1 in any case. (ii) Suppose that n y^ 0 and p is normal. Then, by Theorem 4.4.2, m = n — 1, so Up{Df/f) = —1. Conversely, suppose that Vp[Df/f) = —1, then m = n — 1 < n. By Theorem 4.4.2, m > n if either p G 5 or n = 0, so p must be normal and n ^ 0. (iii) Suppose that p is normal. If n > 0, then m > 0 by Theorem 4.4.2. If n < 0, then m = n — 1 < —1 by Theorem 4.4.2, so m ^ —1. If n = 0, then Pp{Df/f) > —1 by parts (i) and (ii), so Df/f G Op, which impUes that residuep(jD///) = 0 by Theorem 4.4.1. Suppose now that n / 0. By (ii) we have Vp{Df/f) = - 1 , hence Vp{pDf/f) = 0 so pDf/f G Op. By Theorem 4.4.2, we have yp[Df) = n-l and 7rp{p^~'^Df) = niXp{p~'^ f)7ip{Dp). Since p, Dp.,p^'~^Df and p~'^f are all in Op and TTp is a field-homomorphism by Theorem 4.2.1 (p is irreducible), we have residuep(^) = % ( ^ ^ ) = ^ p ( f r ^ ) _

7rp(pi-"Z?/) _ n7rp(p-"/)7rp(I?p) __ TTpip-^f)Trp{Dp) Trp{p-^f)TTpiDp)

4.4 Residues and the Rothstein-Trager Resultant

123

L e m m a 4.4.2. Let p G K[t] be normal irreducible, g E Op and d G K[t] be such that iyp{d) = 1. Then, iesiduep{g/d) = Tip{g/Dd). Proof. Since i'p{d) = 1, Vp{Dd) = 0 by Theorem 4.4.2, so i/p{g/Dd) = i'p{g) > 0, which imphes that g/Dd G Op by Lemma 4.2.L In addition, iyp{pg/d) = 1 + i^p{g) — 1 = J^pig) > 0, so g/d € 7^p, so both iesid\iep{g/d) and 7Tp{g/Dd) are defined. Write d = pq for some q G K[t]^ and let h = gpDq/qDpDd. Then, i^p{q) = i^p{d) — Up{p) = 0 , Vp{Dp) = 0 since p is normal, i/p{Dd) = i/p{d) — 1 = 0 by Theorem 4.4.2, so Up{h) = Vp{g) + 1 + Pp{Dq) > 1. This implies that h e pOp by Lemma 4.2.1, hence that 7Tp{h) = 0 by Theorem 4.2.1. In addition, we have _p_ Dd

1 ^^Dd

pDq ^ qDp_+_pDq_ ^ 1 ^g p qDpDd' ^ qDpDd ^qDp dDp

residue,(|)=7r,^|^j=.r,(/.+

^ )

D

Lemmia 4.4.3. Let q G K\t\ be normal irreducible and f G K{t) be such that z/g(/) = —1. Write f = p + a/d where p^a^d G K[t], d ^^ 0, deg(a) < deg{d) and gcd(a, d) = 1. Then, for any a G K, q I gcd(a — aDd^d) 4=> residueg(/) = a. Proof. Since Vq{f) = —1, we have z/g(a) = 0 and h'q{d) = 1, so Uq{Dd) = 0 by Theorem 4.4.2. This implies that Dd G Oq and that Uq{l/Dd) = 0, hence that 1/Dd G Oq. Furthermore, a^p G Oq and / = {a+pd)/d^ so residueg(/) = TTq{{a +pd)/Dd) = 7Tq{a/Dd) by Lemma 4.4.2. Suppose that q \ gcd(a — aDd, d). Then, 0 = TTq{a — aDd) = T^q{a) — a7Tq{Dd) so

Conversely, suppose that a = residueg(/) = 7Tq{a/Dd). Then, 7rq{a ~ aDd)

= 7rq{a)

- a7Tq{Dd)

= 7rq{a)

so g I a — aDd, hence q \ gcd(a — aDd^ d).

- TTq{ —

)'Kq{Dd)

= 0

D

124

4 The Order Function

We can now state the fundamental property of the Rothstein-Trager result a n t , namely t h a t from any simple function, one can construct a polynomial over K whose nonzero roots in K are exactly the residues of / t h a t are in K. Note t h a t in general, not all the residues of / are in K, expect when K is algebraically closed. T h e o r e m 4 . 4 . 3 . Let f G K{t) be simple w.r.t D, and write f = p -\- a/d where p^a^d £ K[t], d j^O, deg(a) < deg{d), and gcd(a,(i) = 1. Let r = resultantt(a - zDd, d) e K[z] where z is an indeterminate

over K.

(4.5)

Then, for any a € K"",

r{a) = 0 <^=4> residueg(/) = a for some normal irreducible q G K[t]. We call the polynomial

r given by (4-5) the Rothstein-Trager resultant of f.

Proof For any ß G .K*, let rß = resultantt(a — ßDd^d) E if, and aß : K[z] —^ K be the ring homomorphism given by (Jß{z) = ß and (Jß{x) = x for any x E K. Define äß : -?f[^][t] —> K[t] by 'ö'ß{Y^ajP) = Yl^ßi^jWSince 'äß{d) = d, deg^{'äß{d)) = deg^((i), so r{ß) = '^ßij) = :tlc{d)'^^rß for some nonnegative integer mß by Theorem 1.4.3. Recall t h a t / simple means t h a t d is normal, hence squarefree, i.e. t h a t J^pif) ^ —1 for any normal irreducible p G K[t]. Let a G Ü^* be such t h a t r ( a ) = 0. Then, r^ = 0, so deg{g) > 0 by Corollary 1.4.2 where g = gcd(a — aDd^ d). Let then q G K\t] be an irreducible factor of g. Since q \ d and / is simple, q is normal. Also, Vq{d) = 1 since d is squarefree, so a = residueg(/) by L e m m a 4.4.3. Conversely suppose t h a t residueg(/) = a G K* for some normal irreducible q G K[t]. Then, residueg(/) ^ 0, so / ^ Og by Theorem 4.4.1, which imphes t h a t yq{f) = —1. Hence, Uq{d) = 1, so g | gcd(a — aDd.d) by Lemma 4.4.3. Therefore, TQ, = 0 by Corollary 1.4.2, so r{a) = 0. D Let F be a field of characteristic 0, x be an indeterminate over F , and D be t h e derivation d/dx on F{x). Since every irreducible q G F[x] is normal with respect to d/dx., applying t h e above result to ÜT = F , we see t h a t Theorem 4.4.3 and Lemma 4.4.3 respectively prove parts (i) and (ii) of T h e orem 2.4.1. There are similar results relating the order at infinity of an element of K{t) and its derivative. T h e o r e m 4 . 4 . 4 . Let f G K{t) \ {0}. Then, (i) Uoo{Df) > uooif) - max(0,(5(t) - 1). (ii) Ift is nonlinear and Uoo{f) ¥" 0; then equality holds in (i), and

4.4 Residues and the Rothstein-Trager Resultant (Hi) If t is nonlinear and Uooif) = 0; then the strict inequality i.e. VoQ^Df) > 1 — S{t), and

125

holds in (i),

Proof. Write / = a/d where a^d £ K[t], d ^ 0 and gcd(a, d) = 1. Then, Df = {dDa - aDd)/d^, so Voo{Df) = 2deg(d) - deg(6^Da - aDd). Let m = max(0,J(t)-1). (i) By Lemma 3.4.2, deg(D deg{d) — deg(a) — m = ^ooif) -m. (ii) Suppose t h a t t is nonlinear and z^oo(/) 7^ 0- Then m = 6{t) — 1 and Df / 0 by Lemma 3.4.5. Suppose t h a t deg(d) = 0, then deg(a) j^ 0 since Voo{f) ^ 0, so deg{dDa — aDd) = d e g ( D a ) = deg(a) + m by Lemma 3.4.2, which implies t h a t Uoo{Df) = — deg(a) — m = i^ooif) — ^ , hence t h a t Vooit^^Df/f) = 0. Furthermore, lc{dDa — aDd) = lc{dDa) =
^Df\ Y )

^

ddeg(a)lc(a)X(t) d^

d , ^ SWN / .V W X lc(a) = ^ ' ^ S ^ « ) ^ ^ = -i^oo(/)A{t).

Suppose now t h a t deg(a) = 0, then deg(c^) 7^ 0 since 1^00(/) 7^ 0, so deg{dDa — aDd) = deg{Dd) = deg{d)-\-m by Lemma 3.4.2, which implies t h a t Voo{Df) = deg{d) — m = z/oo(/) — ' ^ , hence t h a t yoo{t~^Df/f) = 0. Furthermore, lc{dDa - aDd) = lc{—aDd) = —adeg{d)lc{d)X{t) also by L e m m a 3.4.2. Hence,

t

-j-j

=

^ ^

^

= -deg{d)\{t) =

-u^{f)X{t).

Suppose finally that deg(a) ^ 0 and deg(d) ^ 0. Then, by Lemma 3.4.2, the leading term of dDa — aDd is lc(cf)deg(a)lc(a)A(t)t'*'=s(a)+deg(d)+m -lc(a)deg(d)lc(c^)A(t)t^^s(o)+deg(d)+m = -J/oo(/)lc(a)lc(d)A(i)t^''s(a)+deg(d)+m _

^4 g^

Since I'ooif) ¥= 0; this gives deg(dDa — aDd) = deg(a) + deg{d) + m, hence t^oo{Df) = VooU) ^ "^i so i^oo{t~'^Df/f) = 0. Furthermore, D / \ _

u^{fMaMd)Xit)

* Tj "

HdT'

lc(rf) _

W) ~ "^°°^^^^^*^ •

(iii) Suppose t h a t t is nonlinear and Vooif) = 0- Then m = (5(t) — 1 and deg(a) = deg{d). If Df = 0, t h e n Uoo{Df) = +oo > —m, so suppose t h a t D / ^ 0. If deg(a) 7^ 0, t h e n d e g ( d D a - aDd) < deg(a) + deg(ci) + m by (4.6), so Uoo{Df) = 2deg(d) - deg{dDa - aDd) > -m. If deg(a) = deg{d) = 0, then f e K, so Df G K, which implies t h a t i/ooiDf) = 0 > —m. Hence,

yoo{t-^Dflf)

> 0, so n^{t-^Df/f)

=0.

D

126

4 The Order Function

Exercises Exercise 4 . 1 . Let {k^ D) be a differential field of characteristic 0, t a monomial over k, and / G k{t) be simple. Show that if there are h G k{t) and g G k(t) such that Dg ^ f + h, then / G k[t]. Exercise 4.2. Let {k, D) be a differential field of characteristic 0, t a monomial over /c, and / G k{t), a) Show that if / is the logarithmic derivative of a nonzero element of fc(t), then / is simple and can be written as

where p,a,d G k\t\^ deg(p) < max(l, (5(t)), d / 0, deg(a) < deg((i), gcd(a,d) = 1, and d is normal. Furthermore, all the roots in fc of r = resultantt(a — zDd, d) are integers. b) Show that if / is the logarithmic derivative of a fc(t)-radical, then / is simple and can be written as

where p^a^d G fc[t], deg(p) < max(l, (5(t)), d y>^ 0^ deg(a) < deg{d), gcd{a,d) = 1, and d is normal. Furthermore, all the roots in Ä: of r = resultantt(a — zDd^ d) are rational numbers. Exercise 4.3 (Indicial equation of a linear differential o p e r a t o r ) . Let [k^D] be a differential field of characteristic 0, t a monomial over k^ p G k[t] be normal and irreducible, and / G k{t) be such that i^'pif) < 0. a) Show that Vp{D^f) = i^p{f) — n for any n GN. b) Show that n-l

i=0

for any n G N. c) Let n G N and ao, a i , . . . , a^ G Ä:(t) be such that n > 0 and a^ ^ 0. Let /i = maxo
P{z)=

Yl 0
^p{p-'''^'''\Dpyai)l[{z-j) ^""^

i-Vp{ai)=fj,

and R{z) = resultantt(p, P) G k[z]. Show that either

ek[t]/{p)[z]

4.4 Residues and the Rothstein-Trager Resultant

127

or Pii^pif)) = R[Mf)) = 0. Exercise 4.4. As in Exercises 3.7-3.11, let (k^D) be a differential field of characteristic 0 and {k{t)^Ä) be a differential extension of (k^D) where t is transcendental over k and At = a/h with a^b G k[t] such that 5 7^ 0 and gcd(a,6) == 1. Let p e k[t] be irreducible and normal {i.e. gcd{p^bAp) = 1). Show that for any / £ k{t) \ {0}, jy^iAf) = upif) - 1 -^ 1/^(6) and iyp{Af) > -v^{h) if y^{f) = 0.

if z/p(/) ^ 0

Integration of Transcendental Functions

Having developped the required machinery in the previous chapters, we can now describe the integration algorithm. In this chapter, we define formally the integration problem in an algebraic setting, prove the main theorem of symbolic integration (Liouville's Theorem), and describe the main part of the integration algorithm. From now on, and without further mention, all the fields in this book are of characteristic 0. We also use the convention throughout that deg(O) = — oo.

5.1 Elementary and Liouvillian Extensions We give in this section precise definitions of elementary functions, and of the problem of integrating functions in finite terms. Throughout this section, let k be a di3"erential field and K a differential extension of k. Definition 5.1.1, t G K is a primitive over k if Dt E k, t £ ÜT* is an hyperexponential over k if Dt/t G k. t £ K is Liouvillian over k if t is either algebraic, or a primitive or an hyperexponential over k. K is a Liouvillian extension of k if there are t i , . . . , t^ in K such that K = ^ ( t i , . . . , t^) and ti is Liouvillian over k{ti^..., ti_i) for i m { 1 , . . . , n}. We write t = fa when t is a primitive over k such that Dt ~ a, and t = e-l ^ when t is an hyperexponential over k such that Dt/t = a. Given that t is Liouvillian over /c, we need to know whether t is algebraic or transcendental over k. We show that there are simple necessary and sufficient conditions that guarantee that a primitive or hyperexponential is in fact a monomial over k. L e m m a 5.1.1. If t is a primitive over k and Dt is not the derivative of an element of k, then Dt is not the derivative of an element of any algebraic extension of k.

130

5 Integration of Transcendental Functions

Proof. Let t be a primitive over k^ a = Dt^ and suppose t h a t a is not the derivative of an element of k. Let E be any algebraic extension of fc, and suppose t h a t Da = a for some a G E. Let Tr be t h e trace m a p from k{a) t o k^ n = [k{a) : k]^ and b = Tr{a)/n G k. By Theorem 3.2.4, Db = -D(Tr(a)) n

= -Tr(Da) n

in contradiction with Du ^ a for any u G k.

= -Tr(a) n

= a D

T h e o r e m 5 . 1 . 1 . If t is a primitive over k and Dt is not the derivative of an element ofk, then t is a monomial overk, Const(Ä:(t)) = Const(/c), andS = k (i.e. S^^^ = Sf^ = 9). Conversely, if t is transcendental and primitive over k, and Coiist{k{t)) = Const(Ä;)^ then Dt is not the derivative of an element ofk. Proof. Let t be a primitive over k^ a = Dt^ k be t h e algebraic closure of k, and suppose t h a t a is not the derivative of an element of k. Then, Da ^ a for any a G k by L e m m a 5 . L I , so t must be transcendental over ^, hence it is a monomial over k. Suppose t h a t p G S \k. Let then ß G k he a, root of p. Then, Dß = Dt = a hy Theorem 3.4.3, in contradiction with Da ^ a for any a G k.^ ^o p G k. Conversely, k C. S hy definition. Let c G Const(Ä;(t)). By Lemma 3.4.5, b o t h t h e numerator and denominator of c must be special, hence in A;, so c G /c, which implies t h a t Const(Ä;(t)) C Const(fc). T h e reverse inclusion is given by Lemma 3.3.1, so Const(Ä:(t)) = Const(^). Conversely, let t be a transcendental primitive over k and suppose t h a t Const(Ä;(t)) = Const(fc). If there exists b G k such t h a t Dt — Db^ then c = t — b G Const(Ä;(t)), so c G A: in contradiction with t transcendental over k. Hence Dt is not t h e derivative of an element in k. D T h e o r e m 5 . 1 . 2 . If t is an hyperexponential over k and Dt/t is not a logarithmic derivative of a k-radical, then t is a monomial over k, Const (Ä;(t)) = Const(Ä:); and tS^^^ = Sf^ = { t } . Conversely, if t is transcendental and hyperexponential over k, and Const(fc(t)) = Const(fc)^ then Dt/t is not a logarithmic derivative of a k-radical. Proof. Let t be an hyperexponential over k^ a = Dt/t^ k be t h e algebraic closure of A;, and suppose t h a t a is not a logarithmic derivative of a A:-radicaL We have Dt/t — a and a is not a logarithmic derivative of a fc-radical by Lemma 3.4.8, so t must be transcendental over k^ hence it is a monomial over k since Dt — at. Let p = bt"^ for b Gk and m > 0. Then, Dp = {Db + mab)^^, so p \ Dp, which means t h a t p G S. Let now p G S^^^ and suppose t h a t p has a nonzero root ß G k . Then, Dß/ß = Dt/t = a by Theorem 3.4.3, in contradiction with Da/a ^ a for any a G k . Hence the only root of p in A; is 0, so p = t. We have Sf^ C S^^^ by definition. Conversely, let p G S^^^. Then p = t, so the only root of p in A: is /? = 0. We have Pß = Po = Dt/t = a, which

5.1 Elementary and Liouvillian Extensions

131

is not a logarithmic derivative of a /c-radical, so p G tSf^, which implies that cirr __ oirr

Let c G Const(Ä:(t)). By Lemma 3.4.5, both the numerator and denominator of c must be special, hence c = ht^ foibGk and g G Z. Suppose that b j^ 0 and q ^ 0. Then, 0 = Dc = {Db + qab)t^^ so Db/b = qa^ which implies that a is a logarithmic derivative of a Ä:-radical, in contradiction with our hypothesis. Hence, 6 == 0 or g = 0, so c G A:, which implies that Const(/c(t)) C Const(A:). The reverse inclusion is given by Lemma 3.3.1, so Const(Ä:(t)) = Const(/c). Conversely, let t be a transcendental hyp er exponential over k and suppose that Const(A;(t)) = Const(/c). If there exist b G k* and an integer n / 0 such that nDt/t = Db/b^ then c = f^/b G Const(A:(t)), so c G A; in contradiction with t transcendental over k. Hence Dt/t is not a logarithmic derivative of a Ä:"radical. D In practice, we only consider primitives and hyperexponentials that satisfy the hypotheses of Theorems 5.1.1 or 5.1.2. As we have seen, such primitives and hyperexponentials are monomials that satisfy the extra condition Const{k{t)) = Const(fc). Those monomials are traditionally called Liouvillian monomials in the literature. Definition 5.1.2. t G K is a Liouvillian monomial over kiftis tal and Liouvillian over k and Const(fc(t)) = Const(Ä:).

transcenden-

One should be careful that our definition of monomial in Chap. 3 does not require Const (Ä:(t)) = Const (A:), so it is possible for a monomial in the sense of Chap. 3 to be Liouvillian over k and yet not a Liouvillian monomial in the sense of Definition 5.1.2 (for example log(2) over Q). Theorems 5.1.1 and 5.1.2 can be seen as necessary and sufficient conditions for a primitive or hyperexponential to be a Liouvillian monomial. Furthermore, those theorems describe all the special polynomials in such extensions, and they are all of the first kind. We also have: ^^^^ ' \k[t,t-^,

if Dt/t ek.

^^'^^

The fact that k and k{t) have the same field of constants allows us to refine the relationship between the degree of a polynomial and its derivative in a Liouvillian monomial extension, and to strenghten Theorem 4.4.4. L e m m a 5.1.2. Let t be a Liouvillian monomial over k^ f £ k{t) be such that Df ^ 0, and write f = p/q where p.>q G k[t] and q is monic. If Uoo{f) = 0; then VoQ^Df) > 0. Otherwise, i^oo(/) J^ 0 ct'nd u mn-/^-(^)' ^ooKi^J) | ^ ^ ( j ) + i^

ifDt/t€k orDilc{p))^0, ifDtek andD{lc{p))=0.

Proof. If Voo{f) = O5 then ^^^{Df) > 0 by Theorem 4.4.4, so suppose from now on that 1^00 (/) 7^ 0- Then, n — m / 0 where n = cieg(p) and m = deg{q). We have

132

5 Integration of Transcendental Functions

Df

qDp — pDq

q' hence Voo{Df) = 2m — deg{qDp — pDq), so we need to compute deg{qDp — pDq). Write p = bf^ + r and q = f^ -\- s where b £ k"^ and r, 5 G k[t] satisfy deg(r) < n and deg(s) < m. We treat the primitive and hyp er exponential cases separately. Primitive case: Suppose that Dt — a £ k. Then, Dp = {Db)e + nabe-^

+ Dr

(5.2)

and Dq = maf^-^ + Ds so deg{Dq) < m since deg(Ds) < m by Lemma 3.4.2. Suppose first that Db ^ 0. Then, deg(Dp) = n since deg(Dr) < n by Lemma 3.4.2, so deg(gDp) = m -]- n and deg{pDq) < m + n, which implies that deg{qDp — pDq) = m -\- n^ hence that i/ooiDf) = 2m-

{m^n)

=m-n

=

v^{f).

Suppose now that Db = 0, and write r = ct^^^ + u and s = dt'^~^ + f, where c^d E k and u^v E k[t] satisfy deg{u) < n — 1 and deg(f) < m — 1. We have qDp - pDq = {Dc + n a 6 ) r + ^ - ^ + (n - l ) a c r + ^ - 2 + t"^öix +(dt^-^ + v)Dp - 6(Dd + m a ) r + ^ " ^ ~-(m -~ l ) a 6 d r + ^ - 2 _ 5^n|^^ _ ^^^n-l _^ ^ ) ^ ^

= (Dc - ÖD^f + (n - m)a5) r + ^ - ^ + ((n - l)c - (m - l)6(i) a r + ^ - 2 4_(^t^-i 4. y)Bp + t^Dit - ferDv - ( c ^ - i + ti)I}g. Since n—m ^ 0 and 6 y^ 0, c—bd~\~{n~m)bt ^ k^ so D {c — bd+ {n — m)bt) ^ 0 since Const(Ä;(t)) = Const(Ä:). But D{c-bd

-f (n " m)6t) = Dc - 6D(i + (n - m)ab

since 6 G Const(Ä;), hence Dc — 6Dc^ + (n — m)ab 7^ 0. In addition, (5.2) and Db = 0 imply that deg(Dp) < n, and Lemma 3.4.2 imply that deg(Dtx) < n—l and deg(D^;) < m - l . Hence, {df^-^^v)Dp, f^Du, bf'Dv and {ct''-^+u)Dq all have degrees strictly smaller than n + m — 1, which implies that deg(gDp — pDq) = n + ?7i — 1, hence that Uoo{Df) = 2m — {n + m — 1) = m — n -{-1 = Hyperexponential case: Suppose that Dt/t = a G k. Then, qDp - pDq ={Db + nab)r+'" + f^Dr + sDp - bmae+"' - hf'Ds - rDq = {Db +{n- m)ab) t"+™ + {sDp - rDq + t'^Dr - bt"Ds) .

5.1 Elementary and Liouvillian Extensions

133

Since n - m / 0 and 6 / 0, 6t^-^ ^ k, so D ( 5 ^ - ^ ) 7^ 0 since Const{k{t)) = Const(A:). But D ( 6 ^ " ^ ) = {Db + (n - m)a5) T " ^ , so D6 + (n - m)a6 7^ 0. In addition, deg(Dp) < n, deg(Dg) < 771, deg(i}r) < n and deg(Ds) < m by Lemma 3.4.2, so sDp^ rDq, t^Dr and bt^Ds all have degrees strictly smaller than n^m, which implies that deg{qDp — pDq) = n-\-m, hence that v^{Df) = 2m- (n-i-m) =m-n = Uooif)• Note that when applied to polynomials p G k[t] when t is a Liouvillian monomial over k^ Lemma 5.1.2 implies that d..(Dn) - / ^^S(P), aeg^xyp; - | ^^^^^^ __ ^^

if Dt/t € k or D{lc{p)) ^ 0 , if i:)t G i^ and L^(lc(p)) = 0

whenever Dp ^ 0, and we often use it in this context in the sequel. We now introduce the particular Liouvillian extensions that define the integration in finite terms problem, namely the elementary extensions. Definition 5.1.3. t e K is a logarithm over k if Dt = Db/b for some b E k*. t € K* is an exponential over k if Dt/t = Db for some b £ k. t G K is elementary over k if t is either algebraic, or a logarithm or an exponential over k. t E K is an elementary monomial over k if t is transcendental and elementary over k, and Const(A:(t)) = Const(A;). We write t = log(5) when t is a logarithm over k such that Dt = D5/6, and t = e^ when t is an exponential over k such that Dt/t = b. Since logarithms are primitives and exponentials are hyper exponentials, elementary monomials are Liouvillian monomials and all the results of this section apply to them. Definition 5.1.4. K is an elementary extension of k if there are t i , . . . , t^ in K such that K = k{ti,..., t^) andti is elementary overk{ti^... ,t^_i) fori in { 1 , . . . , n}. We say that f G k has an elementary integral over k if there exists an elementary extension E of k and g E E such that Dg = f. An elementary function is any element of any elementary extension of {G(x),d/dx). We can now define precisely the problem of integration in closed form: given a differential field k and an integrand f £ k, to decide in a finite number of steps whether / has an elementary integral over /c, and to compute one if it has any. Note that there is a difference between having an elementary integral over k and having an elementary antiderivative: consider k = C(a:, ti, t2) where x,ti,t2 are indeterminates over C, with the derivation D given by Dx = 1, Dti = ti and Dt2 = ti/x {i.e. ti = e^ and ^2 = Ei(x)). Then, e^Ei(x) , Ei(x)2 ^ —™-A^ dx = — V^ e k / so e^Ei{x)/x has an elementary integral over k even though its integral is not an elementary function. The two notions coincide only when k itself is a field of elementary functions.

134

5 Integration of Transcendental Functions

Remark that the elementary functions of Definition 5.1.4 include all the usual elementary functions of analysis, since the trigonometric functions and their inverses can be rewritten in terms of complex exponential and logarithms by the usual formulas derived from Euler's formula e-^^^^ = cos(/) + s i n ( / ) \ / ^ . Those transformations have the computational inconvenience that they introduce \/--T, and it turns out that they can be avoided when integrating real trigonometric functions (Sections 5.8 and 5.10).

5.2 Outline and Scope of t h e Integration Algorithm We outline in this section the integration algorithm so that the structure of the remaining sections and chapters will be easier to follow. Given an integrand f{x)dx^ we first need to construct a differential field containing / , and the integration algorithm we describe requires that / be contained in a differential field of the form K = C ( t i , t 2 , . . . ,t^) where C = Const (K), Dti = 1 (i.e. ti = X is the integration variable), and each ti is a monomial over C ( t i , . . . ,ti_i). If the formula for f{x) contains only Liouvillian operations, this requirement can be checked by integrating recursively the argument of each primitive or hyp er exponential before adjoining it^, and verifying using Theorem 5.1.1 or Theorem 5.1.2 that it is a Liouvillian monomial. Another alternative, which is in general more efficient, is to apply the algorithms that are derived from the various structure theorems, whenever they are applicable (Chap. 9). Example 5.2.1. Consider

/•log(x) log(x + 1) log(2x^ + 2x)dx . We construct the differential field K = Q{x,ti^t2jts) Dx = 1,

1 Dti =: - , X

Dt2 =

1 X+ 1

and

with 2a; 4- 1 Dts = -z . X^ -\- X

As we construct K, we integrate at each step and make the following verifications: / (ix ^ Q, so X is a Liouvillian monomial over Q; J dx/x ^ Q(x), so ti is a Liouvillian monomial over Q(x); Jdx/{x + 1) ^ Q(x,ti) so ^2 is a Liouvillian monomial over Q(x,ti) / -T—-dx = ti+t2eQ{x,ti,t2) J X"^ -\- X so ^3 is not a Liouvillian monomial over Q(x,ti,t2), and K is isomorphic as a differential field to Q(c)(a:, ti,^2) where c = ^3 — ti — t2 G Const(ür). ^ A simpler version of the integration algorithm can be used for those verifications, see Sect. 5.12

5.2 Outline and Scope of the Integration Algorithm

135

® Alternatively, applying the Risch structure Theorem (Corollary 9.3.1), we find that the linear equation (9.8) for a = 2x^ + 2x becomes ri

r + 2 __ 2x + 1

X

X+ 1

x'^ -[- X

which has the rational solution ri = r2 = \. This implies that Dt^, is the derivative of an element of K and that c = t^ — ti — t2 £ Const (if). Example 5.2.2. Consider

We construct the differential field K = Q(x, ti, ^2,^3) with Dx = l,

Dti = 2ti,

1 Dt2 = X

/ 1 Dts = { 1 +— \ 2x

and

As we construct K we integrate at each step and make the following verifications: J (ix ^ Q, so X is a Liouvillian monomial over Q; J 2dx ^ log{v)/n for any v G Q{x) and n G Z, so 2 is not the logarithmic derivative of a Q(a:)-radical, which implies that ti is a Liouvillian monomial over Q(a:); J dx/x ^ Q(x,ti), so ^2 is a LiouviUian monomial over Q(x,ti);

/0+i)''"=^°s^"*^) so 1 + 1/(2J:) is the logarithmic derivative of a Q(a:,ti,t2)~radical, so t^ is not a Liouvillian monomial over Q(x,ti,t2), and K is isomorphic as a differential field to Q (x, ti, ^2, y/xti). ® Alternatively, applying the Risch structure Theorem (Corollary 9.3.1), we find that the linear equation (9.9) for 6 = x + ^2/2 becomes

X

2x

which has the rational solution ri = r2 = 1/2. This implies that Dts/ts is the logarithmic derivative of a if-radical, and that c = t1l{xti) G Const(K). Note that the requirement that each ti be a monomial eliminates expressions containing algebraic functions from the algorithm presented here. Although the problem of integrating elementary functions containing algebraic functions is also decidable, the algorithms used in the algebraic function case are beyond the scope of this book [8, 9, 11, 14, 29, 73, 74, 76, 91].

136

5 Integration of Transcendental Functions

Once we have a tower of monomials K = C{ti^... ^tn)-, the algorithms of this chapter reduce the problem of integrating an element of K to various integration-related problems involving elements of C ( t i , . . . , t^_i), thereby eliminating the monomial t^. Since the reduced problems involve integrands in a tower of smaller transcendence degree over C, we can use the algorithm recursively on them, and termination is ensured. In order to avoid writing the full tower of extensions throughout this book, we write K = k{t) where k = C ( t i , . . . ,tn-i) and t = t^ is a monomial over /c, and the task of the algorithms of this chapter is to reduce integrating a given element of k{t) to integration-related problems over k. If t is elementary over k^ then having an elementary integral over k{t) is equivalent to having an elementary integral over A:, so the algorithms we present in this book provide a complete decision procedure for the problem of deciding whether an element of a purely transcendental elementary extension of {C{x)^d/dx) has an elementary integral over C{x). For more general functions, when t is not elementary over A:, it can be proven that if t is either an hyp er exponential monomial or nonlinear monomial over k with <Sf^ = 5^^^, then having an elementary integral over k{t) is equivalent to having an elementary integral over k (Exercise 5.5), so the algorithm is complete for integrands built from transcendental logarithms, arc-tangents, hyper exponentials and tangents. The only obstruction to a complete algorithm for Liouvillian integrands is the case where t is a nonelementary primitive over k: even though we can reduce the problem to an integrand in /c, the problem becomes however to determine whether / € /c has an elementary integral over k{t)^ and although there are algorithms for special types of primitive monomials [6, 21, 22, 52, 53, 94], this problem has not been solved for general monomials (Exercise 5.5f)). As will be seen from numerous examples in this book, the algorithm can still be used successfully on many integrands involving nonelementary monomials. It cannot however always provide a proof on nonexistence of an elementary integral over k{t) when t is a nonelementary primitive over k. The reduction from k(t) to t is also incomplete for general nonlinear monomials, but is complete for tangents and hyperbolic tangents. The general line of the integration algorithm is to perform successive reductions, which all transform the integrand to a "simpler" one, until the remaining integrand is in k (Fig. 5.1): ® The Hermite reduction (Sect. 5.3), which can be applied to arbitrary monomials, transforms a general integrand to the sum of a simple and a reduced integrand; ® The polynomial reduction (section 5.4), which can be applied to nonlinear monomials, reduces the degree of the polynomial part of an integrand; ® The residue criterion (Sect. 5.6), which can be applied to arbitrary monomials, either proves that an integrand does not have an elementary integral over fc(t), or transforms it to a reduced integrand (i.e. an integrand in k{t));

5.2 Outline and Scope of the Integration Algorithm

137

® Reduced integrands are integrated by specific algorithms for each case of LiouviUian or hypertangent monomial (Sect. 5.8, 5.9 and 5.10). Those algorithms either prove that there is no elementary integral over fc(t), or reduce the problem to various integration-related problems over k. Algorithms for solving those related problems are described in Chap. 6, 7 and 8. Except for the last part, the various reductions are applicable to arbitrary monomial extensions.

/em

f = g + h^g simple, h £ k{t)

f = g -\- h^g simple, h G k{t) No elementary integral

No elementary integral fek Fig. 5.1. General outline of the integration algorithm

138

5 Integration of Transcendental Functions

5.3 The Hermite Reduction We have seen in Sect. 2.2 t h a t the Hermite reduction rewrites any rational function as the sum of a derivative and a rational function with a squarefree denominator. In this section, we show t h a t the Hermite reduction can be applied to the normal p a r t of any element of a monomial extension. Let {k, D) be a differential field and t a monomial over k for the next two sections. D e f i n i t i o n 5 . 3 . 1 . For f E k{t), we define the polar multiplicity of / to be

Note t h a t /i(0) = - c o and t h a t / i ( / ) > 0 for any / 7^ 0, since in t h a t case there is always some polynomial p G k[t] for which h'p{f) = 0. Also, the minimum in the above definition can be taken over all the irreducible or squarefree factors of the denominator of / . It is easy to see t h a t for / / 0, / i ( / ) is exactly the highest power appearing in any squarefree factorization of t h e denominator of / (Exercise 5.1). algorithm T h e o r e m 5 . 3 . 1 . Let f G k{t). Using only the extended Euclidean in k[t], one can find g^h^r G k{t) such that h is simple, r is reduced, and f = Dg + /i + r . Furthermore, the denominators of g^h and r divide the denominator of f, and either g — 0 or ß{g) < fi{f). Proof. Let f = f^ -{- f^ ^ f^he the canonical representation of / , and write fn = a/d with a^d G k[t] and gcd(a,d) = 1. We proceed by induction on m = ji{fn)- Let d = did^' • • d^ be a squarefree factorization of d. If m < 1, then either / ^ = 0 or d is normal. In both cases, fn is simple, so g = 0^ h = fn and r = fp -\- fs E k{t) satisfy the theorem. Otherwise, ?7i > 1, so assume t h a t the theorem holds for any nonzero g ~ gp + gn -^ gs with ß{gn) < ^ , and let v = d^ and u = d/v^. Since every squarefree factor of d is normal by the definition of the canonical representation, V is normal, so gcd{Dv^v) = 1. In addition, gcd(ii,i;) = 1 by the definition of a squarefree factorization, so gcd{uDv^ v) = 1. Hence, we can use the extended Euchdean algorithm to find h^c£ k[t] such t h a t = buUv Multiplying both sides by (1 — m)/{uv^)

gives

a

(1 — m)bDv

(1 — m)c

y^yrn

ym

y^ym—1

SO, adding and subtracting Db/v^~^

/ Db

-f cf.

(m-l)bDv\

to the right hand side, we get

(l-m)c~uDb

^

5.3 The Hermite Reduction

139

where po = h/v^~^ and IL» = {{l—m)c—uDh)/{uv^~^). Since the denominator of w divides uv^~^^ w has no special part, so let w = Wp-\-Wn be the canonical representation of w. Since iJ.{w) < m — 1, we have ß{wn) < m — 1, so by induction we can find gi^hi and r i in k{t) such t h a t Wn = Dgi + /i + r i , h is simple, r i is reduced, the denominators of gi^h and r i divide nf"^~\ and /i(^i) < /i(i(;) if^i 7^ 0. Let then ^ = go-^gi and r = / p + ' w ; p + / s + r i , and write e for the denominator of / . Note t h a t d \ e hy the definition of the canonical representation. The denominator of gi divides uv'^~^ and go = b/v'^~^^ so the denominator of g divides d hence e. The denominator of h divides uv^~^ ^ so it divides d hence e. The denominator of w divides d and the denominator of r i divides uv'^~^, so the denominator of r divides e. In addition, / p , lüp, fs and r i are in k{t)^ which is a subring of k{t) by Corollary 4.4.1, so r G /c(t). Finally, we have f = fp + fs -^ fn = fp -i- fs + Dgo + w = fp + fs + Dgo + Wp + Dgi -^h^ri

= Dg +

h^r

which proves the theorem.

D

Although we have used the quadratic version of the Hermite reduction in the above proof, the other versions are also valid in monomial extensions (Exercise 5.2). Instead of splitting a rational function into a derivative and a simple rational function, the Hermite reduction splits any element of k{t) into a derivative, a simple and a reduced element. Thus, it reduces any integration problem to integrands t h a t are the sum of a simple and a reduced element.

H e r m i t e R e d u c e ( / , D)

(* Hermite Reduction - quadratic version *)

(* Given a derivation D on k{t) and / G fc(t), return g^h^r G k{t) such that / = Dg + /i + r, h is simple and r is reduced. *) (/p5 fs, fn) ^— C a n o n i c a i R e p r e s e i i t a t i o n ( / , D) (a,d) <— (numer at or (/n), denominator (/n)) ( d i , . . . , dm) ^— SquarePree((i) for i <— 2 to m such that deg(di) > 0 d o V ^— di

u ^- d/v^ for J 4— i — 1 t o 1 s t e p —1 do (6, c) <~- ExtendedEuclidean('u Dv, v, —a/j) g ^ g + b/v^ a < jc — u Db d <— uv {q,r) <— PolyDivide(a,t£'i;) r e t u r n ( ^ , r/{uv),q + fp + fs)

(* d is monic *)

140

5 Integration of Transcendental Functions

Example 5.3.1. Let k = Q{x) with D = d/dx^ and let t be a monomial over k satisfying Dt = 1 +t^ ^ i.e. t = tan (a:), and consider X — tan(x)

X—t

Since / has no polynomial part and t is normal in /c[t], the canonical representation of / is (/p, /s, /n) = (0,0, / ) so we get a = x — t and d = t'^ = d2 where ^2 = t. We then have: i

V

uj

h

c

a

2 t 1 1 —x xt + 1 —xt and a/uv = —xt/t = —x, so the Hermite reduction returns (—x/t,0,—x), which means that X — taii(x)

tan(x)^

aa: =

X f -—• — x ax

tan(a:)

j

and the remaining integrand is in k{t). The Hermite reduction can also be iterated, yielding a decomposition of / into a sum of higher-order derivatives of reduced and simple elements of k{t) (Exercise 5.3).

5.4 T h e Polynomial Reduction In the case of nonlinear monomials, another reduction allows us to rewrite any polynomial in k[t] as the sum of a derivative and a polynomial of degree less than 5{t). T h e o r e m 5.4.1. If t is a nonlinear monomial, then for any p G k[t], we can find q^r G k[t] such that p = Dq + r and deg(r) < 5{t). Proof We proceed by induction on n = deg(p). If n < (5(t), then g = 0 and r = p satisfy the theorem. Otherwise n > ö{t) so assume that the theorem holds for any a G k[t] with deg(a) < n. Let

(n - 6{t) + l)A(t) go = ct^""^^*^"^"-^, and ro = p — Dqo. Since t is nonlinear and deg(go) > 0, Lemma 3.4.2 implies that deg(jDgo) = deg(go) + <^(0 — 1 = n, and that the leading coefficient of Dqo is {n~6{t)-\-l) cX{t) = lc{p). Hence, deg(ro) < n, so by induction we can find gi, r G k[t] such that ro = Dgi + r and deg(r) < 5{t). Therefore, p = Dqo + ro = Dqo + Dqi + r = Dq + r where q = qo + qi G k[t].

D

5.4 The Polynomial Reduction PolynoniialReduce(p, D)

141

(* Polynomial Reduction *)

(* Given a derivation D on k{t) and p G k[t] where t is a nonlinear monomial over k, return q,r E k[t] such that p = Dq + r, and deg(r) <

m- *)

if deg{p) < ö{t) t h e n r e t u r n ( 0 , p ) m ^ deg(p) - S{t) + 1 qo^{lc{p)/(mX{t)))t^ (q,r) ^- P o l y n o m i a l R e d u c e ( p — Dqo.D) return(go + q,r)

Example 5.4-1- Let k = Q(x) with D = d/dx^ and let t be a monomial over k satisfying D t = 1 + 1 ^ , i e . t = t a n ( x ) , and consider p=l

+ x t a n ( x ) 4-tan(x)^ == 1 + a : t + t^ G Ä;[t].

We have (5(t) = 2, A(i^) = 1, and applying P o l y n o m i a l R e d u c e , we get m = deg(p) — 1 = 1, go = ^7 Dqo = 1 - | - 1 ^ , so p — Dgo = ^t^ which has degree 1. Thus, V+.tan(.) + tan(.n.. = tan(.)4-/.tan(.).. and it will be proven later t h a t the remaining integral is not an elementary function. If tS 7^ /c, i.e. S^^^ y^ 0, then any nontrivial element of S can be used t o eliminate the term of degree 5{t) — 1 from a polynomial. T h e o r e m 5.4.2. Suppose that t is a nonlinear monomial. Let p E k[t] with deg(p) < ö{t), a E k be the coefficient of t^^^^~^ in p, and c = a/X{t). Then,

for any q £ S

\k.

Proof. Let q E S \ k^ then Dq/q £ k[t] and by Lemma 3.4.2, deg{Dq/q) = deg(Dg)—deg(g) = (5(t) —1, and the leading coefficient of Dq is deg(g)lc(g)A(t). Hence Vdeg(g) q J

deg(g)

lc{q)

which implies t h a t the degree of p — c / deg(g) Dq/q is at most ö{t) — 2.

D

142

5 Integration of Transcendental Functions

5.5 Lioiiville's Theorem Given a differential field K and an integrand / G ÜT, if an elementary integral is found, it can be easily proven correct by differentiation. Furthermore, there are usually several ways to find elementary integrals when they exist. Proving t h a t / has no elementary integral is however quite a different problem, since we need results t h a t connect the existence of an elementary integral to a special form of the integrand. The first such result is Laplace's principle [55], which states roughly t h a t we can simpUfy the integration problem by allowing only new logarithms to appear linearly in the integral, all the other functions must be in t h e integrand already^. Liouville was t h e first to state and prove a precise theorem from this observation, first in t h e case of algebraic integrands [57, 58], then for more general integrands [59]. See Chap. IX of [61] for the fascinating history of Liouville's Theorem in t h e 19*^ century. This theorem has become the main tool used in proving t h a t no elementary integral exists for a given function. Furthermore, since it provides an explicit class of elementary extensions to search for an integral, it forms the basis of the integration algorithm. While Liouville used analytic arguments, it is now possible to prove it algebraically in the context of differential fields. Algebraic techiques were first used by Ostrowski [69], who presented a modern proof of Liouville's Theorem, together with an algorithm t h a t reduces integrating in k{t) to integrating in k when t is a primitive monomial over k. The first complete algebraic proof of Liouville's Theorem was then published by Rosenlicht [79] and t h e first proof of the strong version of Liouville's Theorem by Risch, who published it together with a complete integration algorithm for purely transcendental elementary functions [75]. We follow b o t h of them here, first presenting essentially Rosenlicht's proof of the weak Liouville Theorem, and then progressively removing the restrictions on the constant fields, obtaining Risch's proof of t h e strong Liouville Theorem. We remark t h a t Liouville's Theorem has been extended in various directions [17, 71, 81, 86], but those extensions go beyond the scope of this book. Integration algorithms t h a t yield nonelementary integrals [21, 22, 52, 53] are based on such extensions [86]. T h e o r e m 5.5.1 ( L i o u v i l l e ' s T h e o r e m ) . Let K be a differential field and f ^ K. If there exist an elementary extension E of K with C o n s t ( ^ ) = Const(i^) andg £ E such that Dg = f, then there are v G K, ui^... ^u^ G K* and c i , . . . , Cn G C o n s t ( K ) such that

f = Dv + f]c,^.

(5.3)

^". ..la differentiation laissant subsister les quantites exponentielles et radicales, et ne faisant disparaitre les quantites logarithmiques qu'autant qu'elles ont multipliees par des const antes, on doit en conclure que I'integrale d'une function differentielle ne peut contenir d'autres quantites exponentielles et radicales que celles qui sont contenues dans cette fonction..."

5.5 Liouville's Theorem

143

Proof. Write C = Const (K) and let E be an elementary extension of K with Const(£^) = C and g G Ehe such that Dg = f. Then, there are t i , . . . , t ^ ^ •£" such that E = K(ti^... , t ^ ) and each ti is elementary over K{ti^... ,t^_i). We proceed by induction on m. For m = 0, we have E = K, so letting V = g G K, we get / = Dv^ which is of the form (5.3) with n = 0. Suppose now that ?TI > 0 and that the theorem holds for any elementary extension generated by ?7i — 1 elements. Let t = ti and F = K{t). Since K C F C E^ then C C Const(F) C Const(E) = C, so Const(F) = C. In addition, / G F , and E = F ( t 2 , . . . , t ^ ) is an elementary extension of F generated by m — 1 elements, so by induction there are v £ F^ ui^... ^Un E F* and c i , . . . , c ^ € C such that

f = Dv + J2c,:^.

(5.4)

Case 1: t transcendental over K. Then, since Const(F) = C, t is Liouvillian monomial over K by Theorems 5.1.1 and 5.1.2. Let p G K[t] be normal and irreducible. We have i/p{Dui/ui) > —1 by Corollary 4.4.2, hence h'p{Y^^=iCiDui/ui) > —1 by Theorem 4.1.1. Suppose that iyp{v) < 0. Then, Up{Dv) = Up{v)-1 < -1 by Theorem 4.4.2, so i^pif) = mm{i'p{Dv), ~1) < - 1 by Theorem 4.1.1, in contradiction with f E K. Hence I'piv) > 0, so, since this holds for any normal irreducible p, v e K{t). Hence, Dv G K{t) by Corollary 4.4.1. Write now Ui = Wi Yü^iPT/ where Wi G K* ^ each pij G K[t] is monic irreducible, and the e^j's are integers. Then, using the logarithmic derivative identity and grouping together all the terms involving the same pij, we get

1=1

^-^

j=l

where the g^'s are in K[t], monic, irreducible and coprime. Write

3-

Ti

*•

and suppose that one of the g^-'s, say g/^, is normal. We have i^qj^iqk) = 1 and z/q;,(gj) = 0 for j / /c, so i^q^idkDqk/qk) = - 1 and Uq^{djDqj/qj) = 0 by Corollary 4.4.2. This implies that T^qk(^j^k^3-^^31^3) — ^' hence that Vq^ (h) = —1. But qk is normal and Dv G K{t)^ hence z/g^ {Dv) > 0, so z/^^ (/) = — 1, in contradiction with f G K. Hence all the gj's in equation (5.5) are special. Case la: t is a logarithm over K. Then, Dt = Da/a for some a G K* ^ and every irreducible p e K[t] is normal by Theorem 5.1.1, so N = 0 in equation (5.5) and v^ Dv G K[t]. From (5.5) we get Dv = f — g G K. By Lemma 5.1.2, this implies that 01 v = ct-^b where b^c G K and Dc — 0 (otherwise deg(Z}f) > 1). Hence,

144

5 Integration of Transcendental Functions ^,

Da a

v^ ^-^

Dwi w

i=l

which is of the form (5.3). Case lb: t is an exponential over K. Then, Dt/t = Da for some a G K, and the only special monic irreducible p £ K[t] is p = t hj Theorem 5.1.2, so iV = 1 in equation (5.5) and qi = t (with di possibly 0). Hence, diDqi/qi = diDt/t = diDa^ so f = Dw + g where w = v -\- dia E K{t). Suppose that Vt{w) < 0, then Vt{Dw) — Ut{w) < 0 by Theorem 4.4.2 since t G S^^^, so ^t(/) < 0 in contradiction with f £ K. Hence, i^ti^) > 0 so w £ K[t]. By Lemma 5.1.2, Voo{Dw) = Uoo{w)^ so deg{Dw) = deg(tL'), which implies that deg{w) = 0 since / = Dw -{- g £ K. Hence w e K and

Wi

^-^

which is of the form (5.3). Case 2: t algebraic over K. Let Tr : F —^ K and N : F —^ K he the trace and norm maps from F to K and d = [F : K]. Applying Tr to both sides of equation (5.4) we get:

Trif)=Tr{Dv

+ yci^)=Tr{Dv)+yciTr{

— )

since Tr is if-linear and the Q'S are in K. We have Tr{f) = df since f E K, and Tr{Dv) = DiTriv)) and Tr ' ^""'^ DN{ui)

Niu,)

by Theorem 3.2.4, so Ci Dwi

f = Dw + J2

d

which is of the form (5.3) with w = Tr{v)/d

e K and Wi = N{ui) G /f *.

D

Of course, in practice we may have to adjoin new constants in order to compute integrals, as we have seen in Chap. 2. We first show that new transcendental constants are not necessary in order to express an elementary integral T h e o r e m 5.5.2. Let K he a differential field with algebraically closed constant field and f £ K. If there exist an elementary extension E of K and g £ E such that Dg = f, then there are v £ K, i/i,...,ii„ £ K* and c i , . . . , c^ £ Const(ÜT) such that

=1

Uj

5.5 Liouville's Theorem

145

Proof. Suppose that there exist an elementary extension E of K and g £ E such that Dg = f. Write Const (if) = C, Const (£) = C ( a i , . . . , am) for some constants a i , . . . , a ^ in E, and let F = i f ( a i , . . . , a ^ ) . Since C ( a i , . . . ,ay^) C F C E^ C ( a i , . . . , a ^ ) C Const(F) C Const(^), so F and E have the same constant subfield. In addition, f E F and E is elementary over F , so by Theorem 5.5.1, there are v e F, ui,... ,Um £ F* and c i , . . . , c„ G Const(F) such that

f^Dv + Tc,^.

(5.6)

Let X i , . . . , X ^ be independent indeterminates over K. Since the elements of F are rational functions in a i , . . . , a ^ , we can write _ p(ai,...,a^) __ r i ( a i , . . . , a ^ ) _ pi(ai,... ,a^) -y _ __ -^ Q — —;; ^—- ana Ui — — —- [o.i) g(ai,. . . , a ^ ) Si(ai,...,a^J g u a i , . . -,0^771) where p^q^Pi^Qi are in i i r [ X i , . . . , X ^ ] , and Vi^si are in C [ X i , . . . , X ^ ] . In addition, g{ai^..., a ^ ) j^ 0, where

5 = 9 ( n ^ J lf[pA lf[qA

€K[Xi,...,X^].

Replacing v, c i , . . . , c ^ and w i , . . . , ? i ^ by the fractions (5.7) in (5.6), and clearing denominators, we obtain a polynomial h G K[Xi^..., Xm] such that /i(ai,... , a ^ ) = 0. By Lemma 3.3.6 applied to g and 5 = {/i}, there are 6 1 , . . . , 6^ G C such that ^'(^i^ • • • ^ ^m) 7^ 0 a^nd /i(6i,..., bm) = 0. But this implies that

1=1

where p(6i,...,6^) g(6i,..., 6^)

ri(öi,...,ö^) 5^(61, ...,bm)

pi(6i,...,5^) qi{oi, ..•,bm)

Since p,q,pi,qi G K [ X i , . . . , X ^ ] and ri,5i G C [ X i , . . . , X ^ ] , we get w e K, wi^... ,Wn E K* and d i , . . . , d^ ^ C', which proves the theorem. D We can finally remove all the constant restrictions in Liouville's Theorem, showing that for arbitrary constant subfields, v in (5.3) can be taken in K\ and the Ui^s can be taken in K{ci^..., Cn). T h e o r e m 5.5.3 (Liouville's T h e o r e m — Strong version). Let K be a differential field, C = Const(K), and f e K. If there exist an elementary extension E of K and g £ E such that Dg = f, then there are v £ K, C i , . . . , Cyi G C^ and U i , . . . , w^ £ i ^ ( c i , . . . , c„)* such that f = Dvi-}

Ci 1=1

.

146

5 Integration of Transcendental Functions

Proof. Suppose that there exist an elementary extension E of K and g G E such that Dg = / . Since CK is algebraic over K, Const(Cür) = C Pi CK = C by Corollary 3.3.1. Hence, CK has an algebraically closed constant subfield, / € CK^ g E CE", which is an elementary extension of CK^ so by Theorem 5.5.2, there are v £ CK., n i , . . . , i ^ ^ G {CKY and ci,...,Cn G C such that f = Dv+ }

Ci

.

i=l

*

F = K{v^ 1^1,..., tin, c i , . . . , Cn) is finite algebraic over K^ so let Tr;^ : F -^ K be the trace from F to K^ K be the algebraic closure of K and a i , . . . , cr^ be the distinct embeddings of F in if over K. Each aj can be extended to a field automorphism of K over K, and since Trf^ and each cr^ commute with D by Theorem 3.2.4, we have

j=l

j = l i=l

^

so j = l 1=1

with

-*

tu = — Trr^(v) E Ä', rfö^- = — cj^ G K

and

lü^.- = u^^ G K .

In addition, Const(E') = Ü n F = Ü by Corollary 3.3.1, and Ddij = D(c^^/m) = [DciY^/m = 0, so dij G C for each i and j . Let now L = K{dii^... ,djnn) and M = L(t(;ii,... ,tt;^n)- Since L is algebraic over K^ K is the algebraic closure of L. Since M is finite algebraic over L, let Tr^ : M -^ L and N : M —> L he the trace and norm maps from M to L. Since (i^j G L and T r ^ is L-linear, we have 'L l"*^ „.,

i-"ii-L

I ^.^. y - " i i

^(^^^.

by Theorem 3.2.4, so

•

1

•

1

^ Ü

^ ^ - ATK hence

m

n

3 = li=l

T

^

j-^

^'^

which is of the form (5.3) with w G K^ dij G C and Zij = N{wij) in K{dn,...,dmny•

5.6 The Residue Criterion

147

5.6 The Residue Criterion Now that Liouville's Theorem gives us a way of proving that a function has no elementary integral over a given field, we can complete the integration algorithm. For the rest of this chapter, let (/c, D) be a differential field and t a monomial over k. From the Hermite reduction, we can assume without loss of generality that the integrand is given as the sum of a simple and a reduced element of k(t). We have seen in Sect. 2.4 that the Rothstein-Trager algorithm expresses the integral of a simple rational function with no polynomial part as a sum of logarithms. In this section, we show that this algorithm can be generalized to any monomial extension, where it will either prove that a function has no elementary integral, or reduce the problem to integrating elements of k{t). Rothstein had already generalized this algorithm to elementary transcendental extensions in his dissertation [83]. L e m m a 5.6.1. Let f G k{t) be simple. If there are h £ k{t), an algebraic extension E of Coiist(k), v G k{t), c i , . . . , c^ G E^ and i^i,..., n^ G Ek{t) such that

then n

residuep(/) =

^Cijyp{ui) i=l

for any normal irreducible p G Ek[t]. Proof Let / G k{t) be simple, and suppose that there are h G k{t), an algebraic extension E of Const(/c), v G k{t)^ c i , . . . , c„ G £ , and ui^.,. ,Un G Ek{t) such that ^-^ i=\

Ui *

Note that / + /i is simple since h G ^(t). Let p G Ek^ be normal and irreducible. Then, for each i, Vpi^Dui/ui) > —1 and Tesid\iep{Dui/ui) = Pp{ui) by Corollary 4.4.2. Suppose that iyp{v) < 0. Then Vp{Dv) = Vp{v) — 1 < — 1 by Theorem 4.4.2, which implies that I'pif + h) < —1 in contradiction with f -{- h being simple. Hence i'p{v) > 0, so Vp{Dv) > 0, which implies that residuep(D'L') = 0. Furthermore, Vp{h) > 0, so residuep(/i) = 0. Since residucp is E'^-linear, we get residuep(/) = residuep(/) + residuep(/i) = residuep(/ + h) = residuep(D^') + y . Q residuCp ( i=l

\

j = 2 , ^i ^pi'^i) • ^i

J

i=.i

D

148

5 Integration of Transcendental Functions

L e m m a 5.6.2. Suppose that Const(Ä:) is algebraically closed and let f E k{t) be simple. If there exists h G k{t) such that f + h has an elementary integral over k{t), then residuep(/) G Const(A:) for any normal irreducible p G k[t]. Proof. Let C = Const (A:), and suppose C is algebraically closed and that f-\-h has an elementary integral over k{t) where / G k{t) is simple and h G k{t). By Theorem 5.5.1, there are v^ui^... ^Un G k and c i , . . . , c ^ G C such that

Let j> G k[t] be normal and irreducible. By Lemma 5.6.1 we have n

residuep(/) = ^ c ^ iyp{ui) G C. i=l

D

Example 5.6.1. Let Ä: = Q, t be a monomial over k with Dt ^ 1 {i.e. D = d/dt)^ and

Then, / has an elementary integral over k{t): Of — 0

I.

