Formal analysis is the study of formal power series, formal Laurent series, formal root series, and other formal series
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Table of contents :
Preface
Contents
1 Basics of formal power series
2 Calculus of formal power series
3 Applications of formal power series, I
4 Applications of formal power series, II
5 General composition
6 Formal analysis and classical analysis
7 Formal Laurent series
8 Iteration and iterative roots
9 Formal series and general exponents
Bibliography
Index
XiaoXiong Gan Formal Analysis
De Gruyter Studies in Mathematics

Edited by Carsten Carstensen, Berlin, Germany Gavril Farkas, Berlin, Germany Nicola Fusco, Napoli, Italy Fritz Gesztesy, Waco, Texas, USA Niels Jacob, Swansea, United Kingdom Zenghu Li, Beijing, China KarlHermann Neeb, Erlangen, Germany
Volume 80
XiaoXiong Gan
Formal Analysis 
An Introduction
Mathematics Subject Classification 2010 Primary: 13F25, 16W60, 39B12; Secondary: 26E35, 30D05 Author Prof. Dr. XiaoXiong Gan Morgan State University Department of Mathematics 1700 E. Cold Spring Lane Baltimore MD 21251 USA [email protected]
ISBN 9783110597851 eISBN (PDF) 9783110599459 eISBN (EPUB) 9783110598605 ISSN 01790986 Library of Congress Control Number: 2021933954 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2021 Walter de Gruyter GmbH, Berlin/Boston Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com

Dedicated to my mother, Yuemei Xiang, and my father, Yi Gan
Preface At the end of 2017, after my first monograph was published by the Juliusz Schauder Center for Nonlinear Studies, the director of the center, Wojciech Kryszewski, encouraged me to write a “comprehensive book for formal analysis,” the first such book for this special area of mathematics. I accepted the challenge. The earliest power series is believed to be the binomial series invented by young Isaac Newton in 1669. Formal power series attracted mathematicians as early as 1871, when Ernst Schröder introduced them by investigating iterated functions. During this period (1850), Victor Puiseux introduced the socalled Puiseux series, which are generalization of power series that allow for negative and fractional exponents of the indeterminate. The space of formal power series, 𝕏, endowed with linear operation, Cauchy product, and composition, was systematically introduced by Ivan Niven in 1969. This volume presents 𝕏 as a mathematical system. A regular power series has a strong relationship with its coefficients for the variable inside the convergence region if it has a positive radius of convergence. A formal power series is totally determined by its coefficients, if it is denoted by a form of power series, although its variable does not take any value. As sequences, formal power series could be useful for the computer sciences and today’s digital world. We know that ℓp is a popular space of sequences with linear operation and dot product. The sets 𝕏 and 𝕃 are the spaces of formal power series and formal Laurent series, respectively, and they are the spaces of sequences with linear operation and dot product, plus Cauchy product and composition. The formal power series space 𝕏 has more operations and applications than the space ℓp . Mathematicians distinguish between formal and classical analysis. Formal analysis has two characters: formality and analyticity. Formal theory usually uses algebraic approaches to tackle mathematical objects without using the limit or convergence; formal analysis keeps this tradition and, in addition, it does not reject certain kinds of limit or convergence. This volume lays forth the principle of formal analysis that a formal power series f over ℝ or ℂ should maintain all properties when f has a positive radius of convergence. In order to fulfill this principle, we generalize several existing formal series such as formal logarithms. We also introduce some developments of classical analysis, such as the Lagrange inversion problem and boundary convergence problem. The power of formal power series with real exponent is another development of formal analysis that improves the completeness of the space 𝕏. In response to my book proposal, I received a 7page, handwritten review that proved indispensable to me in the development of this work. I was deeply moved by the reviewer’s insights and suggestions, which were as valuable to me as the guidance of my Ph. D. advisor, Karl Stromberg, almost 30 years ago. I am grateful for this reviewer’s recommendations, which drove the research and writing of this book. https://doi.org/10.1515/9783110599459201
VIII  Preface This volume contains the evolution of our work in formal analysis since 2000 and collects the comprehensive works related to formal power series since 1871. Chapter 1 presents the basic algebraic structures of formal power series, including addition, multiplication (Cauchy product), and composition with nonunit formal power series. Here, we also introduce the matrix representations of a formal power series: one matrix representation for the Cauchy product and one matrix representation for the composition of almost units. Formal differentiation has nothing to do with the limit of difference quotient, but it is a very useful tool in studying formal analysis. Chapter 2 introduces many applications of formal differentiation and certain metrics on 𝕏. Formal differentiation returns in late chapters. The Riordan group or Riordan array, a hot topic in recent research, is introduced in Chapter 3. This chapter also collects some wellknown classical theorems that were reproved using formal power series. Chapter 4 features several other applications of formal power series, including real analysis, functional analysis, differential equations, and numerical analysis. The Lagrange inversion problem and other functional analysis problems are introduced in this chapter. Chapter 5 transforms the studies of formal power series into formal analysis, using the general composition theorem. This theorem not only provides a necessary and sufficient condition for composition but also releases the composition of formal power series from the environment of socalled admitting addition. This chapter uses our recently published analytical approach to prove this theorem. The matrix representation for the general composition is also our construction. Chapter 6 further discusses the relationship between classical analysis and formal analysis. First, we tackle a problem that has mystified mathematicians for more than a century: the boundary convergence behavior of power series. A formal analytic point is introduced. We prove that if a power series converges at a boundary point of the convergent region, and this point is formal analytic, then this power series converges at all boundary points. The special Banach spaces ℋp (β) and ℒp (β) are introduced in this chapter. The general formal logarithm Ln works on all formal power series in 𝕏+ (ℝ) with positive constant term, similar to the logarithmic function ln(x) working on all positive real numbers in calculus. This generalization produces many interesting properties for Ln. Several kinds of formal Laurent series existed before the space 𝕃 of formal Laurent series was systematically invented in 2012 by Dariusz Bugajewski and the author. The semiformal Laurent series space ℒs and semireversed formal Laurent series space ℒr are introduced in Chapter 7. We construct ℒs using all results provided in the previous chapters although it is known that ℒs can be generated by the quotients of formal power series in certain sense. We propose a canonical mapping that connects formal Laurent series with the wellknown Lebesgue measurable functions and the Lebesgue integral. The linkage of these two concepts sheds light on the rather complicated multiplication of formal Laurent series and facilitates further investigation of 𝕃. The com
Preface
 IX
position of formal Laurent series remains a challenge for mathematicians. This chapter introduces a composition of formal Laurent series with formal power series. Iteration and iterative roots are closely related to functional equations, and the studies on this subject have a long history. The space of almost unit formal power series forms a group under composition, which yields many interesting results about iteration in this field. Those developments are presented in Chapter 8. The last chapter introduces the power of formal power series with all real exponents, from the root series g 1/n to rational exponents g r , r ∈ ℚ, and finally to the real exponents g s , s ∈ ℝ, where g is a formal power series on ℝ with positive constant term. The formal exponent series is a new subject for the theory of formal power series. In this chapter, we introduce the exponent algorithm for computing formal real exponents, which is a convenient mathematical tool in analysis and in combinatorics. Higherdimensional formal power series warrant further study. But they are outside the scope of this volume and the author’s expertise. As the first comprehensive book on formal analysis, this volume presents to the mathematical world the most complete system of formal power series and formal Laurent series yet. I would like to express my heartfelt thanks to Wojciech Kryszewski, without whom this book would not exist. His mathematical philosophy and vision inspired me to do this work. I would also like to thank my friend, Dariusz Bugajewski. He gave me advice and encouragement throughout this process, and coauthored several papers on formal series with me. Whenever I discuss formal analysis with him and his team at Adam Mickiewicz University in Poznań, Poland, I refresh my mathematical skills and see new mathematical structures. Finally, I extend my gratitude to De Gruyter for giving me this opportunity to contribute to the mathematical world.
Contents Preface  VII 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7
Basics of formal power series  1 Basic algebraic operations  2 Composition of formal power series with nonunit  7 Right distributive law for composition  13 The matrix representation of formal power series  17 Almost unit formal power series  22 Algebraic structure of 𝕏  35 Analytic structure of 𝕏  38
2 2.1 2.2 2.3 2.4 2.5
Calculus of formal power series  49 Formal derivatives of formal power series  49 Formal differential equations and their applications  54 Ultrametric  65 Some topological structure for 𝕏  69 The umbral calculus and formal power series  78
3 3.1 3.2 3.3 3.4
Applications of formal power series, I  85 Riordan group and formal power series  85 Riordan involutions  92 Some subgroups of the Riordan group  93 Cayley–Hamilton theorem and Fermat’s little theorem  96
4 4.1 4.2 4.3 4.4
Applications of formal power series, II  101 Formal power series and some differential equations  101 Functional equations and formal power series  105 Lagrange inversion  115 Difference equations and formal power series  123
5 5.1 5.2 5.3 5.4 5.5 5.6 5.7
General composition  129 Introduction  129 Matrix representation of the general composition, I  132 Coefficients of f n (z)  135 The general composition theorem  141 The general chain rule  150 The general right distributive law  154 Matrix representation of the general composition, II  156
XII  Contents 6 6.1 6.2 6.3 6.4
Formal analysis and classical analysis  161 Introduction to the boundary behavior  161 Boundary behavior and formal analysis  166 Banach spaces ℋp (β) and ℒp (β)  173 General formal logarithm  179
7 7.1 7.2 7.3 7.4 7.5 7.6 7.7
Formal Laurent series  189 Semiformal Laurent series  190 Reversed semiformal Laurent series  199 Basic algebra of formal Laurent series  206 Composition of formal Laurent series and formal power series  217 Canonical mapping and dot product of formal Laurent series  224 Canonical mapping and multiplication of formal Laurent series  226 Topological spaces of formal Laurent series  233
8 8.1 8.2 8.3 8.4
Iteration and iterative roots  245 Conjugacy of formal power series in 𝔻  245 The iteration of formal power series  264 The iterative roots of formal series in 𝔻  269 Schröder’s problem  273
9 9.1 9.2 9.3 9.4
Formal series and general exponents  285 Introduction  286 Formal root series and its algorithm  287 Power of formal power series with rational exponents  298 The real exponents of formal power series  304
Bibliography  315 Index  319
1 Basics of formal power series In 1669, Issac Newton, at 26 years old, wrote that for every real number s, the binomial series 1 + sx +
s(s − 1) 2 s(s − 1) ⋅ ⋅ ⋅ (s − n + 1) n x + ⋅⋅⋅ + x + ⋅⋅⋅ 2! n!
represents the binomial function (1 + x)s for all real numbers, −1 < x < 1 [59]. JosephLouis Lagrange also considered power series in 1797 in his Théorie des fonctions analytiques. The convergence criteria and the connection between power series and the analytic functions was first proved by AugustinLouis Cauchy in 1821 [9]. Brook Taylor and Colin MacLaurin developed the representation theorem involving power series. Without a doubt, the most important part of studies of power series had been the convergence of the series for a certain period. Should we ignore those sequences or series, number sequences or series, and function sequences or series, which are not convergent? The answer is no. Actually Cauchy himself found that the Cauchy product of two convergent series may be divergent. He provided the following example. Example 1.0.1 (p. 72, [95]). Take a0 = b0 = 0,
an = bn =
(−1)n−1 √n
for
n ∈ ℕ.
∞ ∞ Then ∑∞ n=0 an = ∑n=0 bn converges, but ∑n=0 cn diverges, where n
cn = ∑ ak bn−k k=0
for
n = 0, 1, 2, . . . .
Other mathematicians such as Niels Henrik Abel and Ernesto Cesáro in the 19th century showed that divergent series should not be discarded as meaningless or useless (p. 474, [95]). In addition to divergence and convergence, if we bring more operations such as compositions to the sequences or series, we have more reasons to expand our ordinary or conventional studies and theory to a wider range, which is usually called the formal subjects or formal studies. The earliest mention of iterated functions, not yet the formal power series but close to it, could be the paper of 1871 by Ernst Schröder in which the rfold iterated function F r (z) was defined by F 1 (z) = F(z),
F r (z) = F r−1 (F(z)),
which are assumed valid for all value of z discussed [91]. https://doi.org/10.1515/9783110599459001
r ∈ ℕ,
2  1 Basics of formal power series The formal power series is usually written in the form a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ + an x n + ⋅ ⋅ ⋅ which looks like an ordinary power series we mentioned above. That is one reason it is called a power series, but it should not be considered a special kind of the ordinary power series. A significant difference between formal and ordinary power series is that no value is assigned to the symbol x of a formal power series and, therefore, no convergence related with x is concerned there. Ivan Niven systematically set up the space of formal power series in 1969 and at the same time indicated that such formal power series had been investigated by mathematicians before his article [61]. The space of formal power series on a ring S is also considered a set of sequences such as (a0 , a1 , a2 , . . . an , . . . ),
an ∈ S,
n = 0, 1, 2, . . . .
endowed with the linear operation, the Cauchy product, and more interestingly, the composition. These operations will be introduced in Section 1.1 and Section 1.2. We will see that there is no confusion whether the formal power series is considered as the power series or as the sequences, or whether formal power series have a positive radius of convergence or diverge everywhere except at a point (center). Research into formal power series has been getting deeper since 1960. Formal power series are now applied to a wider range of fields. Many fundamental results can be found in the book Applied and Computational Complex Analysis, Volume 1, by Peter Henrici [42], as well as in the books [25, 56, 86].
1.1 Basic algebraic operations A sequence in a set S is considered as a list of members in S in a definite order: s1 , s2 , s3 , s4 , . . . , sn , . . . . The mathematical definition of a sequence in S is a mapping f : ℕ → S such that f (n) = sn for all n ∈ ℕ, where ℕ is the set of all positive integers. Let W be the set of all sequences in a set S, the operations defined in W play important role. For example, if S = ℂ, ℓp is one of such W for 0 < p ≤ ∞, the basic operations defined in ℓp are scalar multiplication and addition, which make ℓp a linear space over ℂ. The other important operation on ℓp is the summation that is defined by n
∞
∑ ak = lim ∑ ak
k=0
n→∞
k=1
1.1 Basic algebraic operations  3
∞ p p and ∑∞ k=0 ak  for any a = (ak ) ∈ ℓ . The form ∑k=0 ak is called a series of complex terms or a series over ℂ. A formal power series could also be considered as a sequence in a set S.
Definition 1.1.1. Let S be a ring and let l ∈ ℕ be given. A formal power series on S is defined to be a mapping from ℕl to S. We denote the set of all such mappings by 𝕏(S), or simply 𝕏. A very brief introduction to the formal power series from N l , l ≥ 2, to S is located in Section 1.6. We always consider the case l = 1 in this book unless we specify otherwise. A formal power series defined in Definition 1.1.1 is a sequence over the ring S. The main reason to call it a formal power series is that the operations endowed to the set 𝕏(S) has a strong relationship with the ordinary power series. We usually write a formal power series f in x by f (x) = f0 + f1 x + ⋅ ⋅ ⋅ + fn xn + ⋅ ⋅ ⋅ ,
or
f = (f0 , f1 , f2 , . . .),
n ∞ or f (x) = ∑∞ n=0 fn x , where {fj }j=0 ⊆ S. In this case, fk is called the kth coefficient of f , k ∈ ℕ ∪ {0}. Two formal power series are equal if and only if all corresponding coefficients of these two formal power series are equal. f is called the zero formal power series if fj = 0 for every j ∈ ℕ ∪ {0}. If f0 = 0, f is called a nonunit, otherwise f is called a unit. We also denote
f (c) = f0 + f1 c + f2 c2 + ⋅ ⋅ ⋅ + fn cn + ⋅ ⋅ ⋅ , if f (c) ∈ S in some sense, although we usually do not assign any value to c. In particular, we denote f0 = f (0). The order of a formal power series f (x) is defined to be the number ord(f ) = min{n  fn ≠ 0, n = 0, 1, 2, 3, . . .}, if f is not the zero formal power series. We define the order of the zero formal power series to be infinity or we say that its order is greater than any positive integer. We denote the set of all nonunits of 𝕏 to be 𝕏0 , that is, 𝕏0 = {f ∈ 𝕏 : 0 < ord(f ) ≤ ∞} or
𝕏0 = {f ∈ 𝕏 : f (0) = 0}.
Definition 1.1.2. Let S be a ring. Let f and g be formal power series in 𝕏(S) such that ∞ n n f (x) = ∑∞ n=0 fn x and g(x) = ∑n=0 gn x , and let r ∈ S. Then g + f , rf , and gf are formal power series in 𝕏(S) and ∞
(g + f )(x) = g(x) + f (x) = ∑ (gn + fn ) x n , n=0
4  1 Basics of formal power series ∞
(rf )(x) = rf (x) = ∑ (rfn ) x n , n=0
and
∞
(gf )(x) = g(x)f (x) = ∑ cn x n , n=0
where n
cn = ∑ gj fn−j , j=0
n = 0, 1, 2, . . . .
It is clear that the sum, the scalar multiplication, and the multiplication are welldefined, that is, g + f , rf , and gf are all in 𝕏. The multiplication gf is also called the Cauchy product of the formal power series f and g, which provides a significant difference between the space 𝕏(S) and the space ℓp when a formal power series is considered as a sequence. If S is commutative, then gf = fg. Equipped with the operations defined in Definition 1.1.2, 𝕏 is a linear vector space with the addition and scalar multiplication as well as 𝕏 is an algebra with the Cauchy product. Definition 1.1.3. Let f ∈ 𝕏(S) be a formal power series over a ring S such that f (x) = f0 + f1 x + f2 x 2 + ⋅ ⋅ ⋅ . The formal power series f is called a constant formal power series if ∞
f (x) = ∑ fn xn , n=0
fn = 0
for all n ∈ ℕ,
or f = (f0 , 0, 0, . . .), where f0 is any element of S. It is clear that the zero formal power series is a constant formal power series such that f0 = 0, or 0 = (0, 0, 0, . . . , 0, . . .). The formal power series I = (1, 0, 0, . . . , 0, . . .) is called the identity formal power series or just identity of 𝕏(S) with respect to the multiplication. We usually write it as ∞
I(x) = ∑ fn xn , n=0
where f0 = 1, fk = 0 for every k ∈ ℕ.
A polynomial over a ring S is considered a formal power series over S. We use the degree of polynomial, deg(f ), to denote the degree of a formal power series f if it is a polynomial, otherwise the degree of a formal power series is infinity. Sometimes people consider that the constant formal power series is a polynomial (formal power series) with deg(f ) = 0. We now can also say that two formal power series in 𝕏(S) are equal if and only if f − g = 0. By Definitions 1.1.2 and 1.1.3, we provide the following proposition without proof.
1.1 Basic algebraic operations  5
Proposition 1.1.4. Let S be a ring. The distributive law holds in 𝕏(S). Or h(f + g) = hf + hg for all f , g, and h in 𝕏(S). We provide another obvious result below. Proposition 1.1.5. Let S be a ring. For any two formal power series f and g in 𝕏(S), we have: (i) ord(f + g) ≥ min { ord(f ), ord(g)}, and (ii) ord(fg) = ord(f ) + ord(g). The following example may help us to understand the formal power series and the operations defined on 𝕏. Example 1.1.6. Let S = ℝ. Suppose that f (z) = 1 + z + z 2 + ⋅ ⋅ ⋅ + z n + ⋅ ⋅ ⋅ , zn z z2 z3 + + + ⋅⋅⋅ + + ⋅⋅⋅, 1! 2! 3! n! h(z) = 1!z + 2!z 2 + 3!z 3 + ⋅ ⋅ ⋅ + n!z n + ⋅ ⋅ ⋅ .
g(z) = 1 +
Then (f + g)(z) = 2 + 2z + (1 +
1 2 1 )z + ⋅ ⋅ ⋅ + (1 + )z n + ⋅ ⋅ ⋅ , 2! n! n
(fh)(z) = z + (2! + 1!)z 2 + ⋅ ⋅ ⋅ + ( ∑ k!)z n + ⋅ ⋅ ⋅ , k=1
ord(fh) = ord(f ) + ord(h) = 0 + 1 = 1. If S is a field, every nonzero element of S has inverse in S. What is 𝕏(S) if S is a field? Definition 1.1.7. Let S be a field, the inverse (or reciprocal) of a formal power series f in 𝕏 = 𝕏(ℂ) is a formal power series, denoted by f −1 , such that ff −1 = I if such an f −1 exists. f −1 is called the inverse of the formal power series f under the multiplication. We also denote f −1 as 1/f . Since a field is a commutative division ring, it follows that f −1 f = ff −1 if f −1 exists. n −1 Theorem 1.1.8. Let f (z) = ∑∞ exists if n=0 fn z be a formal power series in 𝕏(ℂ). Then f and only if f0 ≠ 0. In this case, we write
f −1 (z) = g(z) = g0 + g1 z + g2 z 2 + ⋅ ⋅ ⋅ ,
6  1 Basics of formal power series where g0 =
1 , f0
gn =
−1 n−1 ( ∑ g f ), f0 j=0 j n−j
for every
n ∈ ℕ.
Moreover, the inverse f −1 is unique if it exists. n Proof. Suppose that f (z) = ∑∞ n=0 fn z and f (0) = f0 ≠ 0. We shall construct a formal power series ∞
g(z) = ∑ gn z n , n=0
such that fg = I. Then we write g = f −1 . n Suppose that (fg)(z) = ∑∞ n=0 cn z = I(z). By Definition 1.1.2, n
cn = ∑ gj fn−j , j=0
n = 0, 1, 2, . . . .
Then c0 = 1 yields that f0 g0 = 1, which provides g0 = 1/f0 . c1 = 0 yields that f0 g1 + f1 g0 = 0, and then g1 = −(f1 g0 )/f0 = −f1 /(f02 ). Suppose that g0 , . . . , gn have been computed for some n ∈ ℕ. By n+1
cn+1 = ∑ gj fn+1−j = 0, j=0
we have n
f0 gn+1 = −( ∑ gj fn+1−j ). j=0
Since f0 ≠ 0, it follows that gn+1 ∈ ℂ and gn+1 =
−1 n (∑ g f ). f0 j=0 j n+1−j
The construction is completed. Conversely, suppose that f0 = 0. It is clear that ord(fg) ≥ 1 and hence fg ≠ I. That the inverse is unique comes easily from the formula above and the fact that the product is commutative.
1.2 Composition of formal power series with nonunit  7
Example 1.1.9. Find the inverse of the formal power series: (i) f (z) = 1 + z + z 2 + ⋅ ⋅ ⋅, and (ii) g(z) = 1 + z + 2!z 2 + 3!z 3 + ⋅ ⋅ ⋅ in 𝕏(ℂ). Solution. By Proposition 1.1.4, (1 − z)f (z) = f (z) − zf (z) = (1 + z + z 2 + ⋅ ⋅ ⋅) − z(1 + z + z 2 + ⋅ ⋅ ⋅) = I, then f −1 (z) = 1 − z. We have (i). By Theorem 1.1.8, g −1 exists in 𝕏(ℂ). Denote ∞
g −1 (z) = ∑ bn z n . n=0
Since g −1 g = I, we have b0 = 1 immediately, and also n
∑ bk (n − k)! = 0
k=0
for all n ∈ ℕ.
Then b1 = −b0 = −1 and n−1
bn = −( ∑ bk (n − k)!), k=0
n ∈ ℕ.
Then n−1
g −1 (z) = 1 − z − z 2 − 3z 3 + ⋅ ⋅ ⋅ + ( ∑ bk (n − k)!)z n + ⋅ ⋅ ⋅ . k=0
We know that a commutative ring S is an integral domain if a ⋅ b = 0 in S implies that a = 0 or b = 0. Proposition 1.1.10 ([42]). If S is a field, then the space 𝕏(S) is an integral domain.
1.2 Composition of formal power series with nonunit In addition to the Cauchy product endowed in 𝕏(S), the composition, or another multiplication, is an important operation defined on 𝕏(S), which distinguishes 𝕏(S) from other spaces of sequences significantly. We will see that the composition brings us a lot of interesting results and many applications in the later chapters.
8  1 Basics of formal power series Definition 1.2.1. Let f (z) = f0 + f1 z + f2 z 2 + f3 z 3 + ⋅ ⋅ ⋅ be a formal power series in 𝕏(S). For any n ∈ ℕ, f n = f n−1 ⋅ f is called the nth power of the formal power series f and we write f n (z) = f0(n) + f1(n) z + f2(n) z 2 + ⋅ ⋅ ⋅ + fk(n) z k + ⋅ ⋅ ⋅ ,
(1.1)
where fk(n) is called the kth coefficient of f n . If n = 0, we define that f0(0) = 1, fk(0) = 0 if k ∈ ℕ, or f 0 = I. Let f (z) = f1 z + f2 z 2 + f3 z 3 + ⋅ ⋅ ⋅ be a nonunit. For any k ∈ ℕ, it is obvious that ord(f k ) ≥ k, and hence fn(k) = 0 if n < k, then ∞
f k (z) = ∑ fn(k) z n . n=k
Thus, ∞
n
k=0
k=0
∑ gk fn(k) = ∑ gk fn(k) ∈ S,
n = 0, 1, 2, . . .
for any formal power series g(z) = g0 + g1 z + g2 z 2 + ⋅ ⋅ ⋅. Now, we can define the composition with a nonunit. Definition 1.2.2. Let S be a ring. Let g(z) = g0 + g1 z + g2 z 2 + ⋅ ⋅ ⋅ be any formal power series in 𝕏(S) with deg(g) > 0. Let f (z) = f1 z + f2 z 2 + f3 z 3 + ⋅ ⋅ ⋅ be a nonunit. We define the composition of formal power series g ∘ f , or equivalently g(f ), to be the formal power series ∞
m
∞
g ∘ f (z) = g(f (z)) = ∑ gm (f (z)) = ∑ cn z n , n=0
m=0
where ∞
n
k=0
k=0
cn = ∑ gk fn(k) = ∑ gk fn(k) ∈ S,
n = 0, 1, 2, . . . .
(1.2)
Formula (1.2) ensures socalled admitting addition for the composition g ∘ f . The n formula ∑∞ n=0 gn (f (z)) is also called the admissible sum [61]. We will investigate such admissible sum further in Section 1.7. ∞ n n Theorem 1.2.3. Let f (z) = ∑∞ n=0 fn z , g(z) = ∑n=0 gn z be formal power series over a n 0 ring S and let h(z) = ∑∞ n=1 hn z be a nonunit in 𝕏 (S). Then
(f + g) ∘ h = f ∘ h + g ∘ h.
1.2 Composition of formal power series with nonunit  9
Proof. Since h ∈ 𝕏0 (S), f ∘ h, g ∘ h, and (f + g) ∘ h are elements of 𝕏(S). We show the equality. Write ∞
f ∘ h(z) = ∑ cn z n , n=0 ∞
∞
g ∘ h(z) = ∑ dn z n ,
and
n=0
(f + g) ∘ h(z) = ∑ rn z n . n=0
By (1.2), for n = 0, 1, 2, . . . , n
n
cn = ∑ fk h(k) n ,
n
dn = ∑ gk h(k) n ,
k=0
rk = ∑ (fk + gk )h(k) n .
k=0
k=0
Then cn + dn = rn , n = 0, 1, 2, . . . , which yields (f + g) ∘ h = f ∘ h + g ∘ h. We have defined the equality of two formal power series in Section 1.1. The next theorem is another statement for such equality. The proof is trivial. Theorem 1.2.4. Let f and g be formal power series in 𝕏(S). Then f =g
ord(f − g) ≥ n
⇔
for all
n ∈ ℕ ∪ {0}.
Theorem 1.2.5. Let g be any formal power series in 𝕏(S) and let f be a nonunit. Then ord(g ∘ f ) = ord(g) ⋅ ord(f ). Proof. Since f is a nonunit, it follows that ord(f ) = m ∈ ℕ. It is obvious that the conclusion is true if ord(g) = ∞, or g is the zero formal power series. Suppose that ord(g) = n > 0. Write f (z) = am z m + am+1 z m+1 + ⋅ ⋅ ⋅ ,
g(z) = bn z n + bn+1 z n+1 + ⋅ ⋅ ⋅ ,
where am ≠ 0, bn ≠ 0. We have f (z) = z m (am + am+1 z + am+2 z 2 + ⋅ ⋅ ⋅) = z m ϕ(z), where ϕ(0) = am ≠ 0. Then n
n+1
g ∘ f (z) = bn (f (z)) + bn+1 (f (z)) = bn z
mn
n
(ϕ(z)) + bn+1 z
+ ⋅⋅⋅
m(n+1)
n+1
(ϕ(z))
+ ⋅⋅⋅.
Thus, ord(g ∘ f ) = nm = ord(g) ⋅ ord(f ) by the definition. Finally, suppose that ord(g) = 0. Then g ∘ f (0) = g(0) ≠ 0 and then ord(g ∘ f ) = 0. Then ord(g ∘ f ) = 0 = 0 ⋅ m = ord(g) ⋅ ord(f ).
10  1 Basics of formal power series Definition 1.2.6. Let g(z) = g0 + g1 z + g2 z 2 + ⋅ ⋅ ⋅ be any formal power series in 𝕏(S). The nth partial sum of the formal power series g is the formal polynomial [g]n = g0 + g1 z + g2 z 2 + ⋅ ⋅ ⋅ + gn z n . The following theorem is obvious. Theorem 1.2.7. Two formal power series f and g in 𝕏(S) are equal if and only if [f ]n = [g]n ,
n = 0, 1, 2, 3, . . . .
Corollary 1.2.8. Let f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ be a formal power series in 𝕏(S) where S is a ring. Write (n) (n) 2 (n) k f n (z) = a(n) 0 + a1 z + a2 z + ⋅ ⋅ ⋅ + ak z + ⋅ ⋅ ⋅ ,
for every n ∈ ℕ. Then a(n) is determined only by [f ]k for every nonnegative integer k. In k particular, if ord(f ) = m ∈ ℕ, a(n) =0 k
for all
k < nm.
Proof. It suffices to show that any am , m > k does not contribute to the coefficient a(n) . k m k n It is equivalent to say that the term am z does not contribute to the term of z of f . It is true for n = 2, because a(2) = a0 ak + a1 ak−1 + ⋅ ⋅ ⋅ + ak a0 , k does not contain am , m > k in its expression. Assume the conclusion is true for a(n−1) , then f n = f n−1 f and k a(n) = a(n−1) ak + a(n−1) a1 + ⋅ ⋅ ⋅ + a(n−1) a0 , 1 0 k k does not contain am , m > k in its expression. Thus, a(n) is determined by [f ]k for every n ∈ ℕ. k Finally, suppose that ord(f ) = m ∈ ℕ. Then am ≠ 0 and aj = 0 for 0 ≤ j ≤ m − 1. Then n
f n (z) = (am z m + am+1 z m+1 + ⋅ ⋅ ⋅)
= z nm (am + am+1 z + am+2 z 2 + ⋅ ⋅ ⋅).
Then a(n) = 0 if k < nm. k Corollary 1.2.9. Let g ∈ 𝕏 = 𝕏(S) be given where S is a ring. Let f ∈ 𝕏0 be a nonunit. Then [g ∘ f ]n is determined by [g]n and [f ]n for every n ∈ ℕ.
1.2 Composition of formal power series with nonunit  11
Proof. Write g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ ,
f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ ,
where a0 = 0, and write g ∘ f (z) = c0 + c1 z + c2 z 2 + ⋅ ⋅ ⋅ . (k) Then cn = ∑nk=0 bk a(k) n , n = 0, 1, 2, . . . . By Corollary 1.2.8, aj is determined by [f ]j , 0 ≤ j ≤ n. Then cn is determined by [g]n and [f ]n for every n ∈ ℕ. We complete the proof.
Theorem 1.2.10. Let g ∈ 𝕏(S) and let f and h be nonunits in 𝕏0 (S) where S is a ring. Then g ∘ (f ∘ h) = (g ∘ f ) ∘ h. That is, the composition is associative if the all composed formal power series are nonunits. Proof. Since f , g and h are nonunits, it follows that f ∘ h is a nonunit and, therefore both g ∘ (f ∘ h) and (g ∘ f ) ∘ h exist. Write ∞
[g ∘ (f ∘ h)](z) = ∑ an z n , n=0
∞
[(g ∘ f ) ∘ h](z) = ∑ bn z n . n=0
We show that an = bn for n = 0, 1, 2, . . . . It is clear that f (0) = h(0) = 0, and hence a0 = g(0) = b0 . Let n ∈ ℕ be given. By Corollary 1.2.9, an is determined by [g]n and [f ∘ h]n , same argument yields that [f ∘ h]n is determined by [f ]n and [h]n . Thus, an is determined by [f ]n , [h]n , and [g]n . Similarly, we see that bn is determined by [f ]n , [h]n , and [g]n , too. We write n3
[g]n ∘ ([f ]n ∘ [h]n )(z) = ∑ Ai z i , i=0
where ai = Ai , 0 ≤ i ≤ n. Similarly, we write n3
([g]n ∘ [f ]n ) ∘ [h]n (z) = ∑ Bi z i , i=0
where bi = Bi , 0 ≤ i ≤ n. But [g]n ∘ ([f ]n ∘ [h]n ) = ([g]n ∘ [f ]n ) ∘ [h]n because the composition of polynomials has associative property. This yields that an = bn for all n ∈ ℕ.
12  1 Basics of formal power series
h as
This theorem allows us to write the composition of formal power series g, f , and g∘f ∘h
if both f and h are nonunits. Now we can introduce Definition 1.2.11. Let S be a commutative ring and let f ∈ 𝕏0 (S) be a nonunit. For any integer n ≥ 2, we define f [n] (z) = f ∘ f [n−1] (z) = ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ f ∘ f ∘ ⋅ ⋅ ⋅ ∘ f (z). n
The formal power series f
[n]
is called the nth iterative power of f . We define that f [1] = f .
By Theorem 1.2.10, f [n] ∈ 𝕏(S) if f ∈ X 0 (S). Example 1.2.12. Given formal power series g(z) = 1 + z + z 2 + z 3 + ⋅ ⋅ ⋅ and f (z) = z + 2!z 2 + 3!z 3 + ⋅ ⋅ ⋅ in 𝕏(ℂ). Then ∞
n
g ∘ f (z) = ∑ (f (z)) n=0
is a formal power series in 𝕏(S). Moreover, g ∘ f (z) = 1 + z + (1 + 2!)z 2 + (1 + 2 ⋅ 2! + 3!)z 3 + ⋅ ⋅ ⋅ = 1 + 3z 2 + 11z 3 + ⋅ ⋅ ⋅ .
However, for a very simple formal power series h(z) = 1 + z, it is clear that ∞
g ∘ h(z) = g(1 + z) = ∑ (1 + z)n n=0
does not exist because the sum of the infinitely many constant 1 does not exist. The next theorem is obvious but it is useful in late sections. Theorem 1.2.13. Let g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ be a formal power series in 𝕏(S) where S is a ring. Let f ∈ 𝕏0 be given. Then g ∘ f (z) = b0 + g ∘ f 2
3
where g = b1 z + b2 z + b3 z + ⋅ ⋅ ⋅ = g − b0 , a nonunit of 𝕏(S). Applying Corollary 1.2.8, we have the following. Corollary 1.2.14. Let g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ be a formal power series in 𝕏 = 𝕏(S) where S is a ring. Let f , h ∈ 𝕏0 be given. Then g ∘ f ∘ h = b0 + g ∘ f ∘ h where g = b1 z + b2 z 2 + b3 z 3 + ⋅ ⋅ ⋅ = g − b0 , a nonunit of 𝕏(S).
1.3 Right distributive law for composition
 13
1.3 Right distributive law for composition The right distributive law for composition is a very special property for the formal power series space 𝕏(ℂ). Lemma 1.3.1. Let f and g be formal power series in 𝕏(ℂ) and write f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ ,
g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ .
Let ϕ and ψ be nonunits in 𝕏(ℂ). Then ∞
(f ∘ ϕ) ⋅ (g ∘ ψ) = ∑ cn z n , n=0
and cn = ∑ ak bm p(k,m) , n k+m≤n
(1.3)
where p(k,m) is the nth coefficient of the formal power series (ϕk ψm ) for all k, m ∈ ℕ: n ∞
ϕk ψm (z) = ∑ p(k,m) zn. n n=0
Proof. Let ϕ(z) = q1 z + q2 z 2 + q3 z 3 + ⋅ ⋅ ⋅ ,
ψ(z) = r1 z + r2 z 2 + r3 z 3 + ⋅ ⋅ ⋅ .
Let us denote ∞
ϕk (z) = ∑ qn(k) z n , n=0
∞
ψk (z) = ∑ rn(k) z n n=0
for all k ∈ ℕ. It is obvious that qn(k) = rn(k) = 0,
0 ≤ n < k.
Applying (1.2), we have ∞
n
f ∘ ϕ(z) = ∑ ( ∑ ak qn(k) )z n . n=0 k=0
Similarly, we have ∞
n
g ∘ ψ(z) = ∑ ( ∑ bk rn(k) )z n . n=0 k=0
(1.4)
14  1 Basics of formal power series If we write ∞
(f ∘ ϕ(z)) ⋅ (g ∘ ψ(z)) = ∑ cn z n , n=0
then j
i
cn = ∑ ( ∑ ak qi(k) )( ∑ bm rj(m) ). m=0
i+j=n k=0
(1.5)
On the other hand, ∞
fg(z) = ∑ ( ∑ ak bm )z n , n=0 k+m≤n
where ∑k+m≤n ak bm = ∑nk=0 ak bn−k . We shall show that cn = ∑ ak bm p(k,m) . n k+m≤n
Actually, p(k,m) = ∑ qi(k) rj(m) , n
(1.6)
i+j=n
we have = ∑ ak bm ( ∑ qi(k) rj(m) ) ∑ ak bm p(k,m) n
k+m≤n
i+j=n
k+m≤n
= ∑
∑ ak qi(k) bm rj(m)
i+j=n k+m≤n
j
i
= ∑ ( ∑ ak qi(k) )( ∑ bm rj(m) ) m=0
i+j=n k=0
= cn by formula (1.5).
Lemma 1.3.2. Let f (z) = a0 + a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ be a nonunit in 𝕏0 (ℂ) where a0 = 0. Let us denote (k) (k) 2 f k (z) = a(k) 0 + a1 z + a2 z + ⋅ ⋅ ⋅ ,
for all
k ∈ ℕ ∪ {0}.
Then, for all nonnegative integers k, m, and n, n
(m) a(k+m) = ∑ a(k) n i an−i . i=0
(1.7)
1.3 Right distributive law for composition
 15
Proof. The result is derived from f k+m = f k ⋅ f m . The next formula is called the right distributive law for composition of formal power series. Theorem 1.3.3. Let f and g be formal power series in 𝕏(ℂ) and let ϕ be a nonunit in 𝕏0 (ℂ). Then (f ∘ ϕ) ⋅ (g ∘ ϕ) = (fg) ∘ ϕ. Proof. Let all notation be set as in Lemma 1.3.1 with ϕ = ψ. Applying Lemma 1.3.2 with formula (1.7), we see that formula (1.6) becomes p(k,m) = qn(k+m) . n By formula (1.3), cn = ∑ ak bm p(k,m) = ∑ ak bm qn(k+m) . n k+m≤n
k+m≤n
n ∑∞ n=0 dn z ,
If we write (fg) ∘ ϕ(z) = which suffices to prove that
we shall show that cn = dn for every n ∈ ℕ ∪ {0},
(f ∘ ϕ) ⋅ (g ∘ ϕ) = (fg) ∘ ϕ. Since ∞
n
∞
fg(z) = ∑ ( ∑ ak bn−k )z n = ∑ ( ∑ ak bm )z n , n=0 k=0
n=0 k+m=n
applying (1.2) we have ∞
(fg) ∘ ϕ(z) = ∑ ( ∑ ak bm )ϕn (z) n=0 k+m=n ∞
∞
= ∑ ( ∑ ak bm )(∑ qi(n) z i ) n=0 k+m=n ∞
i=n
i
= ∑ ( ∑ [ ∑ ak bm ]qi(n) )z i , i=0 n=0 k+m=n
and in consequence, i
di = ∑ ( ∑ ak bm )qi(n) = ∑ ak bm qi(k+m) . n=0 k+m=n
k+m≤i
Then dn = ∑ ak bm qn(k+m) = ∑ ak bm p(k,m) = cn . n k+m≤n
k+m≤n
16  1 Basics of formal power series Corollary 1.3.4. For any n ∈ ℕ, the equality f n (ϕ) = (f ∘ ϕ)n is true for any formal power series f and any nonunit ϕ. Proof. Let f = g in Theorem 1.3.3, we have (f ∘ ϕ)2 = f 2 ∘ ϕ. Suppose that f k (ϕ) = (f ∘ ϕ)k ,
k = 1, 2, . . . , n − 1
for some n ≥ 3. Then (f ∘ ϕ)n = (f ∘ ϕ)(f ∘ ϕ)n−1
= (f ∘ ϕ)(f n−1 ∘ ϕ)
= (ff n−1 )(ϕ) = f n (ϕ). Example 1.3.5. Let f (x) = 1 + x + x2 + ⋅ ⋅ ⋅, g(x) = x 2 and ϕ(x) = 1 be three ℝvalued analytic functions. Then both fg ∘ ϕ and f ∘ ϕ are undefined. This example shows that the nonunitness for the formal power series ϕ is important for the right distributive law. Example 1.3.6. Let f (x) = 1 + x + x2 + ⋅ ⋅ ⋅, and ϕ(x) = 1 − x be formal power series in 𝕏(ℝ). Then f (ϕ) is undefined. 1 However, if we consider that f and ϕ are analytic functions such that f (x) = 1−x for x < 1, then we have f (ϕ(x)) =
1 x
for
1 − x < 1.
Examples 1.3.5 and 1.3.6 have showed us the difference between the formal power series and the regular power series. We would like to introduce another example below, which will show that the right distributive law sometimes is working with a unit formal power series ϕ. Example 1.3.7. Let f (x) = 1 + x + x2 + ⋅ ⋅ ⋅, g(x) = x 2 and ϕ(x) = functions. Then
1 2
be ℝvalued analytic
2
1 1 1 1 1 (fg)( ) = ( ) (1 + + 2 + ⋅ ⋅ ⋅) = , 2 2 2 2 2 and 1 1 1 1 1 1 f ( ) ⋅ g( ) = (1 + + 2 + ⋅ ⋅ ⋅) ⋅ = . 2 2 2 2 4 2 Then (fg)(ϕ) = (f (ϕ))(g(ϕ)). Of course, we are interested in the necessary or sufficient condition (or both) for the right distributive law to be satisfied. Readers will see certain developments of such subjects later in this book.
1.4 The matrix representation of formal power series  17
1.4 The matrix representation of formal power series Definition 1.4.1. Let S be a ring and let f ∈ 𝕏(S) be given such that f (z) = f0 + f1 z + f2 z 2 + f3 z 3 + ⋅ ⋅ ⋅ . The infinite triangular matrix f0 [ [0 [ [ A = Af = [ 0 [0 [ [. . [.
f1 f0 0 0 .. .
f2 f1 f0 0 .. .
f3 f2 f1 f0 .. .
... ] . . .] ] . . .] ] . . .] ] ] .. . ]
is called the matrix representation associated with multiplication of the formal power series, or the associated semicirculant matrix of the formal power series f . We define the nth section of the infinite matrix A as the regular matrix f0 [0 [ [ [0 [ An = [ 0 [ [. [. [. [0
f1 f0 0 0 .. . 0
f2 f1 f0 0 .. . 0
f3 f2 f1 f0 .. . 0
... ... ... ... .. . ...
fn−1 fn−2 ] ] ] fn−3 ] ] . fn−4 ] ] .. ] ] . ] f0 ](n×n)
n If we write the matrix representation of the formal power series f (z) = ∑∞ n=0 fn z as A = Af = (aij ), then aij = fj−i for all j ≥ i and aij = 0 if j < i. Let M(𝕏) denote the set of all infinite upper triangular matrices (aij ) over ℂ where
aij = ai−1,j−1 ,
i ≤ j in ℕ and
aij = 0,
i > j.
Define a mapping P: 𝕏(ℂ) → M(𝕏) by P(f ) = Af , then P is a bijective mapping obviously. The mapping P has more properties. Theorem 1.4.2. Let P: 𝕏(ℂ) → M(𝕏) such that P(f ) = Af for every f ∈ 𝕏(ℂ). Let f (z) = ∞ n n ∑∞ n=0 an z and g(z) = ∑n=0 bn z be two formal power series in 𝕏(ℂ). Write P(f ) = A = Af = (aij ),
P(g) = B = Bg = (bij ).
Then (i) P(f + g) = P(f ) + P(g) = A + B ∈ M(𝕏), (ii) P(cf ) = cP(f ) = cA ∈ M(𝕏) for all c ∈ ℂ, (iii) P(fg) = AB ∈ M(ℂ), and (iv) P is a ring isomorphism.
18  1 Basics of formal power series Proof. (i) and (ii) are obvious. Since P(f ) = A = Af = (aij ),
P(g) = B = Bg = (bij ),
it follows that aij = aj−i , bij = bj−i for all j ≥ i, and aij = 0 = bij if j < i. n n Let (fg)(z) = ∑∞ n=0 cn z . Then cn = ∑k=0 an−k bk . Denote P(fg) = C = (cij ),
cij = cj−i
for
j ≥ i,
and cij = 0
if j < i.
Denote AB = D = (dij ). For j ≥ i, we have ∞
j
k=0
k=0
dij = ∑ aik bkj = ∑ aik bkj j
j−i
= ∑ ak−i bj−k = ∑ al b(j−i)−l = cj−i = cij . k=i
l=0
Then P(fg) is a semicirculant matrix and P(fg) = AB = C = (cij ) where cij is the (j − i)th coefficient of the formal power series fg. This is (iii). Thus, P is a homomorphism in the sense of addition and multiplication. It is obvious that P is one to one and onto, then P is a ring isomorphism. It is obvious that two associated semicirculant matrices A and B are equal if and only if An = Bn for all n ∈ ℕ. It is known that the matrix multiplication of two such infinite matrices is still an upper triangular infinite matrix and it is obvious that (AB)n = An Bn for every n ∈ ℕ, where A and B are matrix representations of formal power series f and g, respectively. The next method will help us to compute the inverse or reciprocal of a unit of 𝕏(ℂ). n Corollary 1.4.3. Let f (z) = ∑∞ n=0 an z be a formal power series in 𝕏(ℂ) with a0 ≠ 0 and −1 let P(f ) = A = (aij ). If we denote A = (dij ) and denote An the nth section of A for any n ∈ ℕ, then for every 1 ≤ j ≤ n, we have: (i) djj = a−1 0 ; (ii) dij = −a−1 0 (a1 d(i+1)j + a2 d(i+2)j + ⋅ ⋅ ⋅ + aj−i djj ), 1 ≤ i < j; and (iii) dij = 0 for i > j.
Proof. f −1 exists because a0 ≠ 0, it follows that P(f −1 ) is also semicirculant. We write f −1 (z) = d0 + d1 z + d2 z 2 + ⋅ ⋅ ⋅ ,
1.4 The matrix representation of formal power series  19
and write Af −1 = (dij ). Then dij = 0 if i < j, and dij = dj−i if i ≥ j. By Theorem 1.4.2, we have that A−1 = Af −1 . For every n ∈ ℕ, let A−1 n = (dij )(n×n) = [D1 , D2 , . . . , Dn ], where Dj = [d1j d2j . . . djj 0 0 . . . 0]T is the jth column of the matrix A−1 n , where 1 ≤ j ≤ n. Then a0 [ [0 [ [0 [ [. [ .. [ [0
a1 a0 0 .. . 0
a0 [ [0 [ [0 [ [. [ .. [ [0
a1 a0 0 .. . 0
a2 a1 a0 .. . 0
... ... ... .. . ...
0 d1j [ . ] [.] . [ ] [ an−1 . ] [ .. ] ] ] [ ] [ ] [ an−2 ] [d [0] ] [ (j−1)j ] ] [ ] an−3 ] ] [ ] ] [ [ djj ] = [ 1 ] . ] .. ] [ [ ] 0 ] ] [0] . ] [ [ . ] [.] ] [ ] a0 ]n×n [ [ .. ] [ .. ] [ 0 ]
[0]
Then a2 a1 a0 .. . 0
... ... ... .. . ...
0 aj−1 d1j ] [ ] ] [ aj−2 ] [ d2j ] [0] ] [ ] ] [ [ . ] [.] aj−3 ] ] [ .. ] = [ .. ] . [ ] [ ] .. ] [ ] [ ] . ] ] [d(j−1)j ] [0] a0 ]j×j [ djj ] [ 1 ]
It is a linear equation system with j unknowns d1j , d2j , . . . , djj and this system has unique solution set because a0 ≠ 0. There are many ways to solve this linear system. For example, the last equation provides that djj = a−1 0 . This is (i). (ii) is obvious. Applying the backward method, assume that we have obtained the values of d(i+1)j , d(i+2)j , . . . , djj , by the ith equation a0 dij + a1 d(i+1)j + ⋅ ⋅ ⋅ + aj−i djj = 0, where 0 ≤ i < j, we have dij = −a−1 0 (a1 d(i+1)j + a2 d(i+2)j + ⋅ ⋅ ⋅ + aj−i djj ). This is (ii). We can find all such dij for which i ≤ n, j ≤ n. Since n ∈ ℕ is arbitrary, we completed the proof. If we denote dkj in (ii) by dj−k , i ≤ k ≤ j, we have dm = dj−i = −a−1 0 (a1 dm−1 + a2 dm−2 + ⋅ ⋅ ⋅ + am d0 ), which is the formula in Theorem 1.1.8.
20  1 Basics of formal power series By Corollary 1.4.3 and its proof, we have obtained an algorithm for computing f −1 if the formal power series f is given and f is a unit. We have seen that actually all methods to compute the inverse matrix such as An can be a tool for us to compute f −1 . The next theorem provides another formula for us to do so. The proof of this theorem can be found in (p. 17, [42]) and, therefore, is not listed here. n Theorem 1.4.4. Let f (z) = ∑∞ n=0 an z be a formal power series in 𝕏(ℂ) with a0 ≠ 0 and −1 n let P(f ) = A = (aij ). If we denote f (z) = ∑∞ n=0 bn z , then
a1 a0 (−1)n 0 bn = n+1 0 a0 . . . 0
a2 a1 a0 0 .. . 0
a3 a2 a1 a0 .. . 0
... ... ... ... .. . ...
an−1 an−2 an−3 an−4 .. . a0
an an−1 an−2 an−3 . .. . a1
The above formula is called the Wronski formula, which is a nonrecurrent formula. In addition to applying the matrix representations to calculate the inverse of a formal power series, the matrix representation of formal power series provides a convenient approach for us to compute the nth power of a formal power series, or f n , including provides the long operations of such computations (see p. 175, [52]). For any formal power series f (z) = a0 + a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ over a field S, we denote f n as (n) (n) 2 (n) 3 f n (z) = a(n) 0 + a1 z + a2 z + a3 z + ⋅ ⋅ ⋅ ,
(1.8)
and call a(n) the kth coefficient of f n . We also define that f 0 = I or f 0 = (1, 0, 0, . . . ). The k next proposition is obvious. Proposition 1.4.5. Let f (z) = a0 + a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ be a formal power series over a field S and let n ∈ ℕ be given. Denote the matrix representation of f as a0 [ [0 [ [ A = Af = [ 0 [0 [ [. . [.
a1 a0 0 0 .. .
a2 a1 a0 0 .. .
a3 a2 a1 a0 .. .
... ] . . .] ] . . .] ]. . . .] ] ] .. . ]
Then for any k ∈ ℕ ∪ {0} the kth coefficient of f n , or a(n) is determined by Ak+1 only k where Ak+1 is the (k + 1) × (k + 1) matrix of A as defined in Definition 1.4.1. (n) (n) n Moreover, the coefficients a(n) 0 , a1 , . . . , ak is the first row of the matrix Ak+1 . Applying Theorem 1.4.2 with f = g, we easily obtain the next proposition.
1.4 The matrix representation of formal power series  21
Proposition 1.4.6. Let f (z) = a0 + a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ be a formal power series over a field S. Then for any k ∈ ℕ, T
T
a(1) a(2) 0 [ 0(1) ] [ (2) ] [a ] [a ] [ 1 ] [ 1 ] [ ] [ (2) ] [a ] = [a(1) ] [ 2 ] [ 2 ] [ . ] [ . ] [ . ] [ . ] [ . ] [ . ] (1) (2) a [ak ] [ k ]
a0 [ [0 [ [0 [ [. [ .. [ [0
a1 a0 0 .. . 0
a2 a1 a0 .. . 0
... ... ... .. . ...
ak ] ak−1 ] ] ak−2 ] ] .. ] . ] ] a0 ](k+1)×(k+1)
where AT is the transposed matrix of A. Applying this Proposition (n − 1) times, we have a convenient formula to compute (n) (n) n a(n) , 0 a1 , . . . , ak which, of course, is the first row of A(k+1) . Corollary 1.4.7. Let f (z) = a0 + a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ be a formal power series over a field S and let n ∈ ℕ be given. Then for any k ∈ ℕ, a(n) 0
T
T
a(n−1) 0
[ (n−1) ] [ (n) ] ] [a [a ] [ 1 ] [ 1 ] [ (n−1) ] [ (n) ] ] [a ] = [a [ 2 ] [ 2 ] [ . ] [ . ] [ . ] [ . ] [ . ] [ . ] (n−1) (n) a [ak ] [ k ]
a0 [ [0 [ [0 [ [. [ .. [ [0
a1 a0 0 .. . 0
a2 a1 a0 .. . 0
... ... ... .. . ...
ak ] ak−1 ] ] ak−2 ] . ] .. ] ] . ] a0 ](k+1)×(k+1)
The above proposition provides an algorithm of computing a(n) s. Once we have an k (n) algorithm, we want to know the long operations for computing ak . As usual, we only calculate the number of multiplications and divisions. The next theorem tells us the operation counts of certain algorithms. Theorem 1.4.8. Let f be denoted as in the above corollary and let n, k ∈ ℕ be given. Then (j) the long operations, or the number of multiplications of computing all ai , 0 ≤ i ≤ k, 1 ≤ j ≤ n, is 1 (k + 1)(k + 2)(n − 1). 2 Proof. By Proposition 1.4.5, it is clear that the number of multiplications needed to compute a(2) , 0 ≤ j ≤ k is j 1 1 + 2 + 3 + ⋅ ⋅ ⋅ + k + (k + 1) = (k + 1)(k + 2). 2 By Corollary 1.4.7, the number of multiplications of computing all ai , 0 ≤ i ≤ k, is (j)
also 21 (k + 1)(k + 2) once all ai
(j−1)
have been obtained, 2 ≤ j ≤ n. Thus, the number of
22  1 Basics of formal power series the total multiplications for computing all ai , 0 ≤ i ≤ k, 1 ≤ j ≤ n, is (j)
1 (k + 1)(K + 2)(n − 1). 2
1.5 Almost unit formal power series In the space of formal power series 𝕏, units and nonunits have very different properties. This section introduces the formal power series of a special subset of the nonunit formal power series, socalled delta series or almost unit formal power series. Henrici explained that following a suggestion of Professor I. Kaplansky he used almost units to represent such kind of formal power series [41]. Definition 1.5.1. Let S be a field. If f ∈ 𝕏(S) with ord(f ) = 1, then f is called a delta series, or an almost unit. We denote the set of all delta series by 𝔻(S), or just 𝔻. It is clear that 𝔻 ⊆ 𝕏0 . Definition 1.5.2. Let f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ be a delta series in 𝔻(S) where S is a field. For every n ∈ ℕ, we write (n) (n) 2 (3) 3 f n (z) = a(n) 0 + a1 z + a2 z + a3 z + ⋅ ⋅ ⋅ ,
then a(n) = 0 for 0 ≤ i < n and a(n) n ≠ 0. We now define a matrix i a(1) 1
[ [ 0 [ [ D(f ) = [ [ 0 [ [ 0 [ .. [ .
a(1) 2
a(2) 2 0
0 .. .
a(1) 3
a(2) 3 (3) a3 0 .. .
a(1) 4
a(2) 4 (3) a4 a(4) 4 .. .
... ] . . .] ] ] . . .] ]. ] . . .] ] .. .]
The matrix D(f ) is called the delta associated matrix of the delta series f . The delta series I𝔻 (z) = z is called the identity of composition of 𝔻, we usually denote D(I𝔻 ) = I𝔻 or D(I𝔻 ) = I if there is no confusion. If we write D(f ) = (dij ) in Definition 1.5.2, then dij = a(i) , where djj ≠ 0 and dij = 0 j if i < j for all i, j ∈ ℕ. Please be aware of the difference between D(f ) and the Af = P(f ) in Definition 1.4.1. We extend the delta associated matrix D(f ) = (dij ) to the nonunit associated matrix to any nonunit formal power series f . The significant character for the delta associated matrix is that djj ≠ 0. This character is not necessary for nonunit associate matrix. As we did in Definition 1.4.1, we define that two nonunit associated matrices A and B equal if and only if An = Bn for every n ∈ ℕ.
1.5 Almost unit formal power series  23
The multiplication of finitedimensional matrices is usually not commutative. The delta associated matrices are infinite matrices and they have some very interesting properties. Lemma 1.5.3. Let S be a field. Let f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ ,
g(z) = b1 z + b2 z 2 + b3 z 3 + ⋅ ⋅ ⋅
be nonunit formal power series in 𝕏0 (S). Let D(f ) and D(g) be the nonunit associate matrices for f and g, respectively. Then D(f )D(g) is the nonunit associated matrix of (f ∘ g), or D(f )D(g) = D(f ∘ g). Proof. As we did in (1.8), we write (n) (n) 2 f n (z) = a(n) 0 + a1 z + a2 z + ⋅ ⋅ ⋅ ,
(n) (n) 2 g n (z) = b(n) 0 + b1 z + b2 z + ⋅ ⋅ ⋅
for every n ∈ ℕ, where a0 = b0 = 0. Write D(f )D(g) = (cij ), it is obvious that for any n, m ∈ ℕ such that n ≥ m, (m) (m) (m+1) (n) cmn = a(m) + ⋅ ⋅ ⋅ + a(m) m bn + am+1 bn n bn .
On the other hand, m (m) m+1 m+2 f m ∘ g(z) = a(m) (z) + a(m) (z) + ⋅ ⋅ ⋅ , m g (z) + am+1 g m+2 g
and hence the nth coefficient of f m (g(z)) is (m) (m) (m+1) (n) a(m) + ⋅ ⋅ ⋅ + a(m) m bn + am+1 bn n bn ,
which is cmn . By Corollary 1.3.4, it is the nth coefficient of (f ∘g)m . If we denote D(f ∘g) = (dij ), then dmn is the nth coefficient of (f ∘ g)m . Then we have (m) (m) (m+1) (n) cmn = a(m) + ⋅ ⋅ ⋅ + a(m) m bn + am+1 bn n bn = dmn .
It is trivial that cmn = dmn = 0 if n < m. Thus, D(f )D(g) is the nonunit associated matrix of f ∘ g. Please notice that if ord(f ) = r > 1, then a(n) = 0 for all k < rn. The main diagonal k of D(f ) consists of all zero entries. Without proof, we can easily have the next result. Corollary 1.5.4. Let f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ with a1 ≠ 0. For any n ∈ ℕ, we have n
(D(f )) = D(f [n] ), where f [n] is the nth iterative power of f as in Definition 1.2.11.
24  1 Basics of formal power series The multiplication of two matrices may not be commutable. However, if two square matrices A and B satisfy that AB = I, then A is invertible (p. 172, [65]). Lemma 1.5.5. Let D(f ) be the delta associated matrix of a delta series f and let D(g) be the delta associated matrix of a delta series g. Then D(f )D(g) = I if and only if D(g)D(f ) = I, where I = D(I𝔻 ) is the identity matrix. In this case, we call D(g) the inverse of D(f ) and denote it as D(f )−1 . Proof. It suffices to show that, for every n ∈ ℕ, D(f )n D(g)n = In
if and only if D(g)n D(f )n = In .
If we write f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ ,
g(z) = b1 z + b2 z 2 + b3 z 3 + ⋅ ⋅ ⋅ ,
then we have a1 ≠ 0, a(k) = ak1 and b1 ≠ 0, b(k) = bk1 for k ∈ ℕ. k k Moreover, looking at the determinants we have 1+⋅⋅⋅+n 1+⋅⋅⋅+n b1 ≠ 0, D(f )n D(g)n = D(f )n ⋅ D(g)n = a1 whenever D(f )n D(g)n = In or D(g)n D(f )n = In . We have the conclusion, and hence, we complete the proof. By Theorem 1.4.8, we have the long operations of calculating D(f )n . Corollary 1.5.6. Let f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ with a1 ≠ 0. For any n ∈ ℕ, the long operations for computing D(f )n is 1 (n − 1)n(n + 1). 2 Proof. Taking k = n in Theorem 1.4.8 and considering that a0 = 0, we have the conclusion. Remark 1.5.7. Lemma 1.5.3 and Lemma 1.5.5 indicate a possibility that if f ∈ 𝔻(S) where S is a field, then D(f ) may have the inverse D−1 (f ) and D(f )D−1 (f ) = D−1 (f )D(f ) = I𝔻 = I. Does the inverse of f exist? An answer to this question is to have an isomorphism between f and the matrix D(f ) under the composition. Let us recall the following.
1.5 Almost unit formal power series  25
Definition 1.5.8. A nonempty set G is said to form a group if in G there is defined a binary operation, denoted by ∗, such that: (i) a, b ∈ G implies that a ∗ b ∈ G. (ii) a, b, c ∈ G implies that a ∗ (b ∗ c) = (a ∗ b) ∗ c. (iii) There is an element e ∈ G such that e ∗ a = a ∗ e = a for all a ∈ G. (iv) For every a ∈ G, there exists an element a−1 ∈ G such that a ∗ a−1 = a−1 ∗ a = e. Theorem 1.5.9. Let S be a field. Then 𝔻(S) is a group under the operation of composition of formal power series. Proof. It is clear that the operation of composition is closed in 𝔻(S). We can easily check that the identity of 𝔻(S) is the formal power series I𝔻 (z) = z because f (I𝔻 (z)) = f (z),
I𝔻 (f (z)) = f (z).
The associative property can be obtained by Theorem 1.2.10 and Lemma 1.5.3 because the multiplication of n by n matrices is associative. We now physically show the existence of the inverse of any delta series f in 𝔻(S). Write f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ . Let g(z) = b0 + b1 z + b2 z 2 + b3 z 3 + ⋅ ⋅ ⋅ and g(f (z)) = z. For every n ∈ ℕ, we write (n) (n) 2 (3) 3 f n (z) = a(n) 0 + a1 z + a2 z + a3 z + ⋅ ⋅ ⋅ .
Then a(n) = 0 for 0 ≤ i < n and a(n) n ≠ 0. Let g(f (z)) = z, we have i z = b0 + b1 f (z) + b2 f 2 (z) + b3 f 3 (z) + ⋅ ⋅ ⋅ 2
n
k=1
k=1
(k) 2 (k) n = b0 + (b1 a(1) 1 )z + ( ∑ bk an )z + ⋅ ⋅ ⋅ + ( ∑ bk an )z + ⋅ ⋅ ⋅ . n Then b0 = 0 and b1 = 1/a1(1) = 1/a1 because a1 ≠ 0. Then a(n) n = a1 ≠ 0 for every n ∈ ℕ. Suppose that b0 , b1 , b2 , . . . , bn−1 have been determined for some n ∈ ℕ, put ∑nk=1 bk a(k) n = 0, we have
bn = −
1 (n−1) (b a(1) + b2 a(2) ). n + ⋅ ⋅ ⋅ + bn−1 an an1 1 n
Then g ∈ 𝔻(S) and g is the inverse of f under the composition. By Lemma 1.5.5, we know that such an inverse is commutative. Thus, 𝔻(S) is a group.
26  1 Basics of formal power series Definition 1.5.10. For every delta series f ∈ 𝔻(S) where S is a field, the iterated inverse, or iterative inverse or the reverse of f , is denoted by f [−1] for which f [−1] ∘ f = f [−1] ∘ f [1] = I𝔻 , where I𝔻 is the formal power series I𝔻 (z) = z. The next corollary is obvious which completes the answer to Remark 1.5.7. Corollary 1.5.11. Let f ∈ 𝔻(ℂ) be given. Denote the associate delta matrices of f and f [−1] as D(f ) and D(f [−1] ), then D(f )D(f [−1] ) = I𝔻 . Remark 1.5.12. Let f ∈ 𝕏0 (S) be given where S is a field. If g ∘ f = I𝔻 , then 1 = ord(z) = ord(g ∘ f (z)) = ord(g) ord(f ). If ord(f ) = 1, then ord(g) = 1. If ord(f ) = m > 1, then ord(g) = 1/m ∉ ℕ which is a contradiction. Then g ∘ f = I𝔻 if and only if both g and f are delta series. Corollary 1.5.13. If f ∈ 𝔻(z), such that f (z) = z + ak z k + ak+1 z k+1 + ⋅ ⋅ ⋅ , with a0 = 0, a1 = 1, ak ≠ 0, aj = 0, j = 2, 3, . . . , k − 1. Then f [−1] (z) = z − ak z k + Pk+1 (z), where Pi ∈ 𝕏(ℂ) with ord(Pi ) ≥ i, i ∈ ℕ. In particular, if f (z) = z + ak+1 z k+1 with ak+1 ≠ 0, then f [−1] (z) = z − ak+1 z k+1 + P2k+1 (z). Proof. By Theorem 1.5.9, f [−1] exists. Write f [−1] (z) = b1 z + b2 z 2 + b3 z 3 + ⋅ ⋅ ⋅ . We prove that b1 = 1,
bk = −ak ,
and
bj = 0
for
2 ≤ j < k.
It is trivial that b1 = 1. For 2 ≤ j < k, we may easily verify that (2) a(1) j = aj = ⋅ ⋅ ⋅ = aj
(j−1)
= 0,
1.5 Almost unit formal power series  27
where a(n) is the kth coefficient of f n . By the formula in Theorem 1.5.9, k (2) bj = −(b1 a(1) j + b2 aj + ⋅ ⋅ ⋅ + bj−1 aj
(j−1)
) = 0.
Also, bk = −(b1 a(1) + b2 a(2) + ⋅ ⋅ ⋅ + bk−1 a(k−1) ) k k k
= −(b1 a(1) + 0a(2) + ⋅ ⋅ ⋅ + 0a(k−1) ) = −ak . k k k
So, f [−1] (z) = z − ak z k + Pk+1 (z). Now let f (z) = z + ak+1 z k+1 be given where ak+1 ≠ 0, a1 = 1 and aj = 0 for all other j ∈ ℕ. By the above result, we have f [−1] (z) = z − ak+1 z k+1 + Pk+2 (z). We show that actually ord(Pk+2 ) ≥ 2k + 1. For any n ∈ ℕ, n n j n f n (z) = (z + ak+1 z k+1 ) = z n ∑ ( )ak+1 z jk j j=0
n n = z n + ( )ak+1 z n+k + ( )a2k+1 z n+2k + ⋅ ⋅ ⋅ + ank+1 z n+nk . 1 2
Then n j = ( )ak+1 , a(n) n+jk j
0 ≤ j ≤ n and
a(n) i =0
otherwise.
Applying the recurrence formula for f [−1] in Theorem 1.5.9, −bk+i = b1 a(1) + bk a(k) + bk+1 a(k+1) + ⋅ ⋅ ⋅ + bk+i−1 a(k+i−1) =0 k+i k+i k+i k+i for 1 < i ≤ k. Then ord(Pk+2 ) ≥ 2k + 1. We rename Pk+1 as P2k+1 . Thus, f [−1] (z) = z − ak z k + P2k+1 (z). Example 1.5.14. For a ∈ ℂ, let a a f (z) = z + ( )z 2 + ( )z 3 + ⋅ ⋅ ⋅ , 1 2
28  1 Basics of formal power series where a a(a − 1)(a − 2) ⋅ ⋅ ⋅ (a − k + 1) ( )= , k k!
k = 0, 1, 2, . . . .
Then D(f ) = (amn ), where ma ), n−m
amn = (
n ≥ m,
or (a1 )
1 [ [0 [ [ D(f ) = [ [0 [0 [ [. . [.
(a2 )
(2a ) 1
1
0 0 .. .
1 0 .. .
(a3 )
...
(2a ) 2 3a (1 ) 1 .. .
] . . .] ] ] . . .] ], . . .] ] ] .. .]
and D(f [−1] ) = (bmn ), where bmn =
m −na ( ), n n−m
n ≥ m.
Then 1 −2a 2 1 −3a 3 1 −4a 4 f [−1] (z) = z + ( )z + ( )z + ( )z + ⋅ ⋅ ⋅ . 2 1 3 2 4 3 Solution. We will see in Section 2.2 that f is actually the product of z and formal binomial series Ba , and f n (z) is z n Bna for all n ∈ ℕ. Using the results in Section 2.2, we obtain the D(f ). Next, we verify that 1 [ [0 [ [ [−1] D(f ) = [ [0 [0 [ [. . [.
1 −2a ( ) 2 1
1
0 0 .. .
1 −3a ( ) 3 2 2 −3a ( ) 3 1
1 0 .. .
1 −4a ( ) 4 3 2 −4a ( ) 4 2 3 −4a ( ) 4 1
1 .. .
...
] . . .] ] ] . . .] ], . . .] ] ] .. .]
It is clear that bmn = 0 for n < m because D(f [−1] ) must be a triangular matrix. bmm = (ma ) = 1 and also amm = 1. For n > m, 0 n
n
k=m
k=m n−m
b(k) ∑ a(m) n = ∑ ( k
ma k −na ) ( ) k−m n n−k ma m + p −na ) ( ) p n n−m−p
= ∑( p=0
1.5 Almost unit formal power series  29
= c + c , by the substitution p = k − m, where c =
−na m n−m ma ), ∑ ( )( n p=0 p n−m−p
c =
1 n−m ma −na ). ∑ ( )p( n p=0 p n−m−p
and
Applying Vandermonde’s theorem, n a b a+b )=( ), ∑ ( )( k n − k n k=0
(1.9)
we have m (m − n)a ( ) n n−m m (m − n)a(ma − na − 1) ⋅ ⋅ ⋅ (ma − na − (n − m) + 1) = n (n − m)! ma ma − na − 1 ( ), =− n n−m−1 ma n−m ma − 1 −na c = )( ) ∑( n p=1 p − 1 n−m−p c =
= =
ma n−m−1 ma − 1 −na )( ) ∑ ( n j=0 j n−m−1−j ma ma − na − 1 ( ). n n−m−1
Then c + c = 0. Thus, D(f )D(f [−1] ) = I𝔻 , and hence 1 −2a 2 1 −3a 3 1 −4a 4 f [−1] (z) = z + ( )z + ( )z + ( )z + ⋅ ⋅ ⋅ . 2 1 3 2 4 3 Remark 1.5.15. The group 𝔻(ℂ) is generally not Abelian although each element of 𝔻(ℂ) has a triangular delta associated matrix. If the almost units f and g are denoted by f (z) = a1 z + a2 z 2 + ⋅ ⋅ ⋅ ,
g(z) = b1 z + b2 z 2 + ⋅ ⋅ ⋅ ,
and f (n) and g (n) are denoted as in (1.8), a necessary and sufficient condition for f ∘ g = g ∘ f is (m) (m) (m+1) (n) a(m) + ⋅ ⋅ ⋅ + a(m) m bn + am+1 bn n bn (m) (m) (m+1) (n) = b(m) + ⋅ ⋅ ⋅ + b(m) m an + bm+1 an n an
for all m, n ∈ ℕ with n ≥ m.
30  1 Basics of formal power series Considering 𝕏 or 𝔻 as a linear space or linear algebra, we would like to have bases for these spaces. Theorem 1.5.16. Let g ∈ 𝕏(ℂ) be given. Let f (z) = a0 + a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ with a0 = 0, a1 ≠ 0. There exists a unique sequence {cn }∞ n=0 ⊆ ℂ such that g(z) = c0 + c1 f (z) + c2 f 2 (z) + c3 f 3 (z) + ⋅ ⋅ ⋅ . 0 The sequence {f n }∞ n=0 is called a pseudo basis of 𝕏(ℂ) where we define that f = I, the identity in 𝕏(ℂ) as in Definition 1.1.3.
Proof. Let g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ be a formal power series in 𝕏(ℂ). Let {cn }∞ n=0 ⊆ ℂ be a sequence such that g(z) = c0 + c1 f (z) + c2 f 2 (z) + c3 f 3 (z) + ⋅ ⋅ ⋅ . We prove that such a sequence exists and is unique. As usual, we write (n) (n) 2 (3) 3 f n (z) = a(n) 0 + a1 z + a2 z + a3 z + ⋅ ⋅ ⋅ ,
where a(n) = 0 for 0 ≤ i < n and a(n) n ≠ 0. Then b0 = c0 . i
b1 (1) Next, b1 = c1 a1 = a(1) 1 , which provides that c1 = a1 , where a1 = a1 ≠ 0. Suppose that c0 , c1 , . . . , cn−1 have been computed for some integer n > 1. For n ∈ ℕ, we have n
n−1
k=1
k=1
(n) bn = ∑ ck a(k) = ∑ ck a(k) n + cn an . n Since a(n) n = a1 ≠ 0, it follows that
cn = −
1 n−1 ( ∑ c a(k) ). an1 k=1 k n
We have completed the Mathematical Induction, and hence we have proved the existence of the sequence {cn }∞ n=0 ⊆ ℂ. If we denote the delta associated matrix of f as D(f ), we have D(f )C = B where C = [c1 c2 c3 . . . ]T and B = [b1 b2 b3 . . . ]T are single column matrices, or a(1) 1
[ [ 0 [ [ [ 0 [ [ [ 0 [ .. [ .
a(1) 2
a(2) 2 0 0 .. .
a(1) 3
a(2) 3 a(3) 3 0 .. .
a(1) 4
a(2) 4 a(3) 4 a(4) 4 .. .
...
] [ c1 ] [ b1 ] . . .] c2 ] [ b2 ] ][ ] [ ] ][ [c3 ] [b3 ] . . .] ] = [ ]. ][ ] [ ] ][ [c4 ] [b4 ] . . .] [ ] .] [.] . . .. .] [ . ] [ . ]
1.5 Almost unit formal power series  31
Notice that c0 = b0 and bn , the coefficient of z n of g, is determined only by f , f , . . . , f n ; in practice, we need only to compute (D(f ))n Cn = Bn for n ∈ ℕ, where 1
2
Cn = [c1 c2 c3 . . . cn ]T ,
Bn = [b1 b2 b3 . . . bn ]T .
The determinant (D(f ))n  = ∏nk=1 a(k) ≠ 0 ensures the existence and uniqueness of the k solution of this (n × n) linear equation system. n ∞ Since a(n) = 0 for 1 ≤ i < n and a(n) n ≠ 0, the sequence {f }n=0 is obviously linearly i independent. n ∞ If f (x) = x, the pseudo basis {f n }∞ n=0 = {x }n=0 is called the standard pseudo basis of 𝕏(ℂ).
Corollary 1.5.17. The sequence {f n }∞ n=1 ⊂ 𝔻(ℂ) forms a pseudo basis of 𝔻(ℂ) where f is any delta series in 𝔻(ℂ). For f ∈ 𝕏0 , we know that f ∘f is still a nonunit. This property allows us to compose the nonunit many times. If ord(f ) = 1, or f ∈ 𝔻, we have more interesting results. Definition 1.5.18. In the group 𝔻(S), where S is a field, the formal power series f [n] = f [n−1] ∘ f is also called the nth iterative fold, where f [1] = f . If we write f (z) = a0 + a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ with a0 = 0, a1 ≠ 0, we denote f [n] by [n] [n] 2 f [n] (z) = a[n] 0 + a1 z + a2 z + ⋅ ⋅ ⋅ n where a[n] 1 = a1 ≠ 0.
Since 𝔻(ℂ) is a group, the associative law exists and, therefore, f [m] ∘ f [n] = f [n] ∘ f [m]
(1.10)
for all nonnegative integers n and m, no matter 𝔻(ℂ) itself is commutative or not. Thus, the f [n] is welldefined, and f [n] = f [k] ∘ f [l]
if
k+l=n
for nonnegative integers k and l. Since 𝔻(ℂ) is a group, f [−1] exists for every f ∈ 𝔻(ℂ). This property allows us to extend equation (1.8) to all integers m and n with the definition f [−n] = (f [−1] )
[n]
for all n ∈ ℕ.
For example, f [−2] ∘ f [2] = f [−1] ∘ f [−1] ∘ f [1] ∘ f [1] = I𝔻 = f [−2+2] = f [0] .
(1.11)
32  1 Basics of formal power series The investigation of the iterative fold of formal power series or functionals can be traced to 1871 when Schröder provided some interesting results [91]. Schröder’s contribution to formal theory can be seen in later chapters. To compute the coefficients of the nth iterative fold f [n] for every f ∈ 𝔻(ℂ) is a tough task although such f [n] exists. We provide a recursive formula here. Proposition 1.5.19. Let f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ be a delta series in 𝔻(ℂ), where a0 = 0 and a1 ≠ 0. For every n ∈ ℕ, we write (n) (n) 2 (n) 3 f n (z) = a(n) 0 + a1 z + a2 z + a3 z + ⋅ ⋅ ⋅ ,
as in (1.8), and write f [n] (z) = f1[n] z + f2[n] z 2 + ⋅ ⋅ ⋅ + fk[n] z k + ⋅ ⋅ ⋅ , for all k ∈ ℕ. where f [1] = f = f (1) , or fk[1] = ak = a(1) k It is clear that f1[n] = an1 for all n ∈ ℕ. Let k ≥ 2 be given. For n ≥ 2, if a1 ≠ 1, we have n−1
n−1−j [j] Pk ,
fk[n] = ∑ (ak1 ) j=0
(1.12)
where Pk[0] = ak and Pk = f1 a(1) + f2 a(2) + ⋅ ⋅ ⋅ + fk−1 a(k−1) , k k k [j]
[j]
[j]
[j]
1 ≤ j ≤ n − 1.
(1.13)
If a1 = 1, n−1
fk[n] = ∑ Pk .
(1.14)
[j]
j=0
Proof. Suppose that a1 ≠ 1. By Definition 1.5.2, a1 [ [0 [ [ D(f ) = [ [0 [ [0 [ .. [.
a2
a(2) 2
f1[n−1] [ [ 0 [ [ D(f [n−1] ) = [ [ 0 [ [ 0 [ .. [ .
0 0 .. .
a3
a(2) 3 a(3) 3 0 .. .
a4
a(2) 4 a(3) 4 a(4) 4 .. .
... ] . . .] ] ] . . .] ], ] . . .] ] .. .]
f2[n−1]
f3[n−1]
f4[n−1]
(f2[n−1] )(2)
(f3[n−1] )(2) (f3[n−1] )(3)
(f4[n−1] )(2) (f4[n−1] )(3) (f4[n−1] )(4)
0
0 .. .
0 .. .
.. .
... ] . . .] ] ] . . .] ]. ] . . .] ] .. .]
1.5 Almost unit formal power series  33
j
Since D(f [n] ) = D(f [n−1] ) ⋅ D(f ) or f [n] = f [n−1] ∘ f , and f1 = a1 for every j ∈ ℕ, considering a(k) = ak1 and multiplying the first row of D(f [n−1] ) with the kth column of k D(f ), we have [j]
[n−1] (k−1) fk[n] = f1[n−1] a(1) + f2[n−1] a(2) + ⋅ ⋅ ⋅ + fk−1 aK + fk[n−1] a(k) , k k k
for every integer n ≥ 2. Equation (1.12) is true for n = 2 because a(k) = ak1 and k fk[2] = a1 a(1) + a2 a(2) + ⋅ ⋅ ⋅ + ak a(k) = Pk[1] + (ak1 )Pk[0] . k k k We now assume that (1.12) is true for fk[m] , 2 ≤ m ≤ n − 1 for some n ∈ ℕ, then [n−1] (k−1) fk[n] = f1[n−1] a(1) + f2[n−1] a(2) + ⋅ ⋅ ⋅ + fk−1 aK + fk[n−1] a(k) k k k [n−1] (k−1) = (f1[n−1] ak + f2[n−1] a(2) + ⋅ ⋅ ⋅ + fk−1 ak ) + fk[n−1] ak1 k
n−2 [0] Pk ](ak1 )
= Pk[n−1] + [Pk[n−2] + (ak1 )Pk[n−3] + ⋅ ⋅ ⋅ + (ak1 ) n−1
n−1−j [j] Pk .
= ∑ (ak1 ) j=0
For example, if n = 3, (1.12) provides [2] (k−1) fk[3] = f1[2] a(1) + f2[2] a(2) + ⋅ ⋅ ⋅ + fk−1 ak + fk[2] a(k) k k k 1
= Pk[2] + fk[2] (a1 )k = Pk[2] + (ak1 ) (Pk[1] + ak1 Pk[0] ) 2
2
= Pk[2] + (ak1 )Pk[1] + (ak1 ) Pk[0] = Pk[2] + (ak1 )Pk[1] + (ak1 ) ak . j
It is clear that P2 = a1 a2 , 1 ≤ j ≤ n − 1. Put the values of P2 into (1.12), we obtain the formulas [j]
[j]
2 n−1 f2[n] = an−1 1 a2 (1 + a1 + a1 + ⋅ ⋅ ⋅ + a1 ).
With a little more work, one may have n−1
i
n−2
i
2 n−1 2 2 2 n−2−i f3[n] = an−1 ). 1 a3 ∑ (a1 ) + 2a1 a2 ∑ (a1 ) (1 + a1 + a1 + ⋅ ⋅ ⋅ + a1 i=0
i=0
We would like to indicate that the algorithm of Proposition 1.5.19 actually is followed from D(f [n] ) = D(f [n−1] ) ⋅ D(f ) repeatedly.
34  1 Basics of formal power series Definition 1.5.20. Let f ∈ 𝔻(ℂ) be given and let m ∈ ℕ be given. g ∈ 𝔻(ℂ) is called the nth iterative root or iterated root of f if g [m] = f . It is denoted by g = f [1/m] . Some results about the iterative roots of delta series can be found in early 1980s [92], shortly after Niven [61] systematically introduced the space of formal power series. Some recent developments can be found in [75] and [78]. More details will be seen in Chapter 8. It is clear that the delta associated matrix of almost unit formal power series plays a significant role in computing the composition of almost units. Based on Theorem 1.4.8, we have the following theorem. Theorem 1.5.21. Let f be an almost unit in 𝔻(ℂ) and let n ∈ ℕ be given. Then the long (j) operations of computing D(f )n , or computing ai , 1 ≤ i ≤ n, i ≤ j ≤ n, is n3 − n . 6 Proof. Write f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅, where a0 = 0, a1 ≠ 0. Let A(f ) be the matrix representation of f with multiplication. Then An+1 = (A(f ))n+1 a0 [0 [ [ [0 [ =[ 0 [ [. [. [. [0
a1 a0 0 0 .. . 0
a2 a1 a0 0 .. . 0
a3 a2 a1 a0 .. . 0
... ... ... ... .. . ...
0 an [ ] an−1 ] [0 ] [ [ an−2 ] ] [0 = [0 an−3 ] ] [ [ .. ] ] [. . ] [ .. 0 ] [0
a1 0 0 0 .. . 0
a2 a1 0 0 .. . 0
a3 a2 a1 0 .. . 0
... ... ... ... .. . ...
an an−1 ] ] ] an−2 ] ] . an−3 ] ] ] .. ] . ] 0 ]
If we only compute those nonzero entries, we have T
T
0 0 0 [a(1) ] [ [ ] [ 1 ] [0 [ 0 ] [ ] [ [ (2) ] [ (1) ] [ [a ] [ 2 ] = [a2 ] [0 [ . ] [. [ . ] [ . ] [ .. [ . ] [ . ] [ [ . ] (2) (1) a [an ] [0 [ n ]
a1 0 0 .. . 0
a2 a1 0 .. . 0
... ... ... .. . ...
an ] an−1 ] ] an−2 ] ] .. ] . ] ] 0 ](n+1)×(n+1)
and then T
T
T
a(k) 0 0 [ 0. ] [ . ] [ . ] 0 [ . ] [ .. ] [ .. ] [ [ . ] [ ] [ ] [0 [ (k) ] [ ] [ ] [ [a ] [ ] [ ] [ 0 0 [ k−1 ] ] [ ] 0 = [ (k) ] = [ (k) ] (k−1) ] [ [ [ [a ] a a .. [ ] [ ] [ [ k ] [ k. ] [ k−1 ] [ [ . ] [. . [ ] [ ] [ . ] [ .. ] [ .. ] 0 [ . ] [ (k) (k−1) (k) [an ] [an ] [ an ]
a1 0 0 .. . 0
a2 a1 0 .. . 0
... ... ... .. . ...
an ] an−1 ] ] an−2 ] ] .. ] . ] ] 0 ](n+1)×(n+1)
1.6 Algebraic structure of 𝕏
 35
where 2 ≤ k ≤ n. The number of nonzero multiplications of computing [0, . . . , 0, a(k) , k a(k) , . . . , a(k) n ] is k+1
1 1 + 2 + 3 + ⋅ ⋅ ⋅ + (n − (k − 1)) = (n − k + 1)(n − k + 2). 2 Then the long operations of computing D(f )n , denoted by OPn , is n 1 OPn = ∑ (n − k + 1)(n − k + 2). 2 k=2
Using substitution m = k − 1, we have OPn = =
1 n−1 ∑ (n − m)(n − m + 1) 2 m=1
1 n−1 2 1 n−1 2n + 1 n−1 ∑ (n + n) + ∑ m2 − ∑m 2 m=1 2 m=1 2 m=1
1 1 (n − 1)n(2n − 1) (2n + 1)(n − 1)n = (n3 − n) + − 2 2 6 4 n3 − n . = 6
1.6 Algebraic structure of 𝕏 We have seen that the space 𝕏(S) is a set endowed with the addition, scalar multiplication, multiplication or Cauchy product, and the composition. With the linear operation and the multiplication 𝕏(S) is a ring if S is a ring. We know that 𝕏(S) is an integral domain if S is a field [42]. In such a ring, every element of 𝕏(S) has a matrix representation and the mapping between a formal power series and its matrix representation is an isomorphism by Theorem 1.4.2. As an operation, the composition on 𝕏(S) has different properties. The space 𝔻(ℂ) of all almost units with the composition is a group. Every almost unit formal power series has its delta associated matrix. The mapping from an almost units to its delta associate matrix is an isomorphism with respect to the composition. The next theorem is an extension of Proposition 1.1.10. Theorem 1.6.1. The set 𝕏(S) is an integral domain if and only if S is an integral domain. Proof. Suppose that 𝕏(S) is an integral domain. Let a, b ∈ S be given with ab = 0. Then the Cauchy product of the formal power series f = (a, 0, 0, . . . ) and g = (b, 0, 0, . . . ) is fg = (ab, 0, 0, . . . ) = (0, 0, 0, . . . ). Then f = 0 or g = 0, which means that a = 0 or b = 0. Thus, S is an integral domain.
36  1 Basics of formal power series Now we suppose that S is an integral domain. Assume that there are f and g in 𝕏(S) such that f ≠ 0 ≠ g
but fg = 0.
It is impossible if f or g is a constant formal power series. So we suppose that both f and g are not constant formal power series. Then ord(f ) ∈ ℕ, ord(g) ∈ ℕ, and hence ord(f ) + ord(g) ∈ ℕ but
ord(fg) = ∞.
It is a contradiction to (ii) of Proposition 1.1.5. Thus, 𝕏(S) is an integral domain. Example 1.6.2. Let S = ℤ, the set of all integers, then 𝕏(ℤ) is an integral domain because of Theorem 1.6.1 and that ℤ is an integral domain. We know that ℤ is not a field. The next theorem could be interesting in coding theory. ∞ Theorem 1.6.3. Let (an )∞ n=0 and (bn )n=0 be two nonzero sequences of integers. Then n
∑ ak bn−k ≠ 0
k=0
for some nonnegative integer n. Proof. Assume the conclusion is not true. Then n
cn = ∑ ak bn−k = 0 k=0
for every nonnegative integer n. Write f (x) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅
and
g(x) = b0 + b1 x + b2 x 2 + ⋅ ⋅ ⋅ .
Then f and g are two nonzero formal power series in 𝕏(ℤ), and ∞
g ⋅ f (z) = ∑ cn z n = 0. n=0
So either f = 0 or g = 0 by Theorem 1.6.1 because ℤ is an integral domain. It is a ∞ contradiction to that both (an )∞ n=0 and (bn )n=0 are nonzero sequences in ℤ. We complete the proof. Applying this result to linear algebra, we have the following.
1.6 Algebraic structure of 𝕏
 37
Theorem 1.6.4. Let n ∈ ℕ be given and let (ak )nk=1 ⊆ ℤ be given. Then the equation system aj x1 + aj−1 x2 + ⋅ ⋅ ⋅ + a1 xj = 0,
j = 1, 2, . . . , n
has nontrivial solution x = (x1 , x2 , . . . , xn ) if and only if aj = 0 for 1 ≤ j ≤ n. Proof. Consider the formal power series f (z) = a1 + a2 z + a3 z 2 + ⋅ ⋅ ⋅ ,
g(z) = x1 + x2 z + x3 z 2 + ⋅ ⋅ ⋅ ,
in 𝕏(ℤ), where aj = xj = 0 if j > n. By Theorem 1.6.1, 𝕏(ℤ) is an integral domain. Since fg = 0, it follows that either f = 0 or g = 0. We have the conclusion. In addition to have a matrix representation, a formal power series may have a function representation too. This is different from that an analytic function has a power series expansion. The next example may show us how and why a formal power series may have a function representation. Example 1.6.5. Let f (z) = 1 + z + z 2 + z 3 + ⋅ ⋅ ⋅ be a formal power series in 𝕏(ℂ). By (i) of Example 1.1.9, we know that f −1 = 1 − z. then the function representation of f is f (z) = 1 + z + z 2 + z 3 + ⋅ ⋅ ⋅ =
1 . 1−z
Let us now provide a very brief introduction to the formal power series from ℕp to the ring S, where p ∈ ℕ and p ≥ 2. We choose S = ℂ for our convenience. Definition 1.6.6. For any p ∈ ℕ, the lexicographic order of ℕp is the order defined by (n1 , n2 , . . . , np ) > (m1 , m2 , . . . , mp ) if ni > mi , where i = min{j : nj ≠ mj }. We use the symbol n to represent the ptuple (n1 , n2 , . . . , np ) ∈ (ℕ∪{0})p . If ∑ni=1 ni = k, we call n an index of degree k. We define p
𝒩k = the set of all index in ℕ of degree k.
Sometimes we call n = ∑ni=1 ni the total degree of n, and then we may denote p
𝒩k = {n ∈ (ℕ ∪ {0}) : n = k}.
Definition 1.6.7. Let p ∈ ℕ be such that p ≥ 2. Write X n for the monomial n
n
n
X n = X1 1 X2 2 . . . Xp p ,
38  1 Basics of formal power series where X1 , X2 , . . . , Xp are indeterminates or variables. A formal power series F from ℕp to ℂ is defined to be ∞
F(X) = ∑ ∑{λn X n : n ∈ 𝒩k }, k=0
where λn ∈ ℂ. λn X n is called a term of degree k if n ∈ 𝒩k . We use ℂ⟦X1 , X2 , . . . , Xp ⟧ to represent the set of all formal power series over ℂ in p variables. It is clear that for every k ∈ ℕ ∪ {0}, the kth term ∑{λn X n : n ∈ 𝒩k } in the above definition is a finite linear combination of all monomials n
n
n
λn X n = λn X1 1 X2 2 ⋅ ⋅ ⋅ Xp p
for every n = (n1 , n2 , . . . , np ) ∈ 𝒩k .
For example, F(X) = X1 + 2X2 +
∑
n≥(2,0,...,0)
Xn
is a formal power series in ℂ⟦X1 , X2 , . . . , Xp ⟧ where n ∈ (ℕ ∪ {0})p . Schiling investigated the maximally perfect fields and fields of the formal power series in several variables more than 80 years ago [90]. In 1988, Strouse introduced how to embed the algebra of formal power series in several variables into a Banach algebra [96]. Four years later, Reich provided another investigation in such a topic [74]. The composition of such formal power series could be found in [12].
1.7 Analytic structure of 𝕏 The composition g ∘ f actually consists of the formal power series g and the sequence {f n }∞ n=0 of infinitely many formal power series. Since we invented the composition of the formal power series, we actually have been facing the analytic structure such as g0 + g1 f + g2 f 2 + ⋅ ⋅ ⋅ + gn f n + ⋅ ⋅ ⋅ ,
(1.15)
where g(z) = g0 + g1 z + g2 z 2 + ⋅ ⋅ ⋅ We introduced socalled admitting addition in Section 1.2 in order to treat such analytic structure, especially to treat the composition of formal power series. We called expression (1.15) the admissible sum in Section 1.2, and then we successfully set up the composition of any formal power series with a nonunit. We revisit it now and introduce the following definition [61].
1.7 Analytic structure of 𝕏
 39
Definition 1.7.1. Let {fn }∞ n=1 ⊆ 𝕏(S) be a sequence of formal power series over a ring S and write fn (z) = an,0 + an,1 z + an,2 z 2 + ⋅ ⋅ ⋅ + an,k z k + ⋅ ⋅ ⋅ . The sequence {fn }∞ n=1 is said to be a sequence admitting addition if for any nonnegative integer k there exists an integer N = N(k) such that an,0 = an,1 = an,2 = ⋅ ⋅ ⋅ = an,k = 0 for all n ≥ N. If {fn }∞ n=1 ⊆ 𝕏(S) is a sequence admitting addition, the expression a0 f0 + a1 f1 + ⋅ ⋅ ⋅ + an fn + ⋅ ⋅ ⋅ is called an admissible sum, where ak , k ∈ ℕ ∪ {0}, is the sum of the kth coefficient of all fn . Proposition 1.7.2. Let {fn }∞ n=1 ⊆ 𝕏(S) be a sequence of formal power series over a ring S and write fn (z) = an,0 + an,1 z + an,2 z 2 + ⋅ ⋅ ⋅ + an,k z k + ⋅ ⋅ ⋅ . k ∞ If {fn }∞ n=1 is a sequence admitting addition, then {fn }n=1 is also a sequence admitting addition for all k ∈ ℕ.
Proof. Let k ∈ ℕ be given. For n ∈ ℕ, write (k) (k) 2 (k) m fnk (z) = a(k) n,0 + an,1 z + an,2 z + ⋅ ⋅ ⋅ + an,m z + ⋅ ⋅ ⋅ .
Since {fn }∞ n=1 is a sequence admitting addition, for any r ≥ 0 there is an integer N = N(r) such that an,0 = an,1 = an,2 = ⋅ ⋅ ⋅ = an,r = 0
for
n ≥ N.
for
n ≥ N,
Then, for every k ∈ ℕ, (k) (k) (k) a(k) n,0 = an,1 = an,2 = ⋅ ⋅ ⋅ = an,r = 0
k ∞ because a(k) n,r is determined by an,0 , an,1 . . . , an,r . Thus, {fn }n=1 is also a sequence admitting addition for all k ∈ ℕ.
Proposition 1.7.3. Let {fn }∞ n=1 ⊆ 𝕏(S) be a sequence of formal power series over a ring S and write fn (z) = an,0 + an,1 z + an,2 z 2 + ⋅ ⋅ ⋅ + an,k z k + ⋅ ⋅ ⋅ .
40  1 Basics of formal power series If {fn }∞ n=1 is a sequence admitting addition, then f = a0 f0 + a1 f1 + ⋅ ⋅ ⋅ + an fn + ⋅ ⋅ ⋅ is a formal power series in 𝕏(S). If we write f (z) = c0 + c1 z + c2 z 2 + ⋅ ⋅ ⋅, that is, ck ∈ S,
k = 0, 1, 2, . . . .
Proof. Let k ∈ ℕ ∪ {0} be given. The kth coefficient of fn is an,k for all n. Since the sequence {fn } is a sequence admitting addition, there exists an integer N = N(k) such that an,k = 0 if n ≥ N, or there are only finitely many nonzero kth coefficients of all fn , or N
ck = ∑ an,k ∈ S. n=0
Example 1.7.4. Let f (z) = a1 z + a2 z 2 + ⋅ ⋅ ⋅ be a power series in 𝔻(S). For any n ∈ ℕ, we write fn = f n , the nth power of f and write an,k = a(n) , k = 0, 1, 2, . . . , and an,k = 0 if k k < n. That is, for any k ∈ ℕ ∪ {0}, we have N = N(k) = k + 1 such that for all n ≥ N, an,0 = an,1 = an,2 = ⋅ ⋅ ⋅ = an,k = 0. Then the sequence {fn }, or {f n }, is a sequence admitting addition and the expression g0 + g1 f + g2 f 2 + ⋅ ⋅ ⋅ + gn f n + ⋅ ⋅ ⋅ is an admissible sum for any formal power series g(z) = g0 + g1 z + g2 z 2 + ⋅ ⋅ ⋅. The study of the algebraic structures, including Section 1.1 through Section 1.6 and the admitting addition, is called the formal mathematics by Roman in [83]. He says “By this we mean the limiting processes, such as the convergence of infinite series, play no role.” The core of the definition of admitting addition is to set up a finite set of the kth coefficients of all fn , and then we may avoid the convergence of infinite series, which requires many conditions including the metric or topology. Do we have to avoid it? A slightly more analytical sufficient condition for the admissible sum of a sequence of formal power series is using the limit of the integer sequence {ord(fn )} (p. 4, [83]). We prove it below. Proposition 1.7.5. Let {fn }∞ n=1 ⊆ 𝕏(S) be a sequence of formal power series over a ring S and write fn (z) = an,0 + an,1 z + an,2 z 2 + ⋅ ⋅ ⋅ + an,k z k + ⋅ ⋅ ⋅ . If limn→∞ ord(fn ) = ∞, then {fn }∞ n=1 is a sequence admitting addition.
1.7 Analytic structure of 𝕏

41
Proof. Let k be any nonnegative integer. Since limn→∞ ord(fn ) = ∞, there exists N = N(k) ∈ ℕ such that N > k and n ≥ N implies that ord(fn ) > N. Then, if n ≥ N, an,0 = an,1 = ⋅ ⋅ ⋅ = an,k = ⋅ ⋅ ⋅ = aN,k = 0, which shows that the sequence {fn }∞ n=1 is a sequence admitting addition. Instead of using the concept admitting addition, Henrici used socalled distributive laws of composition [42] to ensure the existence of the composition of a formal power series with a nonunit. If S is a field, such as ℂ or ℝ, with certain metric structure, we may extend the finite addition and admitting addition to more general summation under metric convergence, and maintain all existing properties of formal power series. Consequently, we must face the study of formal power series involving certain convergences. Such a study opens a door for the formal analysis. A formal power series could be considered as a regular power series without assigning any value to the variable in the power series. If a formal power series over a field such as ℂ or ℝ does have a positive radius of convergence, should we consider it as a formal power series or a regular power series? Is there a gap or any conflict when we treat it as a regular power series or as a formal power series? We have to say that there is certain gaps if we do not expand the structure of the space of formal power series. The formal analysis should not have such a gap of conflict. Remark 1.7.6. Let f be a formal power series over ℂ or ℝ. As a regular power series, if f has a positive radius of convergence, then as a formal power series, f should maintain all properties that f has in analysis. We call this the principle of formal analysis. We would like to indicate that this principle actually has been used implicitly so far. For example, the formal power series represented by √1 − 4x2 exists because the Maclaurin series of (1 − 4x 2 )1/2 converges for x < 1/2 [47]. We will know that the formal binomial series Ba is derived from a formal differential equation which will appear in the next chapter, and the formal power series (1−4x2 )1/2 will be the composition B1/2 (−4x2 ). More general, (1 − 4x 2 )1/2 will be the exponent of 1/2 of the formal power series 1 − 4x2 that will be discussed deeply in Chapter 9. As P. Henrici indicated that the significant difference between the regular power series and formal power series is that no value is assigned to the symbol x (p. 9, [42]). We understand such distinction as well as we maintain the smooth connection between these two kinds of power series when a formal power series also represents a regular one with a positive radius of convergence. We start this journey by first extending the
42  1 Basics of formal power series admitting addition to the general summation for a sequence of formal power series with a condition that S is a field with a metric, such as ℝ or ℂ. Definition 1.7.7. Let {fn }∞ n=1 ⊆ 𝕏(S) be a sequence of formal power series over a field S with a metric ρ, and write fn (z) = an,0 + an,1 z + an,2 z 2 + ⋅ ⋅ ⋅ + an,k z k + ⋅ ⋅ ⋅ , for all n ∈ ℕ. Under the usual convergence of the metric space (S, ρ), suppose that m
∞
ck = ∑ an,k = lim ∑ an,k ∈ S, m→∞
n=1
n=1
k = 0, 1, 2, 3, . . . ,
(1.16)
we define the series ∑∞ n=1 fn to be the formal power series f ∈ 𝕏(S) where f (z) = c0 + c1 z + c2 z 2 + c3 z 3 + ⋅ ⋅ ⋅ .
(1.17)
∞ If any of such ck does not exist, we say that ∑∞ n=1 fn does not make sense, or ∑n=1 fn is not a formal power series. If (1.16) holds, the sequence {fn } is said to be formally summable over S. The se∞ ries ∑∞ n=1 fn is called a formal series. If (1.16) holds, the series ∑n=1 fn is also said to be formally summable, and f in (1.16) is called the formal sum of the series.
The element an,k in (1.16) is the kth coefficient of z k of all fn (z), n ∈ ℕ, and ck is the sum of all those kth coefficients, in the sense of certain convergence in the metric space (S, ρ). Remark 1.7.8. Let S be a field with a metric such as ℝ or ℂ. (i) If {fn }∞ n=1 ⊆ 𝕏(S) is a sequence of formal power series admitting addition, then the summation in (1.16) is a finite summation and therefore ck ∈ S for all k. (ii) If f ∈ 𝕏0 (S), then {f n } is a sequence admitting addition and hence g ∘ f ∈ 𝕏(S) for any g ∈ 𝕏(S). The sequence {f n } is formally summable. (iii) In analysis, rearrangement of a series requires some conditions (see p. 445, [95]). A formal power series does not assign any value to its variable, although the series of formal power series in (1.17) looks like the rearrangement ∞
∞
∞
∞
∑ ( ∑ an,k z k ) = ∑ ( ∑ an,k )z k ,
n=1 k=0
k=0 n=1
actually there is only one convergence is required, that is, ∞
∑ an,k ∈ S,
n=1
k = 0, 1, 2, 3, . . . .
∞ (iv) If S is a ring and {fn }∞ n=1 ⊆ 𝕏(S), the existence of ∑n=1 fn returns to whether the condition of admitting addition is satisfied.
1.7 Analytic structure of 𝕏

43
We would like to introduce a theorem (Theorem 3.4, [56]), which shows that the principle of formal power series holds for composition of formal power series. Theorem 1.7.9. Let f ∈ 𝕏(ℂ), h ∈ 𝕏0 (ℂ) be given and denoted by ∞
f (z) = ∑ an z n
and
n=0
∞
h(z) = ∑ bn z n . n=1
Suppose that f is absolutely convergent for z ≤ r with r > 0, and ∞
∑ bn sn ≤ r
n=0
for some s > 0. Let g = f (h) be the formal power series obtained by composition. Then g converges absolutely for z ≤ s, and g(z) = f (h(z)) for
z ≤ s.
Theorem 1.7.10. Let f ∈ 𝕏 = 𝕏(ℂ) be given such that f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ + an z n + ⋅ ⋅ ⋅ . n Let α = limn→∞ √a n . If 0 < α ∈ ℝ, then f is an analytic function on {z ∈ ℂ : z < R}, where R = 1/α, or R = ∞ if α = 0. Moreover, in this case the sum function f is the function representation of the formal power series a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ and
an =
f (n) (0) n!
for every
n ∈ ℕ ∪ {0},
(1.18)
where f (n) is the nth order derivative of the analytic function f . Proof. Since α ≥ 0, Cauchy’s root test for power series ensures that f is an analytic function on {z ∈ ℂ : z < R}. Formula (1.18) is just the nth coefficient of the Maclaurin series of f . These two theorems, both are sufficient condition theorems only, support the principle of formal power series. This principle can help us to find some particular formal power series. For example, the formal power series 1 + z + z2 + z3 + ⋅ ⋅ ⋅ has radius of convergence r = 1 and the sum function f (z) = (1 − z)−1 for all z < 1. Then f −1 (z) = 1 − z. It is the same result in Example 1.6.5.
44  1 Basics of formal power series Example 1.7.11. Let f ∈ 𝕏(ℂ) be f (z) = −z + z 2 − z 3 + ⋅ ⋅ ⋅ + (−1)n z n + ⋅ ⋅ ⋅ . Then f (z) = −z/(1 + z) for z < 1. Applying composition of analytic function to f , we have f ∘ f (z) =
−[−z/(1 + z)] = z = I𝔻 . 1 − z/(1 + z)
Then f [−1] = f . As the composition of formal power series, without considering the convergence, one may check the first several terms of f ∘ f and obtain the same result. However, theoretically the formal power series space is very different from the space of power series with positive radius of convergence although 𝕏 keeps the properties of those convergent series. For example, the composed formal power series such as h in Theorem 1.7.9 needs not to be absolutely convergent on some region, h needs only to be a nonunit. We should be very careful while we are using the Maclaurin series of analytic function to work on formal power series. Example 1.7.12. Considering the formal power series a a Ba (z) = 1 + ( )z + ( )z 2 + ⋅ ⋅ ⋅ 1 2 and the nonunit formal power series f (z) = z + 2!z 2 + 3!z 3 + ⋅ ⋅ ⋅ , then the composition Ba ∘ f (z) is a formal power series, which has no connection to Maclaurin series of any function on ℂ, that is, Ba ∘ f has no function representation. In fact, there exists a theorem for rearrangement of a formal power series admitting addition. ∞ Theorem 1.7.13. Let {ϕn }∞ n=1 ⊆ 𝕏(ℂ) be a sequence admitting addition. Let {ψn }n=1 ⊆ 𝕏(ℂ) be a rearrangement of ϕ s in the sense that given any j there exists a unique k such that ϕj = ψk . Then {ψn }∞ n=1 is also a sequence admitting addition, and
ϕ1 + ϕ2 + ϕ3 + ⋅ ⋅ ⋅ = ψ1 + ψ2 + ψ3 + ⋅ ⋅ ⋅ . Proof. Since {ϕn }∞ n=1 is a sequence admitting addition, we may write ∞
∞
j=1
n=0
∑ ϕj = ∑ an z n ∈ 𝕏(ℂ),
(an )∞ n=0 ⊂ ℂ.
Let m ≥ 0 be an integer. There exists N1 ∈ ℕ such that ϕ1 + ϕ2 + ϕ3 + ⋅ ⋅ ⋅ have the same mth coefficient am .
and
ϕ1 + ϕ2 + ϕ3 + ⋅ ⋅ ⋅ + ϕN1
1.7 Analytic structure of 𝕏
 45
By the condition, for any j, 1 ≤ j ≤ N1 , there exists a unique kj ∈ ℕ such that ϕj = ψkj . Then there exists N2 ∈ ℕ such that N
N
2 {ϕj }j=11 ⊆ {ψk }k=1 .
And then the mth coefficient of ψ1 + ψ2 + ψ3 + ⋅ ⋅ ⋅ + ψN2 is also am . This shows that the sequence {ψn }∞ n=1 is a sequence admitting addition, and ϕ1 + ϕ2 + ϕ3 + ⋅ ⋅ ⋅
and ψ1 + ψ2 + ψ3 + ⋅ ⋅ ⋅
have identical mth coefficient am . Thus, ϕ1 + ϕ2 + ϕ3 + ⋅ ⋅ ⋅ = ψ1 + ψ2 + ψ3 + ⋅ ⋅ ⋅ . An analogous result can be obtained for multiplication of formal power series. If 0 {ϕn }∞ n=1 ⊆ 𝕏 (ℂ) is a sequence admitting addition such that ϕn (x) = an,1 z + an,2 z 2 + an,3 z 3 + ⋅ ⋅ ⋅ , then we say that the related sequence {1 + ϕn }∞ n=1 ⊆ 𝕏(ℂ) is a sequence admitting multiplication. Furthermore, we write ∞
∞
n=1
j=1
∏(1 + ϕn ) = 1 + ∑ aj z j ,
(1.19)
where ar is the coefficient of z r in any finite product ∏nk=1 (1 + ϕk ) with sufficiently large n that ak,j = 0 for 1 ≤ j ≤ r if k > n. A result analogous to Theorem 1.7.13 can be proved as the following. Theorem 1.7.14. If {1 + ϕn }∞ n=1 is a sequence admitting multiplication, so is any rearrangement 1 + ψ1 , 1 + ψ2 , 1 + ψ3 , . . . , as in Theorem 1.7.13, and ∞
∞
n=1
n=1
∏(1 + ϕn ) = ∏(1 + ψn ). The details about the admitting multiplication can be found in [61]. The principle of formal analysis ensures a formal power series to maintain all mathematical properties if this series also has a positive radius of convergence on a field such as ℝ or ℂ. Certain formal power series, such as the multiplicative inverse f −1 or iterative inverse f [−1] , are usually generated by given formal power series with some recursive formulas. Those recursive formulas are generated by the coefficients of the
46  1 Basics of formal power series given formal series without considering the convergence of those given series at all. However, sometimes we may have to check whether those generated formal power series have positive radius of convergence or not. We will see such requests in the other chapters and we have to say that it is not a simple problem. Considering this situation, we would like to recommend the Cauchy’s method of majorants to treat such problems. k Suppose that a formal power series ∑∞ k=0 cx t over ℝ or ℂ is given and needs to be determined whether it has positive radius of convergence. The idea of Cauchy’s k method of majorants is to show that there exists a majorant series ∑∞ k=0 dk t , with ck  ≤ ∞ k dk (so the dk are all positive) and yet ∑k=0 dk t converges in t < ϵ for some ϵ > 0. This k implies the convergence of ∑∞ k=0 ck t by the comparison test. The details of Cauchy’s method of majorants can be found in most analysis books. The next lemma is available in [94]. The power series solution of the differential equation is a popular topic in ODE. Let us consider the IVP for ODE 𝜕x u(x) = f (x, u(x)),
u(x0 ) = u0 .
One can in principle determine u(x) with help of the Taylor series, 1 u(x) = u(x0 ) + 𝜕x u(x0 )(x − x0 ) + 𝜕x2 (x0 )(x − x0 )2 + ⋅ ⋅ ⋅ . 2 We say “in principle” because there is no guarantee that the series converges. Of course, the Cauchy–Kovalevskaya theorem provided some useful results in this problem. May we try in such a way: first finding the formal power series solution (Taylor series or not), and then verifying whether this formal power series convergent or not? If so, we may need the following lemma. For example, in order to find a powerseries solution of the ordinary differential equation x (t) = p(t)x (t) + q(t)x(t) + r(t) where p, q, and r are the analytic functions, ∞
p(t) = ∑ pk (t − t0 )k , k=0
∞
q(t) = ∑ qk (t − t0 )k , k=0
∞
r(t) = ∑ rk (t − t0 )k , k=0
converging at least in t − t0  < ϵ. We suppose ∞
x(t) = ∑ ck t k , k=0
and we obtain the formula (∗) below, a recursive formula for {ck }∞ k=0 .
1.7 Analytic structure of 𝕏

47
Lemma 1.7.15 (Majorant lemma). Suppose ∞
P(t) = ∑ Pk t k , k=0
∞
Q(t) = ∑ Qk t k , k=0
∞
and
R(t) = ∑ Rk t k
and
r(t) = ∑ rk t k
k=0
are majorants of ∞
p(t) = ∑ pk t k , k=0
∞
q(t) = ∑ qk t k , k=0
∞
k=0
respectively, meaning pk  ≤ Pk ,
qk  ≤ Qk ,
and
rk  ≤ Rk .
k Suppose c0 and c1 are given and ∑∞ k=0 ck t is defined by k
(k + 1)(k + 2)ck+2 = ( ∑ pj (k − j + 1)ck−j+1 + qj ck−j ) + rk , j=0
(∗)
k and similarly suppose d0 and d1 are given and ∑∞ k=0 dk t is defined by the analogous equation k
(k + 1)(k + 2)dk+2 = ( ∑ Pj (k − j + 1)dk−j+1 + Qj dk−j ) + Rk . j=0
(∗∗)
Finally, assume c0  ≤ d0 and c1  ≤ d1 . Then ∑ dk t k is a majorant for ∑ ck t k , meaning ck  ≤ dk for every k.
2 Calculus of formal power series This chapter introduces the formal derivatives of formal power series over a ring and the properties related to such derivatives. The important difference between a formal derivative of a formal power series and a regular derivative of a regular power series is that the formal derivative has no relation with the limit of difference quotient. The equations with formal derivatives are called the formal differential equations. The solutions of such formal differential equations are formal power series. The formal exponential series and formal binomial series are solutions of certain formal differential equations, which will be introduced in Section 2.1. The formal logarithm defined on 1 + f , f ∈ 𝕏0 (ℂ) is also introduced in this section. The ultrametric and ultrametric space are introduced in Section 2.3 as preparation for making 𝕏(ℂ) a topological space. As an umbral algebra, the formal power series space 𝕏(ℂ) is closely related to the dual space of ℂ[X] where ℂ[X] is the set of all polynomials over ℂ.
2.1 Formal derivatives of formal power series Definition 2.1.1. Let f (x) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ be a formal power series over a ring S, the formal derivative, or just derivative of f (x) is a formal power series denoted by f (x) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ = a1 + 2a2 x + 3a3 x 2 + ⋅ ⋅ ⋅ , that is, an = (n + 1)an+1 for n = 0, 1, 2, . . . . Let n ∈ ℕ be given. The nth order formal derivative of f is defined by
f (n) (z) = [f (n−1) ] (z). The 0th order formal derivative f (0) is defined by f (0) = f . The following proposition is obvious. Proposition 2.1.2. Let f and g be formal power series in 𝕏(S) and let r ∈ S be given. Then: (i) (f + g) = f + g , and (ii) (rf ) = rf . Proposition 2.1.3. Let S be a field of characteristic zero. Then f ∈ 𝕏(S) is a constant formal power series if and only if f = 0. Proof. If f is a constant series, then f = 0 is obvious. Write f (x) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅. Then f (x) = a1 + 2a2 z + 3a3 z 2 + ⋅ ⋅ ⋅ . https://doi.org/10.1515/9783110599459002
50  2 Calculus of formal power series Conversely, if f is a zero formal power series, (n + 1)an+1 = 0,
for
n = 0, 1, 2, . . . .
Since S is a field of characteristic zero and n + 1 ≠ 0, it follows that an = 0 for all n ∈ ℕ, then f = a0 I. Remark 2.1.4. The condition that S is a field of characteristic zero is necessary for Proposition 2.1.3. Here is an example. Let S = ℤ/2, then 2n = 0 for all n ∈ ℕ. Let f (x) = 1 + x2 + x4 + ⋅ ⋅ ⋅ + x 2n + ⋅ ⋅ ⋅ . Then f (x) = 2x + 4x3 + ⋅ ⋅ ⋅ + 2nx2n−1 + ⋅ ⋅ ⋅ = 0, but f is not a constant formal power series. Corollary 2.1.5. Let f and g be two formal power series in 𝕏(S) where S is a field of characteristic zero, then f =g
if and only if
f = g
and
f (0) = g(0).
Proof. One direction is obvious. We suppose that f = g and f (0) = g(0). Then (f −g) = f − g = 0, and hence f − g = cI for some c ∈ S by Proposition 2.1.3. But f (0) − g(0) = 0 implies that c = 0. Thus, f = g. This corollary will be used very often when S = ℝ or S = ℂ. The derivative of the Cauchy product of two formal power series satisfies the Product Rule, That is, if f and g are two formal power series, then (fg) = f g + fg . We prove this rule below by the operations for formal power series without using anything related to the limit of difference quotient. Proposition 2.1.6. Let f (x) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅
and
g(x) = b0 + b1 x + b2 x 2 + ⋅ ⋅ ⋅
be two formal power series over a ring S. Then (fg) = f g + fg . Proof. Write n
∞
f (x)g(x) = ∑ cn xn
where cn = ∑ ak bn−k ,
n=0
k=0
n = 0, 1, 2, . . . .
Then ∞
∞
n+1
n=0
n=0
k=0
(f (x)g(x)) = ∑ cn xn = ∑ (n + 1)( ∑ ak bn+1−k )xn ,
2.1 Formal derivatives of formal power series  51
and the nth coefficient of (fg) is n+1
cn = (n + 1)( ∑ ak bn+1−k ),
n = 0, 1, 2, . . . .
k=0
On the other hand, ∞
n
f (x)g(x) = ∑ ( ∑ (k + 1)ak+1 bn−k )xn , n=0 k=0 ∞
n
f (x)g (x) = ∑ ( ∑ ak (n − k + 1)bn−k+1 )x n . n=0 k=0
If we write ∞
f (x)g(x) + f (x)g (x) = ∑ dn x n , n=0
we have n
n
dn = ∑ (k + 1)ak+1 bn−k + ∑ ak (n − k + 1)bn−k+1 k=0
k=0
n+1
n
m=1
k=0
= ∑ mam bn+1−m + ∑ ak (n − k + 1)bn−k+1 n
= ( ∑ mam bn+1−m + (n + 1)an+1 b0 ) m=1
n
+ (a0 (n + 1)bn+1 + ∑ ak (n − k + 1)bn−k+1 ) k=1
n
= (n + 1)(a0 bn+1 + an+1 b0 ) + ∑ kak bn+1−k k=1
n
+ ∑ (n + 1 − k)ak bn−k+1 k=1
n+1
= (n + 1)( ∑ ak bn+1−k ) = cn k=0
for n = 0, 1, 2, . . . . Thus, (fg) = f g + fg . The next corollary is a straightforward continuation of Proposition 2.1.6. Corollary 2.1.7. Let n ∈ ℕ with n ≥ 2. For any f , g ∈ 𝕏, we have n n (fg)(n) = ∑ ( )f (k) g (n−k) , k k=0
52  2 Calculus of formal power series where f (k) is the kth order formal derivative of f and f (0) = f . This formula is called the formal Leibniz’s formula for higher order formal derivatives. The Power Rule of differentiation, (f n (x)) = nf n−1 (x) ⋅ f (x), is just an application of the Product Rule. The Chain Rule also works for the composition g ∘ f . There is a beautiful proof for the Chain Rule by Henrici using the Distributive Laws of Composition (p. 40, [42]). We use the nth partial sum of f , [f ]n (see Definition 1.2.11), to prove the Chain Rule below. n Theorem 2.1.8. Let g(x) = ∑∞ n=0 bn x be any formal power series in 𝕏(S), and let f (x) = ∞ n ∑n=1 an x be a nonunit over the ring S, then
(g ∘ f ) (x) = g (f (x))f (x). Proof. Since f is a nonunit, k
f k (x) = [f (x)] = a(k) xk + a(k) x k+1 + a(k) x k+2 + ⋅ ⋅ ⋅ , k k+1 k+2 and ord(f k ) ≥ k for all k ∈ ℕ. We write (g ∘ f ) (x) = c0 + c1 x + c2 x 2 + ⋅ ⋅ ⋅
and
g (f (x))f (x) = d0 + d1 x + d2 x 2 + ⋅ ⋅ ⋅ , and we show that cn = dn for all n ∈ ℕ ∪ {0}. Let n ∈ ℕ be given, then cn is determined only by [g ∘ f ]n+1 = b0 + b1 f + b2 f 2 + ⋅ ⋅ ⋅ + bn f n + bn+1 f n+1 because f m does not contribute to cn , the coefficient of the term x n , if m > n + 1. We denote this finite sum of {f j }n+1 j=0 by [g ∘ f ]n+1 . Also the coefficient dn is only determined by [g (f )]n f (x) = (b1 + 2b2 f + 3b3 f 2 + ⋅ ⋅ ⋅ + (n + 1)bn+1 f n )f (x). But [g ∘ f ]n+1 (x) = b1 f (x) + 2b2 f (x)f (x) + ⋅ ⋅ ⋅ + (n + 1)bn+1 f n (x)f (x) = (b1 + 2b2 f (x) + ⋅ ⋅ ⋅ + (n + 1)bn+1 f n (x))f (x)
= ([g ∘ f ]n (x))f (x). Thus, cj = dj , j = 0, 1, 2, . . . , n.
Recalling the admitting addition in Definition 1.7.1, we introduce the termbyterm differentiation property for the series of formal power series admitting addition below.
2.1 Formal derivatives of formal power series  53
Theorem 2.1.9. Let {fk }∞ k=1 ⊆ 𝕏(S) be a formal power series sequence admitting addition, then
∞
∞
( ∑ fk ) = ∑ fk . k=1
k=1
Proof. Write fk (z) = ak,0 + ak,1 z + ak,2 z 2 + ⋅ ⋅ ⋅ for all k ∈ ℕ. Since the sequence {fk } admitting addition ∞
ck = ∑ ak,n ∈ S n=0
because the summation actually is a finite sum, k = 0, 1, 2, . . . . Then ∞
f (z) = ∑ fk = c0 + c1 z + c2 z 2 + ⋅ ⋅ ⋅ + ck z k + ⋅ ⋅ ⋅ k=1
represents ∑∞ k=1 fk . Then ∞
( ∑ fk ) (z) = f (z) = c1 + 2c2 z + 3c3 z 2 + ⋅ ⋅ ⋅ + (k + 1)ck+1 z k + ⋅ ⋅ ⋅ . k=1
On the other hand, the sequence of all fk is also a sequence admitting addition and then ∑∞ k=1 fk = g for some g ∈ 𝕏(S). Write g(z) = d0 + d1 z + d2 z 2 + ⋅ ⋅ ⋅ + dk z k + ⋅ ⋅ ⋅ . It is clear that ∞
dk = (k + 1)( ∑ ak+1,n ) = (k + 1)ck+1 , n=0
which shows that the kth coefficient of g is equal to the kth coefficient of f for all k. Thus, ∞
∞
( ∑ fk ) = ∑ fk . k=1
k=1
This theorem can simplify the proof of the Chain Rule of Theorem 2.1.8.
54  2 Calculus of formal power series
2.2 Formal differential equations and their applications A formal differential equation is an equation in which one or more of the formal derivatives of one or more formal power series occur, the unknown formal power series themselves may also occur. The following are some examples of formal differential equations. Example 2.2.1. (i) f (x) = f (x) + 3x − 5x3 . (ii) g (x) − 4g(x) = 5x. The solution of formal differential equation is a formal power series. By solving certain formal differential equations, we may obtain some popular and useful formal power series. We know that the solution of the ordinary differential equation f (x) = f (x) is the set of all functions aex where a is a real number. A similar result can be obtained for the formal power series. Definition 2.2.2. Let S be a field. For any a ∈ S, the formal power series 1+
a2 2 a z+ z + ⋅⋅⋅ 1! 2!
is called a formal exponential (power) series and denoted by Ea . It is clear that E0 = I. Proposition 2.2.3. Let g be a formal power series over a field S such that g = ag for some a ∈ S, then g = b0 Ea , where b0 = g(0). Proof. Write g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ Then g (z) = b1 + 2b2 z + 3b3 z 2 + ⋅ ⋅ ⋅ + nbn z n−1 + ⋅ ⋅ ⋅ . Let g = ag, we have bn =
a a a an bn−1 = bn−2 = ⋅ ⋅ ⋅ = b n nn−1 n! 0
for every n ∈ ℕ. Then g(z) = b0 (1 +
a a2 an z+ + ⋅ ⋅ ⋅ + z n + ⋅ ⋅ ⋅) = b0 Ea (z). 1! 2! n!
It is easy to have that b0 = g(0).
2.2 Formal differential equations and their applications  55
Proposition 2.2.4. For arbitrary a and b in ℂ, (i) Ea = aEa ; (ii) Ea Eb = Ea+b ; (iii) Ea−1 = E−a . Proof. Since Ea (z) = 1 + 1!a z +
a 2 z 2!
+ ⋅⋅⋅ +
a n z n!
+ ⋅ ⋅ ⋅, it follows that
a a a + 2 z + 3 z2 + ⋅ ⋅ ⋅ 1 2! 3! a a a = a[1 + z + z 2 + z 3 + ⋅ ⋅ ⋅] = aEa (z). 1! 2! 3!
Ea (z) = 0 +
This is (i). Let F = Ea Eb − Ea+b . Using the product rule and (i), F = Ea Eb + Ea Eb − (a + b)Ea+b
= aEa Eb + bEa Eb − (a + b)Ea+b
= (a + b)F.
Then F = F(0)Ea+b by Proposition 2.2.3. Since F(0) = 0, it follows that F = 0, and then F = 0 by Corollary 2.1.5, or Ea Eb = Ea+b . This is (ii). (iii) is the result of Ea E−a = E0 = I. We introduce another important formal power series. Definition 2.2.5. The formal binomial series over a field S, denoted by Ba , is defined to be a a a Ba (x) = 1 + ( )x + ( )x2 + ⋅ ⋅ ⋅ + ( ) + ⋅ ⋅ ⋅ , 1 2 n where a ∈ S and a ( ) = 1, 0
a a(a − 1) . . . (a − n + 1) ( )= n n!
for all n ∈ ℕ. It is clear that B0 = I because (0n ) = 0 for all n ∈ ℕ, and hence B0 (x) = 1. The formal binomial series should not be considered as a transformation from the classical binomial series invented by I. Newton. Without using convergence, we create the formal binomial series by using the operations for formal power series only. Of course, if x < 1, Ba (x) has all properties for the classical binomial series. Proposition 2.2.6. Let f be a formal power series over a field S and let a ∈ ℝ be given. a If f = 1+x f , then f = a0 Ba
where
a0 = f (0).
56  2 Calculus of formal power series Proof. If a = 0, the conclusion is trivial. We suppose that a ≠ 0. Write f (z) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ and then let (1 + x)f (x) = af (x). We have (1 + x)(a1 + 2a2 x + 3a3 x2 + ⋅ ⋅ ⋅) = a(a0 + a1 x + a2 x 2 + ⋅ ⋅ ⋅). Then a1 + (a1 + 2a2 )x + (2a2 + 3a3 )x2 + ⋅ ⋅ ⋅ + [nan + (n + 1)an+1 ]x n + ⋅ ⋅ ⋅ = a(a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅).
Then nan + (n + 1)an+1 = aan for n = 0, 1, 2, . . . , and hence a−n a−n a−n+1 a =( )( )an−1 n+1 n n+1 n (a − n)(a − n + 1) . . . (a − 1)a a = ⋅⋅⋅ = a0 = ( )a . (n + 1)! n+1 0
an+1 =
Thus, f = a0 Ba where a0 = f (0). Proposition 2.2.7. For arbitrary a and b in ℂ, (i) Ba = aBa−1 ; (ii) Ba Bb = Ba+b ; (iii) B−1 a = B−a . Proof. For (i), we have a a a Ba (z) = 0 + ( ) + 2( )z + 3( )z 2 + ⋅ ⋅ ⋅ 1 2 3 a−1 a−1 2 a−1 3 = a[1 + ( )z + ( )z + ( )z + ⋅ ⋅ ⋅] 1 2 3 = aBa−1 .