,

^

jdt = {l + x/=l) log(l + i^/=l) + (1 - v ^ ) log(l -

t./^).

On the other hand, t^ + 1 is irreducible over Q, but /2t — 2\ residuet2+i(/) = 7rt2+i f —^ j = t+1 which is not a constant. This shows that the hypothesis that the constant field of k be algebraically closed is required in Lemma 5.6.2. If we replace Q by C, then t^ + 1 = (t - ^^.){t + \ / ^ ) , residue^

r^(f)

= TT^ r—r i

7=

and

(

2t ~ 2 \

which are constants. This shows that the hypothesis that p be irreducible is also required in Lemmas 5.6.1 and 5.6.2.

5.6 The Residue Criterion

149

T h e o r e m 5.6.1. Let f G k{t) be simple, and write f = p-\-a/d where p^a^d E k[t], d j^ 0, deg(a) < deg{d), and gcd{a^d) = 1. Let z be an indeterminate over k, r = resultantt(a — zDd^ d) £k[z\^ ^ = '^s^n be a splitting factorization of r w.r.t. the coefficient lifiing njj of D to k[z], and Dga

-£ a^^

(5i

rsia)=0

where go, = gcd(a —aDd, d) E k{a)[t] and the sum is taken over all the distinct roots ofrg. Then, (i) g G k{t), the denominator of g divides d, and f — g is simple. (a) If there exists h G k{t) such that f + h has an elementary integral over k{t), then r^ E k and f — g £ k[t]. (Hi) If there are h G k{t), an algebraic extension E of Coiist{k), v G k{t), ci,...,Cn E E, and ui^... ,Un G Ek{t) such that n

j^

then rg factors linearly over E. Proof (i) Let r^ = cr^^ • • -r^" be the irreducible factorization of r^ in k[z]. Then, g can be rewritten as

E

\-^

Dga

i=lri{a)=0

^"^

For each i, let ki be k{t) extended by all the roots of r^, and a^ be a given root of ri. Since ki is a finitely generated algebraic extension of A:(t), the field automorphisms of ki over k{t) commute with D by Theorem 3.2.4, so we get

j:Tr.

Dga, ai 9ai

by Theorem 3.2.4 where Tri is the trace map from k{t){ai) to k{t). Hence, g G k{t). Furthermore, since g^ \ d for each root a of rg, lcmj.^(^ct)=o{9a) \ d^ so the denominator of g also divides d. Hence the denominator oi f — g divides (i, which implies that f ~ g is simple since d is normal. (ii) Suppose that f + h has an elementary integral over k{t) for some h G k{t), and let k be the algebraic closure of k. By Corollary 3.4.1, t is a monomial over /c, and simple (resp. reduced) elements of k{t) remain simple (resp. reduced) when viewed as elements of k{t). Furthermore f + h has an elementary integral over k{t), so we work with k{t) in the rest of this proof. Let a G A: be any

150

5 Integration of Transcendental Functions

root of r. If a = 0, then Da = 0. Otherwise a ^ 0 and a = residueg(/) for some normal irreducible q G k[x] by Theorem 4.4.3, hence Da = 0 by Lemma 5.6.2. Thus rs{a) = 0 in both cases by Theorem 3.5.2, so rn{(^) 7^ 0 since gcd(rn,r5) = 1. Since this holds for all the roots of r, we have r^ G k. For any a G k^ write g^ = gcd{d^a — aDd). Note that all the irreducible factors of QQ, must be normal, since g^ | d^ which is normal. Let a^ß G A:, and q G k\t\ be a normal irreducible common factor of go^ and gß. Then a = residueg(a/(i) = /3 by Lemma 4.4.3, so gcdi{go,^gß) = 1 when a^ ß. Let now q E.k\t\ be irreducible and normal, and ß = residueg(/). If /3 = 0, then q does not divide d, so q does not divide any g^^ which implies that i'q{g) > 0, hence that residueg(^) = 0 = residueg(/ -- g). If ß ^ 0, then r{ß) = 0 by Theorem 4.4.3, and q \ gp hy Lemma 4.4.3, so rs{ß) = 0 since r^ G k. Since d is squarefree, gß is squarefree, so i'q{gß) = 1. By Theorem 4.4.1, residue^ is /c-linear, so we get residueg(/ - g) = ß -

^

aresidue^ I

r^(a)=0

\ = ß --

\ 9a

y

^

ah'q{goc)

rsia)=0

by Corollary 4.4.2. Since i^q{ga) = 0 foi a j^ ß^ this gives residueg(s) = ß—ß = 0. Since this holds for any normal irreducible q £ k[t] and f ~ g is simple, we have f — g G k[t]^ hence f — g E k[t]. (iii) Suppose that there are h G k{t), an algebraic extension E of Const(/c), V G k{t)^ ci^... ^Cn E E^ and t^i,... ,1^^ E Ek{t) such that

/ + /i = i)t; + y c , i l ^ . i=l

(5.9)

*

Let A; be the algebraic closure of k. As explained in part (ii), we can replace k{t) by k{t) and view (5.9) as an equality in k{t). Let a E k he any root of Vs. By Theorem 4.4.3, a = residuep(/) for some normal irreducible p E k[t]^ so by Lemma 5.6.1 n

a = residuep(/) = 2 . ^i^pi^i) E E. Hence, E contains all the roots of rg in Ä;, so Vg factors linearly over E.

D

Note that since the roots of r^ are all constants by Theorem 3.5.2, g as given by (5.8) always has an elementary integral, namely g=

y^

alog{gcd{d^ a — aDd))

rs{a)=0

which is the Rothstein-Trager formula in the case of rational functions. Part (iii) of Theorem. 5.6.1 applied to the rational function case proves part (iii)

5.6 The Residue Criterion

151

of Theorem 2.4.1, thereby completing the proof of t h a t theorem. As in t h e rational function case, a prime factorization Vg = us^'^ • • • s^ is required, as well as a gcd computation in k{ai)[t] for each i, where a^ is a root of s^. There is no need however to compute the splitting field of Vg. Furthermore, the monic part of Vg always has constant coefficients.

R e s i d u e R e d u c e ( / , D)

(* Rothstein-Trager resultant reduction *)

(* Given a derivation D on k(t) and / 6 k(t) simple, return g elementary over k(t) and a Boolean h G {0,1} such that / — Dg G A:[t] if 6 = 1, or f + h and f -\- h — Dg do not have an elementary integral over k{t) for any h e k{t) ifb = 0. *) d <— denominator(/) (p, a) ^- PolyDivide(numerator(/), c?) (* f = P + a/d *) z ^- a. new indeterminate over k{t) r ^- resultantt((i, a — zDd) {rn.rs)^ SplitFactor(r, KD) us\^ '' • s^ <— factor(rs) (* factorization into irreducibles *) for i ^- 1 t o m d o a <— a I Sii^a) = 0 gi <— gcd(c?, a — aDd) (* algebraic gcd computation *) if Vn ^ k t h e n 6 ^ - 1 else 6 <— 0

Example

5.6.2. Consider

r

21og(x)^ — log(x) — x^ log(x)^ — x'^\og{x)

dx.

Let k = Q(x) with D = d/dx^ and let t be a monomial over k satisfying Dt = 1/x, i.e. t ~ log(x). Our integrand is t h e n

which is simple since t^ — x^t is squarefree. We get d = t^ -- xH,

p = 0,

a = 2t^ - t - x ^

and r = resultantt ( {t^ - x^t, —

V = 4x^(1 -- x'^) (z^ -xz^ V

~t^ + {2xz - l)t + x{z - x)

^ 4

z+ T 4

152

5 Integration of Transcendental Functions

which is squarefree. Then, K^or = -x^{4{5x^

+ 3)z^ + 8x(3x^ - 2)z^ + {5x^ - 3)^ - 2x{3x^ - 2))

so the spHtting factorization of r w.r.t. KJJ is Vs = gcd(r, nor)

=x

Iz.2

1^

and Tn = — = - 4 x ( x ^ - 1)(2; - x) ^ k. Hence, / does not have an elementary integral. Proceeding further we get gi = gcd [ t^ + x^t, —

-t'^ + {2xa - l)t i~ x{a - x))

= t + 2ax

where a^ — 1/4 = 0, so ^ =

^

a log(t + 2 a x ) = - log(t + x) - - log(t -

x).

a\a^-l/4=0

Computing / — Dg we find

r

log(x)+

X

21og(x)^ — log(x) — x^dx = •- 1log / l o g ( x ) + x \ + f dx log(a:)"^ — x^ log(x) 2 \ log(x) l o g ( x ) —xy J log(x) -X

log(x) + X

= ^°Hg§F^J+^'^"^ where Li(x) is the logarithmic integral, which has been proven to be nonelementary since r^ ^ k. W i t h the notation as in Theorem 5.6.1, we have gcd(r5,r^) = 1, so any root a of Ts with multiplicity n is also a root of r with multiplicity n. Since gcd(a,d) = gcd( 0, gcd(d, a — aDd)

= pp^(i?^)(a, t)

where deg^(i?^) = i and Rm is in t h e subresultant P R S of d and a — zDd if deg{Dd) < deg((i), or of a — zDd and d if deg{Dd) > deg{d). T h u s , the Lazard-Rioboo-Trager algorithm is applicable in arbitrary monomial extensions, and it is not necessary t o compute the prime factorization of rg, or t h e g^^s appearing in (5.8), we can use the various remainders appearing in the subresultant P R S instead. As in the case of rational functions, we use a squarefree factorization of Vg = YYi=i ^l ^^ ^P^^^ ^^^ ^^™ appearing in (5.8)

5.6 The Residue Criterion

153

into several summands, each indexed by the roots of qi. We can also avoid computing pp^(jR^), ensuring instead t h a t its leading coefficient is coprime with the corresponding qi. And since multiplying any ga in (5.8) by an arbitrary nonzero element of k{a) does not change the conclusion of Theorem 5.6.1, we can make p p ^ ( Ä ^ ) ( a , t ) monic in order to simplify the answer. This last step requires inverting an element of k[a] and is optional. As in t h e rational function case, it turns out t h a t the leading coefficients of t h e pp^(jR^)(a, t)'s are always invertible in k[a] (Exercise 2.7).

R e s i d u e R e d u c e ( / , D) (* Lazard-Rioboo-Rothstein-Trager resultant reduction *) (* Given a derivation D on k{t) and f £ k{t) simple, return g elementary over k(t) and a Boolean b G {0,1} such that / — Dg G k[t] if 5 = 1, or f + h and f -\- h — Dg do not have an elementary integral over k{t) for any h G k{t) if 6 = 0. *) d ^- denominator(/) (p,a) ^— PolyDivide(numerator(/),d) (* f = p + a/d *) z ^~ a. new indeterminate over k(t) if deg{Dd) < deg{d) t h e n (r, (i^o, i^i, • •., Äg, 0)) ^- SubResultanta;{d, a — zDd) else (r, (Äo,ßi, • • •, i^q,0)) ^- SubResultanta;(a — 2;Dd, 0 do if i = deg{d) t h e n Si i— d else Si ^ Rm where deg^{Rm) = i, 1 < m < q ( A l , . . . , A,) <- SquareFree(lct(Ä)) for j
Example 5.6.3. Consider the same integrand as in example 5.6.2. We have deg{Dd) < deg(d) and t h e subresultant P R S of d and a — zDd is

i

Ri

0

t^ - xH 1 (2 - 3z/x)t'^ + {2xz - l)t + x{z - x) 2 (4x2- - 6)z2 4- 3xz - 2^2 + l)t + x{z - x)(2xz --1) 3 4x^(1 — x^) {z? - xz^ - lz + jx)

The Rothstein-Trager resultant is r = i?3, and its split-squarefree factorization w.r.t. K£) is

154

5 Integration of Transcendental Functions

si = gcdfr, K^r) = x^ { z^

),

rii = — = —4:x(x'^ — l)(z — x) 4:k.

Hence, / does not have an elementary integral. Proceeding further we find that si is squarefree, and the remainder of degree 1 in t in the PRS is ^2 = {{Ax^ - 6)z2 + 3xz - 2x2 _^ ^^^ _^ ^(^ _ ^)(2xz - 1). Since gcd(lct(Ä2),5i) = gcd ({Ax^ - 6)^2+ 3 x z - 2 ^ 2 + 1,^2 iz'^ - »Si = i?2- Evaluating for 2; at a root a of z^ — 1/4 = 0 we get Si{a,t)

g=

Y2 Q!|a2-l/4=0

-

= ~-{{2x'^ - 6ax + l)t + 4ax^ - Sx^ + 2ax)

ce log ( — ( ( 2 ^ ^ - 6ax + l)t + x{4:ax'^ - 3x + 2a)) j ^

1 / (2rc2-3x + l)(t + x ) \ 1 / log .— _ log ' 2 ° V 2 y 2 ^ V

^

(2x2 + 3x + l ) ( t - x ) 2

Computing / — Dg we find 21og(x)2-log(x)-x2 ^ 1 f(2x^-3x-\-l)(log(x)+x) ax = - lor log(x)"^ — x'^ log(2^) 2 \ (2x^ + 3x + l)(log(a:) — x) / 1 __ 6x^-3 \ "^ ' ,log(x) 4x4-5x2 + ly where the remaining integral has been proven to be nonelementary. In fact, it is the integral of a rational function plus the logarithmic integral Li(x). If we had decided to make Si{a^t) monic, we would have obtained 5 i ( a , x ) = - - ( 2 x 2 - 6 a x + l)(t + 2ax) so the integral is then the same as in example 5.6.2.

5.7 Integration of Reduced Functions Prom the results of the previous sections, we are left with the problem of integrating reduced elements of a monomial extension. We use a specialized version of Liouville's Theorem for such elements.

5.7 Integration of Reduced Functions

155

T h e o r e m 5.7.1. Let k be a differential field, t be a monomial over k, C = Coiist(k(t)), and f £ k{t). If there exist an elementary extension E of k(t) and g E E such that Dg = f, then there are v G k{t), c i , . . . ,c„ € C, and ui,...,Un € 5/,(ci,...,c„,)[t]:fc(ci,...,c„,) such that

Proof Suppose that there exist an elementary extension E of k{t) and g E E such that Dg = f. Then, by Theorem 5.5.3, there are v G /c(t), c i , . . . , c^ G C, and ui,... ,Un G k{ci,... ,c^)(t) such that / = Dv -hJ27=i CiD{ui)/ui. Write g = J27=i CiD{ui)/ui. Since g = f — Dv, it follows that g G k{t). Let p G k[t] be normal and irreducible, and q G A:(ci,... ,Cn)[t] be any irreducible factor of p over /c(ci,..., c^). Then, i^p(/) > 0 by Corollary 4.4.1, and Vq{ciDui/ui) > — 1 for each i by Corollary 4.4.2, so z/g(^) > —1. Since this holds for any irreducible factor g of p and g G A;(t), Theorem 4.1.2 implies that Vp{g) > —1. Suppose that yp{v) < 0. Then, Pp{Dv) = i^p{v) - 1 < - 1 by Theorem 4.4.2, which implies that Up{Dv -\- g) < —1, hence that z^p(/) < —1, in contradiction with / reduced. Hence, iyp{v) > 0 for all normal irreducible p G k[t], which means that V G k{t) and Dv G k{t) by Corollary 4.4.1. Write now Ui = Wi YYjLi Pi7 where Wi G A:(ci,..., c^), each pij is a monic irreducible element of fc(ci,... ,c„)[t], and the e^j's are integers. Then, using the logarithmic derivative identity and grouping together all the terms involving the same pij^ we get

/ - ^ ^ + E^^^

+ E^i—

(5.10)

where the g^'s are in /c(ci,... ,Cn)[t], monic, irreducible and coprime. Each Wi is special since it is in A:(ci,... ,Cn). Suppose that qs is normal for some s, Then, Lemma 5.6.1 applied to (5.10) implies that n

residucg^ (/) = ^ i=i

n

CiVq,, {wi) + ^

djVq^ (qj).

j=i

But residueg^(/) = 0 since / G A:(t), and Uq^{wi) = 0 since wi G ^ ( c i , . . . , c„), and lyqX^j) = 0 for j ^ s since the g^'s are coprime. Hence, 0 = dsi'q^{qs) ~ dg^ so ds — 0 whenever g^ is normal. Keeping only the nonzero summands in (5.10), we get that each qj is special, which proves the theorem. D In the case of nonlinear monomials, we have seen that we can always rewrite a polynomial p G k[t] as the sum of a derivative and a polynomial of degree less than ö{t). We then have an analogue of the residue criterion that either proves that such a reduced function does not have an elementary integral, or eliminates the term of degree ö{t) — 1 from its polynomial part.

156

5 Integration of Transcendental Functions

T h e o r e m 5.7.2. Suppose that t is a nonlinear monomial Let f G k{t) and write f = p + d/d where p^a^d G k[t], d ^ 0, deg(p) < 6{t) and deg(a) < deg{d). Let b ^ k be the coefficient oft^^^^~^ inp, and c = b/X{t). If f has an elementary integral over k{t) then Dc = 0. Proof. Let C — Const(A:). Replacing C by its algebraic closure, we can assume without loss of generality t h a t C is algebraically closed. Suppose t h a t / has an elementary integral over k(t). Then, by Theorem 5.7.1, there are v G fc(t), c i , . . . , c^ G C, and i ^ i , . . . , l i ^ G tS such t h a t ""

Till-

f = Dv + V c , ^ ,

(5.11)

By Theorem 4.4.4, ^^^{Dui/ui) > —m and 7roo{t~^Dui/ui) = —y^{ui)\{t) for each i where m = 5{t) — 1. Furthermore, u^Q^a/d) > 0 since deg(a) < deg( —m and 7roo{ct/d) = TToo{t~'^o./d) = 0, which implies t h a t TTooif^f) = b. Suppose t h a t z/oo('^) < 0, then U^Q^DV) < —m by Theorem 4.4.4, so UooiDv + J^^^i CiDui/ui) < —m, in contradiction with ^oo{f) ^ —fT^- Hence z^cx)('^) > 0- K i^oo('^) > 0, then i/oo{Dv) > —ra by Theorem 4.4.4. Otherwise, z^oo(^) = 0 and Poo{Dv) > —m also by Theorem 4.4.4. Hence i/oo{t~^Dv) > 0 in any case, so TToo{t~^Dv) = 0. Multiplying both sides of (5.11) by t"^ and applying TTOO, we get b = n^{t-^f)

hence c = b/\{t)

= J2ci7r^(t-^-^]

=~X]Qi/oo(nOA(t)

= — Yll=i ^* ^oo{ui)^ so Dc = 0.

•

If c is a constant, then Theorem 5.4.2 implies t h a t

has degree at most S{t) — 2 for any q e S \k, so in the case of nonlinear monomials, we are left with reduced integrands with polynomial p a r t s of degree at most 6{t) — 2, provided t h a t we know at least one nontrivial special polynomial. If we know t h a t there are no nontrivial special polynomials, then integrating reduced elements of such nonlinear extensions is in fact easier, and an algorithm for t h a t purpose will be presented in Sect. 5.11. We have now all the necessary tools to complete the integration algorithm. In the following sections, we give algorithms t h a t , given an integrand / in k{t) for a monomial t, either prove t h a t / has no elementary integral over ^ ( t ) , or compute an elementary extension E of k{t) and an element g e E such t h a t / — Dg G k. This process eliminates t from the integrand, t h u s reducing the problem to integrating an element of k, which can be done recursively, i.e. the algorithms of this chapter can be applied to elements of k until we are left

5.8 The Primitive Case

157

with constants to integrate. Note that when t itself is not elementary over fc, then the problems of deciding whether an element of k has an elementary integral over k or over k{t) are fundamentally different, so our algorithms will produce proofs of nonintegrability only if the integrand is itself an elementary function. They can be applied however to much larger classes of functions. It turns out that it will also be necessary to assume that some related problems are solvable for elements of k. Those problems depend on the kind of monomial we are dealing with, so we need to handle the various cases separately at this point. Algorithms for all those related problems will be presented in later chapters.

5.8 The Primitive Case In the case of primitive monomials over a differential field k^ the related problem we need to solve over k is the limited integration problem: recall that the problem of integration in closed form is, given f E k to determine whether there exist an elementary extension E of k and g £ E such that Const(£') is algebraic over Const(A:) and Dg = f. Let wi,... ^w^ G k be fixed. The problem of limited integration with respect to wi,... ^Wn is: given f E k^ determine whether there are g E k and c i , . . . , c^ G Const(A:) such that Dg = f — ciwi — . . . — CnWn, and to compute g and the Q'S if they exist. It is very similar to the problem of integration in closed form, except that the specific differential extension k(JWi,..., JWn) is provided for the integral. We present in this section an algorithm that, with appropriate assumptions on fc, integrates elements of k{t) when t is a primitive monomial over k. We first describe an algorithm for integrating elements of k[t]. T h e o r e m 5.8.1. Let k be a differential field and t a primitive over k. If the problem of limited integration w.r.t. Dt is decidable for elements of k, and Dt is not the derivative of an element of k, then for any p £ k[t] we can either prove that p has no elementary integral over k(t), or compute q G k\i\ such that p — Dq G k. Proof We proceed by induction on m = deg(p). If m = 0, then p E k and q = 0 satisfies the theorem, so suppose that m > 0 and that the theorem holds for any polynomial of degree less than m. Since Dt is not the derivative of an element of k^ t is a, monomial over k, Const{k{t)) = Const(Ä:), and S = k hy Theorem 5.1.1. Thus, Theorem 5.7.1 says that if p has an elementary integral over fc(t), then there are t; Gfc[t],c i , . . . , c ^ G C and ui^... ^Un G /c(ci,...,c„) such that p = Dv + y c i ^

(5.12)

where C = Const (A:), if = Ä:(CI, . . . , c^) is an algebraic extension of Ä:, so t is transcendental over K. Furthermore, Dt is not the derivative of an element of

158

5 Integration of Transcendental Functions

K by L e m m a 5.1.1, so t is a monomial over K and Const{K{t)) = Const(K). Equating degrees in (5.12) we get deg{Dv) = deg(p) = m > 0, so deg(t') < m + 1 by L e m m a 5.1.2, so write p = af^ + 5 and v = ct^^^ + hf^ + w where a^h^c E k^ s^w E k[t]^ deg(s) < m and deg{w) < m. Equating the coefficients of t ^ + ^ and t ^ in (5.12) we get i^c = 0 and a = Dh-\-{m^l)cDt.

(5.13)

Since we can solve the problem of limited integration w.r.t. Dt for elements of k and a £ k^we can either prove t h a t (5.13) has no solution b e k^cE Const(Ä:), or find such a solution. If it has no solution, t h e n (5.12) has no solution so p has no elementary integral over k{t). If we have a solution 6, c, letting go = ct^+^ +6t"^, we get p-Dqo

= {at'^ + s)~{{rn

+ l)cDt + Db)t'^-{mbDt)t'^-^

=:s-{rnbDt)t'^-^

hence deg(p — Dqo) < m. By induction we can either prove t h a t p — Dqo has no elementary integral over k{t)^ in which case p has no elementary integral over k{t)^ or we get qi G k\t] such t h a t p — Dqo — Dqi Gfc,which implies t h a t D p — Dq E k where g = go + ^i •

IntegratePriniitivePolynoiiiial(p, D) (* Integration of polynomials in a primitive extension *) (* Given a is a primitive monomial t overfc,and p E k[t]^ return q E k[t] and a Boolean ß E {0,1} such that p — Dq E k if ß = 1, or p — Dq does not have an elementary integral over k(t) ii ß = 0. *) if p E k t h e n r e t u r o ( 0 , 1 ) a <— lc(p) (* Limitedlntegrate will be given in Chap. 7 *) (6, c) <— L i m i t e d l n t e g r a t e ( a , Dt, D) if (5, c) = "no solution" t h e n return(0,0) m -e- deg(p)

(* a — Dh + cDt *)

(^) ß) ^~ I n t e g r a t e P r i n i i t i v e P o l y n o m i a l ( p — Dqo, D) return(g4-go,/ö)

Example

5.8.1. Consider

/

iog(x)y

log(x)/

where Li(x) = J d x / l o g ( x ) is the logarithmic integral. Let k = Q(x,to) with D = d/dx^ where to is a monomial over Q(x) satisfying DIQ = 1/x^ i.e. to =

5.8 The Primitive Case

159

log(a:), and let t be a monomial over k satisfying Dt = 1/to, i.e. t = Li(x). Our integrand is then

We get 1. a = lc(p) = t o + l/to 2. to + 7- ) - — = to = log(x) = —- (xlog(x) -x)

^

iQ J

ax

ZQ

= D{xto - x)

so (6, c) = LimitedIntegrate(to 4- 1/to, 1/to, D) = {xto — x, 1) 3. qo = ct'^/2 + bt = t^/2 + (xto - x)t 4. p — Dqo = —X E k so the call IntegratePFiinitivePoIynoniial(—x, D) returns (g,/3) = (0,1). Hence,

— —

^ (xlog(x) — x)Li(x) — / xdx

Lifx) -

-

^

X + (Xlog(x) - x)Li(x) - y

.

Putting all the pieces together, we get an algorithm for integrating elements of k{t). T h e o r e m 5.8.2. Let k be a differential field and t a primitive over k. If the problem, of limited integration w.r.t. Dt is decidable for elements ofk, and Dt is not the derivative of an element of k, then for any f G k{t) we can either prove that f has no elementary integral over k(t), or compute an elementary extension E of k{t) and g E. E such that f — Dg G k. Proof Suppose that Dt is not the derivative of an element of /c, then t is a monomial over k and Const{k{t)) = Const(/c) by Theorem 5.1.1. Let / € k{t). By Theorem 5.3.1, we can compute gi^h^r G k{t) such that / = Dgi-{-h~\-r^ h is simple and r is reduced. From /i, which is simple, we compute ^2 ^ k{t) given by (5.8) in Theorem 5.6.1. Note that go =^ gi -i- J g2 lies in some elementary extension of k{t). Let p = h — g2 and q = p + r, then / = Dgo -\- q so f has an elementary integral over k{t) if and only if q has one. If p ^ k[t]^ then p -\- r does not have an elementary integral over k{t) by Theorem 5.6.1, so / does not have an elementary integral over k{t). Suppose now that p G k[t]. We have k{t) = k[t] by (5.1), so r G k[t]^ hence q G k[t]. By Theorem 5.8.1 we can either prove that q has no elementary integral over A:(t), in which case / has no elementary integral over k{t)^ or compute s G k[t] such that q — Ds G fc, in which case / — Dg G k where g = go -{- s. D

160

5 Integration of Transcendental Functions

I n t e g r a t e P r i m i t i v e ( / , D)

(* Integration of primitive functions *)

(* Given a is a primitive monomial t over /c, and / G /c(t), return g elementary over k{t) and ß G {0,1} such that / — Dg ^ k if ß = 1, OT f — Dg does not have an elementary integral over /c(t) if /? = 0. *) {gi ,h,r) ^— H e r i i i i t e R e d u c e ( / , D) (^2,/5)<— ResidueRediice(/i, D) if /3 = 0 t h e n r e t u r n ( p i + p2,0) {Q^ ß) ^~ IntegratePriiiiitivePoIynomial(/i — Dg2 + r, D) retiirn(^i + p2 + q,ß)

5.9 The Hyperexponential Case In the case of hyperexponential monomials over a differential field fe, the related problem we need to solve over k is the Risch differential equation problem: given f^g £ k^ determine whether there exists y £ k such t h a t Dy + fy = g

(5.14)

and to compute y if it exists. It may happen in general t h a t (5.14) has more t h a n one solution in fe, so we first need to examine when this can happen. L e m m a 5 . 9 . 1 . Let (K^D) be a differential field. If there are a^y^z £ K that y j^ z and Dy + ay = Dz + az, then a = Du/u for some u E K*. Proof

Letu

= l/{y-z)

Du-

au =

so a = Du/u.

£K\

Dy-Dz {y-

—

z)^

such

Then, a y - z

=

{Dz + az) - {Dy + ay) -,

{y-

r::

z)^

= 0

D

We present in this section an algorithm t h a t , with appropriate assumptions on A;, integrates elements of k{t) when t is a hyperexponential monomial over k. We first describe an algorithm for integrating elements of k{t). T h e o r e m i 5 . 9 . 1 . Let k be a differential field and t an hyperexponential over k. If we can solve Risch differential equations over k, and Dt/t is not a logarithmic derivative of a k-radical, then for any p E kit) we can either prove that p has no elementary integral over k{t), or compute q G k{t) such that p — Dq E k. Proof Since Dt/t is not a logarithmic derivative of a Ä:-radical, t is a monomial over k, Const{k{t)) = Const(Ä:), and S''' = {t} by Theorem 5.1.2. T h u s k{t) = k[t^t~^] by (5.1), and Theorem 5.7.1 says t h a t if p has an elementary integral over fc(t), then there are v G k{t), c i , . . . , c^ G C, 6 i , . . . , 6^ G k{ci,..., c^), and m i , . . . , m^ G Z such t h a t

5.9 The Hyperexponential Case

.^

Dbit"^'

^

Dt^

P = Dv + J2''^'J^=I^^

^

Dbi

161

,_,,

+ T^'^'''' + ^ ' ' ' ^

i=l

i=l

(5.15)

i=l

where C = Const(A;). K = A:(ci,..., c^) is an algebraic extension of k^ so t is transcendental over K. Furthermore, Dt/t is not a logarithmic derivative of a i^-radical by Lemma 3.4.8, so t is a monomial over K and Const(i^(t)) = Const(ÜT). Since p^v £ k[t^ '^~^]) write p = Xli^m ^*^* ^^^ ^ ~ S i ^ r '^^^* where ai,'?;^ G A;, m , M , r , i ? G Z, m < M and r < ß . Let ]}i = Yfi=m^i^'' ^^ M = 0, then p - i^go = Pi where go = 0 G Ä:(t). If M > 0, then UOOIP) = -M < 0, which implies that Uoo{Dv) — —M < 0, so z^oo('^) = —M by Lemma 5.L2, hence i? = M. Equating the coefficients of t , . . . , t ^ in (5.15) we get ai = Dvi + i—Vi

forl
(5.16)

Since we can solve Risch differential equations over k and ai^Dt/t G k^ we can either prove that (5.16) has no solution Vi G k, or find such a solution*^ If it has no solution for some i, then (5.15) has no solution so p has no elementary integral over k{t). If we have solutions -u^ for 1 < 2 < M, letting QQ = Vit -\-... VMt^ •> we get M

O

M

.

Dt

\

.

^

.

p-Dqo = Y1^'^' '^Yl^^'^' ~Y1 {Dvi + i—Vijt' = Y2aif = pi. i=l

i=m

i=l

^

"^

i=m

If m = 0, then pi £ k so q = qo satisfies the theorem. If m < 0, then ^t{Pi) = —'^ < O7 which implies that Pt{Dv) = —m < 0, so Uti^) = —^ by Theorem 4.4.2 (since t G tS^^^), hence r = m. Equating the coefficients of t - \ . . . , t - ^ in (5.15) we get ai = Dvi + i—-Vi

for m < i < —1.

(5.17)

Since we can solve Risch differential equations over k and ai^Dt/t G A:, we can either prove that (5.17) has no solution Vi G k, or find such a solution. If it has no solution for some i, then (5.15) has no solution, so pi and p have no elementary integrals over k{t). If we have solutions Vi foi m < i < ~1^ letting qi = v_it~^ + ... v^rnt""^ and g = go + gi ^ k{t), we get ~^ ~^ ( Dt \ p - Dq = p^ - Dq^ = ^ a^f + ao - ^ f Dvi + i-—Vi j f = ao e k . Ü

Although this fact is not needed by the algorithm, we remark that Lemma 5.9.1 implies that (5.16) has at most one solution in k.

162

5 Integration of Transcendental Functions

I n t e g r a t e H y p e r e x p o n e n t i a l P o I y n o m i a l ( p , D) (* Integration of hyperexponential polynomials *) (* Given an hyperexponential monomial t over k and p G k[t^t~~^] return q G /c[t, t~^ and a Boolean ß G {0,1} such that p — Dq G fc if /3 = 1, or p — Dq does not have an elementary integral over k{t) if /3 = 0. *) q^O,ß^l for i ^- utip) t o —i^oo(p) such that i 7^ 0 d o a ^- coefficient (p,t*) (* RischDE will be given in Chap. 6 *) V i- HischDEiiDt/t, a) {^ a = Dv-\- ivDt/t if t» = "no solution" t h e n /3 <— 0 else q ^— q + vf return(g,/3)

Example

*)

5.9.1. Consider

/ ((tan(x)^

+ (x + 1) tan(x)2 + tan(x) + x + 2) e*^^^^^ + —

j dx .

Let Ä: = Q ( x , to) with D = d/dx^ where to is a monomial over Q(x) satisfying Dto = 1 + ^07 '^•^- "^0 = t a n ( x ) , and let t be a monomial over k satisfying D t = (1 + to)t, i.e.t=^ gtan(x)^ Q^j, integrand is then

p = (t^ + (x + l)tg + to + X 4- 2) t + We get 1. g z = 0 , /?=: 1 2. z/t(p) = -Uooip) = 1 3. i = 1 4. a = lc(p) - tg 4- (x + l)tg + to 4- X + 2 5. D{tQ + x) + (1 + tg)(to + x) = a, so 1; = R i s c h D E ( l 4- tg, a) ^ to + x 6. q = vt = (to 4- x)t 7.p-Dq = l / ( x 2 4- 1 ) . Hence, / ('(tan(x)^ + (x 4-1) tan(x)2 + t a n ( x ) 4- x 4- 2) e*^"^^) + ~ Y ^ - ) ^^ = (tan(x)+x)e*-(-)+

/ - ^ J x^ 4-1

= (tan(x) 4- x)e*^"^^^ 4- a r c t a n ( x ) . P u t t i n g all the pieces together, we get an algorithm for integrating elements of fc(t).

5.10 The Hypertangent Case

163

T h e o r e m 5.9.2. Let k be a differential field andt an hyperexponential over k. If we can solve Risch differential equations over k, and Dt/t is not a logarithmic derivative of a k-radical, then for any f G k{t) we can either prove that f has no elementary integral over k{t), or compute an elementary extension E of k(t) and g £ E such that f — Dg G k. Proof Suppose t h a t Dt/t is not a logarithmic derivative of a Ä;-radical, t h e n t is a monomial over k and Const{k{t)) = Const(fc) by Theorem 5.1.2. Let / G k{t). By Theorem 5.3.1, we can compute gi.h^r G k{t) such t h a t / = Dgi + / i + T', h is simple and r is reduced. From /i, which is simple, we compute 92 ^ k{t) given by (5.8) in Theorem 5.6.1. Note t h a t go — gi -^ j g2 lies in some elementary extension of k{t). Let p = h — g2 and q = p -\- r, t h e n / = Dgo + g so / has an elementary integral over k(t) if and only if q has one. If p ^ k[t], then p + r does not have an elementary integral over k{t) by Theorem 5.6.1, so / does not have an elementary integral over k{t). Suppose now t h a t p G k[t]. We have k{t) = k[t,t~^] by (5.1), so r G k[t^t~^]^ hence q G k[t,t~^]. By Theorem 5.9.1 we can either prove t h a t q has no elementary integral over k{t)^ in which case / has no elementary integral over fc(t), or compute s G k[tjt~^] such t h a t q — Ds G k, in which case / — Dg G k where g = go + s. D

I n t e g r a t e H y p e r e x p o n e n t i a l ( / , D) (* Integration of hyperexponential functions *) (* Given an hyperexponential monomial t over k and / G /c(t), return g elementary over k{t) and a Boolean ß G {0,1} such that / -- Dg G /c if ß = 1, OT f — Dg does not have an elementary integral over k{t) if /3 — 0. *) (gi ,h,r) <— H e r m i t e R e d i i c e ( / , D) (g2,ß)^~ ResidueReduce(/i,D) if /3 = 0 t h e n return(ö'i + ^'2,0) (g, ß) ^- I n t e g r a t e H y p e r e x p o n e n t i a l P o l y n o i n i a l ( / i — Dg2 + r, D) r e t u r a ( p i + p2 + g, ß)

5.10 T h e Hypertangent Case Tangents and trigonometric functions can be integrated by transforming t h e m to complex logarithms and exponentials, b u t the theory of monomial extensions allows us to integrate t h e m directly without introducing the algebraic number \f-l. We start by defining tangent monomials and computing the special polynomials. Let A; be a differential field and K a differential extension ofk.

164

5 Integration of Transcendental Functions

Definition 5.10.1. Lett e K be such thatt'^-\-l ^Q.t is a hyp er tangent over k if Dt/{t'^ + 1) e k. t is a tangent over k if Dt/(i? -{-I) — Dh for some b e k. t is a hypertangent (resp. tangent) monomial over k if t is a hypertangent (resp. tangent) over k, transcendental over k, and Const(fc(t)) = Const(/c). We write t = tan( J a) when t is a hypertangent over k such that Dt/{t'^ +1) = a, and t = tan(6) when t is a tangent over k such that Dt/itP' + 1) = Db. L e m m a 5.10.1. Let (F^D) be a differential field containing \f--\, a G F be such that a^ + 1 7^ Oj and b = (\/—T — a ) / ( \ / ^ + a). Then, 6 ^ 0 and b - ^

a2 + r

Proof, b y^ 0 since a^ + 1 ^ 0 , and we have

Db h

f^f^~a\ ^f^ K^fÄ^aJ /=4

a

( v ^ + a)2/:iT-a

l + a= D

T h e o r e m 5.10.1. If t is an hypertangent over k and \/'-iDt/(t?' + 1) is not a logarithmic derivative of a k{\/^^)-radical, then t is a monomial over k, Const{k{t)) = Const(Ä;); and any p G S^^^ divides t^ + 1 in k[t]. Furthermore, ^irr _ ^irr^ Convcrscly, if t is transcendental and hypertangent over k, and CoTist[k{t)) = Const(A;)^ then \/^^Dt/(t? + 1 ) is not a logarithmic derivative of a k{\/'^)-radical. Proof. Let t be an hypertangent over k^ a = Dt/{t'^ + 1), and suppose that a \ / ^ is not a logarithmic derivative of a fc(A/^)-radical. Let 0 = /^^t G k{\/'^){t). By Lemma 5.10.1, we have

so 9 is hyperexponential over k{^/'^). Since a \ / ^ is not a logarithmic derivative of a Ä:(\/^)-radical, 2 a \ / ^ is not one either, so by Theorem 5.1.2, 6 is a monomial over k{^/^)^ and Const{k{\/'^){9)) = Const(fc(\/^)). But t = A / ^ ( Ö — l)/(6^ +1), so t is transcendental over Ä:(\/^), hence a monomial over k since Dt = a + at'^. Furthermore, k{^/^){9) = k{^/^){t), so Const(A:(V-l)(t)) = Coiist{k{V-l){0))

= Const(fc(V-l)) =Cn

k{V-l)

by Corollary 3.3.1 where C is the algebraic closure of Const(fc). This implies that Const(Ä;(t)) C C fl k{\/^) H k{t) C k since t is transcendental over k. Hence, Const (fc(t)) C Const (A:). The reverse inclusion is given by Lemma 3.3.1,

5.10 The Hypertangent Case

165

so Const{k{t)) = Const(A:), which implies that Const(/c(t)) = Const(fc) by Lemma 3.3.3. We have D{t^ + 1) = 2tDt = 2at{t^ + 1) so t^ + 1 G 5, hence any factor of t^ + 1 is special by Theorem 3.4.1. Suppose now that p E S^ and let ß G k be any root of p. Dß = aß"^ + a by Theorem 3.4.3, so

D ( ± 1 ^ ^ = J^' ~ ^'^^^^ + 1) - (t - ß){tß' +1 + /3t2 + /3) ßt^lj

(/3t + l)2

,^ """^

^^{ßt^+t + ßH + ß)-{tß^+t ^^

+ ßt^ + ß) _^

(/3t + l)2

which implies that c = {t — ß)/{ßt + 1) G Const (A: (t)) C k. Since t is transcendental over k, {cß — l)t + (c + /3) = 0 implies that cß — 1 = c + ß = 0^ so /?2 + 1 = 0 . Since this holds for every root of p, this implies that every irreducible factor of p divides t^ + 1 in k[t]. We have Sf^ C S^^^ by definition. Conversely, let p e S^^^. Then p divides t^ + 1, so all the roots of p in k satisfy ß'^ = —1. Hence,

Dt-Dß Pß = -j^^

=

t^ + l a^--j=a{t-^ß)

which implies that Pß{ß) = 2aß = ±2^/^a^ which is not a logarithmic derivative of a fc(\/^)-radical, hence not a logarithmic derivative of a Ä;(/3)-radical. Thus, p e <Sf^ which imphes that S'f' = 5^^^ Conversely, let t be a transcendental hypertangent over k and suppose that Const(Ä:(t)) = Const(A:). Then, Const(I(t)) = Const(I) by Lemma 3.3.3. If there exist b G k{^/^y and an integer n > 0 such that r— Dt

Db

then, taking 4-

T+ t

nn

and

c= jj 62

ek{V^){t)

we get Dc

DO

Db

^

r— Dt

^Db

^

SO c E Const(/c(t)) C A: in contradiction with t transcendental over k. Hence, \/'^Dt/(iP' + 1) is not a logarithmic derivative of a ^(y^^^)-radical. D As a consequence, we have k{t) = {f e k{t) such that (t^ -f-1)""/ G k[t] for some integer n > 0} when t is a hypertangent monomial over k. We now present an algorithm that, with appropriate assumptions on k, integrates elements of k{t) when t

166

5 Integration of Transcendental Functions

is a hypertangent monomial over k. Note first t h a t if the polynomial X^ + 1 factors over k^ then \ / ^ G k, so Ä:(t) = k{$) where 0 = (\/—T — t ) / ( v ^ ^ -ft) is a hyperexponential monomial over k. Hence we can use the algorithm for integrating elements of hyperexponential extensions in this case, so we can assume for t h e rest of this section t h a t X ^ + 1 is irreducible over k, in other words t h a t \ / ^ ^ k. Since hypertangents are nonlinear monomials, integrating elements of k[t] is straightforward. T h e o r e m 5 . 1 0 . 2 . Let k be a differential field not containing \/--l, and t an hypertangent over k. If ^/^Dt/{t'^ -j- 1) is not a logarithmic derivative of a k{\f^)-radical, then for any p G k[t] we can compute q G k[t] and c G k such that

P-Dq-c Furthermore,

^^^^

efc.

if Dc ^ 0^ then p has no elementary

integral over k.

Proof. Let a = Dt/{t'^ + 1) G k. Since a \ / ^ is not a logarithmic derivative of a A:(A/^^)-radical, t is a monomial over k^ Const{k{t)) = Const(Ä:), and all the special irreducible polynomials divide t^ + 1 in k[t] by Theorem 5.10.1. Since A / ^ ^ k,t'^ + 1 is irreducible over k, so S'^^^ = {t^ + 1}. Since ö{t) = 2, Theorem 5.4.1 shows how t o compute g , r G k[t] such t h a t p — Dq = r and deg(r) < 1. Write r = at -{-b where a^b £ fc, and let c = a/(2a) G k. Since /i = t^ + 1 G 5 , Theorem 5.4.2 says t h a t deg(r — cDh/h) < 1, hence t h a t

p-Dq-c-j^-^ek. Suppose now t h a t Dc / 0, and t h a t r has an elementary integral over k{t). Then, by Theorem 5.7.1, there are v G k{t)^ c i , . . . , C n G C, 6 i , . . . , 5 n G ^ ( c i , . . . , Cn), and m i , . . . , m^ G Z such t h a t ^ D6 ft^ -f 1)"^'^ ^ at + b = Dv + J2^, 'I, ' =Dv + 2ta^m,c,

"^ Db + Y,c,—^.

(5.18)

If z/oo('^) < 0, then VOQ{DV) = z/oo(t;) — 1 < —1 by Theorem 4.4.4, in contradiction with (5.18), hence z/oo('^) ^ 0^ which implies t h a t UQO{DV) > 0 by Theorem 4.4.4. Let c = a/{2a) G k. Equating t h e coefficients of t in (5.18), we get a = 2a YM=I '^i^ii

so n

c= -— = 7 rriiCi G Const(Ä:) ^-^ 2a i=l

in contradiction with Dc 7^ 0. Hence (5.18) has no solution if Dc ^ 0, which implies t h a t r , and hence p, have no elementary integral over kit). D

5.10 The Hypertangent Case

167

I n t e g r a t e H y p e r t a n g e n t P o l y n o i i i i a I ( p , D) (* Integration of hypertangent polynomials *) (* Given a differential field k such that •\/—1 ^ /c, a hypert angent monomial t over k and p G k[t], return q G k[t] and c G k such that p-Dq~ cD{t^ + l)/{t^ + 1) G /c and p -- Dg does not have
(* deg(r) < 1 *)

Example 5.10.1. Consider /(tan(.)^+.tan(.) + l)d. Let k = Q{x) with D — d/dx^ and and let t be a monomial over k satisfying Dt = 1 -{-t"^, i.e. t = tan(x). Our integrand is then p = t'^ ^xt^l

£ k[t].

We get 1. (g,r) = PolynoniialReduce(t^ + j^t + 1) = {t,xt) 2.a = Dt/{t^ + 1) = 1 3. c = x/2. Since Dc = 1/2 7^ 0, we conclude that / ( t a n ( . f + . t a n ( x ) + l)gii92 ^ ^7 determine whether there are yi,t/2 G /c such that

[DyA.ffi \Dy2)^\h

-f2\(yA^(gi h ) W) {92

and to compute yi and ^2 if they exist. T h e o r e m 5.10.3. Let k be a differential field not containing AZ—T; and t an hypertangent over k. If we can solve coupled differential systems over k, and \/~^Dt/{t'^ -f 1) is not a logarithmic derivative of a k{\/^-i)-radical, then for any p G k{t) we can either prove that p has no elementary integral over k{t), or compute q £ k{t) such that p — Dq G k[t].

168

5 Integration of Transcendental Functions

Proof. Let a = Dt/{t^ + 1) E k. Since ayf~l is not a logarithmic derivative of a ^ ( \ / ^ ) - r a d i c a l , t is a monomial over fc, Const(Ä:(t)) = C o n s t ( ^ ) , and all the special irreducible polynomials divide t^ + 1 in k\t] by Theorem 5.10.1. Since ^/^ ^ A:, t^ + 1 is irreducible over k, so S'^"" = S'{' = {t^ + 1}. Thus, Theorem 5.7.1 says t h a t if p has an elementary integral over /c(t), then there are v G k{t)^ c i , . . . ,Cn € C, 5 i , . . . , 6^ ^ ^(ci, • •. , c ^ ) , and ? n i , . . . , m^ G Z such t h a t p = Dv+

y ^Cj

biit^ + iy

= Dv + 2taY^miCi

+ Y^Ci—-^

= Dv + w

(5.19)

where C = Const(/c), and w = 2ta^^miCi

+ X ^ Q - r - ^ ^ Ä:(CI,.. .,Cn)[t] .

i=i

i=i

*

i f = k{ci^..., Cn) is an algebraic extension of fc, so t is transcendental over ÜT. Furthermore, a^/^ is not a logarithmic derivative of a K ( \ / ^ ) - r a d i c a l by Lemma 3.4.8, so t is a monomial over K and Const(i<^(t)) = C o n s t ( i f ) . We proceed by induction on — z^t2+i(p). K i^t^-^i{p) > 0, then p — Dq G k[t] where q = 0 £ k{t)^ so suppose t h a t m = — z/^2_|_i(p) > 0 and t h a t t h e theorem holds for all h G k{t) with —i^t^-^i{h) < m. Since p G Ä;(t) and m — —i^t^-^iip) ^ ^5 we h a v e p = r / ( t ^ + l)"^ where r G fc[t] and gcd(r, t^ + 1) = 1. Since i't'^-^i{p) = —m < 0, (5.19) implies t h a t i't^_^i{Dv) — —m < 0, hence t h a t 2^*2^1(1;) = —m by Theorem 4.4.2, since t^ + 1 G Si. Thus, v = s/{t'^ +1)"^ where s G k[t] and gcd(5,t^ + 1) = 1. Dividing r and 5 by t^ + 1, we get r = ro(t^ 4-1) + at + 6 and s = 5o(t^ + 1) + ct + d, where ro, so G fc[t], a^h^c^d e fc, at + 6 7^^ 0, and ct + d y^O. From (5.19), we get at + b

ro

(t2 + i ) ^

(t2 4-1)^-1

f

ct-j-d

SQ

yit^ + i)"^

(t2 + 1 ) ^ - 1

tDc 4- ca(t2 + 1) + D d (t2 + 1 ) ^

tDc 4- I)
= , ^ (t2 + l ) ^

tDc^Dd

~ — — (t2 + l ) ^

.

2mat{t^ + l)(ct + d) + DlOo + W (t2 + 1) m+1

ct2 + dt

1

2ma—^r — h ca-r^^ -r T + i^'^o + ^ (t2 + l ) ^ (t2 + 1 ) ^ - 1

^

dt-c

1 - 2m

2ma—7: 7-— + ca—7:—— r + DWQ + it; (t2 + l ) ^ (t2 + 1 ) ^ - 1

where IOQ = so/(t^ + 1)'^""'". Since 1/^2+1 ('w^o) > —'^, ^t'^-\-i{Dwo) > —m by Theorem 4.4.2, so, equating t h e coefficients of (t2 + 1)~"^ we get at -[-h = {Dc — 2mad)t

+ Dd +

2mac

5.10 The Hypertangent Case

169

which implies t h a t

f Dc\ \Dd)

f 0 —2ma\ (c\ ' \2ma 0 ) \d)

(a \h)

'

^^'^^^

Since we can solve coupled differential systems over k and a^h^a £ k, we can either prove t h a t (5.20) has no solution c^d E k, or find such a solution. If it has no solution in /c, then (5.19) has no solution, so p has no elementary integral over k{t). If we have a solution c^d E k, letting go = {ct + (i)/(t^ + 1 ) ^ G k{t), we get u for some u E k[t], so z/t2_j_i(p — Dqo) > —m. By induction we can either prove t h a t p — Dqo has no elementary integral over k{t)^ in which case p has no elementary integral over k{t)^ or we get qi E k{t) such t h a t p—Dqo—Dqi E k[t]^ which implies t h a t p — Dq E k[t] where g = go + ^iD

I n t e g r a t e H y p e r t a n g e i i t R e d u c e d ( p , D) (* Integration of hypertangent reduced elements *) (* Given a differential field k such that ^/^^ ^ /c, a hypertangent monomial t over k and p E /c(t), return q E kit) and a Boolean ß E {0,1} such that p— Dq E k[t] if ß = 1, or p — Dq does not have an elementary integral over k{t) if /3 = 0. *) m <

i^t2+i(p)

if m < 0 t h e n returii(0,1) h^{t^ + 1)> (* /i G k[t] *) (g, r) ^ PoIyDivide(/i, t^ + 1) (* /i = (t^ + l)g + r, deg(r) < 1 *) a ^- coefficient(r, t), b ^- r — at (* r = at + & *) (* CoupledDESystem will be given in Chap. 8 *) (c, d) ^ C o u p l e d D E S y s t e m ( 0 , 2mDt/{t^ + 1), a, b) (* Dc - 2mDt/{t^ + l)d = a,Dd + 2mDt/(f + l)c = 6 *) if (c, d) = "no solution" t h e n return(0,0) go <- (ct + 6^)7(^2 + 1)^ (g, /3) ^- I n t e g r a t e H y p e r t a n g e n t R e d u c e d ( p — Dgo, i^) r e t u r n ( g + go,/3)

Example 5.10.2. Consider sin(x) ^ dx. Let Ä: = Q{x) with I^ = (i/dx, and and let t be a monomial over k satisfying D t = (1 + 1 ^ ) / 2 , i.e. t = t a n ( x / 2 ) . Using the classical half-angle formula, our integrand is then

170

5 Integration of Transcendental Functions siii(a:) 2tan(x/2) 2t/x ^ ^ ~ V ^ x(tan(x/2)2 + 1) ^ IßTl ^ ^^^'

We get Dt/{t^ + 1) = 1/2 and l.m = -iyt2+i{p) = 1 2. h=p{t^ ^l) = 2t/x 3. {q,r) = PolyDivide(2t/x, t^ + 1) = (0,2t/x), so (a, 6) = (2/x,0) 4. Since I}c\ /O - 1 \ / c \ /2/:r

0 J \d)

DdJ ~^\1

\ 0

has no solution in Q(x), CoupledDESystem(0,1,2/x,0) returns "no solution". Hence,

is not an elementary function. Example 5.10.3. Consider tan(a:)^ + tan(x)^ + x'^ tan(x) + 1 , ^^

^

^

\_i

^^

drj^

(tan(x)2 + 1)^ Let k = Q(a:) with D = d/dx, and and let t be a monomial over k satisfying Dt = 1 -\-t'^^ i.e. t = tan(x). Our integrand is then t^^t^+xH p=

+1 ^

. . . G kit) .

We get Dt/{t^ + 1) = 1 and l.m = -Ut2^i{p) = 3 2. h= p{t^ + 1)^ = t^ -^t^ -\-xH^l 3. (g,r) = PolyDivide(/i,t^ + 1) = {t^^xH + 1), so (a,5) = (x^, 1) 4. Since 'I)c\ /O - 6 \ / c \ /x^^ has the solution c = x/18 + 1/6 and tf = 1/108 - x^/G in Q(x), (c,d) = CoupledDESystem(0,6,^2,1) = (x/18 + 1/6,1/108 - xV6). 5. ct^d _ (l + x / 3 ) t - ( x 2 - l / 1 8 ) ^° "" (t2 + 1)^ 6 (^2 + i f P - Dqo

t^ + 5x/18 + 1 5 / 1 8 (t2 + 1)^

5.10 The Hypertangent Case 6. Recursively calling (g,/3) = Dgo), we get ß = 1 and

171

IntegFateHypertangentReduced(p —

_ 5 (IH- x/3) t + 77/12 5(14- x/3) t - 43/6 ^~ 24(t2 + l f ^ 16(t2 + l) 7. 5

Hence, tan(x)^ + tan(a:)^ + x^ tan(x) + 1 . __ (tan(:r)2 + 1)^ (1 + x/3) tan(x) - {x^ ~ 1/18) 5(1 + x/3) tan(x) + 77/12 6 (tan(x)2 4-1)^ 24 (tan(x)2 + 1)^ 5(l + a;/3)tan(:z:)-43/6 5 // XN 16(tan(x)2 + l) 16

/(i+i)..

and the remaining integral is of course x + x^/ö. Putting all the pieces together, we get an algorithm for integrating elements of k{t). T h e o r e m 5.10.4. Let k be a differential field not containing \/^, and t an hypertangent over k. If we can solve coupled differential systems over k, and \r~^Dtl(t^ + 1) 25 not a logarithmic derivative of a k{\/'^)-radical, then for any f G k{t) we can either prove that f has no elementary integral over k{t), or compute an elementary extension E ofk{t) and g G E such that f — Dg G k. Proof. Suppose that y/^Dt/{t'^ + 1) is not a logarithmic derivative of a Ä;(\/^)-radical, then t is a monomial over k and Const(Ä:(t)) = Const(fc) by Theorem 5.10.1. Let / G k{t). By Theorem 5.3.1, we can compute gi,h,r G k{t) such that / = Dgi + h-{-r^ h is simple and r is reduced. From /i, which is simple, we compute g2 G k{t) given by (5.8) in Theorem 5.6.1. Note that go = 9i ^ J 92 lies in some elementary extension of k{t). Let p ^ h — g2 and q = p + r, then / = Dgo -\- q so f has an elementary integral over k{t) if and. only if q has one. If p ^ k[t]^ then p + r does not have an elementary integral over k{t) by Theorem 5.6.1, so / does not have an elementary integral over k{t). Suppose now that p G k[t]. Then p e k{t) so q £ k{t). By Theorem 5.10.3 we can either prove that q has no elementary integral over A;(t), in which case / has no elementary integral over k{t)^ or compute s G k{t) such that u = q — Ds G k[t]^ in which case by Theorem 5.10.2, we compute v G k[t] and c G Const(Ä:) such that u - Dv - cD{t^ + l)/(t^ + 1) G A;. If Dc ^ 0, then ?i, and hence / , have no elementary integral over A:(t), otherwise Dc = 0 so / — Dg G k where g = goi-s + v-{-cJ D{t^ + l)/(^^ + 1) lies in some elementary extension of k[t). D

172

5 Integration of Transcendental Functions

I n t e g r a t e H y p e r t a n g e n t ( / , 19)

(* Integration of hypertangent functions

(* Given a differential field k such that \ / ^ ^ /c, a hypertangent monomial t over k and / G /c(t), return g elementary over k{t) and a Boolean ß G {0,1} such that / — Dg e k if ß = 1, oi f — Dg does not have an elementary integral over k(t) if /? = 0. *) {gi ,h,r) i— H e m i i t e R e d u c e ( / , D) {92, ß) ^- ResidueReduce(/i, D) if /3 = 0 t h e n returii(5fi 4- p2,0) p ^- h — Dg2 + r (gi, ß) ^— IntegrateHypertangentReduced(_p, D) if /3 = 0 t h e n r e t u r n ( p i + ^2 + 9i, 0) (92, c) ^- I n t e g r a t e H y p e r t a n g e n t P o l y n o m i a I ( p — Dqi, D) if Dc = 0 t h e n r e t u r n ( p i + ^ 2 + 91+^2 + clog(^^ + 1), 1) else r e t u r n ( p i + P2 + gi + g2,0)

5.11 The Nonlinear Case with no Specials In the case of nonlinear monomials over a differential field k^ we have seen t h a t we can reduce the problem to integrating reduced elements of the form p + a/d where p^a e k[t]^ d £ S \ {0}, deg(p) < 6{t) and deg(a) < deg{d). Furthermore, Theorem 5.7.2 provides a criterion for nonintegrability, and if an element of tS\ A; is known, allows us to reduce the problem to deg(p) < ö{t) — 1. We address in this section the case S = kj i.e. S^^^ = 0, which corresponds to interesting classes of functions as will be illustrated in the examples. Note t h a t if S^^^ = 0, then k{t) = k[t]^ so as a result of the polynomial reduction (Sect. 5.4), we consider integrands of the form p £ k[t] with deg(p) < S{t). It turns out t h a t if such elements are integrable, then they must be in k. C o r o l l a r y 5 . 1 1 . 1 . Suppose that t is a nonlinear monomial and that S^^^ = 0. Let p G k[t] be such that deg(p) < S{t). If p has an elementary integral over kit), then p £ k. Proof. Let C = Const(A:(t)), p £ k[t] be such t h a t deg(p) < J(t), and suppose t h a t p has an elementary integral over k{t). By Theorem 5.7.1 there are v £ k[t], C i , . . . , C n £ C and ui,...,Un £ <S/,(ci,...,c^)[t]:/c(ci,...,c,.) such t h a t p = Dv + g where g = X^ILi ^^^('^O/'^i- Note t h a t g = p — Dv £ k[t]. Since ^t[t]:k = ^' it follows t h a t 4Tci,...,c„.)[t]:fc(ci,...,c„.) = ^ (Excrcise 3.5), hence t h a t 5/c(ci,...,c,,,)[t]:/c(ci,...,c„,) = Kci,..., Cn). This implies t h a t g £ fc(ci,..., c^). Since g £ k\t]^ we get t h a t g £ k. Suppose t h a t deg(t') > 1, then, deg(p) = deg{Dv + g) = deg{Dv)

= deg('?;) + S{t) - 1 > ö{t)

in contradiction with deg(p) < 6(t). Hence, v £ k^ so p = Dv -{- g £ k. This provides a complete algorithm for integrating elements of k{t).