This is (i). Let F = Ba Bb − Ba+b . By Proposition 2.2.6, we have Ba =
a B . 1+x a
Using this expression and the product rule of formal differentiation, F = Ba Bb + Ba Bb − Ba+b
a b a+b B B + B E − B 1 + x a b 1 + x a b 1 + x a+b a b = (B B − Ba+b ) + (B B − Ba+b ) 1+x a b 1+x a b a+b = F. 1+x
=
2.2 Formal differential equations and their applications  57
By Proposition 2.2.6 again, F = F(0)Ba+b . But F(0) = 0, and hence F = 0, or F is a constant formal power series. Using the fact that ℂ is a field of characteristic zero and that F(0) = 0, we have that F = 0 or Ba Bb = Ba+b . This is (ii). (iii) can be followed from the fact Ba B−a = B0 = I. Calculating Ba Bb explicitly, the nth coefficient of this product is a b a b a b ( )( ) + ( )( ) + ⋅ ⋅ ⋅ + ( )( ). 0 n 1 n−1 n 0 Since Ba Bb = Ba+b , we obtain the Vandemonde’s theorem: n a b a+b )=( ). ∑ ( )( k n − k n k=0
We would like to introduce the J. C. P. Miller formula, which is popular in combinatorics. Theorem 2.2.8. Let f ∈ 𝔻(ℂ) be such that f (z) = b1 z + b2 z 2 + ⋅ ⋅ ⋅ . Let Ba be the binomial series and write Ba ∘ f (z) = c0 + c1 z + c2 z 2 + ⋅ ⋅ ⋅ , then c0 = 1, and for all n ∈ ℕ, cn = =
1 n−1 ∑ [a(n − k) − k]ck bn−k n k=0
1 n ∑ [(a + 1)k − n]cn−k bk . n k=1
Proof. It is obvious that c0 = 1. By Proposition 2.2.7, Ba Bb = Ba+b for all a, b ∈ ℝ and Ba = aBa−1 . Then (Ba ∘ f ) = a(Ba−1 ∘ f )f . Multiplying by (B1 ∘ f ) and applying the right distributive law, we have (B1 ∘ f )(Ba ∘ f ) = a(B1 ∘ f )(Ba−1 ∘ f )f
= a((B1 Ba−1 ) ∘ f )f = a(Ba ∘ f )f .
58  2 Calculus of formal power series Since B1 (z) = 1 + z and (Ba ∘ f ) (z) = c1 + 2c2 z + 3c3 z 2 + ⋅ ⋅ ⋅ , the equality (B1 ∘ f )(Ba ∘ f ) = a(Ba ∘ f )f becomes (1 + b1 z + b2 z 2 + ⋅ ⋅ ⋅)(c1 + 2c2 z + 3c3 z 2 + ⋅ ⋅ ⋅)
= a(c0 + c1 z + c2 z 2 + ⋅ ⋅ ⋅)(b1 + 2b2 z + 3b3 z 2 + ⋅ ⋅ ⋅).
Equating the coefficients of the term z n−1 by using the formula of the Cauchy product, we have bn−1 c1 + 2bn−2 c2 + ⋅ ⋅ ⋅ + ncn = a(cn−1 b1 + 2cn−2 b2 + ⋅ ⋅ ⋅ + nc0 bn ). Then ncn = a(cn−1 b1 + 2cn−2 b2 + ⋅ ⋅ ⋅ + nc0 bn )
− bn−1 c1 − 2bn−2 c2 − ⋅ ⋅ ⋅ − (n − 1)b1 cn−1
= cn−1 b1 (a − (n − 1)) + cn−2 b2 (2a − (n − 2)) + ⋅ ⋅ ⋅ + c1 bn−1 ((n − 1)a − 1) + bn c0 (na − 0) n
= ∑ [k(a + 1) − n]cn−k bk . k=1
We have the second formula for cn . Taking that m = n − k in this second formula, we have cn = = =
1 n ∑ [k(a + 1) − n]cn−k bk n k=1
1 0 ∑ [a(n − m) − m]cm bn−m n m=n−1 1 n−1 ∑ [a(n − k) − k]ck bn−k . n k=0
This is the first formula for cn . The recurrence relation for cn is called the J. C. P. Miller formula. If programmed with care, it permits the computation of the coefficients c1 , c2 , . . . , cn in n2 multiplications and n − 1 divisions [42]. It is well known in calculus that the Maclaurin series 1 1 1 ln(1 + x) = x − x2 + x 3 − x 4 + ⋅ ⋅ ⋅ 2 3 4
has the radius of convergence R = 1. If we consider that g(x) = ln(1 + x) is the function representation of the above formal power series, then g ∘ f ∈ 𝕏(ℝ) for any nonunit f . This paves a way to introduce the formal logarithmic function or formal logarithm [61].
2.2 Formal differential equations and their applications  59
Definition 2.2.9. Let S be a ring. Define a subset of 𝕏(S) as 𝕏1 = {1 + f : f ∈ 𝕏0 (S)} = {f ∈ 𝕏(S) : f (0) = 1}. It is clear that the set 𝕏1 is closed for formal power series multiplication. For any F = 1 + f ∈ 𝕏1 , define 1 1 1 1 L(F) = f − f 2 + f 3 − f 4 + ⋅ ⋅ ⋅ + (−1)n+1 f n + ⋅ ⋅ ⋅ . 2 3 4 n
(2.1)
The mapping L is called a formal logarithmic function or formal logarithm. L is not defined for every element of 𝕏(S), but only for F ∈ 𝕏1 , and 1 1 1 L(F)(z) = f (z) − f 2 (z) + f 3 (z) − f 4 (z) + ⋅ ⋅ ⋅ 2 3 4 for z ∈ S is welldefined because f ∈ 𝕏0 , and L(F) ∈ 𝕏0 , too. The formal logarithm L has most properties of what the Maclaurin series ln(1 + x) has and beyond. Theorem 2.2.10. For all F ∈ 𝕏1 (ℂ),
(L(F)) = F −1 F . Proof. Let F ∈ 𝕏 be given and write F = 1 + f . Then 1 1 1 1 L(F) = f − f 2 + f 3 − f 4 + ⋅ ⋅ ⋅ + (−1)n+1 f n + ⋅ ⋅ ⋅ . 2 3 4 n By Theorem 2.1.9 of termbyterm differentiation and the fact that the sequence {f n } is a sequence admitting addition, (L(F)) = f − ff + f 2 f − f 3 f + ⋅ ⋅ ⋅ (−1)n f n f + ⋅ ⋅ ⋅
= f (1 + f )−1 = F F −1 , because f = F . Theorem 2.2.11. For any two formal power series F and G in 𝕏1 (ℂ), L(FG) = L(F) + L(G). Proof. It is clear that F, G ∈ 𝕏1 (ℂ) implies that FG ∈ 𝕏1 , we have
(L(FG)) = (FG)−1 (FG) = (FG)−1 (G ⋅ F + F ⋅ G )
= F −1 F + G−1 G = (L(F)) + (L(G))
= (L(F) + L(G)) .
60  2 Calculus of formal power series Both L(FG) and L(F) + L(G) are elements in 𝕏0 that yields L(FG)(0) = L(F)(0) + L(G)(0) = 0. Thus L(FG) = L(F) + L(G) by Corollary 2.1.5. Corollary 2.2.12. Let F ∈ 𝕏1 (ℂ) be given and let n ∈ ℕ. Then L(F n ) = nL(F). Proof. The condition F ∈ 𝕏1 (ℂ) implies that F n ∈ 𝕏1 (ℂ). Let G = F in the theorem above and inductively we have the conclusion. Theorem 2.2.11 can be extended to a sequence of formal power series admitting multiplication. 0 Corollary 2.2.13. Let {fn }∞ n=1 ⊆ 𝕏 (ℂ) be a sequence admitting addition. Then
∞
∞
(L(∏ Fn )) = ∑ (L(Fn )) , n=1
n=1
where Fn = 1 + fn , n ∈ ℕ. Proof. By equation (1.19), we write ∞
∞
∞
n=1
n=1
n=1
F = ∏ Fn = ∏(1 + fn ) = 1 + ∑ bn z n = 1 + f (z) for some {bn } ⊆ ℂ where f (z) = k ∈ ℕ,
n ∑∞ n=1 bn z
∈ 𝕏0 (ℂ). Then L(F) ∈ 𝕏0 (ℂ) and for every ∞
L(F) = L(Fk ) + L(∏ Fn ). n=k̸
Applying the Product Rule, for every k ∈ ℕ we have
∞
F = (F1 ⋅ (∏ Fn )) n>1
∞
∞ = F1 ∏ Fn n>1 ∞
∞
+ F1 (∏ Fn ) n>1
= F1 ∏ Fn + F1 (F2 ⋅ ∏ Fn ) n>1
=
∞ F1 ∏ Fn n>1
n>2
+
∞ F1 [F2 ∏ Fn n>2
∞
+ F2 (∏ Fn ) ]. n>2
∞
∞
∞
n=1̸
n=2̸
n>2
= F1 ∏ Fn + F2 ∏ Fn + F1 F2 (∏ Fn ) .
2.2 Formal differential equations and their applications  61
Using the mathematical induction, we may have k
∞
j=1
n=j̸
F = ∑ Fj (∏ Fn ) + (∏ Fj )(∏ Fn ) . j≤k
n>k
for every k ∈ ℕ. We claim that F = ∑∞ k=1 Fk (∏n=k̸ Fn ). ∞ n Since F(z) = 1 + ∑n=1 bn z , it follows that ∞
F (z) = ∑ nbn z n−1 , n=1
where bn is the coefficient of z n in any product ∏m j=1 (1 + fj ) with m sufficiently large. It suffices to show that nbn z n−1 is in ∑∞ F (∏ F ). n n=k̸ k=1 k Let n ∈ ℕ be given. Since {fn }∞ is a sequence admitting addition, it follows that n=1 the sequence {Fk }∞ is a sequence admitting multiplication. Then there is a rn ∈ ℕ k=1 m n large enough that bn z is in any finite product ∏j=1 Fj , m ≥ rn . Then nbn z rn
rn
n−1
is in (∏ Fj ) ,
but
j=1
(∏ Fj ) = j=1
rn F1 (∏ Fj ) j>1
+ F1 (∏ Fj ) j>1
rn
=
F1 (∏ Fj ) j>1
rn
+
rn (F1 F2 )(∏ Fj ) j>2
rn
+ (F1 F2 )(∏ Fj )
2
rn
2
rn
k=1
j=k̸
j=1
j>2
rn −1
rn
rn −1
k=1
j=k̸
j=1
rn
rn
k=1
j=k̸
j>2
= ∑ Fk (∏ Fj ) + (∏ Fj )(∏ Fj ) = ⋅ ⋅ ⋅ = ∑ Fk (∏ Fj ) + ( ∏ Fj )(Frn ) = ∑ Fk (∏ Fj ). Then nbn z n−1 is in ∑∞ k=1 Fk (∏n=k̸ Fn ). Thus, the claim is true. Then r
∞ n F ∑k=1 Fk ((∏j=k̸ Fj ) ∞ Fk = =∑ . ∞ F F ∏j=1 Fj k=1 k
We have
(L(F)) =
∞ ∞ F F = ∑ k = ∑ (L(Fk )) . F k=1 Fk k=1
62  2 Calculus of formal power series 0 Corollary 2.2.14. Let {fn }∞ n=1 ⊆ 𝕏 (ℂ) be a sequence admitting addition. Then ∞
∞
n=1
n=1
L(∏ Fn ) = ∑ L(Fn ), where Fn = 1 + fn , n ∈ ℕ. Proof. We first show that {L(Fn )}∞ n=1 is a sequence admitting addition. k Let gk = ∑∞ f , k ∈ ℕ, then gk ∈ 𝕏0 by Proposition 1.7.2 with ord(gk ) ≥ k for all n=1 n ∞ k ∈ ℕ, which ensures that {gk }n=1 is a sequence admitting addition. Then 1 1 1 g1 − g2 + g3 − g4 + ⋅ ⋅ ⋅ 2 3 4
is an element of 𝕏,
0 or ∑∞ n=1 L(Fn ) ∈ 𝕏 . By Theorem 2.1.9,
∞
∞
( ∑ L(Fn )) = ∑ (L(Fn )) . n=1
n=1
Using Corollary 2.2.13, we have
∞
∞
(L(∏ Fn )) = ( ∑ L(Fn )) . n=1
n=1
Since ∞
L(∏ Fn )(0) = L(1) = 0 n=1
and
∞
∞
n=1
n=1
∑ L(Fn )(0) = ∑ L(1) = 0,
the conclusion is followed. Theorem 2.2.15. Let f ∈ 𝕏0 (ℂ) be given. Then L(E1 (f )) = f where E1 is the formal exponential series Ea with a = 1. Proof. Since f ∈ 𝕏0 , E1 (f ) ∈ 𝕏1 and E1 (f ) = 1 + f +
f2 f3 + + ⋅⋅⋅. 2! 3!
Then (E1 (f )) = f E1 (f ). By Theorem 2.2.10 and the Chain Rule again,
−1
−1
(L(E1 (f )) = (E1 (f )) (E1 (f )) = (E1 (f )) E1 (f )f = f . Then L(E1 (f )) = f by Corollary 2.1.5 because both L(E1 (f )) and f are nonunits or L(E1 (f ))(0) = f (0) = 0.
2.2 Formal differential equations and their applications  63
If f ∈ 𝔻(ℂ), then f is a unit and then (f )−1 ∈ 𝕏(ℂ). Theorem 2.2.16. Let f be an almost unit in 𝔻(ℂ), then −1
(f [−1] ) = (f ∘ f [−1] ) . Proof. By definition, we have that f [−1] ∘ f (z) = z. Then
((f [−1] ) ∘ f )f = I𝔻 ,
and then
−1
(f [−1] ) ∘ f = 1/f = (f ) , because f is a unit. Applying composition with f [−1] in both sides, we have (f )
−1
∘ f [−1] = (f [−1] ) ∘ f ∘ f [−1] = (f [−1] ) .
If A is a unit and B is a nonunit, the right distributive law provides that A−1 ∘ B = (A ∘ B)−1 , which yields that −1
(f ∘ f [−1] )
= (f )
−1
∘ f [−1] = (f [−1] ) .
Corollary 2.2.17. Let f (z) = E1 (z) − 1 =
z z2 z3 + + + ⋅⋅⋅. 1! 2! 3!
Then f [−1] (z) = L(1 + z). Proof. Since f is an almost unit formal power series, it follows that its composition inverse f [−1] exists. Write q = f [−1] , then by Theorem 2.2.16 −1
= (f ∘ q)
−1
= [f ∘ f [−1] + 1]
q = ((E1 − 1) ∘ q)
−1
= (E1 ∘ q)−1 ,
because (E1 − 1) = E1 . Then q = [(E1 − 1) ∘ q + 1 ∘ q]
−1
= (I𝔻 + 1)−1 .
Then q (z) = (z + 1)−1 = 1 − z + z 2 − z 3 + ⋅ ⋅ ⋅ . Then f [−1] (z) =
z z2 z3 z4 − + − + ⋅ ⋅ ⋅ = L(1 + z). 1 2 3 4
64  2 Calculus of formal power series Remark 2.2.18. We have to recognize that there are some confusions after obtaining Theorem 2.2.15 and Corollary 2.2.13: (i) L ∘ E1 makes sense by considering L(E1 ) = L(1 + (E1 − 1)). (ii) Is there any reversion relationship between L and E1 such as L ∘ E1 = I𝔻 ? The next theorem partially answers such questions. Theorem 2.2.19. Let Ea , Ba , and L be the formal exponential series, formal binomial series and formal logarithmic series in 𝕏 = 𝕏(ℂ) and a ∈ ℂ. Then Ea ∘ L = Ba . Proof. For any F ∈ 𝕏1 , L(F) is a nonunit by (2.1). Then Ea ∘ L ∈ 𝕏. Define a formal power series P by P(x) = Ea ∘ L(1 + x). Then P(0) = Ea (0) = 1. Also,
P (x) = (Ea ∘ L(1 + x))(L(1 + x))
= (aEa ∘ L(1 + x))(L(1 + x))
= aP(x)(L(1 + x)) . Since L(1 + x) = 1 − x + x2 − x 3 + ⋅ ⋅ ⋅ = (1 + x)−1 , it follows that (1 + x)P = aP, which is exactly the formal differential equation in Proposition 2.2.6. Then Ea ∘ L = P = P(0)Ba = Ba . We introduced some algebraic structure or algebraic properties of 𝕏 in Chapter 1. This section has added some more properties to 𝕏 by help of the formal differentiation. More algebraic properties such as formal sine and formal cosine can be found in [42] and [61]. A termbyterm differentiation theorem was introduced in the end of Section 2.1 for a sequence of formal power series admitting addition. We provide a general termbyterm differentiation theorem below to close this section. Theorem 2.2.20. Let S be a field with a metric such as ℝ or ℂ. Let {fn }∞ n=1 ⊆ 𝕏(S) be a ∞ sequence of formal power series. If ∑n=1 fn = F ∈ 𝕏(S), then ∞
F = ∑ fn . n=1
2.3 Ultrametric  65
Proof. For every n ∈ ℕ, we write fn (z) = an,0 + an,1 z + an,2 z 2 + ⋅ ⋅ ⋅ + an,k z k + ⋅ ⋅ ⋅ . By Definition 1.7.7, we write F(z) = c0 + c1 z + c2 z 2 + ⋅ ⋅ ⋅ where ∞
ck = ∑ an,k ∈ S n=1
for every k ∈ ℕ ∪ {0}.
Then fn (z) = an,1 + 2an,2 z + ⋅ ⋅ ⋅ + kan,k z k−1 + ⋅ ⋅ ⋅ for all n ∈ ℕ, and then ∞
∞
n=1
n=1
∑ kan,k = k ∑ an,k = kck ∈ S
for k = 0, 1, 2, . . . .
By Definition 1.7.7, ∞
∑ fn ∈ 𝕏(S),
n=1
with kck as the (k − 1)th coefficient of z k−1 . It is clear that the coefficient of z k−1 of F is kck . The termbyterm differentiation rule is obtained. We must say that the termbyterm differentiation rule for the formal differentiation is very different from the classical differentiation.
2.3 Ultrametric Let us give a mathematical definition for the metric space now although we mentioned it a couple of times before. A metric space is a set X together with a metric d : X × X → ℝ such that for all x, y, and z in X: (i) d(x, y) ≥ 0, d(x, y) = 0 if and only if x = y, (ii) d(x, y) = d(y, x), and (iii) d(x, z) ≤ d(x, y) + d(y, z). Such a metric space is usually denoted by (X, d). Definition 2.3.1. Let X be a set. A function d : X × X → ℝ is called an ultrametric if d satisfies (i) and (ii) above and thestrong triangular inequality d(x, z) ≤ max{d(x, y), d(y, z)}. The metric space (X, d) is called an ultrametric space if d is an ultrametric
(2.2)
66  2 Calculus of formal power series It is clear that an ultrametric is a metric. The next proposition is also called isosceles triangle principle. Proposition 2.3.2. Let (X, d) be an ultrametric space and let x, y, z ∈ X be given. If d(x, y) ≠ d(y, z), then d(x, z) = max{d(x, y), d(y, z)}. Proof. This result is equivalent to that the largest and second largest of the numbers d(x, y), d(y, z), and d(x, z) are equal. Assume that d(x, y) ≤ d(y, z) ≤ d(x, z), we show that d(y, z) = d(x, z). Because d is an ultrametric, we have d(y, z) ≤ max{d(y, x), d(x, z)} = d(x, z). Also d(x, z) ≤ max{d(x, y), d(y, z)} = d(y, z). Thus, d(x, z) = d(y, z). Definition 2.3.3. Let (X, d) be an ultrametric space. For any real number r > 0 and a ∈ X, the open ball of radius r with center a is the set Br (a) = {x ∈ X : d(a, x) < r}. The closed ball of radius r with center a is the set Br (a) = {x ∈ X : d(a, x) ≤ r}. Similar to what we have done in a metric space, we define the diameter of a nonempty subset A ⊆ X to be d(A) = sup{d(x, y) : x, y ∈ X}. The distance between two nonempty subsets A and B of X is d(A, B) = inf{d(x, y) : x ∈ A, y ∈ B}. The distance between an element x ∈ X and a nonempty subset A ⊆ X is d(x, A) = d({x}, A). We now introduce some characters of ultrametric space and certain difference between the ultrametric and the usual metric. The details can be found in [89].
2.3 Ultrametric  67
Proposition 2.3.4. Let (X, d) be an ultrametric space. Each ball defined in Definition 2.3.3 is both open and closed. Each point of a ball may serve as a center. Proof. Let (X, d) be an ultrametric space. Let a ∈ X be given and let r be a positive real number. We first show that the formula d(x, y) < r defines an equivalent relation on X. It is obvious that d(x, x) = 0 < r, and d(x, y) < r implies d(y, x) < r. Suppose d(x, y) < r and d(y, z) < r, then d(x, z) = max{d(x, y), d(y, z)} < r. We obtain all three: reflexivity, symmetry, and transitivity. The equivalence classes of this equivalence relation are open. Then Br (a) is the complement of a union of classes, and hence is closed. We now show that Br (a) is open. It suffices to show that Br (b) ⊆ Br (a) for every b ∈ Br (a). Let b ∈ Br (a) be given. For any x ∈ Br (b), d(x, a) ≤ max{d(x, b), d(b, a)} ≤ r, then x ∈ Br (a), and hence Br (b) ⊆ Br (a). Thus, a ball in a ultrametric space is both open and closed. Sometimes we call such a ball clopen. From now on, we will use the notation Br (a) only if (X, d) is an ultrametric space. Next, if b ∈ Br (a) be given, d(a, b) ≤ max{d(a, x), d(x, b)} ≤ r, and then a ∈ Br (b) which, by symmetry, shows that every point of Br (a) could be the center of Br (a). Proposition 2.3.5. Let (X, d) be an ultrametric space and let B1 , B2 be two balls in X. Then either B1 and B2 are ordered by inclusion (B1 ⊆ B2 or B2 ⊆ B1 ), or B1 ∩ B2 = 0. In the latter case, for all x ∈ B1 , y ∈ B2 , d(x, y) = d(B1 , B2 ). Proof. If B1 ∩ B2 = 0, it is obvious that neither B1 ⊆ B2 nor B2 ⊆ B1 . Now suppose that B1 ∩ B2 ≠ 0. Pick a ∈ B1 ∩ B2 . If B1 ⊆ B2 or B2 ⊆ B1 , we are done. Otherwise, we pick an x ∈ B1 \ B2 , y ∈ B2 \ B1 . By Proposition 2.3.4, the element a could be the center of both B1 and B2 and then we have that: (1) d(y, a) > d(x, a) because x ∈ B1 , y ∉ B1 and (2) d(x, a) > d(y, a) because y ∈ B2 , x ∉ B2 . It is a contradiction. Thus, either B1 ⊆ B2 or B2 ⊆ B1 . Finally, suppose B1 ∩ B2 = 0 and let x, x ∈ B1 , y ∈ B2 . Then d(x, x ) < d(x, y) and d(x, x ) < d(x , y).
68  2 Calculus of formal power series By isosceles triangle principle for three elements {x, x , y}, we have d(x, y) = d(x , y). By symmetry, d(x, y) = d(x, y ) for all y ∈ B2 . It shows that the function (x, y) → d(x, y) from B1 × B2 to R for every x ∈ B1 , y ∈ B2 is a constant and we are done. Theorem 2.3.6. Let U be a nonempty open subset of a ultrametric space (X, d). Then there is a partition of U into balls. More specifically, given a sequence r1 > r2 > ⋅ ⋅ ⋅ > 0 such that limn→∞ rn = 0, U can be covered by disjoint balls of the form Brn (a), a ∈ X, n ∈ ℕ. Proof. We want to find a collection {Br (a) : a ∈ U, r ∈ {rn }∞ n=1 , Br (a) ⊆ U} of disjoint balls such that U = ⋃a∈U Br (a). Let a ∈ U be given. We choose the ball Br (a) in such a way: if Br1 (a) ⊆ U, taking r = r1 , otherwise taking r = max{rn : Brn (a) ⊆ U}. The existence of such r is ensured by the openness of U, a ∈ U and limn→∞ rn = 0. If there is some b ∈ U such that Br (b) ⊆ U,
Br (a) ∩ Br (b) ≠ 0 for some
r ∈ {rn },
by Proposition 2.3.5 we have that either Br (a) ⊆ Br (b) or Br (b) ⊆ Br (a). Since r is the largest among such radius belonging to {rn }, it follows that Br (a) = Br (b) in either case. This ensures that the collection {Br (a) : a ∈ U, Br (a) ⊆ U}, where a ∈ U is arbitrary and r ∈ {rn } is the largest such that Br (a) ⊆ U, is disjoint. It is clear that this collection covers U. Finally, we provide an example. Example 2.3.7. Let S be a set. The discrete metric d on S is defined by 1
d(x, y) = {
0
if x ≠ y if x = y
for all x, y ∈ S. One checks that d is also an ultrametric on S.
2.4 Some topological structure for 𝕏

69
2.4 Some topological structure for 𝕏 Definition 2.4.1. A valuation on a field F is a mapping v from F into ℝ such that: (a) v(x) = 0 if and only if x = 0; (b) v(xy) = v(x)v(y); (c) v(x + y) ≤ v(x) + v(y); for all x, y in F. The pair (F, v) is called a valued field. Example 2.4.2. Define 0 x = { 1
if x = 0 if x ≠ 0
on ℝ for every x ∈ ℝ. One checks that (ℝ,  ⋅ ) is a valued field in which  ⋅  is called the trivial valuation. Example 2.4.3. Let ℝ[X] be the set of all polynomials over ℝ. Let d(f ) denote the degree of a polynomial f ∈ ℝ[X] and define d(f ) = ∞ if f is the zero polynomial. Then for any f , g ∈ ℝ[X], (1) d(f + g) ≤ max {d(f ), d(g)}; (2) d(fg) = d(f ) + d(g). Pick a real number ρ > 1. For any f ∈ ℝ[X], define 0 f  = { d(f ) ρ
if f = 0
if f ≠ 0.
Then we have (3) f + g ≤ max {f , g} ≤ f  + g; (4) fg = f  ⋅ g. It is obvious that f  ≥ 0; f  = 0 if and only if f = 0. Except that ℝ[X] is not a field, the mapping  ⋅  behaves like a valuation. The Archimedean property of ℝ states: If x ∈ ℝ, y ∈ ℝ with x > 0, then there is a positive integer n such that nx > y. Simply speaking, this property says that ℕ is unbounded. However, we have just seen in (3) and (4) that v(1 + 1) ≤ v(1) for v(f ) = f  for f ∈ ℝ[X], and then ℕ is bounded under this valuation. Such a valuation is called nonArchimedean valuation. Definition 2.4.4. Let v : 𝕏(ℂ) → [0, ∞) be a mapping such that (a) v(f ) = 0 if and only if f = 0; (b) v(fg) = v(f )v(g);
70  2 Calculus of formal power series (c) v(f + g) ≤ max{v(f ), v(g)}; where f and g are formal power series in 𝕏(ℂ). The mapping v is a nonArchimedean valuation. We have a proposition below which can be verified directly. Proposition 2.4.5. The mapping v : 𝕏(ℂ) → [0, +∞), defined by the formula v(f ) = 2− ord(f )
for
f ∈ 𝕏(ℂ),
is a valuation on 𝕏(ℂ), that is, v satisfies (a), (b) and (c) of Definition 2.4.4. Proposition 2.4.6. Let v be the valuation defined in Proposition 2.4.5. Let d : 𝕏(ℂ) × 𝕏(ℂ) → ℝ be defined as d(f , g) = v(f − g)
for
f , g ∈ 𝕏(ℂ).
Then d is an ultrametric on 𝕏(ℂ). Proof. We show that d satisfies (i), (ii), and the strong triangular inequality (2.2) of Definition 2.3.1. If d(f , g) = 0, then v(f − g) = 0 which occurs if and only if ord(f − g) = ∞, or f − g = 0, we have (i). (ii) is clear because ord(f − g) = ord(g − f ). It is known that ord(f + g) ≥ min{ord(f ), ord(g)} for any formal power series f and g. Then − ord(f + g) ≤ max{− ord(f ), − ord(g)} and hence d(f , g) = 2− ord(f −g) = 2− ord(f −h+h−g) ≤ 2max{− ord(f −h),− ord(h−g)}
≤ max{2− ord(f −h) , 2− ord(h−g) }
= max{v(f − h), v(h − g)} = max{d(f , h), d(h, g)}. Then inequality (2.2) is true. Thus, d is an ultrametric on 𝕏(ℂ). We now define a topology on 𝕏(ℂ) by this ultrametric. Definition 2.4.7. Let f ∈ 𝕏(ℂ) be given and let r > 0 be a real number. The ball in 𝕏(ℂ) centered at f with radius r is defined by Br (f ) = {g ∈ 𝕏(ℂ) : d(f , g) < r}.
2.4 Some topological structure for 𝕏
 71
A subset U ⊆ 𝕏(ℂ) is called an open set if for every f ∈ U there exists ε > 0 such that Bϵ (f ) ⊆ U. The ultrametric topology on 𝕏(ℂ) consists of all open subsets of 𝕏(ℂ) with the ultrametric d defined in Proposition 2.4.6. The space 𝕏(ℂ) with this ultrametric topology is then called an ultrametric topological space and is denoted by (𝕏, d). This ultrametric topological space has some interesting properties. Proposition 2.4.8. The ultrametric topological space (𝕏, d) is complete under the ultrametric d. ∞ j Proof. Let {fn }∞ n=1 ⊆ 𝕏(ℂ) be a Cauchy sequence in (𝕏, d). Write fn (z) = ∑j=0 fn (j)z for every n ∈ ℕ. We construct a formal power series f over ℂ such that
lim d(fn , f ) = 0.
n→∞
Assume that f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅. We first claim that for each j ∈ ℕ ∪ {0}, there exists a nj ∈ ℕ such that fn (j) = fnj (j)
for all n ≥ nj .
(2.3)
Assume for some j the claim fails, then for any nj ∈ ℕ there is some n > nj such that fn (j) ≠ fnj (j), Then ord(fn − fnj ) ≤ j and we have v(fn − fnj ) ≥ 2−j which is a contradiction to that the sequence {fn } is Cauchy. Thus, equation (2.3) is true. By equation (2.3), we can define f by f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ where aj = fnj (j)
for
j = 0, 1, 2, . . . .
We may even pick all nj in the order n0 < n1 < n2 < ⋅ ⋅ ⋅ , and we do so. Then we have ord(fn − f ) ≥ j
for
n ≥ nj ,
v(fn − f ) ≤ 2−j
for n ≥ nj .
and
Now let ϵ > 0 be given. Take j ∈ ℕ such that 2−j < ϵ, then d(fn , f ) = v(fn − f ) ≤ 2−j < ϵ whenever n ≥ nj .
72  2 Calculus of formal power series Proposition 2.4.9. The product of formal power series is a continuous mapping, defined on 𝕏(ℂ) × 𝕏(ℂ). Proof. Let (fn ), (gn ) be sequences of elements of 𝕏(ℂ) such that v(f − fn ) → 0
and
v(g − gn ) → 0
as n → ∞ and f , g ∈ 𝕏(ℂ). We have ord(fg − fn gn ) ≥ min{ord(fg − fn g), ord(fn g − fn gn )}
= min{ord(g) + ord(f − fn ), ord(fn ) + ord(g − gn )}.
So, v(fg − fn gn ) ≤ max{v(g)v(f − fn ), v(fn )v(g − gn )}, and, therefore, fn gn → fg
as n → ∞,
which completes the proof. The topology on the set of formal power series defined above possesses good properties and, therefore, one can use wellknown fixedpoint theorems to examine operators defined on the space under consideration. Example 2.4.10. Let us define the mapping φ : 𝕏(ℂ) → 𝕏(ℂ) by the formula ∞
φ(f (z)) = 1 + ∑ an−1 z n n=1
for
∞
f (z) = ∑ an z n . n=0
n For any f , g ∈ 𝕏(ℂ) (f is as above and g(z) = ∑∞ n=0 bn z ) we have ∞
v(φ(f (z)) − φ(g(z))) = v( ∑ (an−1 − bn−1 )z n ) n=1
∞ 1 1 = v( ∑ (an − bn )z n ) = v(f (z) − g(z)). 2 n=0 2
Since φ is a contraction defined on the complete metric space, by the Banach contraction principle it has a unique fixed point. Now let us find this fixed point. We show that this fixed point is the formal power series g(z) = 1 + z + z 2 + z 3 + ⋅ ⋅ ⋅ + z n + ⋅ ⋅ ⋅ .
2.4 Some topological structure for 𝕏
 73
Let f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ be any formal power series on ℂ. It is clear that φ2 (f )(z) = φ(φ(f ))(z)
= φ(1 + (a0 z + a1 z 2 + ⋅ ⋅ ⋅))
= 1 + z + (a0 z 2 + a1 z 3 + ⋅ ⋅ ⋅ +).
Inductively, we have φn (f )(z) = 1 + z + z 2 + ⋅ ⋅ ⋅ + z n−1 + (a0 z n + a1 z n+1 + ⋅ ⋅ ⋅), for all n ∈ ℕ, n ≥ 2. Then lim φn (f )(z) = 1 + z + z 2 + ⋅ ⋅ ⋅ + z n + ⋅ ⋅ ⋅ = g(z).
n→∞
Please be noticed that the statement in the above example was also proved by the fixedpoint theorem in a different way [70]. Example 2.4.11. Let us define the mapping ψ : 𝕏0 (ℂ) → 𝕏(ℂ) by the formula ∞
∞
n=0
n=0
ψ( ∑ an z n+1 ) = ∑ an z n ,
an ∈ ℂ,
n ∈ ℕ ∪ {0},
for any nonunit f (z) = a0 z + a1 z 2 + ⋅ ⋅ ⋅. Let f (z), g(z) ∈ 𝕏0 (ℂ) be given, we have ord(ψ(f (z)) − ψ(g(z))) = ord(f (z) − g(z)) − 1, so v(ψ(f (z)) − ψ(g(z))) = 21−ord(f (z)−g(z)) = 2v(f (z) − g(z)). Moreover, ψ is a surjection from 𝕏0 onto 𝕏(ℂ). Since (𝕏(ℂ), d) is a complete metric space, the Banach contraction principle may play interesting role for this ultrametric topological space. Let p, q ∈ 𝕏 = 𝔻(ℂ) be almost unit formal power series such that q(z) = z + z 2 + z 3 + ⋅ ⋅ ⋅ ,
and p(z) = z − z 2 + ⋅ ⋅ ⋅ + (−1)n−1 z n + ⋅ ⋅ ⋅ .
We know that the function representations for q and p are q(z) =
z 1−z
and p(z) =
z . 1+z
We may see that actually p(z) = q[−1] (z) by verifying their function representations or by verifying the composition q ∘ p of two formal power series directly. The next example shows us that the formal power series p and q can be obtained by applying the fixed point theory to the complete ultrametric topological space 𝕏(ℂ).
74  2 Calculus of formal power series Example 2.4.12. Let us define the mapping φ and ψ from 𝔻(ℂ) to 𝔻(ℂ) by the formulas ∞
∞
φ(f (z)) = z − ∑ an z n+1
for f (z) = ∑ an z n ∈ 𝔻(ℂ),
ψ(f (z)) = z + ∑ an z n+1
for f (z) = ∑ an z n ∈ 𝔻(ℂ),
n=1 ∞
n=1 ∞
n=1
n=1
respectively. Then each of φ and ψ has its fixed point. Moreover, the unique fixed point p ∈ 𝔻(ℂ) for φ and the unique fixed point q ∈ 𝔻(ℂ) for ψ are almost units such that p(z) = z − z 2 + z 3 − ⋅ ⋅ ⋅ + (−1)n−1 z n + ⋅ ⋅ ⋅ , q(z) = z + z 2 + z 3 + ⋅ ⋅ ⋅ + z n + ⋅ ⋅ ⋅ ,
q ∘ p = I𝔻
or
[−1]
q=p
and
.
Proof. Of course, it is easy to verify that p and q are fixed points for φ and ψ, respectively, if such p and q are given. We suppose that the mappings φ and ψ are given. We first prove the existence of the fixed point for the mapping φ and then construct the fixed point. n For any f , h ∈ 𝔻(ℂ) (f is as above and h(z) = ∑∞ n=1 bn z ), where a0 = 0 = b0 but a1 ≠ 0 ≠ b0 , we have ∞
v(φ(f (z)) − φ(h(z))) = v( ∑ (an − bn )z n+1 ) n=1
∞ 1 1 = v( ∑ (an − bn )z n ) = v(f (z) − h(z)). 2 n=1 2
Since φ is a contraction defined on the complete ultrametric space, by the Banach contraction principle it has a unique fixed point. We show that this fixed point is the formal power series p(z) = z − z 2 + z 3 − ⋅ ⋅ ⋅ + (−1)n−1 z n + ⋅ ⋅ ⋅ . Let f (z) = a1 z + a2 z 2 + ⋅ ⋅ ⋅ be an almost unit on ℂ. It is clear that φ2 (f )(z) = φ(z − (a1 z 2 + a2 z 3 + ⋅ ⋅ ⋅)) = z − z 2 + (a1 z 3 + a2 z 4 + ⋅ ⋅ ⋅ +). Inductively, we have φn (f )(z) = z − z 2 + ⋅ ⋅ ⋅ + (−1)n−1 z n + (a1 z n+1 + a2 z n+2 + ⋅ ⋅ ⋅), for all n ∈ ℕ, n ≥ 2, where φn = φ(φ(⋅ ⋅ ⋅)). Then ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ n
lim φn (f )(z) = z − z 2 + z 3 − ⋅ ⋅ ⋅ + (−1)n−1 z n + ⋅ ⋅ ⋅ = p(z).
n→∞
2.4 Some topological structure for 𝕏
 75
Similarly, we can prove that ψ has the unique fixed point q(z) = z + z 2 + z 3 + ⋅ ⋅ ⋅ . One checks that q ∘ p = I𝔻
or
q = p[−1] .
Definition 2.4.13. Let (G, d) be any ultrametric topological space with ultrametric d. G is called discrete if for any sequences {xj } and {yj } in G, d(x1 , y1 ) > d(x2 , y2 ) > d(x3 , y3 ) > ⋅ ⋅ ⋅ implies lim d(xn , yn ) = 0.
n→∞
The space G is called spherically complete if each nested sequence of balls has a nonempty intersection. Proposition 2.4.14. The space (𝕏(ℂ), d) is discrete and spherically complete, where the ultrametric d is defined as in Proposition 2.4.6. Proof. Let {fj } and {gj } be sequences in 𝕏(ℂ) such that d(f1 , g1 ) > d(f2 , g2 ) > d(f3 , g3 ) > ⋅ ⋅ ⋅ . For any j ∈ ℕ, there exists nj ∈ ℕ such that d(fj , gj ) < 2−nj ≤ d(fj+1 , gj+1 ) because d(fj , gj ) = 2− ord(fj −gj ) , ord(fj − gj ) is a positive integer, and nj < nj+1 for every j ∈ ℕ. Then lim d(fn , gn ) = 0.
n→∞
Proposition 20.2 (p. 52, [89]) proves that (𝕏(ℂ), d) is spherically complete. We omit it here. The Banach contraction fixedpoint theorem is a convenient and efficient tool for the fixedpoint problems. If the contraction constant has the value δ = 1 instead of 0 < δ < 1, the fixed point problems need to be considered in different way [7]. Definition 2.4.15. Let (X, d) be a nonArchimedean metric space with ultrametric d. Let T : X → X be a mapping. Then T is said to be a nonexpansive mapping if whenever x and y are different elements in X, d(Tx, Ty) ≤ d(x, y).
76  2 Calculus of formal power series The mapping T is called contractive if d(Tx, Ty) < d(x, y). The next theorem can be found in [62]. Theorem 2.4.16. Considering the ultrametric topological space (𝕏(ℂ), d) where d is an ultrametric defined in Proposition 2.4.6. If T : 𝕏(ℂ) → 𝕏(ℂ) is a contractive mapping, then T has the unique fixed point. Proof. For any a ∈ 𝕏 = 𝕏(ℂ), define ra = d(a, Ta) and denote the ball Ba = Bra (a). If ra = 0, we are done. So we suppose that ra > 0. Let 𝒜 denote the collection of such balls for all a ∈ 𝕏. By Proposition 2.4.14, the space (𝕏, d) is spherically complete. Then the relation Ba ≤ Bb
if and only if Bb ⊆ Ba
is a partial order. Let 𝒜1 ⊆ 𝒜 be a totally ordered subfamily. From the spherical completeness of 𝕏 (Proposition 2.4.14), we have B : = ⋂ Ba ≠ 0. Ba ∈𝒜1
Let b ∈ B and Ba ∈ 𝒜1 . Then x ∈ Bb implies that d(x, a) ≤ max{d(a, b), d(b, x)} ≤ d(a, Ta) = ra , because b ∈ B implies that b ∈ Ba and then d(a, b) ≤ d(a, Ta), and also the contractive property d(Ta, Tb) ≤ d(a, b) implies d(x, b) ≤ rb ≤ max{d(b, a), d(a, Ta), d(Ta, Tb)} = d(a, Ta) = ra . The above two inequalities ensure that x ∈ Ba , and hence Bb ⊆ Ba . Since Ba ∈ 𝒜1 is arbitrary, it follows that Bb ⊆ Ba for every Ba ∈ 𝒜1 . Thus, Bb is an upper bound in 𝒜 for the subfamily 𝒜1 . By Zorn’s lemma, 𝒜 has a maximal element, say Bz , for some z ∈ 𝕏. We claim that z = Tz. Assume that z ≠ Tz. Since T is contractive, d(Tz, T 2 z) < d(z, Tz),
(T 2 = T ∘ T)
(2.4)
and Tz ∈ BTz ∩ Bz , it follows that BTz ⊆ Bz . But z ∉ BTz by (2.4), so BTz is not a subset of Bz , and this contradicts the maximality of Bz . Therefore, T has a fixed point, which is unique obviously.
2.4 Some topological structure for 𝕏
 77
Theorem 2.4.17. Let (𝕏, d) be a spherically complete nonArchimedean metric space and let T : 𝕏 → 𝕏 be a nonexpansive mapping. Then either T has at least one fixed point or there exists a sphere S of radius r > 0 such that T : S → S and for which d(b, T(b)) = r for each b ∈ S. The proof of the theorem can be found in [62], too. The next example shows how this theorem works for the ultrametric space (𝕏(ℂ), d). Example 2.4.18 ([6]). Let (sn )n∈ℕ be a sequence of real numbers such that sn  < 1, sn ≠ 0 for n ∈ ℕ, and sn → 1 as n → ∞. Let us define the mapping φ : 𝕏(ℂ) → 𝕏(ℂ) by the formula ∞
φ(f (z)) = 1 + ∑ (sn an + 1)z n , n=1
∞
for f (z) = ∑ an z n . n=0
If g(z) ∈ 𝕏(ℂ) with ord(f (z) − g(z)) = 0, then obviously ord(φ(f (z)) − φ(g(z))) ≥ 1 and, therefore, v(φ(f (z)) − φ(g(z))) < v(f (z) − g(z)). If ord(f (z) − g(z)) = k ≥ 1, then ord(φ(f (z)) − φ(g(z))) = k and, therefore, φ is a nonexpansive mapping. Let S(0, r) denotes a sphere centered at 0 with radius r > 0 in 𝕏(ℂ). Let r = 1 and f = (1, 0, 0, . . . ). Then 1 φ(f ) = (1, 1, 0, . . . ) and v(f − φ(f )) = . 2 Now, for any sphere S(0, r), where 0 < r < 1, it is obvious that S(0, r) is either an empty set or there is an integer k such that r = 21k . If r = 21k , then φ(S(0, 21k )) ⊆ S(0, 1). In the case of spheres centered at any other point different from zero one can argue in a similar way. Therefore, in view of Theorem 2.4.16, the mapping φ has at least one fixed point in 𝕏(ℂ). n Moreover, φ has exactly one fixed point f ̄(z) = ∑∞ n=0 ā n z for which ā 0 = 1,
ā k = 1/(1 − sk )
for all k ∈ ℕ.
78  2 Calculus of formal power series
2.5 The umbral calculus and formal power series This section provides an introduction to umbral calculus and its relations with formal power series. We basically adopt the introduction by S. Roman with a slight modification. Readers may visit [83] if you are interested in this subject. Definition 2.5.1. We denote ℂ[X] the set of all polynomials over ℂ and denote ℂ[X]∗ the set of all linear functionals on ℂ[X]. We use the notation ⟨L  p(x)⟩, borrowed from physics, to denote the action of a linear functional L on a polynomial p(x). Proposition 2.5.2. Let L and M be two linear functionals in ℂ[X]∗ . For every p(x) ∈ ℂ[X] we have ⟨L + M  p(x)⟩ = ⟨L  p(x)⟩ + ⟨M  p(x)⟩ and ⟨cL  p(x)⟩ = c⟨L  p(x)⟩ where c ∈ ℂ. That is, ℂ[X]∗ is a linear vector space over ℂ. It is known that a linear functional on ℂ[X] is uniquely determined by the basis of this space. We have the following. Proposition 2.5.3. For any formal power series, ∞
f (t) = ∑
k=0
ak k t ∈ 𝕏(ℂ), k!
(2.5)
there is a Lf ∈ ℂ[X]∗ , determined by f (t), such that ⟨Lf  xn ⟩ = ⟨Lf (t)  xn ⟩ = an ,
n = 0, 1, 2, . . . .
Conversely, given a functional L ∈ ℂ[X]∗ , there exists a formal power series fL ∈ 𝕏(ℂ) such that ⟨L  x k ⟩ k t . k! k=0 ∞
fL (t) = ∑ n
x k ∞ ∗ Proof. It is clear that {x n }∞ n=0 or { n! } forms a basis of ℂ[X]. Let {t }k=0 ⊆ ℂ[X] such that
⟨t k  xn ⟩ = n!δn,k , where δn,k is the Kronecker delta function for which 0 δn,k = { 1
if n ≠ k
if n = k.
2.5 The umbral calculus and formal power series 
Then f (t) = ∑∞ k=0 ak k ∑∞ k=0 k! t and
ak k t k!
79
∈ 𝕏(ℂ) defines a linear functional Lf in ℂ[X]∗ such that Lf (t) = ak k t k! k=0 ∞
⟨Lf (t)  xn ⟩ = ⟨ ∑
∞ n x ⟩ = ∑ ak ⟨t k  x n ⟩ = an , k! k=0
(2.6)
for n = 0, 1, 2, . . . . On the other hand, for every linear functional L ∈ ℂ[X]∗ there exists a formal power series f ∈ 𝕏(ℂ) such that ⟨L  xk ⟩ k t . k! k=0 ∞
f (t) = ∑
Using (2.6), we have ⟨Lf  xn ⟩ = ⟨L  xn ⟩ for all n ∈ ℕ ∪ {0}. We write ⟨L  xk ⟩ k t . k! k=0 ∞
fL (t) = ∑ From now on, for all f ∈ 𝕏(ℂ), ∞
f (t) = ∑
⟨Lf  xn ⟩
k=0
k!
tk
(2.7)
and for all polynomials p(x) ⟨t k  p(x)⟩ k x . k! k≥0
p(x) = ∑
(2.8)
Theorem 2.5.4. Let T : ℂ[X]∗ → 𝕏(ℂ) be a mapping such that T(L) = fL
for all
L ∈ ℂ[X]∗ ,
then T is an isomorphism. Proof. Since {x n }∞ n=0 is a basis of ℂ[X], it follows that L=M
if and only if ⟨L  xn ⟩ = ⟨M  x n ⟩,
n = 0, 1, 2, 3, . . . ,
for L, M ∈ ℂ[X]∗ , and hence L=M
if and only if fL (t) = fM (t).
Then T is bijective. Also ∞
fL+M (t) = ∑
n=0 ∞
= ∑
n=0
⟨L + M  xn ⟩ n t n! ⟨L  xn ⟩ n ∞ ⟨M  x n ⟩ n t +∑ t n! n! n=0
= fL (t) + fM (t), Thus, T is a vector space isomorphism.
80  2 Calculus of formal power series Remark 2.5.5. By the above results, we may conclude: 1. A formal power series f ∈ 𝕏(ℂ) can be considered as a linear functional on ℂ[X]. 2. The set 𝕏(ℂ) can represent an algebra of formal power series, a linear vector space of formal power series, and the vector space ℂ[X]∗ . From now on, in this section we may use f ∈ 𝕏(ℂ) to represent a linear functional directly or denote it as fL or Lf as it appears in Proposition 2.5.3. 3. The equality f (t) = g(t) in 𝕏(ℂ) is equivalent to that they are equal as the linear functionals in ℂ[X]∗ . We have now defined an algebra structure on ℂ[X]∗ , namely, the algebra of formal power series. We may call 𝕏(ℂ) the umbral algebra. Such concept provides us another way to investigate the formal power series. Please be noticed that the multiplication considered in such algebra is the Cauchy product of formal power series. Proposition 2.5.6. Let f , g ∈ 𝕏(ℂ) be given and denote the corresponding linear functional as Lf and Lg , respectively. Then Lf Lg ∈ ℂ[X]∗ and Lfg = Lf Lg such that for n ∈ ℕ ∪ {0}, n n ⟨Lfg (t)  xn ⟩ = ∑ ( )⟨Lf (t)  xk ⟩⟨Lg (t)  xn−k ⟩. k k=0
Proof. By Proposition 2.5.3, we write bk k x , k! k=0
ak k x k! k=0
∞
∞
and g(x) = ∑
f (x) = ∑ ∞
Lf (t) = ∑
k=0
ak k t k!
∞
and Lg (t) = ∑
k=0
and also write
bk k t . k!
Then n
ak bn−k )t n ∈ ℂ[X]∗ . k! (n − k)! k=0
∞
Lf Lg (t) = ∑ ( ∑ n=0
We denote Lfg = Lf Lg . On the other hand, ⟨Lf (t)  xk ⟩
∞
f (t) = ∑
k!
k=0
tk
and
∞
g(t) = ∑
⟨Lg (t)  x k ⟩
k=0
k!
tk .
Then ∞
n
f (t)g(t) = ∑ ( ∑
n=0 k=0
⟨Lf (t)  x k ⟩ ⟨Lg (t)  x n−k ⟩ k!
(n − k)!
)t n
1 n n ( ∑ ( )⟨Lf (t)  x k ⟩⟨Lg (t)  x n−k ⟩)t n n! k n=0 k=0 ∞
= ∑
2.5 The umbral calculus and formal power series 
because (nk ) =
n! , k!(n−k)!
81
it follows that
n n ⟨Lfg (t)  xn ⟩ = ∑ ( )⟨Lf (t)  xk ⟩⟨Lg (t)  x n−k ⟩. k k=0
Applying the Mathematical Induction for the number of formal power series involved in the proposition above, we have the following. Corollary 2.5.7. Let the finite sequence {fk }m k=1 ⊆ 𝕏(ℂ) be given. Then ⟨Lf1 f2 ⋅⋅⋅fm (t)  xn ⟩ = ⟨Lf1 (t)Lf2 (t) ⋅ ⋅ ⋅ Lfm (t)  xn ⟩ n )⟨Lf1 (t)  xi1 ⟩⟨Lf2 (t)  x i2 ⟩ ⋅ ⋅ ⋅ ⟨Lfm (t)  x im ⟩, =∑( i1 , i2 , . . . , im where the sun is extended over all nonnegative integers i1 , i2 , . . . , im such that i1 + i2 + ⋅ ⋅ ⋅ + im = n. Proposition 2.5.8. Let {fk }∞ k=0 ⊆ 𝕏(ℂ) be given with ord(fk ) = k for all k ∈ ℕ ∪ {0}. Then ∞ f = ∑k=0 ak fk ∈ 𝕏(ℂ) and ∞
⟨Lf (t)  p(x)⟩ = ∑ ak ⟨Lfk (t)  p(x)⟩ k=0
for all p ∈ ℂ[X], where (ak )∞ k=0 ⊆ ℂ. The sum on the righthand side is finite. Proof. The condition ord(fk ) = k for all k ≥ 0 ensures that the sequence {fk } is a sequence admitting addition, and hence ∞
f = ∑ ak fk ∈ 𝕏(ℂ) for k=0
(ak )∞ k=0 ⊆ ℂ.
Suppose that deg(p) = n, then Lfk (t)(p(x)) = 0, k > n because ord(fk ) > n. Then n
∞
k=0
k=n+1
⟨Lf (t)  p(x)⟩ = ⟨( ∑ ak Lfk (t) + ∑ ak Lfk (t)) p(x)⟩ n
= ⟨( ∑ ak Lfk )(t) p(x)⟩ + 0 n
k=0
= ∑ ak ⟨Lfk (t)  p(x)⟩ + 0 k=0 ∞
= ∑ ak ⟨Lfk (t)  p(x)⟩. k=0
82  2 Calculus of formal power series k For example, if f ∈ 𝔻(ℂ), set fk = f k and write g(z) = ∑∞ k=0 bk z , we have ∞
⟨Lg∘f (t)  p(x)⟩ = ∑ bk ⟨Lf k (t)  p(x)⟩. k=0
Proposition 2.5.9. Let {fk }∞ k=0 ⊆ 𝕏(ℂ) be given with ord(fk ) = k for all k ∈ ℕ ∪ {0}. If ⟨Lfk (t)  p(x)⟩ = ⟨Lfk (t)  q(x)⟩ for all
k,
then p(x) = q(x), where p, q ∈ ℂ[X]. Proof. The sequence {fk } forms a pseudo basis for 𝕏(ℂ). For all n ≥ 0, there exist constants an,k for which t n = ∑∞ k=0 an,k fk (t). Thus ∞
⟨Lt n  p(x)⟩ = ∑ an,k ⟨Lfk (t)  p(x)⟩ k=0 ∞
= ∑ an,k ⟨Lfk (t)  q(x)⟩ k=0
= ⟨Lt n  q(x)⟩, and then p(x) = q(x) by Theorem 2.5.4 or by (2.8) ⟨Lt k  p(x)⟩ k x . k! k=0 ∞
p(x) = ∑
Theorem 2.5.10. Let f ∈ 𝔻(ℂ) be an almost unit and g ∈ 𝕏(ℂ) be a unit formal power series. Then there exists a unique sequence {sn } of polynomials with deg(sn ) = n satisfying the orthogonality condition ⟨Lgf k (t)  sn (x)⟩ = n!δn,k
(2.9)
for all n, k ≥ 0. Proof. The uniqueness follows from Proposition 2.5.9. We show the existence. For every nonnegative integer k, we write ∞
g(t)f k (t) = ∑ bk,i t i i=k
with bk,k ≠ 0. For every n ∈ ℕ ∪ {0}, we write n
sn (x) = ∑ an,j x j , j=0
2.5 The umbral calculus and formal power series  83
and we are going to determine sn which satisfies (2.9). Now let n ≥ 0 be given and let sn satisfy (2.9), we have n n!δn,k = ⟨Lgf k (t)  sn (x)⟩ = ⟨Lgf k (t) ( ∑ an,j xj )⟩ ∞ n
n
n
j=0
= ∑ ∑ bk,i an,j ⟨Lt i  x j ⟩ = ∑ ∑ bk,i an,j ⟨Lt i  x j ⟩ i=k j=0
i=k j=0
n
= ∑ bk,i an,i i!. i=k
Taking k = n, one obtain n! = bn,n an,n n!, or an,n = 1/bn,n . By successively taking k = n, n − 1, . . . , 0, we obtain a triangular system of equations that can be solved for an,k , 0 ≤ k ≤ n. For example, taking k = n−1, we have n!δn,n−1 = 0, and then n
0 = ∑ bn−1,i an,i i! = bn−1,n−1 an,n−1 (n − 1)! + bn−1,n an,n n!. i=n−1
Then an,n−1 =
−nbn−1,n . bn−1,n−1 bn,n
The sequence {sn (x)} in (2.7) is called the Sheffer sequence for the pair (g(t), f (t)), or that {sn (x)} is Sheffer for (g(t), f (t)). In the next chapter, we will see that the Sheffer sequence for (g(t), t) is the Appell sequence for g(t). The pair (g(t), f (t)) will appear in the Riordan group, which will be discussed in the next chapter.
3 Applications of formal power series, I The formal power series has been applied to many fields, including algebra, combinatorics, analysis, functional analysis, and differential equations. G. Raney investigated the Lagrange inversion formula in 1960 before Ivan Niven systematically introduced the formal power series. A. M. Garsia and S. A. Joni reproved the Lagrange inversion theorem by operator theoretic methods in 1977 [32]. D. Pravica, M. Spurr, L. Reich, and J. Tomaschek applied the formal power series to the differential equations in different ways [66, 98]. A lot of applications of formal power series can be found in the book Applied and computational complex analysis by P. Henrici in which the author also introduces how to use formal power series approach to solve difference equations. This chapter can only collect certain interesting applications and some very recent developments of the formal power series. The Riordan array or Riordan group has a strong relationship with the composition of nonunit formal power series. We introduce some initial work about the Riordan group by L. Shapiro and his team [13, 93]. A complete proof of the existence of the Riordan group we provided in Section 4.2. The Cayley–Hamilton theorem is a popular theorem in algebra, especially in linear algebra, named after Arthur Cayley and Sir William Rowan Hamilton. In Section 4.3, we introduce a new proof of this theorem recently contributed by C. Bernhardt [4] by using the formal power series. Fermat’s little theorem is another classical result which can be proved by using formal power series. In order to make such a proof, we set up some necessary results in Chapter 1 and Chapter 2 that will be cited in the proof. These two classical theorems are introduced in Section 3.4.
3.1 Riordan group and formal power series The Riordan group is a significant application of the formal power series in algebra and combinatorics. The authors of [93] named this group in memory of the mathematician John Riordan, the author of the book Introduction to Combinatorial Analysis [81]. Definition 3.1.1. Let M = (mi,j )i,j≥0 = [M0 , M1 , M2 , . . .] be an infinite matrix m0,0 [ [ m1,0 [ [ M = [ m2,0 [m [ 3,0 [ . . [ .
m0,1 m1,1 m2,1 m3,1 .. .
m0,2 m1,2 m2,2 m3,2 .. .
m0,3 m1,3 m2,3 m3,3 .. .
... ] . . .] ] . . .] ], . . .] ] ] .. . ]
with mij ∈ ℂ for all nonnegative integers i and j, where Mj is the jth column of M. For n each Mj , the formal power series Mj (x) = ∑∞ n=0 mn,j x is called the generating function https://doi.org/10.1515/9783110599459003
86  3 Applications of formal power series, I of the jth column of M, or a formal power series associated with Mj . We also write M[x] = [M0 (x), M1 (x), M2 (x), . . .]. Definition 3.1.2. Let g be a formal power series in 𝕏(ℂ) with g(0) = g0 = 1 and f ∈ 𝔻(ℂ) be such that f0 = 0, f1 = 1, and g(x) = 1 + g1 x + g2 x2 + ⋅ ⋅ ⋅ ,
f (x) = x + f2 x 2 + f3 x3 + ⋅ ⋅ ⋅ .
(3.1)
For each nonnegative integer j, we define formal power series j
Cj (x) = g(x)(f (x)) = g(x)f j (x),
(3.2)
and then let Cj (x) be the generating function of the jth column of an infinite matrix C[x] as M[x] in Definition 3.1.1, in this case we write the infinite matrix C[x] = (g(x), f (x)) = [C0 (x), C1 (x), C2 (x), . . .]
(3.3)
and call the matrix C[x] a Riordan matrix associated with g and f . Let C[x] be a Riordan matrix associated with g and f , and let a = (a0 , a1 , a2 , . . .)T be a column vector of infinitely many entries. Write ∞
A(x) = ∑ an xn , n=0
a formal power series associated with a, applying the usual multiplication of matrices we have a formal power series B(x) = C[x]a = a0 C0 (x) + a1 C1 (x) + a2 C2 (x) + ⋅ ⋅ ⋅ = a0 g(x) + a1 g(x)f (x) + a2 g(x)f 2 (x) + ⋅ ⋅ ⋅
= g(x)[a0 + a1 f (x) + a2 f 2 (x) + ⋅ ⋅ ⋅] = g(x)A(f (x)) where A(f (x)) ∈ 𝕏(ℂ) because f ∈ 𝔻. Proposition 3.1.3. A Riordan matrix is a infinite lower triangular matrix or array. Proof. Let C[x] be a Riordan matrix generated by g and f as in formula (3.3). Then Cj (x) = g(x)(f (x))j . Since ord(g) = 0, ord(f ) = 1, it follows that for all j ≥ 0, ord(Cj ) = ord(g) + ord(f j ) = 0 + j = j. Then Cj [x] = (0, . . . , 0, cj,j , cj+1,j , . . . )T if we write C[x] = (cn,j ). Thus, c0,0 [ [ c1,0 [ [ C = [ c2,0 [c [ 3,0 [ . . [ .
0 c1,1 c2,1 c3,1 .. .
0 0 c2,2 c3,2 .. .
0 0 0 c3,3 .. .
... ] . . .] ] . . .] ], . . .] ] ] .. . ]
3.1 Riordan group and formal power series  87
where c0,0 = 1, c1,0 = g1 , c2,0 = g2 , . . . , if g(x) = 1 + g1 x + g2 x 2 + ⋅ ⋅ ⋅. C[x] is a lower triangular matrix. We now define the multiplication between Riordan matrices. Definition 3.1.4. Let (g(x), f (x)) and (h(x), l(x)) be any two Riordan matrices. A multiplication “∗” of these two Riordan matrices is defined by (g(x), f (x)) ∗ (h(x), l(x)) = (g(x)h(f (x)), l(f (x))).
(3.4)
The next proposition will show that the multiplication “∗” is welldefined and, under this multiplication, the set of all Riordan matrices forms a group. Proposition 3.1.5. The set of all Riordan matrices forms a group under the multiplication of (3.4) with Riordan identity IR = (I(x), x) and group inverse −1
(g(x), f (x))
=(
1
g(f [−1] (x))
, f [−1] (x)),
where f (f [−1] (x)) = f [−1] (f (x)) = x, or f [−1] is the iterated inverse as defined in Definition 1.5.10, I(x) is the formal power series identity defined in Definition 1.1.3. This group is called the Riordan group and is denoted as R, or (R, ∗). We usually denote IR as I if there is no confusion. Proof. The existence of the composition in (3.4) and the iterative inverse is guaranteed by the fact that f ∈ 𝔻(ℂ) (see Section 1.5). We need only to check the existence of the identity Riordan matrix, the inverse Riordan matrix, and the associative property of this multiplication. In fact, (g(x), f (x)) ∗ (I(x), x) = (g(x)(I ∘ f (x)), x(f (x))) = (g(x)I(x), f (x)) = (g(x), f (x)), (I(x), x) ∗ (g(x), f (x)) = (I(x)g(x), f (x)) = (g(x), f (x)). Furthermore, 1 , f [−1] (x)) g(f [−1] (x)) 1 = (g(x) [−1] (f (x)), f [−1] (f (x))) g(f ) 1 = (g(x) [−1] , f [−1] (f (x))) g(f ∘ f (x)) 1 = (g(x) , x) = (I(x), x) = IR , g(x)
(g(x), f (x)) ∗ (
88  3 Applications of formal power series, I and (
1
g(f [−1] (x) =(
, f [−1] (x)) ∗ (g(x), f (x)) 1
g(f [−1] (x))
g(f [−1] (x)), f (f [−1] (x))) = (I(x), x) = IR .
For the associative law, let (g(x), f (x)), (h(x), l(x)) and (r(x), s(x)) be Riordan matrices, then [(g(x), f (x)) ∗ (h(x), l(x))] ∗ (r(x), s(x)) = (g(x)h(f (x)), l(f (x))) ∗ (r(x), s(x)) = (g(x)h(f (x)) ⋅ s(l(f (x))), r(l(f (x)))), and (g(x), f (x)) ∗ [(h(x), l(x)) ∗ (r(x), s(x))] = (g(x), f (x)) ∗ (h(x)s(l(x)), r(l(x))) = (g(x) ⋅ (h ⋅ s ∘ l)(f (x)), (r ∘ l)(f (x))) = (g(x)h(f (x)) ⋅ s(l(f (x))), r(l(f (x)))). Example 3.1.6. The matrix P below is called the Pascal matrix 1 [1 [ [ [1 [ P = [1 [ [ [1 [ .. [.
0 1 2 3 4 .. .
0 0 1 3 6 .. .
0 0 0 1 4 .. .
0 0 0 0 1 .. .
0 0 0 0 0 .. .
... . . .] ] ] . . .] ] . . . .] ] ] . . .] ] .. .]
Let g(x) = 1/(1 − x) = 1 + x + x2 + x3 + ⋅ ⋅ ⋅ be a formal power series on ℝ. We claim that the Pascal matrix P = [P0 , P1 , P2 , . . .] is a Riordan matrix induced by (g(x), f (x)) for some almost unit formal power series f . In fact, P0 (x) = g(x)f 0 (x) is true for any formal power series f . To ensure that P is a Riordan matrix we need to find an almost unit f satisfying both equations (3.1) and (3.2) or, equivalently, Pj (x) = g(x)f j (x)
for all j ∈ ℕ.
n Let f (x) = ∑∞ n=0 an x , then
f (x)g(x) = f (x) We use this formula to construct f .
∞ n 1 = ∑ ( ∑ ai )x n . 1 − x n=0 i=0
(3.5)
3.1 Riordan group and formal power series  89
Since P1 = [0, 1, 2, 3, 4, . . .]T , applying equation (3.2) we have P1 = g(x)f (x) and then obtain the recurrent formula n
∑ ai = n,
n = 0, 1, 2, 3, . . . .
i=0
Then we have a0 = 0, a1 = 1, a2 = 1, and an = 1 for all n ∈ ℕ, or f (x) = x + x2 + x3 + ⋅ ⋅ ⋅ = xg(x). To show that P is a Riordan matrix, we need to show that this f satisfies Pj (x) = g(x)f j (x)
for all j ∈ ℕ ∪ {0}.
(3.6)
It has been proved that (3.6) is true for j = 0 and j = 1. For j > 1, j
g(x)f j (x) = g(x)[xg(x)] = xj g j+1 (x) = x j (1 − x)−(j+1) . Considering the formal binomial series for (1 + z)α , or a a a Ba = 1 + ( )z + ( )z 2 + ⋅ ⋅ ⋅ + ( )z n + ⋅ ⋅ ⋅ 1 2 n with z = −x, where (an ) =
a(a−1)⋅⋅⋅(a−n+1) n!
for all n ∈ ℕ, and α = −(j + 1), we have
−(j + 1) −(j + 1) )(−x) + ( )(−x)2 ⋅ ⋅ ⋅] 1 2 j+1 j + (n − j) n−j = xj [1 + ( )x + ⋅ ⋅ ⋅ + ( )x + ⋅ ⋅ ⋅] 1 n−j j+1 n = xj [1 + ( )x + ⋅ ⋅ ⋅ + ( )x n−j + ⋅ ⋅ ⋅]. 1 n−j
g(x)f j (x) = xj [1 + (
If we write g(x)f j (x) = b0 + b1 x + b2 x2 + ⋅ ⋅ ⋅, we have bi = 0
for 0 ≤ i < j,
and bn = (
n n ) = ( ), n−j j
n ≥ j.
On the other hand, the nth component of Pj is (nj ) for n ≥ j, and equals zero if n < j. Then the formula (3.6) is true. Thus, the Pascal matrix is a Riordan matrix induced by (g(x), f (x)) = (
x 1 , ). 1−x 1−x
90  3 Applications of formal power series, I Example 3.1.7. The matrix B below is another version of Pascal’s triangle and it is also a Riordan matrix: 1 [0 [ [ [2 [ [ B = [0 [6 [ [ [0 [ .. [.
0 1 0 3 0 10 .. .
0 0 1 0 4 0 .. .
0 0 0 1 0 5 .. .
0 0 0 0 1 0 .. .
0 0 0 0 0 1 .. .
... . . .] ] ] . . .] ] 1 1 − √1 − 4x 2 . . .] , ). ]=( ] 2x √1 − 4x 2 . . .] ] . . .] ] .. .]
The verification of this formula is similar to Example 3.1.6 that is left to the readers. Hint: Applying Theorem 1.7.10 for the formal binomial series B1/2 and the formal power series f (x) = −4x2 , the composition B1/2 ∘ f produces √1 − 4x2 as a function
representation of a formal power series, and B−1/2 ∘ f produces (1 − 4x2 )−1/2 .
Example 3.1.8. The Riordan matrix F = (1, z(1+z)) is called the Fibonacci matrix whose row sums are Fibonacci numbers. The inverse of this Riordan matrix is
F −1
1 [0 [ [ [0 [ = (1, z c(−z)) = [0 [ [ [0 [ .. [.
0 1 −1 2 −5 .. .
0 0 1 −2 5 .. .
0 0 0 1 −3 .. .
0 0 0 0 1 .. .
0 0 0 0 0 .. .
... . . .] ] ] . . .] ] , . . .] ] ] . . .] ] .. .]
where c(z) = (1 − √1 − 4z)/2z
(3.7)
is called the Catalan generating function Definition 3.1.9. Two Riordan matrices A and B are called similar if there is a Riordan matrix S such that A = S−1 BS. With the principle of formal analysis and the function representation of formal power series, we introduce an application of formal analysis to mathematical biology. The RNA matrix is a recent application of Riordan array to biology. Proposition 3.1.10. Let s(z) = (1 − z + z 2 − √1 − 2z − z 2 − 2z 3 + z 4 )/2z 2 .
3.1 Riordan group and formal power series  91
s(z) is called an RNA generating function from discrete mathematical biology. The matrix R = (s(z), z s(z)) is called the RNA matrix since the RNA numbers consist the leftmost column of the matrix. Then R is similar to the Pascal matrix P. Proof. Let f (z) = −2z − z 2 − 2z 3 + z 4 , then f ∈ 𝔻(ℂ), and hence we have B1/2 ∘ f (z) = √1 − 2z − z 2 − 2z 3 + z 4 ∈ 𝕏(ℂ). Checking the first several terms of this formal power series we have B1/2 ∘ f (z) = 1 − z − z 2 + ⋅ ⋅ ⋅ . Then s is a unit formal power series with s(0) = 1, which ensures that R = (s(z), z s(z)) is a Riordan matrix. We have C0 = (c(z 2 ), zc(z 2 )) = (
1 − √1 − 4z 2 1 − √1 − 4z 2 , ), 2z 2z 2
where c(z) is the Catalan generating function defined by equation (3.7). One checks that C0 is a Riordan matrix. Performing the composition and inverse function with the function representations of the formal power series involved we have C0−1 = (
1 z , ). 1 + z2 1 + z2
Then we may verify that R = C0−1 PC0 . Theorem 3.1.11. Let D = (dn,k )n,k≥0 be an infinite lower triangular matrix. Then D is a Riordan matrix if and only if there exist two sequences A = (a0 , a1 , a2 , . . .),
Z = (z0 , z1 , z2 , . . .)
with a0 ≠ 0 ≠ z0 such that for all k, n ∈ {0} ∪ ℕ, (i) dn+1,k+1 = ∑∞ j=0 aj dn,k+j , (ii) dn+1,0 = ∑∞ j=0 zj dn,j . The sequences A and Z in Theorem 3.1.11 are called Asequence and Zsequence of the Riordan matrix D, respectively. The above results about the Riordan group can be found in [47] and [93].
92  3 Applications of formal power series, I
3.2 Riordan involutions As a group element, some Riordan matrix may have finite order. Definition 3.2.1. Let M = (g, f ) be a Riordan matrix over ℝ. If M n = M ∗ M ∗ ⋅ ⋅ ⋅ ∗ M = IR = (I(x), x) for some n ∈ ℕ and n is the smallest such integer, M is said to have a finite Riordan order, or just order if there is no confusion. If the order of M is 2, then M is called an involution of the Riordan group. Example 3.2.2. The Riordan matrix 1 [ [0 [ [ M = (1, −z) = [0 [0 [ [. . [.
0 −1 0 0 .. .
0 0 1 0 .. .
0 0 0 −1 .. .
... ] . . .] ] . . .] ], . . .] ] ] .. . ]
(3.8)
is an involution of the Riordan group over ℝ. Example 3.2.3. Let T = (1, −x/(1 + x)) be a Riordan matrix over ℝ. By the formula in Proposition 3.1.5 with g(x) = 1, f (x) = −x/(1 + x) where −x/(1 + x) is the function representation of the formal power series −x(1 − x + x2 − x3 + ⋅ ⋅ ⋅) = −x + x 2 − x 3 + ⋅ ⋅ ⋅ + (−1)n x n + ⋅ ⋅ ⋅ . It is not simple to find f [−1] by using the formula in Theorem 1.5.9. Since f has the function representation, by the principle of formal analysis, we may perform function operation and obtain T −1 = (
1∘
1
f [−1] (x)
, f [−1] (x)) = (1, f [−1] (x)).
Since f [−1] (x) = −x/(1 + x) = f (x) by Example 1.7.11, it follows that T is an involution. Definition 3.2.4. Let M be the Riordan matrix defined in (3.8). A Riordan matrix L is called a pseudoinvolution or is said to have pseudoorder 2 if LM or equivalently ML has order 2. The next example tells us some relations between involution and pseudoinvolution of Riordan group. Example 3.2.5. The identity element IR of Riordan group is a pseudoinvolution but not an involution.
3.3 Some subgroups of the Riordan group
 93
Example 3.2.6. The RNA matrix R = (s(z), zs(z)) defined in Proposition 3.1.10 is not an involution but is a pseudoinvolution. One checks that R−1 = (s(−z), zs(−z)) and
(RM)2 = IR .
Proposition 3.2.7. If a Riordan matrix L is a pseudoinvolution, then Lk is also a pseudoinvolution for all k ∈ ℕ. Proof. Let L be a pseudoinvolution, then (LM)2 = IR or LML = M −1 = M. Assume that Lj , 1 ≤ j ≤ k, is a pseudoinvolution for some k ∈ ℕ. Then Lj MLj = M for 1 ≤ j ≤ k. Then 2
(Lk+1 M) = Lk+1 MLk+1 M = L(Lk MLk )LM = (LM)(LM) = IR . We complete the induction. Proposition 3.2.8. If L = (g(z), f (z)) is a pseudoinvolution of the Riordan group, then L−1 = (g(−z), −f (−z)). The next theorem is obvious. Theorem 3.2.9. A Riordan matrix M = (g(z), f (z)) is an involution of the Riordan group if and only if g(z)g(f (z)) = 1 and f (f (z)) = z.
3.3 Some subgroups of the Riordan group The following problem was mentioned by Shapiro in 2008: Did the set of Riordan matrices of the form (g(z), zg n (z)), n ∈ ℕ ∪ {0}, form a subgroup of (R, ∗)? The answer is positive. For example, we have obtained the Appell subgroup and Bell subgroup when n = 0, 1, respectively. For every unit formal power series g ∈ 𝕏(ℝ), the Riordan matrix (g(z), z) has some interesting properties. Proposition 3.3.1. The subset E = {(g(z), z) : g ∈ 𝕏(ℝ) is a unit} is a subgroup of the Riordan group (R, ∗). E is called the Appell subgroup of (R, ∗). Proof. It is clear that IR = (1, z) ∈ E. For (g(z), z) and (h(z), z) in E, we have (g(z), z) ∗ (h(z), z) = (g(z)h(z), z(z)) = (gh(z), z). Since gh is a unit, it follows that E is closed for Riordan multiplication.
94  3 Applications of formal power series, I Finally, for (g(z), z) ∈ E, applying Proposition 3.1.5 we have −1
(g(z), z)
=(
1 , z) ∈ E, g(z)
because I [−1] (z) = I(z) = z, and g −1 = 1/g is a unit. Thus, E is a subgroup of R. Theorem 3.3.2. The Appell subgroup of the Riordan group is a normal subgroup. Proof. Let A = (g(z), z) ∈ E be given. For any L = (h(z), f (z)) ∈ (R, ∗), we have L−1 ∗ A ∗ L = (h(z), f (z))
−1
∗ (g(z), z) ∗ (h(z), f (z))
1 , f [−1] ) ∗ (g(z), z) ∗ (h(z), f (z)) h(f [−1] (z)) 1 = ( [−1] , f [−1] ) ∗ (g(z) ⋅ h(z), f (z)) h(f (z)) =(
= (g(f [−1] (z)), z) ∈ E, because g(f [−1] (z)) is a unit. Similarly, we have a subset G below is also a subgroup of the Riordan group. Proposition 3.3.3. The subset G = {(1, f (z)) : f ∈ 𝔻(ℝ) is an almost unit} is a subgroup of the Riordan group (R, ∗). G is called the associated subgroup. Example 3.3.4. The Riordan matrix F = (1, z(1 + z)) is called the Fibonacci matrix, where 1 [0 [ [ [0 [ [ F = [0 [0 [ [ [0 [ .. [.
0 1 1 0 0 0 .. .
0 0 1 2 1 0 .. .
0 0 0 1 3 3 .. .
0 0 0 0 1 4 .. .
0 0 0 0 0 1 .. .
... . . .] ] ] . . .] ] . . .] ]. . . .] ] ] . . .] ] .. .]
It is obvious that F is an element of the associated subgroup G. A Riordan matrix is also called a Riordan array. We would like to introduce the stochastic Riordan array.
3.3 Some subgroups of the Riordan group
 95
Definition 3.3.5. A Riordan array (g(z), f (z)) is called a stochastic Riordan array if its row sums equal one. Example 3.3.6. The function representation F(z) = 1/(1−z−z 2 ) is called the generating function for the Fibonacci numbers. Let L = (F(z), −z 2 F(z)), this Riordan array is 1 [1 [ [ [2 [ [ L = [3 [5 [ [ [8 [ .. [.
0 0 −1 −2 −5 −10 .. .
0 0 0 0 1 3 .. .
0 0 0 0 0 0 .. .
0 0 0 0 0 0 .. .
0 0 0 0 0 0 .. .
... . . .] ] ] . . .] ] . . .] ]. . . .] ] ] . . .] ] .. .]
The array L is a stochastic Riordan array. This array is not a proper Riordan array because f (z) = −z 2 F(z) is not an almost unit formal power series, but it is of interest since the Fibonacci numbers appear in the leftmost column. Lemma 3.3.7. A Riordan array L = (g(z), f (z)) is stochastic if and only if f (z) = −g(z) + zg(z) + 1. The proof of the lemma can be found in [47]. The Sheffer sequence {sn (x)} = {n!pn (x)} of polynomials is defined by the formal power series g(t)(f (t))k where g is a unit and f ∈ 𝔻(ℂ) (see Theorem 2.5.10). We now know that g(t)f k (t) is the generating function or the kth column of the Riordan matrix (g, f ). Also, the Sheffer sequence for (g(t), t) is the Appell sequence for g(t). Faà di Bruno’s formula ([19], 1855) is a popular formula in classical analysis. The connection between Faà di Bruno’s formula and Riordan array could be found in [48]. We provide it here briefly without providing proof. Theorem 3.3.8 (Faà di Bruno’s formula). If g and f are functions with a sufficient number of derivatives, then dm g(f (t)) dt m =∑
b
b
b
m! f (t) 1 f (t) 2 f (m) (t) m g (k) (f (t))( ) ( ) ⋅⋅⋅( ) , b1 !b2 ! ⋅ ⋅ ⋅ bm ! 1! 2! m!
where the sum is extended over all different solutions in nonnegative integers b1 , b2 , . . . , bm such that b1 + 2b2 + ⋅ ⋅ ⋅ + mbm = m
and
b1 + b2 + ⋅ ⋅ ⋅ + bm = k.
96  3 Applications of formal power series, I By importing Bell polynomials Bm,k , Faà di Bruno’s formula may have the Bell polynomial version (Riordan’s formula) [48] m dm g(f (t)) = g (k) (f (t))Bm,k (f (t), f (t), . . . , f (m−k+1) (t)), ∑ dt m k=0
where g and f are functions with a sufficient number of derivatives.
3.4 Cayley–Hamilton theorem and Fermat’s little theorem Arthur Cayley and Sir William Rowan Hamilton provided us the Cayley–Hamilton theorem more than 100 years ago. In this short section, we are going to give a proof of this theorem using formal power series [4]. Theorem 3.4.1 (Cayley–Hamilton theorem). Let S be a commutative ring with identity. For any n ∈ ℕ, we denote M(n, n) the set of all n × n matrices over S. Let A ∈ M(n, n) be given and write det(tI − A) = c0 t n + c1 t n−1 + ⋅ ⋅ ⋅ + cn−1 t + cn where det(A) is the determinant of A. Then c0 An + c1 An−1 + ⋅ ⋅ ⋅ + cn−1 A + cn = 0. Proof. For A ∈ M(n, n), we have 1 det(I − tA) = t n det( I − A) = c0 + c1 t + c2 t 2 + ⋅ ⋅ ⋅ + cn t n . t By the Laplace expansion theorem and the definition of adjoint matrix or adjugate matrix (p. 276, [65]), we have det(I − tA)I = (I − tA) adj(I − tA),
(3.9)
where adj(A) denotes the adjugate matrix of A. If we consider I − tA as a formal power series with coefficients in M(n, n) which is a ring, I − At is a unit, and hence its inverse exists. Similar to the proof of the inverse j of the geometric formal power series ∑∞ j=0 z we did in Example 1.1.9, we have ∞
(I − tA)−1 = ∑ Aj t j . j=0
Multiplying (I − tA)−1 to the both sides of equation (3.9), we have ∞
( ∑ Aj t j )(c0 + c1 t + ⋅ ⋅ ⋅ + cn t n ) = adj(I − tA). j=0
3.4 Cayley–Hamilton theorem and Fermat’s little theorem  97
Write adj(I − tA) as a formal power series over M(n, n), that is, ∞
∞
j=0
j=0
( ∑ Aj t j )(c0 + c1 t + ⋅ ⋅ ⋅ + cn t n ) = ∑ Bj t j . Notice that the entries in adj(I − tA) are (n − 1) × (n − 1) cofactors of (I − tA), or are polynomials in t of degree less than or equal to (n − 1), then Bj is a zero matrix if j ≥ n, or ∞
n−1
j=0
j=0
( ∑ Aj t j )(c0 + c1 t + ⋅ ⋅ ⋅ + cn t n ) = ∑ Bj t j . Then the coefficient of t n on the left hand side of the above equation is zero, that is, c0 An + c1 An−1 + ⋅ ⋅ ⋅ + cn−1 A + cn = 0. The other interesting work is a curious proof for Fermat’s little theorem: For p prime and a ∈ ℤ, ap ≡ a (mod p). Of course, there have been many proofs for Fermat’s little theorem and its related theorems that can be found in [36]. We introduce a socalled curious proof which uses the formal power series to prove the Fermat’s theorem [3]. k Lemma 3.4.2. Let f (x) = 1 − x − dx 2 + ∑∞ k=3 ak x be a formal power series in 𝕏(ℚ), with coefficients in ℤ. Then there is a unique way to have ∞
f (x) = ∏(1 − mk xk ), k=1
(3.10)
where {mk }∞ k=1 ⊆ ℤ. k Proof. Taking ϕk (x) = −mk xk in equation (1.19), we know that ∏∞ k=1 (1 − mk x ) is welldefined. Let ∞
f (x) = ∏(1 − mk xk ), k=1
we can easily obtain m1 = 1
and m2 = d.
Let N ∈ ℕ be given with N ≥ 3 and suppose that we have chosen mk ∈ ℤ, 1 ≤ k ≤ N − 1, which satisfy (3.10). Then N−1
N−1
k=1
k=3
∏ (1 − mk xk ) = 1 − x − dx2 + ∑ ak x k + Cx N + g(x),
98  3 Applications of formal power series, I where g is a polynomial, or a formal power series, with ord(g) ≥ N + 1 and deg(g) ≤ N(N − 1)/2, where C is an integer generated by all mk ∈ ℤ, 1 ≤ k ≤ N − 1. Since N
N−1
k=1
k=1
∏(1 − mk xk ) = (1 − mN xN ) ∏ (1 − mk xk ) N−1
= (1 − mN xN )[1 − x − dx 2 + ∑ ak x k + Cx N + g(x)], k=3
its coefficient of the Nth term xN is −mN + C. Equating the Nth term of N
∏(1 − mk x k ) = f (x), k=1
we have mN = C − aN ∈ ℤ that completes the mathematical induction. Thus, there is a unique way to have (3.10). ists.
The formal power series f defined in Lemma 3.4.2 is a unit and, therefore, f −1 ex
k Corollary 3.4.3. Let f (x) = 1 − x − dx2 + ∑∞ k=3 ak x be a formal power series in 𝕏(ℚ), with coefficients in ℤ. Then ∞
f −1 (x) = 1 + x + (d + 1)x 2 + ∑ bk x k k=3
is a formal power series in 𝕏(ℤ). Moreover, ∞
f −1 (x) = (1 + x)(1 + (d + 1)x2 ) ∏(1 − nk x k ), k=3
(3.11)
where {nk } ⊆ ℤ with n1 = −1, n2 = −(d + 1). The proof of the corollary is similar to the proof of the Lemma 3.4.2. Lemma 3.4.4. Let f and f −1 be defined in Lemma 3.4.2 and its corollary, respectively, then ∞ kmk x k N = ∑ x N ∑ msN/s k s 1 − m x k N=1 sN k=1 ∞
−x(L(f (x))) = ∑
(3.12)
where s  N means that there is an integer k such that sk = N, and
−x(L(
∞ 1 N )) = x(L(f (x))) = ∑ x N ∑ nsN/s , f (x) s N=1 sN
where L is the formal logarithm defined in Section 2.2.
(3.13)
3.4 Cayley–Hamilton theorem and Fermat’s little theorem  99
Proof. Both f and f −1 are elements in 𝕏1 , then both L(f ) and L(f −1 ) are formal power series. It is obvious that {1 − mk xk }∞ k=1 is a sequence admitting multiplication. Applying (3.10) and then applying Corollary 2.2.13, we have kmk x k . k k=1 1 − mk x ∞
−x(L(f (x))) = ∑ Notice that
1 = 1 + mk xk + (mk )2 x2k + ⋅ ⋅ ⋅ + (mk )n x nk + ⋅ ⋅ ⋅ . 1 − mk x k For every k ∈ ℕ, we write fk (x) =
∞ kmk xk = fk,n xn ∑ 1 − mk xk n=0
2
= kmk xk [1 + (mk xk ) + (mk x k ) + ⋅ ⋅ ⋅], then fk,n = 0
if jk < n < (j + 1)k,
j ∈ ℕ ∪ {0}.
It is clear that {fk } ⊆ 𝕏(ℂ) is a sequence admitting addition. By Definition 1.7.7, we have (3.12). Similarly, we can have (3.13). By the equations (3.12) and (3.13), we have Corollary 3.4.5. Let the notation be as in Lemma 3.4.4, ∑ msN/s
sN
N N = ∑ ns , s sN N/s s
N ∈ ℕ.
(3.14)
where mk and nk are defined by (3.10) and (3.11). It is easy to know that mk = −nk for odd k, but not for the even k. In number theory, the partition function p(n) represents the number of possible partitions of a nonnegative integer n. For example, p(4) = 5 because the integer 4 has five partitions: 1 + 1 + 1 + 1, 1 + 1 + 2, 1 + 3, 2 + 2, and 4. The example below brings us some connection between Lemma 3.4.2 and the partition functions. Example 3.4.6. Denote the infinite product ∞
(x, x)∞ = ∏(1 − xk ). k=1
100  3 Applications of formal power series, I Then ∞
n (x, x)−1 ∞ = ∑ p(n)x , n=1
where p(n) is the Ramanujan’s partition function. Theorem 3.4.7 (Curious proof of Fermat’s little theorem). For p prime and a ∈ ℤ, p  (ap − a). Proof. Let notation be set the same as in Lemma 3.4.2, Lemma 3.4.4, and their corollaries. Recall that m2 = d, n2 = −(d + 1). Let p be a prime with p > 2, and take N = 2p in (3.14). Then s takes the values 1, 2, p and 2p only. Therefore, ∑ msN/s
sN
N = m2p 2p + m2p p + mp2 2 + m2p 1 s = 2pm2p + pm2p + 2dp + 1.
Similarly, we have ∑ nsN/s
sN
N = −2p ⋅ n2p − p ⋅ n2p + 2(d + 1)p − 1. s
By (3.14), we have 2p ⋅ m2p + p ⋅ m2p + 2d2 + 1 = −2p ⋅ n2p − p ⋅ n2p + 2(d + 1)p − 1. Then p ⋅ (2m2p + m2p + 2n2p + n2p ) = 2((d + 1)p − dp − 1). Thus, p  ((d + 1)p − dp − 1). Then we have a−1 p ∑ ((d + 1)p − dp − 1) = ap − a. d=1
4 Applications of formal power series, II This chapter further introduces some applications of formal power series in analysis and differential equations. We have seen that the formal power series solution of the differential equation f = af is the formal exponential series Ea , and the solution of the equation (1+x)f = af generates the formal binomial series Ba . Note that we do not assign any value to the variable in those equations or in the formal power series and, therefore, we are not concerned with the convergence of the power series solutions. We actually focus on generating the formal power series by means of certain differential equations, not on solving equations. If we focus on solving certain differential equations, and we find the solutions are divergent power series or formal power series, what can we say? Some discussions about the differential equations are collected in Section 4.1, where the solutions of the differential equations are formal power series with some sense of convergence. Most of those developments are contributed by D. Pravica and M. Spurr. The application of formal power series to functional equations can be traced back more than 100 years ago. At that time, the solutions of certain functional equations were simply called the formal solutions. Papers by Schröder and Pfeiffer revealed the formal power series in studies of functional equations a century ago. Readers may find some history and developments of this topic, including the Dhombres functional equation, in Section 4.2. The formal functional equation of the translation equation, which is the recent work of Fripertinger and Reich, is collected in Section 4.2, too. Lagrange inversion formula f (z) = zg(f (z)), where g is given, has attracted many mathematicians. This topic will be discussed in Section 4.3. Finally, Section 4.4 provides an alternative method to solve the linear difference equations. This may suggest the possible applications of formal power series to the numerical methods.
4.1 Formal power series and some differential equations We always try to solve an ordinary differential equation or partial differential equation and obtain an analytic solution. Unfortunately, sometimes we may solve a differential equation but its solution is not analytic. We first introduce an example due to Euler [50]. Example 4.1.1. The equation z2 https://doi.org/10.1515/9783110599459004
d u(z) = u(z) − z dz
102  4 Applications of formal power series, II has the unique power series solution ∞
u(z) = ∑ (k − 1)!z k . k=1
Of course, this solution is not convergent on any region except at the point z = 0. The next example is for partial differential equation. Example 4.1.2. Considering the equation in two variables z1 and z2 and letting z = (z1 , z2 ), then the equation 1 𝜕 𝜕 1 u(z) − (z2 − z1 z22 ) u(z) = u(z) − z1 z2 , (z1 + z12 z2 ) 2 𝜕z1 2 𝜕z2 has the unique power series solution ∞
u(z) = ∑ (k − 1)!(z1 z2 )k , k=1
which, of course, is a formal power series. These kinds of solutions are formal power series solution of differential equations. Example 4.1.3. The equation z 2 ψ (z) + ψ(z/q) = z
(4.1)
is a delaydifferential equation for q > 1. One checks that ∞
ψ(z) = ∑ n! q(n+1)(n+2)/2 (−1)n z n+1 n=0
(4.2)
is a solution of (4.1). Again, ψ is a formal power series solution. We also provide a differential equation similar to (4.1), which has analytic solution. The details of the next example and proposition can be found in [69]. Example 4.1.4. The linear homogeneous equation z 2 y (z) − y(z/q) = 0,
q > 1,
(4.3)
has a solution j
e−q /(qz) y(z) = ∑ j(j−1)/2 j=−∞ q ∞
which is defined and analytic for the positive real part R(z) > 0 [68]. Indeed, j
e−q /z j(j−1)/2 j=−∞ q ∞
y(z/q) = ∑
(4.4)
4.1 Formal power series and some differential equations  103
and j−1
e−q /z −1 y (z) = ∑ j(j−1)/2 (−qj−1 ) 2 z q j=−∞ ∞
j−1
=
e−q /z 1 ∞ ∑ z 2 j=−∞ q(j−1)(j−2)/2 k
1 ∞ e−q /z = 2 ∑ k(k−1)/2 , z k=−∞ q where we make substitution k = j − 1 in the last step. Then j
k
∞ z e−q /z 1 ∞ e−q /z z y (z) − y( ) = z 2 ( 2 ∑ k(k−1)/2 ) − ∑ j(j−1)/2 = 0. q z k=−∞ q j=−∞ q 2
Thus, y is an analytic solution of (4.3). If we replace j by j + ϵ in (4.4), y(z) is still a solution of (4.3). Let m(ϵ) be an absolutely continuous measure on [0, 1] such that m (0) = m (1), and let m∗ be an absolutely continuous measure on ℝ such that m∗ (E + 1) = m∗ (E) for any measurable set E, we have 1
y(z) = ∫ ∑
0 j∈ℤ
e−q
j+ϵ
∞
/(qz)
q(j+ϵ)(j+ϵ−1)/2
t
dm e−q /(qt) dm∗ dϵ = ∫ t(t−1)/2 dt. dϵ dt q
(4.5)
−∞
We now denote ℂ̃ ∗ a punctured neighborhood of z = 0 on the logarithmic Riemann surface. For our convenience, we define logq z ≡ ln z/ ln q
and dθ = {reiθ : r > 0},
where dθ is a ray in the θ direction. Substituting u = qt and assuming that m∗ analytically extends to ℂ̃ ∗ in u, we have 1
y(z) = ∫ e−u/(qz) q− 2 logq (u)(logq (u)−1) darg(z)
A natural example results from setting
dm∗ du
dm∗ du. du
(4.6)
≡ C a normalization constant.
Proposition 4.1.5. Let y(z) be as in (4.6). For z ∈ ℂ such that I(z) ≠ 0 or R(z) > 0 if I(z) = 0, and for some angle θ, f (z) = ∫ dθ
ζ 1 z 1 y( )dξ = ∫ y( )dζ , 1+ξ ξ 1 + ζz ζ dθz
(4.7)
104  4 Applications of formal power series, II where I(z) and R(z) represent the imaginary part of z and real part of z, respectively, and the last integral exists for θ < π with θz = arg(z) − θ. Then 1
ψ(z) = ∫ f (u)du = 0
z 1 ∫ ln(1 + ξ )y( )dξ z ξ
(4.8)
dθ
is a solution of (4.1). Definition 4.1.6. For each R > 0 and subset S ⊆ ℂ, define D(S; R) = {f ∈ 𝕏(S) : f is convergent for z < R}, and then denote Q̃ = D(ℂ̃ ∗ ; R). A qadvanced Gevrey series is a formal power series f ̂ ∈ 𝕏(ℂ) such that there exist positive real numbers A and K for which ∞
f ̂(z) = ∑ an z n , n=0
an  ≤ K An n!qn(n+1)/2 ,
(4.9)
where q > 1. The space of qadvanced Gevrey series, denoted by ℂ[[z]]!q , is the subspace of 𝕏(ℂ) consisting of all q advanced Gevrey series. Let Sθ (ϵ) denote the sector centered on the ray dθ in the θ direction of halfwidth ϵ > 0. Then f (z) is said to have an asymptotic qadvanced Gevrey expansion f ̂(z) in the direction θ if there are positive A, K, R, ϵ so that for all n ∈ ℕ and z ∈ D(Sθ (ϵ); R), n n(n+1)/2 φq,A (ze−iθ )zn , f (z) − fn̂ (z) < KA n!q
(4.10)
m ̂ where fn̂ (z) = ∑n−1 m=0 am z is the nth partial sum of f . In (4.10), 1
2
φq,A (z) = q 2 argq z φ(z) in which the factor φ(z) is given by φ(z) = qp1 (z)+p2 (z)+p3 (z) , where 1 2 p1 (z) = − log2q [1 + (argq z/ logq z) ] 8 − (argq z/ ln q) arctan[argq z/(− logq z)], 1 p2 (z) = − [logq z + logq (− logq z) − 1 + logq A] 2 2
× logq [1 + (argq z/ logq z) ],
p3 (z) = (
1
2 ln2 q
) arctan2 (argq z/ logq z).
(4.11)
4.2 Functional equations and formal power series  105
It can be shown that φq,A ≥ 1 for z sufficiently small. We have the following. Proposition 4.1.7. Let φq,A (z) be defined as in (4.11). There exists R > 0 such that φq,A (z) ≥ 1,
for z < R,
z ∈ ℂ̃ ∗ .
Proof. The factor φ(z) is welldefined for z sufficiently small. The base q > 1 ensures that φq,A (z) > 0. Then 1 arg2q z + p1 (z) ≥ 0 2 for z sufficiently small; p2 (z) ≥ 0 for z sufficiently small; and p3 (z) ≥ 0 is always rue. Then the exponent of (4.11), or 1 arg2q z + p1 (z) + p2 (z) + p3 (z), 2 attains its minimum value of 0 when arg(z) = 0. Thus, there exists R > 0 such that φq,A (z) ≥ 1,
z < R,
z ∈ ℂ̃ ∗ .
Now we are ready to introduce the main theorem its proof can be found in [69]. Theorem 4.1.8. Let y(z) be a solution of the homogeneous equation (4.3) which is analytic on ℂ̃ ∗ , and for each ϵ ∈ (0, π), there are positive real numbers K and A so that, for f (z) defined in (4.7), we have
n−1 f (z) − ∑ m!qm(m+1)/2 (−z)m < KAn n!qn(n+1)/2 φq,A (ze−iθ )zn , m=0
(4.12)
for all θ ∈ [−π + ϵ, π − ϵ], ∀ n ∈ ℕ, and ∀ z ∈ ℂ̃ ∗ with z sufficiently small. Then f (z) is the unique analytic function satisfying (4.12) on ℂ̃ ∗ in a neighborhood of z = 0. More applications of formal power series to the differential equations can be found in [67].
4.2 Functional equations and formal power series The name of formal solution was used in solving some functional equations more than 100 years ago [63]. Actually, those formal solutions are formal power series. The next theorem indicates why we have to face formal power series solutions of functional equations, no matter the solutions are convergent or divergent. The proof of the theorem is relatively long because it takes time to construct a particular formal power series.
106  4 Applications of formal power series, II Theorem 4.2.1. There exists an analytic function f (x) = a1 x + a2 x2 + a3 x3 + ⋅ ⋅ ⋅ ,
a1  = 1, an1 ≠ 1 ∀ n ∈ ℕ,
such that the Schröder’s functional equation ϕ[f (x)] = a1 ϕ(x) has no solution which is analytic about the origin and which has a nonvanishing derivative there, i. e., every formal solution ϕ(x) = c1 x + c2 x2 + c3 x 3 + ⋅ ⋅ ⋅ ,
c1 ≠ 0,
is divergent for all values of x except x = 0. Proof. If ϕ is a solution of the Schröder’s equation, then ord(ϕ) = 1 because otherwise an1 ≠ 1 fails (it is not difficult to see it). So, we may suppose that ϕ(x) = c1 x + c2 x2 + c3 x 3 + ⋅ ⋅ ⋅ ,
c1 ≠ 0
is a solution of the functional equation, then the composition ϕ ∘ f should equal a1 ϕ, or ∞
n
∑ cn (f (x)) = a1 (c1 x + c2 x 2 + c3 x 3 + ⋅ ⋅ ⋅).
n=1 n
Write f (x) =
a(n) 1 x
+
2 a(n) 2 x
3 + a(n) 3 x + ⋅ ⋅ ⋅ for all n ∈ ℕ, then formula (1.2) yields that n
a1 cn = ∑ ck a(k) n , k=1
for all n ∈ ℕ.
If n ≥ 2, n−1
(k) a1 cn − cn a(n) n = ∑ ck an . k=1
n n Since a(n) n = a1 and a1 ≠ 1 ∀ n ∈ ℕ, we have
cn =
1 (n−1) (c a(1) + c2 a(2) ). n + ⋅ ⋅ ⋅ + cn−1 an a1 − an1 1 n
Using the matrix representation formula, a(n) 0
T
a(n−1) 0
T
[ (n) ] [ (n−1) ] [a ] [a ] [ 1 ] [ 1 ] [ (n) ] [ (n−1) ] [a ] = [a ] [ 2 ] [ 2 ] [ . ] [ . ] [ . ] [ . ] [ . ] [ . ] (n) (n−1) a [ k ] [ak ]
0 [ [0 [ [0 [ [. [ .. [ [0
a1 0 0 .. . 0
a2 a1 0 .. . 0
... ... ... .. . ...
ak ] ak−1 ] ] ak−2 ] ] .. ] . ] ] 0 ](k+1)×(k+1)
(4.13)
4.2 Functional equations and formal power series  107
in Corollary 1.4.7, we may obtain all cn recurrently. For example, 2 a(2) 2 = a1 ,
a(2) 3 = 2a1 a2 ,
2 a(2) 4 = 2a1 a3 + a2 ,
a(3) 3
a(3) 4
a(3) 5
=
a31 ,
=
3a21 a2 ,
3a21 a3
...,
3a1 a22 ,
...,
c1 [a1 (1 − a1 )a3 + 2a1 a22 ] , a21 (1 − a1 )(1 − a21 )
....
=
+
(4) 4 3 and a(4) 4 = a1 , a5 = 4a1 a2 , . . . . Then we have
c2 =
c1 a2 , a1 (1 − a1 )
c3 =
Assume that for 3 ≤ k ≤ n with n ≥ 3, ck =
2 k−2 c1 [ak−2 1 (1 − a1 )(1 − a1 ) ⋅ ⋅ ⋅ (1 − a1 )ak + Pk ] k−1 2 ak−1 1 (1 − a1 )(1 − a1 ) ⋅ ⋅ ⋅ (1 − a1 )
,
where Pk = Pk (a1 , a2 , . . . , ak−1 ) is a polynomial of (a1 , a2 , . . . , ak−1 ), then 1 (2) (n) (c1 a(1) n+1 + c2 an+1 + ⋅ ⋅ ⋅ + cn an+1 ) a1 − an+1 1 1 (n) = (c1 an+1 + c2 a(2) n+1 + ⋅ ⋅ ⋅ + cn an+1 ) a1 − an+1 1 c1 an+1 1 = + (c a(2) + ⋅ ⋅ ⋅ + cn a(n) n+1 ). a1 (1 − an1 ) a1 (1 − an1 ) 2 n+1
cn+1 =
Since f is a nonunit, it follows that a(k) n+1 contains only ai , 1 ≤ i ≤ n for 2 ≤ k ≤ n. Taking this consideration for (3) (n) c2 a(2) n+1 , c3 an+1 , . . . , cn an+1 ,
we may denote (n) c2 a(2) c P (a , a , . . . , an ) n+1 + ⋅ ⋅ ⋅ + cn an+1 = n 1 n+1 1 22 , n a1 (1 − a1 ) a1 (1 − a1 )(1 − a1 ) ⋅ ⋅ ⋅ (1 − an−1 1 )
for some polynomial Pn+1 of a1 , a2 , . . . , an . Then, writing 2 n−1 c1 [an−1 c1 an+1 1 (1 − a1 )(1 − a1 ) ⋅ ⋅ ⋅ (1 − a1 )an+1 = , a1 (1 − an1 ) an1 (1 − a1 )(1 − a21 ) ⋅ ⋅ ⋅ (1 − an1 )
we have cn+1 =
2 n−1 c1 [an−1 1 (1 − a1 )(1 − a1 ) ⋅ ⋅ ⋅ (1 − a1 )an+1 + Pn+1 ] an1 (1 − a1 )(1 − a21 ) ⋅ ⋅ ⋅ (1 − an1 )
for all n ∈ ℕ where Pn+1 = Pn+1 (a1 , a2 , . . . , an ) is a polynomial in a1 , a2 , . . . , an .
(4.131 )
108  4 Applications of formal power series, II Next, we set up a sequence {γn }∞ n=1 by γn+1 =
γ1 [α1n−1 (1 − α1 )(1 − α12 ) ⋅ ⋅ ⋅ (1 − α1n−1 )αn+1 + Pn+1 ] α1n (1 − α1 )(1 − α12 ) ⋅ ⋅ ⋅ (1 − α1n )
(4.14)
for all n ∈ ℕ where Pn+1 = Pn+1 (α1 , α2 , . . . , αn ) is a polynomial in α1 , α2 , . . . , αn . To prove the theorem, we want to determine a set of values [ai ] for the αi such that n the ai are the coefficients of a convergent power series f (x) = ∑∞ n=1 an x with a1  = 1, n a1 ≠ 1, and such that the sequence ci (i = 2, 3, . . . ), the corresponding values of γi , are the coefficients of a power series with a zero radius of convergence for every value of γ1 except zero. For any n ∈ ℕ and polynomial Pn+1 = Pn+1 (α1 , α2 , . . . , αn ), we define Fn+1 (α1 , . . . , αn+1 ) = α1n−1 (1 − α1 )(1 − α12 ) ⋅ ⋅ ⋅ (1 − α1n−1 )αn+1 + Pn+1 . Let a be such that am = 1 but ak ≠ 1, 1 ≤ k < m, the primitive mth root of unity. Then Fm+1 (a, α2 , . . . , αm+1 ) = am−1 (1 − a) ⋅ ⋅ ⋅ (1 − am−1 )αm+1 + Pm+1 , and 2 m−1 am−1 ) ≠ 0, 1 (1 − a)(1 − a ) ⋅ ⋅ ⋅ (1 − a
it follows that the coefficient of αm+1 in Fm+1 (a, α2 , . . . , αm+1 ) is not zero. Consequently, we may choose definite values of α2 , α3 , . . . , αm+1 , say a2 , a3 , . . . , am+1 , respectively, such that ai − â i  < δ,
2 ≤ i ≤ m + 1,
where δ > 0 is arbitrary and â i , i = 2, 3, . . . are the coefficients of any convergent power series (e. g., â i = 0, i > m), and such that for some positive ϵ1 , Fm+1 (t, a2 , a3 , . . . , am+1 ) ≠ 0
for
t − a < ϵ1 .
In particular, a2 , a3 , . . . , am+1 , â i may all be taken equal to zero. Let ϵ1 > 0 and ϵ1 ≤ ϵ1 such that no root of unity of order less that m is in the range t − a ≤ ϵ1 . The existence of such a ϵ1 is ensured by the fact that there is only a finite number of such roots of unity. Then F (t, a , a , . . . , a ) m+1 2 3 m+1 ≥ μm+1 , m t (1 − t) ⋅ ⋅ ⋅ (1 − t m−1 )
t − a ≤ ϵ1 ,
for some μm+1 > 0, where μm+1 is as large as desired. By adding the factor (1 − t m ) in the denominator we may find a positive ϵ1 ≤ ϵ1 such that F (t, a , a , . . . , a ) 2 3 m+1 > λm+1 , m m+1 t (1 − t) ⋅ ⋅ ⋅ (1 − t m−1 )(1 − t m ) where λm+1 is as large as desired, for 0 < t − a < ϵ1 and t = 1.
4.2 Functional equations and formal power series  109
Now let p ∈ ℕ with p > m and let b be a primitive pth root of unity and a − b < ϵ1 /2. Again, there exists a positive ϵ2 < ϵ /2 such that for fixed values ai , 2 ≤ i ≤ p + 1 such that ai − â i  < δ, 2 ≤ i ≤ p + 1, and Fp+1 (t, a2 , . . . , ap+1 ) ≠ 0
for t − b < ϵ2 .
Here, a2 , . . . , am+1 are those fixed upon above and, again, in particular, am+2 , am+3 , . . . , ap may all be taken equal to zero. Let ϵ2 ≤ ϵ2 be a positive number such that no root of unity of order less than p is in the range t − b ≤ ϵ2 . Then, as above Fp+1 (t, a2 , a3 , . . . , ap+1 ) ≥ μp+1 , p t (1 − t) ⋅ ⋅ ⋅ (1 − t p−1 )
t − b ≤ ϵ2 ,
for some μp+1 > 0, and there exists a ϵ2 , 0 < ϵ2 ≤ ϵ2 such that Fp+1 (t, a2 , a3 , . . . , ap+1 ) p > λp+1 , t (1 − t) ⋅ ⋅ ⋅ (1 − t p−1 )(1 − t p ) where λp+1 is an arbitrary large number, for 0 < t − b < ϵ2 and t = 1. Again, choose r ∈ ℕ, r > p and let c be a primitive rth root of unity and b − c < ϵ2 /2 and continue as before, and so on. Let us denote m, p, r, . . . as N1 , N2 , N3 , . . . and denote a, b, c as r1 , r2 , r3 , and so on, inductively we construct a increasing sequence ∞ ∞ {Nk }∞ k=1 ⊆ ℕ, a sequence {rk }k=1 ⊆ ℝ and a sequence {ϵk }k=1 ⊆ (0, ∞) such that FN +1 (t, a2 , a3 , . . . , aN +1 ) k > λNk +1 , N k t k (1 − t)(1 − t 2 ) ⋅ ⋅ ⋅ (1 − t Nk )
(4.15)
for 0 < t − rk  < ϵk , t = 1, k ∈ ℕ where rk is a primitive Nk th root of unity and rk+1 − rk  < ϵk /2. In (4.15), each λNk +1 is an arbitrary large number and ai , i = 2, 3, . . . , are the coefficients of a convergent power series. The range t − rk+1  < ϵk+1 , where t = 1, is contained in the range t − rk  < ϵk for k ∈ ℕ, and the roots of unity contained in the ith range are all of an order greater than the Ni th (i ∈ ℕ), where {Ni }∞ i=1 ⊆ ℕ increases indefinitely. By the nested interval principle, there is a unique t, say t = a1 , common to all these ranges and this value can not be a root of unity. Then a1 is in all these ranges with the points t = rk all deleted (0 < t − rk ). Thus, all inequalities in (4.15) hold for this one value of t. Consequently, let αi = ai , i ∈ ℕ, recalling (4.131 ) and the definition of γ1 , we have cNk +1  > γ1  λNk +1
for all k ∈ ℕ.
Recall that each λNk +1 > 0 is as large as desired, say λNk +1 = (k!)k , then the sequence {cNk +1 1/k }∞ k=1 is unbounded if γ1 ≠ 0. Therefore, every formal solution ϕ(x) = c1 x + c2 x2 + c3 x3 + ⋅ ⋅ ⋅ ,
c1 ≠ 0,
110  4 Applications of formal power series, II of the functional equation ϕ[f (x)] = a1 ϕ(x) is divergent for all values of x ≠ 0, if f (x) = a1 x + a2 x2 + a3 x3 + ⋅ ⋅ ⋅ ,
a1  = 1, an1 ≠ 1 ∀ n ∈ ℕ,
is taken as a convergent power series. We complete the proof. A year later the author of Theorem 4.2.1 investigated the functional equation f [f (x)] = g(x) and established the following theorem. Theorem 4.2.2. There is an analytic function g(x) = a1 x + a2 x2 + a3 x3 + ⋅ ⋅ ⋅ ,
a1  = 1, an1 ≠ 1 ∀ n ∈ ℕ,
such that the functional equation f [f (x)] = g(x) has no solution which is analytic about the origin, i. e., every formal solution f (x) = c1 x + c2 x 2 + c3 x 3 + ⋅ ⋅ ⋅ is divergent for all values of x except x = 0. The proof is a construction proof and is not short. The proof can be found in [64] if the readers are interested in. Another interesting functional equation is called Dhombres functional equation [17]: f (zf (x)) = f 2 (x)
(4.16)
which may have either locally analytic or formal solution in the real domain. The generalized Dhombres functional equation is defined as f (zf (z)) = φ(f (z)),
(4.17)
in the complex domain, where φ is given. Let g ∈ 𝕏(ℂ) with ord(g) = k ∈ ℕ and consider that f (0) = w0 with w0 ≠ 0, w0n ≠ 1 for every n ∈ ℕ, and f is not a constant. We introduce a transformed generalized Dhombres functional equation ̃ g(w0 z + zg(z)) = φ(g(z)), ̃ − w0 ). where φ̃ is defined by φ(y) = w0 + φ(y
(4.171 )
4.2 Functional equations and formal power series  111
tion
The other formal functional equation we like to introduce is the translation equaF(s + t, x) = F(s, F(t, x)),
s, t ∈ ℂ,
(4.18)
for Ft (x) = F(t, x) = ∑n≥1 cn (t)xn , t ∈ ℂ. The work we are going to introduce below was introduced by Fripertinger and Reich recently which can be found in [22]. A family {Ft }t∈ℂ which satisfies (4.18) is called iteration group, and neglecting the trivial iteration group, there are two types of such groups, namely iteration group of type I where the coefficient c1 is a generalized exponential function different from 1, and iteration group of type II, where c1 = 1. It is known that for each iteration group of type II there exists an integer k ≥ 2 such that ∞
F(t, x) = x + ∑ cn (t)xn ,
t ∈ ℂ,
n=k
where ck : ℂ → ℂ is an additive function different from 0. Let k ∈ ℕ be given with k ≥ 2, a family (Ft )t∈ℂ is an iteration group of type II if and only if the system satisfies cn (s + t) = cn (s) + cn (t),
k ≤ n ≤ 2k − 2,
c2k−1 (s + t) = c2k−1 (s) + c2k−1 (t) + kck (s)ck (t),
c2k (s + t) = c2k (s) + c2k (t) + kck (s)ck+1 (t) + (k + 1)ck+1 (s)ck (t)
and for n > 2k, cn (s + t) = cn (s) + cn (t) + kck (s)cn−(k−1) (t)
+ (n − (k − 1))cn−(k−1) (s)ck (t) + P̃ n (ck (s), ck+1 (s) . . . , cn−k (s), ck (t), . . . , cn−k (t)),
for all s, t ∈ ℂ, where P̃ n are universal polynomials which are linear in ck (x), ck+1 (x), . . . , cn−k (x). Comparing the coefficients in cn (s + t) = cn (t + s),
n ≥ 2k,
we can prove that there exists a sequence of polynomials {Pn }n≥k so that cn (s) = Pn (ck (s)),
∀ s ∈ ℂ,
n ≥ k,
and F(s, x) = x + ck (s)xk + ∑ Pn (ck (s))xn , n>k
s ∈ ℂ.
112  4 Applications of formal power series, II By the system of {cn }, we have Pn (ck (s) + ck (t)) = Pn (ck (s + t)) = cn (s + t)
= Pn (ck (s)) + Pn (ck (t)) + kck (s)Pn−(k−1) (ck (t))
+ (n − (k − 1))Pn−(k−1) (ck (s))ck (t) + P̃ n (ck (s), . . . , Pn−k (ck (s)), ck (t), . . . , Pn−k (ck (t))),
for all s, t ∈ ℂ, and n ≥ k, where Pj = 0 for j < k and P̃ j = 0 for j ≤ 2k. Replacing ck (s) and ck (t) by independent variables y and z, we have Pn (y + z) = Pn (y) + Pn (z) + kyPn−(k−1) (z) + (n − (k − 1))Pn−(k−1) (y)z + P̃ n (y, . . . , Pn−k (y), z, . . . , Pn−k (z)) for all n ≥ k. Writing G(y, x) = x +yxk +∑n≥k+1 Pn (y)x n as a formal power series over ℂ[y], we deduce from the above formula that G satisfies the formal translation equation of type II, G(y + z, x) = G(y, G(z, x))
(4.181 )
in 𝕏(ℂ[y, z]), or a formal power series over ℂ[y, z], the ring of polynomials in y and z over ℂ. We call G(y, x) a formal iteration group of type II. It also satisfies the condition G(0, x) = x.
(4.182 )
We conclude the formal translation functional equation problem by the next theorem. Theorem 4.2.3. The formal power series F(s, x) = x + ck (s)xk + ∑ Pn (ck (s))x n n>k
on ℂ is a solution of (4.18) if and only if G(y, x) = x + yx k + ∑ Pn (y)x n n>k
is a solution of (4.181 ) and (4.182 ). Another kind of functional equations involving formal solutions or formal power series solutions are equations for generalized logarithmic functions and generalized exponential functions. Definition 4.2.4. The function F satisfying F(x + 1) = eF(x)
4.2 Functional equations and formal power series  113
is called a generalized exponential function, and the function G satisfying G(ex ) = G(x) + 1 is called a generalized logarithmic function. If G ∘ F is defined, we have G(F(x + 1)) = G(eF(x) ) = G(F(x)) + 1. We suggest the readers to visit the last part of Section 2.2 where we discussed the formal logarithm and formal exponential series. These two kinds of functions are important in numerical analysis, where they are used in a new system of computer arithmetic, which has significant advantages over floatingpoint arithmetic, including freedom from overflow and underflow, and a more satisfactory error measure. The details can be found in [100] and [14]. Before closing this section, we feel that we must introduce the Böttcher’s equation, which was invented by the Polish mathematician, Lucjan Böttcher (1872–1937), a student of Sophus Lie. Definition 4.2.5. Böttcher’s equation is the functional equation n
F(h(z)) = (F(z)) , where h is a given analytic function on ℂ with a superattracting fixed point of order n at a, that is, h(z) = a + c(z − a)n + 𝒪((z − a)n+1 ), in a neighborhood of a, with n ∈ ℕ, n ≥ 2. The inverse functional of F is called the Böttcher function. In formal analysis, the iterative inverse F [−1] of the solution F of the Böttcher’s equation is also called the Böttcher function, the function representation of the formal power series F [−1] . If F ∘ h ∈ 𝕏1 , the Böttcher’s equation above can be transformed to (L ∘ F)(h(z)) = n(L ∘ F)(z). Taking ϕ = L ∘ F and a1 = n, we have the Schröder’s functional equation ϕ[f (x)] = a1 ϕ(x), as in Theorem 4.2.1 with different condition for a1 . We introduce and discuss the Böttcher equation in formal analysis, or, we only consider that F ∈ 𝕏(ℂ) and h ∈ 𝕏0 (ℂ). It is clear that h(0) = 0.
114  4 Applications of formal power series, II Proposition 4.2.6. Let h ∈ 𝕏0 (ℂ) be given. If the Böttcher equation n
F(h(z)) = (F(z)) has solutions, then
ord(F)(ord(h) − n) = 0. The proof is trivial. A example with ord(h) = n = 2 was presented in (p. 49, [44]). We introduce it in the sense of formal analysis. Example 4.2.7. Let h(x) = x 2 + 2x4 + 22 x 6 + ⋅ ⋅ ⋅ + 2n x 2n+2 + ⋅ ⋅ ⋅ . The Böttcher function of the Böttcher’s equation G(h(x)) = G2 (x) is the formal power series H(x) = x − x3 + x5 − ⋅ ⋅ ⋅ + (−1)n x2n−1 + ⋅ ⋅ ⋅ . Proof. It is clear that ord(h) = 2. The function representations of h and H are h(x) =
x2 1 − 2x2
and H(x) =
x , 1 + x2
respectively. Then the inverse functional of H is the solution of the quadratic equation xy2 − y + x = 0, or G(x) = H [−1] (x) =
1 ± √1 − 4x 2 . 2x 2
1−4x In order to satisfy that G ∈ 𝕏0 , we take G(x) = 1− 2x . Applying the formal binomial 2 B1/2 composes the formal power series −4x , we have √
1/2 1/2 1 2 (1 − [1 + ( )(−4x2 ) + ( )(−4x2 ) + ⋅ ⋅ ⋅]) 2x 1 2 1 1/2 n = (2x2 + 2x 4 + 4x6 + ⋅ ⋅ ⋅ + ( )(−4x2 ) + ⋅ ⋅ ⋅) 2x n
G(x) =
= x + x3 + 2x 5 + ⋅ ⋅ ⋅ .
Also, using the above calculation we have G(h(x)) = =
(1 − 2x 2 ) − √1 − 4x2 2x2 2 −2x + (2x2 + 2x 4 + 4x6 + ⋅ ⋅ ⋅ + (1/2 )(−4x2 )n + ⋅ ⋅ ⋅) n
2x2 1/2 (−4x 2 )n = x2 + 2x 4 + ⋅ ⋅ ⋅ + ( ) + ⋅⋅⋅. n 2x2
4.3 Lagrange inversion
 115
On the other side, 1/2 (−4x2 )n + ⋅⋅⋅. ) n 2x2
G2 (x) = x2 + 2x4 + ⋅ ⋅ ⋅ + (
Thus, H is the Böttcher function of the equation G(h(x)) = G2 (x). Böttcher’s theorem is another interesting result we would like to introduce. However, this theorem discusses the iteration and iterative roots of formal power series. We will present it to the readers in Chapter 8.
4.3 Lagrange inversion Let g ∈ 𝕏(ℂ) be given, the simplest form of Lagrange inversion formula is f (z) = zg(f (z)).
(4.19)
The existence and uniqueness of the formal power series solution of Lagrange inversion equation are the topics we are going to discuss here. Theorem 4.3.1. The Lagrange inversion formula (4.19) has solution if and only if g ∈ 𝕏(ℂ) is a unit and f is an almost unit in 𝔻(ℂ). If f is a solution of (4.19) for some unit g ∈ 𝕏(ℂ), then f is unique. Proof. If the formal power series f and g satisfy (4.19), then ord(f ) = ord(zg(f (z))) = ord(z) + ord(g(f (z))) ≥ 1, and hence f ∈ 𝕏0 (ℂ). Applying Theorem 1.2.5, ord(f ) = ord(zg(f (z))) = 1 + ord(g) ⋅ ord(f ). Then ord(g) =
ord(f ) − 1 1 =1− . ord(f ) ord(f )
Since ord(g) must be an nonnegative integer and ord(f ) ∈ ℕ, it follows that ord(f ) = 1
and then
ord(g) = 0.
z Now let g be a unit and f be an almost unit in 𝕏(ℂ). Then g(z) is an almost unit. z Therefore, g(z) (f (z)) ∈ 𝔻(ℂ), still an almost unit formal power series. Then the equation
z f (z) (f (z)) = =z g(z) g(f (z))
116  4 Applications of formal power series, II has a solution f (z) = (
z ) g(z)
[−1]
,
which is a solution of (4.19). We know that 𝔻 is a group with composition. The uniqueness of the iterative inz [−1] verse ( g(z) ) yields the uniqueness of f . We are interested in the nontrivial solution f which, of course, is an almost unit. There are many other proofs of Lagrange inversion formula and many constructions of the solution f of the Lagrange inversion equation including the proof by Lagrange himself [53]. For example, the residues proof, the factorization proof [33, 34], and Raney’s proof [72] are some popular results. Our construction proof of (4.19) is simple, all tools we need can be found in previous chapters. We also provide a matrix form for the construction of the nontrivial solution of the Lagrange inversion equation. Theorem 4.3.2. Let g ∈ 𝕏(ℂ) be given such that g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ ,
b0 ≠ 0.
Let f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅, a1 ≠ 0 be a nontrivial solution of the Lagrange inversion equation ∞
f (z) = z ∑ bn f n (z), n=0
(4.20)
which is equivalent to (4.19), then k−1
ak = ∑ bj ak−1 , j=0
(j)
k = 1, 2, . . . ,
(4.21)
where we write (n) (n) 2 (n) f n (z) = a(n) 0 + a1 z + a2 z + a3 + ⋅ ⋅ ⋅ ,
for all n ∈ ℕ ∪ {0}, as we defined in Definition 1.2.1. Proof. Denote f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ where a0 = 0 and a1 ≠ 0. We know that a(n) = 0, k < n for all n ∈ ℕ. Since f is a nonunit, it follows that g ∘ f ∈ 𝕏(ℂ), and then k we write g(f (z)) = c0 + c1 z + c2 z 2 + c3 z 3 + ⋅ ⋅ ⋅ ,
4.3 Lagrange inversion
 117
where, as in (1.2), n
cn = ∑ bk a(k) n ,
n = 0, 1, 2, . . . .
k=0
Suppose that f is a solution of (4.20), then a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ = z(c0 + c1 z + c2 z 2 + c3 z 3 + ⋅ ⋅ ⋅), and then ak = ck−1
for all k ∈ ℕ.
Then a1 = c0 = b0 , a2 = c1 = b1 a(1) 1 = b1 b0 . Assume that a1 , a2 , . . . , ak have been determined. Applying Corollary 1.4.7 with j = 2, 3, . . . , k, consecutively we have a0
(j)
T
a0
(j−1)
T
[ (j−1) ] [a0 [ (j) ] ] [a [a ] 0 [ 1 ] [ [ 1 ] [ (j−1) ] [ [ (j) ] [ 0 ] [a ] = [a [ 2 ] [ [ 2 ] .. [ . ] [ [ . ] [ . ] [ [ . ] [ . ] [. [ . ] (j−1) (j) 0 [ak ] [ [ak ]
a1 a0 0 .. . 0
a2 a1 a0 .. . 0
... ... ... .. . ...
ak ] ak−1 ] ] ak−2 ] ]. .. ] . ] ] a0 ]
Since a0 = 0, the above formula becomes a1
(j)
T
a1
(j−1)
T
[ (j−1) ] [a1 [ (j) ] ] [a [a ] 0 [ 2 ] [ [ 2 ] [ (j−1) ] [ [ (j) ] 0 ] [ [a ] = [a [ 3 ] [ [ 3 ] .. [ . ] [ [ . ] [ . ] [ [ . ] [ . ] [. [ . ] (j−1) (j) 0 [ak ] [ [ak ]
a2 a1 0 .. . 0
a3 a2 a1 .. . 0
... ... ... .. . ...
a1 [ [0 [ (j−1) (j−1) [ 0 = [0, . . . , 0, aj−1 , . . . , ak ] [ [. [ .. [ [0
where 2 ≤ j ≤ k, and a(i) = 0 if j < i. j
ak ] ak−1 ] ] ak−2 ] ] .. ] . ] ] a1 ] a2 a1 0 .. . 0
a3 a2 a1 .. . 0
... ... ... .. . ...
ak ] ak−1 ] ] ak−2 ] ], .. ] . ] ] a1 ]
118  4 Applications of formal power series, II We can obtain all a(i) , 1 ≤ j ≤ k, 1 ≤ i ≤ k. Then, for k ∈ ℕ, j k
ak = ck−1 = ∑ bj ak ∈ ℂ (j)
j=1
is determined. Again, we may have all a(i) , 1 ≤ j ≤ k, 1 ≤ i ≤ k. We complete the j induction, and hence we have the solution f of equation (4.20). Let us introduce the residue proof by Gessel [34]. As usual, we define the residue of a Laurent series f (x) = ∑n∈ℤ fn xn to be res(f ) = f−1 , where (fn )n∈ℤ ⊆ S and S is a field. We call the series f−m z −m + f−m+1 z −m+1 + ⋅ ⋅ ⋅ + f−1 z −1 + f0 + f1 z + f2 z 2 + ⋅ ⋅ ⋅ , n or ∑∞ n=−m fn z for some m ∈ ℕ, a formal Laurent series over S in this section, a terminology used by Henrici [42]. It will be renamed to be the semiLaurent series in Chapter 7.