D

5.11 The Nonlinear Case with no Specials

173

monomial T h e o r e m 5 . 1 1 . 1 . Let k be a differential field and t be a nonlinear over k be such that S^^^ = 0. Then, for any f £ k{t) we can either prove that f has no elementary integral over k{t), or compute an elementary extension E of kit) and g G E such that f — Dg £ k. Proof. Suppose t h a t t is Si nonUnear monomial over k and t h a t S^^^ = 0. Then, Const(^(t)) = Const(Ä:) by Lemma 3.4.5. Let / G k{t). By Theorem 5.3.1, we can compute gi^h^r G k{t) such t h a t / = Dgi + /i + r, h is simple and r is reduced. From /i, which is simple, we compute p2 ^ k{t) given by (5.8) in Theorem 5.6.1. Note t h a t go = gi ^ J g2 lies in some elementary extension of k{t). Let p = h ~ g2 and q = p -^ T^ then / = Dgo + g so / has an elementary integral over k{t) if and only if q has one. If p ^ A:[t], then p + r does not have an elementary integral over k{t) by Theorem 5.6.1, so / does not have an elementary integral over k{t). Suppose now t h a t p G k[t]. We have k{t) = k[t] by (5.1), so r £ k[t]^ hence q £ k[t]. By Theorem 5.4.1 we compute gi, ^2 ^ k[t] such t h a t q = Dqi +^2 ^^nd deg(g2) < ^{t)- We now have f — Dg = q2 where g = go-\-qi. If ^2 ^ k^ then the theorem is proven, otherwise 0 < deg(g2) < ö{t)^ so ^2, and therefore / , have no elementary integral over k D by Corollary 5.11.1.

I n t e g r a t e N o n L m e a r N o S p e c i a l ( / , D) (* Integration of nonlinear monomials with no specials *) (* Given a is a nonlinear monomial t over k with S^^^ = 0, and / G k{t), return g elementary over k{t) and a Boolean ß £ {0,1} such that f — Dg £ k if ß = 1, or f — Dg does not have an elementary integral over k{t) if /3 = 0. *) (gi ,h,r) ^- H e r m i t e R e d u c e ( / , D) (g2, ß) ^- ResidueReduce(/i, D) if /? == 0 t h e n r e t u r n ( p i +^2,0) {qi J ^2) ^— PolynoniiaIReduce(/i — Dg2 + T, D) if q2 £k t h e e ß ^1 else ß ^0 r e t u r n ( p i -\- g2 + qi,ß)

Example 5.11.1. Let z/ G Z be any integer and consider

where Jy{x) is the Bessel function of the first kind of order v. From - ^ — - ~ - J , + i ( x ) + -J,(x)

174

5 Integration of Transcendental Functions

we get

where (/)i,{x) is the logarithmic derivative of Jy{x). Since Jy{x) is a solution of the Bessel equation y"{x) + -y'(x) + f 1 - ^ ) y{x) = 0 X^ J X \

(5.21)

it follows that (j)v{x) is a solution of the Riccati equation y\x) + y{xf + -y{x) + ("l - ^ " j = 0 . X \ X^ J

(5.22)

Let k = Q(:r) with D — d/dx, and let t be a monomial over k satisfying Dt = —t'^—t/x — (1 -z/^/x^), i.e. t = (t)y{x). It can be proven that 5^^^ = 0 in this extension^ so Corollary 5.11.1 implies that t has no elementary integral over A:, hence that / J

""^. • dx = z/log(x) - / (j)y{x)dx Ju{x)

J

where the remaining integral is not elementary over Q(x, (^^^(x)). Example 5.11.2. Let z/ € C be any complex number and consider x'^cj^l + x(^t - i^m - x{x^ + l)cßl - {x^ - z/2)^^ - xV4 dx x2(/)4 + x2(x2 + 2)(^2 _^ ^2 _^ ^4 ^ ^ 6 / 4

where (j)y{x) is the logarithmic derivative of Jjy{x)^ the Bessel function of the first kind of order u. Let k = Q{x) with D = d/dx, and let t be a monomial over k satisfying Dt = —t^ — t/x — (1 — v^jx^)^ i.e. t =• (^y(x). Our integrand is then _ xH^ + Xt^ - vH^ - X{X? + l)t2 - {x^ - Z/2)t - x^/4 •^ ~ ™

xH^ + X2(x2 + 2)^2 -f x2 + X4 + x 6 / 4

""""

and we get 1. Calling (^i,/i,r) = H e r m i t e R e d u c e ( / , D ) we get 1+^2/4 (z/2 + x4/2)t + X^ + X __ 1 ^ i - - t 2 + l+x2/2' ^ - ~ ^2^24.^2 + ^4/2 ' a n d r - t + - . ^ The fact that (5.21) has no solutions in quadratures for z/ G Z (its Galois group is SL2{C)) implies that (5.22) has no algebraic function solution, hence no solution in k. Theorem 3.4.3 then implies that ^S^^"" = 0.

5.12 In-Field Integration

175

2. Calling {g2^ß) = ResidueReduce(/i, D) we get ß = 1 and 1^ 3. We have h-Dg2+r

f.2

= 0,so (gi,^2) = (0,0).

Hence / = Dgi + Dg2^ which means that x'^(i)l + x(t)l - u^cßl - x{x^ + l)(j)l - {x^ - z/2)(/)^ - x^/4 x2(/>4 + x 2 ( x 2 + 2)(l)l + a;2 + x4 +

l+xV4 (/),(x)2 + 1 + ^ 7 2

X^/4

dx

1 log((;^,(x)2 + l + : r V 2 ) . 2

Note that the above integral is valid regardless of whether S^^^ is empty. The above examples used Bessel functions, but in fact the algorithm of this section can be applied whenever the integrand can be expressed in terms of the logarithmic derivative of a function defined by a second-order linear ordinary differential equation. If the defining equation is known not to have solutions in quadratures (for example for Airy functions), then *S^^^ = 0, as explained in note 4 of this chapter.

5.12 In-Field Integration We outline in this section minor variants of the integration algorithm that are used for deciding whether an element of k{t) is either a ® derivative of an element of k{t), ® logarithmic derivative of an element of fc(t), ® logarithmic derivative of a ^(t)-radicaL As we have seen in Sect. 5.2, such procedures are needed when building the tower of fields containing the integrand. Furthermore, they will be needed at various points by the algorithms of the remaining chapters, in particular when bounding orders and degrees. Note that the structure Theorems of Chap. 9 provide efficient alternatives to the use of modified integration algorithms, and in some cases the only complete algorithms for recognizing logarithmic derivatives. Recognizing Derivatives The first problem is, given / G fc(t), to determine whether there exists u £ k{t) such that Du = / , and to compute such an u if it exists. We first perform the Hermite reduction on / , obtaining g G fc(t), a simple h G fc(t), and r G k{t) such that / = Dg -\-h-\-r. At that point, we can prove (see Exercise 4.1) that

176

5 Integration of Transcendental Functions

if / = Du for some u G A;(t), then h G k[t]^ so we are left with integrating h-\-r which is reduced. The algorithms of Sects. 5.7 to 5.11 can then be applied (with a minor modification in the nonlinear case, to prevent introducing a new logarithm), either proving that there is no such w, or reducing the problem to deciding whether an element a £ k has an integral in k{t). If t is a primitive over k^ then it follows from Theorem 4.4.2 and Lemma 5.1.2 that if a has an integral in k{t)^ then a = Dv + cDt where V G k and c G Const(/c), and we are reduced to a limited integration problem in k. Otherwise, 6{t) > 1, and it follows from Theorem 4.4.2 and Lemmas 3.4.2 and 5.1.2 that if a has an integral in A:(t), then a = Dv where v £ k^ and we are reduced to a similar problem in k. When / = Da/a for some a £ Ä:(t)*, then Corollary 9.3.1, 9.3.2 or 9.4.1 provide alternative algorithms: / = Du for u £ k{t) if and only if the linear equation (9.8), (9.12) or (9.21) has a solution in Q. Corollary 9.3.2 also provides an alternative algorithm if / = DhjilP' -f 1) for some h £ k{t)^ i.e. f = arctan(6). It is obvious that the solution u is not unique, but that if / = Du = Dv for u,v £ k{t)^ then u ~v £ Const{k{t)). Recognizing Logarithmic Derivatives The second problem is, given / £ k{t)^ to determine whether there exists a nonzero u £ k{t) such that Du/u = / , and to compute such an u if it exists. We can prove (see Exercise 4.2) that if / = Du/u for some nonzero u £ k{t), then / is simple and that all the roots of the Rothstein-Trager resultant are integers. In that case, the residue reduction produces DQO, _ ^{llrsia)=09a)

__ Dv

where v £ k{t) since the a's are all integers. Furthermore, Theorem 5.6.1 implies that if / = Du/u for u £ k{t), then f — g £ k[t]^ so we are left with deciding whether an element p of k\t] is the logarithmic derivative of an element of k{t). li p = Du/u for u £ k{t)^ then it follows from Exercise 4.2 that deg(p) < max(l, S{t)) and from Corollary 4.4.2 that u = p^^ .. .p^"^ where Pi £ S and e^ £ Z. If t is a primitive over fc, then both p and u must be in k since S = k^ so we are reduced to a similar problem in k. If t is an hyperexponential over fc, then p £ k and u = vt^ foi v £ k* and e G Z, since S^^^ ~ {t}. We are thus reduced to deciding whether p £ k can be written as Dv Dt p= + e-V t foTv £ k* and e G Z. This is a special case of the parametric logarithmic derivative problem, a variant of the Umited integration problem, which is discussed in Chap. 7.

5.12 In-Field Integration 177 If t is a hypertangent over k and \/^^ ^fc,then p = a + bt foi a^b G /c, and u = v{t^ + 1)^forv e k"" and e G Z, since S^^^ = {t^ + 1}. We are thus reduced to deciding whether a -\-bt can be written as a^U=

Dv D(t^ + 1) Dv ^ Dt — +e ; / = — + 2e-^—-t V t^ + 1 V t^ + 1

which is equivalent to ^

and

'u

^ ^ . Z . 2 Dt

The second condition can be immediately verified, while the first is the problem of deciding whether an element of k is the logarithmic derivative of an element of k. When f = Db for some b £ k{t), then Corollary 9.3.1, 9.3.2 or 9.4.1 provide alternative algorithms: / is the logarithmic derivative of a ^(t)-radical if and only if the linear equation (9.9), (9.13) or (9.22) has a solution in Q. The solution u is not unique, but if / = Du/u = Dv/v for u^v G k{t)\{0}, then u/v G Const(fc(t)) (this is the case n = ?n = 1 of Lemma 5.12.1 below). Recognizing Logarithmic Derivatives of

fc(t)-radicals

The third problem is, given / G fc(t), to determine whether there exist a nonzero n G Z and a nonzero u £ k{t) such that Du/u = n / , and to compute such an n and u if they exist. We can prove (see Exercise 4.2) that if nf = Du/u for some nonzero n G Z and u G fc(t), then / is simple and that all the roots of the Rothstein-Trager resultant are rational numbers. In that case, let m be a common denominator for the roots of the Rothstein-Trager resultant. Then, the residue reduction produces V—

g=

y

DQO, 1 ^ ( n r . ( a ) = o C )

a

=

^-

7^^

IDv

=

where v G k{t) since the ma is an integer for each a. Furthermore, Theorem 5.6.1 implies that if / = Du/{nu) for n G Z and u G k{t)^ then / — Dg G fc[t], so we are left with deciding whether an element p of k\t] is the logarithmic derivative of a /c(t)-radical, li p = Du/{nu) for n G Z and u G k{t)^ then it follows from Exercise 4.2 that deg(p) < max(l,(5(t)) and from Corollary 4.4.2 that u = p\^ .. .p^' where pi E S and e^ G Z. If t is a primitive over k^ then both p and u must be in k since S = k, so we are reduced to a similar problem in k. If t is an hyperexponential overfc,then p G k and u = vt^ for v G k* and e G Z, since S^^^ = {t}. We are thus reduced to deciding whether p G k can be written as

178

5 Integration of Transcendental Functions

p=

1 Dv

e Dt +- —

n V n t for V G k* and n, e G Z. This is the parametric logarithmic derivative problem, a variant of the limited integration problem, which is discussed in Chap. 7. If t is a hypertangent over k and ^/^ ^ k, then p = a + 6t for a, 6 G A:, and u = v{t^ + 1)^ for i; G A;* and e G Z, since 5^^^ = {t^ + 1}. We are t h u s reduced to deciding whether a -\-bt can be written as 1 Dv

e Dlt^ + 1)

IDv

2e

Dt

n V

n t ^ + 1

n V

nt^ + l

which is equivalent to Dv

na=

V

^ and

bt^^l ^ - —p^-— G Q . 2 Dt

T h e second condition can be immediately verified, while the first is the problem of deciding whether an element of k is the logarithmic derivative of a Ä^-radical. W h e n f = Db for some b G k{t), then Corollary 9.3.1, 9.3.2 or 9.4.1 provide alternative algorithms: / is the logarithmic derivative of a A:(t)-radical if and only if t h e linear equation (9.9), (9.13) or (9.22) has a solution in Q. The solution (n, u) is not unique, b u t any two solutions are related by t h e following lemma. L e m m a 5 . 1 2 . 1 . Let {K, D) be a differential

field and u^v e K*. If

I Du __ 1 Dv n u

m V

for nonzero n, ?n G Z^ then ^.lcm(n,m)/n /i.lcni(n,7n)/m Proof,

G Const(ü:').

L e t C = tfclcm(n,m)/n/^lcm(n,m)/m_ rpj^^^^^

Dc lcin(n,m) Du \-J—l— — = c n u so c G Const (iC).

lcm.(n,m) Dv , , , f 1 Du ^ ' ^ -lcm(n,m) m v \n u

1

Dv\ =0

m v J D

Exercises E x e r c i s e 5 . 1 . Let A: be a differential field of characteristic 0, t a monomial over k^ and d ^ k\t]\ {0}. Let d =• did\ - • - d^ be a squarefree factorization of d. Show t h a t fi{a/d) < n for any a G k[t]^ and t h a t ß{a/d) = n if and only if gcd(a,
5.12 In-Field Integration

179

Exercise 5.2. Rewrite the proof of Theorem 5.3.1 using Mack's Hnear version of the Hermite reduction instead of the quadratic version. Exercise 5.3. Let fc be a differential field of characteristic 0, t a monomial over fe, and / € Ä:(t)*. Show that using only the extended Euclidean algorithm in k[t], one can find /IQ, ^ i , • • •, /^g and r G k{t) such that q < /i(/), each hi is simple, r is reduced, and f = r -\- ho -\- Dhi + D^/i2 + • - • -^ D^hq. Exercise 5.4 (In-field integration). Let A; be a differential field of characteristic 0 and t be a monomial over k^ Write an algorithm that, given any / G /c(t), returns either g £ k{t) such that Dg = / , or "no solution" if / has no antiderivative in k[t) (see Exercise 4.1). Exercise 5.5 (Generalizations of Liouville's T h e o r e m ) . Let /c be a differential field of characteristic 0, C = Const (A:), / G k{t)^ t be a monomial over fc, and suppose that there exist an elementary extension E of k{t) and g G E such that Dg = / . a) Prove that f = Dv + f]ci^ . -,

(5.23) Ui

has a solution v G k{t)^ c i , . . . , c^ G C, and w i , . . . , t^^ G 'S'ck\t]-'Ck \ i^}b) Prove that if t is a nonlinear monomial over A:, then (5.23) has a solution V G k[t], c i , . . . ,Cn G C, and w i , . . . ,tfcn G %fc[t]:C/c \ {0}c) Prove that if Sf^ = S^^^^ then (5.23) has a solution v G k[t], c i , . . . , c^ G C, and ui,,.. ,Un G Sc^ty:ck\ i^}' d) Prove that if t is a nonlinear monomial over k and Sf^ = tS^^^, then / has an elementary integral over k. e) Prove that if t is an hyperexponential monomial over k^ then / has an elementary integral over k. f) Prove that if t is a primitive monomial over k^ then (5.23) has a solution t; = at + 6, ci,...,Cn G C, and ui^... ^Un G Cfc*, where a G C and b G k. Exercise 5.6. Decide which of the following integrals are elementary functions, and compute those that are elementary. Since the recursive problems involving the procedures L i m i t e d i n t e g r a t e , RischDE and C o u p l e d D E S y s t e m are trivial in these exercises, perform the portions allocated to those procedures by elementary methods. a) / tan(ax)^(ix, b)

a G C* .

180

5 Integration of Transcendental Functions

c) log(a; + a)

dx,

a, 6 G C, a 7^ 6.

d) {x + l)e^' + 1

/ . (e-^) -=.^ - 1

dx

e) / ( l + ^ — + ^ ^ h"^^^' 2-n x^ ^

nGZ,n7^2.

f) 2 + tan(x)2 dx 1 + (tan(x) + x)

(3x -- 2) log(a:)^ + (x - 1) log(x)2 + 2x(x - 2) log(x) + x^ : log(x)^ — 4x^ log(x)^ + 6x^ log(x)^ — 4x^ log(x)'^ + x^ log(2:)2

6 The Risch Differential Equation

We describe in this chapter the solution to the Risch differential equation problem, i.e. given a differential field K of characteristic 0 and f^geK, to decide whether the equation Dy + fy = g

(6.1)

has a solution in K, and to find one if there are some. We only study equation (6.1) in the transcendental case, i.e. when if is a simple monomial extension of a differential subfield k^ so for the rest of this chapter, let k he a. differential field of characteristic 0 and t be a monomial over k. We suppose that the coefficients / and g of our equation are in k{t) and look for a solution y G k{t). The algorithm we present in this chapter proceeds as follows: 1. Compute the normal part of the denominator of any solution. This reduces the problem to finding a solution in k{t). 2. Compute the special part of the denominator of any solution. This reduces the problem to finding a solution in k[t]. 3. Bound the degree of any solution in k[t]. 4. Reduce equation (6.1) to one of a similar form but with f^g £ k[t]. 5. Find the solutions in k[t] of bounded degree of the reduced equation.

6.1 The Normal Part of the Denominator We show in this section that the normal part of the denominator of any solution of a Risch differential equation in a monomial extension is given by an explicit formula in terms of the coefficients of the equation, provided that the equation is adequately preprocessed. We describe first the required preprocessing. Definition 6.1.1. We say that f G k{t) is weakly normalized with respect to t if iesid\iep{f) is not a positive integer for any normal irreducible p G k[t] such that f G Tip.

182

6 The Risch Differential Equation

The motivation behind that definition is the following lemma, which gives a formula for the order of Dy + /?/ at a normal polynomial whenever / is weakly normalized: L e m m a 6.1.1. Let f G k{t) \ {0} be weakly normalized with respect to t, y E k{t)\ {0}^ and p £ k\t\ he normal irreducible. Then, ^p{y) < 0 = » ^p{Dy + fy) = Vp{y) + min(z/p(/), - 1 ) . Proof. Let n = iyp{y)^ m = i'p{f) and suppose that n < 0. Then, Vp[Dy) = n — 1 by Theorem 4.4.2, and i'p{fy) = n -h m by Theorem 4.L1. If m < —1, then min(m,—1) = m and iyp{Dy + fy) = i^pify) by Theorem 4.1.1, so iyp{Dy-\-fy) = n+niin(m, —1). If m > —1, then min(m, —1) = —1 and Vp[Dy-{fy) = iyp{Dy) by Theorem 4.1.1, so yp{Dy + fy) = n + min(?Ti, —1). Suppose now that m = —1. Then h'p{pf) = 0 so / G T^p, and i'p{Dy + fy) > n — 1 by Theorem 4.1.1. By Corollary 4.4.2, Dy/y G IZp and residuep(Dy/i/) = n. Since IZp is a vector space over k and residue^ is a linear map by Theorem 4.4.1, we get Dy/y-\-f G Tip and iesidnep{Dy/y + f) = residuep(I}?//;?/)+residuep(/) = n + residuep(/). Since / is weakly normalized, residuep(/) is not a positive integer, hence residuep(/) 7^ —n, so residuep(Z)y/y + / ) y^ 0. By Theorem 4.4.1, this implies that Dy/y + f ^ Op, hence that Up{Dy/y + / ) < 0, so D h'p[Dy + fy) < n. Therefore, i^p{Dy + fy) = n — 1 = n + min(m, —1). Of course, the next step is, given / G fc(t), testing whether / is weakly normalized with respect to t, and finding an adequate transformation otherwise. The following theorem shows that adding an appropriate logarithmic derivative to any / G k{t) makes it weakly normalized, and gives an explicit change of variable that transforms equation (6.1) into a similar one with a weakly normalized coefficient. T h e o r e m 6.1.1, For any f £ k{t), we can compute q £k\t] such that f = f — Dq/q is weakly normalized with respect to t. Furthermore, for any g^y G k(t), Dy^fy

= g <=^ Dz + fz = qg

where z = qy. Proof Let d = dgdn he SL splitting factorization of the denominator of / , and dn = did^ • • • d^ be a squarefree factorization of d^- Write / = a/di + b/c where a^b^c G k[t] and gcd(di,c) = 1, and let z be a new indeterminate over k and r = resultantt(a — zDdi^di) G k[z]. Let n i , . . . ,715 be all the distinct positive integer roots of r, and q = f | g c d ( a - UiDdi^di)"^'' G k[t] , i=l

We now show that f = f — Dq/q is weakly normalized with respect to t. Let p G k[t] be normal irreducible and suppose that / G IZp. By Corollary 4.4.2,

6.1 The Normal Part of the Denominator

183

Dq/q G Tip and residuep(Dg/g) = iyp{q). Since TZp is a vector space over k and residue^ is a linear m a p by Theorem 4.4.1, we get / G Tip and residuep(/) = residuep(/) — ^p{q). Let p = residuep(/). If p is not a positive integer, t h e n residuep(/) = p — i^p{q) is not a positive integer. Thus, suppose t h a t p is a. positive integer. Then p ^ 0, so f ^ Op hj Theorem 4.4.1, which implies t h a t p I d. Since p is normal, this means t h a t p \ dn^ so i'p{f) = --i^p{dn) < 0. Since / G IZp, we have h>p{pf) > 0, so z^p(/) > —1, hence h'p{f) = —1, so Vp{dn) = 1. This implies t h a t p \ di and gcd{p,d/di) = 1, hence t h a t b/c G Op and a/di G 7?.p. Thus, residuep(6/c) = 0, so p = residuep(a/(ii). Since di is normal, a/di is simple. In addition p E k since p is an integer, hence residuep(a/
^p(g) = y^^niUp{gcd{a

- riiDdi.di))

= Uj .

Hence, residuep(/) = p — Vp{q) = rij — n^ = 0, so / is weakly normalized with respect to t. Let ^, y G /c(t) and z = gy. Then, Dz + fz = qDy + yDg + / g y

qy = g(i^t/ 4- fy)

so Dy + fy = g <=^ Dz + fz = qg.

D

We note t h a t in practice a full squarefree factorization of dn is not necessary, since only di is needed for computing g. The above proof gives an algorithm for weak-normalizing any element of k{t).

WeakNoriiializer(/, D)

(* Weak normahzation *)

(* Given a derivation D on k[t] and / G k(t), return q G k[t] such that / — Dq/q is weakly normalized with respect to t. *) (dn.ds) <— SplitFactor(denominator(/), D) g <— gcd{dn,ddn/dt) d* ^ dn/g di ^ d*/ gcd{d*,g) {a,b) <— ExtendedEuclidean(denominator(/)/di,(ii,numerator(/)) r ^r— resultantt(a — zDdi, di) ( n i , . . . , ns) <— positive integer roots of r r e t u r n ( n j = i gcd(a mDdi.diY')

184

6 The Risch Differential Equation

We can assume now t h a t / is weakly normalized with respect to t in equation (6.1). Then, t h e following theorem gives an explicit formula for the normal part of the denominator of a solution. T h e o r e m 6 . 1 . 2 . Let f G k{t) be weakly normalized with respect to t and g G k{t). Let y G k{t) be such that Dy -\- fy = g. Let d = dgd^ be a splitting factorization of the denominator of f, and e = e^e^ be a splitting factorization of the denominator of g. Let c = gcd((i„,e„) and _

gcd{en,den/dt) gcd{c,dc/dt)

^

^J

Then, (i)

yhek{t).

(Ü)

vh — ^ k{t)

for any q G k[t] \ k such that q \ h.

Proof, (i) Let q = yh E k{t). In order to show t h a t q G k{t), we need to show t h a t i'p{q) > 0 for any normal irreducible p G k[t]. We have z^p(g) = ^p{y) + i^pih) by Theorem 4.1.1. If I'piy) > 0, then iyp{q) > i^p{h) > 0 since h £ k[t]. So suppose now t h a t n = z^p(y) < 0. Case 1: i^p{f) > 0. T h e n , i^p{Dy + fy) = yp{y) — 1 by L e m m a 6.1.1. Since g = Dy + / y , this implies t h a t iyp{g) < 0, hence t h a t p \ e. Since p is normal, gcd(p, Cs) = 1, so Up{en) = -^p{g) = 1 - n. Also, p does not divide d since i'p{f) > 0, so i/p(c) = 0, so iyp{gcd{c^dc/dt)) = 0. Hence iyp{h) = Vp{g 0. so Up{h) = Up{gcd{en,den/dt))

-

Vp{gcd{c,dc/dt))

= (z^p(en) - 1) - {i^p{c) - 1) = -^p{g)

+ ypU) = -^ 1

so i'p{q) = n — n = 0. (ii) Let g G k[t]\k and suppose t h a t q \ h. Let p be any irreducible factor of q in k[t]. Then p | /i, so p | e^, so p is normal with respect to D. In addition, min{iyp{Dy),Up{fy)) < Vp{Dy + fy) = Pp{g) = --i^p{en) < 0, so at least one of h'p{Dy) or i^pify) must be negative. If iyp{f) > 0, then Up{Dy) < 0 or Vp{y) < 0, so iyp{y) < 0 in any case by Theorem 4.4.2. If i^pif) < 0, then p \ d, so p \ c and i'p{h) = {i'p{en) — 1) — (^p(c) — 1) > 0, which implies t h a t Vp{en) > ^p(c) = miii{up{dn)^i^p{en))^ hence t h a t jyp(dn) < i^p{en)^ so Mf) > ^p(^)- Since Up{g) = iyp{Dy + fy) = Vp{f) + Vp{y) by Lemma 6.1.1, we must have i'p{y) < 0 in this case also. Thus, h'p{y) < 0. From the proof of part (i), this implies t h a t Vpiyh) = 0, hence t h a t Up{yh/q) = —i^p{q) < 0, so yh/q ^ k{t). Ü

6.1 The Normal Part of the Denominator

185

C o r o l l a r y 6 . 1 . 1 . Let f G k{t) be weakly normalized with respect to t, g G k{t), and dn^en and h be as in Theorem 6.1.2. Then, (i) For any solution solution of

y G k{t) of Dy -\- fy dnhDq-\-{dnhf

= g, q = yh G k{t) and q is a

— dnDh)q

= dnh^g .

(6-2)

Conversely, for any solution q E k{t) of (6.2), y = q/h is a solution Dy^fy = g. (a) If Dy -\- fy = g has a solution in kit) then e^ | d^h^.

of

Proof, (i) Let y G k{t) he d. solution of Dy -^ fy = g^ and let q = yh. q G k{t) by Theorem 6.1.2, and

Dq+(f-^\q

= hDy + yDh + hfy - yDh = h{Dy + fy) = hg .

Multiplying through by dnh yields d^hDq + {d^hf — dnDh)q = d^K^g^ so g is a solution of (6.2). Conversely, the same calculation shows t h a t for any solution q G k{t) of (6.2), y = q/h is a solution of Dy -^ fy = g. (ii) Suppose t h a t Dy -\- fy = g has a solution in k{t). Then, (6.2) must have a solution q G k{t). T h e denominator of dnf is dg^ which has no normal irreducible factor, so dnf G k{t). Since k[t] C k{t) and k{t) is a differential subring of k{t) by Corollary 4.4.1, this implies t h a t dnhDq+{dnhf—dnDh)q G k{t)^ hence t h a t dnh^g G k{t). Let p G k[t] be any irreducible factor of e^. T h e n p is normal, so we must have Vp[dnh'^g) > 0. Hence, i/p{dnh'^) > —^pid) = iyp{en)- Since this holds for any irreducible factor of e^, we have en \ dnh"^- • The above theorem and corollary give us an algorithm t h a t either proves t h a t a given Risch differential equation has no solution in a given monomial extension, or t h a t reduces the equation to one over k{t).

R d e N o r i n a l D e n o m i n a t o r ( / , g, D) (* Normal part of the denominator *) (* Given a derivation D on k\t] and f,g G k{t) with / weakly normalized with respect to t, return either "no solution", in which case the equation Dy + fy = g has no solution in /c(t), or the quadruplet (a,b,c,h) such that a,h G k[t], b,c G k{t), and for any solution y G k{t) of Dy -\- fy ~ g-, q = yh G k{t) satisfies aDq -j-bq = c. *) (dn^ds) <— SplitFactor(denominator(/), D) (en,es) ^— SplitFactor(denominator(^),D) p <— gcd{dn,en) h ^ gcd{en, den/dt)/ gcd(p, dp/dt) if en }(dnh^ t h e n r e t u r n "no solution" return(
186

6 The Risch Differential Equation

Example 6.1.1. Let /c = Q and let t be a monomial over k satisfying Dt = 1, i.e. D = d/dt, and consider the equation Dy^y

= ^

(6.3)

which arises from the integration of e*/t. We have / = 1 and g = 1/t so: 1. {dr,,ds) = SplitFactor(l,d/dt) = (1,1) 2. (en.es) = SplitFactor(t, d/dt) = (t, 1) 3.p = gcd(l,t) = l 4./i-gcd(t,l)/gcd(l,l) = l Since t J/1, we conclude that (6.3) has no solution in Q(t), hence that J e*/t dt is not an elementary function. Example 6.1.2. Let k = Q{x) with D = d/dx^ and let t be a monomial over k satisfying Dt = 1 +t'^, i.e. t = tan(x). Consider the equation Dy + {t^^l)y

= ^

(6.4)

which arises from the integration of e*^'^^^^/tan(x)^. We have / = t^ + 1 and g = 1/t^ so: 1. (dn.ds) = SplitFactor(t2 + 1,D) = (l,t2 + 1) 2. (6^,6^) = SplitFactor(t2,D) = (^^,1) 3.p = gcd(l,t2) = l 4. /i=::gcd(t2,2t)/gcd(l,l)=t 5. dnh"^ =t^ is divisible by t^ 6. dnh'^g = 1 7. Dr^hf - DnDh = t{t^ + 1) -- {t^ + 1) = (t - l)(t2 + 1) so any solution y £ k{t) of (6.4) must be of the form y = q/t where q G k{t) is a solution of tDq + {t-l){t^ + l)q = l. (6.5)

6.2 T h e Special P a r t of t h e Denominator As a result of Corollary 6.1.1, we are now reduced to finding solutions q £ k{t) of (6.2), which we rewrite as aDq -\-hq = c

(6.6)

where a e k[t] has no special factor, 6, c G k{t), a 7^ 0, and t is a monomial over k. We give in this section algorithms that compute the denominator of any solution in k{t) of (6.6) for specific types of monomials, starting with a result valid for arbitrary monomial extensions.

6.2 The Special Part of the Denominator L e m m a 6 . 2 . 1 . Let t be a monomial a / 0 and i'p{y) 7^ 0. Then,

187

over k, p E S^^^ and a^b^y G k{t)

with

(i) If Up{b) < i^p{a), then Vp{aDy + by) = Vp{b) + Vp{y). (a) If p G <Sf^ and h'p{b) > i'p{a), then Up{aDy + by) = h'p{a) + i'p{y)(Hi) If i/p{b) = i'p{a), then either Vp{aDy + by) = Vp{a) + yp{y), or

for some nonzero (Lemma 3.1.2).

u £ k[t]/{p),

where D* is the induced

derivation

Proof. Since p is irreducible, we have Vp{aDy) = z/p(a) -\-Vp{Dy) and i'p{by) = i^p{b) + ^p{y) by Theorem 4.1.1. Furthermore, Vp{Dy) > i^p{y) by Theorem 4.4.2, which implies t h a t Dy/y G Op. (i) Suppose t h a t h'p{b) < h'p{a). Then, yp{by) = Vp{b) + z^p(y) < i/p(a) + yp{y) < iyp{a) + Vp{Dy) =

i'p{aDy)

which implies t h a t i'p{aDy + by) = iyp{by) = i'p{b) + i'p{y). (ii) Suppose t h a t p G Sf^ and t h a t yp{b) > Vp{a). Then, Up{Dy) = i'p{y) by Theorem 4.4.2, so Vp{aDy) = Typ{a) + i^piDy) = i/p{a) + Up{y) < Up{b) + Up{y) = Vp{by) which implies t h a t Vp{aDy + by) — Vp{aDy) — Vpip) + i^piy)(iii) Suppose t h a t i^pib) = z/p(a) and t h a t a ^ 0. Then, Vpib/a) Vp{a) = 0, so 6/a G Op. Furthermore,

= i^p{b) ~~

iyp{aDy) = i/p{a) + i^piDy) > Vp{b) + Pp{y) = yp{by) so Vp{aDy + by) > i^piby) = i/p{ay). Suppose t h a t Up{aDy + by) > Up{ay). Then, {aDy + by)/ay E pOp^ so 0 = 7r„ (

aDy^by\

ay

^^ (£y_ ^ ^ „ f Dy\ , „ /^ ) =^'D\ h-)=7rp( l4-7r„(-

J

^\y

a

since both Dy/y and b/a are in Op, and TTp is a field-homomorphism. Write now y = p^'p^y^z where z £ Op and Up{z) = 0. Then, Dz G Op by Lemma 4.2.1, and since Dp/p E Op^ we get -7r„

b\ -

(Dy\ = X,p

-—]=^p

y / , ^Dp\

.p . z/p y —

f \

r .Dp

Dz

x^phj)-—^-—

p TTpiDz)

, ,

/Dp\

+ - ^ V ^ = z/pü; TTp ^

since D* o Hp = Wp o D hy Theorem 4.2.1.

+

D*7rp z

fY D

188

6 The Risch Differential Equation

Since a G k[t] and has no special factor in (6.6), this means that z/p(a) = 0 for any p G 5, so Lemma 6.2.1 provides a lower bound for iyp{q) where q e k{t) is a solution of (6.6) in the following cases: (i) If iyp{b) < 0, then z/p(g) G {0, z/p(c) - z/p(6)}. (ii) If Uplb) > 0 and p G cSf ^ then Up{q) G {0, i^p(c)}. For p G tS^^^, once we have a lower bound iyp{q) > n for some n < 0, replacing g by hp'^ in (6.6) yields a(p"D/i + np^'-^hDp) + 6/ip^ = c hence ai:^/i + ( 6 + n a ^ ) h = cp-"".

(6.7)

Furthermore, h E k{t) since g G k{t)^ and h £ Op since i^p(g) > n. Thus we are reduced to finding the solutions h E k{t) nOp of (6.7). Note that cp~" G A:(t) since c G /c(t), and 6 + naDp/p G Ä;(t) since 6 G ^(t), a G /c[t] and p E S. The eventual power of p in the denominators of 6 + naDp/p and cp""' can be cleared by multiplying (6.7) by p^ where N = max(0, ~Up{h),n — z^p(c)), ensuring that the coefficients of (6.7) are also in k{t) O Op. Since all the special polynomials are of the first kind in the monomial extensions we are considering in this section, we only have to find a lower bound for yp{q) in the potential cancellation case, i.e. i^p{h) = 0. We consider this case separately for various kinds of monomial extensions. T h e P r i m i t i v e Case If Dt G k^ then every squarefree polynomial is normal, so k{t) = A:[t], which means that a^b^c £ k[t] and any solution in k{t) of (6.6) must be in k[t]. T h e H y p e r e x p o n e n t i a l Case If Dt/t G k, then k{t) = k[t,t~^], so we need to compute a lower bound on i^tiq) where q £ k{t) is Si solution of (6.6). In order to compute such a bound, we need to be able to decide whether an arbitrary element f of k can be written as Du f = mr]+ (6.8) u for some integer m and u £ k* ^ where rj = Dt/t £ k. As explained in Sect. 5.12, this is the parametric logarithmic derivative problem, a variant of the limited integration problem, which is discussed in Chap. 7. Since the integer m in a solution of (6.8) can appear in the lower bound computation, we first need to ensure that m is the same in all the solutions of (6.8).

6.2 The Special Part of the Denominator

189

L e m m a 6 . 2 . 2 . Let K be a differential field of characteristic 0, and suppose that rj ^ k* is not the logarithmic derivative of a K-radical Then, for f £ K, and any solutions (m^u) and (n^v) in Z x K* of (6.8), we have n = m and vju G Const(iir). Proof

Suppose t h a t (m^u) and (n,i;) are b o t h solutions of (6.8). Then, f = mn H

Du u

z=z nrj -\

Dv V

which implies t h a t Dw = im — n)ri w where w — vju. Since 77 is not the logarithmic derivative of a ilT-radical, t h e above implies t h a t m = n and t h a t Dw — 0. D L e m m a 6 . 2 . 3 . Suppose that t is an hyperexponential over k such that rj = Dt/t is not the logarithmic derivative of a k-radical. Let a € Ä:[t],6, g G kit) he such that g c d ( a , t ) = I, Ut{b) = 0, and i^tiq) / 0- Then, either Ut{aDq^bq)

= Vt{q)

or -_- = vAq) Tj H a(0) u Proof

for some u E k .

Suppose t h a t Vt{aDq + hq) / 1^4(9). Then, Lemma 6.2T impUes t h a t

( h\ aj

, ,

fDt\ \ ^/

D'u ^

for some nonzero u e k[t]/{t)^ where D* is t h e induced derivation. B u t k[t]/{t) i::^ k and D* is an extension of D by L e m m a 3.4.3, so u £ k* and D*u = Du. Furthermore, 7rt{p) = p{0) for any p £ k[t]^ so '^ti — ]=TTt{rj)=rj, which proves the lemma.

iTt{a)=a{0),

and

7Tt{b) = b{0), D

Since t G Sf^ by Theorem 5.L2, Lemmas 6.2.1 and 6.2.3 always provide a lower bound for iyt{Q) where q e k{t) is di solution of (6.6): if Ut{b) ^ 0, then Lemma 6.2.1 provides the bound as explained earlier. Otherwise, i^tib) = 0, so either —6(0)/a(0) = mrj + Du/u for some m G Z and ix G fc*, in which case z/t(g) G {0, m, Utic)}^ or i^tiq) ^ {0, ^'t(c)}. Note t h a t such an m is unique by L e m m a 6.2.2 applied to k. Since S^^^ = { t } , k{t) n Ot = fc[t], so having determined a lower bound for 2/^(^)5 we are left with finding solutions h e k[t] of (6.7).

190

6 The Risch Differential Equation

RdeSpecialDeiioniExp(a, 6, c, D) (* Special part of the denominator - hyperexponential case *) (* Given a derivation D on k[t] and a G k[t], b,c E fc(t) with Dt/t G k, a ^ 0 and gcd(a,t) = 1, return the quadruplet (ä 6, c, /i) such that ä,b,c,h G k[t] and for any solution q ^ k{t) of aDq + bq = c, r = qh E k[t] satisfies äDr + br — c. *) (* the monic irreducible special polynomial *)

p<-t rib ^

T^p{b), Tie ^

i^p{c)

(* n < 0 *) n ^- min(0,nc — min(0,nb)) (* possible cancellation *) if Ub = 0 t h e n (* a = -6(0)/a(0) G /c *) a ^— Reniainder(—5/a,p) if a = mDt/t -\- Dz/z for z E k* and m G Z t h e n n <— min(n, m) (* A^ > 0, for clearing denominators *) N <— niax(0, —nb,n — Uc) return(ap"^, (6 + naDp/p)p^, cp

Example 6.2.1. Let k = Q(a;) with D = d/dx, and let t be a monomial over fc satisfying Dt = t, i.e. t = e^, and consider the equation {f + 2xt + x'^) Dq + (n+-^\t^+(2x-l

+

^]t

• a;2 1 g

4-1 +-

(6-9)

which arises from the integration of

(e^ + x) x^

ef

where x'^ -1 X

1 e^ + X

We have a = t^ + 2xt -f x^, 6 = (1 -f l/x^)^^ + (2x - 1 + 2/x)t + x^, and c = t/x'^ — 1 + 2/x, hence 1. 2. 3. 4. 5.

715 = i^t(&) = 0, nc = z^t(c) =: 0 n = min(05 ric — min(0, njj)) = 0 n5 = 0, so a = --6(0)/a(0) = x^/x^ = - 1 —1 = —Dt/t^ so m = —1 and n = min(n, m) = —1 N = max(0, —n^, n — ric) = 0

Hence, any solution q G k{t) of (6.9) must be of the form q = p/t for p G k[t] satisfying

{t'+2xt+x')Dp+(^^+(^l-iyy='^+(^l-iy.

(6.10)

6.2 The Special Part of the Denominator

191

The Hypertangent Case If Dt/{t^ + 1) € fe and / = 4 ^ k, then the only monic special irreducible is ^^ + 1, so we need to compute a lower bound on z/t2+i(^), where q E k{t) is a solution of (6.6). L e m m a 6.2.4. Suppose that y - T ^ k and that t is an hypertangent over k such that ri\/~^ is not the logarithmic derivative of a k{\/—l)-radical, where T] = Dt/{t'^ + 1) G /c. Let a € k[t],h,q G k{t) he such that gcd(a,t2 + 1) = 1, Vt2j^i{h) — 0 and i^t^-^i{q) ^ 0. Then, either Ut2^i{aDq + bq) = Ut^^iiq) or, extending D to k{y/^) via D\/~—\ = 0 and writing —b{y/^)/a{-\/^) a\/~^ + ß for a^ ß £ k, we have ^^^^)-2.,.+l(9)^^/:Il + ^ a(v—1) u for some u G k{\f^y Proof that

and

2ß = ^ V

=

(6.11)

and v £ k*.

Suppose t h a t Vt'^j^i{aDq + hq) 7^ Vt2j^i{q). Then, Lemma 6.2.1 implies (

h\

^ ^

fD{t^ + l)\ D'^u t2 + l 7 ' U

for some nonzero u G k\t]/{t? ^ 1)^ where D* is the induced derivation. But k[t]l{t'^ -\-l) ^ fc(V^), and ( ^ ( v ^ ) , i ^ * ) is an extension of {k,D) by Lemma 3.4.3, and I)* is the unique extension of D to k{\f-i) by Theorem 3.2.3. Since A / ^ is algebraic over Const(/c), D * y ^ ^ = 0 by L e m m a 3.3.2, so Du = D*u. Furthermore, D(t2 + 1)

^

t2 + 1

Dt t2 + 1

SO we get 6(7)

^

, ,

Dii^

for any 7 G Ä ; ( y ^ ) such t h a t 7^ + 1 ==0. Taking 7 = ^/^ yields the first equality in (6.11). Let a : k[y^'^) —^ k{^/^Ä.) be the automorphism t h a t is the identity on k and t h a t takes \f^ to — A / ^ . Applying 1 + cr to the first equality in (6.11) we get 2ß = {ayf^

+ /3) + {-a^f^:

+ ß)

= (2ut2+i{q)r]V^^^j _ Du

+ ('_2z/,2+i(g)T/v^+^"^

Diu"") __ D{uu'')

U

where v = uu^ G k*.

U^

UU^

_

Dv V

D

192

6 The Risch Differential Equation

Since t^ + 1 E Sf^ by Theorem 5.10.1, Lemmas 6.2.1 and 6.2.4 always provide a lower b o u n d for Ut2^i{q) where q G k{t) is a solution of (6.6): if i^t2+i(^) 7^ 0, t h e n L e m m a 6.2.1 provides the bound as explained earlier. Otherwise, z^t2+i(^) = 0, so either — 6 ( \ / ^ ) / a ( > / ^ ) = mr]^/^ -{- Du/u for some m e Z and u G fc(\/^)*, in which case ^'t2+i(^) € { 0 , m , z/^2^i(c)}, or ^t2+i(9) E {O5 ^t2+i(c)}. Note t h a t such an m is unique by Lemma 6.2.2 applied to k{\/^^). We also remark t h a t the verification of (6.11) implies solving This adjunction of a parametric logarithmic derivative problem over k{y/^). \ / ^ is however t e m p o r a r y since only the integer Tyt^-^i{q) is used from the result, so the algorithm proceeds over k once this bound is determined. Since the necessary condition 2ß = Dv/v is defined over /c, it can be checked first, and V ^ needs t o be introduced only if t h a t condition is satisfied. Since ^irr r= {t^ + 1}, A:(t) n Ot2_|_i = k[t]^ so having determined a lower bound for 1/^2^1 (g), we are left with finding solutions h G k[t] of (6.7). There are analogues of Lemma 6.2.4 and the corresponding algorithm for fields containing \/~^ (Exercise 6.1).

R d e S p e c i a l D e n o m T a n ( a , b, c, D) (* Special part of the denominator - hypertangent case *) (* Given a derivation D on k[t] and a G k[t], b,ce k{t) with Dt/(t^ + l) G k, \ / ^ ^ /c, a 7^ 0 and gcd(a, t^ -h 1) = 1, return the quadruplet (ä, 6, c, h) such that ä, b,c,h G k[t] and for any solution q G k{t) of aDq -\- bq = c, r = qh ^ k[t] satisfies äDr -i-br = c. *) p ^- t"^ + 1 rib ^

^p{b),

(* the monic irreducible special polynomial *) Tic ^

i^p{c)

n ^— min(0, nc — min(0,nb)) (* n < 0 *) if n6 = 0 t h e n (* possible cancellation *) (* a,/? G fc *) a\f^ + ß 0, for clearing denominators *) retnrn(ap^,(b + naDp/p)p ,cp ~"',p~"') Example 6.2.2. Continuing example 6.1.2, let k ~ Q{x) with D = d/dx^ t be a monomial over k satisfying D t = 1 + 1 ^ , i.e. t = tan(a:), and consider the solutions q £ k{t) of (6.5), which arises from the integration of e*^"^^^/ tan(x)^. We have a = t,b={t — l)(t^ + 1) and c = 1, hence 1. rib = i^t^^iib) = l,nc = z^t2+i(&) = 0 2. n = min(0, ric - min(0, rih)) = 0 3. 715 :^ 0, so A^ — max(0, —n^, n — ric) — 0 Hence any solution of g G k{t) of (6.5) must be in k{t) D Ot2^i = k[t].

6.3 Degree Bounds

193

6.3 Degree Bounds As a result of the previous sections, we are now reduced to finding solutions q G k[t] of (6.7), which we rewrite as aDq^hq = c

(6.12)

where a^b^c £ k[t], a 7^ 0, and t is a monomial over k. We give in this section algorithms that compute an upper bound on the degree in t of any solution in k[t] of (6.12) for specific types of monomials, starting with a result valid for arbitrary monomial extensions. L e m m a 6.3.1. Let t be a monomial over k and a^b^q G k[t] with a j^ 0 and deg{q) > 0. Then, (i) If deg(b) > deg(a) +max(0,(5(t) — 1), then deg{aDq + bq) = deg{b) + deg{q). (a) If t is nonlinear and deg(6) < deg(a) + 5{t) — 1, then deg{aDq + bq) = deg(a) + deg(g) + 6{t) - 1. (Hi) If 6{t) > 1 and deg{b) = deg(a) -f- ö{t) — 1, then either deg{aDq + bq) = deg(6) + deg{q) or lc{b) lc(a)

f Dq "" V^^^^*^-'

Proof. (i) We have deg(Dq) < deg(g') + max(0, 5(t) — 1) by Lemma 3.4.2, hence deg[aDq) = deg(a) + deg(Dg) < deg(g) + deg(a) + max(0,5{t) — 1) < deg(g) + deg(6) = deg{bq) which implies that deg(aDg + bq) = deg(5g) = deg(6) + deg(g). (ii) If t is nonUnear, then deg{Dq) = deg(g) + S{t) — 1 by Lemma 3.4.2, hence deg(ajDg) ~ deg(a) + deg(Dg) = deg(g) + deg(a) + ^{t) - 1 > deg(g) + deg(6) - deg(6g) which implies that deg(aDg + bq) = deg{aDq) = deg(a) + deg(g) + S{t) — 1. (iii) If 6{t) > 1, then deg{Dq) < deg{q) + S{t) - 1) by Lemma 3.4.2, hence deg(aDg) = deg(a) + deg(Dg) < deg{q) + deg(a) + S{t) -1 = deg{q) + deg(6) = deg{bq)

194

6 The Risch Differential Equation

which impHes t h a t deg{aDq-\-bq) < deg{b) + d e g ( g ) . Suppose t h a t deg(aDg + bq) < deg(6) + deg{q). Then deg(aDg + bq) < deg(a) + deg(g) + S{t) — 1, so {aDq + bq)/{aqt^^^^~^) G t~^Ooo, which impHes t h a t _ (aDq^bq\ __ ^ " ""^yaqtm-l J-''^

( b \atm-l

Dq \ + qtm-l J

since TTOO is a ring-homomorphism and b o t h b/at^^'^^~^ and Dq/qt^^*'^~^ are in OoQ. Since deg(6) = deg(a) + S{t) — 1, we have /

b

\

lc(6)

lc(6)

and the lemma follows.

D

Lemma 6,3.1 provides an upper bound for deg(g') where q G k[t] is a solution of (6.12) in the following cases: (i) If deg(6) > deg(a) + max(0, Ö{t) - 1), then deg{q) G {0, deg(c) - deg(6)}. (ii) If deg(6) < deg(a) + 6{t) - 1 and ö{t) > 2, then deg{q) G {0, deg(c) - deg(a) + 1 - 6{t)} . As a result, we only have t o consider the cases deg(6) < deg(a) for Louvillian monomials, a n d deg(6) = deg(a) + S{t) — 1 for nonUnear monomials. We consider those cases separately for various kinds of monomial extensions. T h e Primitive Case If D t G k^ then, in order t o compute an upper bound on deg(g), we need t o decide whether an arbitrary element f of k can be written as f = mr]^Du

(6.13)

for some integer m and u e k^ where r] = Dt e k. Note t h a t (6.13) is a limited integration problem in k^ so it can b e solved by applying the algorithm of Chap. 7 t o / and TJ. Since the integer ?72 in a solution of (6.13) can appear in the upper bound computation, we first need t o ensure t h a t m is the same in all the solutions of (6.13). L e m m a 6c3«2. Suppose that t is a primitive over k such that rj = Dt is not the derivative of an element ofk. Then, for f G k{t), and any solutions (m^u) and {n^v) in Z x k of (6.13), we have n = m and v — u £ Const(A;).

6.3 Degree Bounds

195

Proof. Suppose t h a t {m^u) and {n^v) are both solutions of (6.13). Then, / = mr] + Du = riT] + Dv so Dw = {m — n)r] where w = v — u. Since r] is not the derivative of an element of k^ the above imphes t h a t m = n and t h a t Dw = 0. D L e m m a 6 . 3 . 3 . Suppose that t is a primitive over k such that rj = Dt is not the derivative of an element of k. Let a^b^q £ k[t] be such that a 7^ 0^ deg{b) < deg(a)^ and deg(g) > 0. Then, (i)

If deg{b) < deg(a) — 1, then deg{aDq + bq) G {deg(a) + deg(g),deg(a) + deg(g) - 1} .

(ii) If deg{b) = deg(a) — 1, then

either

deg{aDq + bq) € {deg(a) + deg(g), deg(a) + deg(g) - 1} or — -—-— = degfo) T] + Du lc(a) (Hi) If deg{b) = deg(a); then

for some u £ k .

either

deg{aDq + bq) E {deg(a) + deg(g),deg(a) + deg{q) - 1} or lc(fe)

Djlciq))

lc(aD(lc(g)) + 61c(g))

-Ho) ^ ^ ^ ( ^

HÖM^)

,

, ,

^

^

= ^''^'^' + ^ "

for some u £ k. Proof Since deg(g) > 0, deg(Dg) G {deg(g),deg(g) — 1} by L e m m a 5.1.2. (i) If deg(6) < deg(a) - 1, t h e n deg(aDg) = deg(a) +

deg{Dq)

> (deg(6) -h 1) + (deg(g) - 1) = deg(6) + deg(g) = deg{bq) which implies t h a t deg(aDg + bq) = deg{aDq)

G {deg(a) + deg{q), deg(a) + deg(g) - 1} .

(ii) Suppose t h a t deg(5) = deg(a) — 1. If deg(Dg) = deg(g), then deg(aDg) = deg(a) + deg{Dq) = deg{b) + 1 + deg(g) > deg(5) + deg(g) = deg(6g) which implies t h a t deg(ajDg + bq) = deg{aDq) = deg(a) + deg(g). Otherwise, deg(Dg) = deg(g) - 1, so D{lc{q)) = 0 by Lemma 5.1.2, which implies t h a t

196

6 The Risch Differential Equation lc{Dq) = deg{q)r]lc{q) + Dv

where v E k is the coefficient of t^eg(g)--i -^^ ^_ j ^ addition, we have deg(aDg) = deg(a) + deg{Dq) = (deg(6) + 1) + (deg(g) - 1) = deg{b) + deg(g) = deg(5g) which impHes that deg(aDg + bq) < deg(a) + deg(g) — 1. Suppose that deg{aDq + bq) < deg(a) + deg(g) - 1. Then, {aDq + bq)t/{aq) G t-^Ooo, which impHes that

ff{aDq-\-bq)t\ (aDq-{-bq)t\ ___ 0 =

TToo (

— aq

\

JI

=

f ftb tb

tDq\

f tb\

TToo ( — H

I =

~ Va

TToo ( —

ftDq I 4 - TToo (

since ITQQ is a ring-homomorphism, and both tb/a and tDq/q are in deg(6) = deg(a) — 1 and deg{Dq) = deg(g) — 1, we have tb\ I aJ

TToo 1 —

=

lc{tb) lc(a)

~

O^Q.

Since

lc(6) lc(a)

and ftDq\

IcjtDq)

lc{Dq)

degiq)ri\c{q) + Dv

,

. .