Lemma 4.3.3. Let f be a formal Laurent series over ℂ such that ∞
f (x) = ∑ fn xn ,
(4.22)
n=−m
n for some m ∈ ℕ, and let g(x) = ∑∞ n=1 gn x with g1 ≠ 0. Then
res(f (x)) = res(f (g(x))g (x)). Proof. By linearity, it is sufficient to prove the formula when f (x) = xk for k ∈ ℤ, or res(xk ) = res(g k (x)g (x)). It is clear that res(xk ) = 0 if k ≠ −1. Also res(g k (x)g (x)) = res(
d g k+1 (x) ( )) = 0, dx k + 1
because g k+1 does not exist if k < −1, and the residue of obviously if k > −1. If k = −1, then res(x k ) = 1. On the other side, res(g k (x)g (x)) = res( = res(
d (g k+1 (x)) dx
g (x) ) g(x)
g1 + 2g2 x + 3g3 x 2 + ⋅ ⋅ ⋅ ) g1 x + g2 x 2 + g3 x 3 + ⋅ ⋅ ⋅
1 g + 2g2 x + 3g3 x 2 + ⋅ ⋅ ⋅ ) = res( ⋅ 1 x g1 + g2 x + g3 x 2 + ⋅ ⋅ ⋅ = 1,
equals to zero
4.3 Lagrange inversion
because
g1 +2g2 x+3g3 x 2 +⋅⋅⋅ g1 +g2 x+g3 x2 +⋅⋅⋅
 119
is a unit, and g1 + 2g2 x + 3g3 x2 + ⋅ ⋅ ⋅ = 1 + h(x), g1 + g2 x + g3 x2 + ⋅ ⋅ ⋅
for some h ∈ 𝕏0 with ord(h) ≥ 1. In this chapter, the Laurent series we mentioned are the same as what we defined in (4.22). We now introduce the notation [xn ]f (x) which denotes the coefficient of the term of xn in the Laurent series f (x). It is easy to see that [xn ]xk f (x) = [xn−k ]f (x). Theorem 4.3.4. Let R(t) be a formal power series in t over ℂ. Then there is a unique formal power series f ∈ 𝕏(ℂ) such that f (x) = xR(f (x)),
(4.23)
and for any formal Laurent series ϕ(t) and ψ(t) in t, as defined in (4.22), and for any n ∈ ℤ we have 1 n−1 [t ]ϕ (t)Rn (t), n ≠ 0, n [xn ]ϕ(f ) = [t n ](1 − tR (t)/R(t))ϕ(t)Rn (t), [x n ]ϕ(f ) =
∞
ϕ(f ) = ∑ xn [t n ](1 − xR (t))ϕ(t)Rn (t), n=m
ψ(f ) = [t n ]ψ(t)Rn (t), 1 − xR (f ) ψ(f ) [xn ] = [t n ]ψ(t)Rn (t). 1 − fR (f )/R(f ) [xn ]
(4.231 ) (4.232 ) (4.233 ) (4.234 ) (4.235 )
Proof. We show that all formulas from (4.231 ) to (4.235 ) are equivalent in the sense that any one of them is easily derivable from any other. If (4.23) has a solution, then R−1 ∈ 𝕏 by Theorem 4.3.1. Then (4.234 ) and (4.235 ) are equivalent because x = f /R(f ). Taking ψ(t) = (1 − tR (t)/R(t))ϕ(t), the righthand side of (4.232 ) becomes the righthand side of (4.235 ), and the leftside of (4.232 ) becomes the lefthand side of (4.235 ). So, (4.232 ) and (4.235 ) are equivalent. We rewrite (4.234 ) as ψ(f ) = ∑ ([t n ]ψ(t)Rn (t))xn . 1 − xR (f ) n≥m Putting ϕ(f ) =
ψ(f ) 1−xR (f )
in (4.234 ) and using (4.236 ), we obtain (4.233 ).
(4.236 )
120  4 Applications of formal power series, II Since we assume that f is a nontrivial solution of (4.23), it follows that the Lagrange inversion formula is equivalent to x = f /R(f ). Let g(t) = t/R(t), we have x = g(f ), and then g = f [−1] . By (4.232 ), we have [xn ]ϕ(f ) = [t n ](1 − tR (t)/R(t))ϕ(t)Rn (t) = [t n ](1 − g(t)R (t))ϕ(t)Rn (t) = [t n ](1 − g(t)( = [t n ]t = [t n ]
n
t t ) )ϕ(t)( ) g(t) g(t)
tn g (t) ϕ(t) n g(t) g (t)
ϕ(t)g (t) n+1 t . g n+1 (t)
Then we have [xn ]ϕ(f ) = [t −1 ]
ϕ(t)g (t) , g n+1 (t)
(4.237 )
which is equivalent to (4.232 ). ψ(t) Let ϕ(t) = tg (t) , using (4.237 ) we have [xn ]
ψ(f ) ψ(t)g (t) n+1 = [t n ] t fg (f ) tg (t)g n+1 (t) t n ψ(t) ψ(t) = [t n ] n+1 = [t 0 ] n+1 , g (t) g (t)
and then [xn−1 ]
ψ(t) ψ(f ) = [t 0 ] n . fg (f ) g (t)
(4.238 )
Formula 4.238 is equivalent to (4.232 ) and (4.235 ) because the later two formulas are equivalent. If n ≠ 0, (4.231 ) provides n[xn ]ϕ(f ) = [t n−1 ]ϕ (t)Rn (t) = [t n−1 ]ϕ (t)
tn −1 ϕ (t) = [t ] . g n (t) g n (t)
To show that (4.231 ) and (4.232 ) are equivalent, we first show that n[t −1 ]
ϕ(t)g (t) ϕ (t) = [t −1 ] n . n+1 g (t) g (t)
But ϕ (t) ϕ(t)g (t) d ϕ(t) − n n+1 = ( n ), n g (t) dt g (t) g (t)
(4.239 )
4.3 Lagrange inversion
 121
because the coefficient of t −1 in the derivative of any Laurent series is 0, Then (4.239 ) is true. By (4.237 ) and (4.239 ), [xn ]ϕ(f ) = [t −1 ]
ϕ(t)g (t) 1 −1 ϕ (t) = [t ] n n g (t) g n+1 (t)
1 −1 −n [t ]t ϕ (t)Rn (t) n 1 = [t n−1 ]ϕ (t)Rn (t). n
=
This is (4.231 ), and hence (4.231 ) and (4.232 ) are equivalent if n ≠ 0. We continue to show the equivalence of formulas stated in the above theorem. Theorem 4.3.5. Let f (x) and g(x) be compositional inverses. Then for any formal Laurent series ϕ, [xn ]ϕ(f ) = res(
ϕ(f (g))g ϕ(t)g ϕ(f ) ) = res( ) = res( n+1 ). n+1 n+1 x g g
Proof. If f and g are compositional inverses, applying (4.237 ) we have res(
ϕ(f ) ϕ(t)g ϕ(f (g))g n ) = [x ]ϕ(f ) = res( ) = res( ). xn+1 g n+1 g n+1
Some equalities in Theorem 4.3.4 can be proved by using some different approach. We use the mathematical induction to prove (4.232 ) but we suppose that ϕ is a formal power series, not the formal Laurent series. Theorem 4.3.4 is more general. Theorem 4.3.6. Let notation be set as in Theorem 4.3.4. A formal power series f ∈ 𝕏0 (ℂ) satisfies (4.232 ) [xn ]ϕ(f ) = [t n ](1 −
tR (t) )ϕ(t)Rn (t) R(t)
for any formal power series ϕ(t) is equivalent to that f is a solution of the Lagrange inversion formula (4.23). Proof. It is clear that R is a unit. Then (1 − tR /R) is a unit with the constant term 1. Also, we know that f is a nonunit, and hence ϕ(f ) ∈ 𝕏(ℂ). If n = 0, the [x0 ]ϕ(f ) = ϕ(0). On the other side, [t 0 ](1 −
tR (t) )ϕ(t)R0 (t) = 1 ⋅ ϕ(0) = ϕ(0). R(t)
The formula is true. Suppose that (4.232 ) is true for 0 ≤ j ≤ m for some m ∈ ℕ ∪ {0}, or [xj ]ϕ(f ) = [t j ](1 −
tR (t) )ϕ(t)Rj (t), R(t)
0 ≤ j ≤ m.
122  4 Applications of formal power series, II We show that the formula is true for n = m + 1. By linearity, it is enough to prove the formula for ϕ(t) = t k for k ∈ ℕ ∪ {0}. In this case, ϕ(f ) = f k , using the relation f = xR(f ), [xm+1 ]f k = [xm+1 ](f k−1 ⋅ xR(f )) = [xm ]f k−1 R(f ) tR (t) k−1 m )t R (t)R(f ) R(t) tR (t) k m+1 = [t m+1 ](1 − )t R (t). R(t) = [t m ](1 −
The popular Catalan number Cn could be an application of the Lagrange revision equation (4.23). n The Catalan generating function c(x) = ∑∞ n=0 Cn x is defined by 2
c(x) = 1 + x[c(x)] .
(4.2310 )
The quadratic equation has two solutions (1 ± √1 − 4x)/(2x), but only the minus sign gives a nonunit formal power series. We take c(x) =
1 − √1 − 4x , 2x
which is also the formula (3.7). In order to apply Lagrange inversion, we set up f (x) = c(x) − 1,
so that f = x(1 + f )2 .
We may then apply Theorem 4.3.4 to the case R(t) = (1 + t)2 . The equation f = x(1 + f )2 has the solution f (x) =
1 − √1 − 4x − 1 = c(x) − 1 = xc2 (x). 2x
Then (4.231 ) with ϕ(t) = (1 + t)k gives k
[xn ](c(x)) = [xn ](1 + f )k = =
1 n−1 [t ]k(1 + t)k−1 (1 + t)2n n
k 2n + k − 1 ( ), n n−1
for n ∈ ℕ. Since the constant term in ck (x) is 1, we have k
k 2n + k − 1 n ( )x . n n−1 n=1 ∞
ck (x) = [c(x)] = 1 + ∑
Lagrange inversion formula was originally introduced by J. L. Lagrange [53] in 1770 which was then generalized by H. H. Bürmann [54] in the late 18th century. There are also many generalizations and applications of Lagrange–Bürmann formula. An algebraic proof of Lagrange–Bürmann formula by Henrici will be presented in Chapter 7 (Formal Laurent Series), a suitable place for that formula and proof.
4.4 Difference equations and formal power series  123
4.4 Difference equations and formal power series Definition 4.4.1. Let aj ∈ ℂ, 0 ≤ j ≤ k be given for some k ∈ ℕ. For any {sn }∞ n=k ⊆ ℂ, the equation a0 xn + a1 xn−1 + ⋅ ⋅ ⋅ + ak xn−k = sn ,
(4.24)
for n = k, k + 1, k + 2, . . . , and a0 ak ≠ 0, is called a linear difference equation of order k. The equation is called homogeneous if sn = 0, n ≥ k, otherwise it is called inhomogeneous. A sequence {xj }∞ j=0 satisfying (4.24) for all n ≥ k is called a solution of the difference equation (4.24). We always suppose that a0 ak ≠ 0, otherwise we face a reduced (4.24). It is clear that for every given set of k numbers x0 , x1 , . . . , xk−1 there is a unique solution {xn }∞ n=k of the equation (4.24). The equation (4.24) plays a role of recurrence formula. A popular approach of solving the linear homogeneous difference equation is using linear difference operator and the characteristic polynomial that will be introduced below. Definition 4.4.2. Let V be the linear vector space of sequences V = {x = (x0 , x1 , x2 , . . . ) : xn ∈ ℂ, n ∈ ℕ}, that is, V is closed with the linear operation. We denote that (x)n = xn
for all n ∈ ℕ ∪ {0}.
The mapping E : V → V such that Ex = (x1 , x2 , x3 , . . . )
if x = (x0 , x1 , x2 , x3 , . . . )
is called the shifting operator on V. It is easy to verify that E 2 x = E(Ex) and (Ex)n = xn+1 , (E 2 x)n = xn+2 , as usual, E 0 = I or E 0 (x) = I(x) = x. A linear difference operator L is usually denoted by m
L = ∑ ci E i , i=0
ci ∈ ℂ,
i for some m ∈ ℕ ∪ {0}. If p is a polynomial such that p(λ) = ∑m i=0 ci λ , then p is called the characteristic polynomial of L.
Proposition 4.4.3. The difference equation a0 xn + a1 xn−1 + ⋅ ⋅ ⋅ + ak xn−k = sn
124  4 Applications of formal power series, II is equivalent to p(E)x = sn ,
n ≥ k,
k
p(λ) = ∑ an λn−k , n=0
where x = (x0 , x1 , x2 , . . . ). There are several fundamental theorems for solving linear difference equations by using characteristic polynomials, which can be found in all text books of numerical analysis. We introduce a method which solves the linear difference equation by formal power series. We first consider the homogeneous equation a0 xn + a1 xn−1 + ⋅ ⋅ ⋅ + ak xn−k = 0,
(4.25)
n = k, k + 1, k + 2, . . . . Theorem 4.4.4. Let x0 , x1 , . . . , xk−1 be the initial values satisfying equation (4.25). Let A ∈ 𝕏(ℂ) and S ∈ 𝕏(ℂ) be given with A(z) = a0 + a1 z + ⋅ ⋅ ⋅ + ak z k ,
S(z) = s0 + s1 z + ⋅ ⋅ ⋅ + sk−1 z k−1 ,
where sm = a0 xm + a1 xm−1 + ⋅ ⋅ ⋅ + am x0 ,
0 ≤ m ≤ k − 1.
(4.26)
Then A−1 exists and X = A−1 S, where X(z) = x0 + x1 z + x2 z 2 + ⋅ ⋅ ⋅ = (x0 , x1 , x2 , . . . ) is a formal power series and a solution of (4.25). Proof. A−1 ∈ X(ℂ) because a0 ≠ 0. Let X(z) = x0 + x1 z + x2 z 2 + ⋅ ⋅ ⋅ , then the Cauchy product provides k
AX = ( ∑ aj z j )X j=0
= a0 x0 + (a0 x1 + a1 x0 )z + ⋅ ⋅ ⋅ + (a0 x k−1 + ⋅ ⋅ ⋅ ak−1 x0 )z k−1 k
+ (a0 xk + ⋅ ⋅ ⋅ + ak x0 )z k + ( ∑ ai xk+1−i )z k+1 + ⋅ ⋅ ⋅ i=0
= a0 x0 + (a0 x1 + a1 x0 )z + ⋅ ⋅ ⋅ + (a0 x k−1 + ⋅ ⋅ ⋅ ak−1 x0 )z k−1 = s0 + s1 z + ⋅ ⋅ ⋅ + sk−1 z k−1 = S,
4.4 Difference equations and formal power series  125
because ∑ki=0 ai xn−i = 0 for n ≥ k by (4.25). Then X = A−1 S. Both formal power series A and S terminate, and thus have an infinite radius of convergence. They represent analytic function, which happen to be polynomials. Then the formal power series X is a regular power series that represents the rational function x(z) :=
S(z) s0 + s1 z + ⋅ ⋅ ⋅ + sk−1 z k−1 = A(z) a0 + a1 z + ⋅ ⋅ ⋅ + ak z k
near z = 0. Conversely, every power series X that represents a rational function of x of the above form satisfies X = A−1 S, hence AX = S in Theorem 4.4.4, The rational function x(z) of the above form is called a generating function of a solution of (4.25). Henrici also investigated the case if x(z) has distinct poles such as z1 , z2 , . . . , zh , of the respective orders m1 , m2 , . . . , mh , where ∑hi=1 mi = k, and provided a solution for it as well as provided an algorithm for X [42]. Theorem 4.4.5. Let all notation be set as in Theorem 4.4.4 and let x(z) be the corresponding generating function of (4.25). Suppose x(z) has distinct poles z1 , z2 , . . . , zh with orders, or multiplicities m1 , m2 , . . . , mh , m1 + m2 + ⋅ ⋅ ⋅ + mh = k, respectively. Then the coefficient xn of X have the form h
xn = ∑ pi (n)zi−n ,
n ∈ ℕ ∪ {0},
i=1
(4.27)
where pi (n) is a polynomial of degree (mi − 1). Proof. As a rational function, x(z) has poles described above. Suppose the partial fractions of x(z) is h
mj
i=1
j=1
ai,j
x(z) = ∑(∑
(z − zi )j
).
For any zi , 1 ≤ i ≤ h, and 1 ≤ j ≤ mi , applying formal binomial series we have 1 1 = (−1)j j (1 − z/zi )−j j (z − zi ) z i
= (−1)j
k
1
∞ −j z ∑ ( )(− ) j zi zi k=0 k ∞
j + k − 1 1 zk ) j k. k z zi
= (−1)j ∑ ( k=0
i
126  4 Applications of formal power series, II zn zin
Define pi (n) as the coefficient of the term of mi
in x(z), then j+n−1 1 ) j, n z
pi (n) = ∑(−1)j ai,j ( j=1
n
and hence the coefficient of z is
pi (n)zi−n .
i
Thus,
h
xn = ∑ pi (n)zi−n ,
n ∈ ℕ ∪ {0},
i=1
where pi (n) is a polynomial of degree (mi − 1). Example 4.4.6. The Fibonacci sequence {xn }∞ n=0 = {0, 1, 1, 2, 3, 5, 8, 13, . . . } is defined by the difference equation xn − xn−1 − xn−2 = 0, with initial conditions x0 = 0, x1 = 1. Letting all notation be set as in Theorem 4.4.4, we have a0 = 1,
a1 = −1,
a2 = −1,
and s0 = a0 x0 = 0,
s1 = a0 x1 + a1 x0 = 1.
Then A(z) = 1 − z − z 2 ,
and
S(z) = z.
Then the generating function of the sequence X is x(z) =
z . 1 − z − z2
Then the poles of x are z1 =
√5 − 1 , 2
z2 =
−√5 − 1 . 2
Taking partial fractions for x(z), we have z1 z2 1 1 + −1 − 2z1 z − z1 −1 − 2z2 z − z2 1 1 1 1 = + . 1 + 2z1 1 − z/z1 1 + 2z2 1 − z/z2
x(z) =
Applying (4.27), we find xn =
1 1 z −n + z −n 1 + 2z1 1 1 + 2z2 2 n
=
n
1 √5 + 1 1 −√5 + 1 ( ) − ( ) . √5 √5 2 2
4.4 Difference equations and formal power series  127
The method we used in the proof of Theorem 4.4.4 can be applied to solve the nonhomogeneous linear difference equation (4.24) with sn not all zeros. Theorem 4.4.7. Let x0 , x1 , . . . , xk−1 be the initial values satisfying (4.24) a0 xn + a1 xn−1 + ⋅ ⋅ ⋅ + ak xn−k = sn , where {sn }∞ n=k ⊆ ℂ is given. Let A ∈ 𝕏(ℂ) and S ∈ 𝕏(ℂ) be given with A(z) = a0 + a1 z + ⋅ ⋅ ⋅ + ak z k ,
S(z) = s0 + s1 z + ⋅ ⋅ ⋅ + sk z k + ⋅ ⋅ ⋅ ,
where sm = a0 xm + a1 xm−1 + ⋅ ⋅ ⋅ + am x0 ,
m = 0, 1, 2, 3, . . . ,
(4.28)
is the nonhomogeneous terms in (4.24). Then A−1 exists, and hence X = A−1 S, where X(z) = x0 + x1 z + x2 z 2 + ⋅ ⋅ ⋅. Using the sequence representation of X, X = (x0 , x1 , . . . , xk , xk+1 , . . . ) is the solution of (4.24). Proof. Please be noticed that sn is the nonhomogeneous term in (4.24) for n ≥ k. Simply calculating as we did in the proof of Theorem 4.4.4, we obtain AX = S. Since a0 ≠ 0, A is a unit formal power series and hence A−1 exists, then we have X = A−1 S. Remark 4.4.8. The given sequence {sn } play an important role in solving the linear difference equations. 1. If the sequence {sn } grows too fast, for example, sn = n!, the formal power series S diverges everywhere except z = 0, the method in Theorem 4.4.5 is not efficient. 2. If the formal power series S has a positive radius of convergence and its sum function (analytic function) can be obtained, then Theorem 4.4.5 provides an analytical technique to determine xn .
5 General composition Chapters 1 through 4 introduced the theory and structure of the formal power series and many applications including the applications of the composition of formal power series with nonunits. Such composition gives us the right distributive law, iterative powers and iterative inverse of almost units, Lagrange inversion, Riordan group, and many others. Now we face the following challenges: 1. Does the formal power series f in the composition g ∘ f have to be a nonunit? 2. Can we keep the all properties of the composition introduced in Chapter 1 when we eliminate the requirement of the nonunitness for f in the composition g ∘ f ? 3. Is it possible to have a necessary and sufficient condition for the existence of the composition of formal power series? This chapter will answer those questions by setting up a general composition theorem including a sufficient and necessary condition for the existence of the composition of formal power series. The investigation with such a general composition will go deeper and wider in the following chapters. By the investigation of the coefficients of f n (z), the general composition theorem for formal power series was invented in 2002 by the author and N. Knox [30]. That paper provided an algorithm for computing the coefficients of f n (z) as well as a necessary and sufficient condition for the existence of the composition of formal power series. These results will be introduced in Sections 5.3 and 5.4. The general Chain Rule of formal power series will be introduced in Section 5.5. The general right distributive law will be introduced in Section 5.6. The full study of the matrix representation for composition will be split into two parts, Part I and Part II, that will be introduced in Section 5.2 and Section 5.7, respectively. Note that in Section 5.4 we provide an analytical approach to prove the general composition theorem [26] based on a new organization of the coefficients in the multinomial theorem. This new organization about the multinomial theorem could drive further research of this topic in combinatorics.
5.1 Introduction So far, the existence of the composition of formal power series has relied on the admitting addition of a sequence of formal power series. It is a very smart definition. By reviewing the development of the formal series, including formal power series, formal Laurent series with finitely many terms of negative exponents, formal solutions, or other formal subjects, we must say that the study of the formal series actually follows some rules, explicit or implicit. Let us conclude these rules in the beginning https://doi.org/10.1515/9783110599459005
130  5 General composition of the section. We will follow these rules when we make any development for formal power series. Remark 5.1.1. All operations on formal series space such as 𝕏(S) abide the two rules: 1. If a power series f has a positive radius of convergence, then as a formal power series, f should maintain all properties that f has in analysis. It is called the principle of formal analysis introduced in Remark 1.7.6; 2. Distribution rule is honored if an operation runs with infinitely many formal series. Let us see some examples before we go any further. Example 5.1.2. Let S = ℝ. Let ∞
g(x) = ∑ xn , n=0
f (x) = 1 + x.
Then ∞ ∞ n ∞ n n n ∑ (1 + x)n = ∑ ( ∑ ( )xk ) = 1 + ∑ (1 + ∑ ( )xk ). k k n=1 n=0 n=0 k=0 k=1
We cannot, following the distribution rule, even calculate the first coefficient of the n series ∑∞ n=0 (f (x)) . Thus, the composition g(f (x)) does not exist. Example 5.1.3. Let S = ℝ. Let ∞
g(x) = ∑ xn n=0
and
∞
f (x) = ∑ n! x n . n=1
n It is clear that the series ∑∞ n=1 n! x converges nowhere except x = 0. However, the composition g(f (x)), not the composition of functions, is a formal power series by Definition 1.2.2 and ∞
k
n=0
n=0
∈ℝ ck = ∑ bn a(n) = ∑ a(n) k k for every k ∈ ℕ. These two examples indicate some difference between the composition of formal power series and the composition of analytic functions. It is well known that the composition of a formal power series with a nonunit is defined. However, the nonunitness of the composed formal power series is just a sufficient condition for the existence of the composition. Can we obtain a sufficient and necessary condition for the existence of the composition? In Section 2.2, we established the formal logarithmic function (2.1) 1 1 1 L(F) = ln(1 + f ) = f − f 2 + f 3 − f 4 + ⋅ ⋅ ⋅ , 2 3 4
5.1 Introduction  131
where f must be a nonunit. Then f cannot be a nonzero constant formal power series, for example, f = ( 21 , 0, 0, . . . ) does not satisfy (2.1). On the other side, ln(1+ 21 ) should be fine by the principle of formal analysis because the Maclaurin series of ln(1+x) has the radius of convergence r = 1. This example tells us that we do need to extend or replace the restriction of admitting addition by some more general definition for operations of the sequence of formal power series. Therefore, we face challenges which, we know, is not easy at all, and that is why it is not surprised to read a concern Peter Henrici said in his book (p. 35, [42]) that “This (means cn , as above), in general, has no meaning; we have not defined convergence, and if we had we wouldn’t know whether it would take place here.” The formal analysis started its journey almost 20 years ago when the challenge or warning by Henrici was investigated. The general summation of a sequence of formal power series has been a question we must answer before we try to expand or to generalize the existing theory of formal power series. Definition 1.7.7 takes the first step in this road. Here, we begin to study the general composition of formal power series, under the two rules in Remark 5.1.1. Before the general composition of formal power series was established, besides of the admitting addition, we would like to introduce the other condition used to determine the existence of the composition. n Proposition 5.1.4 ([84]). Let f (x) = ∑∞ n=0 fn x be a formal power series over a ring S. If g is a formal power series such that
lim f g n→∞ n
n
n
lim 2− ord(fn g ) = 0, = n→∞
n then the sum ∑∞ n=0 fn g is a formal power series. This series is called the composition of f and g and is denoted by f ∘ g.
The proof of this proposition can be found in Proposition 1.7.5. Clearly, if g is a nonunit or f is a polynomial, then lim f g n→∞ n
n
= 0
is true. This proposition provides a sufficient condition for existence of the composition of formal power series. We now define the general composition of formal power series. Definition 5.1.5 ([30]). Let S be a ring with a metric and let 𝕏 be the set of all formal k power series over S. Let g ∈ 𝕏 be given, say g(x) = ∑∞ k=0 bk x . We define a subset 𝕏g ⊆ 𝕏 to be ∞
∞
𝕏g = {f ∈ 𝕏 f (x) = ∑ ak xk , ∑ bn a(n) ∈ S, k = 0, 1, . . . }, k n=0 k=0
(n) k 0 where f n (x) = ∑∞ k=0 ak x , for every n ∈ ℕ, and f = I.
132  5 General composition It is clear that 𝕏g ≠ 0 because all nonunits are in 𝕏g . Then the mapping Tg : 𝕏g → 𝕏 such that ∞
Tg (f )(x) = ∑ ck x k , k=0
(n) ∑∞ n=0 bn ak , k
where ck = = 0, 1, 2, . . . , is welldefined. We call Tg (f ) the general composition of the formal power series g and f . The composition Tg (f ) is also denoted by g ∘f . We call such composition the general composition of the formal power series because this composition does not require the nonunitness for the composed formal power series. If f is a nonunit, a(n) = 0 for all n > k, and then k k
ck = ∑ bn a(n) ∈ S, k n=0
for all k. The next example shows that the general composition definition really extends the composition with nonunits only and maintains the principle of formal analysis. Example 5.1.6. Let g(x) = 1 + x + x2 + ⋅ ⋅ ⋅ be a formal power series in 𝕏(ℝ) and let f (x) = 1/2, a unit. Then 1 1 g ∘ f (x) = 1 + + 2 + ⋅ ⋅ ⋅ = 2 or = 2I. 2 2
5.2 Matrix representation of the general composition, I Matrix, such as the matrix representation associated with multiplication for any formal power series and the delta associated matrix for an almost unit formal power series, is an efficient tool for both research and applications of formal power series. We now introduce a matrix representation for all formal power series, not only for almost units, with the composition if the composition exists. Definition 5.2.1. Let g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ + bk z k + ⋅ ⋅ ⋅ and f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ + ak z k + ⋅ ⋅ ⋅ be formal power series in 𝕏(S) where S is a ring. For every n ∈ ℕ ∪ {0}, we write (n) (n) 2 (n) k f n (z) = a(n) 0 + a1 z + a2 z + ⋅ ⋅ ⋅ + ak z + ⋅ ⋅ ⋅
as we did in Theorem 1.4.4. We define the composition matrix of f to be the matrix a(0) 0
[ (1) [ a0 [ [ (2) Cf = [ [ a0 [ (3) [ a0 [ .. [ .
a(0) 1
a(0) 2
a(0) 3
.. .
.. .
.. .
a(1) 1 (2) a1 a(3) 1
a(1) 2 (2) a2 a(3) 2
a(1) 3 (2) a3 a(3) 3
...
] . . .] ] ] . . .] ], ] . . .] ] .. .]
5.2 Matrix representation of the general composition, I
 133
and denote it as Cf . If we use the notation, (n) (n) f n = (a(n) 0 , a1 , a2 , . . . )
for k ∈ ℕ ∪ {0},
then a(0) 0 [ (1) [ a0 [ [ (2) Cf = [ [ a0 [ (3) [ a0 [ .. [ .
a(0) 1
a(0) 2
a(0) 3
.. .
.. .
.. .
a(1) 1 a(2) 1 a(3) 1
a(1) 2 a(2) 2 a(3) 2
a(1) 3 a(2) 3 a(3) 3
0
] [f 1 ] . . .] f ] ] [ ] [ 2] [ ] f . . .] ]. ]=[ 3] ] [ f [ ] . . .] ] [.] . .. .] [ . ] ...
If f is an almost unit such that f (z) = a1 z + a2 z 2 + ⋅ ⋅ ⋅, a1 ≠ 0, it is clear that 1 [ [0 [ [ 0 Cf = [ [ [ [0 [ .. [.
0 a(1) 1
0 a(1) 2
0
0 a(1) 3
a(2) 2
0 .. .
... ] . . .] ] ] . . .] ] ] . . .] ] .. .]
a(2) 3
0 .. .
a(3) 3 .. .
or Cf = [
1 0
0 ], D(f )
where D(f ) is the delta associated matrix of f as defined in Definition 1.5.2. We provide a straightforward result below. Proposition 5.2.2. Let notation be set the same as in Definition 5.1.5. Then the composition g ∘ f is defined as T
b0 [ ] [ b1 ] [ ] [ b2 ] [ ] [b ] [ 3] [.] . [.]
a(0) 0 [ (1) [ a0 [ [ (2) [ a0 [ [ (3) [ a0 [ .. [ .
a(0) 1
a(0) 2
a(0) 3
.. .
.. .
.. .
a(1) 1 a(2) 1 a(3) 1
a(1) 2 a(2) 2 a(3) 2
a(1) 3 a(2) 3 a(3) 3
T
] [c0 ] . . .] c1 ] ] [ ] ] [ [ c2 ] . . .] ] , ]=[ ] ] [ c [ 3] . . .] ] [.] . .. .] [ . ] ...
if cn ∈ S, n = 0, 1, 2, . . . . Otherwise, we say that g ∘ f does not exit. Let us look at Example 5.1.6 from another point of view.
134  5 General composition Example 5.2.3. Let g(x) = 1 + x + x2 + ⋅ ⋅ ⋅ be a formal power series in 𝕏(ℝ) and let f (x) = 1/2, a unit. Then T
1 [ ] [1] [ ] [ ] g ∘ f = [1] [1] [ ] [.] . [.] or
1 [1 [2 [ [1 [ 22 [ [1 [ 23 [. . [.
0 0
0 0
0 0
0
0
0
0 .. .
0 .. .
0 .. .
T ... ] [2] . . . ] [0] ] [ ] ] . . .] = [ 0] ] , ] [ [0] ] ] . . .] [ ] [.] .. .. .] [ ]
g ∘ f = 2I𝕏 . Let us see another interesting example. Example 5.2.4. Let g(x) = 1 + x + x2 + ⋅ ⋅ ⋅ be a formal power series in 𝕏(ℝ) and let f (x) = a + x, a unit where 0 < a < 1. Then, for any n ∈ ℕ, n n (a + x)n = ∑ ( )xj an−j , j j=0
and then for 0 ≤ k ≤ n,
n a(n) = ( )an−k . k k
By Proposition 5.2.2, T
1 [ ] [1] [ ] [ ] g ∘ f = [1] [1] [ ] [.] . [.]
1 [a [ [ 2 [a [ [ 3 [a [ .. [.
0 1 (21 )a1
(31 )a2 .. .
0 0 (22 )a0 (32 )a1 .. .
0 0 0
(33 )a0 .. .
T
... (1 − a)−1 [ ] . . . ] [ (1 − a)−2 ] ] ] ] [ . . . ] = [(1 − a)−3 ] , ] ] [ −4 ] ] [ . . . ] [(1 − a) ] ] ] [ .. .. . .] [ ]
because for any k = 0, 1, 2, . . . , k k+1 k+2 2 k+3 3 ck = ( ) + ( )a + ( )a + ( )a + ⋅ ⋅ ⋅ k k k 3 k+1 (k + 2)(k + 1) 2 (k + 3)(k + 2)(k + 1) 3 =1+ a+ a + a + ⋅⋅⋅ 1! 2! 3! −k − 1 −k − 1 −k − 1 )(−a)1 + ( )(−a)2 + ⋅ ⋅ ⋅ =( )(−a)0 + ( 0 1 2 = (1 − a)−(k+1) . Thus, g ∘ f (x) =
1 1 1 + x+ x2 + ⋅ ⋅ ⋅ . 1 − a (1 − a)2 (1 − a)3
5.3 Coefficients of f n (z)
 135
We obtain an interesting formula ∞
n+k n )a = (1 − a)−(k+1) , k
∑(
n=0
k = 0, 1, 2, . . . ,
(5.1)
where a < 1 (Note: it is trivial if a = 0). By (5.1) and let a = 1/2, we have ∞
n+2 1 ) n = 22+1 = 8, 2 2
∑(
n=0
(5.2)
for k = 2. Of course, we may obtain (5.2) by differentiating g twice to have ∞
n+2 n )x , 2
g (x) = 2 ∑ ( n=0
and then ∞
n+2 1 ) n. 2 2
g (1/2) = 2 ∑ ( n=0
On the other side, g (x) = 2(1 − x)−3 and then g (1/2) = 2 ⋅ 8 = 16. Thus, ∞
∑(
n=0
n+2 1 ) n = 8. 2 2
More properties of the matrix representation for composition of formal power series will be discussed in Section 5.7.
5.3 Coefficients of f n (z) A formal power series could be considered a sequence of its coefficients. The composition of formal power series actually is a series of the sequence of formal power series that, eventually, or can only be, determined by their coefficients. We must investigate the coefficients of f n (z) for a formal power series f in order to know the structure of the composition of formal power series. The multinomial theorem plays a significant role in the research on the composition of the formal power series. Before we introduce some results about the multinomial theorem, we would like to indicate that the approach of organizing or constructing the coefficients of f n introduced in this section is different from the original work in [30]. In fact, the formula for the coefficients generated by the multinomial theorem in this section is organized in a way better for students to learn by themselves.
136  5 General composition Let f (x) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ be a formal power series over S and denote f n as in formula (1.8) for every n ∈ ℕ ∪ {0}. Suppose S is a commutative ring and let n ∈ ℕ be given. For any k ∈ ℕ ∪ {0}, the kth coefficient a(n) of f n is totally determined by the k polynomial n
(a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ + ak z k )
because all terms at z t do not contribute anything to a(n) if t > k. Therefore, by the k multinomial theorem, we have n − r0 − ⋅ ⋅ ⋅ − rk−1 r0 r1 n n − r0 r )a0 a1 ⋅ ⋅ ⋅ akk )⋅⋅⋅( a(n) = ∑ ( )( k rk r1 r0
(5.3)
where the sum is extended over all possible nonnegative integers r0 , r1 , . . . , rk such that r0 + r1 + ⋅ ⋅ ⋅ + rk = n and
r1 + 2r2 + 3r3 + ⋅ ⋅ ⋅ + krk = k
(5.4)
where k = 0, 1, 2, . . .. We have seen in Definition 5.1.5 that the composition g ∘ f is determined by all ∞
ck = ∑ bn a(n) , k n=0
k = 0, 1, 2, . . . ,
which shows how important each a(n) is. We must investigate all a(n) completely. Since k k n runs through all nonnegative integers in the series of ck , the existence of such ck is mainly determined by those integer n which is greater than k, and hence we only (n) discuss the case n ≥ k for the convergence of series ∑∞ n=0 bn ak in this section. Lemma 5.3.1. Let f ∈ 𝕏(ℂ) be given with the form f (z) = a0 + a1 z + ⋅ ⋅ ⋅ + ak z k + ⋅ ⋅ ⋅ , where a0 ≠ 0, aj = 0, 1 ≤ j < s and as ≠ 0. We write f n as in (1.8) for every n ∈ ℕ. For every nonnegative integer k, let r0 , r1 , r2 , . . . , rk be nonnegative integers that satisfy formula (5.4). Then: (n) n (i) a(n) 0 = a0 and ak = 0 if 1 ≤ k < s; (ii) rj = 0, 1 ≤ j < s. For k ≥ s, (iii) rs + rs+1 + ⋅ ⋅ ⋅ + rk ≤ ⟦k/s⟧, where ⟦⋅⟧ is the greatest integer function; (iv) r0 ≥ n − ⟦k/s⟧ or n − r0 ≤ ⟦k/s⟧ if n ≥ k; (v) If k = sl + j for some l ∈ ℕ and 1 ≤ j < s, then the tuples (r1 , r2 , . . . , rk ) satisfying (5.4) along with the condition that r1 + r2 + ⋅ ⋅ ⋅ + rk = m
with 0 ≤ m ≤ l
are the same tuples (r0 , r1 , . . . , rk ) satisfying (5.4).
5.3 Coefficients of f n (z)
 137
Proof. Actually we may write f (z) = a0 + as z s + as+1 z s+1 + ⋅ ⋅ ⋅ + ak z k + ⋅ ⋅ ⋅ . (i) is trivial. r r r Every nonzero rj , 1 ≤ j < s, makes a11 a22 ⋅ ⋅ ⋅ akk = 0, and hence each such rj can be excluded from the construction of a(n) in (5.3), that yields (ii). k Putting (ii) into formula (5.4), we obtain r0 + rs + rs+1 + ⋅ ⋅ ⋅ + rk = n,
srs + (s + 1)rs+1 + ⋅ ⋅ ⋅ + krk = k.
(5.5)
Then rs +
k s+1 r + ⋅ ⋅ ⋅ + rk = k/s, s s+1 s
(5.6)
by which (iii) and (iv) follow. Now let a tuple (r0 , r1 , . . . , rk ) satisfying (5.4) be given and suppose that k = sl + j, j ∈ ℕ and 1 ≤ j < s. It is clear that if n is given with n ≥ k, then n − r0 = r1 + r2 + ⋅ ⋅ ⋅ + rk is totally determined by the values of r1 + r2 + ⋅ ⋅ ⋅ + rk , or by the values of rs + rs+1 + ⋅ ⋅ ⋅ + rk in this case. Since 0 ≤ rs + rs+1 + ⋅ ⋅ ⋅ + rk ≤ ⟦k/s⟧ by (iii), we have (v). Before we close the proof, we would like to indicate that if s = 1, then there is no j ∈ ℕ such that 1 ≤ j < 1, ⟦k/s⟧ = k, and l = k, all statements are true.
Definition 5.3.2. Let k ∈ ℕ be given. For any integer m, 0 ≤ m ≤ k, we define Rk,m = ∑
m! r r r a 1 a 2 ⋅ ⋅ ⋅ akk , r1 !r2 ! ⋅ ⋅ ⋅ rk ! 1 2
(5.7)
where the sum is extended over all possible nonnegative integers r1 , r2 , . . . , rk such that r1 + r2 + ⋅ ⋅ ⋅ + rk = m
and r1 + 2r2 + ⋅ ⋅ ⋅ + krk = k.
(5.8)
We define that Rk,m = 0, for our convenience, if Rk,m does not exist or the condition (5.8) fails for those tuples (r1 , r2 , . . . , rk ). For example, Rk,m = 0 if m > k, or R2,0 = 0 because (5.8) does not hold for k = 2, m = 0. The real number Rk,m is called the mth multinomial cofactor of the kth coefficient of the formal power series f n , or just the mth cofactor if there is no confusion. It is easy to check that Rk,k = ak1 if k ≠ 0. If m = 0, then R0,0 = 1 by (5.7). If k ∈ ℕ, then Rk,0 = 0 because (5.8) fails for such a tuple.
138  5 General composition Remark 5.3.3. 1. The multinomial cofactor Rk,m was first introduced in [24]. We are introducing more properties of Rk,m now. 2. The cofactor Rk,m relies on nonnegative integers r1 , r2 , . . . , rk , which are defined in (5.8), and hence Rk,m relies on k and m only and is independent from n. 3. In particular, if there is s ∈ ℕ such that as ≠ 0, a0 ≠ 0 but aj = 0, 1 ≤ j < s, then Rk,m relies on ⟦k/s⟧ only by (iii) of Lemma 5.3.1. Theorem 5.3.4 ([26]). Let s ∈ ℕ, as ≠ 0, and let f (z) = a0 + as z s + as+1 z s+1 + ⋅ ⋅ ⋅ be a formal power series in 𝕏(ℂ) with a0 ≠ 0, aj = 0, 1 ≤ j < s. Then k n a(n) = ( )an−m Rk,m , ∑ k m 0 m=0
for
k ≥ s.
(5.9)
Moreover, if k = sl for some l ∈ ℕ, n n−l l l−1 n n−m (n) a(n) = a = ( )a a + ∑ ( )a Rsl,m , k sl l 0 s m=0 m 0
(5.10)
and such k is unique. Proof. Let n ∈ ℕ be given and let r0 be a nonnegative integer such that r0 ≤ n. We have n − r0 − ⋅ ⋅ ⋅ − rk−1 n − r0 n − r0 − rs ) )⋅⋅⋅( )( rk rs+1 rs (n − r0 )(n − r0 − 1) ⋅ ⋅ ⋅ (n − r0 − rs − ⋅ ⋅ ⋅ − rk + 1) = rs !rs+1 ! ⋅ ⋅ ⋅ rk ! (r + r + ⋅ ⋅ ⋅ + rk )! (n − r0 )! = = s s+1 rs !rs+1 ! ⋅ ⋅ ⋅ rk ! rs !rs+1 ! ⋅ ⋅ ⋅ rk !
(
because r0 + r1 + r2 + ⋅ ⋅ ⋅ + rk = n by formula (5.4). For nonnegative integer tuple (r0 , r1 , . . . , rk ) satisfying (5.4), formula (5.3) provides that n r (n − r0 )! r1 r = ∑ ( )a00 a ⋅ ⋅ ⋅ akk a(n) k r0 r1 ! ⋅ ⋅ ⋅ rk ! 1
k (n − r0 )! r1 n r r = ∑ ( )a00 ∑ a1 ⋅ ⋅ ⋅ akk r r ! ⋅ ⋅ ⋅ r ! 0 1 k n−r =0 0
k
= ∑( m=0
n m! r r )an−m ∑ a 1 ⋅ ⋅ ⋅ akk n−m 0 r1 ! ⋅ ⋅ ⋅ rk ! 1
5.3 Coefficients of f n (z)
 139
k n = ∑ ( )an−m Rk,m , m 0 m=0
where the second summations in line 2 and 3 of the equation array are over tuples (r1 , r2 , . . . , rk ) satisfying (5.8) or r1 + r2 + ⋅ ⋅ ⋅ + rk = n − r0 = m and
1r1 + 2r2 + ⋅ ⋅ ⋅ + krk = k.
We obtain (5.9). Separating m = 0 and m = k from the above summation and using the fact tat Rk,0 = 0 for k ∈ ℕ, we have k−1 n n n−k k a(n) = ∑ ( )an−m 0 Rk,m + ( )a0 a1 . k k m=1 m
If aj = 0, 1 ≤ j < s ≤ k, and as ≠ 0, (iii) of Lemma 5.3.1 yields that Rk,m does not exist if m > ⟦k/s⟧ = l, and hence Rk,m = 0 for m > l by Definition 5.3.2. It is also true by (v) of Lemma 5.3.1 that m takes all nonnegative integers between 0 and ⟦k/s⟧ is equivalent to that Rk,m takes all values generated by the nonnegative integers r0 , r1 , r2 , . . . , rk which satisfy (5.4). Thus, (5.9) becomes l n ( )an−m a(n) = ∑ 0 Rk,m k m m=0 l−1 n n n−m l = ( )an−l 0 as + ∑ ( )a0 Rk,m . l m m=0
To obtain the uniqueness of (5.10) is equivalent to show that Rk,l = als or to show that the tuple (rs , rs+1 , . . . , rk ) = (l, 0, 0, . . . , 0) is the only tuple satisfying (5.8) with r1 = 0 = r2 = ⋅ ⋅ ⋅ = rs−1 and m = l, k = ls. It is clear that the tuple (l, 0, 0, . . . , 0) satisfies (5.8) with m = l. Assume that there is another tuple such that rs = l − b and rs+1 + rs+2 + ⋅ ⋅ ⋅ + rk = b
for some b.
Then rs + rs+1 + ⋅ ⋅ ⋅ + rk = (l − b) + b = l. This tuple satisfies the first equation of (5.8), but if this tuple is put into the second equation of (5.8), sl = k = srs + (s + 1)rs+1 + ⋅ ⋅ ⋅ + krk
140  5 General composition = s(l − b) + srs+1 + rs+1 + srs+2 + 2rs+2 + ⋅ ⋅ ⋅ + [s + (l − 1)s]rk
= s(l − b) + s(rs+1 + ⋅ ⋅ ⋅ + rk ) + rs+1 + 2rs+2 + ⋅ ⋅ ⋅ + (l − 1)srk = sl + [1rs+1 + 2rs+2 + ⋅ ⋅ ⋅ + (l − 1)srk ]. Then 1rs+1 + 2rs+2 + ⋅ ⋅ ⋅ + (l − 1)srk = 0, and hence rj = 0, s < j ≤ k. Thus, b = 0. Formula (5.10) is true and is unique. Corollary 5.3.5. Let f ∈ 𝕏(ℂ) be given with f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ Let f = f − a0 or f = f − f (0), and write f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ . n
Let a(n) k be the kth coefficient of f , n ∈ ℕ, k = 0, 1, 2, . . . . Then a(n) k = Rk,n .
(5.11)
n
Proof. If n > k, then a(n) k = 0 because ord(f ) > k, and Rk,n = 0 by Definition 5.3.2. So (5.11) is true for n > k. If n ≤ k, then a(n) k is determined by equation (5.4) with r0 = 0. Then formula (5.4) is reduced to be formula (5.8) with m = n ≤ k which is Rk,n . Then a(n) k = Rk,n
for all nonnegative integers n, k.
The next lemma provides a relationship between the coefficients of f n and the n coefficients of (f − f (0))n = f . Lemma 5.3.6. Let f be a formal power series in 𝕏(ℂ) with the form f (z) = a0 + a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ . Define f (z) = f (z) − a0 = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ where ak = ak for all k ∈ ℕ and a0 = 0. the kth coefficient of f n and denote by a(n) For any n ∈ ℕ and k ∈ {0} ∪ ℕ, denote by a(n) k k n
the kth coefficient of f , respectively. Then
k n (m) a(n) = ∑ ( )an−m 0 ak . k m m=1
(5.12)
5.4 The general composition theorem
 141
Proof. Applying (5.9) and (5.11), we have k k n n n−m (m) a(n) = ∑ ( )an−m 0 Rk,m = ∑ ( )a0 ak . k m m m=1 m=1
5.4 The general composition theorem A formal power series is a mapping from ℕ to a ring S. If this ring is endowed with a metric, the pointwise convergence of a mapping from the set of formal power series to the ring is welldefined. As we said in the Section 5.1, we want to have a necessary and sufficient condition for the existence of such composition under the two rules. We now begin to establish the general composition theorem of formal power series. The approach used here is different from the previous proofs in [5] and [30]. This approach is called the analytical approach which, by reorganizing the multinomial theorem with the cofactors, is easier for college students to learn by themselves as well as for college professors to teach this subject. Theorem 5.4.1 ([26]). Let f , g ∈ 𝕏(ℂ) be given such that f (x) = a0 + a1 x + ⋅ ⋅ ⋅ + an x n + ⋅ ⋅ ⋅ ,
g(x) = b0 + b1 x + ⋅ ⋅ ⋅ + bn x n + ⋅ ⋅ ⋅ . If deg(f ) ≠ 0. then the composition g ∘ f exists if and only if ∞ n dk = ∑ ( ) bn an−k ∈ ℂ for all k ∈ ℕ ∪ {0}, 0 k n=k
(5.13)
. where (nk ) = n(n−1)⋅⋅⋅(n−k+1) k! If deg(f ) = 0, then the existence of g ∘ f is equivalent to the existence of g(a0 ). Proof. If deg(f ) = 0, then f is a constant formal power series or we may consider that f = a0 , then the existence of g ∘ f is equivalent to the existence of g(a0 ). Now we suppose that deg(f ) ≠ 0. If a0 = 0, formula (5.13) is trivial and the conclusion is true. So we suppose that a0 ≠ 0. Suppose that (5.13) holds. We show that g ∘ f ∈ 𝕏(S), or ∞
ck = ∑ bn a(n) ∈ℂ k n=0
∀ k = 0, 1, 2, . . . ,
where a(n) is the kth coefficient of f n . k Using the sequence {dk } in (5.13) we can define a formal power series H ∈ 𝕏(ℂ) as H(z) = d0 + d1 z + d2 z 2 + ⋅ ⋅ ⋅ .
142  5 General composition Let f (z) = f (z) − a0 = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ , that is, ak = ak for all k ∈ ℕ and a0 = 0, then f is a nonunit formal power series. We n denote by a(n) k the kth coefficient of f for all nonnegative integers k and n. It is clear that H ∘ f exists in 𝕏(ℂ), and hence we write ∞
n
H ∘ f (z) = ∑ dn f (z) = r0 + r1 z + r2 z 2 + ⋅ ⋅ ⋅ , n=0
for some {rn }∞ n=0 ⊆ ℂ. Let k be any nonnegative integer, applying Definition 5.1.5 we have ∞
rk = ∑ dn a(n) k ∈ ℂ. n=0
Since f a nonunit, we have k
(j)
rk = ∑ dj ak j=1
k ∞ n n−j (j) = ∑(∑ ( )bn a0 )ak j j=1 n=j ∞ k k−1 n n n−j n−j (j) = ∑[ ∑ ( )bn a0 + ∑ ( )bn a0 ]ak j j j=1 n=j n=k k ∞ k k−1 n n (j) n−j (j) n−j = ∑( ∑ ( )bn a0 )ak + ∑( ∑ ( )bn a0 )ak j j j=1 n=k j=1 n=j k k−1 ∞ k n n−j (j) n (j) n−j = ∑( ∑ ( )bn a0 )ak + ∑ bn (∑ ( )a0 ak ) j j j=1 j=1 n=j n=k k k−1 ∞ n n−j (j) = ∑( ∑ ( )bn a0 )ak + ∑ bn a(n) k j j=1 n=j n=k
by (5.12), where changing the order of summations is ensured by (5.13) and ∞ ∞ k−1 n n n n−j n−j n−j ∑ ( )bn a0 = ∑ ( )bn a0 − ∑ ( )bn a0 j j j n=j n=j n=k k−1 n n−j = dj − ∑ ( )bn a0 ∈ ℂ. j n=j
5.4 The general composition theorem
n−j
 143
n Since ∑kj=1 (∑k−1 n=j ( j )bn a0 )ak is a finite summation and rk ∈ ℂ, it follows that (j)
∞ k k−1 n (j) n−j = r − ( ∑ ( )bn a0 )ak ∈ ℂ. ∑ bn a(n) ∑ k k j j=1 n=j n=k
If we write k−1
∞
k−1
n=0
n=k
n=0
ck = ∑ bn a(n) + ∑ bn a(n) = ∑ bn a(n) + ĉk k k k for all k = 0, 1, . . . , then ĉk ∈ ℂ, and hence ck ∈ ℂ. Thus, g ∘ f ∈ 𝕏(ℂ). Now we suppose that g ∘ f ∈ 𝕏(ℂ). Then ∞
∈ ℂ for all nonnegative integer k. ck = ∑ bn a(n) k n=0
We show that (5.13) is true. (n) n n n It is clear that d0 = ∑∞ n=0 (0 ) bn a0 = c0 ∈ ℂ because a0 = a0 for all n, which shows that (5.13) is true for k = 0. It is trivial if f is a constant formal power series. We assume that as ≠ 0, aj = 0, 1 ≤ j < s for some s ∈ ℕ. By (5.10) of Theorem 5.3.4 (take k = s, or l = 1) and the definition of Rk,m we have ∞ ∞ n cs = ∑ bn a(n) = ( )bn an−1 ∑ s 0 as . n=1 1 n=0
Since as ≠ 0, it follows that ∞ cs n ∈ ℂ. d1 = ∑ bn ( )an−1 0 = as 1 n=1
Next we suppose that (5.13) is true for d0 , d1 , . . . , dl−1 for some l ∈ ℕ with l ≥ 2. Considand applying equation (5.10), we have ering a(n) ls n n−l l l−1 n n−m = ( )a a + ∑ ( )a Rls,m . a(n) ls l 0 s m=0 m 0 Then ∞
cls = ∑ bn a(n) ls n=0
144  5 General composition ∞ l−1 n n l = ∑ bn [( )an−l a + ( )an−m Rls,m ] ∑ 0 s l m 0 n=0 m=0 ∞ ∞ l−1 n n n−m l = ∑ ( )bn an−l 0 as + ∑ bn [ ∑ ( )a0 Rls,m ]. l m n=0 n=0 m=0 n−m n ∈ ℂ, 0 ≤ m ≤ l − 1. By Remark 5.3.3, Rk,m is By inductive assumption, ∑∞ n=m (m )bn a0 independent from n. So we have ∞ l−1 l−1 ∞ n n−m ∑ bn ( ∑ ( )an−m 0 Rls,m ) = ∑ ( ∑ bn a0 )Rls,m ∈ ℂ. m n=0 m=0 m=0 n=0
Then ∞ l−1 ∞ n l a = c − ( bn an−m ∑ bn ( )an−l ∑ ∑ ls 0 s 0 )Rls,m ∈ ℂ. l n=0 m=0 n=0
Since as ≠ 0, we have that ∞ n dl = ∑ bn ( )an−l l 0 n=0
=
l−1 ∞ n 1 [c − ( ( )bn an−m ∑ ∑ ls 0 )Rls,m ] ∈ ℂ, l m as m=0 n=0
and then (5.13) is true for every l ∈ ℕ. Remark 5.4.2. The general composition theorem or Theorem 5.4.1 reveals: (i) Let f and g be given as in Theorem 5.4.1. If a0 = 0, then formula (5.13) is true, and hence g ∘ f exists by Theorem 5.4.1. This result indicates that Definition 1.2.1 is just a special case for Theorem 5.4.1. (ii) The existence of the composition g ∘ f relies only on the series g and the a0 of f . (iii) The general composition theorem has both necessary and sufficient condition for the existence of the composition of formal power series. (iv) If replacing ℂ in Theorem 5.4.1 by a field S with a metric, the theorem is still true. It is now clear that to ensure the existence of g ∘ f we must check the convergence n−k n of ∑∞ for all k ∈ ℕ ∪ {0} which, unfortunately, means a lot of work. The n=k (k )bn a0 next theorem will help us to avoid checking the convergence infinitely many times. Theorem 5.4.3. Let f , g ∈ 𝕏(ℂ) be given with the forms f (x) = a0 + a1 x + ⋅ ⋅ ⋅ + an x n + ⋅ ⋅ ⋅ n
and
g(x) = b0 + b1 x + ⋅ ⋅ ⋅ + bn x + ⋅ ⋅ ⋅ . n If the series ∑∞ n=0 bn R converges for some real number R > a0 , then g ∘ f exists.
5.4 The general composition theorem
 145
Proof. If deg (f ) = 0, the conclusion is obvious, so we assume that deg (f ) ≠ 0. ∞ n n Consider the power series ∑∞ n=0 bn x . Since ∑n=0 bn R converges, it follows that the kth derivative ∞
∑ n(n − 1) ⋅ ⋅ ⋅ (n − k + 1)bn x n−k
n=k
converges for x < R for every k ∈ ℕ ∪ {0}. Then, for any k ∈ ℕ ∪ {0}, ∞ ∞ n n(n − 1) ⋅ ⋅ ⋅ (n − k + 1) bn x n−k ∑ ( )bn xn−k = ∑ k! k n=k n=k
=
1 ∞ ∑ n(n − 1) ⋅ ⋅ ⋅ (n − k + 1) bn x n−k k! n=k
converges for x < R. Then a0  < R implies that ∞ n ∑ ( )bn an−k 0 k n=k
converges for all k ∈ ℕ ∪ {0}. Theorem 5.4.1 yields the conclusion. Corollary 5.4.4. Let f , g ∈ 𝕏(ℂ) be given with f (x) = a0 + a1 x + ⋅ ⋅ ⋅ + an xn + ⋅ ⋅ ⋅
and
n
g(x) = b0 + b1 x + ⋅ ⋅ ⋅ + bn x + ⋅ ⋅ ⋅ Suppose that a0  < 1 and bn  ≤ M, for all n ∈ ℕ ∪ {0} for some positive number M. Then g ∘ f is welldefined. n Proof. Pick R such that a0  < R < 1, then ∑∞ n=0 bn R converges because bn  ≤ M for ∞ all n ∈ ℕ and ∑n=0 Rn converges. Applying Theorem 5.4.3, g ∘ f exists.
Comparing with Example 1.2.12, the next example will help us to further understand the difference between the formal power series and the regular power series. Example 5.4.5. Let S = ℝ. Let ∞
g(x) = ∑ xn n=0
∞
and f (x) = 0.5 + ∑ n! xn . n=1
Then Corollary 5.4.4 yields that g ∘ f exists. Theorem 5.4.6. Let f , g ∈ 𝕏(ℂ) be given such that f (x) = a0 + a1 x + ⋅ ⋅ ⋅ + an x n + ⋅ ⋅ ⋅ ,
g(x) = b0 + b1 x + ⋅ ⋅ ⋅ + bn x n + ⋅ ⋅ ⋅ ,
146  5 General composition and deg(f ) ≠ 0. Let g (n) denote the nth order formal derivative of g for every n ∈ ℕ. Then the composition g ∘ f exists if and only if there are infinitely many n such that g (n) (a0 ) ∈ ℂ.
(5.14)
Proof. If a0 = 0 the conclusion is trivial and, therefore, we suppose that a0 ≠ 0. Let us first assume that (5.14) holds for every k ∈ ℕ. n n−k , we have Let k ∈ ℕ be given. Since g (k) (x) = (k!) ∑∞ n=k bn (k )x ∞ n g (k) (a0 ) = (k!) ∑ ( )bn an−k = (k!)dk , 0 k n=k
where dk is defined in (5.13). Thus, (5.14) is equivalent to (5.13). Then the theorem is true if (5.14) holds for every k ∈ ℕ. Now we show that (5.14) holds for infinitely many k ∈ ℕ is equivalent to that (5.14) holds for every k ∈ ℕ. This equivalence suffices to complete the proof of the theorem. Suppose that (5.14) holds for infinitely many k ∈ ℕ, but not for all k ∈ ℕ. Then there exists a k ∈ ℕ such that g (k) (a0 ) ∉ ℂ but
g (k+1) (a0 ) ∈ ℂ.
n−(k+1) For g (k+1) (a0 ) = ∑∞ , we write n=k+1 n(n − 1) ⋅ ⋅ ⋅ (n − (k + 1) + 1)bn a0 ∞
g (k+1) (a0 ) = ∑ (m + 1)m ⋅ ⋅ ⋅ (m − k + 1)bn am−k 0 , m=k
by setting m = n − 1, and then apply Abel’s limit theorem (p. 425, [95]) to the series ∞
m−k h(r) = ∑ (m + 1)m(m − 1) ⋅ ⋅ ⋅ (m − k + 1)bn am−k , 0 r m=k
which converges uniformly on [0, 1]. Then 1
1
N
0
m=k
r m−k ]dr ∫ h(r)dr = lim ∫[ ∑ (m + 1)m ⋅ ⋅ ⋅ (m − k + 1)bn am−k 0 0
N→∞
N
= lim ∑ (m + 1)m ⋅ ⋅ ⋅ (m − k + N→∞
m=k
1)bn am−k 0
N
= lim ∑ (m + 1)m ⋅ ⋅ ⋅ (m − k + 2)bn am−k 0 N→∞ ∞
m=k
= ∑ n(n − 1) ⋅ ⋅ ⋅ (n − k + 1)bn an−k−1 , 0 n=k+1
1
∫ r m−k dr 0
5.4 The general composition theorem
 147
1
by setting n = m + 1 in the last step, and hence ∫0 h(r)dr ∈ ℂ. Then ∞
g (k) (a0 ) = ∑ n(n − 1) ⋅ ⋅ ⋅ (n − k + 1)bn an−k 0 n=k
∞
= a0 ∑ n(n − 1) ⋅ ⋅ ⋅ (n − k + 1)bn an−k−1 0 n=k ∞
= a0 ∑ n(n − 1) ⋅ ⋅ ⋅ (n − k + 1)bn an−k−1 + k!bk 0 n=k+1 1
= a0 ∫ h(r)dr + k!bk , 0
is a complex number. It is a contradiction to our assumption that g (k) (a0 ) ≠ ℂ. Thus, (5.14) holds for infinitely many k ∈ ℕ is equivalent to that (5.14) holds for every k ∈ ℕ. Let U ⊆ ℂ be an open region, a ℂvalued function f is analytic on U is equivalent to that f can be represented by a power series on U, and f is differentiable on U infinitely many times. The analyticity is connected with an open region and with the differentiations. Let us introduce the formal analyticity. Definition 5.4.7. Let g ∈ 𝕏(ℂ) be given. A formal power series g is said to be formally analytic at a ∈ ℂ if g (n) (a) ∈ ℂ,
for infinitely many n ∈ ℕ.
Such a point a ∈ ℂ is called a formal analytic point of g. Remark 5.4.8. (i) Theorem 5.4.6 shows that the statement that g (n) (a) ∈ ℂ for all n ∈ ℕ is equivalent to the statement that g (n) (a) ∈ ℂ for infinitely many n ∈ ℕ. (ii) Please be noticed that the derivative here is the formal derivative. (iii) The formal analytic point a does not require anything about the neighborhood of a. The next proposition is obvious. It shows that every formal power series on ℂ is formally analytic at 0. Proposition 5.4.9. Let g ∈ 𝕏(ℂ) be given. Then g is formally analytic at a ∈ ℂ if and only if there is f ∈ 𝕏(ℂ) such that f (0) = a and g ∘ f ∈ 𝕏(ℂ).
148  5 General composition If f is a nonunit, Theorem 1.2.5 provides the equation ord(g ∘ f ) = ord(f ) ⋅ ord(g) for all g ∈ 𝕏. The next two examples will show that Theorem 1.2.5 may not be true for the general composition. Example 5.4.10. Let f (x) = 1 + x and g(x) = x − x2 be formal power series on ℂ. Then g ∘ f (x) = (1 + x) − (1 + x)2 = −x − x 2 . It is clear that ord(g ∘ f ) = 1, ord(g) = 1, ord(f ) = 0, which implies that 1 = ord(g ∘ f ) ≠ ord(g) ⋅ ord(f ) = 0. Example 5.4.11. In 𝕏(ℝ), let f (x) = 1 + x, g(x) = x. Then g ∘ f (x) = (1 + x). Then ord(g ∘ f ) = 0, ord(g) = 1, ord(f ) = 0, which implies that ord(g ∘ f ) = ord(g) ⋅ ord(f ). However, the following property of the order is always true. Proposition 5.4.12. For any nonzero formal power series f and g on ℝ, or ℂ, ord(g ∘ f ) ≥ ord(g) ⋅ ord(f ), if the composition g ∘ f exists. Proof. Since f ≠ 0, g ≠ 0, it follows that ord(f ) and ord(g) are nonnegative integers. If f is a nonunit, Theorem 1.2.5 gives that ord(g ∘ f ) = ord(g) ⋅ ord(f ). If f is a unit and g ∘ f exists, then ord(f ) = 0 and then ord(g ∘ f ) ≥ ord(g) ⋅ ord(f ) = 0, is always true. The space 𝔻 of all almost unit formal power series is a group under the composition. Every almost unit f ∈ 𝔻 has the iterative inverse f [−1] ∈ 𝔻. With the general composition, what kind of formal power series f ∈ 𝕏(ℂ) shall have iterative inverse f [−1] ?
5.4 The general composition theorem
 149
Theorems 5.4.1 and 5.4.6 tell us that the existence of g ∘ f mainly depends on the constant term of f and the coefficients of g. This result directs us to the investigations of some subsets of 𝕏 in which the composition is closed. For example, 𝔻(ℂ) is one of such subsets. It is natural for us to think that Theorem 5.4.1 may help us to further investigate the ordinary power series, the power series with positive radius of convergence. In Definition 1.5.18, we defined f [n] for f ∈ 𝔻(ℂ). If f ∉ 𝔻(ℂ), especially if f is a unit, it is not easy to determine when f [n] exits. However, we have the following result. Proposition 5.4.13. Let f ∈ 𝕏(ℂ) be given with f (z) = a0 + a1 z + ⋅ ⋅ ⋅ + an z n + ⋅ ⋅ ⋅ . Then the iterative power f [2] exists if and only if f (k) (a0 ) ∈ ℂ for every k ∈ ℕ where f (k) is the kth formal derivative of f . In other words, f [2] exists if and only if f is formally analytic at a0 . Moreover, f [2] (0) = f (a0 ). Proof. Taking g = f in Theorem 5.4.1, we have the conclusion. For the formula f [2] (0) = f (a0 ), we apply formula (5.13) and the formula ∞
ck = ∑ bn a(n) . k n=0
Setting k = 0, replacing g by f , and replacing bn by an for every nonnegative integer n, we have the constant term of f [2] , or ∞
∞
n=0
n=0
n f [2] (0) = c0 = ∑ an a(n) 0 = ∑ an a0 = f (a0 ).
We defined the iterative power f [n] for f ∈ 𝔻(ℂ) in Section 1.5. We must say that 𝔻(ℂ) is a very special subset of 𝕏, it is a group. We may generalize the iterative power for f ∉ 𝔻(ℂ) using Theorem 5.4.1. This problem is open. We define a generalized iterative power. Definition 5.4.14. Let S be a field with a metric and let f ∈ 𝕏(S) be given. For any integer n ≥ 2, we define f [n] = f ∘ (f [n−1] ), if the compositions involved exist. Example 5.4.15. We know that f (z) = ez is analytic on ℂ with power series representation ez = 1 +
z z2 z3 + + + ⋅⋅⋅. 1! 2! 3!
150  5 General composition z
If we consider ez as a formal power series, then ee = f [2] (z) is a formal power series (also an analytic function on ℂ) whose constant term is f [2] (0) = f (1) = e. Before closing this section, we realize, by Example 5.4.5 and Example 5.4.10, that we should review all properties about the composition of formal power series and verify the consistency of those properties while the composed series is a unit. We do this in next sections.
5.5 The general chain rule The Chain Rule for composition of formal power series was introduced in Section 2.1 where the composed formal power series must be a nonunit. Theorem 5.4.1 now provides a necessary and sufficient condition for the existence of the composition of formal power series, with it, we need to show that the Chain Rule works for all satisfied formal power series without the requirement of nonunitness for the composed series. In order to do it, not only we have to show that the existence of g(f (x)) implies the existence of g (f (x)), but also we need to show the equality (g ∘ f ) (x) = g (f (x)) ⋅ f (x), that is, the formal power series in two sides are equal. The following two lemmas will help us to obtain the General Chain Rule for the differentiation of formal power series. The details can be found in [23]. Lemma 5.5.1. Let f (x) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ be a formal power series over ℂ. Write ∞
f n (x) = ∑ a(n) xk , k k=0
for every n ∈ ℕ. Then a(n) = k+1
n k ak+1−j ∑ (k + 1 − j)a(n−1) j k + 1 j=0
(5.15)
for all k ∈ ℕ ∪ {0}. Proof. Applying the power rule and the product rule for the formal derivatives of formal power series, we have (f n (x)) = n f n−1 (x)f (x)
∞
∞
= n( ∑ a(n−1) x j )( ∑ (i + 1)ai+1 x i ) j j=0
∞
i=0
k
= n ∑ ( ∑ a(n−1) (k + 1 − j)ak+1−j )xk . j k=0 j=0
5.5 The general chain rule
 151
On the other hand, ∞
(f n (x)) = ∑ (k + 1)a(n) xk . k+1
k=0
The coefficients of the term xk should be equal, k
(k + 1)a(n) = n ( ∑ a(n−1) (k + 1 − j)ak+1−j ). j k+1 j=0
Thus, a(n) = k+1
n k ak+1−j ∑ (k + 1 − j)a(n−1) j k + 1 j=0
for all
k ∈ ℕ ∪ {0}.
∞ n n Lemma 5.5.2. Let g(x) = ∑∞ n=0 bn x and f (x) = ∑n=0 an x be two formal power series over ℂ. For any m ∈ ℕ,
g (m) (f (x)) ∈ 𝕏(ℂ)
if and only if
g(f (x)) ∈ 𝕏(ℂ),
where g (m) is the mth order formal derivative of g. Proof. We first show that g(f (x)) ∈ 𝕏(ℂ) if and only if g (f (x)) ∈ 𝕏(ℂ). Suppose g (f (x)) exists. By Definition 5.1.5 and Theorem 5.4.1, we have ∞ n ∈ ℂ for all k ∈ ℕ ∪ {0}. ∑ ( )(n + 1)bn+1 an−k 0 k n=k
(5.16)
For the existence of g(f (x)), using Theorem 5.4.1 again, we need only show that ∞ n ∈ℂ ∑ ( )bn an−k 0 k n=k
for all k ∈ ℕ ∪ {0}.
(5.17)
If a0 = 0, then formula (5.17) is trivial. We suppose that a0 ≠ 0. Let k ∈ ℕ ∪ {0} be given. Define ∞ n n−k ϕ(x) = ∑ ( )(n + 1)bn+1 an−k . 0 x k n=k
Since (5.16) is true, ϕ(x) converges uniformly on [0, 1] by Abel’s limit theorem. Let ϕm (x) be the partial sum of this power series for m = k, k + 1, . . . The uniform convergence of the power series ϕ(x) on [0, 1] yields that 1
1 m
n n−k dx ∫ ϕ(x) dx = lim ∫ ∑ ( )(n + 1)bn+1 an−k 0 x m→∞ k n=k 0
0
152  5 General composition m n 1 = lim ∑ ( )(n + 1)bn+1 an−k 0 m→∞ k n − k+1 n=k m
n+1 )bn+1 an−k 0 k
= lim ∑ ( m→∞
n=k m
j 1 ∞ j j−k j−k−1 = lim ∑ ( )bj a0 = ∑ ( )bj a0 m→∞ a k k 0 j=k+1 j=k+1 =
1 ∞ j 1 j−k ∑ ( )b a − bk . a0 j=k k j 0 a0
j−k
j Then ∑∞ j=k (k )bj a0 ∈ ℂ. Since k is arbitrary, it follows that formula (5.17) is true for all k, and hence g(f (x)) exists by Theorem 5.4.1. Now suppose that g(f (x)) exists. Then formula (5.17) is true by Theorem 5.4.1. We need only show that (5.16) is true. Notice that for any k ∈ ℕ ∪ {0}, ∞ ∞ n n+1 = (k + 1) ∑ ( ) bn+1 an+1−(k+1) ∑ ( )(n + 1)bn+1 an−k 0 0 k k + 1 n=k n=k ∞
= (k + 1) ∑ ( m=k+1
∞
m ) b am−(k+1) k+1 m 0
m = r ∑ ( ) bm am−r 0 m=r r for every r ∈ ℕ if we write r = k + 1. Thus, formula (5.16) is true. Applying mathematical induction, we can easily have the conclusion that the mth order derivative g (m) (f (x)) exists if and only if g(f (x)) exists. We would like to indicate that there is some other way to prove Lemma 5.5.2, we leave it as an exercise. Notice that formula (5.15) can be applied to any formal power series over a field S with a metric. We now introduce the general Chain Rule for formal power series. ∞ n n Theorem 5.5.3. Let g(x) = ∑∞ n=0 bn x and f (x) = ∑n=0 an x be two formal power series in x over ℂ. Then (g ∘ f ) exists if and only if g ∘ f exists. Moreover,
(g ∘ f ) (x) = g (f (x)) f (x)
(5.18)
if g ∘ f exists. Proof. The first part of the theorem is ensured by Lemma 5.5.2 above. We need only show the equality (5.18), that is, the corresponding coefficients on both sides of (5.18) are equal. Since g ∘ f (x) exists, by Definition 5.1.5, we have ∞
∞
g ∘ f (x) = ∑ ( ∑ bn a(n) )xk , k k=0 n=0
5.5 The general chain rule
 153
(0) where a(n) is the kth coefficient of f n (x), n ∈ ℕ ∪ {0}; a(0) 0 = 1, ak = 0 for every k ∈ ℕ; k and ∞
∈ ℂ for every k ∈ ℕ ∪ {0}. ∑ bn a(n) k
n=0
Then ∞
∞
k=1
n=0
∞
∞
k=1
n=1
(g ∘ f ) (x) = ∑ k( ∑ bn a(n) )xk−1 k = ∑ k( ∑ bn a(n) )xk−1 k ∞
∞
m=0
n=1
m = ∑ (m + 1)( ∑ bn a(n) m+1 )x .
If we write (g ∘ f ) (x) =
∑∞ m=0
m
rm x , then ∞
rm = (m + 1) ∑ bk a(k) m+1 , k=1
m = 0, 1, 2, . . . .
(5.19)
n On the other hand, g (x) = ∑∞ n=0 (n + 1)bn+1 x , and then ∞
∞
g (f (x)) = ∑ ( ∑ (n + 1)bn+1 a(n) )xk . k k=0 n=0
m ∑∞ m=0 cm x ,
We write g (f (x)) ⋅ f (x) = and then apply the Product Rule, formula (5.15) and the convergence of the number series involved, we have m
∞
cm = ∑ ( ∑ (n + 1)bn+1 a(n) j ) ⋅ (m + 1 − j)am+1−j j=0 n=0 ∞
m
n=0
j=0
= ∑ (n + 1)bn+1 ⋅ ∑ (m + 1 − j)a(n) j am+1−j ∞
= ∑ (n + 1)bn+1 ⋅ a(n+1) m+1 ⋅ n=0 ∞
m+1 n+1
= ∑ (m + 1)bn+1 a(n+1) m+1 n=0
for m = 0, 1, 2, . . . . Applying (5.19), we have ∞
cm = ∑ (m + 1)bn+1 a(n+1) m+1 n=0 ∞
= ∑ (m + 1)bk a(k) m+1 = rm , k=1
m = 0, 1, 2, . . . .
154  5 General composition
5.6 The general right distributive law The right distributive law for formal power series says (A ⋅ B) ∘ P = (A ∘ P) ⋅ (B ∘ P) for formal power series A, B, and P provided that the compositions involved exist. It is proved in Section 1.3 that the right distributive law is true if the formal power series P above is a nonunit. The following theorem will remove the requirement of the nonunitness of P and tell us when and how the right distributive law may be true. Lemma 5.6.1. Let A, B and P be formal power series in 𝕏 = 𝕏(ℂ). If A ∘ P ∈ 𝕏 and B ∘ P ∈ 𝕏, then (AB) ∘ P ∈ 𝕏. Proof. Write P(z) = p0 + p1 z + p2 z 2 + ⋅ ⋅ ⋅. Suppose that A ∘ P ∈ 𝕏 and B ∘ P ∈ 𝕏, then A(n) (p0 ) ∈ ℂ,
B(n) (p0 ) ∈ ℂ for all n ∈ ℕ
(∗)
by Theorem 5.4.6, where A(n) is the nth formal derivative of A. By this theorem again, all we need to do is to prove (AB)(n) (p0 ) ∈ ℂ for all n ∈ ℕ. By the formal Leibniz’s Formula for higher order formal derivatives, which is the same as the Leibniz’s Formula in classical analysis, we have n n (AB)(n) = ∑ ( )A(k) B(n−k) k k=0
for every n ∈ ℕ ∪ {0},
where A(0) = A. Then n n (AB)(n) (p0 ) = ∑ ( )A(k) (p0 )B(n−k) (p0 ) ∈ ℂ k k=0
by (∗). ∞ ∞ n n n Theorem 5.6.2. Let A(x) = ∑∞ n=0 an x , B(x) = ∑n=0 bn x and P(x) = ∑n=0 pn x be three formal power series over ℂ. The right distributive law, that is,
(A ∘ P)(B ∘ P) = (AB) ∘ P holds if both A ∘ P and B ∘ P exist.
(5.20)
5.6 The general right distributive law
 155
Proof. We suppose that both A ∘ P and B ∘ P exist. Then (AB) ∘ P ∈ 𝕏 by Lemma 5.6.1. Let H = (A ∘ P)(B ∘ P) − (AB) ∘ P, then, applying the Product Rule and General Chain Rule we have H = (A ∘ P) (B ∘ P) + (A ∘ P)(B ∘ P) − [(AB) ∘ P]P = [A (P)P B(P) + A(P)B (P)P ] − [A (P)B(P)P + A(P)B (P)P ]
= 0.
Then H = cI for some constant c ∈ ℂ for any P ∈ 𝕏 such that A ∘ P and B ∘ P exist. Taking P(z) = z, as an almost unit, we have that H = 0. We complete the proof. Corollary 5.6.3. Let f and g be any two formal power series over ℂ. Then for every m ∈ ℕ, g m ∘ f exists if g ∘ f exists. Moreover, if g ∘ f exists, we have g m ∘ f = (g ∘ f )m
for every
m ∈ ℕ.
(5.21)
Proof. Formula (5.21) is true obviously if m = 1. For m = 2, we need only let A = B = g and P = f in formula (5.20), then the existence of g ∘ f yields (5.21). Assume that the formula is true for m = 1, 2, . . . , k for some integer k ≥ 2. Let A = g k , B = g, P = f in formula (5.20). By the condition and the inductive hypothesis, both g k ∘ f and g ∘ f exist, applying formula (5.20) we have g k+1 ∘ f = (g k g) ∘ f = (g k ∘ f )(g ∘ f ) = (g ∘ f )k (g ∘ f ) = (g ∘ f )k+1 . Thus, formula (5.21) is true for very m ∈ ℕ. Corollary 5.6.4. Let g be any formal power series over ℂ. Then, for every m ∈ ℕ, 𝕏g ⊆ 𝕏g m . Proof. In Definition 5.1.5, we know that 𝕏g ≠ 0. Let f ∈ 𝕏g be given. Then g ∘f ∈ 𝕏, and hence (g ∘f )m ∈ 𝕏 for any m ∈ ℕ. Applying Corollary 5.6.3, we have g m ∘f = (g ∘f )m ∈ 𝕏, which means that f ∈ 𝕏g m . Thus, 𝕏g ⊆ 𝕏g m . Remark 5.6.5. Corollary 5.6.4 can be proved by Theorem 5.4.6. We need only realize that g (n) (a0 ) ∈ ℂ
for all n ∈ ℕ
implies that (g m ) (a0 ) ∈ ℂ (n)
where a0 = f (0) and m ∈ ℕ is given.
for all n ∈ ℕ,
156  5 General composition
5.7 Matrix representation of the general composition, II We have seen in Section 1.5 that the matrix representation of the composition for almost units is a good approach for investigating and applying the composition of formal power series. The matrix has been studied very well in algebra, in combinatorics and in numerical analysis that could be applied to the study of formal power series. Let g and f be elements in 𝕏(ℂ) such that g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ ,
f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ .
By Proposition 5.2.2, we have T
b0 [ ] [ b1 ] [ ] [ ] g ∘ f = [ b2 ] [b ] [ 3] [.] . [.]
a(0) 0
[ (1) [ a0 [ [ (2) [ a0 [ [ (3) [ a0 [ .. [ .
a(0) 1
a(0) 2
a(0) 3
.. .
.. .
.. .
a(1) 1 a(2) 1 a(3) 1
a(1) 2 a(2) 2 a(3) 2
a(1) 3 a(2) 3 a(3) 3
T
] [c0 ] . . .] c1 ] ] [ ] ] [ [ c2 ] . . .] ] , ]=[ ] ] [ [ c3 ] . . .] [ ] .. ] .. .] [ . ] ...
(n) if ck = ∑∞ n=0 bn ak ∈ ℂ, k = 0, 1, 2, . . . . Now let n ∈ ℕ be given and let
g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ ,
f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ .
∞ (n) k (n) k n Writing g n (z) = ∑∞ k=0 bk z and f (z) = ∑k=0 ak z , we have
b(n) 0
[ (n) ] [b1 ] ] [ [ (n) ] n ] b g ∘f =[ [ 2 ] [ (n) ] [b3 ] ] [ .. [ . ]
T
a(0) 0
[ (1) [ a0 [ [ (2) [ a0 [ [ (3) [ a0 [ .. [ .
a(0) 1
a(0) 2
a(0) 3
a(2) 1
a(2) 2
a(2) 3
a(1) 1
a(3) 1 .. .
a(1) 2
a(3) 2 .. .
a(1) 3
a(3) 3 .. .
c0(n)
...
T
] [ (n) ] [ ] . . .] ] [c1 ] ] [ (n) ] [ ] . . .] ] = [c2 ] , ] [ (n) ] [ ] . . .] ] [c3 ] . .. . ] [ .. ]
(j)
(n) if ck(n) = ∑∞ j=0 bj ak ∈ ℂ, k = 0, 1, 2, . . . . Then
b(0) [ 0 [ b(1) [ 0 Cg ⋅ Cf = [ [ b(2) [ 0 [ . . [ .
b(0) 1
b(0) 2
.. .
.. .
b(1) 1 b(2) 1
Then we have the following.
b(1) 2 b(2) 2
a(0) ][ 0 [ (1) . . .] ] [ a0 ][ [ (2) . . .] ] [ a0 ][ . .. . . ][ . ...
a(0) 1
a(0) 2
.. .
.. .
a(1) 1 a(2) 1
a(1) 2 a(2) 2
...
] . . .] ] ]. . . .] ] .. ] .]
5.7 Matrix representation of the general composition, II
 157
Proposition 5.7.1. Let f , g ∈ 𝕏(ℂ) be given with the form as above. If g ∘ f ∈ 𝕏(ℂ), then T
Cg ⋅ Cf = (g 0 ∘ f , g 1 ∘ f , g 2 ∘ f , . . . ) .
(5.22)
Proof. By the general right distributive law or Theorem 5.6.2, g ∘ f ∈ 𝕏(ℂ) implies g n ∘ f ∈ 𝕏(ℂ) ∀ n ∈ ℕ, and g 0 ∘ f ∈ 𝕏(ℂ) is obvious. Applying the above expression of Cg ⋅ Cf , we have (5.22). Theorem 5.7.2. If f , g ∈ 𝕏(ℂ) and g ∘ f ∈ 𝕏(ℂ), then Cg ⋅ Cf = Cg∘f . Proof. Let g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ ,
f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ .
Since g ∘ f ∈ 𝕏(ℂ), we may write g ∘ f (z) = c0 + c1 z + c2 z 2 + ⋅ ⋅ ⋅ , (n) where cK = ∑∞ n=0 bn ak , k = 0, 1, 2, . . . . By Definition 5.2.1, we have
Cg∘f
c0(0) [ (1) [ c0 [ [ (2) =[ [ c0 [ (3) [ c0 [ .. [ .
c1(0)
c2(0)
c3(0)
.. .
.. .
.. .
c1(1) c1(2) c1(3)
c2(1) c2(2) c2(3)
c3(1) c3(2) c3(3)
...
] . . .] ] ] . . .] ], ] . . .] ] .. .]
where (g ∘ f )n = (c0(n) , c1(n) , c2(n) , . . . ),
n = 0, 1, 2, . . . .
We may write T
Cg∘f = ((g ∘ f )0 , (g ∘ f )1 , (g ∘ f )2 , . . . ) .
(5.23)
Since g ∘ f ∈ 𝕏(ℂ), it follows by Theorem 5.6.2, or the general right distributive law, that (g ∘ f )n = g n ∘ f
for all
Thus, the equations (5.22) and (5.23) yield Cg ⋅ Cf = Cg∘f .
n ∈ ℕ ∪ {0}.
158  5 General composition The next corollary can be proved easily by using general right distributive law and the mathematical induction. Corollary 5.7.3. Let n ∈ ℕ be given and n ≥ 2, If f [j] ∈ 𝕏(ℂ) for 1 ≤ j ≤ n, then Cf [n] = Cnf . Let A : 𝕏(ℂ) → ℳ be a mapping defined by A(f ) = Cf , where ℳ is the set of all infinite matrices on ℂ, we have the following. Corollary 5.7.4. The mapping A : 𝕏(ℂ) → ℳ is an injective ring homomorphism under the composition on 𝕏(ℂ) and the matrix multiplication on ℳ. Proof. It is clear that no matrix such as Cf can correspond to two different formal power series because the second row of Cf contains the coefficients of f . This proves that A is injective. Considering that both 𝕏(ℂ) and ℳ are rings. It is obvious that A(f + g) = A(f ) + A(g)
for all f , g ∈ 𝕏(ℂ).
Theorem 5.7.2 shows that for the composition on 𝕏(ℂ) and the matrix multiplication on ℳ, A(g ∘ f ) = Cg∘f = Cg Cf = A(g)A(f ), for all g, f ∈ 𝕏(ℂ) if g ∘ f ∈ 𝕏(ℂ). Thus, A is a homomorphism. Remark 5.7.5. 1. The general composition matrix Cf works for the composition of all formal power series on ℂ, including the units, nonunits, or almost units. 2. The mapping from an almost unit to its associated delta series is an isomorphism but the similar mapping for the general composition is a homomorphism. We provide the matrix approach for the solution of the Lagrange inversion formula (4.19). Let us recall the delta associated matrix D(f ) defined in Section 1.5 for an almost unit and suppose that f is a nontrivial solution of (4.19). We extend such matrix to a general case or to a full version. Theorem 5.7.6. Let g ∈ 𝕏(ℂ) be given such that g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ ,
b0 ≠ 0.
If f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ is a nontrivial solution of the Lagrange inversion equation (4.19) f (z) = zg(f (z)) or
f (z) = g(f (z)). z
5.7 Matrix representation of the general composition, II
 159
We use the second equation and write A = BCf , where A = (a1 , a2 , a3 , . . . ), B = (b0 , b1 , b2 , . . . ), and 1 [ [0 [ [ 0 C = Cf = [ [ [ [0 [ .. [.
0 a(1) 1
0 a(1) 2
0
a(2) 2
0 .. .
0 .. .
0 a(1) 3
a(2) 3
a(3) 3 .. .
0 a(1) 4
a(2) 4
a(3) 4 .. .
... ] . . .] ] ] . . .] , ] ] . . .] ] .. .]
is the composition matrix representation of f . Ak = (a1 , a2 , a3 , . . . , ak ) can be inductively computed by Ak = Bk Ck , or T T 1 b0 a1 ] [0 [ [ ] [ b1 ] [ [a2 ] ] [ [ [ ] ] [ [ [a3 ] [ ] = [ b2 ] [0 [ . ] [. [.] [ .. ] [ . [ .. ] ] [. [ [ ] b a [ k−1 ] [0 [ k]
0 a(1) 1 0 .. . 0
0 a(1) 2
a(2) 2 .. . 0
0 ] a(1) k−1 ] ] (2) ] ak−1 ] .. ] ] . ]
... ... ... .. .
.
a(k−1) k−1 ]k×k
...
Proof. In the above matrix equality, we have k−1
(j)
ak = ∑ bj ak−1 , j=0
k = 1, 2, . . . ,
which is (4.21). Actually, we may also write f (z) = 0 + a1 x + a2 x2 + ⋅ ⋅ ⋅, and denote f = (0, a1 , a2 , . . . ), the sequence representation of f , and then denote f k = (0, . . . , 0, a(k) , a(k) , . . . ), which k k+1 0 2 is the (k + 1)th row of C. Since f = 1 + 0z + 0z + ⋅ ⋅ ⋅, it follows that f0 [f 1 ] [ ] ] Cf = [ [f 2 ] . [ ] .. [.] Example 5.7.7. Let f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ where a0 = 0 and a1 ≠ 0. Then 2 a(2) 2 = a1 ,
a(2) 3 = 2a1 a2 ,
2 a(2) 4 = 2a1 a3 + a2 ,
3 a(3) 3 = a1 ,
2 a(3) 4 = 3a1 a2 ,
2 2 a(3) 5 = 3a1 a3 + 3a1 a2 ,
3 a(2) 6 = 2a1 a5 + 2a2 a4 + a3 ,
a(2) 5 = 2a1 a4 + 2a2 a3 ,
160  5 General composition 2 3 a(3) 6 = 3a1 a4 + 6a1 a2 a3 + a2 ,
4 a(4) 4 = a1 ,
3 a(4) 5 = 4a1 a2 ,
3 2 2 a(4) 6 = 4a1 a3 + 6a1 a2 ,
3 2 3 a(4) 7 = 4a1 a4 + 12a1 a2 a3 + 4a1 a2 .
(k) For g(z) = 1 + z + z 2 + z 3 + ⋅ ⋅ ⋅, or bn = 1, n ∈ ℕ ∪ {0}, since a(0) 0 = 1 and a0 = 0, k ∈ ℕ, (4.21) becomes k−1
(j)
ak = ∑ ak−1 , j=0
k ∈ ℕ.
Then we have a1 = 1 ⋅ a(0) 0 = 1,
a2 = a(1) 1 = 1,
(2) 2 a3 = a(1) 2 + a2 = 1 + 1 ⋅ 1 = 2, 3
a4 = ∑ a3 = a3 + 2a1 a2 + a31 = 2 + 2 ⋅ 1 ⋅ 1 + 13 = 5, (j)
j=1 4
a5 = ∑ a4 = a4 + (2a1 a3 + a22 ) + (3a21 a2 ) + a41 (j)
j=1
= 5 + (2 ⋅ 1 ⋅ 2 + 12 ) + (3 ⋅ 12 ⋅ 1) + 14 = 14, k−1
(j)
ak = ∑ ak−1 , j=1
∀ k ∈ ℕ, k ≥ 2.
Then the solution of Lagrange inversion equation with the given formal power series g(z) = 1 + z + z 2 + z 3 + ⋅ ⋅ ⋅ is f (z) = z + z 2 + 2z 3 + 5z 4 + 14z 5 + ⋅ ⋅ ⋅ .
6 Formal analysis and classical analysis By the principle of formal analysis, if a formal power series over ℂ has a positive radius of convergence, then this series, when it joins the formal power series activities, must keep all properties it has in the classical analysis, such as the properties of convergence, the analyticity of the sum functions, the term by term differentiation and integration, the topology involved, and so on. Of course, a formal power series can be considered as a sequence that represents the basis of socalled digital subjects, could be used in the fields of computer sciences or coding sciences. The two rules, the general composition theorem, and the formal analytic points provide a smooth connection between formal analysis and classical analysis. The boundary behavior of the convergence of ordinary or regular power series was a hot topic around 100 years ago but this study ended without resolution. D. Bugajewski and the author provided a partial conclusion to this problem by using the general composition of formal power series [27], or the formal analytic points. This work will be introduced in Section 6.2. The umbral calculus by S. Roman and Ultrametric calculus: An introduction to padic analysis by W. H. Schikhof introduce some fundamental concepts for the space of formal power series. B. Yousefi, K. Hedayatian, and other mathematicians contribute a lot to the Banach space ℋp (β) that will be introduced in Section 6.3 [40, 101]. The formal logarithm introduced in Chapter 2 [61] is defined on formal power series g with the form g = 1 + β where β ∈ 𝕏0 . Section 6.4 will generalize the formal logarithm to be defined on 𝕏+ (ℝ) = {f ∈ 𝕏(ℝ) : f (0) > 0}. This generalization can connect the formal logarithm with the popular Maclaurin series ln(1 + x) without any gap. This general formal logarithm is defined on 𝕏+ (ℝ) and the logarithmic function ln(x) is defined on ℝ+ = {r ∈ ℝ : r > 0}. Are they similar?
6.1 Introduction to the boundary behavior The boundary behavior of a power series, or an analytic function, has been a challenge since the radius of convergence of a power series was investigated by d’Alembert and Cauchy about 200 years ago. The discussion on this topic was very hot during the end of the 19th century and the beginning of the 20th century when many wellknown mathematicians joined the investigation and provided their contributions. n In 1911, Russian mathematician, N. Lusin, constructed a power series ∑∞ n=0 cn a with radius of convergence r = 1 and cn → 0 as n → ∞, and the series diverges at every z ∈ ℂ, z = 1. In 1912, a Polish mathematician, W. Sierpiński, produced a power series with radius of convergence r = 1, which converges at the point z = 1 but diverges https://doi.org/10.1515/9783110599459006
162  6 Formal analysis and classical analysis at every other point on the unit circle (pp. 119–120, [79]). More examples can be found in [55], which was published in 1916. In his book, Complex Analysis, S. Lang put a Warning on page 60 of [56], which says “Let r be the radius of convergence of the (power) series f (z). Nothing has been said about possible convergence if z = r”. R. Burckel also says in his book [8] that “All sorts of behaviors of power series ∑ cn z n are possible on its circle of convergence.” The investigation of the boundary behavior of power series or analytic functions has never been stopped although it was slowed down after 1916 when people somehow agreed with no conclusion for this topic, and that is why there has always been a warning in the section involving this topic in the books of analysis. In order to know more about the boundary convergence behavior, the following definition is introduced. n Definition 6.1.1. Let f (z) = ∑∞ n=0 an z ∈ 𝕏(ℂ) be a formal power series with the radius of convergence R > 0, or has a convergence disk B = BR (0) = {z ∈ ℂ : z < R}. A point ω on the boundary
𝜕B = {z ∈ ℂ : z = R} is called a singular point of f if there is no holomorphic function f ̂ in any neighborhood U of ω, which satisfies f ̂U∩B = f U∩B . If every point of 𝜕B is a singular point of f , 𝜕B is called the natural boundary of f and B is called the region of holomorphy of f . It is known that the set of singular points of f on 𝜕B is always closed and can be empty. The next proposition shows how strange the boundary behavior could be. The following example and proposition can be found in (p. 151, [79]). Example 6.1.2. Let g ∈ 𝕏(ℂ) be such that ∞
n
g(z) = ∑ z 2 = z + z 2 + z 4 + z 8 + ⋅ ⋅ ⋅ . n=0
Let n ∈ ℕ be given and let ζ be a 2n th root of unity, then limg(tζ ) = ∞. t↑1 Proof. It is easy to know that radius of convergence for g is 1. For any n ∈ ℕ, n
n−1
g(z 2 ) = g(z) − (z + z 2 + ⋅ ⋅ ⋅ + z 2 ), and then 2n g(z ) ≤ g(z) + n
for z < 1.
(6.1)
6.1 Introduction to the boundary behavior
 163
q
For all q ∈ ℕ and real number t such that ( 2√2 )−1 < t < 1, q
n q 1 g(t) > ∑ t 2 > (q + 1)t 2 > (q + 1), 2 n=0
then lim g(t) = ∞. t↑1
n
Take ζ such that ζ 2 = 1, then n
n
n+1
n=1
g(tζ ) = tζ + t 2 ζ 2 + ⋅ ⋅ ⋅ + t 2 ζ 2 + t 2 ζ 2 2 2
= (tζ + t ζ + ⋅ ⋅ ⋅ + t
2n−1 2n−1
ζ
2n
+ ⋅⋅⋅
n+1
n−1
) + t + t2 ζ 2
+ ⋅⋅⋅.
Then 2 2 2n−1 2n−1 2n 2n+1 2n−1 + ⋅⋅⋅ g(tζ ) ≥ −tζ + t ζ + ⋅ ⋅ ⋅ + t ζ + t + t ζ n
n+1
≥ −n + (t 2 + t 2 2n
n+2
+ t2
+ ⋅ ⋅ ⋅)
= g(t ) − n. Thus, limg(tζ ) = ∞. t↑1 Let Gn be the set of all nth roots of unity, n = 0, 1, 2, . . . , let ∞
G = ⋃ Gn , n=0
∞
and H = ⋃ G2n n=0
are subgroups of 𝜕B1 (0), with H ⊆ G. The density theorem says that both H and G are dense in 𝜕B1 (0). Proposition 6.1.3. Let all notation be set as in the example above. Then the boundary of the unit disc 𝜕B1 (0) is the natural boundary of g. Proof. For every 2n th root of unity ζ ∈ H, limt↑1 g(tζ ) = ∞ shows that ζ is a singular point of g. Since H is dense in 𝜕B1 (0) and the singular points constitute a closed set, the claim follows. Corollary 6.1.4. The unit disc B = B1 (0) is the region of holomorphy of the function −n 2n h(z) = ∑∞ n=0 2 z and h is continuous on B = B ∪ 𝜕B. Proof. It is clear that h absolutely converges for every point on 𝜕B and then h is absolutely convergent in B. Thus, h is continuous in B together with its boundary 𝜕B.
164  6 Formal analysis and classical analysis Recalling the function g in Example 6.1.2 and h in Corollary 6.1.4, we have zh (z) = g(z). We would be interested in the question: what is the relationship of natural boundary between the functions h and h ? A recent development on boundary behavior is worth of being introduced here. It is a certain kind of universal result. The proof of the proposition below can be found in [11]. k ∞ Proposition 6.1.5. Let {ζk }∞ k=1 ⊆ 𝜕B where B = B1 (0). For every k ∈ ℕ, let {zn }n=1 ⊆ B such that
lim z k n→∞ n
= ζk .
Then there exists a function f which is analytic on B and continuous on B and has the following property: for any sequence (wk )k∈ℕ ⊆ ℂ there exists an increasing sequence (nj )j∈ℕ ⊆ ℕ such that for any k ∈ ℕ, lim f (k) (znkj ) = wk .
j→∞
The set of such functions, denoted by U((ζk ), (znk )), is a dense Gδ subset of the set of all functions that are analytic on B and continuous on B. A wellknown result is that if a power series has a positive radius of convergence then all derivatives of this power series have the same radius of convergence. We should also remember that the derivative of a formal power series has nothing to do with socalled the sum function of power series and has nothing to do with the limit of difference quotient, although there is some relationship between the formal derivan tives and the derivatives of the regular power series. For example, if g(x) = ∑∞ n=0 n!x is considered as a regular power series over ℝ, it converges at x = 0 only and, therefore, the derivative of g does not make any sense. However, if we consider g as a formal power series, then ∞
g (k) (x) = ∑
n=k
n! n!x n−k (n − k)!
is welldefined for all k ∈ ℕ. Therefore, we must be very careful when we deal with the convergence of a power series and when we deal with the derivatives of formal power series. The convergence of a power series on the boundary points of the region of convergence is a very special topic for mathematicians. Such convergence actually not only relies on the given power series but also relies on the derivatives of the power series. Let us see the example below.
6.1 Introduction to the boundary behavior
 165
n
(−1) n Example 6.1.6. Let g(x) = ∑∞ n=0 n x and h = 1 be formal power series on ℝ. Then g 1 is analytic on B1 (0) = {x ∈ ℝ : x < 1} and g(h) = g(1) ∈ ℝ but g(−1) = ∑∞ n=1 n ∉ ℝ. ∞ n−1 Moreover, g (h) = g (1) = ∑n=0 (−1) ∉ ℝ.
Let g ∈ 𝕏(ℂ) be analytic in B = B1 (0) and g ∈ C ∞ (B). We may write g(z) = b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ . We know that g(w) = b0 + b1 w + b2 w2 + ⋅ ⋅ ⋅ ∈ ℂ for every w ∈ B, but we do not know whether g(z) ∈ ℂ if z ∈ 𝜕B or if z = 1. We also understand that f is analytic at a point w ∈ ℂ means that f is differentiable infinitely many times in an open region centered at w. A boundary point does not have such advantages. If we consider that g is a formal power series and g (n) , n ∈ ℕ, is the nth order formal derivative of g, then the outcome could be different. Recalling Definition 5.4.7 and Theorem 5.4.6 about the formal analytic points we may find: 1. a formal analytic point does not require a neighborhood of the point in which g is differentiable infinitely many times; 2. if a is a formal analytic point, then g (n) (a) ∈ ℂ for all n ∈ ℕ is equivalent to that g (n) (a) ∈ ℂ for infinitely many n ∈ ℕ. For boundary convergence problem of power series, can formal power series contribute anything? We close this section by providing a classical example for a power series over ℂ. Example 6.1.7. Let zn n=1 n ∞
g(z) = ∑
be a power series over ℂ. It is well known that the radius of convergence of this series is 1. Actually zn = − Log(1 − z) ∀ z < 1. n=1 n ∞
g(z) = ∑
(∗)
where Log is the principal logarithm. The right side − Log(1 − z) of (∗) is continuous on Xδ = {z ∈ ℂ : z ≤ 1, z − 1 ≥ δ} for each 0 < δ < 2, so is the left side. This example shows that actually this power series g converges at every boundary point of its region of convergence except z = 1.
166  6 Formal analysis and classical analysis
6.2 Boundary behavior and formal analysis This section provides a test for checking the boundary convergence of a power series by connecting it to the formal analytic point. Definition 6.2.1. Let g ∈ 𝕏(S) be given, where S is a metric field. We denote by r(g) the radius of convergence of g and by I(g) the interval, or disc, or ball of convergence of g. As usual, 0 ≤ r(g) ≤ +∞. We denote by 𝕏(S) the set of all convergent power series over S centered at 0 with positive radius (not infinity) of convergence. We also define the composition domain or formal analytic domain of g to be the set D(g) = {a ∈ S : g (n) (a) ∈ S
for infinitely many
n ∈ ℕ},
where g (n) is the nth order formal derivative of g and g (0) = g. It is clear that D(g) ≠ 0 for every g ∈ 𝕏 = 𝕏(S) because at least 0 ∈ D(g). Recalling the set 𝕏g defined in Definition 5.1.5, now we define a similar set 𝕏g = D(g) + 𝕏0 (the Minkowski sum of D(g) and all nonunits in 𝕏), which is also said to be the gcomposition subset of 𝕏, where D(g) is considered as a subset of 𝕏 consisting of all formal analytic points of g. Let us emphasize that the termwise differentiation of series is an interesting topic in analysis. There are some general results about the termwise differentiation of a n power series f (x) = ∑∞ n=0 an x ∈ 𝕏 with x < r = r(f ). If r is a positive real number, n−1 then the existence of f (r) = ∑∞ is not a trivial question. Some results about n=0 nan r the termwise differentiation at the endpoints of the interval of convergence could be found in [95]. We are interested in those power series or analytic function g in 𝕏(ℝ) or 𝕏(ℂ). Example 6.1.2 has showed us how complicated the boundary behavior could be for g ∈ 𝕏(ℂ). Example 6.1.6 tells us how the derivative of a power series may affect the convergence at the endpoints of the interval of convergence, or at the boundary points of the region of convergence. In particular, if we seek such convergence of termwise differentiation for power series not only for once, but for the infinitely many times, what can we say? No matter whether g ∈ 𝕏 or g ∈ 𝕏(ℂ), the formal derivative g (n) always exists for all nonnegative integer n. Therefore, we can test every point of ℂ for whether it is a formal analytic points formal analytic points. For any g ∈ 𝕏(ℂ), it is obvious that if z ∈ ℂ with z < r(g), then z is an analytic point and also a formal analytic point of g. We are interested in testing the boundary point of g whether it is a formal analytic point. Therefore, we must investigate D(g).
6.2 Boundary behavior and formal analysis  167
Lemma 6.2.2. Let g ∈ 𝕏(ℝ) and a ∈ ℝ be given. Then a ∈ D(g)
if and only if −a ∈ D(g).
n Proof. Let g(x) = ∑∞ n=0 bn x . The conclusion is true obviously if a = 0. We now suppose that a ≠ 0. Without loss of generality, we assume that
a ∈ D(g)
but
−a ∈ D(g) fails.
By Theorem 5.4.1, Formulas (5.13) and (5.14), ∞ n ∑ ( )bn an−k ∈ ℝ k n=k
for every k ∈ ℕ ∪ {0}
but n )b (−a)n−k0 ∉ ℝ k0 n
∞
∑ (
n=k0
n−k0 n for some k0 ∈ ℕ. Then ∑∞ diverges which means that there are inn=k0 (k0 )bn (−a) finitely many n ∈ ℕ such that
1 n )bn  ⋅ an−k0 > 2 . k0 n
(
Then there exists a sequence of positive integers (nk ), nk ≥ k0 , such that 1 nk )bnk  ⋅ ank −k0 > 2 k0 nk
( Then
for every k ∈ ℕ.
1 1 nk (nk − 1) ⋅ ⋅ ⋅ (nk − k0 + 1)bnk ank > 2 , nk ak0 ⋅ k0 !
so
bnk  ⋅ ank >
ak0 ⋅ k0 ! ak0 ⋅ k0 ! . > k +2 n3k (nk − 1) ⋅ ⋅ ⋅ (nk − k0 + 1) nk0
Hence 1 n ( k )bnk  ⋅ ank −3 ak0 k0 + 3 n (n − 1) ⋅ ⋅ ⋅ (nk − k0 − 2) = k k ⋅ bnk  ⋅ ank (k0 + 3)!ak0 +3 > =
nk (nk − 1) ⋅ ⋅ ⋅ (nk − k0 − 2) ak0 ⋅ k0 ! ⋅ k +2 (k0 + 3)!ak0 +3 nk0
k +2 1 1 ⋅ nk (1 − ) ⋅ ⋅ ⋅ (1 − 0 ). 3 nk nk (k0 + 1)(k0 + 2)(k0 + 3) ⋅ a
168  6 Formal analysis and classical analysis Then lim (
k→∞
nk )b  ⋅ ank −3 = ∞. k0 + 3 nk
Therefore,
n )b  ⋅ an−3 = 0 k0 + 3 n
lim (
n→∞
fails, and hence
n )b ⋅ an−3 = 0 k0 + 3 n
lim (
n→∞
fails. Then n )b an−k0 ∈ ℝ k0 + 3 n
∞
∑ (
n=k0
fails. It is a contradiction to our assumption. Thus, −a ∈ D(g). If g ∈ 𝕏(g) in Lemma 6.2.2 has a positive radius of convergence and the formal analytic point a ∈ D(g) is the endpoint of the interval of the convergence, the lemma actually provides a test for the boundary behavior of such a power series. We can extend this result from 𝕏(ℝ) to 𝕏(ℂ). Corollary 6.2.3. Let g ∈ 𝕏(ℂ) and a ∈ ℂ be given. If a ∈ D(g), then z ∈ D(g) for all z ∈ ℂ with z = a. Proof. Taking z = aeiθ for some real number θ and applying the similar approach as in the proof of Lemma 6.2.2, we can obtain the conclusion. It is clear that D(g) ⊆ I(g) for every g ∈ 𝕏 = 𝕏(ℂ). It is also clear that z ∈ ℂ with z < r(g) implies that g (n) (z) ∈ ℂ for every n ∈ ℕ and, therefore, z ∈ D(g). By these facts and the results above, we can prove the following theorem. Theorem 6.2.4. The composition domain D(g) is convex and balanced for every formal power series g ∈ 𝕏 = 𝕏(ℂ). Proof. Let g ∈ 𝕏 be given and let r = r(g). If r = 0 or r = +∞, then D(g) = {0} = I(g) or D(g) = ℂ = I(g), respectively, so the conclusion is obvious. Now, let 0 < r < +∞ and let 0 < t < 1 be given. Suppose that a0 , c0 ∈ D(g). If a0  < r or c0  < r, then ta0 + (1 − t)c0  < r, and hence ta0 + (1 − t)c0 ∈ D(g). Further, if a0  = c0  = r, then ta0 + (1 − t)c0  ≤ r and, therefore, Corollary 6.2.3 yields that ta0 + (1 − t)c0 ∈ D(g). It means that D(g) is always convex. Next, suppose that α ∈ ℂ with α ≤ 1. If a0  < r or α < 1, then αa0  < r and, therefore, αa0 ∈ D(g). In the case that a0  = r and α = 1, we have αa0  = r. Applying Corollary 6.2.3 again, we obtain the conclusion. Thus, D(g) is balanced and the proof is completed.
6.2 Boundary behavior and formal analysis  169
The absolute convergence on the boundary of the region of convergence is a more interesting topic for the study of power series. Theorem 6.2.5. Let g ∈ 𝕏(ℂ) be given such that r = r(g) < ∞. If a ∈ ℂ is a formal analytic point of g with a = r, then the power series g (k) converges absolutely on the closed disk {z ∈ ℂ : z ≤ r}, for every k ∈ ℕ. Proof. The case that r = 0 is trivial. We suppose that r > 0. Let a be a formal analytic point of g with a = r, it follows that g (k) (a) ∈ ℂ for every k ∈ ℕ ∪ {0}. Then g (k) (r) ∈ ℂ for every k ∈ ℕ by Corollary 6.2.3. It suffices to show that g (k) (r) converges absolutely for every k ∈ ℕ. Assume that n )bn r n−k0 = +∞ k0
∞
∑ (
n=k0
for some k0 ∈ ℕ. Then there are infinitely many n ∈ ℕ such that n 1 )bn r n−k0 > 2 . k0 n
(
Further, by similar reasoning as in the proof of Lemma 6.2.2, n ) b r n−k0 −3 = 0 k0 + 3 n
lim (
n→∞
fails which contradicts that n ) b r n−k0 −3 ∈ ℂ. k0 + 3 n
∞
∑ (
n=k0
Thus, the power series g (k) converges absolutely on the boundary of the closed disc {z ∈ ℂ : z ≤ r}, and hence it converges on this closed disk. We know that every power series g ∈ 𝕏(ℂ) is continuous on the open disc D = Br(g) (0) = {z ∈ ℂ : z < r(g)}. It may happen if r(g) < ∞, however, that g(z0 ) is defined for some z0  = r(g) but g is not continuous on D ∪ {z0 }. We provide a sufficient condition for the uniform continuity of a power series on closed disc D below. Lemma 6.2.6. Let g ∈ 𝕏(ℂ) be given such that 0 < r = r(g) < ∞. If z0 ∈ ℂ is a formal analytic point of g with z0  = r, then the power series g (k) is uniformly continuous on the closed disc D = {z ∈ ℂ : z ≤ r} for every k ∈ ℕ ∪ {0}.
170  6 Formal analysis and classical analysis Proof. Let z ∈ ℂ be such that z = r. Since g (k) (z0 ) ∈ ℂ for every k ∈ ℕ ∪ {0}, it follows that g (k) (z) ∈ ℂ for every k ∈ ℕ ∪ {0} by Corollary 6.2.3. Let h ∈ ℂ be such that 0 < h < r and z − h ≤ r. Then ∞ n n g(z − h) − g(z) = ∑ bn [(z − h) − z ] n=0 ∞
≤ ∑ bn h[(z − h)n−1 + (z − h)n−2 z + ⋅ ⋅ ⋅ + (z − h)z n−2 + z n−1 ] n=0 ∞
≤ ∑ bn h[z − hn−1 + z − hn−2 z + ⋅ ⋅ ⋅ + z − hzn−2 + zn−1 ] n=0
∞
≤ h ∑ bn nr n−1 . n=0
By Theorem 6.2.5, g converges absolutely at r which means that ∞
∑ bn nr n−1 ∈ ℝ.
n=0
Letting h → 0, we deduce that g is continuous on the closed disc D. Therefore, g is uniformly continuous on D. Let us notice that since the convergence of g (k+1) on the closed disc D is absolute, we can similarly prove that g (k) is uniformly continuous on D for every k ∈ ℕ. By means of the results above, we introduce some properties of gcomposition subset of 𝕏(ℂ). Corollary 6.2.7. For any g ∈ 𝕏 = 𝕏(ℂ), the gcomposition subset 𝕏g of 𝕏 is convex and balanced. Proof. It is clear that 𝕏0 (ℂ) is convex and balanced. Theorem 6.2.4 shows that D(g) is convex and balanced. Thus, 𝕏g = D(g) + 𝕏0 is convex and balanced. The next result describes the relation between 𝕏g and 𝕏g , where g ∈ 𝕏(ℂ). Corollary 6.2.8. Let g ∈ 𝕏(ℂ) be given. Then: (i) 𝕏g ⊆ 𝕏g , (ii) 𝕏g ≠ 𝕏g if and only if there exists some nonzero a ∈ ℂ such that a = r(g),
g(a) ∈ ℂ
but
g (k) (a) ∉ ℂ for some
k ∈ ℕ.
Proof. (i) follows from Theorem 5.4.1 and formula (5.14). For (ii). suppose that there exists a nonzero a ∈ ℂ such that a = r(g),
g(a) ∈ ℂ but
g (k) (a) ∉ ℂ
for some k ∈ ℕ.
6.2 Boundary behavior and formal analysis  171
Let f = a. Then f ∈ 𝕏g but f ∉ 𝕏g by the fact that g(f ) = g(a) ∈ ℂ but a ∉ D(g). Then 𝕏g ≠ 𝕏g . Conversely, we suppose that 𝕏g ≠ 𝕏g . Then there exists a formal power series f ∈ 𝕏g but f ∉ 𝕏g by (i) above. Denote f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ . If deg(f ) > 0, then g (k) (a0 ) ∈ ℂ for all k ∈ ℕ ∪ {0} by Theorem 5.4.6 and hence f ∈ 𝕏g . Therefore, deg(f ) = 0
or
f = a0 ∈ ℂ.
If a0 = 0, then f = a0 = 0 ∈ 𝕏g which contradicts to that f ∉ 𝕏g . So a0 ≠ 0. Since a0 = f ∈ 𝕏g , it follows that g(f ) = g(a0 ) ∈ ℂ and then r(g) > 0. If 0 < a0  < r(g), then g (k) (a0 ) ∈ ℂ for all k ∈ ℕ ∪ {0} by termwise differentiation for power series, and hence f ∈ 𝕏g . Then a0  = r(g) because g(a0 ) = g(f ) ∈ ℂ. Finally, it is clear that g (k) (a0 ) ∉ ℂ for some k ∈ ℕ because otherwise f ∈ 𝕏g by Theorem 5.4.1. Thus, f ∈ 𝕏g but f ∉ 𝕏g if and only if there exists a nonzero a ∈ ℂ such that a = r(g),
g(a) ∈ ℂ but
g (k) (a) ∉ ℂ for some k ∈ ℕ.
Proposition 6.2.9. Let g ∈ 𝕏 = 𝕏(ℂ) be given. Then: (i) 𝕏g = 𝕏0 if and only if r(g) = 0. (ii) 𝕏g = 𝕏 if and only if r(g) = +∞. (iii) h, g ∈ 𝕏 with r(h) < r(g) implies that D(h) ⊆ D(g), and 𝕏h ⊆ 𝕏h ⊆ 𝕏g ⊆ 𝕏g . (iv) It is not always true that h, g ∈ 𝕏 with r(h) = r(g) implies that 𝕏h = 𝕏g . (v) 𝕏 = ⋃g∈𝕏 𝕏g . Proof. (i) follows from Corollary 6.2.8 and Theorem 5.4.1. Properties (ii) and (v) are obvious. (iii) is yielded by Theorem 5.4.3 and Corollary 6.2.8. For (iv) considering 1 n x n n=1 ∞
h(x) = ∑
1 n x . 2 n=1 n ∞
and g(x) = ∑
It is clear that a formal power series 1 ∈ 𝕏g but 1 ∉ 𝕏h .
172  6 Formal analysis and classical analysis 1 n Example 6.2.10. (a) 𝕏(ℂ) = 𝕏ez where ez = ∑∞ n=0 n! z . ∞ 1 n (b) Let ga (x) = ∑n=0 an x . This geometric series has the radius of convergence r(g) = a if a > 0. Then applying Theorem 5.4.3, we have ∞
𝕏(ℝ) = ⋃ 𝕏ga = ⋃ 𝕏gk . a>0
k=1
1 n ∑∞ n=1 n2 x ,
(c) Let g(x) = as in the proof of Proposition 6.2.9 (iv). Observe that 1 ∈ 𝕏g , however, 1 ∉ D(g). Notice also that deg(1) = 0. Now we provide a necessary and sufficient condition under which an operator Tg , where g ∈ 𝕏(ℂ) or g ∈ 𝕏(ℂ), maps 𝕏g or 𝕏g into itself. n Proposition 6.2.11. Let g = ∑∞ n=0 bn z ∈ 𝕏(ℂ) be given. Then Tg maps 𝕏g into itself if and only if g(a) ∈ D(g) for every a ∈ D(g), where Tg (f ) = g ∘ f .
Proof. Let f ∈ 𝕏g = D(g)+𝕏0 be given. Then f = a+f , for some a ∈ D(g) and f ∈ 𝕏0 . By n Definition 5.1.5, if g ∘f is welldefined, the constant term of g ∘f is c0 = ∑∞ n=0 bn a = g(a) 0 and, therefore, g ∘ f ∈ {c0 } + 𝕏 . Thus g ∘ f ∈ 𝕏g
if and only if g(a) ∈ D(g).
If the formal power series g in the above proposition does not have a positive radius of convergence, the composition mapping Tg may still map 𝕏g to itself but needs an additional condition. Corollary 6.2.12. Let g ∈ 𝕏(ℂ) be given. Then Tg maps 𝕏g into 𝕏g if and only if g(a) ∈ D(g) for every a ∈ D(g), and g(g(a)) ∈ ℂ for every
a ∈ 𝕏g \ 𝕏g .
Proof. The conclusion is a consequence of Proposition 6.2.11 and (ii) of Corollary 6.2.8. The boundary convergence problem landed to inconclusive after the intensive investigations and discussions a little more than one hundred years ago, as we described in Section 6.1. Since then, mathematicians slowed down the pace but have never stopped tackling this problem, and continued to make contributions, such as Example 6.1.2 and Proposition 6.1.5. We are providing a formal analytic theorem or a boundary behavior test by using the set of formal analytic points. Let us first denote by 𝜕E the boundary of any subset E ⊆ ℂ. Theorem 6.2.13. Let g ∈ 𝕏(ℂ) be given with a positive radius of convergence r and a region of convergence I(g). If g is not continuous on I(g) = I(g) ∪ 𝜕I(g), then there exists an integer k ∈ ℕ and a point w with w = r such that g (k) (w) ∉ ℂ.
6.3 Banach spaces ℋp (β) and ℒp (β)
 173
If there exists a formal analytic point w ∈ D(g) with w = r, then g is continuous on the closed disk I(g). Proof. Write B = Br (0) = {z ∈ ℂ : z < r}. It is clear that I(g) = Br (0) = B ∪ 𝜕B. Suppose that g is not continuous on I(g). Since g is analytic on B, and hence g is continuous at every point of B, then there must exist w ∈ ℂ with w = r and g is not continuous at w. If g (n) (w) ∈ ℂ for all n ∈ ℕ ∪ {0}, then w is a formal analytic point of g and then z ∈ D(g)
for all z ∈ 𝜕I(g),
and then g is uniformly continuous on I(g) by Lemma 6.2.6. It is a contradiction. Thus, there exists a k ∈ ℕ ∪ {0} such that g (k) (w) ∉ ℂ. If there exists a formal analytic point w ∈ D(g) with w = r, Lemma 6.2.6 ensures the conclusion. Please be noticed that although I(g) = B, generally I(g) ≠ B.
6.3 Banach spaces ℋp (β) and ℒp (β) The set 𝕏(ℂ) is a linear space and a linear algebra with Cauchy product. A natural challenge for mathematicians is whether we may develop a normed space for 𝕏(ℂ) or for some subspaces of 𝕏(ℂ). We introduce such a normed space ℋp (β), a subspace of 𝕏(ℂ), based on [101] and [103]. Definition 6.3.1. Let β: ℕ ∪ {0} → (0, ∞) be a mapping from nonnegative integers to (0, ∞) such that β(0) = 1. For a real number 1 ≤ p < ∞ and for a formal power series f ∈ 𝕏(ℂ) such that f (z) = f0 + f1 z + f2 z 2 + ⋅ ⋅ ⋅, the mapping ‖ ⋅ ‖β,p : 𝕏(ℂ) → ℝ ∪ {∞} is defined by ∞
p
p
1/p
‖f ‖β,p = ( ∑ β (n)fn  ) n=0
,
where we define that ∞a = ∞ for a > 0. We usually write ‖f ‖β,p as ‖f ‖β if there is no confusion. We define the subset p
ℋ (β) = {f ∈ 𝕏(ℂ) : ‖f ‖β < ∞}.
It is easy to verify that ‖⋅‖β is a norm on ℋp (β), and then the subset ℋp (β) becomes a normed space under this norm. A significant character of such a normed space is that every formal power series f in ℋp (β) has relationship with the sequence {β(n)}∞ n=0 , which is called the βsequence.
174  6 Formal analysis and classical analysis n ̂ ̂ Proposition 6.3.2. Let fk (z) = ∑∞ n=0 fk (n)z ∈ 𝕏(ℂ) be given such that fk (n) = δk (n) for every nonnegative integer k, where δk (k) = 1 and δk (n) = 0 for all n ≠ k. Then {fk }∞ k=0 forms a base for ℋp (β), 1 ≤ p, and ‖fk ‖β = β(k).
Proof. Actually fk (z) = z k for every k ∈ ℕ ∪ {0}, which shows that the sequence {fk }∞ k=0 forms a base for 𝕏(ℂ), of course, also for ℋp (β). By the definition of δk (n), ∞
1/p
p ‖fk ‖β = ( ∑ β (n)̂fk (n) ) p
n=0
∞
1/p
p = ( ∑ β (n)δk (n) ) p
= β(k).
n=0
Proposition 6.3.3. For 1 ≤ p < ∞, the space ℋp (β) is a Banach space under the norm ‖ ⋅ ‖β . Proof. It is not difficult, by using Minkowski’s Inequality, for us to prove that ℋp (β) is a linear vector space. We need only show that every Cauchy sequence in ℋp (β) converges in ℋp (β). Let 1 ≤ p < ∞ be given and let {fk } ⊆ ℋp (β) be a Cauchy sequence. For each k ∈ ℕ, write ∞
fk (z) = ∑ ̂fk (n)z n , n=0
and define ak = (ak,0 , ak,1 , ak,2 , . . .) by ak,n = β(n)̂fk (n) for every
n ∈ ℕ ∪ {0}.
Let us also define ∞
1/p
p
.
‖ak ‖p = ( ∑ ak,j  ) j=0
Now let k, m ∈ ℕ be given. For any sequences ak and am we have ∞
p
1/p
‖ak − am ‖p = ( ∑ ak,n − am,n  ) n=0 ∞
1/p
p = ( ∑ β(n)̂fk (n) − β(n)̂fm (n) ) n=0
= ‖fk − fm ‖β . This equality shows that {fk } ⊆ ℋp (β) is equivalent to (ak ) ⊆ ℓp under the corresponding norms. Minkowski’s It is well known that the space ℓp is a Banach space under the norm ‖ ⋅ ‖p . Therefore, ℋp (β) is a Banach space under the norm ‖ ⋅ ‖β .
6.3 Banach spaces ℋp (β) and ℒp (β)
 175
Remark 6.3.4. In the proof of Proposition 6.3.3, we have seen that given a formal n p p ̂ power series f (z) = ∑∞ n=0 f (n)z ∈ H (β) we can find a sequence af ∈ ℓ such that af = (an )∞ n=1
with
an = β(n)̂f (n)
for all
n ∈ ℕ.
Conversely, for any a = (an ) ∈ ℓp , let ̂f (n) = an /β(n) for every nonnegative integer n, we have ∞
f (z) = ∑ ̂f (n)z n ∈ ℋp (β). n=0
One checks that the mapping Tβ : ℋp (β) → ℓp , 1 ≤ p < ∞ by Tβ (f ) = af is an isomorphism between these two linear spaces. Definition 6.3.5. For 1 ≤ p < ∞, the mapping Mz : ℋp (β) → 𝕏 defined by ∞
∞
n=0
m=1
Mz (f )(z) = ∑ ̂f (n)z n+1 = ∑ ̂f (m − 1)z m , n p p ̂ for every f (z) = ∑∞ n=0 f (n)z in ℋ (β), is called a shifting operator on ℋ (β).
We know that the space ℋp (β) relies on the β sequence {β(n)}∞ n=0 . We want to know more about this sequence. Proposition 6.3.6. Mz is a bounded operator if and only if the sequence β(k + 1) } β(k) k=0 ∞
{ is bounded.
Proof. Let fk (z) = z k , k = 0, 1, 2, . . .. By Proposition 6.3.2, we know that {fk } forms a basis of ℋp (β) and ‖fk ‖β = β(k) for every nonnegative integer k, Suppose that the sequence {
β(k+1) } β(k)
β(k + 1) ≤B β(k)
is bounded, say that
for every k ∈ ℕ ∪ {0},
n p ̂ for some real number B > 0. Then for any f (z) = ∑∞ n=0 f (n)z in ℋ (β) with ‖f ‖β ≤ 1, 1/p
∞
p p ‖Mz f ‖β = ( ∑ ̂f (n) [β(n + 1)] ) n=0
p 1/p
p β(n + 1) p = (∑̂f (n) [β(n)] [ ] ) β(n) 1/p
p ≤ B(∑̂f (n) βp (n)) so ‖Mz ‖ < ∞.
= B ⋅ ‖f ‖β ≤ B,
176  6 Formal analysis and classical analysis Conversely, suppose that ‖Mz ‖ < ∞. It is clear that Mz fk = fk+1 for every nonnegative integer k where {fk } is a basis of ℋp (β). Then ‖fk+1 ‖β = ‖Mz fk ‖β ≤ ‖Mz ‖ ⋅ ‖fk ‖β , or β(k + 1) ≤ ‖Mz ‖ ⋅ β(k),
k = 0, 1, 2, . . . ,
β(k + 1) ≤ ‖Mz ‖ < ∞, β(k)
k = 0, 1, 2, . . . .
and hence
When a power series has been investigated, the next topic for mathematicians is usually the investigation of the Laurent series. Before we set up the space of the formal Laurent series, which will be done late in the next chapter, we introduce how to extend the subspace ℋ(β) to a subspace of Laurent series which is also connected with the βsequence. Definition 6.3.7. Let {β(n)}n∈ℤ be a βsequence where β(n) > 0 for all n ∈ ℤ with β(0) = 1, and let {f ̂(n)}n∈ℤ be a sequence in ℂ. Write ∞
f (z) = ∑ f ̂(n)z n = ⋅ ⋅ ⋅ + f ̂(−1)z −1 + f ̂(0) + f ̂(1)z + ⋅ ⋅ ⋅ n=−∞
without assigning any value to z. f is called a formal Laurent series. Then for any 1 ≤ p < ∞ we define a mapping ‖ ⋅ ‖ from the set of all such formal Laurent series to [0, ∞] by 1/p
p ‖f ‖ = ‖f ‖β,p = ( ∑ f ̂(n) β(n)p ) n∈ℤ
.
We usually write ‖ ⋅ ‖β,p as ‖ ⋅ ‖β if there is no confusion. We define ℒp (β) by p n ℒ (β) = {f (z) = ∑ f ̂(n)z : ‖f ‖β < ∞}. n∈ℤ
A similar argument in the proof of Proposition 6.3.3 yields the proposition below. We omit the proof. Proposition 6.3.8. For 1 ≤ p < ∞, the space ℒp (β) is a Banach space under the norm ‖ ⋅ ‖β .
6.3 Banach spaces ℋp (β) and ℒp (β)
 177
The shifting operator Mz can be extend to work on ℒp (β), too. A similar statement as Proposition 6.3.6 is also true for Mz on ℒp (β). It is obvious that Mz (f ) ∈ ℒp (β) for every f ∈ ℒp (β), and then Mzn = Mz (Mzn−1 ) is welldefined for every n ∈ ℕ. Actually Mzn (f ) = Mz n (f ) = ∑ f ̂(k)z k+n , k∈ℤ
for every n ∈ ℕ, where Mz0 = I or Mz0 (f ) = f . Then 1/p
n p p Mz (f )β = ( ∑ f ̂(k) β(k + n) ) k∈ℤ
.