^ „

where u = v/lc{q) G k. (iii) Suppose that deg(6) = deg(a). If deg(Dg) = deg(^) — 1, then deg(aDg) = deg(a) + deg(Dg) = deg(6) + deg(g) - 1 < deg(6) + deg(g) = deg(6g) which imphes that deg(aDg + bq) = deg(6<3') = deg(a) -h deg(g). Otherwise, deg{Dq) = deg(g), which imphes that deg(aDg) = deg(a) + deg{Dq) = deg{b) + deg(g) = deg{bq) hence that deg(aDg + bq) < deg(a) 4- deg(g). Suppose that deg{aDq + bq) < deg(a) + deg(g). Then, {aDq + bq)/{aq) £ t~^0^, which imphes that faDq + bq\ \ aq J

[b \a

Dq\ q J

[ b\ \aj

(Dq \ q

since TTOO is a ring-homomorphism, and both b/a and Dq/q are in (9oo- Since deg(6) = deg(a) and deg{Dq) = deg(g), we have

..^l"-\ = 'M and ^aj which imphes that

lc(a)

,^fD,\_HDq)_DiHq)) \ QJ

lc{q)

lc(g)

6.3 Degree Bounds

- 1 ^ =^

197

(6.14)

ic(a) u where u = lc(g). Write p = u~^q G k[t]. Then, deg(p) = deg(g) and aDq -j-bq — aD{up) + bup = ADp + Bp where A = ua and B = aDu + bu. Note that deg(yl) = deg(a), and deg(B) < deg(A), since (6.14) imphes that the coefficient of t^^^^^^ in B is 0. Suppose first that deg(ß) < deg(^) — 1. Then, (i) imphes that deg{ADp + Bp) e {deg(A) + deg(p), deg(y4) + deg{p) - 1} hence that deg(aDg -h bq) G {deg(a) + deg(g), deg(a) + deg(g) - 1} . Suppose finally that deg{B) — deg(^) — 1. Then, (ii) implies that either deg(ADp + Bp) £ {deg{A) + deg(p), deg{A) + deg(p) - 1} or

HB)

^ ,,

"kM) =^^gW^ + ^^ for some v E k. Noting that deg(p) = deg(g), deg(74) = deg(a), and lc{A) = lc(a)lc(g) completes the proof. D Lemmas 6.3.1 and 6.3.3 always provide an upper bound for deg(g) where q G k[t] is a, solution of (6.12): if deg(6) > deg(a), then Lemma 6.3.1 implies that deg(g) € {0,deg(c) — deg(6)}. If deg(6) < deg(a) — 1, then Lemma 6.3.3 implies that deg(g) G {0,deg(c) — deg(a),deg(c) — deg(a) + 1}. If deg(6) = deg(a) — 1, then either —lc(6)/lc(a) = mr] + Du for some m G Z and u £ k/in which case deg(g) € {0,m,deg(c) — deg(a), deg(c) — deg(a) + 1}, or deg(g) E {0,deg(c) — deg(a),deg(c) — deg(a) + 1}. Note that such an m is unique by Lemma 6.3.2. Finally, if deg(6) = deg(a), then either —lc(6)/lc(a) = Du/u for some u e k* and —lc{aDu + bu)/{ulc{a)) = mr] + Dv for some m G Z and v £ k, in which case deg(g) G {0,m,deg(c) — deg(a),deg(c) — deg(a) + 1}, or deg(g) G {0,deg(c) — deg(a),deg(c) — deg(a) + 1}. We can compute such an li by a variant of the integration algorithm (Sect. 5.12). Although it is not unique, if —lc(6)/lc(a) = Du/u = Dv/v for u,v £ k*^ then u = cv for some c G Const(fc) by Lemma 5.12.1, which implies that lc{aDu + bu) lc{a)u

lc{acDv + bcv) lc{a)cv

c {lc{aDv + bv) clc{a)v

lc{aDv + bv) lc{a)v

so the solution we use does not affect the bound m, which is unique by Lemma 6.3.2.

198

6 The Risch Differential Equation

R d e B o u n d D e g r e e P r i m ( a , 6, c, D) (* Bound on polynomial solutions - primitive case *) (* Given a derivation D on k[t] and a, 6, c G k[t] with DtG A; and a ^ 0, return n G Z such that deg(g) < n for any solution q e k[t] of aDq-\-bq = C. *) da ^ deg(a), 4 ^ deg(6), dc ^ deg(c) if db > da t h e n n <— max(0, dc — <^b) elsen <— max(0, dc-da -hi) (* possible cancellation *) if db = da — I t h e n a ^ -lc(6)/lc(a) if a = mDt + D^ for z E k and ?n G Z t h e n n '^- max (n^m) (* possible cancellation *) if c?6 = da t h e n a < lc(6)/lc(a) if a = Dz/z for z ^ k* t h e n /? ^ -lc(ai:>z + bz)/(z lc(a)) if /3 = mDt + Dw for w G k and m G Z t h e n n ^ max(n, m) return n

Example 6.3.1. Let k = Q(x,to) with D = d/dx^ where to is a monomial over Q(a:) satisfying Dto/to = 1/x^, i.e. to = exp(—1/x), and let t be a monomial over k satisfying Dt = 1/x, i.e.t = log(x). Consider t^Dy - (^i" \x^

+ -\y XJ

= {2x- l)t^ + ^ 2 ± ^ i 3 - h±^t^ x 2x

+ ^t

(6.15)

which arises from the integration of

Theorem 6.1.2 gives /i = 1, so any solution in /c(t) must be in k{t) = k\t]. We have a = t^^h = —t^jx^ — 1/x and . X 4 to 4- a:^ Q to + 4x^ 9 c = (2x - l)t^ + -iL__ t^ - -^^ 1^ + a:t X

1x

hence 1. da = deg(a) = 2, 4 = deg(5) = 2, 4 = deg(c) = 4 2. n = max(0,4 — 4 + 1) = 3 3. 4 is equal to 4 , so a) a = -lc(6)/lc(a) = l/rr^ b) a is equal to Dii^^^ji^., so i. /5 = -lc(ai)to + 6to)/(tolc(a)) = - 1 / x ii. /3 is equal to —Dt, so n = max(n, — 1) = 3 So any solution in fc[t] of (6.15) must have degree at most 3.

6.3 Degree Bounds

199

In the specific case where D = d/dt^ then Du = 0 for any u G k^ so in particular, —lc(6)/lc(a) is not of the form Du/u for u e. k. This yields a simpler form of Lemma 6.3.3 for t h a t case, together with a simpler algorithm. C o r o l l a r y 6 . 3 . 1 . Suppose that t is transcendental over k and that D = Let a^h^q E k\t] he such that a 7^ 0 and deg(g) > 0. Then,

d/dt.

(i) If deg{b) > deg(a) — 1 then, deg{aDq + bq) = deg(6) + deg(g). (ii) If deg{h) < deg(a) — 1, then deg{aDq + hq) — deg(a) + deg(g) — 1. (Hi) If deg{b) = deg(a) — 1, then either deg{aDq + bq) = deg(5) + deg(g), or lc{b) lc{a)

R d e B o u n d D e g r e e B a s e ( a , 6, c) (* Bound on polynomial solutions - base case *) (* Given a,b,c G k[t] with a 7^ 0, return n G Z such that deg(g) < n for any solution q ^ k[t] of adq/dt + bq = c. *) da ^ deg(a), db <— deg(6), dc ^ deg(c) n ^- niax(0, dc — max((i6, da — 1)) if db = da — 1 t h e n m < lc(5)/lc(a) if m G Z t h e n n <— max(0, m, dc — db) return n

(* possible cancellation *)

Example 6.3.2. Let /c = Q and let t be a monomial over k satisfying D^ = 1, i.e. D = d/dt^ and consider t h e equation Dy - 2ty = 1 which arises from t h e integration of e~* dt. Theorem 6.L2 gives h = 1, so any solution in k{t) must be in k{t). Lemma 6.2.1 shows t h a t Ut{y) > 0 for any solution, hence any solution in k{t) must be in k[t]. We have a = c = 1 and b = -~2t, hence 1. da = deg(a) = 0, 4 = deg(5) = 1 , 4 = deg(c) = 0 2. n = m a x ( 0 , 4 — m a x ( 4 7 ^a — 1)) = 0So any solution in k{t) must be in k = Q. Since t is transcendental over Q, —2ty 7^ 1 for any y € Q, which implies t h a t

/ is not an elementary function.

e-*\

200

6 The Risch Differential Equation

The Hyperexponential Case L e m m a 6 . 3 . 4 . Suppose that t is an hyperexponential over k such that r] = Dt/t is not the logarithmic derivative of a k-radical. Let a^b^q G k[t] be such that a ^ 0, deg(6) < deg{a), and deg(g) > 0. Then, (i) If deg{b) < deg(a), then deg{aDq + bq) = deg(a) + deg(g). (ii) If deg{b) = deg(a)^ then either deg{aDq + bq) = deg(6) + deg(g), or lc(a)

^^SW^+

j^^^)

•

Proof Since deg(g) > 0, we have deg(Dg) = deg(g) by L e m m a 5.1.2. (i) If deg(5) < deg(a), then deg(ajDg) = deg(a) + deg{Dq) > deg{b) + deg(g) = deg{bq) which imphes t h a t deg{aDq + bq) = deg(aDg) = deg(a) + deg(g). (ii) Suppose t h a t deg(6) = deg(a) and deg{aDq-j-bq) ^ deg(6) + d e g ( g ) . Since 6{t) = 1, L e m m a 6.3.1 impUes t h a t

lc(6)

fDq

lc(a)

\ q HDq)

= 1

D{lc{q))+deg{q)r,\ciq)

^=

H^)

, J3(lc(g))

=^^^('^)^ + n ^ K ^ D

Lemmas 6.3.1 and 6.3.4 always provide an upper bound for deg(g) where q £ k[t] is a solution of (6.12): if deg(6) > deg(a), t h e n Lemma 6.3.1 implies t h a t deg(g) £ {0,deg(c) — deg(6)}. If deg(6) < deg(a), then Lemma 6.3.4 implies t h a t deg(g) G {0,deg(c) — deg(a)}. Finally, if deg(6) = deg(a), then either —lc(fe)/lc(a) = mr] -{- Du/u for some m EIJ and ix € Ä:*, in which case deg(g) G { 0 , m , d e g ( c ) — deg(6)}, or deg(g) G {0,deg(c) — deg(6)}. Note t h a t such an m is unique by Lemma 6,2.2. R d e B o u n d D e g r e e E x p ( a , 6, c, D) (* Bound on polynomial solutions - hyperexponential case *) (* Given a derivation D on k[t] and a^b^c G k[t] with Dt/t G k and a 7^ 0, return n £ IL such that deg(g) < n for any solution q G k[t] of aDq -\-bq = c. *) da ^ deg(a), dh ^ deg(6), dc ^ deg(c),n ^ max(0, dc — max(ci5, da)) (* possible cancellation *) if da = db t h e n a ^ -lc(&)/lc(a) if a = mDt/t + Dz/z for z G k* and m G Z t h e n n ^— max(n, m) return n

6.3 Degree Bounds

201

Example 6.3.3. Continuing example 6.2.1, let k = Q(x) with D = d/dx, t b e a monomial over k satisfying Dt = t^ i.e. t = e^^ and consider the solutions in k[t] of (6.10). We have a = t^ + 2xt + x^, 6 = c = t'^/x'^ + {2/x - l)t, hence 1. 2. 3. 4.

da= db = dc = 2 da = db^ so a = —lc(6)/lc(a) = —Ijx^ n = max(0, d^ — max((i5, (ia)) = 0 — 1/x^ cannot be written in the form vn -h Dzjz

foimE'Z

and z G Q{x)

Hence any solution p G k[t] of (6.10) must be of degree 0, i.e. in Q{x). The Nonlinear Case L e m m a 6 . 3 . 5 . Suppose that t is a nonlinear monomial over k, and let a,b,q G k[t] be such that a 7^ 0, deg(6) = deg(a) + S{t) — I, and deg(g) > 0. Then, either deg{aDq + bq) = deg(6) + deg(g), or -|^=deg(g)A(i). ic(a) Proof. Suppose t h a t deg{aDq + bq) 7^ deg(6) + deg(g). Then, Lemma 6.3.1 implies t h a t lc(6) _ / Dq lc(a) " ^ ^ \qtm-i) Furthermore, lc{Dq) deg{q)X{t).

\ _ lc(Dg) ~~ lc{qt^(t)-^)

= deg{q)lc{q)X{t)

_ ~

IcjDq) \c{q) '

by L e m m a 3.4.2, so —lc(5)/lc(a) = D

Lemmas 6.3.1 and 6.3.5 always provide an upper bound for deg(g) where q G k[t] is a solution of (6.12): if deg(6) 7^ deg(a) -{-6{t) — 1, then Lemma 6.3.1 provides t h e bound as explained earlier. Otherwise, either --lc(6)/lc(a) = mX{t) for some m G Z, in which case deg(g) G {0,m,deg(c) — deg(6)}, or deg(g) G {0,deg(c) - d e g ( 6 ) } . R d e B o u n d D e g r e e N o n L i n e a r ( a , 6, c, D) (* Bound on polynomial solutions - nonlinear case *) (* Given a derivation D on k[t] and a,b,c G k[t] with de g{Dt) > 2 and a 7^ 0, return n G Z such that deg(g) < n for any solution q G k\t] of aDq -^bq = c. *) - deg(Dt), A <-- lc{Dt) da ^ deg(a), db ^ deg(6), dc ^ deg(c),(5 < n <— max(0, dc — max{da -{-6 —.db)) 1 if db = da + 6 — 1 t h e n (* possible cancellation *) m^-lc(6)/(Alc(a)) if ??i G Z t h e n n <— niax(0, m, dc - db) return n

202

6 The Risch Differential Equation

Example 6.3.4- Continuing examples 6.1.2 and 6.2.2, let k = Q(x) with D = d/dx, t be a monomial over k satisfying D t = 1 + t"^, i-e. t = tan(a:), a n d consider the the solutions q £ k[t] of (6.5). We have a = t, b = (t — l)(t^ + 1) and c = 1, hence 1. da = deg(a) = 1 , 4 = deg(6) = 3, dc = deg(c) = 0 2.6 = 6{t) = 2, A = l c ( l + t 2 ) = l 3. n = max(0, dc — max((ia + (5 — 1, (if,)) = 0 4. 4 / ö^a + (^ - 1 Hence any solution q e k[t] of (6.5) must be of degree 0, i.e. in Q{x).

6.4 T h e S P D E Algorithm We are now reduced to finding solutions q G k[t] of (6.12) and we have an upper bound n on deg(g). We present here an algorithm of Rothstein [83] t h a t either reduces equation (6.12) to one with a == 1, or proves t h a t it has no solutions of degree at most n in k[t]. This algorithm is based on the following theorem. T h e o r e m 6 . 4 . 1 . Let a^b^c e k[t] with a ^0 and gcd(a, 6) = 1. Let z^r G k[t] be such that c = az + br and either r = 0 or deg(r) < deg(a). Then, for any solution q G k[t] of aDq -\-bq = c, h — [q ~ r)/a £ k\t], and h is a solution of aDh + (6 + Da)h = z - Dr .

(6.16)

Conversely, for any solution h G k\t\ of (6.16), q = ah -\- r is a solution aDq -{~bq = c.

of

Proof. Let q Gk\t] be a solution of aDq -\-bq = c. Then, aDq -\- bq = az ^br, so b{q — r) = a{z — Dq)^ so a \ b{q — r ) . Since gcd(a, 6) == 1, this implies t h a t a I g — r, hence t h a t h = {q — r)/a £ k[t]. We then have: n r^ NT fDq~Dr aDh + (6 + Da)h = a —^=

(q-r)Da\ ^ f

biq ~-r)-{-(q + ~-~^^i—~L )^

r)Da L

Dq-Dr^^^'' a (az 4- br) — bq

= -i

^—i

_ jj'p -I

bq ~ br ^—„_

^ = 2 — Dr .

a a Conversely, let h E k[t] be a solution of (6.16), and let q ~ ah-\-r.

Then,

aDq -\-bq = a^Dh + ahDa + aDr + abh + br = a{aDh + (6 + Da)h) + aDr + br = a{z — Dr) + aDr + br = az -}- br = c. D

6.4 The SPDE Algorithm

203

Theorem 6.4.1 reduces (6.12) to (6.16), which is an equation of t h e same type. However, if (6.12) has a solution q of degree n, then the corresponding solution h of (6.16) must have degree at most n — deg(a) since q = ah -\- r and deg(r) < deg(a). Thus, if deg(a) > 0 and gcd(a, 6) = 1, we can use Theorem 6.4.1 to reduce the degree of the unknown polynomial. T h e hypothesis t h a t gcd(a,6) = 1 is not a restriction: if (6.12) has a solution in k[t], then c E (ci^b), so g = gcd(a, b) must divide c, in which case we can divide a, b and c by ^f in order to get an equivalent equation with gcd(a, b) = 1. Note t h a t this step reduces the degree of a. If gcd(a, b) / c , we can conclude t h a t (6.12) has no solution in k[t]. We can repeat this until either we have proven t h a t (6.12) has no solution of degree at most n in fc[t], or until deg(a) = 0 i.e. a G Ä:*, at which point we divide t h e equation by a and we get an equation of the type (6.12) with a = 1. This is the SPDE^ algorithm of Rothstein [82, 83].

S P D E ( a , 6, c, D, n)

(* Rothstein's SPDE algorithm *)

(* Given a derivation D on k[t], an integer n and a^b^c e k[t] with a / 0, return either "no solution", in which case the equation aDq -j-bq = c has no solution of degree at most n in k[t], or the tuple (b,c,m,a, ß) such that b,c,a,ß G k[t], m G Z, and any solution q G k[t] of degree at most n of aDq-\-bq — c must be of the form q = ah^ ß, where h G k[t], deg(h) < m and Dh-\-bh = c. *) if n < 0 t h e n if c == 0 t h e n r e t u r n ( 0 , 0 , 0 , 0 , 0 ) else r e t u r n "no solution" g ^ gcd(a, b) if g )fc t h e n r e t u r n "no solution" a ^ a/g, b ^ b/g, c ^ c/g if deg(a) = 0 t h e n r e t u r n ( 5 / a , c/a^ '^,1,0) (r, z) <— E x t e n d e d E u c l i d e a n ( 6 , a, c) (* br + az = c, deg(r) < deg(a) *) u^ SPDE(a,6+I>a,^-i^r,D,n-deg(a)) if II = "no solution" t h e n r e t u r n "no solution" (b,c,m,a,ß) ^— u (* The solutions of (6.16) are h = as -\- ß where Ds -\-bs = c '^) {"^ ah + r = aas -\- aß -\- r *) return(6, c, m, aa, aß -{- r)

Example 6.4-1. Continuing examples 6.1.2, 6.2.2 and 6.3.4, let k = Q{x) with D = d/dx^ t be a monomial over k satisfying Dt = 1 + t^, i.e. t = t a n ( x ) , and consider the the solutions in k[t] of (6.5). We have a = t, b = (t — l)(t^ + 1) = t^— t ^ + t — l , c = : l and n = 0 from example 6.3.4, hence 1. ^ = gcd(a,6) = 1 2. (r,2) = E x t e n d e d E u c l i d e a n ( t ^ - t^ + 1 - l , t , 1) = {~l,t^ ^Special Polynomial Differential Equation

- t + 1)

204

6 The Risch Differential Equation

3. b + Da = t^ - t'^ + t - 1 + Dt = t^ -^ t 4:. z- Dr = t^ - t ^ l 5. recursive call, SPDE(t, t^ + t, t^ - t + 1, D, - 1 ) : a) —1 < 0 and t^ — t + 1 / 0, so return "no solution" Thus (6.5) has no solution in k[t], hence it has no solution in k{t). This implies that (6.4) has no solution in k{t), hence that ptan(a:)

tan(x)^

• dx

is not an elementary function. Example 6.4-2. Continuing examples 6.2.1 and 6.3.3, let k = Q{x) with D = d/dx^ t be a monomial over k satisfying Dt = t, i.e. t = e^^ and consider the solutions in k[t] of (6.10). We have a = t'^-\-2xt-\-x'^, b = c = t^/x'^ + [2/x-l)t and n = 0 from example 6.3.3, hence 1. g = gcd[a,h) = 1 2. (r, z) = ExtendedEuclidean(6, a, c) = (1,0) 3. 6 + Da = {2x'^ + l)t^/x'^ + (2x2 _|_ ^ ^ 2)t/x + 2x A. z~Dr = 0 5. recursive call, SPDE(t, b + Da, 0, D, - 2 ) : a) - 2 < 0 so return (0,0,0,0,0) ß,b = c — m = a = ß = 0

7. return (0,0,0,0,1) Thus any solution in k[t] of degree at most 0 of (6.10) must be of the form Oh -\-1 = 1 where Dh = 0. It follows that p = 1 is a solution of (6.10). Going back to example 6.2.1, this implies that q = 1/t is a solution of (6.9), hence that

Example 6.4-3. Continuing example 6.3.1, let k = Q(x,to) with D = d/dx, where to is a monomial over Q{x) satisfying Dto/to = l/x^, i.e. to = exp(—1/x), t be a monomial over k satisfying Dt = 1/x, i.e.t = log(x), and consider the solutions in k[t] of (6.15). We have a = t^, b = —t^jx^ — 1/x, X and n = 3 from example 6.3.1, hence

2x

1. ö' = gcd(a,6) = 1 2. (r, z) = ExtendedEuclideaii(6, a, c)

= f-x2i,(2:.-l)t2 + ^ ^ - ^ i ± i ^ V

X

2x

6.4 The SPDE Algorithm

b-{-Da =

205

^2 2 1 ^+ -t- -

X 2x 5. recursive call, SPDE(ai = t^, 6i = & + Da^ ci = z — Dr^ D, 1): a) g = gcd(ai,6i) = 1 b) {ri^zi)=

ExtendedEuciidean(6i,ai,ci) =: ( J:^+

2 '

2^2

c)

6i+Dai =

t^ 4 1 ^ + - t - •-,zi -i:^ri = 0

d) recursive call, SPDE(t2, -t^/x'^ + 4t/x - 1/x, 0, D, - 1 ) : _ i. - 1 < 0 so return (0,0,0,0,0) e) hi = ci = mi = ai = ßi = ^ _ f) return (0,0,0,0, x^ + to/2) 6.fe= c = m = a = 0, /3 = x 2 + to/2 7. return (0,0,0,0, (x^ + tQ/2)i^ - x^t) Thus any solution in k[t] of degree at most 3 of (6.15) must be of the form O/i + (x^ + to/2)t2 ~ x^t where Dh = 0. It follows that

is a solution of (6T5), hence that

I

-l/x

0(3.)ei/iog(x)+iA^^ = ((^--1+xA

log(x)2 - x^\og{x)\

gi/iosW+i/x

where 0(x) = (2x - 1) log(x)^ +

g-i/x_|_^ e"V^+4x^ ^-^ log(x) 2x

X ^oz{x)

Example 6.4-4- Let A: = Q and t be a monomial over k satisfying Dt = 1, i.e. D = d/dt^ and consider the solutions in k[t] of arbitrary degree n of (t^ + t + 1) i^g - (2t + l)g = ^t^ + ^t4 + t^ ^ t^ + 1.

(6.17)

We have a = t^ + t + 1, 5 = - 2 t - 1, and c = t^/2 + 3t4/4 + t^ - t^ + 1 so: 1. p = gcd(a,6) = 1

206

6 The Risch Differential Equation

2. (r, z) = ExtendedEuclidean(6, a, c) = (5t/4, t^/2 + tV4 + t/4 + 1) 3. 6 + Da = - 2 t - 1 + D(t2 + t + 1) = 0 4. recursive call, SPDE(t2 + t + 1, 0, t^/2 + t^/i + t/4 - 1/4, D, n - 2): a) ö' = gcd(t2 + t + 1, 0) = t^ + t + 1 b) ^ I c so a = 1, 6 = 0, c == t/2 - 1/4 c) deg(a) = 0, so return (6, c, m, a, /?) = (0, t/2 — 1/4, n — 2,1, 0) 5. return (0,t/2 - l / 4 , n - 2,t^ + t + l,5t/4) so any solution q G A;[t] of degree at most n of (6.17) must be of the form g = ( t ^ + t + l)/i + 5t/4 where h ^ k[t] has degree at most n — 2 and satisfies Dh=-t-~. 2

4

(6.18) ^ ^

6.5 T h e Non-Cancellation Cases We are now reduced to finding solutions in k[t] of the following equation: Dq-^bq==c

(6.19)

where 5, c G Ä;[t] and t is a monomial over k. Furthermore, we have an upper bound n on deg(g). We describe in this section an algorithm that can be used in any monomial extension whenever the leading terms of Dq and bq do not sum to 0. Sufficient conditions for this are either D = d/dt^ or deg(6) > max(0, 6{t) — 1), or t is nonlinear and either deg(ö) j^ Ö{t) — 1 or deg(6)/A(t) is not a negative integer. Since there is no cancellation between the leading terms of Dq and bq in those cases, we call them the non-cancellation cases. L e m m a 6.5.1. Let b^q G k[t] with q ^ 0. (i)

Suppose that b j^ 0. If D = d/dt or deg(6) > max(0, J(t) — 1), then the leading monomial of Dq + bq is lc(6)lc(g)t^^^-^^)+^"^(^) .

(a) If deg{q) > 0, deg(6) < 6(t) - I, and either 6{t) > 2 or D = d/dt, then the leading monomial of Dq + bq is deg(g)lc(g)A(t)t^^^"(^)+'^(*)-^ (Hi) If6{t) > 2, deg(6) = 3{t) - 1, deg(g) > 0 and deg(g) ^ -lc(5)/A(t), then the leading monomial of Dq + bq is (deg(g)A(t) +lc(ö))lc(g)t^^s-^^)+^(*)-i.

6.5 The Non-Cancellation Cases

207

Proof, (i) Suppose t h a t 6 ^ 0 . If D = d/dt, then deg{Dq) < deg(g) < deg(g) + deg(6) = deg{bq) so deg{Dq + bq) = deg{b) + deg{q) and the leading coefficient of Dq + bq is t h e leading coefficient of 6g, which is t h e product of the leading coefficients of b and q. If t is an arbitrary monomial and deg{b) > m = max(0, 5{t) — 1), then, by Lemma 3.4.2, deg{Dq) < d e g ( g ) + m , so deg{Dq) < deg(g) + deg(6) = deg{bq). Hence, deg(Dg+6g) = deg{bq) = deg(6)+deg(g) as previously, and t h e leading coefficient of Dq + bq is t h e leading coefficient of 6g, which is t h e product of the leading coefficients of b and q. (ii) Suppose t h a t deg(g) > 0, deg(ö) < 6{t) — 1, and either 6{t) > 2 or D = d/dt. If 5{t) > 2, then deg(Dg) = deg(g) + 6{t) - 1 by Lemma 3.4.2, so deg{Dq) > deg(g) + deg(6) = deg{bq). Hence, deg{Dq 4- bq) = deg(jDg) = deg{q)-i-5{t) — l^ and t h e leading coefficient of Dq-\-bq is t h e leading coefficient of Dq, which is deg(g)lc(g)A(t) by Lemma 3.4.2. If D = d/dt, t h e n ö{t) = 0, so deg(6) < 0 which implies t h a t 5 = 0, so deg{Dq-\-bq) = deg{Dq) = deg{q) — 1, and the leading coefficient of Dq + bq is t h e leading coefficient of Dq which is deg(g)lc(g)A(t) since X{t) = L (iii) Suppose t h a t 5{t) > 2, deg(6) = ö{t) — 1, deg(g) > 0 and deg(g) j^ - l c ( 6 ) / A ( t ) . Then, deg{Dq) = deg{q) + 5{t)--l by Lemma 3.4.2, so deg{Dq) = deg{bq). T h e leading coefficient of Dq is deg{q)lc{q)X{t) by L e m m a 3.4.2, and the leading coefficient of bq is lc(6)lc(g). Since deg{q)X{t)+lc{b) ^7^ 0 by hypothesis, we get t h a t the leading coefficient of Dq + bq is (deg(g)A(t) -h lc(5))lc(g) D and the degree of Dq -h bq is deg(g) + 6{t) — 1. Lemma 6.5.1 yields t h e following algorithms for finding t h e solutions of equation (6.19) whenever one of its hypotheses is satisfied. W h e n d e g ( 5 ) is L a r g e E n o u g h Suppose t h a t 6 7^ 0, and t h a t D = d/dt or deg(6) > max(0, ^(t) — 1). Then, for any solution q £k\i\\ {0} of Dq + bq = c, we must have deg(g) + deg(6) = deg(c), so deg(g) = deg(c) — deg(6) and lc(6)lc(g) = lc(c). This gives the leading monomial ut^ of any such g, and replacing q by uf^ + /i in (6.19), we get

Diuf") +Dh + but"" + bh = c so Dh^bh

=

c-D{ut'')-but''

which is an equation of t h e same type as (6.19) with t h e same b as before. Hence the hypotheses of part (i) of Lemma 6.5.1 are satisfied again, so we can repeat this process, but with a bound of n — 1 on deg(/i). This bound will decrease at every pass through this process, guaranteeing termination.

208

6 The Risch Differential Equation

P o I y R i s c h D E N o C a n c e I l ( 6 , c, D n)

(* Poly Risch d.e. - no cancellation *)

(* Given a derivation D on fc[t], n either Mi integer or +oo, and 6, c G k[t] with 6 7^ 0 and either D = d/dt or deg(6) > max(0,8{t) -1) , return either "no solution", in which case the equation Dq -\- bq = c has no solution of degree at most n in k[t], or a solution q G k\b] of this equation with deg(g) < n. *) q^O while c ^ 0 do m <— deg(c) — deg(6) if n < 0 or m < 0 or m > n t h e n r e t u r n "no solution" V ^ (lc(c)/lc(5)) t^ q^ q + p n ^- m — 1 c ^^ c— Dp — hp return q

Example 6.5.1. Let k = Q{x) with D = d/dx, and let t be a monomial over k satisfying Dt = 1 -^t^^ i.e. t = tan(a::), and consider the equation Dy + {t^ + 1)^ = t^ + (x + l)t^ + 1 + X + 2

(6.20)

which arises fom the integration of (tan(x)^ + (x + 1) tan(x)^ + tan(x) + x + 2) e*^"^^) . Theorem 6.1.2 gives h = 1, so any solution in k{t) must be in k{t). Lemma 6.2.1 shows that I't^-^iiy) > 0 for any solution, hence any solution in k{t) must be in k[t], so looking for solutions in k[t] of arbitrary degree we get: 6 = t^ + 1, •2, n = +00 and r ^{xi-l)r +t•

t

xr + x-\-l

so y = t -i- X is Si solution of (6.20), hence / (tan(x)^ + (x -h 1) tan(x)^ + tan(x) + x + 2)e*^"^^^dx = (tan(x) + x)e tan(a;) W h e n deg(6) is Small E n o u g h Suppose that deg(6) < 6{t) — 1 and either D = d/dt, which implies that 6 = 0, or 6{t) > 2. Let q e k[t] be a solution of Dq + bq = c. If deg(g) > 0, then deg(g) + ö{t) — 1 = deg(c), so deg(g) = deg(c) + 1 — 6{t) and deg(g)lc(g)A(t) = lc(c). This yields the leading monomial ut^ of g, and

6.5 The Non-Cancellation Cases

209

replacing q by uf^ + /i in the equation yields a similar equation with a lower degree bound on its solution. If g E /c, then: if 6 € fc*, then Dq + bq e k, so either c G /c, in which case we are reduced to solving a Risch differential equation of type (6.1) over A;, or deg(c) > 0 and (6.19) has no solution in k, hence in k[t]. If deg(6) > 0, then the leading term of Dq + bq is qlc{b)t^^^^^\ so either deg(c) = deg(6), in which case q — lc(c)/lc(5) is the only potential solution, or deg(c) ^ deg(6) and (6.19) has no solution in k^ hence in k\i\.

P o l y R i s c h D E N o C a n c e l 2 ( 6 , c, D, n)

(* Poly Risch d.e. - no cancellation

(* Given a derivation D on /c[t], n either an integer or +oo, and b^c £ k[t] with deg(6) < 5{t) — 1 and either D = d/dt or S{t) > 2, return either "no solution", in which case the equation Dq -\- bq = c has no solution of degree at most n in k[t], or a solution q G k[t] of this equation with deg(g) < n, or the tuple (/i,5o,co) such that h G k[t], bo,co G k, and for any solution q G k[t] of degree at most n of Dq -j- bq = c, y = q — h is a solution in k of Dy + 6oy = CQ. *) while c ^7^ 0 d o if n = 0 t h e n m -^r- Q else m ^- deg(c) — 6{t) + 1 if n < 0 or m < 0 or m > n t h e n r e t u r n "no solution" if m > 0 t h e n p <- (lc(c)/(m A(t))) t"" else if deg(5) ^ deg(c) t h e n r e t u r n "no solution" if deg(6) = 0 t h e n return(g, 6, c) p ^ lc(c)/lc(6) q ^q + p n <— m — 1 c <— c — Dp — bp return q

(* m = 0 *)

Example 6.5.2. Continuing example 6.4.4, let Ä: = Q, t be a monomial over k satisfying Dt = 1, i.e. D = d/dt, and consider the solutions h G k[t] of arbitrary degree n of (6.18). We get c = t/2 — 1/4, n = +oo and

m

P

q

n

c

2 t'/A t'/A 1 -1/4 1 - t / 4 tV4 - 1 / 4 0 0 SO h = t^/A — t/A is a solution of (6.18). Going back to example 6.4.4, this implies t h a t g = (t^ + t + 1) {\t^ - \t)-\-\t = \t^^t is a solution of (6.17). This example illustrates t h a t in the case 6 = 0, t h e algorithm P o l y R i s c h D E N o C a n c e l 2 is computing exactly an integral of c, taking into account the degree

210

6 The Risch Differential Equation

constraints. Using the integration algorithm for t h a t purpose would not be more efficient. W h e n ö{t) > 2 a n d d e g ( 6 ) = ö{t) - 1 In t h a t case, we have cancellation only when deg{q) = —lc(6)/A(t), which implies in particular t h a t —lc(6)/A(t) is a positive integer. Let q e k[t] he SL solution of Dq -\- bq = c. If deg(g) > 0 and deg(g) 7^ - l c ( 6 ) / A ( t ) , then deg(g) + 6{t) - 1 = deg(c), so deg{q) = d e g ( c ) + 1-(^(t) and (deg(g)A(t) + lc(6))lc(g) = lc(c). This yields t h e leading monomial uf^ of g, and replacing q by uf^ -i- h in the equation yields a similar equation with a lower degree bound on its solution. We can repeat this as long as the new degree bound is greater t h a n —lc(5)/A(t), or until we have a complete solution if —lc(5)/A(t) is not a positive integer. If q e k, then the leading t e r m of Dq + bq is qlc{b)t^^*'^~^^ so either deg(c) = (5(t) — 1 , in which case q = lc(c)/lc(6) is the only potential solution, or deg(c) ^ ^(t) — 1 and (6.19) has no solution in A:, hence in k[t].

P o l y R i s c h D E N o C a n c e l 3 ( 6 , c, D, n)

(* Poly Risch d.e. - no cancellation *)

(* Given a derivation D on k[t] with S(t) > 2, n either an integer or +00, and 6,c G k[t] with deg(5) = S(t) — 1, return either "no solution", in which case the equation Dq -j-bq = c has no solution of degree at most n in k[t], or a solution q G k[t] of this equation with deg(g) < n, or the tuple (h^m^c) such that h G k[t], m G Z, c e k[t], and for any solution q E k[t] of degree at most n of Dq -h bq = c, y = q -- h is a solution in k[t] of degree at most m of Dy -\-by = c, *) q<-0 if -lc(6)/A(t) G N t h e n M <- -~-lc(6)/A(t) else M ^ - 1 while c / 0 do m ^ max(M, deg(c) - 6{t) + 1) if n < 0 or ? n < 0 or m > n t h e n r e t u r n "no solution" u ^ mX{t) + lc(6) if li = 0 t h e n r e t u r n ( g , m, c) if m > 0 t h e n p ^— (lc(c)/u) f^ else if deg(c) ^ 6{t) — 1 t h e n r e t o r n "no solution" p ^ lc(c)/lc(6) q^r-q-^-p

n <~ m — 1 c <— c — Dp — bp return q

(* m -: 0 *)

6.6 The Cancellation Cases

211

Example 6.5.3. Let k = Q{x) with D = d/dx^ and let t be a monomial over k satisfying Dt = 1 + 1 ^ , i.e. t = tan(a;), and consider the equation Dy-h{l-

t)y = t^ +t^-

2xt - 2x

(6.21)

which arises from the integration of (tan(x)^ + tan(x)2 - 2xtan(x) - 2x) e^-iog(i+tan(a:)2)/2 Theorem 6.1.2 gives /i = 1, so any solution in k(t) must be in k{t). Since —6(\/^) = A / ^ — 1 and —2 is not the logarithmic derivative of an element of A:, Lemma 6.2.4 implies that ^'t2+i(y) ^ 0 for any solution, hence any solution in k(t) must be in A:[t], so looking for solutions in k\t] of arbitrary degree we get: 6 = 1 — t, c = t"^ + 1 ^ — 2xt — 2x, n = +oo, M = 1 and m u P q n c 2 1 t^ t^ 1 -2{x + l)t--2x 1 0 so any solution of (6.21) must be of the form y = t'^ -\- q where q £ k[t] is a. solution of degree at most 1 of Dq + i l - t)q = ~2{x + l)t - 2x .

(6.22)

6.6 The Cancellation Cases We finally study equation (6.19) whenever the non-cancellation cases do not hold, i.e. in one of the following cases: 1. 5{t) 2, deg(ö) = 5{t) - 1, and deg(g) = -lc(6)/A(t). We present in this sections algorithms for the above cases for specific types of monomials. T h e Primitive Case If Dt G /c, then 6{t) = 0, so the only cases not handled by Lemma 6.5.1 are b = 0 01 b e k*. If b = 0^ then (6.19) becomes Dq = c for c G k[t], which is an integration problem in k[t]^ and deciding whether it has a solution in k[t] can be done by the in-field integration algorithm (Sect. 5.12), so suppose now that b e k*. If 6 = Du/u for some u £ k* ^ which can also be checked by a variant of the integration algorithm (Sect. 5.12), then (6.19) becomes Dq-\- qDu/u = c, i.e. D{uq) — uc which is as earlier an integration problem in k[t].

212

6 The Risch Differential Equation

If h is not of the form Du/u for some u e k*^ then D(lc(g)) + blc{q) / 0, so the leading monomial of Dq + bq is (D(lc(g)) + &lc(g))t^®^^^\ This implies t h a t deg(g) = deg(c), and t h a t lc{q) is a solution in A;* of Dy + by = lc(c)

(6.23)

which is a Risch differential equation in k. If it has no solution in k^ then (6.19) has no solution in k[t]. Otherwise, Lemma 5.9.1 implies t h a t it has a unique solution which must then be lc(g). This gives the leading monomial yt^®^^^^ of any solution g, and as earher, replacing q by yt^^^^^) + /i in (6.19) yields an equation of the same type with a lower degree b o u n d on its solution, and a lower degree right hand side.

P o i y R i s c h D E C a n c e l P r i m ( 6 , c, D, n) (* Poly Risch d.e., cancellation - primitive case *) (* Given a derivation D on k[t], n either an integer or +oo, b ^ k and c ^ k[t] with Dt £ k and 6 ^ 0 , return either "no solution", in which case the equation Dq -\- bq = c has no solution of degree at most n in k[t]^ or a solution q E k[t] of this equation with deg(q) < n. *) if 6 = Dz/z for z ^ k* t h e n if zc = Dp for p e k[t] and deg(p) < n t h e n return(p/2;) else r e t u r n "no solution" if c = 0 t h e n r e t u r n 0 if n < deg(c) t h e n r e t u r n "no solution" while c 7^ 0 d o m <— deg(c) if n < m t h e n r e t u r n "no solution" 5 ^ RischDE(6, lc(c)) {^ Ds + bs = lc(c) *) if s = "no solution" t h e n r e t u r n "no solution" q^q + sf^ n -^r— m — 1 c^ c~ bsf^ — D{sf^) (* deg(c) becomes smaller *) return q

The Hyperexponential Case If Dt/t = T] E k^ t h e n 6{t) = 1, so the only cases not handled by Lemma 6.5.1 are 5 = 0 or 5 G fc*. If 6 = 0, then (6.19) becomes Dq = c for c E k[t]^ which is an integration problem in k[t]^ and deciding whether it has a solution in k[t] can be done by a variant of the integration algorithm (Sect. 5.12), so suppose now t h a t b e k*.

6.6 The Cancellation Cases

213

If 6 = Du/u + mrj for some u G k* and m G Z^ then (6.19) becomes Dq-\-{Du/u-}-mr])q = c, i.e. D{uqV^) = ucV^ which is an integration problem in k{t), and deciding whether it has a solution in kif) can be done by a variant of the integration algorithm (Sect. 5.12). Suppose finally t h a t h is not of the form Du/u + mr] for some u E k'^ and m G Z. T h e n D(lc(g)) + deg(g) ?7lc(g) + 61c(g) j ^ 0, so the leading monomial of Dq^bq is (D(lc(g)) + deg((7)77lc(g) + 61c(g))t^^g(^). This implies t h a t deg((7) = deg(c), and t h a t lc(g') is a solution in /c* of Dy + (6 + deg(g) r])y = lc(c)

(6.24)

which is a Risch differential equation in k. If it has no solution in k^ then (6.19) has no solution in k[t]. Otherwise, Lemma 5.9.1 implies t h a t it has a unique solution which must then be lc(g). This gives the leading monomial yt^®^^^^ of any solution g, and as earUer, replacing q by yt^^^^^^ + /i in (6.19) yields an equation of the same type with a lower degree bound on its solution, and a lower degree right hand side.

P o l y R i s c h D E C a n c e l E x p ( 5 , c, D, n) (* Poly Risch d.e., cancellation - hyperexponential case *) (* Given a derivation D on k[t]^ n either an integer or +00, h G k and c G k[t] with Dt/t G k and 6 ^ 0 , return either "no solution", in which case the equation Dq -\-bq = c has no solution of degree at most n in k[t], or a solution q G k[t] of this equation with deg(g) < n. *) if 6 = Dz/z + mDt/t for z G k* and m EZ t h e n if czt"^ = Dp for p ek{t) and q = p/izt"^) G k[t] and deg(g) < n t h e n Feturn(g) else r e t u r n "no solution" if c = 0 t h e n r e t u r n 0 if n < deg(c) t h e n r e t u r n "no solution" g <- 0 while c^ 0 do m <— deg(c) if n < m t h e n r e t u r n "no solution" s ^ RischDE(6 + mDt/t, lc(c)) (* Ds + {b-i- mDt/t)s = lc(c) *) if s = "no solution" t h e n r e t u r n "no solution" q^q + st"^ n ^~ m — 1 c ^ c - bst"^ - Dist"^) (* deg(c) becomes smaller *) return q

214

6 The Risch Differential Equation

T h e Nonlinear Case If 6{t) > 2, then we must have deg{b) = 6{t) — 1 and lc(6) = —n\{t) where n > 0 is the bound on deg(g). There is no general algorithm for solving equation (6.19) in this case. If however <S^^^ ^ 0, then the following can be done: for p G ^S^^^, applying TTp to (6.19) and using the fact that D*o7Tp = iTpoD where D* is the induced derivation on k[t]/{p) (Theorem 4.2.1), we get D''q*+TTp{b)q* =7Tp{c)

(6.25)

where q* = 7Tp{q). Assuming that we have an algorithm for solving (6.25) in k[t]/{p), we can then solve (6.19) as follows: if (6.25) has no solution in k[t]/{p)^ then (6.19) has no solution in k[t]. Otherwise, let g* G k[t]/{p) be a solution of (6.25), and let r £ k[t] be such that deg(r) < deg(p) and TTp{r) = g*. Note that TTp{Dr + hr) = 7rp(c), so p | c — Dr — br. In addition, 7Tp{q) = 7rp(r), so h = (^q — r)/p £ k[t] and we have

c = Dq + bq=p(Dh+

(b+—]h]

-{-Dr + br

so /i is a solution in k[t] of degree at most deg(g) — deg(p) of Dh^(b+^)h='-^'-^' \ P J P

(6.26)

which is an equation of type (6.19), but with a lower bound on the degree of its solution. There are cases when (6.25) can be solved, for example if there exists p £ S^^^ with deg(p) = 1. Then, k[t]/{p) :^ k^ so (6.25) is a Risch differential equation in k. Another possibility is if S^^^ n Const(Ä:)[t] ^ 0, in which case taking p = t — a where a is a constant root of an irreducible special, we get k[t]/{p) c::L k{a)^ so (6.25) is a Risch differential equation in k{a). This is the case when t is an hypertangent monomial with a = ±^/^. Taking p = t — a can also be done with a not constant, but (6.25) is then a Risch differential equation in a nonconstant algebraic extension of A:(t), and no algorithms are known for such curves when t is a nonlinear monomial. Although the techniques of [11, 73] are probably generalizable to such curves, they would not yield a practical algorithm in their current form. T h e H y p e r t a n g e n t Case If Dt/it? -\- 1) = T] £ k^ then 5{t) = 2, so the only case not handled by Lemma 6.5.1 is 6 = 6o — niqt where bo £ k and n > 0 is the bound on deg(g). In such extensions, the method outlined above provides a complete algorithm: if / = 4 £ k, then S''' = {t - A/=T,^ + v ^ } , and (6.25) is simply a Risch differential equation over k.

6.6 The Cancellation Cases

215

If y ^ ^ k, then taking p = t"^ + 1 £ 5^^^ (6.25) becomes i^g* + (6o - n 7 ] v ^ ) g * = c ( v ^ )

(6.27)

where D is extended to k[t]/{p) :^ k{y/^) by Dy/^ = 0. One possibility is to view (6.27) as a Risch differential equation in fc(\/^) and to solve it recursively. If it has no solution in ^ ( \ / ^ ) , then (6.19) has no solution in k[t]. Otherwise, if u-\- vy^^ is a solution of (6.27) with u^v E k^ t h e n letting r = u-^vt^ h = {q — r)/p is a solution in k[t] of degree at most n — 2 of (6.26). It is also possible however t o avoid introducing V ^ by considering the real and imaginary parts of (6.27): writing q* = u -{- vy^^^^ we get

where CQ -{- Cit is the remainder of c by t^ + 1. This is the coupled differential system introduced in Sect. 5.10. If it has no solution in k, then (6.19) has no solution in k[t]. Otherwise, if (u^v) € A;^ is a solution of (6.28), then letting r = u -\- vt^ h = [q — r)/p is a solution in Ä:[t] of degree at most n — 2 of (6.26).

PolyRischDECancelTan(6o, c, D, n) (* Poly Risch d.e., degenerate cancellation - tangent case *) (* Given a derivation D on /c[t], n G Z, 6o G A: and c £ k[t] with Dt/{t^ + 1) G /c, ^/—l ^ k and n > 0, return either "no solution", in which case the equation Dq + {ho — ntDt/(t^ -\- \))q = c has no solution of degree at most n in k[t]^ or a solution q G k\t] of this equation with deg(g) < n. *) if n = 0 t h e n if c G A: t h e n if 6o 7^ 0 t h e n r e t u r n RischDE(6o, c) else if / c = g Gfct h e n return(g) else r e t u r n "no solution" else r e t u r n "no solution" p <-- t^ + 1 (* the monic irreducible special polynomial *) 7] <— Dt/p

(^ t = t a n ( / 77) *)

(c, cit + Co) ^- PolyDivide(c,p) (* c{y/~^) = c i \ / ^ - | - co *) (* CoupledDESystem will be given in Chap. 8 *) {u,v) <— CoupledDESysteni(6o, —n77,co,ci) (* Du -f bou + nriv = co, Dv — niqu + hov = ci *) if (u^v) = "no solution" t h e n r e t u r n "no solution" if n == 1 t h e n return(w + vt) r = u-\- vt (* this division is always exact *) c ^- (c— Dr — (bo — nrjt)r)/p h ^ PolyRischDECancelTan(6o, c,D,n~- 2) if h = "no solution" t h e n r e t u r n "no solution" retnrn{ph + r)

216

6 The Risch Differential Equation

Example 6.6.1. Continuing example 6.5.3, let k = Q(x) with D = d/dx^ t be a monomial over k satisfying Dt = 1 -\-1"^^ i.e. t = tan(x), and consider the solutions q G k[t] of degree at most 1 of (6.22). We have 6 = 1 — t, 6o = 1, c = —2{x + l)t — 2x, n = 1 and: l . p = t2 + l 2. 77 = Dt/p = 1 3. (c, cit + Co) = PolyDivlde(-2(x + l)t - 2x,p) = (0, -2{x + l)t - 2x) 4. Since

has the solution t^ = 0 and v = — 2x, (ti, v) = C o u p l e d D E S y s t e m ( l , - 1 , -2x, -2{x + 1)) = (0, -2x). Thus, q = —2xt is a solution of degree at most 1 of (6.22). Going back to example 6.5.3, this implies that y = t'^ — 2xt is a solution of (6.21), hence that / (tan(x)^ + t^ii{xf

- 2xtan(x) - 2x) e^-iog(i+tan(x)2)/2 ^^ _ (tan(x)2 - 2xtan(x)) e--i°g(i+*-"(-)')/2 .

Note that the above can also be written as /'

(tan(x)"^ + tan(a:)^ — 2xtan(a::) — 2x) co^{x)e^dx = (tan(a:)'^ — 2xtan(x)) cos(x)e^ .

Exercises Exercise 6.1. Prove the following analogue of Lemma 6.2.4 for fields containing A / ^ : let A: be a diflFerential field of characteristic 0 containing \ / ^ , t be a hypertangent over k such that 7 / \ / ^ is not the logarithmic derivative of a /c-radical, where 77 = Dt/{t^ + 1) G /c Let a € Ä:[t], 6, g G A:(t) be such that gcd(a, t^ + 1) = 1, i/^_y3i(^) = z^t+v'^(^) ~ ^' ^^^ e = ±.1 and suppose that ^t-e^f^i^) 7^ 0- Then, either Pt^,^fzi{aDq + hq) = Ut_^^y^{q) or

6(eV^) a(ev—1) for some u G k*.

^

Du W '

Parametric Problems

We describe in this chapter solutions to several integration-related problems involving parameters. Those problems arise as subproblems in the integration algorithm: the limited integration problem, which arises from integrating polynomials in a primitive extension (Sect. 5.8) and the parametric logarithmic derivative problem, which arises from recognizing logarithmic derivatives (Sect. 5.12) and from bounding orders and degrees of solutions of the Risch differential equation (Sects. 6.2 and 6.3). The common thread between those problems is that they ask whether there exists constants for which a given parametric differential equation has a solution in a given differential field.

7.1 The Parametric Risch Differential Equation We present first the classical parametric problem, namely the parametric Risch differential equation, which is a Risch differential equation where we replace the right hand side g e K hj the linear combination YALI ^i9i ^i^^ Qi ^ K. The problem is then to determine all the constants Q G Const(ÜT) for which the equation m

Dy + fy = Y.''^9i

(7-1)

i=l

has a solution in ÜT, and of course to compute such solutions. This problem, which does not arise when we integrate only transcendental elementary functions, shows up in the integration of nonelementary functions, or in integration in terms of nonelementary functions [6, 21, 22, 52, 53]. In addition, the problem of limited integration can be seen as a special case of this problem. Note that the set of constants ( c i , . . . , c^) for which (7.1) has a solution in K forms a linear subspace of Const(if)'^. This motivates the following formal definition of the parametric Risch differential equation problem: given a differential field K of characteristic 0 and / , ^ i , . . . , gm G K^ to compute / i i , . . . , /i^ € if, a matrix A with m ~{- r columns and entries in Const(Ü") such that (7.1) has

218

7 Parametric Problems

a solution c i , . . . , C772 £ Const(ÜT) and y e K if and only if y = Ylj=i

^j^j

where < i i , . . . , (i^ G Const(i^) and A{ci,..., c ^ , ( i i , . . . , ^^^i
de /dt)

gcd(c, dc/dt)

^ ^ ^

Then, yh E k{t). Proof. Let q = yh £ k{t). In order t o show t h a t q £ fc(t), we need t o show t h a t ^P{Q) > 0 foi" any normal irreducible p £ k[t]. We have i^p(g) = Up{y)-^iyp{h) by Theorem 4.1.1. If z^p(y) > 0, t h e n z^p(^) > I'pih) > 0 since h £ k[t]. So suppose now t h a t n = iyp{y) < 0. Let g = J^^i <^i9i ^ Ht) and fi = mini /i. In addition, eg £ k[t] since e is a least common multiple of the denominators of t h e ^r^'s, so z^p(e) + Typ{g) = yp{eg) > 0, which implies t h a t Up{e) > —h'p{g). Case 1: Up{f) > 0. Then, i'p{Dy + fy) = Vp{y) - 1 by Lemma 6.1.1. Since g = Dy + fy, this implies t h a t i^pig) < 0, hence t h a t p \ e. Since p is normal, gcd(p, Cs) = 1, so Up{en) = i^p{e) > -yp{g) = l - n . Also, p does not divide d since i'p{f) > 0, so i/p(c) = 0, so i/p(gcd(c, dc/dt)) = 0. Hence iyp{h) = h'p{gcd{en^den/dt)) = Up{en) — 1 > - n , so yp{q) = n + ^p(^) > n — n = 0. Case 2: i'p{f) < 0. Then, Vp{g) = Vp{Dy + fy) = iyp{f) + n by L e m m a 6.1.1, so n = ^p{g) — ^p[f)' Since n < 0, this implies t h a t Vp{g) < ^p{f) < 0, hence t h a t p I d and p | e. As above, since p is normal, gcd(p, «i^) == gcd(p, e^) = 1, so Up{dn) = -i^pif) < --^p{9) > ^pien)' Thus, Up{c) = min{up{dn),jyp{en)) = i^p{dn) = -yp{f) > 0, so

7.1 The Parametric Risch Differential Equation Up{h) = Up{gcd{en,den/dt))

-

219

Vp{gcd{c,dc/dt))

= (z^p(en) - 1) - (^p(c) - 1) z/p(e) + Vp{f) > -iyp{g) + iyp{f)

-n,

so i^p(g) = n + z^p(/i) > n — n = 0.

D

gi^...^gm C o r o l l a r y 7 . 1 . 1 . Let / € k{t) be weakly normalized w.r.t. t, in k{t), and d^^en and h he as in Theorem 7.1.1. Then, for any solution c i , . . . , c ^ £ Const(fe) and y G k{t) of Dy + fy = YlZ^i ^i9i, q = yh G k{t) and q is a solution of m

dnhDq + {dnhf - d^Dh) q = Y^ Ci{dnh'^gi) •

(7.2)

i=l

Conversely, for any solution c i , . . . , c ^ G Const(A:) and q E k{t) y = q/h is a solution of Dy -\- fy = Y^^i ^i9i-

of

(7.2),

Proof. Let c i , . . . , c ^ G Const(A:) and y G k{t) be a solution of Dy -\- fy = g^ and let q — yh. q G k{t) by Theorem 7.1.1, and Dq+(f~~ ^

—

]q = hDy + yDh^

hfy - yDh = h{Dy + fy) =

/

hj^agi. 7= 1

Multiplying through by dnh yields dnhDq+{dnhf — dnDh)q = dnh'^ Y^iLi ^i9i^ so g is a solution of (7.2). Conversely, the same calculation shows t h a t for any solution c i , . . . , c ^ G Const(/c) and q e k{t) of (7.2), y = q/h is a solution of Dy + fy=-YT=i^i9i' n The above theorem and corollary give us an algorithm t h a t reduces a given parametric Risch differential equation to one over k{t).

P a r a i i i R d e N o r m a I D e n o m m a t o r ( / , 5 ^ 1 , . . . ^Qm^D) (* Normal part of the denominator *) G k{t) with / weakly (* Given a derivation D on k[t] and f^gi,...,gm normalized with respect to t, return the tuple (a,5, G^i,. . . , Gm^h) such that a,h e k[t], b G k{t), Gi,...,Gm G k{t), and for any solution c i , . . . ,Cm G Const(/c) and y G k{t) oiDy + fy^ YH=I ^^^^' Q = yh G k{t) satisfies aDq i- bq = ^^iCiGi. *) (dn,ds) <— SplitFactor(denominator(/), JD) (en, es) <— SplitFactor(lcm(denominator(c?i),..., denominator(prn)),!)) p ^ gcd{dn,en) h ^ gcd{en,den/dt)/gcd{p,dp/dt) Teturn{dnh,dnhf — dnDh.dnh^gi, • • • ,dnh'^gm,h)

220

7 Parametric Problems

T h e Special P a r t of t h e D e n o m i n a t o r As a result of Corollary 7.1.1, we are now reduced to finding solutions c i , . . . , c^ G Const(/c) and g G k{t) of (7.2), which we rewrite as m

aDq -^bq = ^

CiQi

(7.3)

i=l

where a G k[t] has no special factor, b G k{t)^ gi^...^gm G k{t)^ a / 0, and t is a monomial over k. Since ^p{Yl^i^i9i) ^ ^^i min(0,mini 0 and p G <Sf, then z/p(g) > m.iii{0,mini n for some n < 0, replacing q by /ip"" in (7.3) yields a{p''Dh + np^'-^hDp) + fe/iK = X]c^^i i=l

hence

aDh -f ( 6 + n a ^ J ^ = ^^i

{diP"'') •

('^•4)

Furthermore, h G A:(t) since g G fc(t), and h G Op since i^p(g) > n. Thus we are reduced to finding the solutions c i , . . . , c^ G Const(fc) and h G k{t) H Op of (7.4). Note that b + naDp/p G A:(t) since 6 G Ä:(t), a G Ä:[t] and p £ S. The eventual power of p in the denominator of 5 + naDp/p can be cleared by multiplying (6.7) by p^ where N = max(0, ~z/p(6)), ensuring that the coefficients of the left hand side of (7.4) are also in k{t) H Op. Since all the special polynomials are of the first kind in the monomial extensions we are considering in this section, we only have to find a lower bound for i^p(g) in the potential cancellation case, i.e. Ppib) = 0. We consider this case separately for various kinds of monomial extensions. T h e P r i m i t i v e Case If Dt G k., then every squarefree polynomial is normal, so k{t) = k\t\., which means that a^b £ k\t\ and any solution in g G fc(t) of (7.3) must be in k\t]. T h e H y p e r e x p o n e n t i a l Case If Dt/t = T] £ k^ then k{t) = fc[t,t~-^], so we need to compute a lower bound on i^tiQ) where c i , . . . , c ^ G Const(fc) and q G k{t) is a solution of (7.3).