Theorem 6.3.9. If Mz is bounded, then for every k ∈ ℕ, β(k + n) k }. Mz β = sup{ β(n) n∈ℤ
Proof. Let k ∈ ℕ be given. Denote by Bk = sup{ n∈ℤ
β(k + n) }, β(n)
no matter Bk if finite or infinity. We are going to show that Bk is bounded if ‖Mz ‖ is bounded. Suppose that ‖Mz ‖ is bounded, it follows that B1 is a positive real number by the similar proof of Proposition 6.3.6 (extend ℕ to ℤ). Then for every k ∈ ℕ given, β(n + k) β(n + k) β(n + k − 1) β(n + 1) = ⋅ ⋅⋅⋅ , β(n) β(n + k − 1) β(n + k − 2) β(n) for all n ∈ ℤ, which shows that Bk ≤ Bk1 < ∞, because sup{ n∈ℤ
β(n + j) β(m + 1) } = sup{ } = B1 , β(n + j − 1) β(m) m∈ℤ
for every j ∈ ℤ, then Bk is bounded. Let k ∈ ℕ be given, then for f ∈ ℒp (β) with ‖f ‖β ≤ 1, we have 1/p
p k p Mz f β = ( ∑ ̂f (n) [β(n + k)] ) n∈ℤ
p 1/p
p β(n + k) p = (∑̂f (n) [β(n)] [ ] ) β(n)
178  6 Formal analysis and classical analysis p 1/p
β(n + k) p p }] ) ≤ (∑̂f (n) [β(n)] [sup{ β(n) n∈ℤ p ≤ Bk (∑̂f (n) βp (n))
1/p
= Bk ⋅ ‖f ‖β ≤ Bk < ∞.
So ‖Mzk ‖ ≤ Bk < ∞ or ‖Mzk ‖ is bounded. It is not difficult to verify that the sequence {z n }n∈ℤ forms a base for ℒp (β). For the basis {fj }j∈ℤ where fj (z) = z j , we may easily obtain that ‖fj ‖β = β(j) as we did in the proof of Proposition 6.3.2. Then for every k ∈ ℕ we have k k+j Mz (fj )β = z β = β(j + k)
for all j ∈ ℤ.
Then β(n + k) = Mzk (fn )β ≤ Mzk ⋅ ‖fn ‖β = Mzk β(n), and hence sup{ n∈ℤ
β(n + k) } ≤ Mzk . β(n)
Thus, ‖Mzk ‖ = Bk for all k ∈ ℕ if ‖Mz ‖ is bounded. Actually the relationship between ‖Mz ‖ and one is bounded or not.
β(n+k) β(n)
could be closer, no matter any
Corollary 6.3.10. For any shifting operator Mz on ℋp (β) or ℒp (β), β(k + n) k }, Mz β = sup{ β(n) n∈ℤ
for all k ∈ ℕ. Proof. We have just proved that the conclusion is true if ‖Mzk ‖β is bounded. As we did in the proofs of the above theorems, β(n + k) k β(n + k) ≤ Mz ≤ sup{ }, β(n) β(n) n∈ℤ for every k ∈ ℕ. Thus, the conclusion is true for either ‖Mzk ‖β is bounded or k Mz = ∞
or
sup{ n∈ℤ
β(n + k) } = ∞. β(n)
p
Another interesting operator on ℒ (β) is the multiplication operator of a formal power series φ ∈ ℒp (β), which is denoted by Mφ . The set of multipliers ℒp∞ (β) is defined by p
p
p
p
ℒ∞ (β) = {φ ∈ ℒ (β) : φℒ (β) ⊆ ℒ (β)}.
6.4 General formal logarithm
 179
Theorem 6.3.11 ([102]). Let 2 ≤ p < ∞. Then we have: (i) If φ ∈ ℒp∞ (β), then φω ∈ ℒp∞ (β) and ‖Mφω ‖ = ‖Mφ ‖ for all ω ∈ 𝜕U, where U ⊆ ℂ is the unit disk. (ii) Let φ ∈ ℒp∞ (β). If ω → 1, then Mφω → Mφ in the strong operator topology. (iii) If g is a continuous complex valued function on 𝜕U and dλ = dω/2π is the normalized Lebesgue measure on 𝜕U, then the operator ∫𝜕U φω g(ω)dλ defined by ( ∫ φω g(ω)dλ)f = ∫ g(ω)Mφω fdλ 𝜕U
𝜕U
is in ℒp∞ (β) and ∫ φω g(ω)dλ ≤ ‖Mφ ‖ ∫ gdλ. ∞ 𝜕U
𝜕U
The proof is set as an exercise. More properties of ℒp (β) and ℋp (β) can be found in [101–103].
6.4 General formal logarithm We introduced the formal logarithmic function L in Section 2.2 provided by Niven [61]. L is a transformation on 𝕏, which transforms a formal power series in 𝕏1 to a formal power series in 𝕏0 . Niven at the same time reminded the readers to be aware that L is defined only for F ∈ 𝕏1 (ℂ), where 𝕏1 = {1 + f : f ∈ 𝕏0 } = {F ∈ 𝕏 : F(0) = 1}. For every F = 1 + f ∈ 𝕏1 we define 1 1 1 L(F) = L(1 + f ) = f − f 2 + f 3 − f 4 + ⋅ ⋅ ⋅ , 2 3 4 which is formula (2.1). We know that the formal logarithm has brought us many useful and interesting results including L(FG) = L(F) + L(G),
and
L(E1 (f )) = f ,
for all F, G ∈ 𝕏1 , and f ∈ 𝕏0 . Although a formal power series f can be considered as a sequence, there should not have any conflict for f to be considered as a power series, no matter it has positive radius of convergence or not. The principle of formal analysis says: all properties of a convergent power series should be maintained when it is considered as a formal power
180  6 Formal analysis and classical analysis series. If so, the formal logarithm L defined in (2.1) faces some challenges. For example, if f (z) = 21 , a nonzero constant formal power series, then f ∉ 𝕏0 or F = 1 + f ∉ 𝕏1 , and hence L(1 + f ) = L(1 + 21 ) is not defined by (2.1). However, in Calculus, the MacLaurin series for ln(1 + x) does provide that 2
3
4
1 1 1 1 1 1 1 1 ln(1 + ) = − ( ) + ( ) − ( ) + ⋅ ⋅ ⋅ = ln 3 − ln 2. 2 2 2 2 3 2 4 2 The formal logarithm defined in (2.1) needs to be generalized. Actually the condition F ∈ 𝕏1 for L(F) is just a sufficient condition for the existence of the formal logarithm. So, it would be better if we may expand 𝕏1 , and it will be perfect if we obtain a necessary and sufficient condition for the existence of a formal logarithm. Definition 6.4.1. Let hL ∈ 𝕏 = 𝕏(ℝ) be such that 1 1 1 hL (z) = z − z 2 + z 3 − z 4 + ⋅ ⋅ ⋅ . 2 3 4 Define the subset 𝕏+ = 𝕏+ (ℝ) ⊆ 𝕏 by 𝕏+ = {f ∈ 𝕏(ℝ) : f (0) > 0}. The general formal logarithm or just formal logarithm on ℝ is defined to be the mapping Ln : 𝕏+ → 𝕏 such that Ln(g) = ln(c) + hL ( gc ), or Ln(g) = ln(c) + [
2
3
4
1 g 1 g g 1 g − ( ) + ( ) − ( ) + ⋅ ⋅ ⋅], c 2 c 3 c 4 c
(6.2)
for all g = g(0) + g ∈ 𝕏+ (ℝ), where g = g − g(0) ∈ 𝕏0 , c = g(0) > 0. The formal power series hL is called the generating series for Ln. The function ln is the natural logarithmic function on ℝ. The series Ln(g) is called the formal logarithmic series of g. Proposition 6.4.2. Let g = cI ∈ 𝕏+ (ℝ) be a positive constant formal power series, then Ln(g) = ln(c)I, where ln is the natural logarithmic function on ℝ. Proof. It is clear that g = g − g(0) = 0. by (6.2), Ln(g) = ln(c) + [0 − 0 + 0 − 0 + ⋅ ⋅ ⋅] = ln(c)I. Following this proposition, we have Ln(g(0)) = ln(g(0))I.
(6.3)
6.4 General formal logarithm
 181
The mapping T : 𝕏(ℂ) → ℂ by T((c, 0, 0, . . . )) = c, from the set of all constant formal power series cI of 𝕏(ℂ) to ℂ, is obviously an isomorphism under the linear operation and Cauchy product on 𝕏, therefore, if there is no confusion, we may write (c, 0, 0, . . . ) as c. In sense of this discussion, we may easily obtain Ln(I) = Ln(1) = ln(1) = 0.
(6.4)
Theorem 6.4.3. Let g ∈ 𝕏+ (ℝ) be given. Then (Ln(g)) = g /g.
(6.5)
Proof. Write g = g(0) + g = c(1 + gc ) where c = g(0). Then 2
(Ln(g)) = [ln(c) +
2
= =
3
4
1 g 1 g g 1 g − ( ) + ( ) − ( ) + ⋅ ⋅ ⋅] c 2 c 3 c 4 c
3
h g g g [1 − ( ) + ( ) − ( ) + ⋅ ⋅ ⋅] c c c c g 1 c 1+
g c
= g /g,
=
g g
because g = g . Once we obtain the formal derivative of Ln, we may apply its properties to solve many problems. Theorem 6.4.4. If g, f ∈ 𝕏+ (ℝ), then Ln(fg) = Ln(f ) + Ln(g).
(6.6)
Proof. Since f , g ∈ 𝕏+ (ℝ), it follows that fg ∈ 𝕏+ (ℝ). Let P = Ln(fg) − (Ln(f ) + Ln(g)), then P = (Ln(gf )) − (Ln(f ) + Ln(g))
= =
(gf ) − (f /f + g/g ) gf
f g + fg f g + fg − = 0. fg fg
Then Ln(gf ) = Ln(g) + Ln(f ) + cI𝕏 for some constant formal power series cI. Since f (0)g(0) = fg(0), (6.3) yields that c = 0, and hence, Ln(gf ) = Ln(g) + Ln(f ).
182  6 Formal analysis and classical analysis For any g ∈ 𝕏+ (ℝ), we have that g −1 ∈ 𝕏+ (ℝ), too. Corollary 6.4.5. If g ∈ 𝕏+ (ℝ), then Ln(g −1 ) = − Ln(g). Proof. Applying (6.6) to the equation gg −1 = I. It is defined that g −n = (g −1 )n for all n ∈ ℕ if g −1 ∈ 𝕏. Corollary 6.4.6. Let n ∈ ℤ and let g ∈ 𝕏+ (ℝ) be given. Then Ln(g n ) = n Ln(g). Proof. For n = 0, Ln(g 0 ) = Ln(I) = 0 = 0 Ln(g) by (6.4). Since g ∈ 𝕏+ (ℝ), it follows that both g n and g −n in 𝕏+ (ℝ) for all n ∈ ℕ. Taking f = g in (6.6), we have Ln(f 2 ) = 2 Ln(f ). Inductively, we have the conclusion for n ∈ ℕ. Also n
Ln(g −n ) = Ln((g −1 ) ) = n Ln(g −1 ) = (−n) Ln(g). The formal root series and formal rational exponent series will be discussed systematically in Chapter 9. We briefly introduce some of those formal power series now in order to introduce an important property of Ln. If f , g ∈ 𝕏 and f n = g for some n ∈ ℕ, we say that f is the nth formal root series of g and we denote f by g 1/n . It is known [29] that if g ∈ 𝕏+ (ℝ), then g 1/n ∈ 𝕏+ (ℝ) for all n ∈ ℕ. As for g −n , we also define 1/n
g −1/n = (g −1 )
.
In order to define g m/n for g ∈ 𝕏+ (ℝ), we need only the following. Corollary 6.4.7. Let f , g ∈ 𝕏+ (ℝ) be given. For every n ∈ ℕ, (fg)1/n = f 1/n ⋅ g 1/n . Proof. Let a = f 1/n , b = g 1/n . Then an = f , bn = g, and hence fg = an bn = (ab)n . Then ab = (fg)1/n , or (fg)1/n = f 1/n ⋅ g 1/n .
6.4 General formal logarithm
 183
Taking f = g, we have (f 2 )1/n = (f 1/n )2 . Continuing this process, we have (f m )
1/n
m
= (f 1/n ) .
For r = m/n, m, n ∈ ℕ we now define that 1/n
g r = (g m )
for g ∈ 𝕏+ (ℝ). Similarly, if r = −s < 0, we define s
g r = (g −1 ) . Theorem 6.4.8. Let g ∈ 𝕏+ (ℝ) and let r ∈ ℚ be given, then we have Ln(g r ) = r Ln(g). Proof. It has been proved in Corollary 6.4.6 for r ∈ ℤ. We suppose that r = 1/n, n ∈ ℕ. Let g r = g 1/n = a, then an = g, and Ln(g) = Ln(an ) = n Ln(a). Then Ln(g 1/n ) = Ln(a) =
1 n
Ln(g). If r = m/n, m, n ∈ ℕ, using Corollary 6.4.5 we have
Ln(g m/n ) = Ln((g m )
1/n
)=
m 1 Ln(g m ) = Ln(g) = r Ln(g). n n
If r = −m/n, m, n ∈ ℕ, we have m/n
Ln(g −m/n ) = Ln((g −1 )
)=
m m Ln(g −1 ) = − Ln(g) = r Ln(g). n n
Remark 6.4.9. The general formal logarithm Ln has the following properties: 1. Ln(1 + x) performs as its counterpart the Maclaurin series for ln(1 + z) in Calculus if x < 1. Here, we consider that Ln(1 + x) does not assign value to x but we may treat x as some formal power series including certain constant formal power series. 2. If g = 1 + β ∈ 𝕏1 where β ∈ 𝕏0 is a nonunit, then 1 1 1 Ln(1 + β) = L(1 + β) = β − β2 + β3 − β4 + ⋅ ⋅ ⋅ 2 3 4 3.
where L is the formal logarithm introduced in Section 2.2. It will be proved in the end of this section that the mapping Ln is injective.
Theorem 6.4.10. Let f ∈ 𝕏(ℝ) be given. Then Ln(Ea (f )) = af where Ea is the formal exponential series.
184  6 Formal analysis and classical analysis Proof. It is obvious that Ln ∘ Ea (f ) ∈ 𝕏 because Ea (f ) ∈ 𝕏+ (ℝ). Write f (0) = c, then Ea (f (0)) = 1 + ac +
a2 2 a3 3 c + c + ⋅ ⋅ ⋅ = eac > 0. 2! 3!
By (6.5), (Ln(Ea (f )) = (Ea (f )) (Ea (f )) = (Ea (f )) aEa (f )f = af .
−1
−1
Then Ln(Ea (f )) = af because Ln(Ea (f (0))) = Ln(eac ) = ac = af (0). Corollary 6.4.11. Taking a = 1 in the above theorem, we have Ln(E1 (f )) = f . It looks like that Ln plays the role of iterative inverse E1[−1] on 𝕏+ (ℝ). Is it true? Let us investigate what Ea ∘ Ln is. Theorem 6.4.12. Let Ea be the formal exponential series with a ∈ ℝ. Then a
Ea ∘ Ln(g) = (g(0)) Ba (
g(x) ), g(0)
(6.7)
for g ∈ 𝕏+ (ℝ) where g = g − g(0) ∈ 𝕏0 , and Ba is the formal binomial series a a Ba (z) = 1 + ( )z + ( )z 2 + ⋅ ⋅ ⋅ . 1 2 Proof. If a = 0, (6.7) is trivial. We suppose that a ≠ 0. Write Ea ∘ Ln = P and denote g(0) = b, we have (P(g)) = Ea (Ln(g))(Ln(g)) = aEa (Ln(g))
g g = (aP(g)) g g
because g = g , it follows that (g(0) + g)(P(g)) = ag P(g).
(∗)
Writing P(g) in the form P(g) = (Ea ∘ Ln)(g(0) + g) = c0 + c1 g + c2 g 2 + ⋅ ⋅ ⋅ , for some {cn }∞ n=0 ⊆ ℝ. Then a
c0 = Ea (Ln(0)) = Ea (ln(g(0))) = ea ln(g(0)) = (g(0)) = g a (0).
6.4 General formal logarithm
 185
If g = 0, a zero series, (6.7) is true now. We suppose that g ≠ 0: P (g) = g (c1 + 2c2 g + 3c3 g 2 + ⋅ ⋅ ⋅). Applying (∗) and using b = g(0) > 0, we have (b + g)[c1 + 2c2 g + 3c3 g 2 + ⋅ ⋅ ⋅] = a[c0 + c1 g + c2 g 2 + ⋅ ⋅ ⋅]. Since g ≠ 0 and 𝕏(ℝ) is an integral domain [42], it follows that ncn + b(n + 1)cn+1 = acn
for all n ∈ ℕ.
Then a − n a − (n − 1) a−n c = ⋅ cn−1 b(n + 1) n b(n + 1) n (a − n)(a − n + 1) = cn−1 = ⋅ ⋅ ⋅ b2 (n + 1)n (a − n)(a − n + 1)(a − n + 2) ⋅ ⋅ ⋅ (a − 1)a c0 = bn+1 (n + 1)n(n − 1) ⋅ ⋅ ⋅ 2 ⋅ 1 a(a − 1) ⋅ ⋅ ⋅ (a − (n + 1) + 1) = c0 , bn+1 (n + 1)!
cn+1 =
or cn =
1 a ( )c . bn n 0
Thus, 2
3
a g(x) a g(x) a g(x) c + ( )( ) c0 + ( )( ) c0 + ⋅ ⋅ ⋅ P(g) = c0 + ( ) 2 b 3 b 1 b 0 2
3
a g(x) a g(x) a g(x) = c0 (1 + ( )( ) + ( )( ) + ( )( ) + ⋅ ⋅ ⋅) 1 b 2 b 3 b g(x) a = (g(0)) Ba ( ). g(0) Corollary 6.4.13. Let notation be set the same as those in Theorem 6.4.12. We have E1 ∘ Ln(g) = g, for all g ∈ 𝕏+ (ℝ). Proof. If a = 1, then (an ) = 0 for all integer n ≥ 2, and hence E1 ∘ Ln(g) = g(0)B1 (
g(x) g(x) ) = g(0)(1 + ) = g. g(0) g(0)
186  6 Formal analysis and classical analysis The following corollary is the formula (1.7–9) of [42]. Corollary 6.4.14. Let notation be set the same as those in Theorem 6.4.12. We have Ea ∘ Ln(1 + g) = Ba (g). Proof. Let g(0) = 1 in (6.7). Combining Corollary 6.4.11 and Corollary 6.4.13, we have the following. Theorem 6.4.15. E1 and Ln are composition inverse each other on the space 𝕏+ (ℝ), or Ln[−1] = E1
E1[−1] = Ln
and
E1 ∘ Ln(g) = Ln ∘E1 (g) = g
for all
on
or
𝕏+ (ℝ),
g ∈ 𝕏 (ℝ). +
Remark 6.4.16. It is known that the exponential function ex and the natural logarithmic function ln x are iterative inverse each other on all positive real numbers, not on ℝ. Very similar, Ln and E1 are inverse each other under composition on 𝕏+ (ℝ) only. Theorem 6.4.17. The formal logarithm Ln : 𝕏+ (ℝ) → 𝕏(ℝ) is injective. Proof. Let f ∈ 𝕏+ (ℝ) be given such that f (x) = f0 + f1 x + f2 x2 + ⋅ ⋅ ⋅
and
Ln(f ) = 0.
We show that f = I. k 0 Write f = c + f with c = f0 , f = ∑∞ k=1 fk x ∈ 𝕏 . Then Ln(f ) = ln(c) + [
2
3
4
f (x) 1 f (x) 1 f (x) 1 f (x) − ( ) + ( ) − ( ) + ⋅ ⋅ ⋅] = 0. c 2 c 3 c 4 c
Then c = 1. For our convenience, we rewrite f = 1 + f and
1 2 1 3 1 4 Ln(f ) = f (x) − f (x) + f (x) − f (x) + ⋅ ⋅ ⋅ = 0. 2 3 4 n
Then f1 = 0 because ord(f ) ≥ n for all n ∈ ℕ. Suppose that fj = 0, 1 ≤ j ≤ n for some n ∈ ℕ. Then f (x) = fn+1 xn+1 + fn+2 x n+2 + ⋅ ⋅ ⋅ . k
It follows that f (x), k > 1, does not contribute anything to the term of x n+1 in the formal power series Ln(f ). Or, the term of xn+1 of Ln(f ), which is zero, is only determined by the term of xn+1 of f (x). So, fn+1 = 0, which completes the induction. Thus, f = I.
6.4 General formal logarithm
 187
We have seen the relationship between formal logarithmic series and formal exponential series. We want to see more properties about Ea . By the general Cauchy product for the multiplication of any two series, it is reasonable to set up the Cauchy product of two formal power series each of them is a composition with some formal power series. Definition 6.4.18. Let f , g ∈ 𝕏(S) where S is a field with a metric such as ℝ or ℂ. Let P, Q ∈ 𝕏(S) be such that P(x) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ ,
and Q(x) = b0 + b1 x + b2 x 2 + ⋅ ⋅ ⋅ .
If P(f ) ∈ 𝕏(S), Q(g) ∈ 𝕏(S), then the Cauchy product P(f )Q(g) is a formal power series in 𝕏(S) such that P(f )Q(g) = c0 (x) + c1 (x) + c2 (x) + c3 (x) + ⋅ ⋅ ⋅ where n
cn (x) = ∑ (ak f k (x))(bn−k g n−k (x)), k=0
for all n ∈ ℕ ∪ {0}. If P = Q = Ea , a ∈ ℝ, in the above definition, we have a very useful result. Theorem 6.4.19. Let f , g ∈ 𝕏(ℝ) be given. For any a ∈ ℝ, Ea (f ) ⋅ Ea (g) = Ea (f + g). Proof. By the general composition theorem or by Theorem 5.4.3, Ea (f ), Ea (g) and Ea (f + g) are all welldefined formal power series in 𝕏(ℝ) and a2 a3 a f + f2 + f3 + ⋅⋅⋅, 1! 2! 3! 2 a a a3 Ea (g) = 1 + g + g 2 + g 3 + ⋅ ⋅ ⋅ . 1! 2! 3! Ea (f ) = 1 +
For any k ∈ ℕ ∪ {0}, by Definition 6.4.18 we define the formal power series Pk = Pk (f i , g j ) =
∑
0≤i,j≤k,i+j=k
k ai i aj j aj ak−j k−j f g = ∑ fj g , i! j! j! (k − j)! j=0
then ∞
Ea (f ) ⋅ Ea (g) = ∑ Pk . k=0
188  6 Formal analysis and classical analysis Also, k
aj j ak−j k−j f g j! (k − j)! j=0
Pk = ∑ =
k! ak k ak k k j k−j f j g k−j = ∑ ∑ ( )f g k! j=0 j!(k − j)! k! j=0 j
=
ak (f + g)k . k!
Thus, ak (f + g)k = Ea (f + g). k! k=0 ∞
Ea (f ) ⋅ Ea (g) = ∑
7 Formal Laurent series Pierre Alphonse Laurent (1813–1854) had difficulty publishing his work on the series that would come to bear his name. In 1843, AugustinLouis Cauchy reported Laurent’s work to the French Academy: Cauchy’s theorem on the representability via power series of holomorphic functions in disk is valid in annuli, if the series allows negative power of (z − c). Unfortunately, Cauchy failed to convince his colleagues to accept and publish Laurent’s work. Twenty years later, in 1863, the widow of P. A. Laurent had Laurent’s work published. Karl Weierstrass proved the same theorem in 1841. And like Laurent’s, Weierstrass’ work was not published until later, in 1894. The Laurent series is usually considered as a natural extension of power series in classical analysis and, therefore, the formal Laurent series is supposed to be a natural extension of formal power series. For formal Laurent series, we should be keeping the same or similar operations endowed for formal power series while we extend the series to have infinitely many terms with negative exponents. Considered as a sequence, a formal Laurent series is denoted as (. . . , c−2 , c−1 , c0 , c1 , c2 , . . . ). Unfortunately, the extension task has not been fulfilled smoothly. The difficulty in establishing the space of formal Laurent series is how to establish the multiplication and composition and, at the same time, make sure that these two operations stay in the space of Laurent series the same as they stay in 𝕏 although the outcome will be quite different. A semiformal Laurent series is a formal Laurent series with finitely many terms with negative exponents. Such formal Laurent series had been working very well before the general formal Laurent series was invented in 2011, and we believe it will continue to play a strong role in many mathematical fields. Section 7.1 and Section 7.2 introduce the semiformal Laurent series space ℒs and the reversed semiformal Laurent series space ℒr including their applications in the Lagrange–Bürmann formula, Riordan array, and Diophantine approximation problem. Section 7.3 introduces the basics of formal Laurent series space 𝕃, which was invented in 2011 by the author and Dariusz Bugajewski [28]. Section 7.4 provides the composition of a formal Laurent series with a formal power series. Many properties of formal power series, such as right distributive law and the chain rule, are proved to be working with the composition of formal Laurent series. Section 7.5 and Section 7.6 introduce a canonical mapping that connects certain classical Lebesgue measurable functions to the formal Laurent series. Under such connection, the dot product of formal Laurent series can be represented by the multiplication of the corresponding Lebesgue measurable functions, and the multiplication https://doi.org/10.1515/9783110599459007
190  7 Formal Laurent series of formal Laurent series can be represented by the convolution of the corresponding Lebesgue measurable functions. Section 7.7 introduces two topological spaces of formal Laurent series. One is an ultrametric topological space established by Bugajewski and the author, and the other is a Banach space induced by the norm ‖ ⋅ ‖β , the same norm introduced in Section 6.3. Yousefi and Soltani contributed a lot for this Banach space.
7.1 Semiformal Laurent series While the formal power series was developing successfully in both theory and applications, the formal Laurent series became an interesting topic for mathematicians to investigate. Such investigation has not gone very smoothly if our goal is to set up the formal Laurent series similar to the regular Laurent series in real and complex analysis. The challenge we met was not what Cauchy faced in the French Academy, it was the mathematical structure of formal Laurent series itself. Let us first see an example below. Example 7.1.1. Consider the expression ∑ xn = ⋅ ⋅ ⋅ + x−2 + x −1 + 1 + x + x 2 + ⋅ ⋅ ⋅ .
n∈ℤ
A formal Laurent series is supposed to be like this expression. If we denote this expression as f (x) and run the multiplication, as we did for formal power series, then f 2 (x) = f (x)f (x) is not defined, simply every coefficient of each x k in f (x)f (x) is not defined under the usual multiplication because x −n x n+k = x k for all n ∈ ℕ. A problem in this example is caused by the fact that this expression, or formal Laurent series, has infinitely many negative power terms. In his popular book [42], P. Henrici introduced a formal Laurent series H as H(z) = cm z m + cm+1 z m+1 + cm+2 z m+2 + ⋅ ⋅ ⋅ ,
cm ≠ 0,
(7.1)
over ℂ, where m is a fixed integer (positive, negative, or zero). Henrici also pointed out that if the formal Laurent series of (7.1) has infinite number of nonzero terms with negative exponents, and it is therefore not clear that the product of any two such series exists (p. 219, [42]). This chapter will answer Henrici’s question late. In this section, all formal Laurent series are defined as in (7.1), or are semiformal Laurent series. Such formal Laurent series have many interesting properties and have many applications. With all respect, we still call H of (7.1) a formal Laurent series or semiformal Laurent series to distinct it from the real formal Laurent series, which should allow the
7.1 Semiformal Laurent series  191
existence of infinitely many nonzero terms with negative exponents. Let us give the semiformal power series a clear mathematical definition although we had introduced it in (4.22). Definition 7.1.2. Let S be a field and let m ∈ ℤ be any integer, a semiformal Laurent series H over S is defined as H(z) of (7.1). We write cn = 0 if n < m, and then we may write H(z) = ∑n∈ℤ cn z n . If L1 (z) = cm z m + cm+1 z m+1 + cm+2 z m+2 + ⋅ ⋅ ⋅ ,
and L2 (z) = dk z k + dk+1 z k+1 + dk+2 z k+2 + ⋅ ⋅ ⋅ ,
then aL1 , L1 + L2 and L1 L2 are semiformal Laurent series such that aL1 (z) = ∑ (acn )z n , n∈ℤ
a ∈ S,
(L1 + L2 )(z) = ∑ (cn + dn )z n , n∈ℤ
(L1 L2 )(z) = ∑ rn z n , n∈ℤ
rn = ∑ ci dj = i+j=n
where ∑
i+j=n,i≥m,j≥k
ci dj ,
n ∈ ℤ,
(7.2)
if all rn ∈ ℂ, otherwise we say that L1 L2 does not exist. Please be noticed that there are only finitely many nonzero terms with negative exponents in the summation. The formal derivative of a semiformal Laurent series H(z) = ∑ cn z n is defined by H (z) = ∑(n + 1)cn+1 z n , or cn = (n + 1)cn+1 ,
(7.3)
if we write H (z) = ∑ cn z n . Similar to the residue of a Laurent series in analysis, we define the residue of a semiformal Laurent series to be the coefficient of z −1 , for example, c−1 is the residue of H(z) above, and we usually write c−1 = res(H). Finally, we denote ℒs = ℒs (ℂ) = {all semiformal Laurent series over ℂ}.
The zero element of ℒs is the series such that each coefficient of the series is zero, and the identity element of ℒs is the series whose each coefficient is zero except the constant term is 1.
192  7 Formal Laurent series The formal Laurent series defined in (7.1), although has only a finite number of nonzero terms with negative exponents, is very interesting, and has many applications. It is obvious that such formal Laurent series includes the set of formal power series as its subset. Certain results related to such formal Laurent series were presented in [42]. The most important character of the semiformal Laurent series is the next theorem. Theorem 7.1.3. With the operations defined in Definition 7.1.2, ℒs is a field. Proof. It is clear that we need only show that every nonzero semiformal Laurent series H has the multiplication inverse, or reciprocal, H −1 . Let fk (z) = z k ∈ ℒs , k ∈ ℤ, it is obvious that fk−1 = z −k . More generally, let H(z) = cm z m + cm+1 z m+1 + cm+2 z m+2 + ⋅ ⋅ ⋅ , where cm ≠ 0, and m ∈ ℤ is any integer (negative, positive, or zero). Then H(z) = z m (cm + cm+1 z + cm+2 z 2 + ⋅ ⋅ ⋅) = z m P, where P(z) = cm + cm+1 z + cm+2 z 2 + ⋅ ⋅ ⋅ is a unit formal power series. Then P −1 ∈ 𝕏(ℂ), and hence P −1 z −m ∈ ℒs . We have (P −1 z −m )H = P −1 z −m z m P = 1. Thus, H −1 = P −1 z −m . Corollary 7.1.4. If a semiformal Laurent series has expression H(z) = cm z m + cm+1 z m+1 + cm+2 z m+2 + ⋅ ⋅ ⋅ , where cm ≠ 0, then there exists a sequence {dn } ⊆ ℂ such that H −1 (z) = d−m z −m + d−m+1 z −m+1 + d−m+2 z −m+2 + ⋅ ⋅ ⋅ , where d−m ≠ 0 and dn = 0, n < −m. The proof is trivial. Proposition 7.1.5. A semiformal Laurent series is a formal derivative of a semi Laurent series if and only if its residue is zero. Proof. Let H ∈ ℒs be such that H(z) = cm z m + cm+1 z m+1 + cm+2 z m+2 + ⋅ ⋅ ⋅ . By formula (7.3), c−1 = (−1 + 1)c−1+1 = 0 ⋅ c0 = 0. Conversely, let M(z) = ∑ bn z n ∈ ℒs such that b−1 = 0. We may set a sequence {cn }n∈ℤ ⊆ ℂ such that
cn =
1 b , n n−1
n ∈ ℤ,
n ≠ 0,
and let c0 be arbitrary, then M = H , where H(z) = ∑ cn z n .
7.1 Semiformal Laurent series  193
Example 7.1.6. For any k ∈ ℤ, ∞
n
2
∞
( ∑ z ) = ∑ (n − 2k + 1)z n . n=k
n=2k
Solution. Applying the formal geometric series and then applying the formal binomial series we have ∞
n
2
k
∞
n
2
H(z) = ( ∑ z ) = (z ∑ z ) n=k
n=0 ∞
= z 2k (1 − z)−2 = z 2k ∑ ( n=0
−2 )(−z)n n
∞
= ∑ nz n+2k−1 n=1 ∞
= ∑ (m − 2k + 1)z m . m=2k
The set 𝕏(ℂ) is not a field because the nonunits don’t have the multiplication (Cauchy product) inverse. The set ℒs (ℂ) is a field and it is obvious that 𝕏(ℂ) ⊆ ℒs (ℂ). This property is very helpful for mathematicians. We must indicate that there is another way to introduce ℒs (ℂ) as a field and 𝕏(ℂ) ⊆ ℒs , especially for the readers who are familiar with the abstract algebra. Some details can be found in [42] or [79]. We provide the following remark for this purpose. Remark 7.1.7. (i) 𝕏(ℂ) is an integral domain by Proposition 1.1.10. It is well known that any integral domain can be imbedded in a field (Theorem 3.6.1 of [43]). (ii) Roughly speaking, the field we seek in (i) above should be the set of all quotients f /g where f , g ∈ 𝕏(ℂ) including those g ∈ 𝕏0 (ℂ) (defined by Theorem 7.1.3). When such structure is carried through for the integral domain 𝕏(ℂ), it yields a field that is isomorphic to the field of formal Laurent series ℒs . For example, the semiformal Laurent series in (7.1) H(z) = cm z m + cm+1 z m+1 + cm+2 z m+2 + ⋅ ⋅ ⋅ , with m ∈ ℤ, m < 0 can be written as H(z) = z m (cm + cm+1 z + cm+2 z 2 + ⋅ ⋅ ⋅) =
h(z) , ϕ(z)
where h(z) = cm + cm+1 z + cm+2 z 2 + ⋅ ⋅ ⋅ and ϕ(z) = z −m are both formal power series in 𝕏(ℂ). It is clear that the representation l = f /g above are nonunique but the formal series ϕ is uniquely determined by each H and so is h/ϕ.
194  7 Formal Laurent series (iii) By (ii), for any ϕ ∈ 𝕏(ℂ) we have that H ∘ ϕ(z) =
f (ϕ(z)) ∈ ℒs (ℂ) g(ϕ(z))
if and only if f ∘ ϕ ∈ 𝕏(ℂ), g ∘ ϕ ∈ 𝕏(ℂ) and g ∘ ϕ ≠ 0. (iv) We also call ℒs (ℂ) a quotient field in the sense of imbedding (the minimal field in all such fields). However, our quotient field is a little different from the popular v quotient field Q(𝒜) defined in page 131 of [79] where ∑∞ 0 av z is a convergent power series for every ∑v≥m av z v ∈ Q(𝒜). We promised by the end of Section 4.3 that we would introduce the proof of Lagrange–Bürmann formula when we are ready. Let us start with the Schur–Jabotinski theorem [42]. Theorem 7.1.8. Let f ∈ 𝔻(ℂ) be such that f (z) = a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ , and for every k = ±1, ±2, ±3, . . . , write ∞
n f k = ∑ a(k) n z ,
(7.4)
n=k
where f −1 is the multiplication (Cauchy product) inverse of f in ℒs , and f −k = (f −1 )k , k ∈ ℕ. If q = f [−1] , the composition inverse of f , we write q(z) = b1 z + b2 z 2 + b3 z 3 + ⋅ ⋅ ⋅ , and for all m ∈ ℕ ∞
n qm = ∑ b(m) n z , n=m
then b(m) n =
m (−n) a , n −m
n ≥ m.
(7.5)
Proof. Let A = D(f ) be the delta associated matrix of f a(1) 1
[ [ 0 [ [ D(f ) = [ [ 0 [ [ 0 [ .. [ .
a(1) 2
a(2) 2 0
0 .. .
a(1) 3
a(2) 3 (3) a3 0 .. .
a(1) 4
a(2) 4 (3) a4 a(4) 4 .. .
...
] . . .] ] ] . . .] ], ] . . .] ] .. .]
7.1 Semiformal Laurent series  195
and let B = (bmn ), where bmn = 0 for n < m and bmn =
m (−n) a , n −m
n ≥ m.
We claim that B = A−1 by showing that B is a right inverse of A. Let AB = C = (crs ). Then C is also an upper triangular matrix with crs = 0 if s < r. Then (r) −r r crr = a(−r) −r ar = a1 a1 = 1,
for all r ∈ ℕ.
For s > r, s s 1 s k = ∑ ka(r) a(−s) . crs = ∑ a(r) b = ∑ a(r) ( )a(−s) −k k −k k ks k s s k=r k=r k=r
Considering the formal power series r (r) r+1 (r) r+2 f r (z) = a(r) + ar+2 z + ⋅⋅⋅, r z + ar+1 z
and the semiformal Laurent series −s (−s) −s+1 −s+2 f −s (z) = a(−s) + a(−s) + ⋅⋅⋅. −s z + a−s+1 z −s+2 z
Then r−1 r (r) r+1 + ⋅⋅⋅. (f r (z)) = ra(r) + (r + 1)a(r) r z r+1 z + (r + 2)ar+2 z
We write (f r (z)) f −s = ∑n∈ℤ pn z n , then (7.2) yields that (r) (−s) (−s) (r) (−s) p−1 = ra(r) r a−r + (r + 1)ar+1 a−r−1 + ⋅ ⋅ ⋅ + sas a−s ,
or, s
res[(f r ) f −s ] = ∑ ka(r) a(−s) . k −k
k=r
Then scrs = res[(f r ) f −s ].
Also we have (f r ) f −s = rf r−1 f f −s = rf r−s−1 f =
r (f r−s ) . r−s
By Proposition 7.1.5, res((f r−s ) ) = 0, and hence crs = 0, for s > r. Then we have AB = I.
196  7 Formal Laurent series In the above theorem, actually B = D(f [−1] ) by the expression of qm or (f [−1] )m . So, without using (7.5), AB = I is obviously true. The point of this theorem is to connect the entries of B or the coefficients of f [−1] to the coefficients of f −1 . Theorem 7.1.9 (Lagrange–Bürmann expansion, [42]). Let P ∈ 𝔻(ℂ) be given such that P(z) = a1 z + a2 z 2 + ⋅ ⋅ ⋅ , and write Q = P [−1] , the composition inverse of P. Then for any R ∈ 𝕏(ℂ) 1 res(R P −n )z n , n n=1 ∞
R ∘ Q(z) = R(0) + ∑
where P −1 is the multiplication inverse of P in ℒs and P −n = (P −1 )n . Proof. Let Q(z) = P [−1] (z) and let R ∈ 𝕏(ℂ) be given. Writing Q(z) = b1 z + b2 z 2 + ⋅ ⋅ ⋅
and
R(z) = c0 + c1 z + c2 z 2 + ⋅ ⋅ ⋅ ,
and then we write R ∘ Q(z) = d0 + d1 z + d2 z 2 + ⋅ ⋅ ⋅ . As we did in the proof of Theorem 7.1.8, we write A = D(P) = (amn ) = (a(n) m ),
B = D(Q) = (bmn ).
It is obvious that d0 = c0 , and n
dn = ∑ ck b(k) n , k=1
n ∈ ℕ,
where b(m) n = bmn =
m (−n) a , n −m
n ≥ m,
and bmn = 0 otherwise, by Theorem 7.1.8. Then n k 1 n dn = ∑ ck a(−n) = ∑ ck ka(−n) . −k −k n n k=1 k=1
Using the same approach in the proof of Theorem 7.1.8, we can readily obtain dn =
1 res[R P −n ]. n
Thus, 1 res(R P −n )z n . n n=1 ∞
R ∘ Q(z) = R(0) + ∑
(7.6)
7.1 Semiformal Laurent series  197
Corollary 7.1.10. Let P ∈ 𝔻(ℂ) be an almost unit and denote Q = P [−1] . Then, for any R(z) = c0 + c1 z + c2 z 2 + ⋅ ⋅ ⋅ in 𝕏(ℂ), ∞
(R ∘ Q) (z) = ∑ res(R P −n−1 )z n . n=0
Proof. Taking the formal derivative both sides of (7.6). Corollary 7.1.11. Let P ∈ 𝔻(ℂ) be an almost unit and denote Q = P [−1] . Then, for any S ∈ 𝕏(ℂ), ∞
(S ∘ Q)Q = ∑ res(SP −n−1 )z n . n=0
Proof. By the Chain Rule (5.20), (R∘Q) = (R ∘Q)Q . Let S = R . Since R ∈ 𝕏 is arbitrary, it follows that R , or S, is also arbitrary. The full version of the Lagrange–Bürmann theorem covers the convergent power series or Laurent series, or the representations of analytic functions. The principle of formal analysis guarantees that the Lagrange–Bürmann theorem and its corollaries still work for analytic case. We introduce the analytic version of Lagrange–Bürmann theorem without proof (p. 101, [42]). We denote vectors in a Banach space by the bold letters. A Banach algebra is a complete linear space and closed for the multiplication defined on it. Theorem 7.1.12. Let X be a Banach algebra. Let f and h be analytic at 0 ∈ X, f (0) = 0. Let F and H be the series representations of f and h, respectively, and let ρ > 0, τ > 0 such that h(W) = H(W) for
‖W‖ < τ.
and f (Z) = F(Z),
f (Z) < τ
for ‖Z‖ < ρ.
Let f be the analytic function defined by f (Z) = F(Z), ‖Z‖ < ρ, where F is the formal derivative of F. Let f (0) ≠ 0 and let G = F [−1] . Then for ‖Z‖ < ρ the following expansions hold: 1 res(HG−n )Zn , n n=1 ∞
h(f (Z)) = h(0) + ∑ ∞
h(f (Z))f (Z) = ∑ res(HG−n−1 )Zn . n=0
198  7 Formal Laurent series W. Jones and W. Thron [49] in 1979 investigated the relationship between a semiformal Laurent series and a sequence {Rn (z)} of meromorphic functions. Let H(f ) be the Laurent expansion of f if f is a function meromorphic at the origin and if H is a semiformal Laurent series as in (7.1), they proved that a necessary and sufficient condition for a sequence {Rn (z)} to correspond to some formal Laurent series H is that lim λ(H − H(Rn )) = ∞,
n→∞
where λ is the function defined on the set ℒs such that λ(H) = ∞
if H = 0;
and
λ(H) = m if
H ≠ 0,
where m is a fixed integer as in (7.1). This λ is similar to the order we defined for a formal power series. We would like to let readers know that the formal Laurent series defined in (7.1) has been also applied to Riordan group [39]. The most significant condition in equation (7.1) is that m ∈ ℤ is fixed. In order to make a formal Laurent series to be or to look like the one Cauchy introduced to the French Academy in 1843, we must take off the restriction for m. Let us close this section by briefly introducing the Puiseux series which has close relation with ℒs (ℂ). Puiseux series was first introduced by Issac Newton in 1676 [60] and rediscovered by Victor Puiseux in 1850 [71] and, therefore, it is also called the Newton–Puiseux series. For example, the series z −2 + 2z −1/2 + z 1/3 + 5z 8/3 + ⋅ ⋅ ⋅ is a Puiseux series in z. Definition 7.1.13 (Newton–Puiseux series). Let 𝕂 be a field of characteristic 0 such as ℝ or ℂ. A Puiseux series ψ in variable z is a power series η(z 1/n ), where η ∈ 𝕏(𝕂) and n ∈ ℕ. For a fixed n ∈ ℕ, we denote the set of all such series as 𝕂⟦z 1/n ⟧ which is a ring. The set of all fractions is denoted by 𝕂((z 1/n )) which is a field, very much like the construction of ℒs . We also define 𝕂 z 1/ℕ = ⋃ 𝕂 z 1/n , ?
?
n∈ℕ
?
?
the local 𝕂algebra of Newton–Puiseux series in the variable z. A Puiseux series f (z) on ℂ is usually denoted by ∞
f (z) = ∑ cv z v/n v=v0
7.2 Reversed semiformal Laurent series  199
1/n where v0 ∈ ℤ, {cv }∞ )) are v=v0 ⊆ ℂ, n ∈ ℕ. The addition and multiplication on ℂ((z defined as
(z −1 + 2z −1/2 + z 1/3 + ⋅ ⋅ ⋅) + (z −5/4 − z −1/2 + 2 + ⋅ ⋅ ⋅) = z −5/4 + z −1 + z −1/2 + 2 + ⋅ ⋅ ⋅ and (z −1 + 2z −1/2 + z 1/3 + ⋅ ⋅ ⋅) ⋅ (z −5/4 − z −1/2 + 2 + ⋅ ⋅ ⋅) = z 9/4 + 2z −7/4 − z −3/2 + z −11/12 + 4z −1/2 + ⋅ ⋅ ⋅ .
Finally, we present the following. Theorem 7.1.14 (The Newton–Puiseux theorem). Any monic reduced polynomial f (x, y) on 𝕂 of degree n ∈ ℕ has n roots in 𝕂⟦z 1/ℕ ⟧. If f is irreducible, then those roots are precisely the series of the form: ψρ = η(ρ ⋅ z 1/n ), where ψ = η(z 1/n ) ∈ 𝕂⟦z 1/n ⟧ is any one of them and ρ ∈ 𝕂∗ varies among the multiplicative subgroup Gn of (𝕂∗ , ⋅) of nth root of 1 ∈ 𝕂∗ . The proof of this theorem and more details about Puiseux series can be found in [31].
7.2 Reversed semiformal Laurent series In Section 7.1. we saw that every element of ℒs is a formal Laurent series which has only finitely many terms with negative exponents. The space ℒs has many applications. We now introduce another kind of formal Laurent series which have only finitely many terms with positive exponents. Definition 7.2.1 ([39]). For all fn ∈ ℂ, n ∈ ℤ, the reversed semiformal Laurent series with the order m is defined by ∞
f (z) = ∑ fn z −n , n=m
for some m ∈ ℤ.
(7.7)
The set of all such reversed formal Laurent series is denoted by ℒr . The order of such f ∈ ℒr is defined by ord(f ) = min{n ∈ ℤ : fn ≠ 0}. The set of all reversed formal Laurent series of order m is denoted by Σm . The addition, scalar multiplication and the Cauchy product (under distribution law) are defined as they were defined in ℒs .
200  7 Formal Laurent series Actually, f (z) ∈ ℒr if and only if f (z −1 ) ∈ ℒs . For example, f (z) = z 2 + z + 1 + z −1 + z −2 + ⋅ ⋅ ⋅ is a reversed semiformal Laurent series of order (−2). Proposition 7.2.2. (i) If f ∈ Σ0 , then 1/f (z −1 ) is a unit formal power series in 𝕏(ℂ). (ii) If h ∈ Σ−1 , then 1/h(z −1 ) ∈ 𝔻(ℂ). Proof. Let f ∈ Σ0 be given. We may write f (z) = a0 + a1 z −1 + a2 z −2 + ⋅ ⋅ ⋅ ,
a0 ≠ 0.
Then g(z) = f (z −1 ) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ is a unit in 𝕏(ℂ), and hence 1/f (z −1 ) = g −1 (z) ∈ 𝕏(ℂ). This is (i). Now let h ∈ Σ−1 be given with h(z) = b−1 z + b0 + b1 z −1 + b2 z −2 + ⋅ ⋅ ⋅ . Then h(z −1 ) = b−1 z −1 + b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ , is an element in ℒs . Applying Corollary 7.1.4, we have 1/h(z −1 ) = d1 z + d2 z 2 + d3 z 3 + ⋅ ⋅ ⋅ , −1 for some {dn }∞ n=1 ⊆ ℂ, d1 ≠ 0. Thus, 1/h(z ) ∈ 𝔻.
We define the Riordan array associated with formal Laurent series. Definition 7.2.3. Let f (z) ∈ Σ0 and g(z) ∈ Σ−1 be given. As we did in Definition 3.1.1, let D = (dn,k ) be a infinite matrix with entries dn,k = [z n ]
k
1 1 ( ) , f (z −1 ) g(z −1 )
(7.8)
where n, k ≥ 0, and [z n ]h(z) is the nth coefficient of the formal Laurent series h. The matrix D is called the Riordan array associated with the formal Laurent series f and g, and denoted by [f (z), g(z)] = (1/f (z −1 ), 1/g(z −1 )) = (d(z), h(z)),
(7.9)
where d(z) = 1/f (z −1 ) is a unit in 𝕏(ℂ) and h(z) = 1/g(z −1 ) ∈ 𝔻(ℂ) by Proposition 7.2.2.
7.2 Reversed semiformal Laurent series  201
Similar to formula (3.2), the generating function of the kth column of D is k
1 1 ( −1 ) . −1 f (z ) g(z ) We define the set R to be R = {[f (z), g(z)] : f ∈ Σ0 , g ∈ Σ−1 }. Theorem 7.2.4 ([39]). Define ⊙ : R × R → R by [f1 (z), g1 (z)] ⊙ [f2 (z), g2 (z)] = (d1 (z), h1 (z)) ∗ (d2 (z), h2 (z)), for all [fj (z), gj (z)] ∈ R, where dj (z) = 1/fj (z −1 ), hj (z) = 1/gj (z −1 ), j = 1, 2, and “∗” is the multiplication of Riordan group defined in (3.4). Then R forms a group. Proof. Let [fj (z), gj (z)] ∈ R be given, j = 1, 2. Then [f1 (z), g1 (z)] ⊙ [f2 (z), g2 (z)] =( =(
1 1 1 1 , )∗( , ) f1 (z −1 ) g1 (z −1 ) f2 (z −1 ) g2 (z −1 ) 1 1 1 , ) −1 −1 f1 (z ) f2 (g1 (z )) g2 (g1 (z −1 ))
= [f1 (z)f2 (g1 (z)), g2 (g1 (z))].
Then R is closed under the operation ⊙. It is easy to verify that ⊙ satisfies the associative law. Next, we show that the element [1, z] ı̇ s the identity of R. In fact, [f (z), g(z)] ⊙ [1, z] = (1/f (z −1 ), 1/g(z −1 )) ∗ (1, z)
= (1/f (z −1 ), 1/g(z −1 )) = [f (z), g(z)].
Finally, we show that [f (z), g(z)]
−1
=[
1 , g [−1] (z)], f (g [−1] (z))
where g [−1] (z −1 ) is the composition inverse of g(z −1 ) in terms of z −1 , that is, g [−1] (g(z −1 )) = g(g [−1] (z −1 )) = z −1 . Then [f (z), g(z)] ⊙ [
1 , g [−1] (z)] f (g [−1] (z))
= (1/f (z −1 ), 1/g(z −1 )) ∗ (f (g [−1] (z −1 )), =( Thus, R is a group.
1 ) g [−1] (z −1 )
1 1 f (g [−1] (g(z −1 ))), [−1] ) = [1, z]. −1 f (z ) g (g(z −1 ))
202  7 Formal Laurent series The group R is called the Riordan group associated with Laurent series. Sheffertype polynomial sequence is an interesting subject in Riordan group theory ([38],[37]). We mentioned this sequence when we closed Section 3.3 without providing the details. Let us do it now. Let A ∈ 𝕏(ℂ) be given such that A(z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ ,
an ≠ 0,
n = 0, 1, 2, . . . ,
then a generalized Sheffertype polynomial sequence {pn } associated with Riordan matrix (d, h) is defined by d(z)A(xh(z)) = p0 (x) + p1 (x)z + p2 (x)z 2 + ⋅ ⋅ ⋅ ,
(7.10)
where (d(z), h(z)) ∈ R, or d is a unit and h is an almost unit formal power series. Similarly, a generalized Sheffertype polynomial sequence {pn } associated with reversed Laurent series could be defined. n Definition 7.2.5. Let f ∈ Σ0 and g ∈ Σ−1 be given, and let A(z) = ∑∞ ∈ 𝕏(ℂ) n=0 an z be given with an ≠ 0 for all n. A generalized Sheffertype polynomial sequence {pn } associated with Laurent series generated by [f , g] ∈ R with respect to A is defined by ∞ x 1 A( ) = pn (x)z n , ∑ f (z −1 ) g(z −1 ) n=0
(7.11)
∞ x 1 A( ) = ∑ pn (x)z −n . f (z) g(z) n=0
(7.111 )
or equivalently
The generalized Sheffertype polynomial sequence {pn } associated with Laurent series has many applications in combinatorics. For example, if an = 1/n!, or A(z) = ez , formula (7.11) defines the classical Sheffertype polynomial sequence. If a0 = 1, an = 1/n, n ∈ ℕ, or A(z) = 1 − ln(1 − z), then (7.11) produces the sequence {pn } which is called the Dirichlet polynomial sequence. More details can be found in [39]. We now start to introduce another application of the reversed formal Laurent series. In number theory, the study of Diophantine approximation deals with the approximation of real numbers by rational numbers. It is named after Diophantus of Alexandria [1, 20]. Definition 7.2.6. x ∈ ℝ is said to be very well approximable if there exists ϵ > 0 such that p −(2+ϵ) x − < q q
for infinitely many
p ∈ ℚ. q
7.2 Reversed semiformal Laurent series  203
x ∈ ℝ is called a Liouville number if the above inequality holds for all ϵ > 0. We denote by 𝒲 the set of all well approximable numbers. Let x be an irrational real number, the irrationality exponent of x, denoted by μ(x), is defined by p μ(x) = sup{μ : x − < q−μ q
for infinitely many
p ∈ ℚ}. q
A wellknown result for such approximation is Dirichlet’s theorem, which says p −2 x − < q q
for infinitely many
p ∈ ℚ. q
Definition 7.2.7. For any prime number q > 1, let Fq be a field of q elements and denote by Fq ((z −1 )) the set of all reversed formal Laurent series over Fq such that ∞
Fq ((z −1 )) = {x = x(z) = ∑ cn z −n : m ∈ ℤ, m < 0, {cn } ⊆ Fq }. n=m
Denote by Fq [z] the set of all polynomials on Fq , and denote by Fq (z) the set of all rational functions on Fq . −n For each x = ∑∞ ∈ Fq ((z −1 )), define n=m cn z 0
[x] = ∑ cn z −n , n=m
and call it the integer part of x. Actually [x] is a polynomial. The degree of x is defined by deg x = − inf{n ∈ ℤ : cn ≠ 0}, and deg(0) = −∞. Proposition 7.2.8 ([57]). The mapping ‖ ⋅ ‖ : Fq ((z −1 )) → [0, ∞) by ‖x‖ = qdeg x , for every x ∈ Fq ((z −1 )) satisfies: (i) ‖x‖ ≥ 0; and ‖x‖ = 0 if and only if x is a zero formal Laurent series in Fq ((z −1 )); (ii) ‖xy‖ = ‖x‖ ⋅ ‖y‖; (iii) ‖αx + βy‖ ≤ max{‖x‖, ‖y‖} for α, β ∈ Fq ; (iv) For all α, β ∈ Fq , α ≠ 0, β ≠ 0, if ‖x‖ ≠ ‖y‖, then ‖αx + βy‖ = max{‖x‖, ‖y‖}. The mapping ‖ ⋅ ‖ is a nonArchimedean norm on the field Fq ((z −1 )). It is called the Dnorm, or just a norm, on Fq ((z −1 )).
204  7 Formal Laurent series Proof. It is trivial that ‖αx‖ = ‖x‖ if α ≠ 0. We may call (iv) the isosceles triangle principle for the norm ‖ ⋅ ‖ as we did for the ultrametric in Proposition 2.3.2. (i), (ii), and (iii) are obvious. We only show (iv). We assume that ‖x‖ < ‖y‖ and we show that ‖x + y‖ = ‖y‖. By (iii) and ‖x‖ < ‖y‖, ‖x + y‖ ≤ ‖y‖ = ‖y + x − x‖ ≤ max{‖x + y‖, ‖x‖} ≤ ‖x + y‖, which yields (iv). If we denote ℒs as ℒs (z), then Fq ((z −1 )) could be considered as ℒs (z −1 ) except that the field for ℒs (z −1 ) is Fq . That is, for each H ∈ ℒs , H(z) = cm z m + cm+1 z m+1 + cm+2 z m+2 + ⋅ ⋅ ⋅ ,
m ∈ ℤ,
x = x(z) = H(z −1 ) ∈ Fq ((z −1 )) if {cn } ⊆ Fq , or x = cm z −m + cm+1 z −(m+1) + cm+2 z −(m+2) + ⋅ ⋅ ⋅ ,
m ∈ ℤ.
The proposition bellow expresses some useful properties that can be easily verified. Proposition 7.2.9. Let all notation be set as in Definition 7.2.7, then: (i) The set Fq ((z −1 )) is a field. (ii) The norm ‖ ⋅ ‖ is a nonArchimedean norm on the field Fq ((z −1 )). (iii) The field Fq ((z −1 )) is locally compact and complete under the metric ρ for which ρ(x, y) = ‖x − y‖ for all
x, y ∈ Fq ((z −1 )).
(iv) Define Iq = {x ∈ Fq ((z −1 )) : ‖x‖ < 1}, then ∞
Iq = {x = ∑ cn z −n : {cn } ⊆ Fq }. n=1
Definition 7.2.10. Let β ∈ Fq ((z −1 )) with ‖β‖ > 1, where Fq ((z −1 )) is defined as in Definition 7.2.7. The transformation Tβ on Iq , defined by Tβ x = βx − [βx],
for every x ∈ Iq ,
is called the βtransformation. Define ϵ1 (x) = [βx] and ϵn (x) = ϵ1 (Tβn−1 x), n ≥ 2, then x=
ϵ (x) ϵ1 (x) ϵ2 (x) + 2 + ⋅⋅⋅ + n n + ⋅⋅⋅, β β β
(7.12)
for every x ∈ Iq . Formula (7.12) is called the βexpansion of x in base β. ϵ1 (x), ϵ2 (x), ϵ3 (x), . . . are called the βdigits of x. We denote by (ϵ1 (x), ϵ2 (x), ϵ3 (x), . . . ) the βexpansion of x.
7.2 Reversed semiformal Laurent series  205
For any y ∈ Fq ((z −1 )), we have that deg(y − [y]) ≤ −1, and then ‖Tβ x‖ = βx − [βx] < 1. Then Tβn x ∈ Iq for all n ∈ ℕ and ϵn (x) < ‖β‖ for all n ∈ ℕ. Scheicher proved the next lemma and theorem. Lemma 7.2.11 ([87]). Let z, w ∈ Fq ((z −1 )) with ‖z‖ = ‖w‖ where ‖ ⋅ ‖ is defined in Proposition 7.2.8, then n n n−1 z − w ≤ ‖z − w‖ ⋅ ‖z‖
for all
n ∈ ℤ.
Proof. The statement is trivial for n = 0. Let n ∈ ℕ be given, we have n n−1 n n−2 n−1 z − w = ‖z − w‖z + z w + ⋅ ⋅ ⋅ + w ≤ ‖z − w‖(max{z n−1−j wj : 0 ≤ j ≤ n − 1}) = ‖z − w‖ ⋅ ‖z‖n−1 .
Also, −n −n n n −n −n z − w = w − z z w ≤ ‖w − z‖‖w‖n−1 z −n w−n = ‖z − w‖ ⋅ ‖z‖−n−1 . Theorem 7.2.12. The polynomial sequence {ϵn } over Fq is a βexpansion of x ∈ Iq if and only if ‖ϵn ‖ < ‖β‖
for all
n ∈ ℕ.
Now let Fq [z] be the set of all polynomials over Fq . Let β ∈ Fq [z] with ‖β‖ > 1 and let J(β) be a proper subset of the set {ϵn ∈ Fq [z] : ‖ϵn ‖ < ‖β‖}, with at least two elements. Define ϵn , ϵn ∈ J(β)}. n n=1 β ∞
𝒦J(β) = {x ∈ Iq : x = ∑
Let ϕ : Fq [z] → {qn , n ∈ ℤ} be a nonincreasing function such that ϕ(Q) ≥ ϕ(P)
whenever ‖Q‖ ≤ ‖P‖,
Q, P ∈ Fq [z].
206  7 Formal Laurent series Finally, we set
P
𝒦(ϕ) = {x ∈ Iq : x − < ϕ(Q) Q
for infinitely many
P }, Q
where P, Q ∈ Fq [z]. We have the following theorem. Theorem 7.2.13. Let β, J(β), 𝒦J(β) , and 𝒦(ϕ) be defined as above. Let ϕ : Fq [z] → {qn , n ∈ ℤ} be a function such that the induced function ‖Q‖2 ϕ(Q) is nonincreasing and lim ‖Q‖2 ϕ(Q) = 0.
‖Q‖→∞
Then for every γ such that 0 < γ < 1/‖β‖, the set (𝒦(ϕ) \ 𝒦(γϕ)) ∩ 𝒦J(β) is uncountable. The details of the proof of Theorem 7.2.13 can be found in [58].
7.3 Basic algebra of formal Laurent series Now we begin to introduce the real or general formal Laurent series, which allows infinitely many terms to have negative exponents, as the Laurent series in analysis. Algebraically, the addition, scalar multiplication and Cauchy product on the formal Laurent series will be the same as these operations for regular Laurent series that has nonempty annulus of convergence. The semiformal Laurent series space ℒs and the reversed formal Laurent series space ℒr actually just form certain subsets of the general formal Laurent series space. Definition 7.3.1 ([28]). A formal Laurent series on ℂ is defined to be a mapping from ℤ to ℂ. A formal Laurent series g in z from ℤ to ℂ is usually denoted by ⋅ ⋅ ⋅ + b−n z −n + ⋅ ⋅ ⋅ + b−1 z −1 + b0 + b1 z + b2 z 2 + ⋅ ⋅ ⋅ ,
(7.13)
or by g(z) = ∑n∈ℤ bn z n , where bn ∈ ℂ for every n ∈ ℤ. ∞ n −n The series ∑∞ are called the regular part and the principal n=0 bn z and ∑n=1 b−n z + part of g, and they are denoted by g and g − , respectively. The complex conjugate of a Laurent series g(z) = ∑n∈ℤ bn z n is defined to be the series g(z) = ∑n∈ℤ bn z n where bn is the complex conjugate of bn , n ∈ ℤ. We define the reverse of the formal Laurent series g by 1 ̆ g(z) = g( ) = ∑ b̆ n z n , z n∈ℤ
(7.14)
where b̆ n = b−n for every n ∈ ℤ. Finally, we denote by 𝕃(ℂ), or simply by 𝕃, the set of all formal Laurent series over ℂ.
7.3 Basic algebra of formal Laurent series  207
Definition 7.3.2. Let g(z) = ∑n∈ℤ bn z n and f (z) = ∑n∈ℤ an z n be formal Laurent series over ℂ and c ∈ ℂ. Then g + f and cg are formal Laurent series in 𝕃 and (L1) (g + f )(z) = ∑n∈ℤ (bn + an )z n , (L2) (cg)(z) = ∑n∈ℤ cbn z n . Now we define the zero formal Laurent series to be the series g with bn = 0 for every n ∈ ℤ. In this chapter, we will understand that ∑n∈ℤ bn ∈ ℂ for a sequence (bn )n∈ℤ ⊆ ℂ if ∞ and only if both ∑∞ n=0 bn ∈ ℂ and ∑n=1 b−n ∈ ℂ. It is clear that 𝕃, with (L1) and (L2), is a linear vector space over ℂ. It is not clear whether 𝕃 is generally an algebra under the usual multiplication. Example 7.1.1 provided a suspicion for such consideration. Can we define a multiplication for formal Laurent series as we did for formal power series? For the semiformal Laurent series introduced in (7.1), the answer is yes. For the formal Laurent series introduced in Definition 7.3.1, yes is not the answer as we have seen in the Example 7.1.1. However, the product of certain nonformal Laurent series, P. Henrici called them in [42] but we call them the regular Laurent series, or the product of Laurent expansions of the analytic functions g and f on an annulus, is welldefined. If g(z) = ∑n∈ℤ bn z n and f (z) = ∑n∈ℤ an z n are Laurent expansions of two analytic functions on the annulus Ar,s = {z ∈ ℂ : r < z < s}, then (fg)(z) = ∑n∈ℤ cn z n and cn = ∑ bm an−m m∈ℤ
for every
n ∈ ℤ,
(7.15)
where each cn , the coefficient of the Laurent expansion of gf is calculated by cn =
1 f (t)g(t) dt, ∫ 2πi t n+1
(7.16)
Γρ
where r < ρ < s and Γρ is the circle {t = ρeiθ : 0 ≤ θ ≤ 2π}. This result, such as (7.15), about the product of the regular Laurent expansions of analytic functions indicates a possible product of formal Laurent series defined in Definition 7.3.1, although we clearly can not introduce the product of formal Laurent series in that way. We simply may not be able to apply the Cauchy integral formula (7.16) because our formal Laurent series are not necessary the Laurent expansion of analytic functions on some annulus. Let us begin our considerations with the following example.