7.1 The Parametric Risch Differential Equation

221

Since t G Sf^ by Theorem 5.1.2, Lemmas 6.2.1 and 6.2.3 always provide a lower bound for z^t(g): if z^t(6) ^ 0, then L e m m a 6.2.1 provides the bound as explained earlier. Otherwise, Utib) = 0, so either —&(0)/a(0) = srj-\-Du/u for some 5 G Z and u £ k"^, in which case ut{q) > min

0,5, min

{Pt{gi))

l
Otherwise, ut{q) > min

0, min {iyt{gi)) l
Note t h a t such an s is unique by L e m m a 6.2.2 applied to k. Since S^^^ = { t } , k{t) D Of — k[t], so having determined a lower bound for i't{q), we are left with finding solutions c i , . . . , c ^ G Const(/c) and h G k[t] of (7.4).

P a r a m R d e S p e c i a l D e n o m E x p ( a , b , g i ^ . . . ,gm,D) (* Special part of the denominator - hyperexponential case *) (* Given a derivation D on k[t] and a G k[t], b G k{t) and p i , . . . ,pm G k{t) with i9t/t G /c, a y^ 0 and gcd(a,t) = 1, return the tuple (ä, 6 , ^ , . . . , p ^ , /i) such that ä,b,h G fc[t], pT,... ,'g^ G /c(t), and for any solution c i , . . . ,Cm G Const(/c) and g G k{t) of aDg + 6g = X^I^i CiQi, r = qh E k[t] satisfies äDr -\-br = YlTLi ^i9i- *) p <— t (* the monic irreducible special polynomial *) m ^ ^p{b), He ^ mini 0 , for clearing denominators *) r e t u m ( a p ^ , (6 + naDp/p)p''^gip""-^,... ,p^p^-"-,p-")

T h e Hypertangent Case If Dt/{t^ -i-l) = T] £ k and \ / ^ ^ fc, t h e n the only monic special irreducible is t^ -f 1, so we need to compute a lower bound on z^t2+i( min

0 , s , min l<^<m

{ut^^iiOi))

222

7 Parametric Problems

Otherwise, ^t^+i{q) > min ( 0 y

min {i^t^-^i{gi)) l<*<7n

Note t h a t such an s is unique by Lemma 6.2.2 appHed t o fc(\/^). The remarks made in the nonparametric case about adjoining \ / ^ temporarily remain valid in this case. Since S^^^ = {t^ + 1}, A:(t) fl C^t^+i = k[t]^ so having determined a lower bound for z^t2+i(^), we are left with finding solutions c i , . . . , c ^ € Const(Ä:) and h G k[t] of (7.4).

P a r a m R d e S p e c i a l D e n o m T a n ( a , 6, ^ i , . . . , prn, -Ö) (* Special part of the denominator - hypertangent case *) (* Given a derivation D on k[t] and a E k[t],b ^ k{t) a n d p i , . . .,prn € k{t) with Dt/{t^ + 1) ek, x / ^ ^ /c, a 7^ 0 and gcd(a,t^ + 1) = 1, return the tuple (ä, 6, ^ , . . . , öv^, /i) such that ä,b,h £ k[t], 'gi,. .. , ^ G/c(t), and for any solution c i , . . . , c^ G Const(A:) and g G k(t) of aDg + 5g = X^^^ CiQi, r = qh £ k[t] satisfies äDr -\-br = YlT^i ^^Qi- *) p <- t^ + 1 (* the monic irreducible ;special polynomial *) rih ^ ^p{b), Tic ^ miiii 0, for clearing denominators *) r e t u r n ( a p ^ , (b + naDp/p)p^, gip^""",..., gmP^"'' .p-n

T h e Linear Constraints on t h e Constants As a result of the previous paragraphs, we are now reduced to finding solutions c i , . . . , c ^ G Const(/c) and g G /c[t] of (7.4), which we rewrite as: aDq -i-bq = CiQi + . . . -f

(7.5)

where a^b £ k\t]^ 9i- - - ^gm ^ k(t)^ a 7^ 0, and t is a monomial over k. In addition, dividing (7.5) by gcd(a, b) if needed, we can assume without loss of generality t h a t gcd(a, 6) = 1. We show t h a t if some gi is not in k[t]^ then we can obtain linear constraints on t h e Q ' S , and reduce (7.5) to a similar equation with the right hand side in k[t].

7.1 The Parametric Risch Differential Equation

223

L e m m a 7 . 1 . 1 . Let a^h.q G k\t\, gi,... ^Qm G k{t) and c i , . . . , c ^ € Const(A:) be such that aDq + bq = cigi + . . . + Cmgm- Let di be the denominator of gi for 1 < i < m, d = l c n i ( ( i i , . . . , d^), cind g i , . . . , qm-,'^ii • • • 5^m ^^ such that dgi = dqi + r^ and either ri = 0 or deg(r^) < deg{d) for each i. Then, (7.6)

y ^ Ciri = 0

and

(7.7)

aDq -\-hq = ciqi + . . . + Cmqm •

Proof. Since gi — qi ^ ri/d for each i, we obtain from (7.5) t h a t \--\m

m

^ ^ ^ ^ - ^ = aDq + bq -J^c^q^ G k[t] . i-l

Since deg (X^I^i ^*^*) < deg((i), it follows t h a t Yl^i hence t h a t aDq -\- bq = Yl^i ^i^i-

^i^i must be equal to 0, ^

Equating the coefficients of the powers of t on b o t h sides of (7.6) yields a homogeneous system of linear equations for the Q ' S , i.e. a matrix M with coefficients in k such t h a t M

C2

(7.8)

0.

L i n e a r C o n s t r a i n t s ( a , 6, ^ i , . . . , gm, i^) (* Generate linear constraints on the constants *) (* Given a derivation D on k(t), a^b G k[t] and gi,... ,gm G fc(t), return 9i,." •, ^m G fc[t] and a matrix M with entries in k{t) such that for any solution G Const(fc) andp Gfc[t]of aDp-\-bp = cig\-\-.. .+c mgm 5

( c i , . . . , Cm) is a solution of Mx = 0, and p and the Q satisfy aDp + bp = Ciqi + .. . + Cmgm. *)

d 4- lcm(denominator(pi),..., denominator(ö'rn)) for i ^— 1 t o m d o {qi^ri) <- PolyDivide(d^i,c^) (* dgi = qid-\-ri *) if ri = ... ~ rm = ^ t h e n A^ = —1 else A'" <— max(deg(ri),.. ., deg(rm)) for i <-- 0 t o A/ d o for j ^— 1 t o m d o Mij <— coefficient(rj,t*) r e t u r n ( g i , . . . ,qm,M)

224

7 Parametric Problems

Example 7.1.1. Let /c = Q, t be a monomial over k satisfying Dt = 1, i.e. D = d/dt^ and consider the equation 2t3 + 3t + l 1 , 1 i^F = c , - ^ , - - ^ + c . ^ 4 - C 3 ^ .

,^^. (7.9)

We have a = 1, 5 = 0, ^i = (2t3 + 3t + l ) / ( t 2 - l ) , ^2 = 1/(^-1), Qs = 1/(^ + 1) and: l.d = lcm(t2 - 1, t - 1, t + 1) = t^ - 1 2. rf^i = 2t^ + 3t + 1 = 2td + 5t + 1, d^2 = * + 1, dgs = t - 1, so gi = 2t, ^2 = ^3 = 0, ^1 = 5t + 1, r2 = t 4-1 and ra = t - 1. 3. Equation (7.6) becomes ci{5t-\-1) 4-C2(t + 1) + C3(t — 1) = 0, which yields the linear system C2 I = 0

(7.10)

which has the solution space (ci, C2, C3) = (A, —3A, —2A) for any A G Q. 4. Replacing ci, C2 and C3 by the above solution in (7.9) yields

which is now a parametric Risch differential equation with polynomial right-hand side. Since we are interested only in the constant solutions of (7.8), we need to reduce it to an equivalent system with coefficients in Const(fc). An algorithm for this reduction is provided in the following lemma. L e m m a 7,1.2. Let (K^D) be a differential fields A be a matrix with coefficients in K, and n be a vector with coefficients in K. Then, using only elementary row operations on A and u, we can either prove that Ax = u has no constant solution, or we can compute a matrix B and a vector v, both with coefficients in Const(Ü"); such that the constant solutions of Ax = u are exactly all the constant solutions of Bx = v. Furthermore, if n = 0, then v = 0. Proof Let C = Const(iC), and write Ri for the i^^ row of A, and aij for the j ^ ^ entry of Ri. By applying the usual Gaussian elimination, we can compute an equivalent system in row-reduced echelon form, so suppose that A is in that form. If all the entries of A are in C, let B = A and v = u. Otherwise, let j be the smallest index such that the j * ^ column of A has a non-constant entry, and let i be such that aij ^ C. Then, Doij / 0, so we add the row Rm+l

DRi

f Dan

Düij

\ Düij

Da. Doij

7.1 The Parametric Risch Differential Equation

225

at the b o t t o m of A, and the entry tXm+i = Dui/Daij at the b o t t o m of u . By our choice of j , the first nonzero entry in jRm+i is a 1 in column j , so we add adequate multiples of Rm^i to all t h e other rows to ensure t h a t a^j = 0 for i — 1 . . . m. We now have a new m a t r i x A and a new vector ü with one more row, b u t with only constant entries in columns 1 through j . Repeating this, we eventually obtain a matrix B and a vector v such t h a t all the entries of B are in C By construction, v = 0 if u = 0. Since we have only added extra rows to A and performed elementary row operations to A, any solution of Bx = V must be a solution of Ax = u. Conversely, let x be a constant solution of Ax = u. In order for x to satisfy Bx = v, it only has to satisfy Rm-^ix = tim+i, where Äm+i is the extra row added in the reduction step. But, _ {Daii)xi

+ . . . + {Dair)xr

__ D [anxi

Düij

+ . . . + airXr) Düij

_ D{Rix)

__ Dui

Düij

Düij

so X is a constant solution of Bx = v, which implies t h a t the constant solutions of Ax = u are exactly the constant solutions of Bx = v. Note t h a t if v has a nonconstant entry, then Bx = v has no constant solution, implying t h a t Ax = u has no constant solution. D

C o n s t a i i t S y s t e m ( M , u, D) (* Generate a system for the constant solutions *) (* Given a differential field (K, D) with constant field C, a matrix A and a vector u with coefficients in if, returns a matrix B with coefficients in C and a vector v such that either v has coefficients in C, in which case the solutions in C of Ax = u are exactly all the solutions of Bx = v, or V has a nonconstant coefficient, in which case Ax = u has no constant solution. *) {A, u) ^- R o w E c h e l o n ( ^ , u) m 4— number of rows of A while A is not constant d o j ^- minimal index such that the j ^ ^ column of A is not constant i ^- any index such that aij ^ C, Ri 4- i^^ row of A D{Ri)/D{aij),u m-f-1 ^~" D{u i)/D{ai_^•) Rm +i = for s 4— 1 t o ?Ti d o ^

J^s

Us ^

its

Us

A^ - A\jRm+l,VL retiirn(A, u)

OJSJJ^m-\-l

Ö'Sj^m-l-l

<- UU Um+1

(* vertical concatenation *) (* ve

7 Parametric Problems

226

Example 7.1.2. Let k

X, and consider the system (7.12)

and u

Ax = u where A • 1. RowEchelon(yl, u) yields A:

and

3,i = 2,i?2 = (0

1

x-1 x-i-l

3. lid

DR. = (0 0 D ( ( x - l ) / ( x + l))

1),U4=

Dli2

D{{x-l)/{x

+ l))

0.

4. Adding (i?4,i^4) to (A, u) yields

A:

0

and

0

Vo 0 5. Finally, adding ~{x — 1)/{x -\~ 1)R4 to R2 yields (I A--

0 0

0^ 0 O X 1 0 0 0

and

Vo 0 1/ which both have constant entries. The above constant system has the unique solution

which is thus the unique constant solution of (7.12). Note that (7.12) has a one-dimensional affine space of solutions over Q(x), namely

+w

for any w € Q(x).

Using Lemmas 7.1.1 and 7.1.2, we can produce a constant homogeneous linear system for the Q'S. If its kernel has dimension 0, then the only solution of 7„5 is g = ci = . . . = c^ = 0 . Otherwise, a basis of its kernel allows us to express some of the Q'S in terms of others, thereby decreasing m and reducing the problem to solving equation (7.7).

7.1 The Parametric Risch Differential Equation

227

Degree B o u n d s As a result of Lemma 7.1.1, we are now reduced to finding solutions c i , . . . , Cm in Const(/c) and q G k[t] of (7.7) where a,5,gi,... ^qm G k[t]^ a / 0, and t is a monomial over k. Since deg(^^-j^ c^g^) < maxi deg(a) + max(0, ö{t) - 1), then deg(g) < max(0, max (deg(gi)) - deg(6)). l
(ii) If deg(6) < deg(a) + ö{t) - 1 and 6{t) > 2, then deg(g) < max(0, max {deg{qi)) — deg(a) + 1 — 6{t)). l
As a result, we only have to consider the cases deg(6) < deg(a) for Louvillian monomials, and deg(5) = deg(a) + 5{t) — 1 for nonlinear monomials. We consider those cases separately for various kinds of monomial extensions. T h e Primitive Case If Dt = T] E k^ then Lemmas 6.3.1 and 6.3.3 always provide an upper bound for deg(g) as in the nonparametric case: if deg(6) > deg(a), then Lemma 6.3.1 implies that deg(g) > max(0, max (deg(gi)) — deg(6)). l<2<m

If deg(6) < deg(a) — 1, then Lemma 6.3.3 implies that deg(g) > max(0, max (deg(gi)) - deg(a) + 1). l
If deg(6) = deg(a) — 1, then either —lc(6)/lc(a) = srj -}- Du for some 5 G Z and u G k^ in which case deg(g) > max(0, 5, max (deg(gi)) - deg(a) + 1), l
or

deg(g) > max(0, max (deg(gi)) - deg(a) + 1). l<'i<m

Note that such an s is unique by Lemma 6.3.2. Finally, if deg(6) = deg(a), then either

JM-£^

and - M ^ ^ ^ = . , + z^.

lc(a) u u[c{a) for some u G k*^ v E k and s G Z, in which case deg(g) > max(0, s, max (deg(g^)) - deg(a) + 1), l<'i<m

228

7 Parametric Problems

or deg(g) > max(0, max (deg(g^)) — deg(a) + 1 ) . l
We can compute such a u by a variant of the integration algorithm (Sect. 5.12). As explained in t h e nonparametric case, although u is not unique, Lemmas 5.12.1 imply t h a t the choice of u does not affect lc{aDu + bu)/{ulc{a)), so s is unique by Lemma 6.3.2.

P a r a m R d e B o u n d D e g r e e P r i i i i ( a , b , q i , . . . ,qm,D) (* Bound on polynomial solutions - primitive case *) (* Given a derivation D on k[t] and a,h,qi,.. . ,qm ^ k[t] with Dt e k and a y^ 0, return n G 1^ such that deg(q) < n for any solution c i , . . . , Cm E Const (A:) and q ^ k[t] of aDq + bq = Y^^i Ciqi. *) da ^ deg(a), 4 ^ deg(6), dc ^ maxi da t h e n n ^— max(0, dc — db) else n ^— max(0, dc ~ da ~\- 1) if db — da — 1 t h e n (* possible cancellation *) a ^ -lc(6)/lc(a) if a = sDt + Dz for z G k and s G Z t h e n n ^- max(n, s) if db = da t h e n (* possible cancellation *) a ^ -lc(6)/lc(a) if a = Dz/z for z ^ k* t h e n ß < lc{aDz + bz)/{z lc(a)) if /3 = sDt + Dw; for t/; G A: and s E Z t h e n n ^— max(n, s) return n In the specific case where D = d/dt^ Corollary 6.3.1 yields a simpler algorithm, as in t h e nonparametric case.

P a r a m R d e B o u n d D e g r e e B a s e l a,b,qi,... ^Qm) (* Bound on polynomial solutions - base case *) (* Given a,b^qi,.. . ,qm G k[t] with a j^ 0, return n G Z such that deg(g) < n for any solution c i , . . . , Cm Gk and q G k[t] of a^ + bq *) - maxi<^< m(d€ m{q^)) da ^ deg(a), db ^ deg(6), dc
Example 7.1.3. Let A: = Q, t be a monomial over k satisfying Dt = 1^ i.e. D d/dt^ and consider t h e parametric Risch differential equation

7.1 The Parametric Risch Differential Equation Dy-\-y

= cit^

^C2,

where N is an arbitrary positive integer. We have so by Theorem 7.1.1, any solution y € k{t) must have a = b = 1, so da = djy = 0^ dc = max(A/', 0) max(0, —1)) = A^, which implies t h a t any solution at most A^.

229 (7.13)

f = 1, gi =t^ and ^2 = 15 be in k{t) = k[t]. We t h e n = N and n = max(0, N — y E k[t] of (7.13) has degree

T h e Hyperexponential Case If Dt/t = T] E: k^ then Lemmas 6.3.1 and 6.3.4 always provide an upper b o u n d for deg(g) as in the nonparametric case: if deg(6) > deg(a), then Lemma 6.3.1 implies t h a t deg(g) > max(0, m a x (deg(gi)) — deg(6)). l
If deg(5) < deg(a), then Lemma 6.3.4 implies t h a t deg(g) > max(0, m a x (deg(gi)) - d e g ( a ) ) . l<'i
At last, if deg(a) = deg(5), then either —lc(6)/lc(a) = sr] + Du/u 5 G Z and w G fc*, in which case

for some

deg(g) > max(0, 5, m a x (deg(g^)) — deg(6)), l
or deg(g) > max(0,maxi<^<-^(deg(gi))—deg(6)). Note t h a t such an s is unique by L e m m a 6.2.2.

P a r a m R d e B o u n d D e g r e e E x p ( a , b ^ q i , . . . ,qm,D) (* Bound on polynomial solutions - hyperexponential case *) (* Given a derivation D on k[t] and a, 6,gi,.. ., gm € k[t] with Dt/t G k and a ^ 0, return n G Z such that deg(g) < n for any solution c i , . . . , Cm. G Const(/c) and q £ k[t] of aDq -}-bq = YlTl^i ^^^i- *) da ^ deg(a), 4 ^ deg(6), dc ^ ma,Xi 0 *) if da ~ db t h e n (* possible cancellation *) a ^ -lc(6)/lc(a) if a = sDt/t + Dz/z for z E k* and s E Z t h e n n ^— max(n, 5) return n

T h e Nonlinear Case If 6{t) > 2, then Lemmas 6.3.1 and 6.3.5 always provide an upper bound for deg(g) as in the nonparametric case: if deg(6) ^ deg(a) + S{t) — 1,

230

7 Parametric Problems

then Lemma 6.3.1 provides the bound as explained earlier. Otherwise, either —lc(5)/lc(a) = sX{t) for some 5 € Z, in which case deg(g) > max(0, 5, m a x (deg(g^)) - deg(6)), l
or deg(g) > max(0,maxi<^<^(deg(gi)) - deg(6)).

P a r a m R d e B o u n d D e g r e e N o i i L i n e a r ( a , b , q i , . . . ,qm,D) (* Bound on polynomial solutions - nonlinear case *) (* Given a derivation D on k[t] and a,b,qi,... ,qm ^ k[t] with deg{Dt) > 2 and a ^ 0, return n G Z such that deg(g) < n for any solution c i , . . . , Cm G Const(Ä:) and q G k[t] of aDq + hq = f^i ciqi. *) da ^ deg(a), 4 ^ deg(6), (ic ^ maxl<^<m(deg((?^)) 6 <— deg(Dt), A ^- lc(I}t), n ^- max(0, (ic — max((ia + 5 — 1,
The Parametric SPDE Algorithm We are now reduced to finding solutions c i , . . . , c ^ in Const(fc) and q G k\t] of (7.7) and we have an upper bound n on deg(g). Theorem 6.4.1 and t h e S P D E algorithm of Sect. 6.4 generalize to the parametric case. T h e o r e m 7.1.2. Let a, 6, g i , . . . , g^n ^ ^[^] with a / 0 and gcd(a, h) = 1. Let zi^ •. • ^ Zjn^ri^... ^Tm E k\t] he such that for each i, qi = azi + bri and either ri = 0 or deg(r^) < deg(a); and let r = Yl^i^i'^iThen, for any solution c i , . . . , c ^ G Const(/c) and q £ k[t] of aDq -\~ bq =^ YllLi ^iQi^ P = (^ ~ '^)/<^ ^ k[t], and p is a solution of aDp + (6 + Da)p = ci{zi - Dri) + . . . + Cm{zm - Drm) • Conversely, for any solution c i , . . . ^Cm G Const(^) and p G k[t] of q = ap -i- r is a solution of (7.7).

(7.14) (7A4),

Proof. Let c i , . . . , c ^ G Const(Ä:) and q e k[t] be a solution of (7.7). Then, m

aDq -\-bq = a Y ^ CiZi + br i=i

so b{q — r) = ^(Yl^i^i^i ~ T)q) , which implies t h a t so a \ b{q — r). But gcd(a, 6) = 1, so a I g — r and p =:^ [q — r)/a G k[t]. We then have:

7.1 The Parametric Risch Differential Equation ,, ^ , [Dq-Dr aDp + ^ + Da)p = a -^ \ a

(q-r)Da\ ^ ^^ a^

J

b(q - r)-{-{q + -^ ^ ^a

231 r)Da ^

a

Conversely, let c i , . . . , c ^ G Const(Ä:) and p G A:[t] be a solution of (7.14), and let q = ap + r. Then, aDq -{- bq = o?Dp + apDa + aDr + a6p H- 5r = a (ai:^p + (6 + Da)p) + a D r 4- br a V ] Q(zi — Dri) + a D r + br i=l i=l m

m

m

a D

Theorem 7T.2 reduces (7.7) to (7.14), which is an equation of the same type. If the coefficients a and b of the new equation have a nontrivial gcd, we divide it by t h a t gcd, obtaining an equation of type (7.5) and reapply the hnear constraints algorithm, obtaining a new equation of type (7.7). However, in all cases if (7.7) has a solution q of degree n, then the corresponding solution of the new equation must have degree at most n — deg(a) since q ~ ap + r and deg(r) < deg(a). Thus, if deg(a) > 0, we can use Theorem 7.1.2 and Lemma 7.1.1 to reduce the degree of the unknown polynomial. We can repeat this until deg(a) = 0 i.e. a G fc*, at which point we divide t h e equation by a and we get an equation of type (7.7) with a = 1.

P a r S P D E ( a , 6, g i , . . . , g^, D, n)

(* Parametric SPDE algorithm *)

(* Given a derivation D on /c[t], an integer n and a,b,qi,. .. ,qm G k[t] with deg(a) > 0 and gcd(a,5) = 1, return {ä,b,^,... ,'q^,ri,... ,rm,fi) such that for any solution c i , . . . , Cm G Const(fc) and q ^ k[t] of degree at most n of aDq + bq = ciqi + . . . + Cmqm, P = {q ~ ciri — . . . — Cmrm)/a Cmq-, has degree at most n and satisfies aDp -{- bp = ciqi i- .. . -\-•m^m • ) for i <— 1 t o m d o (* bri -\- azi = g^, deg(ri) < deg(a) *) {vi^ Zi) ^— E x t e n d e d E u c l i d e a n ( 6 , a, qi) r e t u r n ( a , 6 + Da.zi — Dn,... ,Zm — Drm.ri,... ,rm,n~ deg(a))

232

7 Parametric Problems

Example 7.1.4- Let k = Q{x) with D = d/dx^ t be a monomial over k satisfying Dt = 1/x, i.e. t = log(x), and let us search for a polynomial solution of arbitrary degree n of q = Cix — C2xt.

tDq

(7.15)

X

We have a = t^ b = —1/x^ m = 2, gi = x and g2 = ~xt so: 1- (^i,^i) = E x t e n d e d E u c l i d e a n ( — 1 / x , t , a : ) = (—x'^,0), (^25 Z2) = E x t e n d e d E u c l i d e a n ( — 1 / x , t, —xt) = (0, —x) 2. b-i- Da =

\- Dt =

h - = 0, zi - Dri = 2x, zi - DT2 = - x .

X

X X

So (7.15) is reduced to iDp -= 2c\x — C2X where p = (g + ciX'^)/t G k\t] has degree at most n — 1. We have gcd(a, b) = t in the above equation, so it becomes 2x

X

which is of type (7.5). Calling L i n e a r C o n s t r a i i i t s ( l , 0 , 2 x / t , —x/t) gives: 1. d = lcm{t,t) = t 2. ( g i , r i ) = P o I y D i v i d e ( 2 a : , t ) = (0,2^:), (92,^2)= PolyDivide(-a:,t) = ( 0 , - x ) 3. iV = m a x ( d e g ( r i ) , d e g ( r 2 ) ) = 1 4. ' 0 0 2x —X so we are reduced to the equation Dp = 0 and the linear constraints M{ci,C2)'^ = 0. Calling C o n s t a n t S y s t e m ( M , 0 ) yields the constant system

0 0 2

0 \ / c, 0 \C2 -1

which has t h e 1-dimensional solution space (ci, C2) = (A, 2A). Since Dp — 0 has the 1-dimensional solution space p = c for any c G Q, we get the 2-dimensional solution space {q ~ ßt — Xx'^^ ci = A, C2 = 2A) of (7.34). Given the formal definition of the parametric Risch differential equation problem, this solution space is represented by g = dihi -\- 0^2/12 where hi = t^ h2 = x^ and /Cl

2 -1 1 0

0

0 \ \c,

0 ij

\d2

= 0.

7.1 The Parametric Risch Differential Equation

233

T h e Non-Cancellation Cases We are now reduced to finding solutions c i , . . . , c^ G Const(A;) and q E k[t] of the following equation: m

i=l

where 6,gi,... , g ^ G k[t] and t is a monomial over k. Furthermore, we have an upper bound n on deg(g). As in the nonparametric case, Lemma 6.5.1 provides algorithms for all the non-cancellation cases. W h e n deg(6) is Large Enough Suppose that 6 7^ 0, and that either D = d/dt or deg(6) > max(0, (5(t) — 1). Then, for any solution q = y^f^-j-.. .-\-yo G k[t] of (7.16), Lemma 6.5.1 implies that deg{Dq + bq) < n + deg(5) and equating the coefficients of t^+^®s(fc) on both sides yields m

lc(6) y„ = ^

ci coefficient(gi, ^+'^''^(6)) _

Replacing qhy h + X^^j CiSint"' in (7.16), where .» =

coefficient (gi,i"+d«'g(f')) ^^^ ^

, efc,

^^^^^ (7.17)

we get m

m

m

^ i=l

Cibsir^r + bh = ^

i=l

Ciqi i=l

which is equivalent to m

Dh+bh = Y,ci {qi - Disi^n - bs.nn i=l

which is an equation of the same type as (7.16) with the same b as before. Hence the hypotheses of part (i) of Lemma 6.5.1 are satisfied again, so we can repeat this process, but with a bound of n — 1 on deg(/i). Note that although b remains the same, the right side of (7.16) changes at every pass, so we must recompute the g^'s that appear in (7.17). The bound on deg(g) will decrease at every pass through this process, guaranteeing termination. After finishing the case n = 0, we get that any solution q G k[t] of the initial equation with deg(g) < n must be of the form q = Yl^i ^i^i where hi — Xl?=o ^^3^^ ^ ^WReplacing q by that form in (7.16) with the original g^'s yields y ^ Ci{(ii - Dhi - bhi) = 0 . i=l

234

7 Parametric Problems

T h e left side is an element of /c[t], so setting all its coefficients to 0 yields a homogeneous linear system of the form M ( c i , . . . ,c^)"^ = 0, where M has entries in k. The same system can also be obtained from the last g^'s when n = 0 and the equation Yl^i ^i{Qi ~ Dsio — bsio) = 0, and this is how it is obtained in the algorithm below. By Lemma 7.1.2, we can compute an equivalent system of the form A{ci^... ^ Cm)^ = 0 where A has entries in Const(Ä:). The solution of the initial problem is then q — YlT=i ^^^^ where t h e additional equations di = Ci ioi 1 < i < m are added to A, i.e. an m x 2m block of the form /I 0

0 0

-1 0

0 -1

0 \ 0 (7.18)

Vo

0

1 0

•••

"•-

0

- 1 /

is concatenated to the b o t t o m of A, as well as a zero block to its right. T h e final system of linear constraints is then A{ci^..., c ^ , < i i , . . . , dmY' = 0.

ParamPolyRischDENoCancell(6, (* Parametric Poly Risch d.e. - no cancellation *) (* Given a derivation D on /c[t], n G Z and 6 , g i , . . . , g^n G k[t] with 6 7^ 0 and either D — djdt or deg(6) > max(0,5{t) — 1), returns hi,.. . ,hr G k[t] and a matrix A with coefficients in Const(fc) such that if c i , . . . , Cm G Const(fc) and q G k[t] satisfy deg{q) < n and Dq+bq — YlTLi ^iQi then q = J2^j=i ^j^j where di,... ,dr G Const(/c) and A{ci,..., Cm,c^i, • • •, dr)^ = 0. *) db ^ deg(6), bd <— lc(6) for i ^- 1 to m do hi ^~ 0 while n > 0 do for i -e- 1 t o m do Si ^— coefficient{qi,t'^'^'^^)/bd hi -^ hi + Sif^ qi^qiD{sir) - bsir n •^- n — 1 (* The remaining linear constraints are X^^i c^gi = 0 *) if qi = ... = qm = 0 t h e n dc < 1 else dc ^— maxi
7.1 The Parametric Risch Differential Equation

235

Example 7.1.5. Continuing example 7.1.3, let /c — Q, t be a monomial over k satisfying Dt = \^ i.e. D = d/dt^ iV be a positive integer, and consider the solutions y G k[t] of degree at most N of (7.13). We have 6 = 1, m = 2, qi = t^ and q2 = 1. Then, 1. db = 0, bd = 1, hi = h2 = 0 2.n = N,Si = 1, hi = t ^ , qi = -Nt^-\ 3.n = N-l,si = -iV, hi=t^ - Nt^-\

52 = 0, /i2 = 0, 92 = 1 qi = N{N - l ) t ^ - 2 ,

52 = 0, /l2 = 0, 92 = 1

4. . . .

It is easy to prove by induction that after r steps through the loop (r < N) we get r-l

/ii = X l ( " l ) ' ' Nit^-^,qi

= {-ly

i V ^ t ^ - ^ /i2 - 0 and 92 = 1

j=0

where N^ = YllZo ( ^ ~~ 0- Thus, after N iterations we get n = 0, N-l

hi= J2{~iy

Nit^-^,

j=o

qi = (—1)^ iV^, /i2 = 0 and 92 = 1- The last iteration then gives 1.5i = (~-l)^iV^ 2. /ii = hi + 51 = E , l o ( - l P ' ^^^'^'"'' 3. gi = (^1)^ N^ -Dsi--si=0 4. 52 = 1, h2 = I, q2 =0

Proceeding with the algorithm, we get 1. dc = --1, M and A are 0 by 0 matrices 2. Ueq = 0

A \o

0 -1 1 0

0 ^1

So the algorithm returns the above matrix, /i2 = 1 and N

hi = ^ h i y Nlt^~^

= t^ - Nt^-^

+ N{N - l)t^""2 + . . . + (-l)^iV!,

The general solution of (7.13) is y = dihi + d2 where

236

7 Parametric Problems

W h e n d e g ( 6 ) is S m a l l E n o u g h Suppose t h a t deg(6) < ö{t) — 1 and either D = djdt^ which impUes t h a t 5 = 0, or b(t) > 2. Let q = y^f^ + . . . + 2/0^ k[t] be a solution of (7.16). If n > 0, t h e n Lemma 6.5.1 implies t h a t deg(Dg + bq) < n + ö{t) — 1 and equating the coefficients of t^+^(*)-i on b o t h sides yields m

nX{t)yn

= ^c,coefficient(g,,r+'^(*^-^).

Replacing g by /i + YllLi ^iSint^ in (7.16), where n A(t) we get m

which is an equation of the same type as (7.16) with the same h as before. Hence the hypotheses of part (ii) of Lemma 6.5.1 are satisfied again, so we can repeat this process, but with a bound of n — 1 on deg(/i). Note t h a t although h remains the same, the right-hand side changes at every pass, so we must recompute t h e g^'s t h a t appear in (7.19). The bound on deg(g) will decrease at every pass through this process, until we reach n = 0, i.e. we are looking for solutions q = yo e k. At this point, the algorithm proceeds differently for deg(6) > 0 and for b E k. If deg(6) > 0, then equating the coefficients of t^^^^^^ on b o t h sides yields m

lc(6) yo = X ^ Q coefficient(g^, t^^s(b)) ^ i=l

SO any solution yo ^ k must be of the form yo = Y^iLi ^i^io where coefficient (g^,t^^s(^)) lc(6) This imphes t h a t any solution q e k[t] of the initial equation with deg(g) < n must be of the form q = J^^i ^i^i where n

j=o

Replacing q by t h a t form in (7.16) with the original g^'s yields

7.1 The Parametric Risch Differential Equation

237

m

"^CiiQi-

Dhi-bhi)

=0.

i=l

As we have seen earlier, this can be converted to a homogeneous system of the form A{ci^... ^Cm)^ = 0 where A has entries in Const(Ä^). The solution of the initial problem is then q = YlT^i ^i^i where the additional equations di = Ci for 1 < i < m are added t o A as earlier. The final system of linear constraints is then A ( c i , . . . , c ^ , d i , . . . , dm)^ = 0. If 6 G k, then any solution yo ^ ^ of (7.16) satisfies m

Dyo-}-byo = Y.Ciq,{0).

(7.20)

i=i

This is a parametric Risch differential equation of type (7.1) over k. Assuming t h a t we can solve such problems over k, we obtain / i , . . . , / r G k and a m a t r i x B with coefficients in Const(/c) such t h a t any solution yo E k of (7.20) is of the form r

where di^... ,dj G Const(A;) and - B ( c i , . . . , C m , d i , . . . , d r ) ^ = 0. This implies t h a t any solution q £ k[t] of the initial equation with deg(g) < n must be of the form q = X]^=i ^jfj + X^I^i ^i^i where n

3=1

Replacing q by t h a t form in (7.16) with the original qis yields m

r

Y, Ciiqi - DK - bhi) - Y. ^3 (Dfj +&/,) = 0 . In a similar way t h a n in the previous cases, this can be converted to a homogeneous system of the form A{ci,..., c ^ , d i , . . . , d^)^ = 0 where A has entries in Const(Ä:). The solution of the initial problem is then r

m

j=l

i=l

where the additional equations B{ci,..., c ^ , d i , . . . , d^)^ = 0 are added to A, as well as the equations ei = Ci foi 1 < i < m. The final system of linear constraints is then A ( c i , . . . , c ^ , d i , . . . , d^-, e i , . . . , e^)"^ = 0.

238

7 Parametric Problems

ParamPolyRischDENoCancel2(6,gi,. ..,qm,D,n) (* Parametric Poly Risch d.e. - no cancellation *) (* Given a derivation D on k[t], n G Z and 6 , g i , . . . , g ^ G k[t] with deg(6) < S{t) — 1 and either D — d/dt or 6{t) > 2, returns hi,... ,hr E k[t] and a matrix A with coefficients in Const(A:) such that if c i , . . . ,Cm G Const (A;) and g G A:[t] satisfy deg(<3') < n and Dq-\-hq = ^^^^ Qg^ then g = S^=i <^i^i where di,... ,dr G Const(Ä:) and A ( c i , . . . , Cm, ( i i , . . . , dr)^ — 0. *) for i ^— 1 t o m d o /ij ^— 0 while n > 0 do for 2 ^- 1 t o m d o Si^ coefficient(gi,r+'^-^)/(nA) hi^hi + Sif, qi^qiD{sir) n -^— n — 1 if deg(6) > 0 t h e n for i ^- 1 t o m do Si^ coefficient(gi,t^"s(^))/lc(6) hi ^~ hi -\- Si, qi ^

bsir

qi — Dsi — bsi

if q^ =z , _ = q^ = 0 t h e n dc < 1 else dc ^ mdiXi
1 t o m d o Ai-^n^gA

^

1^ A^-^-neg,m+^ <

1

r e t u r n ( / i i , . . . ,hm,A) else (* 6 G A; *) P a r a m R i s c h D E ( 6 , g i ( 0 ) , . . . ,gm(0)) (/i, ...Jr,B)^ if gi = . . . = Qm = 0 t h e n if Dfi + hfi = ... = Dfr + bfr = 0 t h e n 4 < 1 else
1 t o m d o Ai-^neq,i

^

1, ^ i + n e g , m + r + i <

1

r e t u r n ( / i , . . . , /^, / i i , . . . , / i ^ , A)

Example 7.1.6. Continuing example 7.1.1, let /c = Q, t be a monomial over k satisfying Dt = 1^ i.e. D = d/dt^ and consider the solutions p G A:[t] of (7.11). We have a = 1^ b = 0 and gi = 2x, so the degree bound algorithm for the base case yields an upper bound of n = 2 on deg(p). Then, 1. (5 = 0, A - : 1, /ii = 0 2. 5i = 2/2 = l^hi= t2, gi = 0, n = 1

7.1 The Parametric Risch Differential Equation

239

3. si = 0, hi =t^, qi=0,n = 0 4. 5i = 0, /ii = t^, gi = 0, n = - 1 At this point, since b e k, we recursively find the solutions y G k of Dy = 0. This returns fi = l and the linear constraint di = X^ i.e. B = {1 —1 )• We then have 1. gi = Dfi — 0 so dc = —1^ hence M and A are 0 by 0 matrices 2.A = AUB = {1 -1) 3. Ueq = 1 4.

^~\1

0

-1

So the algorithm returns the above matrix, / i = 1 and hi — t^. T h e general solution of (7.11) is p = di -i- eit^ where

0.

Going back to example 7.1.1, Since we had (ci, 02,03) = (A, —3A,--2A), the general solution of (7.9) is p = di + eit^ where 3 1 0

0

I 2 0

1

0

1 0 \1 0

0 - 1 0 0

0 \ / ^ ^ \

0 0 -1

C2

= 0. C3

\ej Since the above constraints imply t h a t di = ei — ci, the general solution can also be simplified t o p = c i ( l 4-1^) subject to the constraints (7.10). W h e n 5{t) > 2 a n d d e g ( 5 ) = ö{t) - 1 In t h a t case, we have cancellation only when deg(g) = —lc(6)/A(t), which implies in particular t h a t —lc(6)/A(t) is an integer between 1 and our degree bound n. Let q = ynt^-^.. .yo G k[t] be a solution of (7.16). If n / --lc(5)/A(t), then Lemma 6.5.1 implies t h a t deg(Dg + 6g) < n + S{t) — 1 and equating the coefficients of t'^+^^s(b) QJ^ b o t h sides yields m

(nX{t) + lc(&)) y^ = Y^Ci coefficient^, t " + * ( * ) - i ) . i=i

Replacing q by h + YllLi CiSm^""'"''^*^"'^ in (7.16), where coefficient(gi,t"+*(*)-i) nA(t)+lc(6)

ek,

(7.21)

240

7 Parametric Problems

we get m

Dh+bh = j2ci {Qi - D{sinn - bsinn i=l

which is an equation of the same type as (7.16) with the same b as before, but with a bound of n — 1 on deg(/i). As long as the degree bound is not equal to —lc(6)/A(t), the hypotheses of part (iii) of Lemma 6.5.1 are satisfied again, so we can repeat this process until either we have finished the case n = 0, or we reach the case n = —lc{b)/X{t). If we have finished the case n == 0, then any solution q G k[t] of the initial equation with deg(g) < n must be of the form q = Y^iLi ^i^i where n 3=0

Replacing q by that form in (7.16) with the original g^'s yields m

^

Ci{qi - Dhi - bhi) = 0 .

i=l

As we have seen earlier, this is equivalent to homogeneous system of the form yl(ci,..., c^)-^ = 0 where A has entries in Const(^). The solution of the initial problem is then q = Yl^i ^i^i where the additional equations di = Ci foi 1 < i < m are added to A as earlier. If we reach the case n = —lc{b)/X{t) > 0, then the algorithm of the next section on the cancellation cases produce / i , . . . , /^ G k[t] and a matrix B with coefficients in Const (A;) such that any solution q e k[t] of degree at most n must be of the form

= Yldjfj 3=1

where di,..,,dj G Const(Ä:) and B{ci,...,c^,di,..., dr)^ ~ O.This implies that the solutions of the initial equation must be of the form r

m

where n

hi=

Y^

Sijt^

Gk[t].

j = l-lc{b)/X{t)

Replacing q by that form in (7.16) with the original g^'s yields m

r

Y, cMi - Dhi - bhi) - J2 '^i^^fi

+bfj)=0.

7.1 The Parametric Risch Differential Equation

241

As we have seen earlier, this is equivalent to homogeneous system of the form A{ci^..., c^, ( i i , . . . , dr)"^ = 0 where A has entries in Const(Ä;). The solution of the initial problem is then r

m

j=i

i=i

where the additional equations B ( c i , . . . , c^, d i , . . . , d^^)-^ = 0 are added to A, as well as the equations ei = Ci for 1 < i < m. If deg(g) > 0 and deg(g) ^ -lc(6)/A(t), then deg(g) + 6{t) - 1 = deg(c), so deg(g) = deg{c) + l-5{t) and (deg(g)A(t)+lc(6))lc(g) =lc(c). This yields the leading monomial ut^ of g, and replacing q by ut^ + /i in the equation yields a similar equation with a lower degree bound on its solution. We can repeat this as long as the new degree bound is not equal to —lc{b)/X{t). If g E ^, then the leading term of Dq + bq is glc(ö)t^^*^~^, so either deg(c) = (5(t) — 1, in which case q = lc(c)/lc(6) is the only potential solution, or deg(c) j^ 6{t) — 1 and (6.19) has no solution in k^ hence in k[t]. T h e Cancellation Cases We finally study equation (7.16) whenever the non-cancellation cases do not hold, i.e. in one of the following cases: 1. 6{t) 2, deg(6) = 6{t) ~ 1, and deg(g) = -lc(6)/A(t). T h e Liouvillian Case If D ^ d/dt and Dt G k or Dt/t G k^ then S{t) < 1, so the only case not handled by Lemma 6.5.1 is b E k. Then, for any solution q = ynt^ + . • • +2/o ^ k[t] of (7.16), Lemma 5.1.2 implies that deg{Dq -\-bq) < n and equating the coefficients of t^ on both sides yields m

Dyn + byji = \ ^ Ci coef!icient(g4, t^)

(7.22)

i=i

if Dt Gfc,and Dyn + ib + n— j y^ = ^

Q coefficient(g^, ^'')

(7.23)

if Dt/t G k. Both (7.22) and (7.23) are parametric Risch differential equations of type (7.1) over k. Assuming that we can solve such problems over k^ we obtain / i ^ , . . . , frn.n ^ ^ and a matrix An with coefficients in Const(A:) such that yn is of the form

242

7 Parametric Problems

^ ^ ^^ / ^ 3=1

^jnjjn

where din,...,
r

Dh + bh = Yl ^^^^ - E ^.n(i^(/,nf") - ^/,nt") j=l

^=l

which is an equation of the same type as (7.16) with the same b as before. Hence, we can repeat this process, but with a bound of n — 1 on deg(/i). Note t h a t although b remains the same, the right side of (7.16) changes at every pass, so we must recompute t h e g^'s t h a t appear in (7.22) or (7.23). Note also t h a t the number of undetermined constants in the right side increases at each step. The bound on deg(g) will decrease at every pass through this process, guaranteeing termination. After finishing the case n = 0, we get t h a t any solution q ^ k[t] of the initial equation with deg(g) < n must be of t h e form n

g= ^

Ti

^

djihji

where

hji = fjif

.

(7.24)

i=0 j = l

Replacing g by t h a t form in (7.16) with the original g^'s yields m

i=l

n

Ti

i=0 j = l

As we have seen in the non-cancellation homogeneous system of the form A{ci,..., has entries in Const (A:). T h e solution of by (7.24) where the additional equations are added to A for 0 < i < n.

cases, this can be converted to a c ^ , du^..., dr.^^n)^ = 0 where A the initial problem is then given Ai[ci,..., c^, dn^...,

T h e Nonlinear Case If 6{t) > 2, then we must have deg(6) = 6{t) — 1 and lc(6) = —nX(t) where n > 0 is the bound on deg(g). As in the nonparametric case, there is no general algorithm for solving equation (7.16) in this case. If however S^^^ ^ 0, then projecting (7.16) to ^[t]/(p) for p G S^^^ can be done as in the nonparametric case. Since ^[t]/(p) is a finite algebraic extension of A:, Const(A:[t]/(p)) is a finite algebraic extension of Const(/c) by Corollary 3.3.1, so let 5 i , . . . , bg be a vector space basis for Const(/c[t]/(p)) over Const(A;). Now, with D* being the induced derivation on A:[t]/(p), we get m

D*g* + 7rp(6)g* = ^

Ci^p(g,)

(7.25)

7.1 The Parametric Risch Differential Equation

243

where g* = TTp{q). Assuming that we have an algorithm for solving (7.25) in k[t]/{p)^ we obtain / i i , . . . , / i ^ in k[t]/{p) and a matrix B with coefficients in Const{k[t]/{p)) such that any solution in k[t]/{p) of (7.25) must be of the form q* = Yl^j=i^j^j where di,...^dr € Const(^[t]/(p)) and B{ci,... ^Cm.di^... ,dr)^ = 0. Expanding formally the constants d i , . . . , d ^ and the entries of B with respect to the basis 6 i , . . . , 5s we obtain a matrix A with coefficients in Const(fc) such that the system JB(CI, . . . , c^, ^i, • • •, d^)'^ = 0 is equivalent to A{ci^..., c^, d u , . . . , drs)^ = 0 where s

dj = ^ d j i k

for

1 <j


1=1

and any solution in k[t]/{p) of (7.25) must now be of the form r

/ s

\

j=i

\i=i

J

r

s

j=i 1=1

where hji = bihj G k[t]/{p). For each j and /, let rji G k[t] be such that deg(r^7) < deg(^) and 7Tp{rji) = hß, and let r

s

j=i 1=1

We have deg{u) < deg(p) and (7.26) implies that 7rp{q) = 7Tp{u), hence that h = {q ~ u)/p e k[t]. Replacing q hj ph ^ u in (7.16) we get ^

Ciqi = Dq^bq=p(Dh+

(b^

— ]h]

+Du^bu

so /i is a solution in k[t] of degree at most deg(g) — deg(p) of Dh+(b+^y=

^ - ^ '''' - ^^-^

^ ' - ^ '-'^'^^-' + '^-'^ .

(7.27)

Write now qi = pq~ + qi and DTJI + bvji = prj! + fj] where g^i,f7i G Ä:[t], deg(g^) < deg(p) and deg(fj]) < deg(p). The right hand side of (7.27) becomes IT=1 ^i^i - E i = i E L i djijDrji +fer^-Q__ ~

V^ v ^

7

, E i = i ^*^* ~ Z ^ j = i 2^1=1

E ^^^^" E E ^^^-^^^- +

Since 7rp{u) = g* is a solution of (7.25), we have

v~

^ji^ji

'

244

7 Parametric Problems

(

m

\

I m

Y^ CiQi - Du-bu\ i=^l

= ^p J

r

s

X^ Q$i - X ] ^ \i=l

djif^i

j = l 1=1

Since deg(gj) < deg(p) and deg(fj]) < deg(p), it follows that m

r

s

so (7.27) becomes FJ \

^^

r

s

Dh ^

^

i=l

3 = 1 1=1

which is an equation of type (7.16), but with a lower bound on the degree of its solution. Repeating this process until the lower bound becomes negative, and grouping all the linear constraints obtained at each step yields a complete solution of the initial parametric problem. The remarks made in the nonparametric case about when (7.25) can be solved, for example when we can find an element of degree one, or an element with constant coefficients, in S^^^, remain valid in the parametric case. T h e H y p e r t a n g e n t Case If Dt/iiP' + 1) = ?7 G Ä:, then 5{t) = 2, so the only case not handled by Lemma 6.5.1 is 5 = 6o — nr]t where bo G k and n > 0 is the bound on deg(g). In such extensions, the method outlined above provides a complete algorithm: taking p = t'^ + 1 G cS^^^ (7.25) becomes m

Dq- + (bo - n77x/^)g* = ^

Qg,(v^)

(7.28)

i=l

where D is extended to k[t]/{p) ~ k{y^^) by D ^ / ^ = 0. One possibility is to view (7.28) as a parametric Risch differential equation in k{\/~^) and to solve it recursively. After expanding the result with respect to the basis { 1 , \ / ^ } (only if \ / ^ ^ A:), we obtain / i i , . . . , /i,. G k{-s/^) and a matrix A with entries in Const(A:) such that all the solutions in k{^/^) of (7.28) must be of the form q* = X]j=i ^j^j where dj e Const(/c) and A ( c i , . . . , c^, c^i,..., dr)^ = 0. Write hj = hjo -\- hjwf—l for each j with hjo^hji G k. Then, letting Vj = hjo + hjit G k[t]^ qi he the quotient of qi by p and f] be the quotient of DTJ + bvj by p, /i = (g — YJT=I ^j'^j)/P ^^ ^ solution in k[t] of degree at most n-2of m

r

Dh + {bo — {n — 2)rj t)h = \ ^ ciql — 2_\ djfj. i=i

j=i

7.2 The Limited Integration Problem

245

If V —1 ^ /c, it is possible to avoid introducing y —1 by considering t h e real and imaginary parts of (7.28): writing q* = u -{-1'^/^, we get

n:)-{X :)(:)=Ec.(::;)

(->

where qio-hqnt is the remainder of qi by t^ + 1. This is the parametric version of t h e coupled differential system introduced in Sect. 5.10, and the algorithm of C h a p . 8 can be generalized to the parametric case, in a manner similar t o what is done for the Risch differential equation in this chapter.

7.2 The Limited Integration Problem We describe in this section a solution to the limited integration problem, ,Wm ^ K^ t o i.e. given a differential field K of characteristic 0 and f^wi,... decide whether there are constants c i , . . . , c ^ G Const(K) such t h a t f = Dv + CiWi + . . . + CmWm

(7.30)

has a solution v E K^ and to find one such solution if there are solutions. As we have seen in Chap. 5, this problem arises from integrating polynomials in primitive extensions. There are several possible approaches to this problem: ® If all the Wi^s are logarithmic derivatives of elements of K^ then t h e existence of a solution of (7.30) implies t h a t / has an integral in an elementary extension of K, so equation (7.30) can be seen as an elementary integration problem, and the algorithm of Chap. 5 can be used, followed by a step t h a t a t t e m p t s to rewrite the resulting integral in terms of the Wi^s. ® Equation (7.30) can be considered a parametric Risch differential equation for V and can be solved by the algorithm of Sect. 7.1. T h e first approach is applicable only when integrating elementary functions, since the only primitive monomials appearing in the integrand are t h e n logarithms, and it is in fact the approach originally taken by Risch [75] and in most computer algebra systems and texts [32, 28, 39]. How to rewrite an elem e n t a r y integral in terms of the Wi^s is however never made explicit^ and adds new difficulties and complexities to the algorithm. The second approach is applicable for arbitrary w;^'s, so it allows arbitrary primitives in the integrand. Furthermore, algorithms for integrating in terms of some nonelementary functions, like Erf, Ei, Li and dilogarithms [21, 22, 52, 53], first produce candidate special functions and then solve the limited integration problem for those special functions. Because of those advantages, we essentially use t h a t m e t h o d ^In [75], Risch only has the following remark about the hypothesis that we can integrate elements of k: ^ ' . . .we assume t h a t the simpler v a r i a n t s , which occur when some of the Ci and Vi are given, have been e s t a b l i s h e d . ' \

246

7 Parametric Problems

here. However, the parametric Risch differential equation algorithm is made simpler by the fact that cancellation at the poles of v (including infinity) cannot occur since only Dv appears in the equation and no multiple of -u, so bounding orders and degrees is significantly easier. We present in this section an simplified version of the algorithm of Sect. 7.1 that takes advantage of this fact. We only study equation (7.30) in the transcendental case, i.e. when K is a simple monomial extension of a differential subfield fc, so for the rest of this section, let /c be a differential field of characteristic 0 and t be a monomial over k. We assume in addition that Const(Ä:(t)) = Const (A:). We suppose that the coefficients / and i ^ i , . . . , w^ of our equation are in k{t) and look for solutions c i , . . . ,Cm G Const(^) and v £ k(t). Because of the special form of equation (7.30), Theorem 7.1.1 can be strengthened to yield not only the normal part of the denominator, but also its special part whenever Sf^ = cS^^^, and the degree bound whenever t is either a Liouvillian or nonlinear monomial. Theorem 7.2A, Let v^ f^wi,... ,Wm G k(t) and ci^... ,Cm € Const(Ä:) be such that f = Dv + ciwi -|- . . . -f Cm^m- Let d = dgdn he a splitting factorization of the denominator of f, and e^ = es^iSn^i be splitting factorizations of the denominators of the wi^s. Let c = lcm((in, e^,i,. • •, en,m); /is = lcm(4,es,i,...,es,m); and

K = gcd [c, I Then, (i) vhn ek{t), (ii) IfS'f"^ = 5^^^ then vh^hs G k[t]. (Hi) If t is nonlinear or Liouvillian over k, then either i/ooi^) = 0 or l^ooiv) > m i n (l^oo(/), Z^oo('^l), • • • , ^ o o ( ^ m ) ) + 5{t) - 1 .

Proof, (i) Let q = vhn G k{t). In order to show that q G k{t)^ we need to show that Typ{q) > 0 for any normal irreducible p G k[t]. We have z^p(g) = h'p{v) + iyp{hn) by Theorem 4.1.1. If I'piv) > 0, then z^p(g) > i^p{hn) > 0 since hn G k[t]. So suppose that n — Vp(y) < 0 and let w = ciWi-{-.. . + c^t(;^. Then h'p{Dv) = n — 1 by Theorem 4.4.2, so Vp{f — w) = n — 1^ which implies that ^p{f) < n — 1 01 Up{wi) < n — 1 for some i. Hence p^~^ \ c, so i^p(c) > 1 — n, which implies that i'p{hn) — Vp{c) — \ > —n, hence that z/p(g) = n-{-Up{hn) > 0. (ii) Suppose that Sf^ = S^^^ and let q = vhnhg G k{t). Since hs E k[t] C kit)., and we have from (i) that vhn G k{t), we get that q G k{t)^ so in order to show that q G A:[t], we need to show that i'p{q) > 0 for any q G S^^^. We have iyp{q) = I'piy) + Vp{hn) + i^p{hs) by Theorem 4.1.1. If i'p{v) > 0, then I^P{Q) ^ ^p{hn)-^^p{hs) > 0 since h^ hs G k[t]. So suppose that n = i'p{v) < 0 and let w = ciWi + ... -\- CmW^i- Since Sf^ = S^^^^ p G Sf^, so i/p{Dv) = nhj

7.2 The Limited Integration Problem

247

Theorem 4.4.2, so z^p(/ — w) = n^ which implies that z/p(/) < n or Vp{wi) < n for some i. Hence p~'^ \ hs^ so Vp{hs) > —n, which implies that Up[q) = n + Up{hn) + yp{hs) > i^piK) > 0 . (iii) Let /i = min (i/oo(/), z^oo(^i),•••, ^cx)(^m))- Then UooiDv) = Voo{f Y^=i^i^i) ^ A^ by Theorem 4.3.1. Suppose now that z/oo('^) 7^ 0 and that t is either nonlinear or Liouvillian over k. If t is nonlinear, then v^{Dv) = i^oo(^)—<^(^)+l by Theorem 4.4.4. If t is hyperexponential over k^ then 6{t) = 1, so i'oo{Dv) = z^oo('^) = ^oo(^) — ^(^) + 1 by Lemma 5.1.2. If t is primitive over k, then 6{t) = 0 and Voo{Dv) € {i^ooC'^), ^cx)('^) + 1} by Lemma 5.1.2. Hence z/oo(-ö'y) < ^oo('^) — 5{i) + 1 in all cases, so z^cx)(^) > ^oo{Dv) + 5{t) - 1 > /i 4- (^ - 1. D

Corollary 7.2.1. Let f,wi,... Then,

^Wm^c.hn and hg be as in Theorem 7.2.1.

(i) For any solution c i , . . . , c^ e Const(Ä:) and v G k{t) of (7.30), q = vhn £ k{t) and q is a solution of m

hnDq - qDhn = hlf-Y.

^^^'^^ •

(^•^^)

2=1

Conversely, for any solution with q G k{t) of (7.31), v — q/hn yield a solution of (7.30). (a) If cSp ~ S^^^, then for any solution Ci,... , c ^ G Const{Ä:) and v G k{t) of (7.30), p = vhnhs G k[t] and p is a solution of Dhn^hn-~]p

= hlhsf ~ Y, <^ihlhsWi.

(7.32)

In addition, if t is nonlinear or Liouvillian over k, then either deg(p) = deg{hn) + deg{hs),

deg{p) < deg(/i^)+deg(/i5) + l-(^(t)-min(z/oo(/),^oo(^i),--.,z^oo(^m)) • Conversely, for any solution with p E k[t] of (7.32), v = p/{hnhs) solution of (7.30).

yield a

Proof, (i) Let v G k{t)^ c i , . . . ,Cm be a solution of (7.30), and let q — vhnq e k{t) by Theorem 7.2.1, and

248

7 Parametric Problems n , V^ ^
Dhn . ^ Q-j;2~ + 1^ ^^'^^ -

Multiplying through by /i^ yields (7.31). Conversely, t h e same calculation shows t h a t for any solution with q £ k{t) of (7.31), v = q/hn yield a solution of (7.30). (ii) Let V G k{t)^ c i , . . . , c ^ be a solution of (7.30), a n d let p = vh^hg. Since ^irr ^ ^irr^ ^ ^ ^[^] ^^y Theorem 7.2.1, and ^ J = Dv^y

Dp

Dhn

Dhs

CiWi = — — - P T Ö I - - P-r-r2

,

^

+ Z ^ ^^^^ •

Multiplying through by /i^/is yields (7.32). Conversely, t h e same calculation shows t h a t for any solution with p E k[t] of (7.32), v = q/{hnhs) yield a solution of (7.30). Suppose additionally t h a t t is either nonlinear or Liouvillian over k^ a n d let /i = mm{uoo{f),iyoo{wi),... ,iyoö{wm))- We have i^ooiv) = ^oo{v/{Khs)) = deg{hn)^deg{hs)-deg{p). So if i^oo('^) = 0, t h e n deg(p) = deg(/in) + deg(/is). And if £^oo('^) 7^ 0, then i^oo('^) > /i + (^(t) — 1 by Theorem 7.2.1, which implies t h a t deg(p) < deg{hn) + deg{hs) + 1 - 5{t) - /i. D This gives us an algorithm t h a t reduces a limited integration problem t o ^'irr one over k{t)^ or k\i\ if S^^ = <S"

L i m i t e d I n t e g r a t e R e d u c e ( / , w;i,..., WmyD) (* Reduction to a polynomial problem *) (* Given a derivation D on k{t) and f,wi,... ,Wm G k{t)^ return (a, 6, /i, iV, p, t»!,..., Vm) such that a,b,h G fc[t], iV G N, ^, - ^ i , . . . , Vm G fc(t), and for any solution v G fc(t), ci,...,Cm G C of f — Dv + ciwi + ... CmWm, p = vh £ k{t), and p and the Q satisfy aDp -\- hp — p + cif 1 + . . . + Cm'i^m- Furthermore, if «Sf^ == «S^'^'^, then p G A;[t], and if t is nonlinear or Liouvillian over fc, then deg(p) < A^. *) (dn^ds) ^- SplitFactor(denominator(/), D) for 2 <— 1 t o m d o (en,i,es,i) ^— S p lit Fact or (denominator (t^i), D) C ^ - l c m ( d n , en,l, . . . , en,m)

/in ^— gcd(c, dc/dt) a<~ hn.bi Dhn,N if Sr = S''' t h e n

^0

/15 <— l c m ( ( i s , e s , i , . . . ,es,m)

a ^-- hnhs,b <

Dhn — hnDhg/hg

/i < - m i n ( l / o o ( / ) , i ^ o o ( ' W ^ l ) , . - . ,i^oo('W^m))

N <- deg(/in) + deg(/i,) + max(0,1 - (5(t) - fi) r e t u r n ( a , 6, a, N, ahnf, —ahnWi,..., —ahnWm)

(* exact division *)

7.2 The Limited Integration Problem

249

Example 7.2.1. Let k = Q(x) with D = d/dx and let t be a monomial over k satisfying Dt = 1/x, i.e. t = log(a:), and consider the limited integration problem ^=Dv + cr^ (7.33) which arises when asking whether J x/ log{x)'^dx is expressible in terms of x,log(x) and Li(a:^). We have / = x/t"^ and wi = x/t^ so: 1. 2. 3. 4. 5. 6. 7. 8. 9.

(dn.ds) = SplitFactor(t2,I}) = ( t ^ l ) (en,i,e5,i) = SplitFactor(t,D) = (t,l) c = lcm{t^,t) =t^ K =gcd{t^,2t/x) =t a = t,b = -Dt = -l/x hs = l c m ( l , l ) = 1 /i = min(2,1) = 1 N = deg(t) + deg(l) + max(0,1 - jj.) = 1 ahnf = X, —ahnWi = —xt

so any solution ci G Const(A;) and v G k{t) of (7.33) must be of the form V = q/t where q e k[t] has degree at most one and is a solution of tDq

q = X ~ cixt.

(7.34)

X

In the case of Liouvillian or hypertangent monomials, we are reduced to finding solutions c i , . . . , c^ G Const(/c) and p G k[t] of (7.32), which we rewrite as m

aDp + bp=: go + Y2 ^^di

(7.35)

where a, 6 G k[t]^ a ^ 0 and go^...,gm G k{t). In addition, we have an upper bound n on deg(p) by Corollary 7.2.1. This is equivalent to looking for solutions Co, c i , . . . , c^ G Const(fc) and p G k[t] of m

aDp~\-bp = Y^Cigi

(7.36)

with the additional constraint CQ = 1. Since (7.36) is an equation of type (7.5), we can use the algorithms of the previous section to find all its solutions. This produces / i i , . . . , /i^ G k[t] and a matrix A with coefläcients in Const(A:) such that any solution of (7.36) must be of the form p = Yl^i=i^j^j where ( i i , . . . , (ir ^ Const(/c) and A{co,..., c^, ( i i , . . . , dr)^ — 0. If this linear system has no solution with CQ = 1, then (7.35) and the original hmited integration problem have no solution. Otherwise any solution with CQ = 1 yields a solution of (7.35), hence of the original limited integration problem. Several modifications can be made to the parametric Risch diflFerential equation algorithm for the equations that arise from limited integration problems: if the hnear constraints algorithm produces a 0-dimensional nuUspace,

250

7 Parametric Problems

then there are no solutions with CQ = 1 and we can stop. If it produces a 1-dimensional nullspace, then there is a unique solution with CQ = 1, so replacing c i , . . . , c ^ by t h a t unique solution in (7.35) yields a nonparametric problem to which the S P D E algorithm of Sect. 6.4 is applicable. It is also possible to replace c i , . . . , c ^ by t h a t unique solution in (7.30) and apply t h e in-field integration algorithm of Sect. 5.12, but t h a t would imply recomputing the denominator of v. Finally, we can use the upper bound on deg(p) provided by Corollary 7.2.1 rather t h a n recomputing it. Example 7.2.2. Continuing example 7.2.1, let k = Q{x) with D = d/dx and let t be a monomial over k satisfying Dt = 1/x, i.e. t = log(x), and consider the solutions q G k[t] with degree at most 1 of (7.34). Solving t h a t equation is equivalent to finding a solution with CQ = 1 of the parametric Risch differential equation tDq

q = CQX •— cixt. X

T h a t equation was solved in example 7.1.4, the general solution being {q = jit — Xx , Co = A, ci = 2A). Setting Co == 1, we find t h a t (7.34) has a 1-parameter solution space, namely ci = 2 and q = jit — x'^ for any /i € Const(A:). This means t h a t t h e solutions of (7.33) are ci = 2 and v = q/t = /i — x"^/t for any fi G Const(Ä:) {v is of course defined up to an additive constant). As a consequence, we get xdx

x^

/ log(x)2

log(a:;

+ 2Li(a:2).

7.3 The Parametric Logarithmic Derivative Problem. We describe in this section a solution to the parametric logarithmic derivative problem, i.e. given a differential field K of characteristic 0, an hyperexponential monomial 0 over K and / G ÜT, to decide whether there are integers n,m G Z with n ^ 0 such t h a t / Dv ^ DO nf = -—+m-— V

(7.37)

U

has a solution v E K^ and to find one such solution if there are solutions. As we have seen in Chap. 5, this problem arises from determining whether elements of K{9) are logarithmic derivatives of elements of K{0) or logarithmic derivatives of K(ö)-radicals. We can thus assume t h a t we are able to determine recursively whether elements of K are logarithmic derivatives of if-radicals. Even though equation (7.37) is very similar to (7.30), the limited integration algorithm of Sect. 7.2 is not directly applicable to this problem. However, because the unknown constants are restricted to be integers, the structure

7.3 The Parametric Logarithmic Derivative Problem

251

theorems of Chap. 9 provide a complete solution to this problem whenever they are applicable. In fact they provide the only known complete solution, but we present first a variant of the linear constraints algorithm, which is often able to yield a unique potential solution for m/n, thereby solving the problem. This method is not a complete algorithm since it may fail to determine mjn^ in which case we must revert to the structure theorems and their associated algorithms. As previously, we only study equation (7.37) in the transcendental case, i.e. when K is o, simple monomial extension of a differential subfield A:, so for the rest of this section, let Ä: be a differential field of characteristic 0 and t be a monomial over k. We assume in addition that Const (A; (t)) = Const (/c). We suppose that the coefficients / and D6 jO of our equation are in k{t) and look for solutions n^m e Z and v e k{t). Lemma 7.3.1. Let u^v^w E k{t) and c^c E Const(A:) be such that v j^ 0, c yi^ 0, and u=c

\-cw .

(7.38)

V

Write u — p -{- a/d and w = q + h/e where p^ g, a, 6, (i, e G k[t], d ^ {), e / 0^ gcd(a,(i) = gcd(5, e) = 1^ deg(a) < deg{d) and deg(6) < deg(e). Then, deg{p - cq) < max(0, 5{t) - 1).

(7.39)

Furthermore, let I = l^lg be a splitting factorization of I = lcm((i, e), and l^ be the deflation of In (Definition 1.6.2). Then, lu — clw = 0

{mod Isl~).

Proof. Since u = cDv/v + cto, it follows that u — cw = cDv/v, hence that Uoo{u - cw) = z/oo ( c — j = z^oo ( — ) - ~ max(0,6{t) - 1) by Theorem 4.4.4. Since u — cw = p — cq + a/d — c5/e, either deg(p — cq) = 0, or UQO{U — CW) = — deg(p — eg), in which case deg(p — cq) < max(0, ö{t) — 1). Since I = lcm((i, e), we have lu — clw € k[t]. Let p G S^^^ be any special irreducible factor of ^5. From u ~ cw = cDv/v we get f Dv\ fDv\ Up[u — cw) = Vp I c—- \ = Up I

^ I > 0

by Theorem 4.4.2. Therefore, Up{lu—clw) = i'p{l)-\--Up{u—cw) > iyp{l) = i^pih) since p is special. Since this holds for every irreducible factor of Ig^h \ lu — clw. Let p G k[t] be any irreducible factor of l~. From u — cw — cDv/v we get f-Dv\ fDv\ i/p{u — cw) = Vp f c — j = z/p I I > —1

252

7 Parametric Problems

by Corollary 4.4.2. Therefore, iyp{lu - clw) = Up{l) + iyp{u - cw) > Up{l) - 1 = z/p(/n) ~1 = i^p{ln) since p \ In- Since this holds for every irreducible factor of ^~, l~ \ lu — clw. Finally, lsl~ \ lu — clw since gcd(^5,^~) = 1. D Given u^w E k{t)^ Lemma 7.3.1 either proves t h a t (7.38) has no solution V e k{t) and c^c G Const(Ä:), or it yields a unique candidate c G Const(Ä:) for the solution in the following cases: ® If deg(g) > max(0, S{t) — l): then equating all t h e terms ofp — cq of degree higher t h a n max(0, S{t) — 1) to 0 yields an over deter mined linear algebraic system for c. If this system has no solution in Const(/u), then (7.38) has no solution, otherwise we get a unique candidate for c. ® If deg(p) > max(0,(5(t) — 1) > deg(g): t h e n (7.39) is never satisfied, so (7.38) has no solution. ® If deg(^s/^) > 0, let then r G k[t] be the remainder of lu — clw modulo Isln- If ^ is identically 0, then lu = Iw (mod /s^~), which implies t h a t l^u G k[t] and l^w G k[t] where l^ is the squarefree part of Z^, hence t h a t d and e are normal, in contradiction with deg(/5^~) > 0. Therefore r is not identically 0, so equating all its coefficients to 0 yields an overdetermined linear algebraic system for c. If this system has no solution in Const(fc), then (7.38) has no solution, otherwise we get a unique candidate for c. This in t u r n s yields a method for solving (7.37): given / and Ö, applying Lemma 7.3.1 to u ~ f and w = DO/9, we can either prove t h a t (7.38) has no solution with c G Q, in which case (7.37) has no solution, or get a unique candidate for c = m/n, or fail to get information about c if none of the above conditions is satisfied. If we get a unique candidate c G Q, write c = M/N where M , iV G Z, iV > 0 and gcd(M, N) = 1. T h e n for any solution of (7.37), we must have n = QN and m = QM for some nonzero integer Q, which implies t h a t QNf = Dv/v + QMDO/O, hence t h a t Nf - MDO/O is the logarithmic derivative of a A:(t)-radical, something t h a t we can test recursively. Note t h a t if 6 is an exponential over k{t), t h e n DO/6 = Drj for some 7] G k{t). If Uooiv) < 0^ "then UooiDrj) < - max(0, (5(t) - 1) by Theorem 4.4.4, so deg(g) > max(0, J(t) — 1) and the above m e t h o d succeeds. If i^pir]) < 0 for any normal irreducible p G k[t], then i/p{Drj) < — 1 by Theorem 4.4.2, so p | /~ and the above method succeeds. If iyp{r]) < 0 for any special p G cSf^, then Up{Drj) < 0 by Theorem 4.4.2, so p \ Is and the above method succeeds. Thus, the only cases where the above method can fail when 0 is an exponential over k{t) and S^^^ = Sf^ is if ?7 G k, i.e. 0 is an exponential over k. In a similar fashion, we see t h a t if / = Dg for some g G k{t), then the above method succeeds when S^^^ = Sf^, unless gek. Thus if S^^^ = cSf^, / = Dg and 0 is an exponential over k{t), then t h e above method fails only if f e k and 0 is an exponential over A;, in which case an analysis similar to the one made in Sect. 5.12 shows t h a t for any solution of (7.37), v must be in A:*

7.3 The Parametric Logarithmic Derivative Problem

253

if D t Gfc,in which case we are reduced to solving a similar problem over fc, or V must be of the form v = wt^ for w e k* and an integer q if Dt/t e k. In t h a t latter case, we are reduced to solving an equation of the form nDg =

Dw w

DO Dt + m - — + q-9 t

7.40)

where b o t h 0 and t are exponentials over k. Lemma 7.3.1 can be generalized t o an arbitrary number of w^s (Exercise 7.1) and applied to (7.40). This process stops when we reach the constant field, since Dw = 0 at t h a t point, and (7.40) becomes a linear algebraic equation with integer unknowns. In practice however, it is preferable to use the structure theorems if Lemma 7.3.1 fails t o produce c the first time around.

P a r a m e t r i c L o g a r i t h m i c D e r i v a t i v e ( / , 0, D) (* Parametric Logarithmic Derivative Heuristic *) (* Given a derivation D on A:[t], / G k{t) and a hyperexponential monomial 6 over k(t)^ returns either "failed", or "no solution", in which case nf = Dv/v + mD6/9 has no solution v G k{ty and n^m ^Z with n 7^ 0, or a solution (n,m,v) ofthat equation. *) w ^ DO/e (* It; G k{t) *) d <— denominator(/), e ^- denominator(lo) (p, a) <— PolyDivide(numerator(/), B t h e n s 4— solve( coefficient(p, f) = c coefficient(g, f)^B + 1 < i < C) if s = 0 or s ^ Q t h e n r e t u r n "no solution" (* s G Q *) N ^r— numerator(s), M <— denominator(s) if Q{Nf - Mw) = Dv/v for some Q G Z and v G k{t) with Q 7^ 0 and V y^ 0 t h e n return(QiV, QM, 1;) else r e t u r n "no solution" if deg(p) > B t h e n r e t u r n "no solution" (* deg(g) < B *) I <— lcm((i, e) (Injs)^ SplitFactor(/,D) Z^

Is gcd(ln,dln/dt)

(* 2 == / ^ C *)

if z G /c t h e n r e t u r n "failed" (* n = If (mod hlü) *) {ui,ri) ^ PolyDivide(^/,^) (1^2,^2) ^- PolyDivide(/tü,z) (* r2 = /ic» (mod Islü) *) s ^s— solve( coefficient(ri,r) = c coefficient(r2, t*),0 < ^ < deg(^)) if s = 0 or s ^ Q t h e n r e t u r n "no solution" (* s G Q *) M -e— numerator(s), N ^- denominator(s) if Q{Nf - Mw) = Dv/v for some Q G Z and 1; G k{t) with Q 7^ 0 and V -^0 t h e n return((3A^, QM, f) else r e t u r n "no solution"

254

7 Parametric Problems

Example 7.3.1. Let k = Q(x) with D = d/dx^ t be a monomial over k satisfying Dt = 1/x and 6 be an exponential monomial over k{t) satisfying DO = —6/{xt^)^ i.e. t = log(x) and 0 = e^/iogC^)^ and consider the parametric logarithmic derivative problem 5t2_|_t-6

Dv

DO

,^^^,

for n^m £ Z and i; € Ä:(t). We get l.w = DO/0 = -l/{xt^) 2. (p, a) = PolyDivide(5t2 + t - 6, 2x^2) = (5/(2x), t - 6) 3. (g,6) = PolyDivide(-l,j:t2) = (0,-1) 4. B = max(0,deg^(l/x) -1) = 0,C = max(deg^(5/(2x)),deg^(0)) = 0 5. Since deg(p) < B and deg(g) < B, I = lcm(2xt^,a:t^) = xt"^ 6. (InJs) = SplitFactor(xt2,D) = ( x t ^ l ) 7. z = gcd^(a:t2,2xt) =t 8. (?ii,ri) = PolyDivide(5/2t2 + t / 2 - 3 , t ) = (5/2^ + 1/2,-3) 9. (^2,^2) = P o I y D i v i d e ( - l , t ) = (0,-1) 10. s = solve(-3 = - c ) = 3, so M = 3 and iV = 1 IL f --3w = (5t + l)/(2tx) Using the algorithm of Sect. 5.12, we find that / — 3K; is the logarithmic derivative of a /c(t)-radical, namely

^•^

^

tx

xH

so (7.41) has the solution n = 2iV = 2, m = 2M = 6 and K = xH. Note that it has in fact no solution with n and m coprime. Example 7.3.2. Let k = Q and t be a monomial over k satisfying Dt = 1^ 0 be an exponential monomial over k{t) satisfying DO = 0, i.e. D = d/dt and 0 = 6^^ and consider the parametric logarithmic derivative problem Dv DO 11 = — + m — V

6

,

7.42

,

for m G Z and v G k{t). Even though this problem is trivial, it arises from bounding the degree in 0 of the solutions q G k{t)[0] of ,^ X r. / 9 X 234662231 1255151 (^ + 2t + 1)D, - (llt^ + 22. + 10), = - ^ ^ ^ ^ t + ^ ^ ^

(7.43)

which itself arises from computing the nontrivial integral^ [ 2581284541e^ + 1757211400 i/(e*+i)-iot . J 39916800e3t + 119750400e2* + 119750400e* + 39916800^ *' This example is attributed to M. Rothstein in [34].

7.3 The Parametric Logarithmic Derivative Problem

255

Despite its triviality, (7.42) is not solved by Lemma 7.3.1 because both 11 and DO/Q do not involve t. Since t is a primitive over k, (7.42) has a solution with V G k{t) if and only if it has a solution with v G k. Kt this point, Dv = 0, so (7.42) becomes 11 — TTI^ whose solution yields the degree bound 11 on the solutions of (7.43). There happens to be a solution of degree 11, and the above integral is an elementary function. The structure theorems of Chap. 9 provide an efficient alternative to solving (7.37): suppose first that / has an elementary integral over K^ which turns out always to be the case in the parametric logarithmic problems that arise from the integration of elementary functions. Let then F be an elementary extension of K{9) and g E F he such that / = Dg. Then, if (7.37) has a solution with n 7^ 0, we get , Dv De DivO"^) nj = — + m— V U VU'" which implies that / = Dg is the logarithmic derivative of an F-radical. If F = C[x){ti,... ,tn) where C = Const(i^), Dx = 1, and each ti is either algebraic, or an elementary or real elementary, or a nonelementary primitive monomial over C{x){ti^... ,t^_i), then it can be proven (Chap. 9) that / is the logarithmic derivative of an F-radical if and only if there are r^ G Q such that

J2r,Dt, + J2n^=f

(7.44)

where E = {i G { 1 , . . . , n} s.t. ti exponential monomial over C{x){ti^...,

t^_i)}

and L = {i G { 1 , . . . , n} s.t. ti logarithmic monomial over C ( x ) ( t i , . . . , tj_i)} . Finding the rational solutions of (7.44) can be done by considering it a system of one linear equation for the r^'s with coefficients in F , then applying Lemma 7.1.2 to obtain a system with coefficients in C and the same constant solutions. Assuming that we have a vector space basis containing 1 for C over Q, projecting that system on 1 yields a system with coefficients in Q and the same rational solutions as (7.44). This method is also applicable to equations of the form (7.40) with an arbitrary number of terms in the right hand side, since the existence of a solution implies that / is the logarithmic derivative of an F-radical.

Exercises Exercise 7.1. Prove the following generalization of Lemma 7.3.1: let ^ be a differential field of characteristic 0, t be a monomial over k with Const(A;(t)) =

256

7 Parametric Problems

Const (/c), V^WQ^ ... ,Wn £ k{t) and c i , . . . , c^, c G Const (A:) be such t h a t v ^ 0^ c 7^ 0, and Wo = C V

h > CiWi . ^-^

Write Wi — pi^ai/di for 0 < i < n where pi^ai^di 1 and deg(a^) < deg((i^). Then,

deg

Po - ^

G /c[t],
QPi I < max(0, 8{t) - 1 ) .

Furthermore, let I = Inh be a splitting factorization of I = lcm((io, • • • ^^n), and /~ be the deflation of In- Then, IWQ ~ 2_^ Cilwi = 0

(mod Igl^ ) .

i=l

E x e r c i s e 7 . 2 . Solve the parametric logarithmic derivative problems (7.41) and (7.42) using the structure theorem approach.

8 The Coupled Differential System

We describe in this chapter the solution to the coupled differential system problem, i.e. given a differential field K of characteristic 0 and / i , /2, ö'i, P2 in K, to decide whether the system of equations

has a solution in K x K, and to find one if there are some. It turns out that (8.1) is not really a second order equation, but the coupled system for the real and imaginary parts of a Risch differential equation. Indeed, suppose that {yi^y2) E K X K is Si solution of the slightly more general system

for an arbitrary a G ConstD{K). writing y = yi -\- y2 y/ä we have

Then, since D^/a = 0 by Lemma 3.3.2,

Dy + (A + f2V^)y = Dyi + D(y2)\/^+ (/i + /2v^)(yi +t/2v^) = Dyi + fiyi + 0/2^2 + {Dy2 + /2I/1 + fiy2)Va = 91 -i-92\/ä which implies that y is a solution in K{y/a) of the Risch differential equation Dy + (/i + f2Vä)y = 91+ g2V^ •

(8.3)

Conversely, if y/a ^ K and y = y-^ -\- y2y^ satisfies (8.3) for ^1,2/2 ^ K^ then the above calculation shows that Dyi + fiyi + af2y2 + {Dy2 + Ayi + fiy2)Vä = gi+ 92^/^ hence that Dyi + fiyi 4-a/2?/2 = 9i and i^2/2 + /2?/i 4-/i?/2 = ^2, since {1, y^} is a vector space basis for K{^/E) over K. Therefore, (2/1,^2) is a solution

258

8 The Coupled Differential System

of (8.2). Since coupled differential systems are generated by the integration algorithm only when \ / ^ ^ K^ the above remarks yield a trivial algorithm for finding the solutions in K x K of (8.1): find the solutions y G K{y/'^) of Dy + (/i + f2\/--l)y = gi -\- ö'2\/—T, and let yi and y2 be the real and imaginary parts of y. While this approach is feasible, it has the following inconvenient: when K = k{t) and t is a monomial over k^ the Risch differential equation algorithms of Chap. 6 can generate Risch differential equations to be solved recursively over k. If we adjoin \f^ to K, then the equations to be solved recursively are over k(y/^)^ which means that any other hypertangent monomial in k has to be rewritten in terms of complex hyper exponentials. This cause the eventual solutions yi, ^2 to be in a differential field isomorphic to K rather than in K itself, something that we would like to prevent. In order to avoid this problem, we present in this chapter direct algorithms corresponding to the cancellation cases of the Risch differential equation. When K = k{t) and t is a monomial over /c, these algorithms generate recursively coupled systems of the form (8.1) over k rather than Risch differential equations over fc(v^). As in Chaps. 6 and 7, we only study (8.2) in the transcendental case, i.e. when K is a simple monomial extension of a differential subfield Ä:, so for the rest of this chapter, let A: be a differential field of characteristic 0 and t be a monomial over k. We assume in addition that Const (A:(t)) = Const (A;). We suppose that the coefficients /i,/2,5'i and ^2 of our system are in fc(t), that a G Const(A:) and ^ a ^ k{t), and look for solutions (^1,^2) ^ k{t) x k{t) of (8.2). In the first stage, we let / = /^ + /2\/ä, g = Qi + g2y^ and apply the algorithms of Chap. 6 to the Risch differential equation Dy-\-fy = ^ up to and including the SPDE algorithm. At that point, either we have proven that (8.3) has no solution in k{y^){t)^ in which case (8.2) has no solution in k{t) x fc(t), or we have computed 6, c, d^a^ ß G k{y^)[t] such that any solution in k{^/a){t) of (8.3) must be of the form y = {aq + ß)/d where q G k{^/ä)[t] is a solution of (6.19), i.e. Dq-\-bq = c. Furthermore, we have an upper bound n on deg(g). Although it may have been necessary to solve various problems over k{^/a) recursively during the reduction of (8.3) to (6.19), those problems only occur when we compute various bounds on the poles of y, so after those integer bounds are computed, we are again computing in k[^/a){t)., even though we may have used isomorphic fields during the computation. If we are in one of the non-cancellation cases of Sect. 6.5, then we can apply the corresponding algorithms to the reduced equation Dq -\- bq = c since they do not generate any recursive problem over k{^/ä). Thus, in the non-cancellation cases, we can either prove that (8.3) has no solution in k{y/ä){t), or compute such a solution, thereby solving (8.2). We can therefore assume for the rest of this chapter that we are in one of the cancellation cases of Sect. 6.6, i.e. in one of the following cases: 1. ö{t) < 1, 6 G k{y/E) and D ^ d/dt, 2. 6{t) > 2, deg(6) = S{t) - 1, and deg(g) =

-lc{b)/X{t).

.1 The Primitive Case

259

By our previous remarks, (6.19) is equivalent to

Dq2

+

ah

qi

(8.4)

.12

where q = qi +g2\/ä, b = bi + Ö2\/ä, c = Ci +C2\/ä and gi,g2, &i,^2,ci, C2 are in k[t]. Since y ^ ^ k{t), deg(g) = max(deg(gi),deg(g2))5 so deg(gi) < n and deg(g2) ^ '^- In addition, deg(6) = max(deg(6i), deg(62))5 so 6 G k{y/a) if and only iibi Ek and 62 € k.

8.1 The Primitive Case If Dt E /c, then 5{t) = 0, so the only cancellation case for (8.4) is 5i, 62 ^ ^• If 61 = 62 = 0, then (8.4) becomes Dqi = c\ and Dg2 = ^2 for ci,C2 in k\f\, which are integration problems in Ä:[t], and deciding whether they have solutions in k\f\ can be done by the in-field integration algorithm (Sect. 5.12), so suppose now that 61 G Ä:* or 62 ^ ^*If b\ + 62 Vö- = Du/u for some n G k{y/äy, which can also be checked by a variant of the integration algorithm (Sect. 5.12), let then ui

au2

U2

Ui

where u = ui -\- U2\fä with u\^U2 G k. Letting p == pi + p = (i^igi + 01^2^2) + {u2q\ + uiq2)\ß

(8.5) P2A/Ö

we have

= {ui -f W2\/ä)(gi + g2\/ä) = uq

so Dp = D{uq) = liDg + ^Dti = uDq + ii6g == u{Dq H- 6g) = i^c. This implies that (8.4) becomes

f Dpi\ ^ fui au2\ / c i \ \Dp2J \U2 Ui J \C2j which is, as earlier, a pair of integration problems in k[t]. The change of variable (8.5) is invertible since Ul

CLU2

U2

Ui

aun

(Ui

+ U2y/ä){ui ~ U2\fa) ^ 0 .

Hence, solutions of the integration problems yield solutions gi,g2 of (8-4). Note that a necessary condition for 61 + 62 y ^ = Du/u is 26i = Dv/v for some V E k*^ and that condition can be tested in k rather than k{^/ä). If bi+b2\/ä is not of the form Du/u for some u G k{y/äY^ then D(lc(g)) + b\c{q) ^ 0, so the leading monomial of Dq + bq is (D(lc(g)) + 61c(g))t^^s^^l This implies that deg(g) = deg(c), and that lc(g) is a solution in k{y/äY of (6.23), i.e. Dy -\-by = lc(c). Since deg(g) = max(deg(gi), deg(g2)) and deg(c) = max(deg(ci),deg(c2)), we get that deg(gi) < n and deg(g2) < ^

260

8 The Coupled Differential System

where n = max(deg(ci), deg(c2)). Furthermore, lc(g) = yi + y 2 v ^ and lc(c) = zi + Z2 y/a where ?/i, 1/2, ^i and Z2 are the coefficients of t^ in gi, g2 7 ci and C2. Therefore (6.23) is equivalent to

(DyA \Dy2j^\b2

(h

ab2\fyi\_fz, h ) \ y 2 )

\z2

which is a coupled differential system in k. If it has no solution in k^ then (8.4) has no solution in k\t]. Otherwise, since it is equivalent to (6.23), Lemma 5.9.1 implies t h a t it has a unique solution y 1,2/2, which must be the coefficients of f^ in qi and ^2- Replacing each qi by yif^ + hi in (8.4), we get (Dhi\ {Dh2j

(hi ^ \b2

ah2\(hi\_(ci~-D{yit^)\__(hi 61 ; ^ 2 / ~ \C2-D{y2nj

{b2

ah^ h J

(yit^ \y2t^

which is a system of the same type as (8.4) with the same bi and 62 as before, but a bound of n — 1 on deg(/ii) and deg(/i2). We can therefore repeat this process, decreasing the bound each time until we have solved (8.4).

C o u p l e d D E C a n c e l P r i m ( a , 61,62, ci, C2, D, n) (* Cancellation - primitive case *) (* Given a derivation D on k[t], n either an integer or +cx), a G Const(A;), ^1, ^2 ^ k and ci, C2 G k[t] with Dt G /c, ^/ä ^ k(t) and 61 7^ 0 or 62 7^ 0, return either "no solution", in which case the system (8.4) has no solution with both degrees at most n in k[t], or a solution gi, 92 G k[t] X k[t] of this system with deg(gi) < n and deg(g2) < n. *) if 6 = Dz/z for z G k{^/Ey t h e n {^ zi,Z2 E k *) z = zi -\- Z2\/ä if zici + a2;2C2 = Dpi and Z2C1 -\- z\C2 = Dp2 for pi,p2 G k[t] with deg(pi) < n and deg(p2) < n t h e n return((2;ipi - az2P2)/(^i - az^), {zip2 - Z2Pi)/{zj - azf)) else r e t u r n "no solution" if ci — 0 and C2 = 0 t h e n r e t u r n (0,0) if n < max(deg(ci), deg(c2)) t h e n r e t u r n "no solution"
8.2 The Hyperexponential Case

261

8.2 The Hyperexponential Case If Dt/t = T] e k, then 5{t) = 1, so t h e only cancellation case for (8.4) is 61,62 e ^• If bi = b2 = 0, then (8.4) becomes Dqi = ci and Dq2 = C2 for ci,C2 in k[t], which are integration problems in /c[t], and deciding whether they have solutions in k\i\ can be done by the in-field integration algorithm (Sect. 5.12), so suppose now t h a t hi G Ä:* or b2 G k*. If bi + b2^/ä — Du/u + mr] for some u G k{y^y and ??i G Z, let t h e n Pi\_fui

au2\

(qit'^\

V2)-\U2

u,){q2t-)

.^^x ^'-'^

where u = ui -i- U2\fä with wi, 1^2 ^ ^- We have Uiq2f^)^/ä p = Pl + P 2 \ / ä = {uiqif^ + au2q2f^) + {u2qit'^ + = (^xi + t^2V^)(gi + g2v^)t'^ = ^ g t ^

I ) p = D(2xgt^) = (^iDg + qDu + mr]uq)e'

= u{Dq + 6g)t^ = i^ct^

which implies t h a t (8.4) becomes Dpi\ Dp2j

_ (^i ~ \U2

au2\ f cif ui J \c2t-

which is a pair of integration problems in k{t)^ and deciding whether t h e y have solutions in k{t) can be done by a variant of the integration algorithm (Sect. 5.12). As in the primitive case, t h e change of variable (8.6) is invertible, so solutions of the integration problems yield solutions of (8.4). Note t h a t a necessary condition for 61 + 6 2 \ / ä = Du/u-\-mr] is 26i = Dv/v-\-2mr] for some V G /c*, and t h a t condition can be tested in k rather than k{^/ä)^ yielding a unique potential candidate for the integer m. Suppose finally t h a t 61 + 62^/^ is not of the form Du/u + mrj for some u G k{^Y and m G Z. Then D{\c{q)) + deg(g) ?7lc(g) + b\c{q) / 0, so t h e leading monomial of Dq + bq is (D(lc(g)) + deg(g) ry lc(g) + 61c(g)) t^^^C^). This implies t h a t deg(g) = deg(c), and t h a t lc(g) is a solution in k{^/äy of (6.24), i.e. Dy-\-{b-\-deg{q)r])y = lc(c). Since deg(c) = max(deg(ci),deg(c2)) and deg(g) = max(deg(gi),deg(g2)), we get t h a t deg(gi) < n and deg(g2) < ^ where n = max(deg(ci),deg(c2)). Furthermore, lc{q) = yi + ^ 2 \ / ß and lc(c) = zi + Z2\^ where ^1,^2, ^1 and Z2 are the coefficients of f^ in gi, ^2, ci and C2. Therefore (6.24) is equivalent to

262

8 The Coupled Differential System Dyi\ Dy2 J

f bi-\-717] V ^2

ab2 \(y^\ h+rir]J\y2j

= (^'^ \Z2

which is a coupled differential system in k. If it has no solution in k^ then (8.4) has no solution in k[t]. Otherwise, since it is equivalent to (6.24), L e m m a 5.9.1 implies t h a t it has a unique solution yi,y2, which must be the coefficients of f^ in qi and g2- As in the primitive case, replacing qi by yif^ + hi and q2 by 2/2^"' + ^2 in (8.4) yields a system of the same type with a lower degree bound on its solutions, and a lower degree right hand side.

C o u p l e d D E C a n c e l E x p ( a , bi,b2,ci,C2, D,n) (* Cancellation - hyp er exponential case *) (* Given a derivation D on k[t], n either an integer or +oo, a G Const(Ä;), hi,b2 ^ k and ci,C2 G k[t] with Dt/t G k, ^/a ^ k{t) and &i 7^ 0 or 62 7^ 0, return either "no solution", in which case the system (8.4) has no solution with both degrees at most n in k[t], or a solution ^1,^2 ^ k[t] x k[t] of this system with deg(gi) < n and deg{q2) < n. *) ifb = Dz/z 4- mDt/t for 2; G /c(^/ä)* and m ^ Z t h e n 2; = 21 + 2:2^/0 (* 2:1, 2:2 G fc *) if (2:1 ci + az2C2)t'^ = Dpi and (2:201 + 2:102)^'^ = Dp2 for pi,P2 G fc(t) then gi ^ (^ipi - a2;2P2)t~"^/(2;i -

azl)

92 ^ (^iP2 - Z2Pi)t~'^/{zl - azl) if gi G A:[t] and g2 G k\t] and deg(gi) < n and deg(j92) < ^' Tt h e n return(gi,g2) else r e t u r n "no solution" else r e t u r n "no solution" if ci = 0 and C2 = 0 t h e n r e t u r n (0,0) if n < max(deg(ci),deg(c2)) t h e n r e t u r n "no solution" gi ^ 0, g2 <- 0 while ci 7^ 0 or C2 7^ 0 d o m ^— max(deg(ci), deg(c2)) (* m becomes smaller at each pass *) if n < m t h e n r e t u r n "no solution" (si,S2) ^ C o u p l e d D E S y s t e m ( 6 i + mDt/t,62, coefficient(ci,f^), coefficient(c2,t"^)) if (51,52) = "no solution" t h e n r e t u r n "no solution" qi ^ qi -f sit"^, q2^

q2 + 82^^

n ^~ m — 1 Ci^CiC2^C2-

D(5it"^) - {biSi + ab2S2)f^ D ( 5 2 t ^ ) - (^251 + 6152)^"

r e t u r n (91,^2)

8.3 The Nonlinear Case

263

8.3 The Nonlinear Case If S{t) > 2, then t h e only cancellation case for (8.4) is max(deg(öi), deg(62)) = 6{t) — 1 and lc(6) = —nX{t) where n > 0 is the bound on deg(g). Since lc(6) = ßi -\- ß2\/ä where ßi and /?2 are the coefficients of t^^^'^~^ in bi and 62, and A(t) E k^ we must have ß2 = 0 and ßi = —nX{t), i.e. deg(6i) = 6{t) — 1, deg(62) < S{t) — 1 and lc{bi) = —nX{t). As in the Risch differential equation case, there is no general algorithm for solving the system (8.4) in this case. If however S^^^ / 0, then projecting (8.4) t o k[t]/{p) for p E S^^^ can be done: with D* being the induced derivation on k[t]/{p), applying Tip to (8.4) we get ^*^n

. (Mh)

^*^2y

^ fM^i)\

rrp{^)7Tp{b2)\ (Qt\

\Mh)

TTpih) J

\QV

\M^2)J

(g^)

^' ^

where ql = 7Tp{qi) and g | = 7rp{q2)- Assuming t h a t we have an algorithm for solving (8.7) in k[t]/{p)^ we can then solve (8.4) as follows: if (8.7) has no solution in k[t]/{p)^ then (8.4) has no solution in k[t]. Otherwise, let (g|, gl) ^ k[t]/{p) X k[t]/{p) be a solution of (8.7), and let r i , r 2 G k[t] be such t h a t deg(ri) < deg(p), deg(r2) < deg(p), 7rp(ri) = ql and 7rp(r2) = g | . Note t h a t TTp{Dri + bivi + a62r2) = 7rp(ci), and 7rp{Dr2 + b2ri + 6ir2) = ^p{c2), so p \ ci — Dri — biri ~ab2r2 and p\ C2 — Dr2 — ^2^1 ~-bir2. In addition, 7rp{qi) = 7Tp{ri) and 7Tp{q2) = 7rp(r2), so hi = {qi - ri)/p G k[t], /12 = (g2 " ^ 2 ) / ^ G A:[t] and we have

C2J

\Dq2J

\b2

-pyiDhJ

bi )

+i

\q2

b,

6i +

p

^JU2

SO (/ii, /12) is a solution in /c[t] x fc[t] of

DhA^fh

+f

Dh2j^\

62 1 f[ci P

ab2 b^ +

\fh, ^jyh2

— Dri\ C2 ~ Dr2 J

[bi ab2\ (ri V &2 bi I \r2

which is a coupled system of type (8.4), but with a lower bound on the degree of its solutions since deg(/ii) < deg(gi) — deg(p) and deg(/i2) < deg(g2) — deg(p). As was the case for Risch differential equations, there are cases when (8.7) can be solved, for example when we can find an element of degree one, or an element with constant coefficients, in S^^^. Although y ^ ^ k{t), it may happen t h a t ^y7^p{a) G k[t]/{p), in which case two new difficulties arise:

264

8 The Coupled Differential System

® p is then reducible over k{^)^ so we must use an irreducible factor p of p in k{y^)[t] rather than p. ® (8.7) is not equivalent to a Risch differential equation over k[t]/{p) anymore, so we must revert to solving (6.25), taking care to generate coupled systems over k recursively, rather than Risch differential equations over

fc(v^). An example of those difficulties is provided by the hypertangent case with a = - 1 and <S^^^ = {t^ + 1}.

8.4 The Hypertangent Case If Dt/{t^ -h 1) = ?7 G /c, then 5{t) = 2, so the only cancellation case for (8.4) is b2 £ k and bi = bo — nr]t^ where b^ E. k and n > 0 is the bound on deg(g). In such extensions, the method outlined above is not applicable for a = - 1 , since 5^^^ = {t^ + 1}, and yfÄ. e k[t]/{t'^ + 1). Since ^/^ ^ k{t) by assumption, (8.4) is equivalent to (6.19), in this case Dq + (6o - nr]t + 62V-1)^ = ci + C2^ and we can use the method of Sect. 6.6: taking

(6.25) becomes Dg* + {bo + (62 - nr/)/=4)g* - c i ( v ^ ) + where D is extended to k{^f~i) Writing q* = yi + y2\/~-^

and

by D\f~i

(8.9)

C2(A/^)V^

= 0 and g* = g ( v ^ ) G

k{^/-l).

Ci(v^) + C2(\/^)\/^ = ^i+2:2\/^

where yi^y2^zi, Z2 G k^ (8.9) is equivalent to Dyi \ , f bo Dy2j ' \b2-nr]

nrj-b2\ fyi\ bo )\y2)

__ f zi V^2/

^""

which is a coupled differential system in /c. If it has no solution in k^ then (8.9) has no solution in k{^/~—l)^ which implies that (8.4) has no solution in k\t]. Otherwise, if yi^y2 E k x k is di solution of (8.10), letting r = yi + ?/2A/--T, h = {q — r)/{t — A / ^ ) is a solution in k{\/~^)\t] of degree at most n — 1 of (6.26), which in this case becomes Dh + (bo - (n - l)j?i + (62 + V)V^)

h

t-^J-l

ci - zi + nri{yit + 7/2) + (c2 - 2:2 + nr}{y2t - j / i ) ) \ / ^

8.4 The Hypertangent Case

265

where the right hand side is an exact quotient in k{\/'^)[t]. Repeating this process we either prove t h a t (8.4) has no solution in k[t], or obtain a solution q e k{^/^)[t] of (6.19). Writing q = Qi + q2V^ with qi,q2 £ k[t], we get t h a t gi,g2 is a solution of (8.4).

CoupledDECancelTan(6o, ^2, ci, C2, D, n) (* Cancellation - tangent case *) (* Given a derivation D on k[t], n either an integer or +oo, 6o, &2 G A: and ci, C2 e k[t] with Dt/{t'^ -\-l) =r] e k, v ^ ^ ^ k{t) and 6o 7^ 0 or 62 7^ 0, return either "no solution", in which case the system Dqi \ . fbo ~ nr]t Dq2 J \ b2

-62 \ f qi\ 60 + nr]t) \q2)

^ f d \C2

has no solution with both degrees at most n in k\t]^ or a solution ^1,^2 ^ k\t] X k\t] of this system with deg{qi) < n and deg(g2) < n. *) if n = 0 t h e n if ci G A: and C2 Gfct h e n r e t u r n CoupledDESystem(5o, 62,01,02) else r e t u r n "no solution" T] ^ Dt/{t^ + 1) (* t = t a n ( / 77) *) c i ( v ^ ) + C 2 ( \ / ^ ) \ / ^ = ^1 + 2 ; 2 \ / ^ (* 2:1,2:2 G A; *) (51,52) <— CoupledDESysteni(6o,62 — n?7, ^1,2:2) if (si,S2) — "no solution" t h e n r e t u r n "no solution" c ^ (ci - zi-{- nr}{sit + S2) + (c2 - 2:2 + nri{s2t Sl))^/-i)/p c =
Example 8.4-1- Let A: = Q(x) with D = d/tix, and let t be a monomial over k satisfying D t = 1 + 1 ^ , i.e. t = t a n ( x ) , and consider the coupled system DyA.fO Dy2j^{Ax

-4x\fyA_(-{t^-2t 0 J \ y 2 j - \

+ 8x^-l)/{t^ + 2{1 - 2x) / (t^ + 1)

l)\ J

^^'''^

which arises from computing

/

(tan(x)2 - 2 t a n ( x ) -f- 8^^ - 1) tan(x2) + 4x - 2 dx. (tan(x)2 + l)(tan(a;2)2 + i )

(8.12)

T h e system (8.11) is equivalent to the Risch differential equation r-r (t2-2t + 8x2-l) + 2(2x-l)/=T Dy + 4xy^/^ = -^ — ^, T ^ ^ t^ + 1

,^ _ , 8.13

266

8 The Coupled Differential System

over k{y^^){t). Since 4 x \ / ^ is weakly normalized w.r.t. t and the denominator of the right hand side of (8.13) is special, Theorem 6.1.2 implies that any solution in k{^/^){t) of (8.13) must be in k{^/^){t). With a = l,b = 4 X A / ^ and e z= ±1^ we have

'^^^-'K-i.V^i

«(eV=T)

which is not of the form 2mev^-4 + Du/u for m € Z and u G k{^/^y, so J^t-e^/^iiDy + 4.xyy/-l) = i/^_,^/3T(?/) for e = ±1 and any y G k{^/-i){t) by Lemma 6.2.1 (see also Exercise 6.1). Hence, i'i__^^j-z:i{y) > — 1 for any solution y G k{\/^^)[t) of (8.13), which implies that any such solution must be of the form y = g/(t^ + 1) where q G k{^/~^)[t]. Making that substitution in (8.13) we obtain Dq + {Ax^f^. - 2t)q = -{f - 2t + Sx^ - 1) + 2(1 - 2x)^f-i

(8.14)

which is equivalent to the system Dqi\ Dq2j^\

f-2t 4x

-4x\ -2tJ

fqA __ f-t^ + 2t - Sx^ -}-1 \q2j ~~ \ 2(1 - 2x)

.15)

With a = 1, 6 = 4 x / ^ - 2t and c = {-t^ + 2t - 8^^ + 1) + 2(1 - 2x) we have lc(6) lc(a)

2 > deg(c) - deg(6)

so any solution q G k{y/^)\t] of (8.14) has degree at most 2 by Lemma 6.3.5. Since deg(6) = 1 and lc(5) = —2, we are in the cancellation case of this section. Applying C o u p l e d D E C a n c e l T a n to 6o = 0, 62 = 4x, ci = - t 2 + 2t--8x^ + 1, C2 = 2(1 — 2x) and n = 2, we get 2. T] = Dt/lt^ + 1) = 1 3. c i ( v ^ ) + C 2 ( V ^ ) A / ^ = 2(1 - 4^2) + 4(1 - x ) v ^ so zi == 2(1 - 4x^) and Z2 = 4(1 ~ x) 4. Since (DsA

(

0

VD.2;'^ V2(2x-i)

2(l-2x)\ / 5 i \

0

; 1^2;

/2(l-4x2)

V 4(1-x)

has the solution si = —1 and S2 = 2x + 1, (si,52) = C o u p l e d D E S y s t e m ( 0 , 4 x - - 2 , 2 - 8 x 2 , 4 - 4 x ) = ( - l , 2 x + l) 5. c=

t2-2(2x + l ) t v ^ ~ 4 x - l ,^ ^, r-T ^ ^-j=—— = - t + (4x + 1 ) \ / ^

so di = —t and d2 = 4x -\-l

8.4 The Hypertangent Case

267

6. recursive call, CoupledDECancelTan(0,4x + 1, - t , 4x + 1, D, 1): a) c i ( - \ / ^ ) H- C2{y/—l)\/^ = 4 r c \ / ^ , so zi = 0 and Z2 = 4a: b) Since

[DsAfO \Ds2J ^\ix

-Ax\ fsA 0 y ^^2;

^fO \^x^

has the solution si = 1 and 52 = 0, (51, S2) = CoupledDESystem(0,4x,0,4a:) = (1,0) c) c = ( - t + t + (4J: + 1 - 4x - l ) v ^ ) / ( t - v ^ ^ ) = 0, so (^1 = ^2 = 0 d) CoupledDECancelTan(0,4a: + 2,0,0,1), 0) returns (0,0), so we re=turn hit + /i2 + 5i == 1 and /12t — hi -j- S2 = 0 7. We obtain (/ii, /i2) = (1,0) from the recursion, so we return the following solution of (8.15): qi = hit + /i2 + "51 = t — l

and

q2 = /i2^ — /ii + 52 = —1 + 2x + 1 = 2x

We conclude that a solution of (8.11) is y^

=

t-1 t2 +I 1l

4.9

^ ^^^

^2

^"

=

2x t2 + l

hence that (tan(x)2 - 2tan(x) + Sx^ - 1) tan(x2) + 4x - 2 (jnf zzz (tan(x)2 + l)(tan(x2)2 + l) (tan(x) — 1) tan(x2) + 2x f tan(x) — 1 , (tan(x)2 + l)(tan(x2)2 + l) ^ J '^tan(x)2 + T "^ and the latter integral does not involve tan(x2) anymore. Applying the algorithm of Sect. 5.10 to it, which involves solving a coupled differential system over Q(x), we find

/

tan(a:) — 1 , ^tan(:c)2 + l

x{l — x) tan(x)2 + (1 — 2x) tan(x) — ^2 — j : — 1 2(tan(x)2 + 1)

which yields a complete formula for (8.12).

9 Structure Theorems

We present in this chapter proofs of the various structure theorems that were used in Chap. 7 for solving the parametric logarithmic derivative problem. Although they are used in the integration algorithm, the main application of structure theorems is to determine algebraic dependencies between functions.

9.1 T h e Module of Differentials We first need to slightly generalize the concept of derivation we used previously. Definition 9.1.1. Let S C R be commutative rings and M be an R-module. An iS-derivation of R into M is a map D : R —^ M such that for any x^y G R: (i) D{x + y) = Dx + Dy. (a) D{xy) = xDy + yDx. (Hi) Do = 0 for any c E S. Note that a derivation of R in the sense of Chap. 3 is an ^-derivation of R into R for any subring S of Consti3(i?), in particular for S = Z. The usual properties of derivation (Theorem 3.1.1) are easily generahzed to 5-derivations. T h e o r e m 9.1.1. Let S C R be commutative rings, M be an R-module, and D : R -^ M be an S-derivation. Then, (i) D{cx) = cDx for any c £ S and x E R. (a) If R is a field, then X yDx — xDy y

y'^

for any x,y e R, yj^O. (Hi) Dx^ = nx'^ ^Dx for any x E R\ {0} and any integer n > 0 (any intege r n if R is a field).

270

9 Structure Theorems

(iv) Logarithmic

derivative

identity:

D{u\K..ul-)

—— = ei

U-^' . . . Un

for any ui^...

Dui Ui

Dun

h . . . + en

Un

^Un G R* and any integers e i , . . . , en-

(v) ^ DP{Xi,...,Xn)

= ^

dP -^{Xi,

for any : c i , . . . , x^ € R and any polynomial

...

,Xn)DXi

P with coefficients

in S.

Proof The proofs are similar t o the proofs of the corresponding statements in Theorem 3.1.1 and are left as exercises. D Let ß be a commutative ring and ^R be the free fi-module generated by the symbols Ox for all x G R. Its elements are all the finite sums ^ ^ aiöxi with üi^Xi E R. For any subring S of jR, let ^R/S be the submodule of $R generated by 6{x -\- y) — 5x — 5y and 5{xy) — x8y — y5x for all x^y £ R and ÖC for all c G S, and let ÜR^S be the quotient module ^R/^R/SIt is easily checked t h a t the m a p dR^s ' R ~^ ^R/s that sends x € -R to the equivalence class of 5x is an ^'-derivation of R into OR/SThe pair {OR^SJ^R^S) is called the module of S-differentials of R^ and we omit the subscript on d when the context is clear. The 5-differentials of R are all the finite sums ^2% ^idxi with a^, x^ € fi, subject to the relations d{x -\- y) = dx -\- dy^

d{xy) = xdy + ydx

for all x^y G R

and (ic = 0 for all c G S. If S Q T C R are commutative rings, then ^R/s is a submodule of ^R/T- This implies the existence of a canonical projection TT : ÜR/S -^ ^R/Ti which is the surjective Ä-linear m a p given by TT {J2i O^idR/s Xi) = Yli ClidR/T Xi. We first show t h a t QR/S is a universal object, i.e. t h a t every S'-derivation can be factored as in t h e following diagram:

L e m m a 9 . 1 . 1 . Let S C R be commutative rings^ M be an R-nnodule, and D : R -^ M be an S-derivation. Then, there is a unique R-linear map D : ^R/s -^ M such that D = Dd.

9.1 The Module of Differentials

271

Proof. Since ^R is a free Ä-module, let D : <^R —> M be the i?-linear map given by D{6x) = Dx for all x G R. Since D is an 5'-derivation, D{6c) = Dc = 0 for all c E S. Furthermore, D{6{x -{-y) — 5x — 6y) = D{x + 2/) — Dx — Dy = 0 and D{S{xy) — x6y — yöx) = D{xy) — xDy — yDx = 0 for any x^y G -R, which implies that ^R/S Q kerD, hence that D induces an i?-linear map D : ÜR/S -^ M. For any x £ Rwe have Ddx = D5x = Dx, so D = Dd. Suppose that Di and D2 are both i?-linear maps from QR/S into M such that D = Did = D2d. Then, since any uj G f^R/s is a finite sum of the form LÜ = J2i didxi with ai^^Xi G ß , we get by linearity of Di and D2 that

hence that D is unique.

D

Any ring homomorphism induces a skew-linear map on the differentials such that the following diagram commutes:

aT/S'^

Q R/S-

R L e m m a 9,1.2. Let T and S C R be commutative rings, and a \ R —^ T he a homomorphism. Then, there is a unique map a* : QR/S ~^ ^T/S'^ such that (i)

{uj + r]Y* = a;^* + r]''* for any uj,r] e

(a)

{XUJY

=• x^uj^

for any ÜÜ G OR^S

QR/S

o.nd x G R.

(Hi) cr*d = da Proof. Let ä : ^R —^ # T be the map given by

r^aiSxA

=Y,a15{x1)

for all finite sums with ai^Xi E R. W is well-defined since $R is free over R. Furthermore, W is an abelian group homomorphism by definition. Since a is a homomorphism, 1^ = 1, so aS = Öa. In addition we have {6cY = S{c^) G ^T/S'^ for all c G /S. Furthermore,

{5{x + y)-6x-

6yf = S{{x + yY) - Six") - Siy") = 5{x-+y-)-S{x-)-6{y'')

e^r/S"

272

9 Structure Theorems

and {5{xy) - x5y - ySxf

= 5{{xyY) - x''5{y'') - y''5{x'') = Six-^y-) - x^5{y'^) ~ y'^5(x'')

6 ^T/S-

for any x^y e R^ which implies that {^R/SY ^ ^T/s^i hence that ä induces an abehan group homorphism a* : ÜR/S -^ ^T/S^ that satisfies

{Ta^dxA

=^
Furthermore, a*d = da since ad — da. Let x G R and cu £ f^R/Si ^tnd write (JO = ^^aidxi with ai^Xi G R. Then,

{XUJY^

= I Y^ üixdxi I

= Y,{aixYd{x^)

= x^ ^

a,""^«) =

X'^CÜ^*

.

Suppose that ai and (J2 are both maps from OR/S into ÜT/S^' that satisfy the lemma. Then, since any a; G ^R/S is a finite sum of the form u = J2i ^idxi with tti^Xi e R, we get UJ

'^=Y.a^d{ X,

=

UJ-

hence that a* is unique.

D

In a similar manner, we can show that a derivation on R induces a skewderivation on the differentials such that the following diagram commutes:

Ü R/S'

R

D* ^

D

^R/S

R

L e m m a 9.1.3. Let [R^D) he a differential ring and S C R be a differential subring. Then, there is a unique map D* : ÜR/S -^ ÜR/S such that (i)

D*(a; + 77) = D'^uj + D*r] for any LÜ^TJ e

ÜR/S

(a) D*{xuj) = {Dx)üj + xD^'uj for any uj E ÜR/S and x E R. (Hi) D'^d = dD Proof Let D : $R —> ^ ^ be the map given by

9.1 The Module of Differentials D I y . diSxi

= 2.{Dai)öxi

273

-(- aiS{Dxi)

for all finite sums with ai^Xi £ R. D is well-defined since $R is free over R. Furthermore, D is an abelian group homomorphism by definition. Since D is a derivation, Dl = 0, so D5 = 6D. Since 5 is a differential subring, DS C ^S', so D{5c) = 5{Dc) G ^R/s for all c G 5. Furthermore, 'D{6{x + y) -Sx~

6y) = 6D{x + y) - öDx - 6Dy = ö{Dx + Dy) - 5Dx - 5Dy

£

^R/S

and D{6{xy) — x6y — ySx) = 6D{xy) — {Dx)6y — x5Dy — {Dy)6x — yÖDx = 6{xDy + yDx) — 6{xDy) + {6{xDy) - {Dx)6y -6{yDx)

x5Dy)

+ {6{yDx) - {Dy)Sx -

= {6{xDy + yDx) - 6{xDy) + {S{xDy) - {Dy)6x -

y5Dx)

6{yDx))

xöDy)

+ [S{yDx) - {Dx)6y - y5Dx)

G ^R/S

for any x,y e R^ which implies that D^R/S ^ ^R/S^ hence that D induces an abehan group homorphism D* : QR/S —^ ^R/S that satisfies ^ * ( A2^^^^^

I "^ '^{Dai)dxi

+

aid{Dxi).

Furthermore, D*d = dD since Dd = (ijD. Let x G R and cj G OR^SI ^^^ write a; = Yli^idxi with ai,x^ G Ä. Then, D*(xcc;) = D* y ^ ajxdxj j = V^(Da^x)(ix4 -h aixd{Dxi) = V^ ai{Dx)dxi 4- x{Dai)dxi + aixd{Dxi) = {Dx)uj + xD*u . i

Suppose that Di and D2 are both maps from QR/S into i7ß/5 that satisfy the lemma. Then, since any u G OR/S is a finite sum of the form uo = ^ ^ a^dx^ with ai^Xi G i?, we get Diuj = 2^{Dai)dxi i

+ aiDi{dxi) = 2_\{Dai)dxi + aid{Dxi) i

= ^ ( I ^ a i ) d x i 4- aiD2{dxi) = D2UJ i

hence that D* is unique.

D

274

9 Structure Theorems

L e m m a 9.1.4. Let S C R be commutative rings, [O^is^d) S-differentials of R, and B C. R. Then,

he the module of

(i) d{S[B]) and {d})}}y^B generate the same suhmodule of Q^js over R. (ii) If R and S are fields, then d[S(B)) and {dl)\i)^ß generate the same submodule of QR/S over R. Proof For any S C QR/S^ we write R{S) for the submodule of OR/S generated by S over R. (i) Let p e S[B]. Then p = P ( x i , . . . , x ^ ) where Xi e B and P e S[Xi,..., Xn] is a polynomial with coefficients in 5. Therefore,

i=l

*

by Theorem 9.1.1, so R{d{S[B])) C R{d{B)). Since B C S[B], R{d{B)) C R{d{S[B])) so both submodules are equal. (ii) Suppose that R and S are fields, let x G S{B) and write x = p/q where p,qe S[B] and q^O.By Theorem 9.1.1, p qdp-pdq ax = a- = q q^

1 p = ~ dp -—7: dq . q q^

Since dp and dq are in the span of {db}}y^s by (i), we conclude that R{d{S{B))) C R{d{B)). Since B C S{B), R{d{B)) C R{d{S{B))) so both submodules are equal.

D

We now determine the dimension of QR/S over R when R and S are fields. L e m m a 9.1.5. Let k C K be fields of characteristic 0 and B C K be algebraically dependent over k. Then, {db}b^ß is linearly dependent over K. Proof. Since B is algebraically dependent over A:, there a r e i C ] ^ , . . . , Xfi € B and a polynomial P € k[Xi,..., Xn] \ {0} such that P ( x i , . . . , Xn) = 0. Let Q be a nonzero polynomial with coefficients in k and of minimal total degree such that Q ( x i , . . . ,Xn) = 0. Applying d and Theorem 9.1.1 we get 0 = dO = dQ{xi,...,Xn)

= y^

^^{xi,...,Xn)dxi.

Since Q ^ k and k has characteristic 0, dQ/dXi^ is not identically 0 for some io- By minimality of the total degree, (öQ/öXio)(xi,..., x^) / 0, which implies that d x i , . . . , dx^ hence {db}h^ß^ are linearly dependent over K. D T h e o r e m 9.1.2. Let k C K be fields and B C K be algebraically independent overk. If K is separable algebraic overk{B), then {db}ij^ß is a basis for Qx/k over K.

9.1 The Module of Differentials

275

Proof. Let x G K^ then x is separable algebraic over k{B)^ so let P G k{B)[X] be its minimal irreducible polynomial and write P = J2T=o^J'^'^ where ao, • • •, CLm ^ k{B). Since d is a Ä:-derivation of K into O^/k we get

(

m

\

771

V ^ ßj 3;-^ I = V ^ x^ düj + j üj x^ ~^dx j=0 J j=0 m

m

= 2 , oc^duj + dx y^ jajX^~^ . j=0

Let a = YlT=iJ^j^'^~^

j=l

^ ^ - Since x is separable over k{B), a / 0, so dx = — y

—daj a

j=0

is in t h e subspace of Ox/k generated by d a o , . . . , da^^ hence in t h e subspace generated by d{k{B)). By Lemma 9.L4, this implies t h a t dx is in the subspace of Ox/k generated by {db}h£ß. Since this holds for any x G K , {db}beB generate Ox/kSuppose t h a t X]?=i ^jdxj = 0 for some a i , . . . , a^ G K and xi^... ^Xn G B. Since B is algebraically independent over /c, d/dxi^..., d/dx^ are derivations on A:(B) by Theorem 3.2.2. Those derivations can be extended t o derivations of K by Theorem 3.2.3. Since k C ConstDi{K), each d/dxi is a fc-derivation of K into K , so let D i , . . . , Dn be the induced ÜT-linear maps from Ü^/k i^^o i f given by L e m m a 9 . L I . Applying Di and Lemma 9 . L I we get

y^ajdxj

I = y^ajDi{dxj)

= / . % ' ^ ~ ^ =" ^i

which implies t h a t c i x i , . . . ^dxn are linearly independent over if, hence t h a t {(i6}5^ß is linearly independent over K. D As a consequence, the dimension of ü^/k over K is exactly t h e transcendence degree of K over k. Another consequence is t h a t in characteristic 0, algebraically independent elements yield linearly independent differentials. C o r o l l a r y 9 . 1 . 1 . Let k C K be fields of characteristic 0. Then, B C K is algebraically independent over k if and only if {db}beB ^ ^K/k ^^ linearly independent over K. Proof. Let ^ be a transcendence basis of K over k containing B. Since the fields have characteristic 0, K is separable over k{A)^ so {dh]i)^j^^ and therefore {db}heß^ is hnearly independent over K by Theorem 9.L2. Conversely, if {db]i)^ß is linearly independent over if, then B must b e algebraically independent over k by Lemma 9.L5. D

276

9 Structure Theorems

Corollary 9.1.2. Let k C K he fields and t G K be transcendental over k. If K is separable algebraic over k{t), then ox dx = -r- dt dt in ÜK/k foT any x E K, where d/dt is the derivation on K that maps t to 1 and every element of k to 0. Proof Note that d/dt is uniquely defined on K by Theorems 3.2.2 and 3.2.3. Since t is transcendental over k and K is separable algebraic over k{t)^ dt is a basis for Ox/k over K by Theorem 9.1.2, so let D : ÜT —> K he the map given by dx = {Dx)dt for every x E K. Since d is a ^-derivation we have D{x + y)dt = d{x ^ y) = dx + dy = {Dx)dt + {Dy)dt and D{xy)dt = d{xy) = xdy + ydx = {xDy)dt + {yDx)dt for any x^y G K^ which implies that D is a derivation on K. Since Dc = 0 for any c e k and Dt = 1, we have D = d/dt by unicity of the differential extension. D Let k C K C L he fields of characteristic 0. The restriction to K of the /c-derivation di/}. : L -^ i^L//c is a A:-derivation of K into Oi/k^ so by Lemma 9.LI, it induces a K-linear map d : Ox/k "^ ^L/k such that d^/k = d dx/k' Let B be a transcendence basis of K over k. Then, {dK/k^}b£ß is a basis of ÜK/k over K by Theorem 9.1.2. In addition, {dLikb}heB is linearly independent over L by Corollary 9.1.1. Since d d^/k^ = ^L/fc^ foi" ^W b G B^ this implies that d is injective, hence that it is an embedding of Ox/k into ^L/k-

9.2 Rosenlicht's Theorem We prove in this section a fundamental theorem of Rosenlicht, itself a generalization of a result of Ax [4] on Schanuel's conjecture for differential fields, that is used to prove the various structure theorems later. From now on, let all fields in this chapter have characteristic 0. We start with an analogue of Theorem 3.2.4 for differentials: the trace map in algebraic extensions induces a linear trace on the differentials. L e m m a 9.2.1. Let k C K be fields, E a finitely generated algebraic extension of K, and Tr : E -^ K and N : E —^ K be the trace and norm maps from E to K. Then, there is a K-linear map Tr* : Üß/k ~^ ^K/k 5iic/i that Tr* d ~ dTr and TrU — ]= - — ^

\a J

N{a)

for any a G E\

9.2 Rosenlicht's Theorem

277

Proof. Let K be the algebraic closure of K and cJi,..., cr^ be the distinct embeddings of E in K over K. Note that k^'^ = k for each i since ai is the identity on A:, so let a* : Üß/k ~~^ ^'K/k ^^ ^^^ induced map given by Lemma 9.L2. Define Tr* : Üß/k ~^ ^~K/k ^^ -^^* ~ Y^l=i ^i • ^i^^^e a^ is the identity on K^ a* is a K-linear map by Lemma 9.L2, so Tr* is if-linear. We have n

n

/ ^

\

Tr*{da) = Y^{da)< = ^ c ^ ( a ^ ' 0 = ^ 1 X^a^M = d{Tr{a)) i=i

for any a G E^ so Tr* d = dTr. Furthermore,

for any a G E*. Let >B be a transcendence basis for K over k. Then, {db}beB is a basis for Ox/k over if by Theorem 9.L2. But E is algebraic over if, so ;B is a transcendence basis for E over k and {db}beB is a basis for üß/k over E by Theorem 9.1.2. Write then u G i^^;//^ as a; == YlbeB^^db where the üb are in E and only finitely many of them are nonzero. Then, n i=l

n ^=l b£B

beB

which is in the image of O^/k under the natural embedding Oj^^k ~^ ^l
eÜK/k-

(9-1)

Ui

Then, cj = 0 <=4> ui,... ^Un^v are all algebraic over k 4=^ dui = ... = dun = dv = 0. Proof. Note that Corollary 9.1.1 implies that any x G if is algebraic over k if and only if dx = 0. Suppose first that ui^... ^Un^v are all algebraic over k. Then, dui = ... = du^ = dv = 0 so UJ = {). Conversely, suppose that uj — Q and that one of the i^^, say ni, is transcendental over k. Let B be a transcendence basis of if over k containing ui and E — k(B \ {i^i}). K is then algebraic over E{ui), so F = E{ui){u2^. •. .u^v) C if is a finitely generated algebraic extension of E{ui). Identifying Qp/k with its image under

278

9 Structure Theorems

the embedding Qp/k -^ ^K/k mentioned at the end of the previous section, we can consider uj and the differentials appearing in (9.1) as being elements of Qp/k- Let Tr : F -^ E{ui) and N \ F —^ E{ui) be the trace and norm maps from F to E(ui) and Tr"" : Op/k —^ ^E{ui)/k be the induced £'(ni)-linear map given by Lemma 9.2.1. Applying Tr* and Lemma 9.2.1 to (9.1) we get n

,

0 = Tr'' (0) = Tr"" (oj) = Tr"^ (dv) + Y " Q Tr* — 1=1

^(rp ( ^^ , \^ dN{ui) dui ^ = d{Tr{v))+ } Ci =dw + mci h}

dvi Ci —

in ÜE{ui)/ki where w = Tr{v) E E{ui)^ rn =[F : E{ui)\ > 0 and Vi — N{ui) G E{ui) for 2 < i < n. Applying the canonical projection TT : OE{ui)/k —^ ^E{ui)/E to the above, we get n

J

,

dE(ui)/EUi

^

0 = dE(ut)/E^ + ^ c i —^^-^

+ Z_^^i —

^^

i=2

^

i=2

•

^*

By Corollary 9.1.2, dE{m)/EX = {Ox/dui)dE(ui)/EUi

^

dE(m)/EVi

for any x e E{ui), so

^

^

Since ui is transcendental over E, dE(ui)/E^i 7^ O5 so the above implies that m ui

'^-^ Ci dvi ^ Vi dui

dw dui '

Note that 2x1 is a monomial over E with respect to d/dui and that ?ii is normal and irreducible as an element of ^[t^i]. Furthermore, the left hand side of (9.2) is simple by Corollary 4.4.2, so dw/dui must be simple too, which implies that Uu^{dw/dui) > —1. But Vui{dw/dui) ^ —1 by Corollary 4.4.2, so lesidiMeu^iydw/dui) = 0 by Theorem 4.4.1. Applying Corollary 4.4.2 and residue^^j to (9.2) we get

(

n

m

rx

^ - ^ Ci avi

\

n

\

\-y

. .

ci f- > ,—-K— =mci + } Uu^{vi)Ci where i/u^ is the order function at ui. Since m is a positive integer, the above is a contradiction with c i , . . . , c^ linearly independent over Q, Therefore u i , . . . , 1^^ are all algebraic over k^ so dui = ... = dun = 0, which impHes that 0 = CO =^ dv^ hence that v is also algebraic over k. D L e m m a 9.2»3. Let (K^D) he a differential field, k C K be a differential sub field and D* : O^/k ~^ ^K/k ^^ the induced skew-derivation given by Lemma 9.1.3. For any u^v G K, if u and v are algebraically dependent over Constp(A:)^ then D*{udv) = d{uDv) in Üx/k-

9.2 Rosenlicht's Theorem

279

Proof. Suppose that u and v are algebraically dependent over C = ConstD{k), and let P G C[X^ Y] be a nonzero polynomial of minimal total degree such that P{u^v) = 0 . Applying D we get 0 = D{P[u,v))

dP = j^^.v)Du

dP + ^{u,v)Dv

(9.3)

dP + ^{u,v)dv

(9.4)

by Theorem 3.1.1. Applying d we get dP 0 = d{P{u,v)) = —{u,v)du

by Theorem 9.1.1. Using (9.3) and (9.4), we obtain dP

dP

\

fdP

\

fdP

dP \ f dP - — {u,v)Dv\ (-^K^)^^ f dP

dP

\

If •ö^{u,v) = 0, then ^ is identically 0 by the minimality of P, which implies that P £ C[Y], hence that v is algebraic over C, i.e. that dv = 0 and Dt? = 0 by Lemma 3.3.2. Therefore, {Du)dv = {Dv)du = 0. Similarly, {Du)dv = {Dv)du = 0 if §^iu,v) = 0. If §^{u,v) 7^ 0 and ^{u,v) / 0, then (9.5) implies that {Du)dv — {Dv)du. Using that equality together with Lemma 9.1.3 we get D'^iudv) = {Du)dv + uD*{dv) = {Dv)du + ud{Dv) = d{uDv). D

L e m m a 9.2.4. Let (K^D) be a differential field, k C K be a differential sub field and D* : O^/k -^ ^K/k ^e the induced skew-derivation given by Lemma 9.1.3. Then, for any v G K, w i , . . . , i i ^ £ K* and c i , . . . , c ^ G ConstD(A:)^ d \

Dv 4- 2_\ ^i ~ \ = D* \dv ^ V^ Ci — - \ i=i ^W \ i=i ^^ J

in Qx/k -

Proof Since d is Ä:-linear, we have

''(o''+i:=.^)=''(^«)+|:«.<'(^)^ Since UiU~ — 1 = 0, it^ and u~ are algebraically dependent over Consti:)(Ä;), so d{Duj/uj) = D'^{duj/uj) by Lemma 9.2.3. In addition, d{Dv) = D*{dv) by Lemma 9.1.3, so

280

9 Structure Theorems

<^[D- + tc.^j=^*idv) + ±c.D*(^^ D

L e m m a 9.2.5. Let {K,D) be a differential field, k C K be a differential subfield, D* : f2x/k ~^ ^K/k ^6 ihe induced skew-derivation given by Lemma 9.1.3 and cji,... ,cj„ G OK/U ^^ such that D*Ui = 0 for each i. //a;i, • • • ,^n are linearly dependent over K, then they are linearly dependent over Consto{K). Proof. Suppose that c^i,..., cj^ are linearly dependent over K. Let a i , . . . , a^ in K be not all 0, and such that X^ILi ^*^^ ~ ^ with the number of nonzero aiS minimal over all such linear combinations. We can assume without loss of generality that ai ^ 0, and dividing by ai if needed, that ai = 1. Applying D* we get

(

n

\

n

n

i=2

/

i=2

i=2

Since ai = 1, the above is a linear combination of the Ui with one less nonzero a^, so by minimality we must have Doi = 0 for 2 < i < n, hence a i , . . . , a„ E Const 1)(Ü"). Therefore c j i , . . . , a;^ are linearly dependent over Const/:)(Ü^). D T h e o r e m 9.2.1 (Rosenlicht [81]). Let (K^D) be a differential field, k be a differential subfield of K with Const x)(if) = Coiistjy{k), and let vi,... ^Vn E K and w i , . . . , i i ^ G Ü"*. If there are constants Cij in Consti:)(^) such that D

^0

Dvi + 7 Cij—-^ •^ n

Gk

for 1 < i < n

then either k(u\^..., Um^vi,... ^Vn) has degree of transcendence at least n over k, or a;i,..., cj^ ci^e linearly dependent over Const£>(A:); where X-^

dUj

Ui = dvi + 2^ Cij — j=i

^

G ÜK/k •

^^

Proof. Let C = Consti:)(i^) = Constp(Ä:), E = k{ui^... ^Um,vi, • • • jV^) and P* : ÜK/k -^ ^K/k be the induced skew-derivation given by Lemma 9.L3. Let Xi = Dvi + Y^iLi^ij^^j/^^j^ ^^^ suppose that x^ G ^ for 1 < i < n. Then dxi = 0 for each i by the definition of üx/kt so D*a;j = dxi = 0 by Lemma 9.2.4. Suppose that Ü;I, . . . ,Ü;^ are linearly independent over C. Then, they are lin= early independent over K by Lemma 9.2.5, so considered as elements of Oß/k^ they are linearly independent over E. This implies that the dimension of Üß/k over E is at least n, hence by Theorem 9.1.2 that E has transcendence degree at least n over k. D

9.2 Rosenlicht's Theorem

281

Corollary 9.2.1. Let (K^D) be a differential fields k C K be a differential subfield with Const/:) (ÜT) = Const/) (k) and letvi,... ,Vn £ K and u i , . . . , t^^ E K* be such that ßy^ 1 ^ j^ for 1 < i < n. Ui

Then, either k{ui^..., Un-, f i , . . . , Vn) has degree of transcendence at least n over k, or there are c i , . . . ,c^ G Constj^(A:) not all zero, and e i , . . . , e^ € Z not all zero, such that X^ILi ^i^i ^^^ HlLi ^T ^^^ ^^^^ algebraic over k. Proof. Suppose that the degree of transcendence of k{ui^..., w^, '^i, • • •, '^n) over k is stricly less than n. Taking Cij = ~\ li i = j and 0 if i ^ j , we see that i^i,..., w^,t?i,... ,i;^ satisfy the hypothesis of Theorem 9.2.1. Therefore, uji,... .uOn are hnearly dependent over Consta(/c) where LOi = dvi — dui/ui G Ox/k- Let then c i , . . . ,c^ € Constjr)(/c) be not all zero and such that Y27=i^i^i ~ ^- Since Const/:) (Ä;) contains Q, it is a vector space over Q, so there are &i,..., 6^ G Const/:)(/c) linearly independent over Q, and rriij G Z not all zero such that Q = X]^=i '^ijbj. We then have n

n

0 = ^

CiiOi = ^

2=1

n Cidvi

i=l

-

^

r ^

J ^ij^j

—-

i=l j =l


/

3= 1

'^i

By Lemma 9.2.2, this imphes that Yll=i ^i^i i^ algebraic over fc, and that n r = i ^i ^'^ ^^ algebraic over k for each j . Since at least one of the m^j's is nonzero, this proves the corollary. D Corollary 9.2.2. Let C be a field, x be transcendental over C, (K^D) be a differential extension of {C{x), d/dx) with Const/) (ÜT) = C, and vi^... ^Vn G K and i t i , . . . , u^ G K* be such that Dvi

Du-

eC

forl
Ui

Then, either C{x){ui^...,?i„,'Ui,..., Vn) has degree of transcendence at least n over C(x), or there are e i , . . . , e^ G Z not all zero such that YYi=i ^T ^ ^• Proof Let UQ = 1^ VQ = x and suppose that the degree of transcendence of C{x){ui^... ,Un^vi^... ,Vn) ovor C{x) is stricly less than n. Then the transcendence degree of C{uo^ui^... ^Un, VQ^VI, ... ,Vn) over C is stricly less than n + 1, so by Theorem 9.2.1, a;o,... ,Ü;„ are linearly dependent over C where uji = dvi — dui/ui G Ox/C' Let then CQ, . . . , c^ G C be not all zero and such that X^iLo ^^^^ ~ ^' Note that (JJQ = dx is linearly independent over K by Corollary 9.1.1, so there must be some io > 0 such that QQ ^ 0. Expressing each Ci as Ci = J2]=i ^ij^j with rriij G Z as in the proof of Corollary 9.2.1,

282

9 Structure Theorems

we get in a similar way that Yfi=o ^i '^ ~ Yl7=i ^i '^ ^^ algebraic over C for each j . Since C = Constp(A"), Lemma 3.3.2 implies that n f ^ i '^i ^''^ ^ ^ ^^^ each j . Since QO / 0, m^o^ ^ 0 for some j , which proves the corollary. D

9.3 The Risch Structure Theorems Recall (Definition 5.1.4) that a differential field (K, D) is an elementary extension of (A:, D) ii K = k(ti^..., tn) with each ti elementary over k{ti,..., t^-i). In that case we define the following index sets: Ex/k = {i € { 1 , . . . , n} such that ti transcendental over /c(ti,..., t^_i) and Dti/ti = Dai^ai e k{ti,...

,ti_i)}

(9.6)

and ^K/k = {i € { 1 , . . . , n} such that ti transcendental over fc(ti,..., t^-i) and Dti = Dai/ai^ai G k{ti,... ,ti_i)*} . (9-7) Note that the cardinality of E^/k U Lj^^j^ is exactly the transcendence degree of K over k and that it is at most n. In addition, if Consti:>(Ä') = Const/) (Ä:), then deg^. (Dt^) is 0 when ti € Lj^ß and 1 when ti € -B^/zc (since Dti ^ 0), so £i^/fc n Lxjk = 0Lemma 9.3.1. Let (K^D) be an elementary extension of (k^D) satisfying Consti:)(i^) = Consti:)(fc) = C. Write K = k{ti^... ,tn) with each ti elementary over k{ti^... ,t^_i)^ and let E^/k ^'^d L^/k ^^ given by (9.6) and (9.7) respectively. If there are integers e^ G Z such that

n ^^n «^ e^ where ti = log(a^) for i G L^/k^ then e^ = 0 for all i in Ex/k U E^/kProof Suppose that Si ^ 0 for some i and let then j = max{i G E^/k U ivj^//^ such that e^ 7^ 0} ,

^ ^ n *^' n ^^' ^^^ ^ ' "^ x^ ^^^^ "^ s ^^*^ where t^ = log(a^) for i G I/^/A: and ti = exp(ai) for i G Ex/k- Note that Da/a = Dß by the logarithmic derivative identity. If j G Ex/k^ then t^-^a G C, in contradiction with tj transcendental over k(ti^... ,tj_i) since a G fc(ti,... ,tj_i).

9.3 The Risch Structure Theorems If j e Lx/ki

0 =

283

t h e n a^-^a G C, so

\^. "^^ = e j ^ + — = ejDtj a -^ a ttj a

+Dß = D{ejtj

+ ß)

which implies t h a t ejtj + /3 € C, in contradiction with tj transcendental over ,t^--i). k{ti,... , t j _ i ) since ß G k{ti,... T h e o r e m 9 . 3 . 1 ( R i s c h [77]). Let C he a field, x he transcendental over C, and (ÜT, D) he an elementary extension of {C{x)^d/dx) with Constx)(ii") = C Write K = C ( x ) ( t i , . . . , t ^ ) with each ti elementary over C{x){ti^... ,t^_i)^ and let Ex/c{x) ^'^d Lx/cix) ^^ given hy (9.6) and (9.7) respectively. If there are v E K and u E K"" such that Dv = Du/u, then there are r^ € Q such that

where ti = exp(a^) for i G

Ex/c{x)-

Proof. Let ui = ti and Vi = ai for i G Exic{x)i ^ ^ LK/C(X)I

^ = Ex/c{x)

U LK/C{X)I

^i = ^i ^^^ ^i = ti for

'^ be t h e cardinality of / , and F =

C{x){u^ tJ, {ui}i^i^ {vi\i^i). Since t h e degree of transcendence of K over C{x) is exactly ?7i, t h e degree of transcendence of F over C{x) is at most m, hence strictly less t h a n m + 1 . Since Dv — Du/u == 0 G C and Dvi — Dui/ui = 0 e C for each i G I, Corollary 9.2.2 implies t h a t there are integers e and {e^}^^/, not all zero, such t h a t u^ Yliei ^T ^ ^- Note t h a t e / 0 by Lemma 9.3.L We then have

eDv 4-

y^

= D lev -^

^

e^Dt^ +

e^ti +

y_]

e^Da^

^

eiai

which implies t h a t

^^ ^ X^ ^*^* "^ X] ^^^^ ^ ^ and dividing by e proves t h e theorem.

D

As consequences we get algorithms for determining whether new logarithms or exponentials over a differential field are monomials over t h a t field having t h e same constants.

284

9 Structure Theorems

Corollary 9.3.1. Let C^x^K^Ex/c{x) a G K"" and b e K. Then,

^'^d Lx/c{x)

^^ o.s in Theorem 9.3.1,

(i) Da/a is the derivative of an element of K if and only if there are r^ E Q such that ^ "c:^ Dti Da ,^ .

E

^GLx/c(,'c)

*€-EK/C(X-)

(ii) Dh is the logarithmic derivative of a K-radical if and only if there are r^ G Q such that

Y,

nDti+ Yl

n^=Db.

(9.9)

Proof, (i) Suppose that Da/a = Dv for some v E K. By Theorem 9.3.1 there are r^ E Q such that

where ti = exp(a^) for i G EK/C{X)- Applying D yields (9.8). Conversely, if there are r^ E Q satisfying (9.8), then

Yl ^'^' "^ Yl ^^^'

"" ^

is the derivative of an element of K. (ii) Suppose that nDb = Du/u for some integer n 7^ 0 and u E K*. By Theorem 9.3.1 applied to ^' = n5, there are r^ E Q such that

^^ "^ X! ^*^* "^ X^ ^^^^ ^ ^ where ti = exp(ai) for i E Ex/c{x)- Applying D and dividing by n yields (9.9). Conversely, if there are r^ E Q satisfying (9.9), then

where ti = log(a^) for i E Lx/c(x)- Putting the r^'s over a common denominator e 7^ 0, we get ^. eDb = where

\r^ y

Dai a

h

\-^ y

Dti Du Si —— = —

"= n <' n *i' ex*D

9.3 The Risch Structure Theorems

285

The algorithms follow from Corollary 9.3.1 and Theorems 5.1.1 and 5.1.2: let (K^D) be given explicitely as an elementary extension of {C{x)^d/dx) where C = Const]j{K) and suppose t h a t the sets Ex/c{x) ^^d Lx/c{x) ^^^ known (those can be computed by applying the algorithm t o ^ 1 , ^ 2 , . . . , t ^ in t h a t order). Let a E ÜT* and let t in a differential extension of K be such t h a t t = log(a), i.e. Dt = Da/a. If (9.8) has a solution r^ G Q, then it provides v E K such t h a t Dv = Da/a^ hence c = t — v G Consti^(K(t)) and K{t) = ( 7 ( c ) ( t i , . . . , t n ) . Otherwise, Da/a is not the derivative of an element of K by Corollary 9.3.1, so t is a monomial over K and Consti:)(iir(t)) = Const/:)(if) by Theorem 5.1.1. Let b £ K and let t in a differential extension of K be such t h a t t = exp(6), i.e. Dt/t = Db. If (9.9) has a solution r^ G Q, then it provides a nonzero integer e and u E K* such t h a t eDb = Du/u, hence c = t^/u G Const]j{K{t)) and K{t) is algebraic over C ( c ) ( t i , . . . , t„) since t^ = cu. Otherwise, Db is not t h e logarithmic derivative of a ilT-radical by Corollary 9.3.1, so t is a monomial by Theorem 5.1.2. over K and ComtD{K{t)) = ConstD{K) To determine whether (9.8) and (9.9) have solutions in Q, we compute a linear system with coefficients in C and t h e same constant solutions by L e m m a 7.1.2. Assuming^ t h a t we have a vector space basis B containing 1 for C over Q, projecting t h a t system on 1 yields a linear system with coefficients in Q for the r^'s. Risch also gave a real version of his structure theorem, which is applicable t o towers of logarithms, exponentials, arc-tangents and tangents over a real constant field. Recall (Definition 5.10.1) t h a t t is a tangent over k if Dt/{t^ 41) = Da for some a G fc. In a similar fashion, we say t h a t t is an arc-tangent over k if Dt = Da/{a^ + 1) for some a £ k such t h a t a^ + 1 ^ 0, and t h a t t is real elementary over k if t is either algebraic, a logarithm, an exponential, an arc-tangent or a tangent over k. We say t h a t (i^, D) is a real elementary extension of {k^D) if K = ^ ( t i , . . . , ^ ^ ) with each ti real elementary over In t h a t case, in addition to the index sets Ex/]^ and Lj^/^ k{tiy... ^ti-i). defined by (9.6) and (9.7), we introduce the following index sets: Tx/k

= {^ ^ {I5 • • • 5 ^ } such t h a t ti transcendental over and Dti/{t^^ + 1) = Düi.ai

G k{ti,...,

fe(ti,...,

Vi)}

t^_i) (9.10)

and Ax/k

= {^ G { 1 , . . . , n ) such t h a t ti transcendental over k{ti^..., and Dti = Dai/{af

+ 1), a^ G k{ti,...,

ti^i)}

.

t^_i) (9.11)

Note t h a t the cardinality of Ej^/k U L^/k U T^/k U A^/k is exactly the transcendence degree of K over k and t h a t it is at most n. In addition, if ^This may cause undecidability problems in general, but we usually compute in cases where C is a finitely generated extension of Q and such a basis is available. The integration algorithm requires an explicitly computable constant field.

286

9 Structure Theorems

Consti:)(iir) = ConstD{k), then deg^.(Dt^) is 0 when ti G Lj^^j^ U Ax/ti 1 when ti G Ej^/^ ^^nd 2 when ti £ Tx/ki so the sets AL/]^ U Li^ß, Ej^ß ^iid ^K//c a^i'e disjoints. It can be shown t h a t A^ß H Li^ = 0 when A / ^ ^ L (Exercise 9.3), so t h e four index sets are disjoints. L e m m a 9 . 3 . 2 . Let K be a fields X be an indeterminate irreducible. Then, p irreducible over K[X]/{q)

and p^q £ K[X] be

<^==^ q irreducible over

K[X]/{p)

Proof. Let K be the algebraic closure of K and a^ ß £ K he such t h a t p{a) = q{ß) = 0. Then, K[X]/{q) :^ K{ß) and K[X]/{p) :^ K{a). We proceed by a degree argument in t h e following diagram:

K{a,ß)

K{a)

K{ß)

We have [K{a) : K] = deg(p) and [K{ß) : K] = deg(g) since p and q are irreducible over K. Up is irreducible over K{ß), then [K{a,ß) : K{ß)] = deg{p), so [K{a,ß)

: K] = [Kia,ß)

: Kiß)][Kiß)

: K] = d e g ( p ) d e g ( g ) .

But [K{a,ß)

: K] = [K{a,ß)

: K{a)][K{a)

: K] = [K{a,ß)

: i ^ ( a ) ] deg(p)

which implies t h a t [K{a^ß) : K{a)] = deg(g), hence t h a t q is irreducible over K{a). T h e converse follows by symmetry. D L e m m a 9 . 3 . 3 . Let C be a field, x be transcendental over C, and (K^ D) be a real elementary extension of {C{x),d/dx) with A / ^ ^ K and Const£)(i^) = FurC. Then, i^(\/—1) is an elementary extension of {C{^/—l){x),d/dx). = C{^/^), and the elementary tower from C{^/^){x) thermore, ConstA{E) can be chosen so that Lj^(^rri)/c(y^:^) — ^K/c ^ A^/c ^'^d to K{^/^) ^K(^^:T)/C(^/=T) = ^K/C U TK/C Proof. Note t h a t C o n s t i ^ ( Ü ^ ( V ^ ) ) = C{^fÄ) by L e m m a 3.3.4. We write K = C ( a : ) ( t i , . . . ^tn) where each ti is real elementary over C ( x ) ( t i , . . . , t i _ i ) = and proceed by induction on n. If n = 0, then K = C{x) so K{\/'^) C ( \ / ^ ) ( x ) . Suppose now t h a t n > 0 and t h a t t h e lemma holds for k =

9.3 The Risch Structure Theorems

287

(7(x)(ti,... , t n - i ) . Then, Consta(/c(V^)) = C ( A / ^ ) , k{y^^) is an elementary extension of (C{^/^){x)^d/dx), and the elementary tower can be chosen so that L^^yzr[)/c{yriri) = L^/c U Ak/c and E^(^^^/zri)/c{./^) = ^k/c U Tj^/cIn addition, K = k(t) where t = t„ is real elementary over K. Case 1: t is transcendental over k. Then, t is transcendental over k{\/^Ä). Case la: t is a logarithm over k. Then, t is a logarithm over k{\/^Ä)^ so K{^/^) = k{^/^){t) is elementary over fc(^/^), hence over C{^/^). Furthermore, ^K(^^:T)/C(^/^) = {'^}^Lk(V^)/C{V^)

= {n}ULk/cUAk/c

=

LK/C^AK/C

and ^K(V'=T)/C(y=T) = ^/c(y=T)/C(y=T) = -^fc/C U T/e/c = EK/C U T ^ / C •

Case lb: t is an exponential over k. Then, t is an exponential over A:(v^^), so K{\/'^) = A;(\/^)(t) is elementary over k{\/~—l)^ hence over C{^/^). Furthermore, ^K(^/=l:)/C(^/=T) = {^}U£^fc(yi:T)/c(v'irT) = {n}UEk/c^Tk/c

=

EK/C^TK/C

and ^K(V'=T)/C(y^) = i/e(v/IT:)/C(y^) = ^/c/C U A/,/c = I/i^/C U AK/C •

Case Ic: t is an arc-tangent over k. Let then 6 = 2^^/^^ G ii!^(\/^). We have Dt = Da/{a^ -h 1) where a Gfcsatisfies a*^ + 1 ^^^ 0, so a^ + 1

6

where b = (a — \ / ^ ) / ( a + \ / ^ ) G Ä:(\/^)*, so ^ is a logarithm over which implies that i ^ ( ^ / ^ ) = k{y^^){9) is elementary over k{^/^), over C ( \ / ^ ) . Furthermore,

fe(^/^), hence

and ^i^(^/^)/c(^/=T) = Ej^(^^jzri)ic{yr-i) = ^/c/c U Tfc/c = EK/C U T ^ / C • Case Id: t is a tangent over k. Let then 6 = ( y ^ —t)/(A/^4-t) G if(\/—T). We have Dt = {1? + 1)-Da for some a G Ä:, so = —-

e

/=T-t

—=

/=T + t

= —

-— = i J ( 2 a v - l )

^^ + 1

288

9 Structure Theorems

so 6 is an exponential over ^ ( \ / ^ ) , which impUes that K{^/^) = is elementary over k{^/^)^ hence over C{y/^). Furthermore, ^K{V^)/c{V^)

= {'^}^Ek{V^)/c{V^)

= {n}^Ek/c^Tk/c

=

k{^/^){0)

EK/C^TK/C

and LKi./^)/ciV^)

= ^fc(y=:T)/c(V=T) = ^k/c U Ak/c = LK/C U AK/C -

Case 2: t is algebraic over k. Let then X be an indeterminate and p G k[X] be the minimal irreducible polynomial for t over k. Since ^ / ^ ^ k{t)^ p remains irreducible over k{\/^) by Lemma 9.3.2, so K{\/^) = k{y/^){t) c::^ k{\/^)[X]/{p) is algebraic, hence elementary, over k{\/^). Furthermore, ^K(^/Z:T)/C(V/:=T)

= E^^^zrT)/c(^:-i) = Ek/c U Tk/c =

EK/C

U TK/C

and EK{./^)/C{^T)

= ^/c(v'=T)/c(x/=ö[) - ^k/c U yl/c/c - LK/C U A K / C • D

We can now prove a shght generalization of the real Risch structure Theorem to arbitrary constant fields. T h e o r e m 9.3.2 (Risch [77]). Let C he a field, x he transcendental over C, and (K^D) he a real elementary extension of{C{x)^d/dx) with Consta) (if) = C and \ / ^ ^ K. Write K = C ( x ) ( t i , . . . ,t^) with each ti real elementary over C{x){ti,...,ti__i), and let EK/C{X), LK/C{X), TK/C{X) and AK/C{X) be given by (9.6), (9.7), (9.10) and (9.11) respectively. (i) If there are v G K and u E ilT* such that Dv = Du/u, then there are ri £ Q such that

y^

X]

^^^^ ^

X]

^^^^ ^ ^

where ti = exp(ai) for i G EK/C{X)(a) If there are u^v £ K such that Dv = Du/{u^ + 1); then there are r^ G Q such that

where ti = tan(ai) for i G Txic{x)' Proof. Let F — C ( v ^ ) . By Lemma 9.3.3, E = i^(\A-T) is elementary over {F{x),d/dx), Coii^tD{E) = F , LE/F{X) = LK/C{X) ^AK/C{X) and EE/F{X) = EK/C{X)

UTX/C(X)-

9.3 The Risch Structure Theorems

289

(i) Suppose that Dv = Du/u ioi v £ K and u G -ff*. By Theorem 9.3.1 appUed to E, there are r^ G Q such that

v+ Yl ^^^^ + Yl ^^^^ ^ C{y^ where Oi = exp(a^) for i €

EE/F{X)-

0 = Dt;+

Yl ^iD^i+ YJ ^^^^^ Uuj

E = Dv+

Yl

+

Differentiating, we get

\—^

Uurj

riDti + 2V^

}_^ r , - - + 2A/-l

Yl }_^

''i^^i r,:

Since y ^ ^ if, we get that

^^LK/C(X)

^^EK/C{X)

hence that i£LK/c(x)

*€£^K/c(x)

where t^ = exp(ai) for i G Ex/c{x)(ii) Suppose that Di; = Du/{y? + 1) for u^v £ K. Then, D{2v^J^)

= 2DvV^

= 2 A / ^ - ^"^

^2 + 1

^^ 2

where z = {u — \/~—i)l{u + \/—I), so by Theorem 9.3.1 appHed to E, there are r^ G Q such that

i^LE/F{x)

where 6i = exp(ai) for i G

EE/F{X)'

0 - 2Dv^f-i + ^-^ I£EK/C{X)

Y Dti

i^EE/F{x)

Differentiating as in part (i), we get

^^^^^ + 2^/"^ ^— ,^-^ f'^TK/cix)

Y^ ^^^^^ Dti

290

9 Structure Theorems

Since \ / ^ ^ iC, we get t h a t DU

Dv+ Yl ^iDU+ Y.

'tf + 1

hence t h a t

where ti = tan(a^) for i G Tx/c{x)-

D

As consequences we get algorithms for determining whether new logarithms, exponentials, tangents and arc-tangents over a real differential field are monomials over t h a t field having t h e same constants. C o r o l l a r y 9 . 3 . 2 . Let C,X,K,EK/C{X),LK/C{X),TK/C{X)

in Theorem 9.3.2, a £ K"" and h e K. (i)

Da/a is the derivative such that

^

Y"^

riDtiiderivative

J2

. ^.

'''T^ = ~-

^^'^^^

of a K-radical

if and only if there are

(9.13)

of an element of K if and only if there are

^

nDu+

-c-^

Y.

Dti

Db

. , - ^ = ^-_-.

*GTx/C(a;)

(iv) \/~lDb is the logarithmic derivative there are r^ G Q such that

ieAK/dx)

Da

*G-BK/C(.T)

(Hi) Dh/ifP" + 1) is the derivative ri £ Q such that

i€Ai^/C(x)

Y

Dti

nDti+ Yl n^ = Dh.

i€Lj^/c(x)

E

be as

of an element of K if and only if there are r^ G Q

E (a) Db is the logarithmic ^i ^ Q such that

and AK/C{X)

Then,

(9.14)

*

of a K(y^^^)-radical

i^Tx/dx)

/^ . .X

if and only if

*

Proof. T h e proofs of parts (i) a n d (iii) are similar t o t h e proof of part (i) of Corollary 9.3.1, using Theorem 9.3.2 instead of 9.3.1, while t h e proofs of part (ii) is similar t o t h e proof of part (ii) of Corollary 9.3.1. (iv) Suppose t h a t n\/^Db = Dw/w for some integer n 7^ 0 a n d w in i i r ( \ / ^ ) * , and write w = y -{- z\PÄ. where y,2; G ÜT. If y = 0, t h e n z 7^ 0 and Dw/w = Dz/z = ny^^^Db^ which implies t h a t Dz = Db = 0, hence

9.3 The Risch Structure Theorems

291

that (9.15) is satisfied with r^ = 0 for each i, so suppose from now on that y ^ 0. We then have, ^nV-lDb

=

Dw Dy-\-^f^Dz = • j=^w y + zy —1

=

yDy + zDz yDz - zDy r-+ V^l y^ + 2^ y^ + z^

so yDy + zDz = 0, which imphes that c = y'^ + z"^ G C*. Let then W = w'^/c and u —

1 — W I—-—iV—1 1+ W 1 + yz

c — w ^^-j^

y'^-^yzy^-l

yy-^z^-1

c-y^^z^-2yz^ri

z2;\/^4-2/

We have W = ( \ / ^ — u)/{y/^ Du u^ -\-l

r-i

z

-eK. y

+ -u), so by Lemma 5.10.1,

DW W

Dw w

Dc c

^

r—r..

which imphes that nDh = Du/{v? + 1). By Theorem 9.3.2 appHed to t' = nh, there are r^ G Q such that n6 +

^

Titi +

^

r^a^ € C

where ti = tan(ai) for i £ Tx/c{x)- Applying D and dividing by n yields (9.15). Conversely, if there are r^ G Q satisfying (9.15), then Do

DU K/C(x)

SO putting the r^'s over a common denominator e / 0 and multiplying by 2ydY^ we get Z_-/

' "^

a? + 1

^—^

t? + 1

where t^ = arctan(ai) for i G Aj^/(7(a.). Let 6^ = ( \ / ^ — a 4 ) / ( \ / ^ + a^) G ii:(v^^)* and 6>i = {^/^~-ti)/{y^-hti) G Ü ^ ( v ^ ) * . By Lemma 5.10.1, Dbi/bi = 2^rÄ.Dail{a\ + 1) and DQilQi = 2^fÄ.Dti|(t\ + 1), so

2e.f^\Db= Yl iGA,K/C{x)

where

i^ö.,

^^T^+

E ieTj K/Cix)

9i

^= n ^' n ^r^ eif(v^r.

w

(9.16) G

292

9 Structure Theorems

T h e algorithms follow from Corollary 9.3.2 and Theorems 5.1.1, 5.1.2 and 5.10.1: let (K^D) be given explicitely as a real elementary extension of {C{x)^d/dx) where C = ConstD{K)^ \ / ^ ^ K, and suppose t h a t t h e sets EK/C(X),LK/C{X),TK/C{X) and AK/C{X) are known (those can be computed by applying t h e algorithm t o t i , ^ 2 , . . . , t „ in t h a t order). Let a G K* and let t in a differential extension of K be such t h a t t = log(a), i.e. Dt = Da/a. If (9.12) has a solution r^ G Q, then it provides v e K such and K{t) = C ( c ) ( t i , . . . , t ^ ) . t h a t Dv = Da/a, hence c = t—v G Const]j{K{t)) Otherwise, Da/a is not t h e derivative of an element of K by Corollary 9.3.1, so t is a monomial over K and Constjr:)(iir(t)) = Constjj{K) by Theorem 5.1.1. Let b E K and let t in a differential extension of K be such t h a t t = exp(6), i.e. Dt/t = Dh. If (9.13) has a solution r^ G Q, then it provides a nonzero integer e and u G K* such t h a t eDh — Du/u.^ hence c == t^/u G Const£)(iC(t)) and K(t^ is algebraic over C ( c ) ( t i , . . . , t^) since t^ = cu. Otherwise, Dh is n o t t h e logarithmic derivative of a if-radical by Corollary 9.3.1, so t is a monomial over K and Constr>{K{t)) = ConstD(i^) by Theorem 5.1.2. Let b £ K and let t in a differential extension of K be such t h a t t = arctan(6), i.e. Dt = Db/{b'^ + 1). If (9.14) has a solution n € Q, then it provides v e K such t h a t Dv = Db/{b^ + 1), hence c = t - v G ConstD{K{t)) and K{t) = C ( c ) ( ^ i , . . . , t ^ ) . Otherwise, D6/(6^4-l) is not t h e derivative of an element of i f by Corollary 9.3.2, so t is a monomial over K and ConstJT)(if (t)) = ConstD{K) by Theorem 5.1.1. Let b G K and let t in a differential extension of K be such t h a t t = tan(6), i.e. D t / ( t ^ + l ) = Db. If (9.15) has a solution r^ G Q, then it provides a nonzero = Dw/w. integer e and w G K ( \ / ^ ) * given by (9.16) such t h a t 2e^/^Db Let v^=T + t and

c=-GKw-mr. w

Using L e m m a 5.10.1 we get Dc D0 DW ^ r—- Dt — = e— = 2eV^c 9 w t^ + 1

^ r-~^^ ^ 2ey^Db =0

so c G Const£){K{\/^){t)) and 9 is algebraic over C{c, ^/^){tl^... ,t^) since 9^ = cw. Since t = (^ — 1 ) A / ^ / ( Ö + 1), this implies that K{t) is algebraic over C(c, A / ^ ^ ) ( t i , . . . , t^), hence over C ( c ) ( t i , . . . , t^). It is actually possible to compute the minimal polynomial for t over C ( c ) ( t i , . . . ,t„) directly from the solution of (9.15) without introducing \ / ^ , see [10] for details. Otherwise, if (9.15) has no solution in Q, then ^f-lDb is not the logarithmic derivative of a if ( \ / ^ ) - r a d i c a l by Corollary 9.3.2, so t is a monomial over K and ConstD{K{t)) = ConstD{K) by Theorem 5.10.1.

9.4 The Rothstein-Caviness Structure Theorem

293

9.4 The Rothstein-Caviness Structure Theorem Rothstein and Caviness [84] have generahzed the Risch structure theorem by allowing arbitrary primitives instead of logarithms in the tower of extensions. Since a hyperexponential extension can be embedded in an exponential extension of a primitive extension, this yields a structure theorem applicable t o arbitrary Liouvillian extensions. In order to avoid having logarithms cancel with primitives, it is necessary to introduce the restriction t h a t for a primitive t over a field F , either t is explicitely given as a logarithm over F^ oi Dt does not have an elementary integral over F. D e f i n i t i o n 9 . 4 . 1 . Let {K,D) he a differential extension of {k,D). t e K is nonsimple primitive over k if Dt G k and Dt does not have an integral in any elementary extension ofk. Note t h a t Theorem 5.1.1 implies t h a t if t is nonsimple primitive over k^ then t is transcendental over k and Consti^(Ä:(t)) = Consti:>(Ä:). D e f i n i t i o n 9 . 4 . 2 . {K^D) is a log-explicit Liouvillian extension of (k^D) if t^) and for i G { 1 , . . . , n } ^ there are ^ i , . . . , t„ in K such that K = k{ti,..., over either ti is elementary over k{ti^... , t ^ _ i ) ; or ti is nonsimple primitive k{ti,. . . , t i _ i ) . T h e o r e m 9.4.1 ( R o t h s t e i n & C a v i n e s s [84]). Let C be a fields x he transcendental over C, and {K^ D) he a log-explicit Liouvillian extension of {C{x)^d/dx) with ConstoiK) = C. Write K = C ( x ) ( t i , . . . , t ^ ) with each ti either elementary or nonsimple primitive over C{x){ti,... ^ti-i), and let EK/C{X) ^"^d Lj^^c(x) ^^ given hy (9.6) and (9.7) respectively. If there are V £ K and u G -ftT* such that Dv = Du/u, then there are r^ G Q such that

where ti = exp(a^) for i G

Ex/c{x)'

Proof. We proceed by induction on the number /i of nonsimple primitives among t i , . . . ,tn. If /^ = 0, t h e n K is elementary over C{x) and the result follows by the Risch structure Theorem. Suppose t h a t /i > 0 and t h a t the theorem holds for any log-explicit Liouvillian extension of {C{x),d/dx) with constant field C and at most /i — 1 nonsimple primitives. Let io be the largest index such t h a t ti^ is nonsimple primitive over C ( x ) ( t i , . . . , t^Q_i), t = t^^, k — C(x)(t\^... ^ti^-^i) and Oi = ti^j^i for i G { 1 , . •. , m } where m = n — io^ Then, fc is a log-explicit Liouvillian extension of {C{x)^d/dx) with at most /i—l nonsimple primitives, and K = k{t){9i^..., 6m) is elementary over k{t) by the maximality of i^. Let Ui = Oi and vi = Oi for i G Ex/k{t)i '^i = ^i ^^^d Vi = 6i for i G LK/k{t), UQ = 1, vo =t, I = EK/k{t) U L^/kit)^ P be t h e cardinality of / , and F = k{uo^ vo)(w, v^ {ui}i^j^ {viji^j). Since the degree of transcendence

294

9 Structure Theorems

of K over k{t) — k^uo^vo) is exactly p, and the degree of transcendence of k{uo^vo) over fc is 1, the degree of transcendence of F over k is at most p + 1, hence strictly less t h a n p + 2. Since Dv — Du/u = 0 e k, DVQ — DUQ/UQ = Dt £ k and Dvi — Dui/ui = 0 G A: for each i G I^ Theorem 9.2.1 implies t h a t the elements uj = du/u — dv^ ÜOQ = dt and tOi = dui/ui — dvi of fix Ik are linearly dependent over C, so let c, c o , ! ^ } ^ ^ / G C be not all zero such t h a t cijj + coc/t + XliG/ ^^'^^ "^ ^- Dividing by c if c 7^ 0, we can assume t h a t c € {0,1}. Since C is a vector space over Q, there are 5 i , . . . , 6^ E C linearly independent over Q, and rriij such t h a t Ci = Xl^=i^^*i^i- We can assume without loss of generality t h a t 61 = 1, hence t h a t c = c6i, since c G {0,1}. We then have 0 = c^dt + ccj + ^

auji

(9.17)

— cbi

== co
\ , / . '^ij^j —~

iei 3=1

=:dicot + J2^jyj

iei j=i

*

-YJ^^-^

where

and öij = 1 if i = j and 0 if i 7^ j . By Lemma 9.2.2, this implies t h a t w = Cot + Xl^=i ^jVj is algebraic over Ä:, and that Zj is algebraic over k for each j . We also have — ^ = c(5ij cdi^ —

^ TTiij h' y^

^ = c5ijDv + 2 ^ rriijDvi = Dt/^

(9.18)

for each j , so = coDt + Y^bjDyj

Dw~-Y^bj—^ 3=1

^^

3=1

- ^ b j - ^ j=i

=

coDt.

^^

Applying the trace from E = k{w^ z i , . . . , z^) into k and Theorem 3.2.4, we get co[E : k]Dt = D{Tr{w))

-•

^ 6 , ^ ^ ^

where Tr and N are the trace and norm maps respectively. Since Dt has no elementary integral over k by hypothesis, the above implies t h a t CQ = 0, hence t h a t w = ^ ^ ^ 1 bjyj is algebraic over k. Let F = E{y2^..., yr)- Since each yj is either a logarithm or algebraic over E by (9„18), F is elementary over E, hence

9.4 The Rothstein-Caviness Structure Theorem

295

elementary over A:, and therefore log-explicit Liouvillian over {C[x)^d/dx) with at most /i — 1 nonsimple primitives. Furthermore, Constp(F) = C since F C ÜT, so applying the induction hypothesis to yi = w — Yl\=2 ^jyj ^ ^ ^^^ 21 G F , we get that there are r^ G Q such that

where t^ = exp(a^) for z G Ep/c(x)- Since F is algebraic over k and each y^ is either a logarithm or algebraic over E by (9.18), Ep/c{x) = Ek/c{x) and r

r

where fj = 0 whenever y^ is not a logarithmic monomial, Vj otherwise. In addition, yi = cv -\- J^^^j mnVi, so (9.19) becomes r

= ^'^ '^ Yl ^*^^ "^" X] ^^^^ "^ X] ^*^ '* ^ ^ where fJ = mn + Xlj=2 ^ * i ^ ^ Q- Furthermore,

X^n'^i^

X]

n^^i-f

X] ^^^*

where 6^^ = exp(?7^) for i G Ex/k{t)i so CV+

X] i^Lk/c(x)

^^*^+

X] i^Ek/c{x)

^^^^+

X^ ^^LK/k{t)

^^^^+

X]

^^^^

GC.(9.20)

i^EK/k{t)

Since {6i}i^i is a transcendence basis for K over A;(t), {d6i}i^i is a basis for ^K/k{t) over ÜT by Theorem 9.1.2. For i G Ex/k(t) we have Ü;^ = dOi/Oi — drji where Oi = exp(77^) and rji G k{t){9i,... ,6i^i). For i G Lx/k{t) we have a;^ = drji/rji — dOi where 9i = log{rji) and r/i G k{t){9i,... ,9i-i). In both cases, 77^ G k{t){9i,... ,6^i_i) implies that that dr/^ is the K-span over K of {(i^j}j^jj
296

9 Structure Theorems

Corollary 9.4.1. Let C^x^K,Ex/c{x) a £ K"" and b £ K. Then,

^'^d Lx/c{x)

^^ o^s in Theorem 9.4-1,

(i) Da/a is the derivative of an element of K if and only if there are r^ G Q such that

(ii) Dh is the logarithmic derivative of a K-radical if and only if there are ri G Q such that

^

riDti+ Yl

n^=Dh.

(9.22)

The proof and corresponding algorithms are exactly the same than for Corollary 9.3.1 and the algorithms following it. In the cases arising from the integration algorithm, we can always ensure that the differential field containing the integrand is a log-explicit Liouvillian extension of its constants by applying recursively the integration algorithm to primitives. For the general case, Rothstein and Caviness also proved that any Liouvillian extension of a differential field can be embedded in a log-explicit Liouvillian extension [84].

Exercises Exercise 9.1. Prove Theorem 9.LL Exercise 9.2. Let i? be a commutative ring of characteristic 0. Show that the left Ä-module Ü{R) of all the derivations on R is isomorphic to the module Hom^(i7/^/^,A) of all the Ä-linear maps from ß ß / z into R. Exercise 9.3. Let C be a field of characteristic 0, x be transcendental over (7, and (K, D) be a real elementary extension of {C{x),d/dx) with Con.stjy{K) = C. Suppose that there are a, 6 in Ü" such that 6^ + 1 7^ 0, Da/a is not the derivative on an element of K, and Da a

Dh 62 _!_ 1 •

Show that \f-i G C. Conclude that if C is a real field, then the index sets LK/C{X)

and AK/C{X)

are disjoint.

10 Parallel Integration

We describe in this chapter an alternative approach to integration, also based on Liouville's Theorem, that attempts to avoid the recursive nature and associated computational cost of the algorithms presented in the previous chapters. This method, introduced by Risch and Norman in a verbal presentation at the SYMSAC'76 conference, attempts to handle all the generators of the differential field containing the integrand "in parallel", i.e. all at once rather than considering only the topmost one. As such, it has been called either the "new Risch algorithm" [67], the "Risch-Norman algorithm" [37, 38, 42] or the "parallel Risch algorithm" [30, 31, 35] in the literature. It is in fact heuristic rather than algorithmic, as it can fail both in theory and practice to compute elementary antiderivatives. However, its relative ease of implementation, speed and satisfactory rate of success make it an attractive alternative for designers of computer algebra systems. As a result, it is present in several systems, either as a replacement or preprocessor for the complete integration algorithm. The general idea behind parallel integration is to avoid the recursive nature of the integration algorithm by viewing the differential field containing the integrand as a field of multivariate rational functions over its constants. Recall from Sect. 5.2 that our integrand / belongs to a differential field of the form K = C ( t i , . . . ,t^) where C = Const(ir) and each ti is transcendental^ over C ( t i , . . . ,ti_i). By the strong Liouville Theorem (5.5.3), if the integral of / is elementary over K, then there are v G K, c i , . . . , c^ algebraic over C and H i , . . . ^Um G i^(ci,... ,c^)* such that

f = Dv + J2c,^^.

(10.1)

Since i; is a quotient of multivariate polynomials in t i , . . . , t^, and the i^^'s can be assumed without loss of generality to be polynomials in t i , . . . ,ty^ (logarithmic derivative identity), the parallel method consists in making educated ^We do not require in this chapter that the t^'s be monomials or that Dti = 1.

298

10 Parallel Integration

guesses for the Ui^s and the denominator of ?;, as well as for the degree of its numerator. This reduces the problem of solving equation (10.1) to finding the Q'S and the constant coefficients of the numerator of t;, and this can be done with elementary linear algebra. If the linear equations for the unknown constants have a solution, then an integral of / is found. On the other hand, the nonexistence of a solution does not always imply that / does not have an elementary integral over K^ as it could just mean that the guess was wrong. There are several published variants of the parallel approach, differing in the details of the guess or the approach used to solve for the unknown constants, but all share the property that there are functions having elementary antiderivatives that cannot be found with that variant. As a consequence, parallel integration is a convenient heuristic, that is significantly easier to implement than a complete integration algorithm. To turn it into an algorithm for a class of integrands requires proving that the given guess catches all the functions in that class having elementary antiderivatives. This remains an open problem, although partial results have appeared for logarithmic integrands in [35] and are extended here to more general differential fields.

10.1 Derivations of Polynomial Rings We describe first several properties of monomial extensions that hold in the more general setting of derivations of multivariate polynomial rings. Let F be a field, t i , . . . ,tn independent indeterminates over JP, and D be a derivation of the polynomial ring R = F [ t i , . . . ,t^]. Since F is a field, i? is a unique factorization domain, so its elements have greatest common divisors and least common multiples (see Sect. 1.1 and Exercise 1.16). Algorithms for computing gcd's and Icm's, as well as squarefree and multivariate factorizations can be found in computer algebra textbooks^ such as [39, 97] so we make use of them without further description. The notions of content and primitive part (see Sect. 1.6) can be extended to similar notions with respect to each variable ti by writing R = Ri[ti] where Ri = F [ t i , . . . ,t^_ 1,^^4-1,... ,tn]. We denote them by content (p, ti) and pp(p, ti) respectively, and say that p E Ris primitive with respect to ti if content(p, t^) e R* = F*. We first remark that Lemma 3.4.4 remains valid in this setting. L e m m a 10.1.1. Let p i , . . . , p ^ G F[ti,... ,tn] be such that gcd{pi^pj) = 1 for i j^ j , and let p = YflLi PT ^here the Ci ^s are positive integers. Then,

gcd{p,Dp) = n ^ ^ " '

llgcd{pi,Dpi

\i=i

Proof. The proof is exactly the same than the one of Lemma 3.4.4. There are also several newer and improved algorithms in the research literature.

10.1 Derivations of Polynomial Rings

299

By analogy with t h e monomial case, we say t h a t p E R is normal with respect to D if gcd(p, Dp) = 1, and special with respect to D if p \ Dp. Since Theorem 3.4.1 is a consequence of Lemma 3.4.4, it is not surprising t h a t it remains valid here. T h e o r e m 10.1.1. (i) Any finite product of normal and two by two relatively prime polynomials is normal Any factor of a normal polynomial is normal. (a) Any finite product of special polynomials is special. Any factor of a special polynomial is special. Proof. T h e proof is exactly the same t h a n t h e one of Theorem 3.4.1, using D Lemma 10.1.1 instead of Lemma 3.4.4. We say t h a t p = PsPn is a splitting factorization of p if Ps->Pn ^ FL^ Ps is special and every squarefree factor of Pn is normal. We proceed to show t h a t , given t h e additional step of extracting content, t h e splitting factorization algorithms of Sect. 3.5 can be generalized t o R. T h e o r e m 1 0 . 1 . 2 . Let p G , F [ t i , . . . ,tn] be primitive some j . Then,

with respect to tj for

gcd(p. Dp) gcd{p,dp/dtj) is the product of all the coprime special irreducible factors of p. (ii) If in addition p is squarefree, then p = PsPn ^s a splitting factorization p, where Ps = gcd(p. Dp) and pn = p/ps-

of

Proof. Let q G R he any irreducible factor of p. Since p is primitive with respect t o tj, q does not divide content(p,tj), so*^ dq/dtj ^ 0. It follows t h a t deg^.(dq/dtj) < deg^.(g), hence t h a t gcd{q,dq/dtj) = 1. T h e proof is then exactly t h e same t h a n the one of Theorem 3.5.1, using Lemma 10.1.1 and Theorem 10.1.1 instead of Lemma 3.4.4 and Theorem 3.4.1. D The hypothesis t h a t p be primitive with respect t o tj is really needed, as the following example illustrates. Example 10.1.1. Consider the derivation D on Q[X, Y] given by DX = 1 and DY = y , and let p =^ XY + X^Y\ Then, gcd(p. Dp) gcd{p,dp/dX)

_ ^ ^ -^ Y

^^^

gcd(p, Dp) gcd{p,dp/dY)

_ Y X

while t h e product of the special irreducible factors of p is Y. ^Recall that all fields are of characteristic 0 from Chap. 5 onwards.

300

10 Parallel Integration

For any p G F [ t i , . . . , t ^ ] \ F , define the main variable of p^ denoted mamVar(p) t o be tj where j is the highest index such t h a t deg^.(p) > 0. Writing p = content(p,tj)pp(p, tj) where tj = mainVar(p), Theorem 10.1.2 shows t h a t the algorithm S p l i t F a c t o r can be applied to p p ( p , t j ) , yielding its splitting factorization pp{p,tj) = qsQu- Recursively computing a splitting factorization of content(p,tj), which has one variable less t h a n p, yields content(p, tj) = hghn^ and it follows t h a t p = PsPn is a splitting factorization of p where Ps = Qshs and p^ = Quhn-

S p lit Fact or (p, D)

(* Splitting Factorization *)

(* Given a derivation D on jP[ti,...,tn] and p G i^[ti,...,tn], return {Pn,Ps) G F[ti,..., tn]'^ such that p = PnPs, Ps is spccial, and each squarefree factor of pn is normal. *) if p £ F t h e n r e t u r n ( l , p ) t ^— mainVar(p) c <— content(p, t), q <— p/c

(*
{hn.hs) ^— SplitFactor(c, D) S <— gcd(q, Dg)/gcd(g, dq/dt) if deg^(S') = 0 t h e n return(/inP, hs) {qn.qs) ^ SplitFactor(g/5, D) YetVLYn{hnqn, Shsqs)

(* exact division *) (* exact division *)

Example 10.1.2. Applying S p l i t F a c t o r t o the polynomial p G Q[X, Y"] of example 10.1.1 we get: 1. mainVar(p) = Y 2. content(p, Y) = X,q = p/content(p, Y) = Y + XY'^ 3. recursive call, S p l i t F a c t o r ( X , 19): a) mainVar(X) = X b) c o n t e n t ( X , X ) = 1 c) recursive call, S p l i t F a c t o r ( 1 , D ) returns (1,1). d) gcd{X,DX) 1, so return {X, 1) gcd{X, dX/dX) 4. hn = X,hs

= 1. gcd(g, Dq) gcd{q, dq/dV)

^ g c d ( y + X F ^ Y + Y^ + 2XY^) ~

gcd(y + X y 2 , 1 +

^

2XY)

6. q/S = l-\- XY 7. recursive call, S p l i t F a c t o r ( l + XY, D) returns (1 + XY, 1) So a splitting factorization of p is XY

+ X'^Y'^ = pnPs = ( X +

X'^Y)Y.

10.2 Structure of Elementary Antiderivatives

301

The algorithm S p l i t S q u a r e f r e e F a c t o r can be generalized t o j P [ t i , . . . , t-^j in a similar fashion, by extracting the content of each squarefree factor as shown below. Another alternative would be to compute a squarefree factorization of the primitive part of p, then separately compute a splitting factorization of its content and recombine b o t h factorizations. SplitSquarefreeFactor(p, D)

(* Splitting Squarefree Factorization *)

(* Given a derivation D on F[ti,.. ., tn] and p e F[ti,... ,tn], return {Ni,...,Nm) and {Si,...,Sm) in F [ t i , . . . ,tn]^ such that p = {NiNi • • • N^){SiS2 • • • S^) is a splitting factorization of p and the Ni and Si are squarefree and coprime. *) {pi,...,Pm) ^ Squarefree(p) for 2 ^— 1 t o m do t <— niainVar(pi) Ci <— content(pi,t) Qi ^

Pi/Ci

{Mi,Ti)^ SplitFactor(ci,J9) Si i- gcd{qi,Dqi) Ni ^ qi/Si r e t u m ( ( M i A r i , . . . , MmNm), (TiS'i,.. • 5

(*Q

(* Qi =PP{Pi,t) *) is already squarefree *) (* exact division *)

-^m^m))

Let now K = F{ti,..., t^) be t h e quotient field of F [ t i , . . . , t^] and let D be a derivation on K such t h a t DF C F. As F [ t i , . . . ^t^] is not necessarily closed under D , we define the denominator of K with respect to D, denoted den£)(ii^), to be the least common multiple of the denominators of D t i , . . . , Dtn- Since D = KJJ i- Yl7=ii-^^^)^/^^^ (^^^ Exercise 3.6) where KD is the derivation on F [ t i , . . . , t^] given by

KD

fX^aetr---t^H=X](^ae)tr...t^

D = deii]j{K)D is a derivation of j P [ t i , . , . , t^], so we can use splitting factorizations with respect to D.

10.2 Structure of Elementary Antiderivatives We now return to integrating elements of a differential field of the form K = F ( t i , . . . , t^) where DF C F and each ti is transcendental over C ( t i , . . . , t^_i). As seen at the end of the previous section, D = deii]j{K)D is a derivation of F [ t i , . . . , t „ ] . It turns out t h a t a key property of normal polynomials (see Theorem 4.4.2) remains valid in the multivariate case.

302

10 Parallel Integration

L e m m a 1 0 . 2 . 1 . Let p G F[ti^ • • • ,tn] be irreducible and normal with respect to D. If p^ divides the denominator of f £ K* for some m > 0, then p'^^^ divides the denominators of Df and Df. Proof. Let p be irreducible and normal with respect t o D. Suppose t h a t p^ divides the denominator of / G ÜT* for some m > 0 and write / = a/bp^ where s >m, a,b E F[ti,..., t„] and p divides neither a nor b. Then, -=-

bpDa — apDb — sabDp

Since p does not divide a, b and Dp^ p does not divide sabDp. As it divides bpDa — apDb, it follows t h a t p does not divide the numerator of Df, hence t h a t p^^-*- divides the denominator of Df. Since s > m, p^+i ^^jg^ divides it. Since Df = denD{K)Df, the denominator of Df divides the denominator of D / , whence p ^ + ^ also divides the denominator of Df. D We can now describe the general shape of an elementary integral. We write C for Constjj{K) and C for its algebraic closure. T h e o r e m 1 0 . 2 . 1 . Suppose that ConstD(i^) C F, and let f = a/d e K"" where a^d E F [ t i , . . . , t ^ ] are coprime. Let d = dgdn be a splitting factorization of d w.r.t D, and 11^=1 ^^^ ^^^ Y\.k=i'PT ^^ respectively the squareof d^ and its irreducible factorization in C F [ t i , . . . , t n ] . free factorization / / / has an elementary integral over K, then there are b, s G F[ti,... ,tn], wi,...,Wr_e CF"", a i , . . . , a ^ , / 3 i , . . . , / 3 i , 7 i , . . . , 7 ^ G C and s i , . . . , 5 ^ irreducible in ( 7 F [ t i , . . . , tn] such that 5, s i , . . . , s ^ are special with respect to D, and

!-4j^^^*t^.^*tß.^^t-,.^^

(102 .)

Proof Suppose t h a t / has an elementary integral over K. By the strong Liouville Theorem (5.5.3), there are ^' G JFC, C i , . . ., c^ G C, and i ^ i , . . . , w^ G K{ci,..., Or)* such t h a t

f = Dv + J2^'^ ^ Ui As in the proof of Liouville's Theorem, write Ui — ^iTliPi^ where Wi G J ^^j F ( c i , . . . , c^)*, each pij G F ( c i , . . . , C7.)[ti,..., t^] is monic irreducible and the Cij^s are integers. Using the logarithmic derivative identity and grouping together all the terms involving the same pij we get Dvj

v-^ Dwi

/ = ^« + EOTr^ + E^^ Vj

10.2 Structure of Elementary Antiderivatives

303

where (j E C ( c i , . . . , c^) and Vj G -F(ci,..., c^)[ti,..., t^] are coprime monic irreducibles. Multiplying through by den£f{K) yields

j= l

-^

^=l

Let p E F [ t i , . . . , t^] be an irreducible normal factor of the denominator of v, and let 1/ > 0 be such that p^ divides that denominator. By Lemma 10.2.1, p^^^ divides the denominator of Dv. The denominator of J2j Cj{-^'^j)/'^j ^^~ vides lcmj('Uj), which is squarefree, so it is squarefree. Therefore, the factor p^'^^ is not canceled by the sum of logarithmic derivatives, so p^+-^ divides the denominator of deii]j{K)f. But the denominator of deiLiD{K)f divides the denominator of / , so p^+i | dn^ which implies that the normal part of the denominator of v divides 0^=2 ^7" ' hence that

where 6, s G F[ti^ • • • ,^n] and s is special with respect to D. Suppose finally that some Vi is normal with respect to D. Then, the denominator of dDvi/vi is Vi, which is coprime to the denominator of J2j^i Cj{Dvj)/vj. If Vi does not divide d^^ then Vi does not divide the denominator of i;, so it does not divide the denominator of Dv, whence Vi must divide the denominator of deiiD{K)f, which divides d, in contradiction with Vi not dividing dn- Therefore each Vi is either special with respect to D or a factor of d^. Breaking up V . (j{Dvj)/vj into a sum over special irreducibles and a sum over irreducible factors of dn yields (10.2). D Theorem 10.2.1 gives parts of an elementary integral of / . We use it to compute integrals when F = Consti:)(il"), in which case Yl^jiDwi/wi = 0, so that term can be removed from (10.2) and the remaining unknown parts are the numerator of v and several special polynomials. For general derivations, the special polynomials are not known, although the requirement that F = Constjj{K) implies that there can be only finitely many coprime irreducible special polynomials (see Exercise 3.6 and [27, 85, 93]). When K = F{ti,..., tn) is a tower of nested monomial extensions over F , the situation is easier and we can describe all the special polynomials. Remark first that if F is closed under D and each ti is a monomial over F{ti,... ,ti_i), then F ( t i , . . . ,ti__i) and F{ti,... ,^i_i)[ti] are closed under D, so the notion of special with respect to D is well defined for elements of F ( t i , . . . ,ti-i)[ti]. For any nonzero p £ F ( t i , . . . , ti_i)[t^], write p'^ G F [ t i , . . . , t^] for the product of p by the least common multiple of the denominators of its coefficients with respect to t^. We introduce the notation

5]^^^ = {p+ for p e U 5]?^,,,...,,,„,)[,,]:^(,, l
,,:^,)} C F\h,...,

g .

304

10 Parallel Integration

W h e n the ti are all primitive, hyperexponential or hypertangents, then S^^^.p is finite and consists exactly of the ti t h a t are hyperexponential and of t h e factors of 1 + 1 | for the tj t h a t are hypertangents (see Chap. 5). T h e o r e m 1 0 . 2 . 2 . Let K = F{ti^ • • • ^t^) where DF C F and each ti is a monomial over F{ti^... , t ^ _ i ) . Then, the irreducible special polynomials for D are exactly all the members of S^^.p and all the irreducible factors of deii]j{K). Proof Let p e F[ti,..., t^] be an irreducible factor of deiiD{K) and tj be its main variable. Then, p G E[tj] where E = F ( t i , . . . , t j _ i ) , so Dp € E[tj] since tj is a monomial over E. We can then write Dp = a/b where a € - F [ t i , . . . ,tj] and b G F[ti,..., t j _ i ] , which implies t h a t p )[b. We have bDp = bdenD{K)Dp

= deiiD {K)a

and p I dei[ijj{K), so p \ bDp. Since p is irreducible, it follows t h a t p \ Dp^ hence t h a t p is special with respect to D. Let now p G '^^^•p 3iid tj be its main variable. Then, p = q~^ for some q G *^fL,].ß, where E = F ( t i , . . . , t j _ i ) . Since p = aq with a e E* a n d q is irreducible in £[^j], it follows t h a t p is irreducible in E[tj]^ hence in j P [ t i , . . . ,tj_i][tjj by Gauss' Lemma (see Exercise L15). In addition, a \ Da in E[tj], so p I Dp in E[tj]^ which implies t h a t p \ Dp in E[tj^... , t ^ ] , hence t h a t p\ Dp in F [ t i , . . . ,t^_i][tj-,... , t n ] . Conversely, let p € F [ t i , . . . , t n ] be an irreducible special with respect t o D , and tj be its main variable. Then, Dp G ^ [ t j , . . . ,tn] where E = F ( t i , . . . , t j _ i ) . Since p is special with respect to D , p | den]j{K)Dp in £ ' [ t j , . . . , tn]. Suppose t h a t p does not divide deiiD{K) in F [ t i , . . . , t^]. T h e n p does not divide den£)(K) in E[tj^... , t n ] , so p | Dp in - B [ t j , . . . , t n ] . As p and Dp are both in E[tj]^ this implies t h a t p divides Dp in £^[tj], hence t h a t p is special with respect to D . Write p = g+ where q G £^[tj] is p divided by its leading coefficient with respect to tj. T h e n q = ap foi a £ E* is monic, irreducible, and special with respect to D since Dq = pDa + aDp = q

fDa \ a

\

Dp\ -] . p J

Therefore, q G S^^^.

D

As a consequence, there can be special polynomials with respect to D even when S^^^.p is empty, as in the case of nested primitive extensions, b u t their irreducible factors are limited to the factors ofdeii£f{K). Consider for example K = Q ( x , t ) with Dx = 1 and Dt = Dq/q for some q G Q[x]. Since deni:>(ür) is the squarefree part of g, every irreducible factor of q is special and can appear as a logand in integrals, which explains the origin of the log(x + 1) t e r m in the following integral from [35]:

/

log(x^ -l)dx

= {x-

1) log(x^ - 1) + 21og(x + 1) - 2x .

10.3 The Integration Method

305

It also explains the "anomalous" log(H-tan'^ 0) terms reported in [31] that can arise when integrating expressions involving tan^, since 1 + tan^ 9 is special. As a consequence of Theorem 10.2.2, the equation (10.2) can be refined when K he a tower of nested monomial extensions of F. The Si can be taken to be the irreducible factors (over CF) of deii]j{K) and of the elements of '^W-.F^ while s is of the form

^=

n

^'' n ^'^

p\denD{K)

(10.3)

q€S%]p,

where ßp > 0, e^ > 0 and the first product is taken over all the monic irreducible factors of deiiD{K) over F rather than CF. Those exponents can be bounded in some differential fields, including nested logarithmic extensions, as will be shown in Sect. 10.4.

10.3 The Integration Method The parallel method now follows from the results of the previous section, and it can be applied to multivariate rational function fields with arbitrary derivations as long as Const/:) (X) = F. It consists in making educated guesses for the s^Si and pi in (10.2), bounding the degree of the numerator b of v^ and finally solving linear equations for the coefficients of b and the unknown constants ai^ßi. Since s is special with respect to D, it must be of the form s = HpP^^ where p ranges over all the monic irreducible special polynomials and all but finitely many gp are 0. Since no bound is known for e^, the common guess in the literature is to take the highest e such that p^ \ d. This has the added advantage that s = ds and that no factoring of dg is needed. It also sounds reasonable, since any higher exponent, or the appearance in s of a special irreducible factor not dividing d would require cancellation in Dv in order to disappear. Such cancellation can however happen, as illustrated in the following example, so the guess s = ds turns this procedure into a heuristic that might fail to compute an elementary antiderivative. Example 10.3.1. Consider ,2

J

(e^ (e^ 4- x) x) a:^

We have K = Q(x,ti,t.2) with Dx = 1, Dti = ti {i.e. ti = e^), and ^^

(x2 + l)tl + {2x^ - ^2 + 2x)ti + x^ x^(x + ti)^

i.e. t2 = e(^'-i)/^+V(e^'^+^). Therefore, denoiK) = x^{x + h)^ and the polynomials X and x-j-ti are special with respect to deii]j{K)D. Our integrand is then

306

10 Parallel Integration

^~

(ti - X^ + 2x)t2 X2(x + ti)2

whose denominator is d = x^{x -\- ti^^ which is special. It turns out however (see example 6.4.2) that

h where the special denominator s = ti does not divide d. Therefore, the integral of / cannot be computed by the parallel method with the guess s = ds. Next come the monic special irreducible 5^. When ÜT is a tower of nested primitive, hyp er exponential and hypertangent monomials over F, then Theorem 10.2.2 provides a finite exhaustive list of all of them, namely the factors of deii]j{K)^ together with the hyperexponential t^, and the factors of 1 + 1 | for the hypertangent tj. According to Theorem 10.2.1, one should really take their irreducible factors in F[ti,... ,t^] (recall that F = Consti:)(K))) rather than in F [ t i , . . . , t„], as illustrated by the following example. Example 10.3.2. Let K = Q(x, t) with Dx = 1 and Dt = 2x/{x^ - 2) {i.e. t = log(x2 — 2)). Then, deiiD{K) = x^ — 2 is special with respect to den£){K)D^ and irreducible in Q[x,t]. However,

which shows that the irreducible factors of deni:)(ii") in Q[x,t] must be considered. Similarly, the pi should be the irreducible factors of d^ in F [ t i , . . . , t^^] rather than in F [ t i , . . . ,t^] as illustrated by dx

/

1

1

-X) 2^/2: l0g(A/2 ^' '

-2

p log(A/2 + X) .

2A/2

There are however arguments for factoring d^ and denjj (K) only over F rather than F when K is a tower of nested primitive, hyperexponential and hypertangent monomials. Since the parallel approach can fail to find integrals no matter what we choose, we must switch back to the complete integration algorithm if it does not find an integral. So it is useful only when it is significantly faster than the complete algorithm. While there are algorithms for computing irreducible factorizations in F [ t i , . . . ,^n] (called absolute factorizations)^ their cost can actually be higher than the full integration algorithm. In addition, this factorization could compute algebraic extensions of F that are not necessary to compute the integral, as in the case of

/

^

dx = log(x2 - 2).

10.3 The Integration Method

307

For those reasons, both approaches appear in the Hterature: those that report on actual implementations ([37, 38]) factor over F , while those that analyze the method ([30, 35]) factor over F. We finally need to bound the degree of the numerator b of v. When K is a tower of nested monomial extensions, Lemma 3.4.2 suggest the following natural guess, which is the one used in the literature: ^*'^^-\

max(deg,,(a),degt_(rf))

ii5{ti)>0.

^

''

where / = a/d. Once we have bounds bi on deg^. (6) we write 61

E ...£u,,...,„4'...c

(10.5)

ii=0

where the Ui^...i^ are undetermined constants (in F not F). Substituting all our guesses in the right hand side of (10.2), equating it with / and clearing denominators yields an equation of the form 61


bn

m

t

••• X^ Wii-i„,^n-i,,. +'^airi-^Y^ßiWi in=0

i=l

(10.6)

i=l

where q and the qi^...i.^ are in F [ t i , . . . ,tn], and the r^ and wi are either in F [ t i , . . . , t^] or F [ t i , . . . , t^], depending upon the choice made earlier. Equating the coefficients of the same monomials on both sides of (10.6) yields an inhomogeneous linear system for the unknown constants w^^...^^, a^ and ßi. There are a couple of variants as to how to solve that linear system: while the original approach [30, 37, 67] equates first the monomials of highest total degrees and solve for the t/ii,...i,„ of lowest weight in the obtained equation, [38] reports that directly solving the whole system appears to perform at least as well in practice. The parallel method can easily be adapted to differential fields that are not nested monomial extensions, since Theorem 10.2.1 is valid in such fields. In that case, since there is no complete integration algorithm in case of failure of the parallel approach, one should use irreducible factors over F rather than F for the candidate si and pi of (10.2). In addition, the degree bounds (10.4) should be replaced by deg(5) < 1 + deg(a) + max(0, deg(denj:)(Ä')) — max (deg(Dt^))) l
that arises from assuming no cancellation of total degree (deg means total degree). Finally, the form (10.5) for h should be replaced by

il + ...+i„<deg(b)

308

10 Parallel Integration

Another problem in such extensions is that the special polynomials with respect to D are not known. One should start by taking all the irreducible factors over F of the special part of denD{K) (which is not always special, see example 10.3.4). It is possible in addition to look for special polynomials of low total degree by taking p to be a polynomial of fixed degree with unknown constant coefficients. The condition p \ Dp translates into a system of nonlinear algebraic equations for the coefficients of p, which can be solved using algebraic techniques [39]. This computation depends only on the field X, so specials can be precomputed and stored for some non-monomial extensions. A finite, potentially empty, set S of monic irreducible specials can be found in that way, and used as a replacement for S^^^.p in the non-monomial case. Finally, the guess s = dg should be replaced by

s = ds

Yl P

in order to catch potential cancellations outside the factors of dg. Example 10.3.3. Consider

2 — log(a:)^ Let K = Q(x,t) with Dx = 1 and Dt = 1/x {i.e. t = log(x)). We have denoiK) = x, which is special with respect to deiiD {K)D. Since t is a primitive monomial over Q(x), <S]^^Q is empty, so S = {x}. Our integrand is

--t2

with numerator a = 1 — t and denominator d = x^ — t^, which is normal and factors as d = piP2 = {x — t){x i-t). We find the degree bounds bi = b2 = 3, so the candidate integral is y ^ UijxH^ + a i log(x) + ßi log(x -t) -\- ß2 log(x + t). Equating its derivative to / yields a system of 23 equations for the 13 unknowns ai^ßi,ß2^Uij. Solving that system yields the general solution

/

f = at-h'u — a log(a:) — - log(x — t) -h - log(a: + t)

where a and u are arbitrary constants. Choosing a = u = 0 yields f

1 - lOg(x)

,

1 1

/

1

/

XN

1 1

/

1

/ XX

J x2-log(x)2'^'^ " 2 ^^^"^ ^ ^^^""^^ ~ 2 ^^^"^ ~ ^^^""^^'

10.3 The Integration Method P a r a l l e l l n t e g r a t e ( / , D)

309

(* Parallel Integration *)

(* Given a derivation D on K = F{ti,..., tn) such that Consta(i^) = F, return either an elementary integral of / , or "failed", in which case it is unknown whether / has an elementary integral over K. *) (* The first 4 steps need to be done once per field K *) h <— lcm(denominator(Dtl),..., denominator(Dtn)) if each ti is a monomial t h e n hs •^— h else (hn.hs)^ SplitFactor(/i, D) A ^r- I -\- max(0,deg(/i) — maxi
qirr

else S ^— FindSpecials(/iD) S ^ S U IrreducibleFactors(/is)

(* exhaustive search for low degree *) (* over F or F *)

(* The remaining steps are done for each integrand *) a ^— numerator(/), d <— denominator(/) {dn,ds)^ SplitFactor(rf, D) ( d i , . . . , (ie) ^- SquareFree(c?n) { p i , . . . ,pt} ^- IrreducibleFactors(c?i • - • de) (* over F or F *) if each U is a primitive, hyperexponential or hypertangent monomial t h e n Vs ^

ds

for i ^— 1 t o n d o bi ^ max(deg^. (a), deg^. (d)) if deg^. (Dti) = 0 t h e n h <-h-\~ 1 else Vs <- ds Ilpes,pXd, P Ö <— deg(a) -i- A

(* for general fields *) (* total degree *)

Solve the inhomogeneous linear system for the UiQ...i,^, as and ßi obtained by clearing denominators and equating coefficients of equal monomials in / = Dv + ^^^gasDs/s + Yli=i ßi^Pi/Pi- ^^ ^^ ^^^ ^° solution t h e n r e t u r n "failed" else r e t u r n v + J2ses^^ log("S) + Yll=i A log(pO-

Example 10.3.4- We illustrate the applicability of the parallel approach to non-monomial extensions by integrating an expression involving t h e Lambert W function. T h a t function is defined (see [24]) as the solution of W{x)e^^''^

=x.

Consider

/

x{l + W{x^)Y

(10.7)

310

10 Parallel Integration

Taking logarithmic derivatives on both sides of (10.7) yields

dw{x) ( dx

1 ^ _ 1

V

^(^)

whence dW{x) W{x) dx x{l + W{x)) and W{x) is not a monomial over C(x). By the chain rule, dWix'^) _ dx

W{x^)

_

X2(l + W ( x 2 ) )

2W{x'^) x{l

+ TF(x2))

so let K = Q{x,t) with Dx = 1 and Dt = 2t/{x + xt) {i.e. t = W{x^)). We have denD{K) = x -\- xt and A = \ — max(0, deg(x + xt) — max(deg(x + xt),deg(2t))) = 1. Computing a splitting factorization oi x -{- xt yields the special part x and normal part 1 + t. Searching for specials by computing denD{K)Dp/p where p = coo + CIQX + coit + cnxt + C2ox'^ + co2^^, we find that the only monic irreducible specials w.r.t. D of total degree at most 2 are X and t, so we let S = {t}. Our integrand is

x^ + (x^ + 2)t •^ ~ X + 2a:t + xt2 with numerator a = x^ + (x^ + 2)t and denominator (i = x + 2xt + xt^. Proceeding with parallel integration, we obtain 1. 2. 3. 4. 5.

(dn.ds) = SplitFactor(d, D) = (1 + 2t + t^, x) SquarePree(
So the candidate integral is

Equating its derivative to / yields a system of 22 linear equations for the 18 unknowns ai^a2,ß,Uij. Solving that system yields the general solution / /

^ ^ ^

2alog(x) + alog(t) + log(l + t)

where a and u are arbitrary constants. Choosing a = u = 0 yields /

a;(l + iy(x2))2

^ ^ 2 ; = - - — — + l o g ( l + PF(x2)). 2W{x-^)

The denominator W{x'^) appearing in the integral does not appear in the integrand, but is one of specials predicted by Theorem 10.2.1.

10.4 Simple Differential Fields

311

10.4 Simple Differential Fields We describe in this section how Theorem 10.2.1 can be improved in a class of differential fields that includes nested logarithmic extensions, as well as more general ones. For general derivations, special polynomials do not always satisfy Lemma 10.2.1, for example p = e^ in Q(x,e^). We are interested in differential fields where every irreducible special divides deui:» and satisfies Lemma 10.2.1. Throughout this section, {K,D) is a differential field of the form K = F ( t i , . . . ,tn) where DF C F and each ti is transcendental over F ( t i , . . . ,t^_i). Definition 10.4.1. We say that an irreduciblep G F[ti,..., t^] is simple with respect to D if p^ divides the denominator of f G K* for some m > 0 implies that p^^^ divides the denominator of Df. We say that (K^D) is simple if every irreducible special of F[ti^ • • • ,tn] divides deiiD{K) and is simple. A differential field with no nontrivial special polynomial is clearly simple. This is the case for {C{x), d/dx) but also for towers of nested monomial extensions when S^^.p is empty and deiiDiK) = 1, as in the following example. Example 10.4-1- Consider K = Q{x^t) with the derivation D given by Dx = 1 and Dt = X — t'^. As explained in note 4 of Sect. 5.11, it can be proven using differential Galois theory that the differential equation y'{x)-\-y{xy' = x has no algebraic function solution, and Theorem 3.4.3 then imphes that ^Qjx)it]-Q(x) is empty. Since deiiD{K) = 1, Theorem 10.2.2 implies that every irreducible p G Q[x,t] is normal, hence that (K^D) is simple. The monomial t in this example corresponds to the logarithmic derivative of an Airy function solution of y''{x) = xy{x). A key property of simplicity, is that it is preserved by primitive extensions, provided that no new constant is introduced and that den/^ is squarefree, which is equivalent to requiring that the denominators of the Dti are all squarefree. L e m m a 10.4.1. Let K = F ( t i , . . . ,tn) where DF C F and each ti is transcendental over F ( ^ i , . . . ,t^_i). Suppose that Const(ür) C F, that E = F ( t i , . . . , t n - i ) is closed under D, and that Dt^ G E. If E is simple and den-D{K) is squarefree, then K is simple. Proof Since Dtn G £', the denominators of all the Dti are in F [ t i , . . . ,tn-i], so denoiK) G F [ t i , . . . ,tn-i]- Let D = denD{K)D and p G F [ t i , . . . ,t„] be irreducible and special with respect to D. Since deii]j{K) e E*^ p divides Dp in E[tn]^ so p G F [ t i , . . . ,tn-i] by Theorem 5.1.1, which implies that denD{E)Dpe F [ t i , . . . ,tn_i]. Let /i = deni)(K)/deni)(E) G F [ t i , . . . ,t^^i]. Since p \ Dp = h{deiiD{E)Dp) and p is irreducible, it must divide h or deiiD{E)Dp. If p I /i, then p \ deiiD{K). Otherwise, p \ deiijj{E)Dp^ which means that p is special with respect to dejiD{E)D^ hence that p \ denoiE)

312

10 Parallel Integration

since E is simple. Therefore, every irreducible of jP[ti,... ,tn] that is special with respect to deiiD {K)D divides deii]j{K). Let now / G K* be such that p'^ divides the denominator of / for some m > 0, and write / = a/bp^ where s >m^ a^b e F[ti,...,t^] and p divides neither a nor b. Then, den n (K)

1—

bDa - oDb -

sab^

which shows that at most p^+i divides the denominator of {deii]j{K)/p)Df. Write a = ^ ^ a^t^ where a^ G F[ti^..., t„_i]. Since p E jP[ti,..., t^-i] does not divide a, it cannot divide all the a^, so the set / = {i s.t. p / a^} is not empty and we can write a = ä -\- pa^ where p does not divide ä = X^^^j a^t^ and ä = Yli0{^i/p)'^li ^ ^[^i? • • • ? ^n]- Similarly, since p does not divide 6, we can write b = b -\- pb^ where 6, 6 G -F[ti,..., t^] and p does not divide b. We have Da = Da + P-Dä + äDp = Da + p I Da + ä — ) = Da

V and similarly 1^6 = Db

(mod p ) ,

PJ

(mod p). Therefore,

_ — 'Dp — -Dp 6Da - aDb - sab—^ = bDä - äDb - säb—^ (mod p). (10.9) P P Let a = deg^^(d), ß = deg^^ (6) and A, ß G F [ t i , . . . ,tn-i] be the leading coefRcients with respect to tn of d and 6 respectively. Since Dt^ G .E and denD(iC) G F [ t i , . . . ,tn-i], Dt^ = deni^(il^)Dt„ G F [ t i , . . . ,tn_i], which implies that Wa - äDb - sab^ where q G F[ti,...,

= (BDA - ADB - sAB—j

t^+^ + q

(10.10)

t^] and deg^^ (g) < a + /?• Let now

From the definitions of d and 6, p divides neither A nor D, so p^ divides the denominator of ^. If p | denp(£)Dp, then p is simple since E is simple. Otherwise, p is normal with respect to deiijj{E)D, so p is simple by Lemma 10.2.1 applied to E. So p^+^ must divide the denominator of Dg in any case. Since p I deni:)(K), which is squarefree, p does not divide denjn> (i<")/p, so p^+^ divides the denominator of {denD{K)/p)Dg. We have denniK) ^_ l - _ _ BDA-ADB -Dg = -Dg +1 p B'^p'

SAB'^

10.4 Simple Differential Fields

313

so PKBDA-ÄDB-SAB-

P It follows then from (10.9) and (10.10) that — — 'Dp p YbDa — aDb — sab— p and (10.8) implies that p^~^^ divides the denominator of {den]j{K)/p)Df, which divides the denominator of Df. Therefore, p^+^ divides the denominator of Df. Since s > m^ pm+i ^^^^ divides it, so p is simple, which implies that K is simple. D We now have a large class of simple differential fields, namely nested primitive extensions with denjj squarefree. T h e o r e m 10.4.1. Let K = F ( t i , . . . , t n ) where DF C F, ConstD(ü:') C F and each ti is a primitive monomial over F ( t i , . . . ,t^_i). / / deiiD{K) is squarefree, then [K^ D) is simple and the denominator of Df/f is squarefree for any f G ÜT*. Proof. We first prove by induction on n that (K^D) is simple. For n = 1, we have deiiD{F{ti)) = 1 since Dti G F. In addition, Theorem 5.1.1 imphes that SpL i.p is empty, so Theorem 10.2.2 implies that every irreducible in F[ti] is normal, whence F{ti) is simple. Suppose now that n > 1 and that the towers with n — 1 generators are simple when den/^ is squarefree. Let E = F ( t i , . . . ,tn_i). Since denjj{E) \ deiiD{K)^ deiiD{E) is squarefree, so E is simple by the induction hypothesis, and K is simple by Lemma 10.4.1. We now prove the last assertion, first for an irreducible p G F [ t i , . . . , t^]. Let S be the set of indices i such that p divides the denominator of Dti. Note that 1 ^ 5 since Dti G F , so let m be n if 5 is empty, min(5) — 1 otherwise. Then, p G F[ti,..., tjn] and p does not divide the denominators of D t i , . . . , Dtmj so p does not divide h — deni:)(F(ti,... , t ^ ) ) . Since hDp G F\t\,... , t ^ ] , the denominator of Dpjp divides ph., which is squarefree, so the denominator of Dpjp is squarefree. Let now / G Ä'* and write f 8iS f = YliPT where the pi are irreducible and the e^ are integers. By the logarithmic derivative identity, Df f

^ "^

Dpi Pi

Since the denominator of each Dpi/pi is squarefree, so is the denominator of

Df/f.

a

The condition deii]j{K) squarefree is satisfied in particular by nested logarithmic extensions of a single primitive extension, which are therefore simple.

314

10 Parallel Integration

Corollary 10.4.1. Let K = F ( t i , . . . ,tn) where DF C F, ConstD{K) C F, ti is a primitive monomial over F and each ti is a logarithmic monomial over F ( t i , . . . , t^__i) for 2 < i < n. Then, denD{K) is squarefree and (K^D) is simple. Proof. We first prove by induction on n that deii]j{K) is squarefree. Since Dti G F , den2)(F(ti)) = 1, which is squarefree. Suppose now that n > 1 and that deiiD{E) is squarefree, where E = F ( t i , . . . ,tn-i). Let / G £'* be such that Dtn = Df/f. Since denD{E) is squarefree, the denominator of Df/f is squarefree by Theorem 10.4.1 applied to E. Therefore, denz){K)^ which is the least common multiple of denD{E) and the denominator of Df/f is also squarefree. Once deii]r){K) is squarefree, (K^D) is simple by Theorem 10.4.1. D

We can finally refine Theorem 10.2.1 in the case of simple differential fields with deiij:){K) squarefree. As earlier, we write C for the algebraic closure of Const D{K). Theorem 10,4.2, Suppose that Const£)(i
Proof. Suppose that / has an elementary integral over K. Then, by Theorem 10.2.1, there are c, 5 E F [ t i , . . . ,t^], « i , . . . ,ag,/3i,... ,/?t,7i,... ,7r> G C, wi,... ^Wr G C F * and ui,... ,Uq irreducible in C F [ t i , . . . , t^] such that s, 1^1,..., liq are special with respect to D, and

\ sTl

cd-

Ui

Pi

Wi

where d^ = Yl7=i^nj ^^ ^^^ squarefree factorization of d^. Since (K^D) is simple, the Ui must divide denD(K), so we can take q = m and Ui = 5^, since 5 i , . . . , 5^ are all the special irreducible factors of den]j{K). Let now P ^ F [ t i , . . . ^tn] be an irreducible factor of s and let 1/ > 0 be such that p^ divides s. Since s is special, so is p by Theorem 10.1.1, so p \ denD{K)Dp. Therefore, the denominator of Dp/p divides den]j{K), which implies that the denominator of Dp/p is squarefree. Since (K^D) is simple, p is simple, so

10.4 Simple Differential Fields

315

pi^+i divides the denominator of D{c/sYl^=2^rij )• ^^ ^^^ denominator of Dp/p is squarefree, the factor p^+^ is not canceled by the sum of logarithmic derivatives, so p^~^^ \ dg. Therefore, s \ Yi^j=2^iJ where dg = 0^=1 <^sj i^ the squarefree factorization of ds, which implies that / can be written in the form (10.11). D As earlier, when F = Constp (Ü!'), ^ ^ jiDwi/wi = 0, so that term can be removed from (10.11). The algorithm P a r alle l l n t e g r a t e is easily modified for simple differential fields with den/:^ squarefree: the set S consists only of the irreducible factors of /ig, we compute the squarefree factorization of d rather than of dn, as well as the irreducible factorization of dn- Finally, the term Vg can be ignored in the candidate integral.

References

1. S.K. Abdali, B.F. Caviness, and A. Pridor. Modular polynomial arithmetic in partial fraction decomposition. In Proceedings of the 1977 MACSYMA Users Conference, pages 253-261. NASA Pub. CP-2012, 1977. 2. A. Akritas. Elements of Computer Algebra with Applications. Wiley Interscience, New York, 1989. 3. E. Artin and O. Schreier. Algebraische Konstruktion reeller Körper. Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg, 5:83-115, 1926. 4. J. Ax. On Schanuel's conjectures. Annals of Mathematics, 93:252-268, 1971. 5. J. Baddoura. A conjecture on integration in finite terms with elementary functions and poly logarithms. In Proceedings of ISSAC'94, pages 158-162. ACM Press, 1994. 6. J. Baddoura. Integration in Finite Terms with Elementary Functions and Dilogarithms. PhD thesis, MIT, Mathematics, 1994. 7. T. Becker and V. Weispfenning. Gröbner bases: a computational approach to commutative algebra, volume 141 of Graduate Texts in Mathematics. SpringerVerlag, New York, 1993. 8. L. Bertrand. Calcul Symbolique des Integrales Hyperelliptiques. These de doctor at, Universite de Limoges, Mathematiques, 1995. 9. L. Bertrand. Computing a hyperelliptic integral using arithmetic in the jacobian of the curve. Applicable Algebra in Engineering, Communication and Computing, 6:275-298, 1995. 10. M. Bronstein. Simplification of real elementary functions. In G. Gönnet, editor, Proceedings of ISSAC'89, pages 207-211. ACM Press, 1989. 11. M. Bronstein. On the integration of elementary functions. Journal of Symbolic Computation, 9(2):117-173, February 1990. 12. M. Bronstein. The Transcendental Risch Diff"erential Equation. Journal of Symbolic Computation, 9(l):49-60, January 1990. 13. M. Bronstein. A unification of Liouvillian extensions. Applicable Algebra in Engineering, Communication and Computing, l(l):5-24, 1990. 14. M. Bronstein. The Risch differential equation on an algebraic curve. In S. Watt, editor. Proceedings of ISSAC'91, pages 241-246. ACM Press, 1991. 15. M. Bronstein and B. Salvy. Full partial fraction decomposition of rational functions. In Proceedings of ISSAC'93, pages 157-160. ACM Press, 1993.

318

References

16. W.S. Brown and J.F. Traub. On Euclid's algorithm and the theory of subresultants. Journal of the ACM, 18:505-514, 1971. 17. B.F. Caviness and M. Rothstein. A Liouville theorem on integration in finite terms for line integrals. Communications in Algebra, 3:781-795, 1975. 18. P.L. Chebyshev. Sur I'integration des differentielles qui contiennent une racine carree d'un polynome du troisieme ou du quatrieme degre. Journal de Mathematiques Pures et Appliquees (2^™-^ serie), 2:1-42, 1857. 19. P.L. Chebyshev. Sur I'integration de la differentielle {x + A)dx / \/x^ + ax^ + hx^ + ex +
References

319

37. J. Fitch. User-based integration software. In Proceedings of SYMSAC^Sl, pages 245-248. ACM Press, 1981. 38. K. Geddes and L. Stefanus. On the Risch-Norman integration method and its implementation in MAPLE. In Proceedings of ISSAC'89, pages 212-217. ACM Press, 1989. 39. K.O. Geddes, S.R. Czapor, and G. Labahn. Algorithms for Computer Algebra. Kluwer Academic PubHshers, Boston, 1992. 40. P. Gianni and B.M. Trager. Square-free algorithms in positive characteristic. Applicable Algebra in Engineering, Communication and Computing, 7:1-14, 1996. 41. G.H. Hardy. The Integration of Functions of a Single Variable. Cambridge University Press, Cambridge, England, 1916. 42. S.H. Harrington. A new symbolic integration system in reduce. The Computer Journal, 22(2):127-131, 1979. 43. E. Hermite. Sur I'integration des fractions rationelles. Nouvelles Annales de Mathematiques (2^"^^ serie), 11:145-148, 1872. 44. E. Horowitz. Algorithms for Symbolic Integration of Rational Functions. PhD thesis. University of Wisconsin, Madison, 1969. 45. E. Horowitz. Algorithms for partial fraction decomposition and rational function integration. In Proceedings of SYMSAM'71, pages 441-457. ACM Press, 1971. 46. D.J. Jeffrey. Integration to obtain expressions valid on domains of maximurn extent. In Proceedings of ISSAC'93, pages 34-41. ACM Press, 1993. 47. D.J. Jeffrey. Rectifying transformations for the integration of rational trigonometric functions. Journal of Symbolic Computation, 24(5):563-574, November 1997. 48. D.J. Jeffrey, G. Labahn, M. von Mohrenschildt, and A.D. Rich. Integration of the signum, piecewise and related functions. In Proceedings of ISSAC'97, pages 324-330. ACM Press, 1997. 49. D.J. Jeffrey and A.D. Rich. The evaluation of trigonometric integrals avoiding spurious discontinuities. ACM Transactions on Mathematical Software, 20:124135, 1994. 50. D.J. Jeffrey and A.D. Rich. Recursive integration of piecewise-continuous functions. In Proceedings of ISSAC'98, pages 290-294. ACM Press, 1998. 51. I. Kaplansky. An introduction to differential algebra. Hermann, Paris, 1957. 52. P. Knowles. Integration of a class of transcendental Liouvillian functions with error-functions, part I. Journal of Symbolic Computation, 13:525-543, 1992. 53. P. Knowles. Integration of a class of transcendental Liouvillian functions with error-functions, part II. Journal of Symbolic Computation, 16:227-241, 1993. 54. S. Lang. Algebra. Addison Wesley, Reading, Massachusetts, 1970. 55. P.S. Laplace. Theorie Analytique des Probabilites. V^ Courcier, Paris, 1820. 56. D. Lazard and R. Rioboo. Integration of rational functions: Rational computation of the logarithmic part. Journal of Symbolic Computation, 9:113-116, 1990. 57. J. Liouville. Premier memoire sur la determination des integrales dont la valeur est algebrique. Journal de VEcole Poly technique, 14:124-148, 1833. 58. J. Liouville. Second memoire sur la determination des integrales dont la valeur est algebrique. Journal de VEcole Poly technique, 14:149-193, 1833. 59. J. Liouville. Memoire sur I'integration d'une classe de fonctions transcendantes. Journal für die reine und angewandte Mathematik, 13:93-118, 1835.

320

References

60. R. Loos. Generalized polynomial remainder sequences. In B. Buchberger, G.E. Collins, and R. Loos, editors, Computer Algebra, Symholic and Algebraic Computation^ pages 115-137. Springer-Verlag, New York, 1982. 61. J. Luetzen. Joseph Liouville 1809-1882, volume 15 of Studies in the history of mathematics and physical sciences. Springer-Verlag, New York, 1990. 62. D. Mack. On rational integration. Research Report UCP-38, Dpt. of Computer Science, University of Utah, 1975. 63. J. Mafik. A note on integration of rational functions. Mathem^atica Bohemica, 116:405-411, 1991. 64. B. Mishra. Algorithmic Algebra. Texts and Monographs in Computer Science. Springer-Verlag, New York, 1993. 65. J. Moses. Symbolic integration: the stormy decade. Communications of the ACM, 14:548-560, 1971. 66. T. Mulders. A note on subresultants and a correction to the Lazard-Rioboo— Trager formula in rational function integration. Journal of Symbolic Computation, 24(l):45-50, 1997. 67. A.C. Norman and P.M.A. Moore. Implementing the new Risch integration algorithm. In A. Visconti, editor. Proceedings of the 4* International Colloquium on Advanced Computing Methods in Theoretical Physics, pages 99-110, 1977. 68. M.W. Ostrogradsky. De integration des fractions rationelles. Bulletin de la Classe Physico-Mathematiques de VAcademic Imperiale des Sciences de St. Petersbourg, IV: 145-167,286-300, 1845. 69. A. Ostrowski. Sur I'integrabilite elementaire de quelques classes d'expressions. Commentarii Mathematici Helvetici, 18:283-308, 1946. 70. J.C. Piquette. A method for symbolic evaluation of indefinite integrals containing special functions or their products. Journal of Symbolic Computation, 11:231-249, 1991. 71. N.V. Rao. A generahzation of Liouville's theorem on elementary functions. Preprint, University of Toledo, Ohio, 1994. 72. R. Rioboo. Quelques aspects du calcul exact avec des nombres reels. These de doctorat, Universite de Paris 6, Informatique, 1991. 73. R. Risch. On the integration of elementary functions which are built up using algebraic operations. Research Report SP-2801/002/00, System Development Corporation, Santa Monica, CA, USA, 1968. 74. R. Risch. Further results on elementary functions. Research Report RC-2402, IBM Research, Yorktown Heights, NY, USA, 1969. 75. R. Risch. The problem of integration in finite terms. Transactions of the American Mathematical Society, 139:167-189, 1969. 76. R. Risch. The solution of problem of integration in finite terms. Bulletin of the American Mathematical Society, 76:605-608, 1970. 77. R. Risch. Algebraic properties of the elementary functions of analysis. American Journal of Mathematics, 101:743-759, 1979. 78. J.F. Ritt. Integration in Finite Terms. Columbia University Press, New York, 1948. 79. M. Rosenlicht„ Liouville's theorem on functions with elementary integrals. Pacific Journal of Mathematics, 24:153-161, 1968. 80. M. Rosenlicht. Integration in finite terms. American Mathematical Monthly, 79:963-972, 1972. 81. M. Rosenlicht. On Liouville's theory of elementary functions. Pacific Journal of Mathematics, 65:485-492, 1976.

References

321

82. M. Rothstein. Aspects of Symbolic Integration and Simplification of Exponential and Primitive Functions. PhD thesis, University of Wisconsin, Madison, 1976. 83. M. Rothstein. A new algorithm for the integration of exponential and logarithmic functions. In Proceedings of the 1977 MACSYMA Users Conference, pages 263-274. NASA Pub. CP-2012, 1977. 84. M. Rothstein and B.F. Caviness. A structure theorem for exponential and primitive functions. SIAM Journal on Computing, 8:357-367, 1979. 85. M.F. Singer. Liouvillian first integrals of differential equations. Transactions of the American Mathematical Society, 333:673-687, 1992. 86. M.F. Singer, B.D. Saunders, and B.F. Caviness. An extension of Liouville's theorem on integration in finite terms. SIAM Journal on Computing, 14:965990, 1985. 87. J. Slagle. A Heuristic Program that Solves Symbolic Integration Problems in Freshman Calculus. PhD thesis. Harvard University, 1961. 88. R. Tobey. Algorithms for Antidifferentiation of Rational Functions. PhD thesis, Harvard University, 1967. 89. B.M. Trager. Algebraic factoring and rational function integration. In Proceedings of SYMSAC'76, pages 219-226, 1976. 90. B.M. Trager. Integration of simple radical extensions. In Proceedings of EUROSAM'79, volume 72 of Lecture Notes in Computer Science, pages 408-414, 1984. 91. B.M. Trager. On the integration of algebraic functions. PhD thesis, MIT, Computer Science, 1984. 92. B.L. van Der Waerden. Algebra. Frederic Ungar Publishing, New York, 1970. 93. J.-A. Weil. Constantes et polynömes de Darboux en algebre differentielle : application aux systemes differentiels lineaires. These de mathematiques, Ecole Poly technique, 1995. 94. M. Weileider. Integration elementarer Funktionen durch elementare Funktionen, Errorfunktionen und logarithmische Integrale. Doctoral thesis, LMU München, Mathematik, 1990. 95. D.Y.Y. Yun. On square-free decomposition algorithms. In Proceedings of SYMSAC'76, pages 26-35, 1976. 96. D.Y.Y. Yun. Fast algorithm for rational function integration. In Proceedings of IFIP'77, pages 493-498. North Holland, 1977. 97. R. Zippel. Effective Polynomial Computation. Kluwer Academic Publishers, Boston, 1993.

Index

Airy function Bessel function

175

integral 245 monomial 133 ExtendedEuclidean extension elementary real 285

173,174

CanonicalRepresentation 103 coefficient lifting 80,103,149 ConstantSystem 225 content 25,298 CoupledDECancelExp 262 CoupledDECancelPrim 260 CoupledDECancelTan 265

factorization irreducible 4 prime 4 split-squarefree 101,153 splitting 99,100,102,103,149,182, 184,218,246,251,300 squarefree 28, 39, 41, 42, 45, 51, 54, 71,101,138,152,182 F u l l P a r t i a l F r a c t i o n 58

deflation 26,28,43,251 derivation 75 induced 78 differential 270 extension 79 field 75 ring 75 elementary extension 133,282,293 function 133 integral 133 monomial 133 error function 245 Euclidean algorithm 10 extended 10 division 8 polynomial remainder sequence Euclidean 10 exponential extension 133

11-14

Gauss' Lemma 26,33 Grobner basis 53 HalfExtendedEuclidean 11,13 Hermite reduction 40, 41,43, 71,138, 175 HermiteReduce 40, 41, 44,139 Hubert's Nullstellensatz 8, 90 HorowitzOstrogradsky 46 hyp er exponential extension 129, 293 hypertangent extension 164

22

indicial equation 126 IntegrateHyperexponential 163 IntegrateHyperexponentialPolynomial 162 IntegrateHypertangent 172

324

Inde

IntegrateHypertangentPolynomial 167 IntegrateHypertangentReduced 169 IntegrateNonLinearNoSpecial 173 IntegratePrimitive 160 IntegratePrimitivePolynomial 158 IntegrateRationalFunction 52,59 IntegrateRealRationalFunction 70 IntRationalLogPart 48,51, 54 Lambert W function 309 LaurentSeries 57 Icm 33 LimitedlntegrateReduce 248 LinearConstraints 223 Lionville's t h e o r e m 142, 144, 145,155 Liouvillian extension 129 log-explicit 293 monomial 131 logarithmic derivative identity 76,104, 270 of a radical 97 extension 133 integral 152,154,158,245 133 monomial LogToAtan 63 LogToReal 69 m o n o m i a l 91 arctangent 285 elementary 133 real 285 exponential 133, 282, 285 hyperexponential 129,160,164,166, 1 7 6 , 1 7 7 , 1 8 8 , 200, 212, 220, 229, 250,258,261 hypertangent 164,177,178,191, 214,221,244,258,264 Liouvillian 131 logarithmic 133, 282, 285 nonlinear 9 1 , 9 5 , 1 2 4 , 1 4 0 , 1 4 1 , 1 5 6 , 172,201,214,229,242,263 primitive 129,157,176,177,188, 194,211,220,227,245,259 nonsimple 293 tangent 164,285

N e w t o n - L e i b n i z - B e r n o u l l i algorithm 36,58 n o r m a l polynomial 92 Parallellntegrate 309 ParametricLogarithmicDerivative 253 ParamPolyRischDENoCancell 234 ParainPolyRischDENoCancel2 238 ParamRdeBoundDegreeBase 228 ParamRdeBoundDegreeExp 229 ParamRdeBoundDegreeNonLinear 230 ParamRdeBoundDegreePrim 228 ParamRdeNormalDenominator 219 ParamRdeSpecialDenomExp 221 ParamRdeSpecialDenomTan 222 ParSPDE 231 p a r t i a l fraction decomposition 15, 17, 36,39,47,54 PartialFraction 15, 17 polar multiplicity 138 PolyDivide 8 polynomial remainder sequence 22 E u c h d e a n 22 fundamental theorem 23 primitive 22 subresultant 23, 49-51,152 PolynomialReduce 141 PolyPseudoDivide 9 PolyRischDECancelExp 213 PolyRischDECancelPrim 212 PolyRischDECancelTan 215 PolyRischDENoCancell 208 PolyRischDENoCancel2 209 PolyRischDENoCancelS 210 primitive extension 129,245 monomial 129,157,176,177,188, 194,211,220,227,245,259 nonsimple 293 part 25,298 polynomial 25,298 polynomial remainder sequence 22 quotient

field

7, 33, 4 3 , 1 0 7 , 1 1 6 , 1 1 7

RdeBoundDegreeBase 199 RdeBoundDegreeExp 200 RdeBoundDegreeNonLinear

201

Index RdeBoundDegreePrim 198 RdeNormalDenominator 185 RdeSpecialDenomExp 190 RdeSpecialDenomTan 192 real

field 65 integration 59,163, 177, 178,191, 214,221,244,264 structure theorem 288 reduced function 103 Remainder 115 residue 118,147 ResidueReduce 151,153 Risch difi^erential equation 181 parametric 217 Risch structure theorem 283 real 288 Rosenlicht's theorem 280 Rothstein-Caviness theorem 293 Rothstein-Trager resultant 47, 67, 72, 124,149,176,177 similarity 22,50 simple function 103 SPDE 203

325

special polynomial 92 of the first kind 98 SplitFactor 100,300 SplitSquarefreeFactor 102,301 Squarefree 29, 32 squarefree factorization 28, 39, 41, 42, 45, 51, 54,71,101,138,152,182 polynomial 28 SubResultant 24 Sylvester matrix 18, 20 tangent extension monomial

164 164

value map at a point 112 at infinity 116 ValueAtInfinity 118 weak normalization 181 WeakNormalizer 183 Wronskian 71,88