208  7 Formal Laurent series Example 7.3.3. Let g(z) = ∑n∈ℤ z n and f (z) = ∑n∈ℤ 2−n z n . Then g is not analytic on any nonempty annulus, and hence we cannot apply the product formula (7.16) to obtain (7.15) for the regular Laurent series to this g. However, the product gf is a formal Laurent series under the dot product or the usual multiplication (Cauchy product), which will be seen soon. Let us introduce the dot product of formal Laurent series. Definition 7.3.4. Let g(z) = ∑n∈ℤ bn z n and f (z) = ∑n∈ℤ an z n be two formal Laurent series over ℂ. We define the dot product of g and f as g ⋅ f = ∑ bn an , n∈ℤ
if ∑n∈ℤ bn an ∈ ℂ; otherwise we say that the dot product g ⋅ f does not exist. We also define the pdot product of g and f as (g ⋅ f )p = ∑ bpn apn ,
0 < p < +∞,
n∈ℤ
if the series on the right side is convergent in ℂ. It is obvious that the dot product is commutative, if it exists. For all g(z) = ∑n∈ℤ bn z n ∈ 𝕃, we denote by DP(g) the set DP(g) = {f ∈ 𝕃 : f ⋅ g ∈ ℂ}. We see that DP(g) ≠ 0 for every g ∈ 𝕃 because the zero formal Laurent series is in the DP(g). Every polynomial is in DP(g), too. We provide a nontrivial example below. Example 7.3.5. Let g(z) = ∑n∈ℤ bn z n be given. We define an = 2−n
1 bn  + b−n  + 1
for every n ∈ ℤ.
Then f (z) = ∑n∈ℤ an z n ∈ DP(g) because ∑ an bn  ≤ ∑ 2−n .
n∈ℤ
n∈ℤ
Now, we introduce the shifting mapping. Definition 7.3.6. Let k ∈ ℤ be given. The kthshifting mapping Sk : 𝕃 → 𝕃 is defined by Sk (f )(z) = ∑ an−k z n n∈ℤ
for every
f (z) = ∑ an z n ∈ 𝕃. n∈ℤ
Let g(z) = ∑n∈ℤ bn z n ∈ 𝕃 be given. We define the reverseshifting set of g as 𝕃(g) = {f ∈ 𝕃  Sk (f ̆) ∈ DP(g)
for every
k ∈ ℤ}.
7.3 Basic algebra of formal Laurent series  209
Let us recall the multiplication operator Mz in Definition 6.3.5. Mz can be extended to 𝕃 and can be understood as the particular shifting mapping S1 : 𝕃 → 𝕃 on particular subset of 𝕃 because S1 (f )(z) = ∑ an−1 z n = ∑ an z n+1 = zf (z) = (Mz f )(z), n∈ℤ
n∈ℤ
n
if f (z) = ∑n∈ℤ an z ∈ ℋ ⊆ 𝕃. Similarly, Sk (f )(z) = z k f (z)
for every k ∈ ℤ.
By Definition 7.3.6, f ∈ 𝕃(g) if and only if ∑ bm ak−m ∈ ℂ for every k ∈ ℤ,
m∈ℤ
because it is equivalent to g ⋅ Sk (f ̆) ∈ ℂ, for every k ∈ ℤ. Actually g ⋅ Sk (f ̆) is just the ck in (7.15), the kth coefficient of the product of socalled nonformal Laurent series which requires that both f and g are analytic on an annulus. Let us prove the following proposition. Proposition 7.3.7. Let g(z) = ∑n∈ℤ bn z n and f (z) = ∑n∈ℤ an z n be two formal Laurent series on ℂ. Then f ∈ 𝕃(g) if and only if g ∈ 𝕃(f ). Proof. In fact, ̆ g ⋅ Sk (f ̆) = ∑ bm ak−m = ∑ bk−n an = f ⋅ Sk (g), n∈ℤ
m∈ℤ
for every k ∈ ℤ, what completes the proof. Now we are going to define the multiplication of any two formal Laurent series. Definition 7.3.8 ([28]). Let g(z) = ∑n∈ℤ bn z n and f (z) = ∑n∈ℤ an z n be two formal Laurent series on ℂ. We define the product of f and g, denoted by fg, to be (L3) fg(z) = ∑k∈ℤ dk z k , if dk = ∑m∈ℤ bm ak−m ∈ ℂ for every k ∈ ℤ. The identity of formal Laurent series under this multiplication is defined to be I𝕃 (z) = ∑ ek z k , k∈ℤ
e0 = 1, en = 0
for all other n ∈ ℤ.
We usually denote I𝕃 = I if there is no confusion. Please be noticed that the condition dk ∈ ℂ for every k ∈ ℤ in Definition 7.3.8 is equivalent to that f ∈ 𝕃(g). Actually dk is the series formed by all coefficients of kth term determined by g(z)f (z) through the distributive law and 𝕃(g) is the set of all f ∈ 𝕃 such that fg ∈ 𝕃. By Proposition 7.3.7, fg = gf for every f ∈ 𝕃(g) and, therefore, the product of formal Laurent series is commutative if it exists.
210  7 Formal Laurent series Remark 7.3.9. (i) Let f and g be formal Laurent series defined in Definition 7.3.1. If b−n = 0 = a−n for every n ∈ ℕ, then the formal Laurent series g and f become the formal power series and each dk is reduced to be the coefficient of the Cauchy product of these two formal power series. One can also check that g(z)f (z) = f (z) if g = I. Therefore, we may say that the multiplication on 𝕃 is the Cauchy product. (ii) Plainly, 𝕏(ℂ) ⊆ 𝕃(ℂ), ℒs ⊆ 𝕃 and ℒr ⊆ 𝕃. All those subspaces maintain (L1), (L2), (L3), and the dot product. Remark 7.3.10. If g and f are analytic on an nonempty annulus, then the coefficients of the regular Laurent expansion of fg have the same forms as in (L3). This shows that the principle of formal analysis works well here. The following two examples show that the existence of the dot product of two formal Laurent series does not have relationship with the existence of the product defined by (L3). It is an open question whether and how these two operations are related. Example 7.3.11. Let f (z) = ∑n∈ℤ an z n and g(z) = ∑n∈ℤ bn z n such that a0 = b0 = 1, and a−n = n,
b−n = n−3 ,
an = 1,
bn = n−2
for every n ∈ ℕ.
Then ∞ 1 n ∈ ℝ, + ∑ 3 2 n=1 n n=1 n ∞
f ⋅g =1+ ∑ so f ∈ DP(g). However,
1 ∞ −3 +∑n n=1 n n=1 ∞
d0 = ∑ bm a0−m = 1 + ∑ m∈ℤ
does not exists, so f ∉ 𝕃(g). Example 7.3.12. Let f (z) = ∑n∈ℤ an z n and g(z) = ∑n∈ℤ bn z n such that a0 = b0 = 1, and a−n = n,
b−n = n−2 ,
an = 1,
bn = n−3 ,
for every n ∈ ℕ.
Then −∞
∑
m=−k−1
bm ak−m =
−∞
∑
m=−k−1
m−2 ,
and ∞
∑ bm ak−m =
m=k+1
∞
∑ m−3 (m − k),
m=k+1
7.3 Basic algebra of formal Laurent series  211
and hence for every k ∈ ℤ,
dk = ∑ bm ak−m ∈ ℝ m∈ℤ
or f ∈ 𝕃(g). However, ∞
∞
n=1
n=1
1 ∞ 1 +∑ 3 n=1 n n=1 n ∞
f ⋅ g = 1 + ∑ a−n b−n + ∑ an bn = 1 + ∑ does not exist, so f ∉ DP(g). Proposition 7.3.13. Let f , g ∈ 𝕃 be given. Then: (i) 𝕃(g) ≠ 0; (ii) if f ∈ 𝕃(g) then αf ∈ 𝕃(g) for every α ∈ ℂ; (iii) if f , h ∈ 𝕃(g) then f + h ∈ 𝕃(g).
Proof. If we denote by E(P) the set of all polynomials over ℂ, then E(P) ⊆ 𝕃(g), what proves (i). Items (ii) and (iii) are obvious. Now, let us provide a nontrivial example of the product of formal Laurent series. Example 7.3.14. Let g(z) = ∑n∈ℤ bn z n be given. We define an = 2−n
1+
1
for every
∑2n i=−2n bi 
n ∈ ℤ,
and define f (z) = ∑n∈ℤ an z n . We shall show that Sk (f ̆) ∈ DP(g) for every k ∈ ℤ. Let k ∈ ℤ be given. We have g ⋅ Sk (f ̆) = ∑ bm ak−m = ∑ bm 2−k−m m∈ℤ
m∈ℤ
1+
1
b  ∑2k−m i=−2k−m i
.
If m > 2k, we have 0 < m < 2(m − k) ≤ 2k − m. Then bm ak−m  = bm  2−k−m
1+
1
b  ∑2k−m i=−2k−m i
< 2−k−m ,
so ∞
∞
m=2k
m=2k
∑ bm ak−m  < ∑ 2−k−m < +∞.
If m < −2k, we have m < 0 and m + 2k < 0, so −2k − m ≤ −2(k − m) = m + m − 2k < m + m + 2k < m < 0.
212  7 Formal Laurent series Then bm ak−m  = 2−k−m
bm 
1 + ∑2k−m b  i=−2k−m i
< 2−k−m ,
and hence −∞
∑
m=−2k
bm ak−m 
1 and 1/p + 1/q = 1; (ii) 𝕃∞ ⊆ 𝕃(g) if p = 1; (iii) 𝕃1 ⊆ 𝕃(g) if p = +∞; (iv) if ϕ ∈ 𝕃p and ψ ∈ 𝕃q , where 1/p + 1/q = 1, then ϕψ ∈ 𝕃(g) for every g ∈ 𝕃1 . Proof. Suppose that 1 < p < +∞. Let (an )n∈ℤ be a sequence in ℓq and let us write f (z) = ∑n∈ℤ an z n . Then 1/q
( ∑ am q ) m∈ℤ
< +∞
1/p
( ∑ bm p )
and
< +∞.
m∈ℤ
Applying Hölder’s inequality, we get 1/p
∑ bm ak−m  ≤ ( ∑ bm p ) m∈ℤ
m∈ℤ
1/p
= ( ∑ bm p ) m∈ℤ
1/q
( ∑ ak−m q ) m∈ℤ
1/q
( ∑ an q ) n∈ℤ
which proves (i). Now suppose that p = 1. Let (an ) ∈ ℓ∞ be given and let f (z) = ∑n∈ℤ an z n . Then an  < M for every n ∈ ℤ for some M > 0, and we have ∑ bm ak−m  ≤ ∑ Mbm  = M( ∑ bm ) < +∞
m∈ℤ
m∈ℤ
m∈ℤ
for every k ∈ ℤ, and then f ∈ 𝕃(g). This proves (ii). The proof of (iii) is similar to the proof of (ii). Now, let ϕ(z) = ∑ an z n ∈ 𝕃p , n∈ℤ
ψ(z) = ∑ cn z n ∈ 𝕃q . n∈ℤ
We show that ϕψ ∈ 𝕃(g) where g ∈ 𝕃1 . By (i), it is clear that ϕψ ∈ 𝕃. Put ϕψ(z) = ∑k∈ℤ dk z k , where dk = ∑m∈ℤ am ck−m for every k ∈ ℤ. Applying Hölder’s inequality, we get ∑ dk bn−k  = ∑ ∑ am ck−m ⋅ bn−k  m∈ℤ k∈ℤ k∈ℤ ≤ ∑ ( ∑ am ck−m ) ⋅ bn−k  k∈ℤ m∈ℤ
1/p
≤ ∑ [( ∑ am p ) k∈ℤ
m∈ℤ
1/q
( ∑ ck−m q ) m∈ℤ
] ⋅ bn−k 
214  7 Formal Laurent series
= ∑ (‖ϕ‖p ‖ψ‖q ) ⋅ bn−k  = ‖ϕ‖p ‖ψ‖q ‖g‖1 < +∞. k∈ℤ
It means that ϕψ ∈ 𝕃(g). Theorem 7.3.16. The set 𝕃1 is a linear algebra. In particular, f k ∈ 𝕃1 for every f ∈ 𝕃1 and k ∈ ℕ. Proof. It is obvious that 𝕃1 is a linear space over ℂ. We show that 𝕃1 is closed in terms of the multiplication of formal Laurent series. Let f (z) = ∑n∈ℤ an z n and g(z) = ∑n∈ℤ bn z n be any two formal Laurent series in 𝕃1 . We know that f ∈ 𝕃∞ for all f ∈ 𝕃p , p ≥ 1. Then fg ∈ 𝕃 by Theorem 7.3.15 (ii). We show that fg ∈ 𝕃1 . Let fg(z) = ∑n∈ℤ dk z k , where dk = ∑m∈ℤ am bk−m for every k ∈ ℤ. Then ∑ dk  = ∑ ∑ am bk−m ≤ ∑ ∑ am bk−m  m∈ℤ k∈ℤ m∈ℤ k∈ℤ k∈ℤ = ∑ am  ∑ bk−m  = ‖f ‖1 ‖g‖1 < +∞. m∈ℤ
k∈ℤ
Thus, fg ∈ 𝕃1 , and hence 𝕃1 is a linear algebra. Taking f = g and repeating the above process ktimes, we get f k ∈ 𝕃1 for every f ∈ 𝕃1 and k ∈ ℕ. Remark 7.3.17. Obviously, every sequence (bn )n∈ℤ such that lim b n→±∞ n
∈ℂ
is bounded, and thus g = ∑n∈ℤ bn z n ∈ 𝕃∞ . We conclude by Theorem 7.3.15 (iii) that 𝕃1 ⊆ 𝕃(g) if (bn ) is bounded, which provides a wide range of formal Laurent series that have products. If we have noticed that many sequences converge in some spaces but their corresponding formal Laurent series converge nowhere such as g(z) = ∑n∈ℤ z n or f (z) = ∑n∈ℤ n1 z n , we may realize how far away we have been from the analyticity. We met a difficulty in Example 7.3.3 where the formal Laurent series g is nowhere analytic. However, with the definition (L3), we can provide the following example. Example 7.3.18. Let f (z) = ∑ an z n = ∑ 2−n z n n∈ℤ
n∈ℤ
and
g(z) = ∑ bn z n = ∑ z n , n∈ℤ
n∈ℤ
that is, an = 2−n and bn = 1 for every n ∈ ℤ. As we know, the formal Laurent series g is not analytic on any nonempty annulus. However, gf ∈ 𝕃 by Theorem 7.3.15, because g ∈ 𝕃∞ and f ∈ 𝕃1 . If we write gf (z) = ∑k∈ℤ dk z k , then we have m=−1
∞
−∞
m=0
dk = ∑ am bk−m = ∑ am bk−m + ∑ am bk−m m∈ℤ
7.3 Basic algebra of formal Laurent series  215
∞
+∞
m=1
m=0
= ∑ 2−m + ∑ 2−m = 1 + 2 = 3, for every k ∈ ℤ. Therefore, we have g(z)f (z) = ∑ dk z k = ∑ 3z n = 3g(z). n∈ℤ
k∈ℤ
Proposition 7.3.19. For any f , g ∈ 𝕃(ℂ): ̆ (i) f ∈ 𝕃(g) if and only if f ̆ ∈ 𝕃(g); (ii) f ∈ 𝕃(g)̆ if and only if f ̆ ∈ 𝕃(g); (iii) f ∈ 𝕃(g) if and only if f ∈ 𝕃(g). Proof. Let g(z) = ∑n∈ℤ bn z n and f (z) = ∑n∈ℤ an z n . By (L3), f ̆ ∈ 𝕃(g)̆ if and only if ∑ b̆ m ă k−m = ∑ b−m a−(k−m) = ∑ bm a−k−m ∈ ℂ
m∈ℤ
m∈ℤ
m∈ℤ
for every k ∈ ℤ. This is equivalent to that ∑m∈ℤ am bj−m ∈ ℂ for every j ∈ ℤ, or f ∈ 𝕃(g). This proves (i). ̆ Since f ̆ = f for all f ∈ 𝕃, (ii) is a special case of (i). Similarly, we can prove (iii). Theorem 1.6.1 tells us that the space of the formal power series over ℂ is an integral domain. We have already seen in Example 7.3.18 that (f (z)−3)g(z) = 0 but neither (f −3) nor g is the zero formal Laurent series. The next proposition will provide more details about 𝕃(g) when we pursue properties of an integral domain. Proposition 7.3.20. Let g(z) = ∑n∈ℤ bn z n be a formal Laurent series. Then (i) zg(z) = g(z) if and only if bn = b for every n ∈ ℤ for some b ∈ ℂ; (ii) if Pn (z) = ∑nk=0 pk z k is a polynomial of degree n and g = ∑n∈ℤ bz n , b ∈ ℂ, then n
(Pn (z) − ∑ pk )g(z) = 0; k=0
n
(iii) let g(z) = ∑n∈ℤ bz and f (z) = ∑n∈ℤ an z n be given. If ∑n∈ℤ an = A ∈ ℂ converges absolutely, then (f (z) − A)g(z) = 0. Proof. Supposing that zg(z) = g(z), it follows that bn = bn−1 for every n ∈ ℤ. Let us write b = b0 . We have b = bn for every n ∈ ℤ. Conversely, suppose that g(z) = ∑n∈ℤ bz n . Then zg(z) = ∑ bz n+1 = ∑ bz n = g(z). n∈ℤ
This proves (i).
n∈ℤ
216  7 Formal Laurent series Now, by (i), we can easily obtain the equality z k g(z) = g(z) for every k ∈ ℕ, if bn = b for every n ∈ ℤ. Then (pk z k − pk )g(z) = 0
for 0 ≤ k ≤ n,
and we get n
(Pn (z) − ∑ pk )g(z) = 0. k=0
Finally, fg ∈ 𝕃 by (iii) of Theorem 7.3.15. Writing fg(z) = ∑k∈ℤ dk z k , we have dk = ∑ bn ak−n = ∑ bak−n = bA. n∈ℤ
n∈ℤ
Hence f (z)g(z) = ∑ dk z k = ∑ bAz k = A ∑ bz n = Ag(z). k∈ℤ
n∈ℤ
k∈ℤ
Proposition 7.3.21. Let b ∈ ℂ \ {0} and (an )n∈ℤ ⊆ ℂ, write g = ∑ bz n , n∈ℤ
f (z) = ∑ an z n . n∈ℤ
Then f (z)g(z) = 0
if and only if
∑ an = 0.
n∈ℤ
Proof. It is clear that f (z)g(z) = 0 if and only if ∑ bm an−m = 0
m∈ℤ
for every n ∈ ℤ.
Then ∑m∈ℤ bm an−m = 0 is equivalent to ∑ ban−m = 0
m∈ℤ
or
∑ an = 0.
n∈ℤ
Remark 7.3.22. If f ∈ 𝕃1 , then f ∈ 𝕃(f ) by Theorem 7.3.16. Since 𝕃∞ ⊆ 𝕃(f ), by (ii) of Theorem 7.3.15, the above two propositions show that 𝕃(f ) is not an integral domain for every f ∈ 𝕃1 . The space 𝕃 of formal Laurent series maintains all properties of the regular Laurent series and contains many interesting subspaces such as 𝕃(g), 𝕃p , ℒs , ℒr , Fq ((z −1 )), and 𝕏. Similar to the formal power series space 𝕏, the formal Laurent series space 𝕃 enjoys all basic algebraic operations, such as addition, scalar multiplication, and Cauchy product. We have not seen the composition, the most special operation the space 𝕏 has. That is the next section.
7.4 Composition of formal Laurent series and formal power series  217
7.4 Composition of formal Laurent series and formal power series The composition of formal power series is a very interesting issue, which also significantly distinguishes itself, in view of the sequences, from the regular sequence space such as ℓp . Some applications of the composition have been developed rapidly in many fields including the differential equations, combinatorics, number theory, numerical analysis, Riordan arrays, and so on. We have seen those applications in the previous chapters. Certain kind of composition of formal Laurent series in ℒs with formal power series were discussed in [42], where the formal Laurent series has a finite principal part and the composed formal power series must be nonunit. Except for the analytic functions on an annulus, the composition of the general formal Laurent series had not been established by 2012. In the previous sections, we have seen that the multiplication for formal Laurent series is not a simple issue. The composition involving formal Laurent series could be more complicated. For example, composition of formal Laurent series must face the existence of the inverse or reciprocal of the formal Laurent series and the uniqueness of such existence. Such existence and uniqueness are guaranteed in the space of formal power series, as we discussed in Chapter 5. Let us consider the following examples. n n + Example 7.4.1. Let f (z) = ∑∞ n=1 z and g(z) = ∑n∈ℤ z . We know that g (f (z)) is a formal − + power series but we do not know what g (f (z)) is where g is the regular part of g and g − is the principal part of g.
Example 7.4.2. Let f (z) = 1 − z and g(z) = ∑n∈ℤ z n . Obviously, (1 − z)k is a polynomial n for all k ∈ ℕ and (1−z)−1 = ∑∞ n=0 z is a welldefined formal power series. We can easily find out that ∞
g + (f (z)) = ∑ (f (z))
n
n=0
is not defined. Example 7.4.1 shows that we must have the inverse (of multiplication) of some kind of formal Laurent series before we establish a composition of Laurent series. Example 7.4.2 shows that we may not have the composition even if the composed formal power series is a unit or a polynomial. These two examples tell us that generally it is impossible to have a composition of a formal Laurent series with a nonunit and also may not be possible to have a composition of a formal Laurent series with a unit. This is a significant difference to the composition of formal power series although a formal power series can be considered as a formal Laurent series whose coefficients of all terms with negative exponents are zeros. By means of the generalized composition established in Chapter 5, however, we are able to establish the composition of a formal Laurent series with a formal power series.
218  7 Formal Laurent series In the next definition, we keep the notations 𝕏g and Tg defined in Definition 5.1.5 and extend the composition to formal Laurent series g. j Definition 7.4.3. Let f (z) = ∑∞ j=0 aj z be a unit formal power series on ℂ. If n ∈ ℕ ∪ {0}, (n) k we write f n (z) = ∑∞ k=0 ak z as we did in Definition 1.2.1. Denote ∞
f −1 (z) = ∑ a(−1) zk , k k=0
and then write ∞
n
f −n (z) = [f −1 (z)] = ∑ a(−n) zk k k=0
for every n ∈ ℕ, where f −1 is the inverse formal power series of f such that the Cauchy product f −1 (z)f (z) = 1. Since f is a unit, f −n is welldefined for all n ∈ ℕ. Similarly, as we did in the Definition 5.1.5, let g(z) = ∑n∈ℤ bn z n be a formal Laurent series and extend 𝕏g to be ∞
𝕏g = {f (z) = ∑ an z n ∈ 𝕏(ℂ) : ∑ bn a(n) ∈ℂ k n∈ℤ
n=0
∀ k ∈ ℕ ∪ {0}}.
If 𝕏g ≠ 0, for example, if g is a formal power series or g is a Laurent expansion of an analytic function, we define Tg : 𝕏g → 𝕏 by ∞
Tg (f )(z) = ∑ ck z k k=0
where ck = ∑ bn a(n) ∈ ℂ for every k ∈ ℕ ∪ {0}. k n∈ℤ
The mapping Tg (f ) is called the composition of the formal Laurent series g with the unit formal power series f on ℂ, or the composition of g with f . The omposition Tg (f ) is also denoted by g ∘ f or k
(g ∘ f )(z) = ∑ bk [f (z)] . k∈ℤ
Now, we prove the following. Theorem 7.4.4. Let g(z) = ∑k∈ℤ bk z k be a formal Laurent series over ℂ such that b−k ≠ k 0 for some k ∈ ℕ, and let f (z) = ∑∞ k=0 ak z be a formal power series on ℂ such that a0 ≠ 0. Then g(f (z)) is a formal power series, that is, ∞
g(f (z)) = ∑ cn z n n=0
with
cn ∈ ℂ for every
n ∈ ℕ ∪ {0},
7.4 Composition of formal Laurent series and formal power series  219
if and only if both ∞ n ∈ ℂ and ∑ ( )b−n ak−n 0 k n=k
∞ n ∈ℂ ∑ ( )bn an−k 0 k n=k
for every k ∈ ℕ ∪ {0}. Proof. Let us write g = g − + g + , where g − is the principal part of g and g + is the regular part of g. By Theorem 5.4.1, g + ∘ f exists if and only if ∞ n ∈ ℂ for every k ∈ ℕ ∪ {0}. ∑ ( ) bn an−k 0 k n=k
Since a0 ≠ 0, it follows that f −1 (z) = we write
1 f (z)
is a welldefined formal power series. If
∞
n f −1 (z) = ∑ a(−1) n z , n=0
then a(−1) = 0
1 , a0
a(−1) =− 1
a1 , a20
a(−1) = 2
a21 − a0 a2 a30
,
....
Then we may write ∞
g − (f (z)) = ∑ b−n (f (z))
−n
n=1
∞
n
= ∑ b−n (f −1 (z)) . n=1
By Theorem 5.4.1, g − (f (z)) exists if and only if ∞ n n−k ∈ ℂ for every ∑ ( ) b−n (a(−1) 0 ) k n=k
k ∈ ℕ ∪ {0}
which is equivalent to ∞ n ∈ ℂ for every k ∈ ℕ ∪ {0}. ∑ ( ) b−n ak−n 0 k n=k
Example 7.4.5. Let ∞
∞
n=1
n=0
g(z) = ∑ (2z)−n + ∑ z n
and f (z) =
3 ∞ + ∑ n!z n . 4 n=1
Then ∞
g − (z) = ∑ (2z)−n n=1
and
∞
g + (z) = ∑ z n . n=0
220  7 Formal Laurent series It is clear that f diverges everywhere except at z = 0, and then g ∘ f does not make any sense in the classical complex analysis. However, (g − ∘ f )(z) ∈ 𝕃 and
(g + ∘ f )(z) ∈ 𝕃
by Theorem 5.4.3, because g − converges for z > 1/2 and g + converges for z < 1. Then (g ∘ f ) ∈ 𝕃. The above example is a particular case of the following. Proposition 7.4.6. Let r, s ∈ ℝ ∪ {+∞} with 0 ≤ r < s and let g(z) = ∑k∈ℤ bk z k be a Laurent series, which converges on the annulus Ar,s = {z ∈ ℂ : r < z < s}. If f ∈ 𝕏(ℂ) with f (0) = a0 ∈ Ar,s , then g ∘ f ∈ 𝕃. −n Proof. Let us write g = g − + g + . Then g − (z) = ∑∞ converges for all z such that n=1 b−n z ∞ + n z > r, or 1/z < 1/r, and g (z) = ∑n=0 bn z converges for all z such that z < s. Since a0 ∈ Ar,s , it follows that 1/a0  < 1/r. Applying Theorem 5.4.3 to g − and f −1 , we have that ∞
∞
n=1
n=1
n
(g − ∘ f ) ∈ 𝕃(z) = ∑ b−n f −n (z) = ∑ b−n (f −1 (z)) ∈ 𝕏(ℂ). Similarly, because a0  < s, applying Theorem 5.4.3 to g + and f , we obtain (g + ∘ f ) ∈ 𝕃. Thus, (g ∘ f )(z) ∈ 𝕃. Proposition 7.4.7. Let g(z) = ∑k∈ℤ bk z k be a formal Laurent series over ℂ, and let k f (z) = ∑∞ k=0 ak z be a formal power series on ℂ such that a0 ≠ 0. If there exists some m ∈ ℕ such that bk = 0 for every k ≤ −m, then g ∘ f ∈ 𝕏 if and only if g + ∘ f ∈ 𝕏. Proof. Since bk = 0 for every k ≤ −m, it follows that m−1
g − (z) = ∑ b−i z −i . i=1
Then m−1
(g − ∘ f )(z) = ∑ b−i f −i (z) ∈ 𝕏 i=1
is always true because f (0) = a0 ≠ 0. Thus, g ∘ f ∈ 𝕏 if and only if g + ∘ f ∈ 𝕏. This proposition has introduced the composition g ∘ f , where g ∈ ℒs and f ∈ 𝕏(ℂ). For the reversed semiformal Laurent series g ∈ ℒr such as ∞
g(z) = ∑ cn z −n , n=m
m ∈ ℤ,
7.4 Composition of formal Laurent series and formal power series  221
we have the following corollary whose proof is very similar to the proof of Proposition 7.4.7. Corollary 7.4.8. Let g ∈ 𝕃(ℂ) such that ∞
g(z) = ∑ cn z −n ,
m ∈ ℤ,
n=m
and let f ∈ 𝕏(ℂ) be given with f (0) = a0 ≠ 0. Then g ∘ f ∈ 𝕃 if and only if ∞ n ∈ ℂ for all ∑ ( )b−n ak−n 0 k n=k
k ∈ ℕ.
Now, we are going to introduce the formal derivatives of a formal Laurent series. Definition 7.4.9. Let g(z) = ∑n∈ℤ bn z n be a formal Laurent series over ℂ. Then the formal derivative of g is a formal Laurent series defined by g (z) = ∑ (bn z n ) = ∑ nbn z n−1 .
n∈ℤ
n∈ℤ
The higher order derivatives of g are defined recursively. It is obvious that g = (g − ) + (g + ) . Corollary 7.4.10. Let g(z) = ∑k∈ℤ bk z k be a formal Laurent series over ℂ, and let f (z) = k ∑∞ k=0 ak z be a formal power series on ℂ such that a0 ≠ 0. Then (g ∘ f )(z) is a formal n power series, that is, (g ∘ f )(z) = ∑∞ n=0 cn z with cn ∈ ℂ for every n ∈ ℕ ∪ {0}, if and only if (k)
(g + ) (a0 ) ∈ ℂ
and (ğ + ) (a−1 0 ) ∈ ℂ for every (k)
k ∈ ℕ ∪ {0}.
where g (k) is the kth order formal derivative of g. Proof. By Theorem 5.4.6, (g + )(k) (a0 ) ∈ ℂ for every k ∈ ℕ ∪ {0} is equivalent to ∞ n ∈ ℂ for every k ∈ ℕ ∪ {0}. ∑ ( ) bn an−k 0 k n=k n Further, let us note that ğ + (z) = ∑∞ n=1 b−n z . Thus ∞
−1 (ğ + ) (a−1 0 ) = ∑ n(n − 1) ⋅ ⋅ ⋅ (n − k + 1)b−n (a0 ) (k)
n−k
n=k
for every k ∈ ℕ ∪ {0} which is equivalent to ∞ n ∈ ℂ for every k ∈ ℕ ∪ {0}. ∑ ( )b−n ak−n 0 k n=k
It is enough to apply Theorem 7.4.4 to complete the proof.
∈ℂ
222  7 Formal Laurent series The right distributive law is an important property for formal power series if the composition exists. The next theorem shows that the right distributive law is also true for formal Laurent series under certain conditions. Theorem 7.4.11. Let g, h ∈ 𝕃(ℂ) be given and let f ∈ 𝕏(ℂ) be given with f (0) = a0 ≠ 0. If g ∘ f ∈ 𝕏(ℂ),
h ∘ f ∈ 𝕏(ℂ),
and
gh ∈ 𝕃(ℂ),
then the right distributive law holds, that is, (gh) ∘ f = (g ∘ f )(h ∘ f ). Proof. Let g, h ∈ 𝕃(ℂ) be given and let f ∈ 𝕏(ℂ) be a unit with f (0) = a0 ≠ 0. Suppose that g ∘ f ∈ 𝕏(ℂ),
h ∘ f ∈ 𝕏(ℂ),
and
gh ∈ 𝕃(ℂ).
Write g = g − + g + , h = h− + h+ , where g − is the principal part of g and g + is the regular part of g, h− is the principal part of h and h+ is the regular part of h. Then gh = (g − + g + )(h− + h+ ) = g − h− + g + h− + g − h+ + g + h+ , and (gh) ∘ f = (g − h− + g + h− + g − h+ + g + h+ )(f )
= g − (f )h− (f ) + g + (f )h− (f ) + g − (f )h+ (f ) + g + (f )h+ (f ).
Since g ∘ f ∈ 𝕏(ℂ) and h ∘ f ∈ 𝕏(ℂ), it follows by Theorem 7.4.4 that all eight compositions in the righthand side exist in 𝕏. On the other hand, (g ∘ f )(h ∘ f ) = (g − (f ) + g + (f ))(h− (f ) + h+ (f ))
= g − (f )h− (f ) + g + (f )h− (f ) + g − (f )h+ (f ) + g + (f )h+ (f ).
Thus, (gh) ∘ f = (g ∘ f )(h ∘ f ). For any g ∈ 𝕃(ℂ) and any unit f ∈ 𝕏(ℂ), if g ∘ f ∈ 𝕏, what could be possible Chain Rule for the formal differentiation? We should answer two questions: 1. does g (f ) exist? 2. does the equality (g(f )) (z) = g (f (z))f (z) hold?
7.4 Composition of formal Laurent series and formal power series  223
Theorem 7.4.12. Let g ∈ 𝕃(ℂ) and let the unit f ∈ 𝕏(ℂ) be given. Then (g ∘ f ) ∈ 𝕏(ℂ) if and only if g ∘ f ∈ 𝕏(ℂ). Moreover, (g ∘ f ) (z) = g (f (z))f (z),
(7.17)
if g ∘ f ∈ 𝕏. Proof. By Corollary 7.4.10, it is clear that (g ∘ f ) ∈ 𝕏(ℂ) if and only if g ∘ f ∈ 𝕏(ℂ). We need only prove (7.17). Suppose that g ∘ f ∈ 𝕏(ℂ) and write g = g − + g + such that n
−∞ ∞ ∞ 1 g − (z) = ∑ bn z n = ∑ b−n z −n = ∑ b−n ( ) . z n=−1 n=1 n=1
Let h = g − . We show that (h ∘ f ) (z) = h (f (z))f (z). ∞
h(f (z)) = ∑ b−n ( n=1
n
1 ) . f (z)
Since f is a unit, 1/f = ϕ ∈ 𝕏(ℂ) for some formal power series ϕ. Then, by the general Chain Rule (5.20) we have ∞
n
∞
(h(f (z))) = [ ∑ b−n (ϕ(z)) ] = ∑ b−n n(ϕ(z))
n=1
∞
= ∑ b−n n( n=1 ∞
n−1
1 ) f (z)
n=1
n=1
ϕ (z)
−1 f (z) f 2 (z)
−(n+1)
= ∑ (−n)bn (f (z))
n−1
,
and the other side is ∞
h (f (z))f (z) = ∑ (−n)bn (f (z)) n=1
f (z).
−(n+1)
Thus, (g − ∘ f ) (z) = (g − ) (f (z))f (z). It is clear that g + and f are formal power series which satisfy the Chain Rule (5.20). We complete the proof. Remark 7.4.13. In view of the composition g ∘ f , where g ∈ 𝕃 and f ∈ 𝕏, we have: (i) if g ∈ 𝕏, then Theorem 7.4.4 is reduced to Theorem 5.4.1; (ii) If g ∈ 𝕃 and g ∉ 𝕏, then f must be a unit. If we consider g ∘ f where both g and f are formal Laurent series, what can we say?
224  7 Formal Laurent series
7.5 Canonical mapping and dot product of formal Laurent series In this section, we are going to set up a mapping, which will be called the canonical mapping from 𝕃 = 𝕃(ℝ) to a subset M0 of the Lebesgue measurable functions on ℝ, and investigate this mapping from the point of view of dot product in 𝕃 and multiplication in M0 . We would like to indicate that choosing 𝕃 = 𝕃(ℝ) is just for our convenience, that is, all results in this section will remain true if 𝕃 = 𝕃(ℂ) and M0 = M0 (ℂ). Let ξI be the characteristic function of the interval I ⊆ ℝ, that is, 1
ξI (t) = {
if t ∈ I,
0
if t ∉ I.
Definition 7.5.1. We define M0 = M0 (ℝ) = { ∑ an ξ(n,n+1] : (an )n∈ℤ ⊆ ℝ}, n∈ℤ
the subset of extended ℝvalued step functions on ℝ. let
It is clear that the functions belonging to M0 are Lebesgue measurable. Moreover, f = ∑ an ξ(n,n+1] , n∈ℤ
g = ∑ bn ξ(n,n+1] n∈ℤ
be given in M0 , then obviously f p = ∑ an p ξ(n,n+1] n∈ℤ
for every
0 < p < +∞;
(f ⋅ g)(x) = f (x) ⋅ g(x) = ∑ an bn ξ(n,n+1] (x) n∈ℤ
(7.18) (7.19)
for every x ∈ ℝ. Definition 7.5.2. We define a mapping A: 𝕃 → M0 by A(f ) = ∑ an ξ(n,n+1] n∈ℤ
for
f (z) = ∑ an z n ∈ 𝕃. n∈ℤ
The mapping A is said to be the canonical mapping. Usually, we use notation ̃f = A(f ). Now, we define the set 𝕄p as 𝕄p = Lp (ℝ) ∩ M0 ,
0 < p ≤ +∞,
where Lp (ℝ) is the classical Banach space of functions integrable in the Lebesgue sense with pth power on ℝ and L∞ (ℝ) denotes the Banach space of essentially bounded functions on ℝ. Now, let us state some obvious results in the following.
7.5 Canonical mapping and dot product of formal Laurent series  225
Lemma 7.5.3. A is an isomorphism with respect to the addition and scalar multiplication defined in 𝕃 and M0 . By Lemma 7.5.3, the inverse of A, denoted by A−1 , is welldefined on M0 . Lemma 7.5.4. Let 1 ≤ p ≤ +∞. Then the spaces 𝕃p (ℝ) and 𝕄p are isometric (shortly: 𝕃p ≅ 𝕄p ) under the canonical mapping A. Proof. By Lemma 7.5.3, A is isomorphism. It is obvious that A(𝕃p (ℝ)) = 𝕄p . Indeed, suppose that 1 ≤ p < +∞ and let ϕ = ∑ an ξ(n,n+1] ∈ 𝕄p . n∈ℤ
By (7.18), we have 1/p
‖ϕ‖Lp = (∫ ϕp )
= ( ∑ an p )
1/p
n∈ℤ
ℝ
< +∞.
So, for f (z) = ∑n∈ℤ an z n , we have 1/p
( ∑ an p ) n∈ℤ
= ‖f ‖𝕃p < +∞,
and A(f ) = ϕ. The case p = +∞ is also clear, because A(f )L = sup{an  : n ∈ ℤ} = ‖f ‖𝕃∞ , ∞
n
for every f (z) = ∑n∈ℤ an z ∈ 𝕃∞ , and we denote A(𝕃∞ ) = 𝕄∞ . Lemma 7.5.5. Let g(z) = ∑n∈ℤ bn z n and f (z) = ∑n∈ℤ an z n be formal Laurent series in 𝕃(ℝ). Then f ∈ DP(g)
if and only if ̃f g̃ ∈ L1 (ℝ).
Moreover, f ⋅ g = ∫ ̃f g̃ dx, ℝ
where f ⋅ g is the dot product of the formal Laurent series f and g. Proof. Under the canonical mapping, ̃f = ∑ a ξ n (n,n+1] n∈ℤ
and g̃ = ∑ bn ξ(n,n+1] . n∈ℤ
226  7 Formal Laurent series By equation (7.19), ̃f g̃ (x) = ( ∑ a ξ n (n,n+1] (x))( ∑ bn ξ(n,n+1] (x)) n∈ℤ
n∈ℤ
= ∑ an bn ξ(n,n+1] (x) n∈ℤ
for every x ∈ ℝ, and then f ∈ DP(g) if and only if ̃f g̃ is Lebesgue integrable, in this case, ∫ ̃f g̃ dx = ∑ an bn = f ⋅ g. n∈ℤ
ℝ
Corollary 7.5.6. Let g(z) = ∑n∈ℤ bn z n ∈ 𝕃p , 1 ≤ p ≤ +∞. Then: (i) 𝕃q ⊆ DP(g), if p > 1 and 1/p + 1/q = 1; (ii) 𝕃∞ ⊆ DP(g), if p = 1; (iii) 𝕃1 ⊆ DP(g), if p = +∞. Proof. Let p > 1, 1/p + 1/q = 1, and let f ∈ 𝕃q. Then ̃f ∈ Lq (ℝ), and ∫ ̃f g̃  dx ≤ ‖̃f ‖q ⋅ ‖g̃ ‖p ℝ
by Hölder’s inequality. Then (i) is true by Lemma 7.5.5. If f ∈ 𝕃∞ and p = 1, then ∫ ̃f g̃  dx ≤ ‖̃f ‖∞ ∫ g̃  dx ≤ ‖̃f ‖∞ ⋅ ‖g̃ ‖1 , ℝ
ℝ
and again we have (ii) by applying Lemma 7.5.5. The proof of (iii) is also straightforward. The most important character of the canonical mapping A is its role in the multiplication or Cauchy product of formal Laurent series.
7.6 Canonical mapping and multiplication of formal Laurent series The multiplication of formal power series, or Cauchy product, is a welldefined operation. The similar multiplication of formal Laurent series, however, is very different and even very difficult to define. Some existence theorems and verifications of the existence of multiplication of formal Laurent series were introduced in the previous sections of this chapter. This section provides another approach to express such multiplication by using certain existing results in classical analysis.
7.6 Canonical mapping and multiplication of formal Laurent series  227
It is known that given two ℝvalued functions f and g on ℝn , n ∈ ℕ, we may define a new function, f ∗ g, by f ∗ g(x) = ∫ f (t)g(x − t) dt ℝn
if the righthand side is welldefined. f ∗ g is called the convolution of f with g. Readers may find many results about the convolution in [95] if they are interested in. We provide a popular result about the convolution below as a lemma. Lemma 7.6.1. Let f ∈ L1 (ℝn ), g ∈ M(ℝn ) where n ∈ ℕ and M represents the set of all Lebesgue measurable functions on ℝn , suppose that g is bounded. Then the formula f ∗ g(x) = ∫ f (x − y)g(y) dy ℝn
defines the convolution f ∗ g that is bounded and uniformly continuous on ℝn . Moreover, if A, B ⊆ ℝn , f vanishes on ℝn \ A, and g vanishes on ℝn \ B, then f ∗ g vanishes on ℝn \ (A + B), where A + B = {a + b : a ∈ A, b ∈ B}. Also, f ∗ g(x) = ∫ f (t)g(x − t) dt ℝn
for all x ∈ ℝn . The next lemma describes some properties of the convolution of functions belonging to M0 . Lemma 7.6.2. Let ϕ and ψ be any two functions in M0 . Then the convolution ϕ∗ψ exists on ℝ if and only if ϕ ∗ ψ exists on ℤ, or (ϕ ∗ ψ)(k) ∈ ℝ
for every
k ∈ ℤ.
Moreover, if t ∈ ℝ is such that k < t ≤ k + 1 for some k ∈ ℤ, then (ϕ ∗ ψ)(t) = (t − k) (ϕ ∗ ψ)(k + 1) − (t − k − 1) (ϕ ∗ ψ)(k).
(7.20)
Proof. The implication in one direction is obvious. Now, let ϕ = ∑n∈ℤ an ξ(n,n+1] and ψ = ∑n∈ℤ bn ξ(n,n+1] be any two functions in M0 . Suppose that ϕ ∗ ψ exists on ℤ, that is, (ϕ ∗ ψ)(k) = ∫ ϕ(x)ψ(k − x) dx ∈ ℝ for every
k ∈ ℤ,
ℝ
we show that ϕ ∗ ψ exists on ℝ. It suffices to show that formula (7.20) is true.
228  7 Formal Laurent series Let k ∈ ℤ be given, we have n+1
(ϕ ∗ ψ)(k) = ∑ ∫ ϕ(x)ψ(k − x) dx n∈ℤ n
n+1
= ∑ ∫ an ψ(k − x) dx = ∑ an bk−(n+1) . n∈ℤ
n∈ℤ n
Let t ∈ ℝ. Then k < t ≤ k + 1 for some k ∈ ℤ. For n < x ≤ n + 1, we have ϕ(x)ψ(t − x) = an ψ(t − x) and k − (n + 1) < t − (n + 1) < t − x < t − n ≤ k − (n − 1). Bisecting the interval (k − (n + 1), k − (n − 1)], we have: (j) for t − x such that k − (n + 1) < t − (n + 1) ≤ t − x ≤ k − n: ψ(t − x) = bk−(n+1) ; (jj) for t − x such that k − n < t − x < t − n ≤ k − (n − 1): ψ(t − x) = bk−n . Substituting u = t − x and applying (j) and (jj) above, we have n+1
t−n
∫ ϕ(x)ψ(t − x) dx = an n
∫
ψ(u) du
t−(n+1) k−n
= an
∫
t−n
ψ(u) du + an ∫ ψ(u) du
t−(n+1)
k−n
= (k − t + 1)an bk−(n+1) + (t − k)an bk−n
= an [(t − k)bk−n − (t − k − 1)bk−(n+1) ].
Applying formula (7.21) we have n+1
(ϕ ∗ ψ)(t) = ∑ ∫ ϕ(x)ψ(t − x) dx n∈ℤ n
= (t − k) ∑ an bk−n − (t − k − 1) ∑ an bk−(n+1) n∈ℤ
n∈ℤ
= (t − k) (ϕ ∗ ψ)(k + 1) − (t − k − 1) (ϕ ∗ ψ)(k), this is equation (7.20).
(7.21)
7.6 Canonical mapping and multiplication of formal Laurent series  229
Corollary 7.6.3. The convolution ϕ ∗ ψ considered in Lemma 7.6.2 is a continuous function on ℝ. Proof. It is a simple consequence of formula (7.20). Now we establish a relation between the multiplication of formal Laurent series and the convolution of certain Lebesgue measurable functions. Theorem 7.6.4. Let g(x) = ∑n∈ℤ bn xn ∈ 𝕃(ℝ). The following are true: (i) Let f ∈ 𝕃(g) be given. If we write f (x) = ∑ an xn n∈ℤ
and
fg(x) = ∑ dk x k , k∈ℤ
then dk−1 = ̃f ∗ g̃ (k)
for every
k ∈ ℤ.
(ii) f ∈ 𝕃(g) if and only if the convolution ̃f ∗ g̃ exists on ℝ. (iii) If f ∈ 𝕃(g), then we have ̃ = ∑ (̃f ∗ g̃ )(k + 1) ξ fg (k,k+1] , k∈ℤ
(7.22)
̃ = A(fg). where fg Proof. Let f ∈ 𝕃(g) be given by f (x) = ∑n∈ℤ an xn . Then ̃f can be considered as ϕ and g̃ be considered as ψ in Lemma 7.6.2, respectively. Notice that fg(x) = ∑k∈ℤ dk x k , where dk = ∑ an bk−n ∈ ℂ n∈ℤ
for every k ∈ ℤ.
(i) is followed by (7.21) in Lemma 7.6.1. Let k ∈ ℤ be given. Suppose that n < x ≤ n + 1 for some integer n, then k − n − 1 ≤ k − x < k − n. Therefore, we have ̃f (x) = an and g̃ (k − x) = bk−n−1 = b(k−1)−n
almost everywhere.
Actually, the equality above is true for all k − x except possibly k − x = k − n − 1, which does not affect the Lebesgue integral of g̃ (k − x). Then ̃f (x)g̃ (k − x) = a b n (k−1)−n ,
if n < x < n + 1,
and then we have ∫ ̃f (x)g̃ (k − x) dx = ∑ an b(k−1)−n ∈ ℂ, ℝ
n∈ℤ
or ̃f ∗ g̃ exists on ℤ. By Lemma 7.6.2, ̃f ∗ g̃ exists on ℝ.
230  7 Formal Laurent series Conversely, suppose that ̃f ∗ g̃ exists on ℝ. (i) yields that (̃f ∗ g̃ )(k) = dk−1 ∈ ℂ for every k ∈ ℤ, where dk is the kth coefficient of the multiplication fg(x) as in (L3). Thus f ∈ 𝕃(g). We have (ii). Finally, f ∈ 𝕃(g) implies that fg(x) = ∑k∈ℤ dk x k ∈ 𝕃, where dk is as in (L3). Using (i), we have ̃ = A( ∑ d xk ) = ∑ d ξ fg k k (k,k+1] k∈ℤ
k∈ℤ
= ∑ (̃f ∗ g̃ )(k + 1) ξ(k,k+1] , k∈ℤ
so formula (7.22) is true, what completes the proof. Now, we introduce a functional multiplication on M0 . Definition 7.6.5. Let ϕ, ψ ∈ M0 be given. The functional multiplication of ϕ and ψ is defined to be the function ϕ ⊗ ψ of the form ϕ ⊗ ψ = ∑ (ϕ ∗ ψ)(k + 1)ξ(k,k+1] , k∈ℤ
provided the convolution ϕ ∗ ψ exists. Remark 7.6.6. We can summarize the properties of the canonical mapping A: 𝕃 → M0 as follows: (i) A(αf + βg) = αA(f ) + βA(g) = α̃f + βg̃ , for any f , g ∈ 𝕃, and α, β ∈ ℂ; (ii) A(fg) = ̃f ⊗ g̃ = A(f ) ⊗ A(g), where f ∈ 𝕃(g). For ϕ, ψ ∈ M0 , ϕ∗ψ, and ϕ⊗ψ may be different because ϕ∗ψ is always continuous on ℝ by Corollary 7.6.3 but ϕ ⊗ ψ may not. The theorem below claims that the integrals of ϕ ∗ ψ and ϕ ⊗ ψ are equal, if ϕ, ψ ∈ 𝕄1 . Theorem 7.6.7. Let ϕ, ψ ∈ 𝕄1 be given. Then both ϕ ∗ ψ and ϕ ⊗ ψ are integrable in the Lebesgue sense on ℝ and ∫(ϕ ∗ ψ) dt = ∫(ϕ ⊗ ψ) dt. ℝ
ℝ
7.6 Canonical mapping and multiplication of formal Laurent series  231
Proof. Since ϕ, ψ ∈ L1 (ℝ), it follows that ϕ ∗ ψ ∈ L1 (ℝ). Then formula (7.20) yields that k+1
∫(ϕ ∗ ψ) dt = ∑ ∫ (ϕ ∗ ψ)(t) dt k∈ℤ k
ℝ
k+1
= ∑ ∫ [(t − k) (ϕ ∗ ψ)(k + 1) − (t − k − 1) (ϕ ∗ ψ)(k)] dt k∈ℤ k
1 1 = ∑ [ (ϕ ∗ ψ)(k + 1) + (ϕ ∗ ψ)(k)] 2 2 k∈ℤ = ∑ (ϕ ∗ ψ)(k + 1). k∈ℤ
This also proves that ϕ ⊗ ψ is integrable on ℝ by Definition 7.6.5. Then we have ∫(ϕ ⊗ ψ) dt = ∑ (ϕ ∗ ψ)(k + 1) = ∫(ϕ ∗ ψ) dt. ℝ
k∈ℤ
ℝ
Corollary 7.6.8. Let ψ ∈ M0 be given. Setting up 𝕄(ψ) by 𝕄(ψ) = {ϕ ∈ M0 : ϕ ∗ ψ is continuous on ℝ}. Then 𝕄(ψ) = A(𝕃(A−1 (ψ))). In other words, by using Lemma 7.5.3 with ψ = g̃ , 𝕄(g̃ ) = A(𝕃(g)) for every
g ∈ 𝕃.
Proof. Write ψ = ∑n∈ℤ bn ξ(n,n+1] . Let ϕ = ∑ an ξ(n,n+1] n∈ℤ
be a Lebesgue measurable function in 𝕄(ψ). Then A−1 (ψ)(z) = ∑ bn z n , n∈ℤ
A−1 (ϕ)(z) = ∑ an z n . n∈ℤ
Therefore, A−1 (ψ)(z)A−1 (ϕ)(z) = ( ∑ bn z n )( ∑ an z n ) n∈ℤ
k
n∈ℤ
= ∑ dk z , k∈ℤ
where dk = ∑n∈ℤ an bk−n ∈ ℂ for every k ∈ ℤ, because ϕ ∈ 𝕄(ψ) and (ϕ ∗ ψ)(k + 1) = ∑ an bk−n ∈ ℂ for every n∈ℤ
k∈ℤ
232  7 Formal Laurent series by formula (7.21). Then A−1 (ϕ) ∈ 𝕃(A−1 (ψ)), so ϕ ∈ A(𝕃(A−1 (ψ))), what proves that 𝕄(ψ) ⊆ A(𝕃(A−1 (ψ))) or 𝕄(g̃ ) ⊆ A(𝕃(g)) for any g ∈ 𝕃. Now, we check that A(𝕃(g)) ⊆ 𝕄(g̃ ) for every g ∈ 𝕃. Let g ∈ 𝕃 be given. For any ϕ ∈ A(𝕃(g)), there exists f ∈ 𝕃 such that ϕ = A(f ) = ̃f and f ∈ 𝕃(g). By Theorem 7.6.4 and Corollary 7.6.3, we have ϕ ∈ 𝕄(g̃ ), and hence A(𝕃(g)) ⊆ 𝕄(g̃ ). Thus, 𝕄(g̃ ) = A(𝕃(g))
for every g ∈ 𝕃.
By Theorem 7.6.4, by using some wellknown facts in analysis concerning the convolution operator. We prove Theorem 7.3.15 again below. Theorem 7.6.9. Let g(z) = ∑n∈ℤ bn z n ∈ 𝕃p , where 1 ≤ p ≤ +∞. Then: (i) 𝕃q ⊆ 𝕃(g), if p > 1 and 1/p + 1/q = 1; (ii) 𝕃∞ ⊆ 𝕃(g), if p = 1; (iii) 𝕃1 ⊆ 𝕃(g), if p = +∞; (iv) if ϕ ∈ 𝕃p and ψ ∈ 𝕃q , where 1/p + 1/q = 1, then ϕψ ∈ 𝕃(g)
for every
g ∈ 𝕃1 .
Proof. Since g ∈ 𝕃p , it follows that g̃ ∈ A(𝕃p ) = 𝕄p . Taking f ∈ 𝕃q , we have ̃f ∈ A(𝕃q ) = 𝕄q . By the property of the convolution, ̃f ∗ g̃ is a continuous function, and hence f ∈ 𝕃(g) by Theorem 7.6.4. This is (i). Let g ∈ 𝕃1 . For any f ∈ 𝕃∞ we have that ̃f ∈ A(𝕃∞ ) = 𝕄∞ . By the property of the convolution, ̃f ∗ g̃ is a continuous function, so again by Theorem 7.6.4, f ∈ 𝕃(g). This is (ii). Now let g ∈ 𝕃∞ and let f ∈ 𝕃1 . Then ̃f ∈ A(𝕃1 ) = 𝕄1 . The convolution ̃f ∗ g̃ is a continuous function, so using Theorem 7.6.4, f ∈ 𝕃(g). We have proved (iii). ̃ ∈ A(𝕃 ) = 𝕄 and ψ ̃ψ ̃ ∈ A(𝕃 ) = 𝕄 . Then ϕ ̃ ∈ L (ℝ). Since g ∈ 𝕃 , For (iv), let ϕ p p q q 1 1 ̃ ̃ it follows that g̃ ∈ A(𝕃1 ) = 𝕄1 . Therefore, g̃ ∗ (ϕψ) ∈ L1 (ℝ), so by Theorem 7.6.4, ϕψ ∈ 𝕃(g). Theorem 7.6.10. 𝕃1 with the multiplication is a linear algebra. ̃ ψ ̃ ψ ̃∗ψ ̃ ∈ A(𝕃 ). Then ϕ, ̃ ∈ L (ℝ), so ϕ ̃ ∈ L (ℝ), what means, by Proof. Take ϕ, 1 1 1 ̃⊗ψ ̃ exists and Theorem 7.6.4 and Definition 7.6.5, that the functional multiplication ϕ ϕψ ∈ 𝕃1 . The next corollary describes the character of the subspace 𝕃1 .
7.7 Topological spaces of formal Laurent series  233
Corollary 7.6.11. For any f , g ∈ 𝕃1 (ℝ), we have ‖fg‖1 = A(fg)1 = ‖̃f ⊗ g̃ ‖1 . Proof. Let f , g ∈ 𝕃1 (ℝ) be given such that f (x) = ∑ an xn , n∈ℤ
g(x) = ∑ bn x n . n∈ℤ
Then f ∈ 𝕃(g) and fg(x) = ∑ dk xk ∈ 𝕃(ℝ), k∈ℤ
where dk = ∑n∈ℤ an bk−n ∈ ℝ for every k ∈ ℤ. Therefore, by Theorem 7.6.10, ‖fg‖1 = ∑ ∑ an bk−n . n∈ℤ k∈ℤ By Theorem 7.6.4 and Definition 7.6.5 we have ̃ = ∑ (̃f ∗ g̃ )(k + 1) ξ ̃ ̃. fg (k,k+1] = f ⊗ g k∈ℤ
Notice that formula (7.21) provides (̃f ∗ g̃ )(k + 1) = ∑ an bk−n . n∈ℤ
Applying these formulas, we get ‖fg‖1 = A(fg)1 = ‖̃f ⊗ g̃ ‖1 .
7.7 Topological spaces of formal Laurent series We defined a metric topology on the space of formal power series 𝕏(ℂ) with the ultrametric d induced by the valuation v on 𝕏 × 𝕏 in Section 2.4. We now extend those concepts to the formal Laurent series space 𝕃. Let f (z) = ∑n∈ℤ an z n be a formal Laurent series over ℂ. We define the order of f by ord(f ) = min{n : an ≠ 0}. As in the case of formal power series we also put ord(f ) = +∞,
if an = 0
for every n ∈ ℤ.
234  7 Formal Laurent series Remark 7.7.1. It is easy to see that the property ord(f + g) ≥ min{ord(f ), ord(g)}, which was proved in Section 2.4 for formal power series. It holds for formal Laurent series, too. However, the next two brief examples show that the property ord(fg) = ord(f ) + ord(g), in Proposition 1.1.5 is not necessarily true for the formal Laurent series. Example 7.7.2. Let f (z) = ∑n∈ℤ an z n , g(z) = ∑n∈ℤ bn z n be two formal Laurent series such that a1 = 1,
an = 0
for n ∈ ℤ \ {1},
b−1 = 1,
and
bn = 0
for n ∈ ℤ \ {−1}.
Obviously, we have ord(f ) = 1,
ord(g) = 1,
d0 = ∑ bm a−m = 1 m∈ℤ
and but dk = ∑ bm ak−m = 0 if k ≠ 0. m∈ℤ
So ord(fg) = 0 and, therefore, ord(fg) < ord(f ) + ord(g). Example 7.7.3. Let f (z) = 1 − z,
g(z) =
1 + 1. z
Then ord(f ) = 0 = ord(g) and
ord(fg) = 1,
which shows that ord(fg) > ord(f ) + ord(g). We define an ultrametric on 𝕃(ℂ). Definition 7.7.4. The mapping d : 𝕃(ℂ) × 𝕃(ℂ) → [0, 1], defined by the formula d(f (z), g(z)) = 2−ord(f (z)−g(z)) is an ultrametric on 𝕃(ℂ).
for any f , g ∈ 𝕃(ℂ),
7.7 Topological spaces of formal Laurent series 
235
It is not difficult to verify that d satisfies Definition 2.3.1. By Remark 7.7.1, for the formal Laurent series f and g we have − ord(f + g) ≤ max{− ord(f ), − ord(g)}, which yields that d(f , g) = 2− ord(f −g) = 2− ord(f −h+h−g) ≤ 2max{− ord(f −j),− ord(h−g)} = max{d(f , h), d(h, g)},
for any h ∈ 𝕃(ℂ), and hence the mapping d, defined in Definition 7.7.4 is an ultrametric on 𝕃(ℂ). The detail of the proof is similar to Proposition 2.4.6. Proposition 7.7.5. The ultrametric defined in Definition 7.7.4 is complete. Proof. Let (fn ) be a Cauchy sequence of elements of 𝕃(ℂ). Write fn (z) = ∑ bn (k)z k , k∈ℤ
n ∈ ℕ.
We shall construct a formal Laurent series f (z) = ∑n∈ℤ bn z n in 𝕃(ℂ) such that limn→∞ d(fn , f ) = 0. There exists an N0 ∈ ℕ such that d(fn , fm ) < 1
for all integers n, m ≥ N0 .
This means that ord(fn −fm ) ≥ 1, or bn (0) = bN0 (0) if n ≥ N0 . We define that b0 = bN0 (0), then d(fn , b0 ) ≤ 2−1 < 1
for n ≥ N0 .
There exists an N1 ∈ ℕ such that N1 > N0 , and d(fn , fm )
0 be given. Among the above chosen positive integers Nn , there exists a positive integer Nk such that 2−Nk < ε. Therefore, for n ≥ Nk we have ord(fn −f ) ≥ Nk +1, so d(fn , f ) = 2−ord(fn −f ) ≤ 2−(Nk +1) < ε. It means that fn → f as n → ∞, what proves our claim. Remark 7.7.6. It appears that unfortunately the product (L3) defined in Definition 7.3.8 is not continuous, if one considers the topology induced by the ultrametric defined in Definition 7.7.4. Indeed, let E = {(f , g) : f , g, fg ∈ 𝕃(ℂ)}. The mapping (f , g) → fg, defined on E, is not continuous. For example, let fn (z) = ∑ bnk z k , k∈ℤ
gn (z) = ∑ ank z k , k∈ℤ
n ∈ ℕ,
where 1,
bnk = {
0,
k = 1 or k = n,
1,
ank = {
otherwise,
0,
k = −1 or k = −n, otherwise,
and let f (z) = ∑k∈ℤ bk z k , g(z) = ∑k∈ℤ ak z k , where 1,
bk = {
0,
k = 1,
1,
otherwise,
ak = {
0,
k = −1,
otherwise.
Obviously, we have ord(f − fn ) → +∞
and
ord(g − gn ) → +∞
(n → +∞),
so d(f , fn ) → 0
and d(g, gn ) → 0
as n → +∞.
7.7 Topological spaces of formal Laurent series 
237
Let n ∈ ℕ be given and write fn gn = ∑k∈ℤ dnk z k , fg = ∑k∈ℤ dk z k , where dnk = ∑ anm bnk−m m∈ℤ
and dk = ∑ am bk−m . m∈ℤ
It is clear that d0 = 1, and dn0 = ∑ anm bn−m = an−1 bn1 + an−n bnn = 2. m∈ℤ
So ord(fg − fn gn ) = 0 and, therefore, d(fn gn , fg) = 1 for n ∈ ℕ, what means that fn gn does not approach fg, as n → +∞. As in the case of the space of formal power series we can also use wellknown fixed point theorems to deal with some operators defined on the space of formal Laurent series. The next example is an extension of Example 2.4.10. Example 7.7.7. Let us define the mapping φ: 𝕃(ℂ) → 𝕃(ℂ) by the formula ∞
φ(f (z)) = ∑ a−n+1 z −n + n=2
∞ 1 + 1 + ∑ an−1 z n , z n=1
n
for f (z) = ∑n∈ℤ an z ∈ 𝕃(ℂ). Let g(z) = ∑n∈ℤ bn z n ∈ 𝕃(ℂ). We have ord(φ(f (z)) − φ(g(z))) ∞
∞
n=2
n=1
= ord( ∑ (a−n+1 − b−n+1 )z −n + ∑ (an−1 − bn−1 )z n ) = ord(z( ∑ (an − bn )z n )) n∈ℤ
= ord(f (z) − g(z)) + 1, so
1 d(φ(f (z)), φ(g(z))) ≤ d(f (z), g(z)). 2
Thus φ is a contraction defined on the complete metric space and, therefore, it possesses a unique fixed point. We should remember another topological space formed by a subset of 𝕃, denoted by ℒp (β). The Banach space ℒp (β) and its formal power series version ℋp (β) were introduced in Section 6.3 although at that moment we did not know the formal Laurent series space 𝕃 very well. Now it is suitable for us to introduce some more properties of ℒp (β). Let us first recall some necessary definitions. Let β: ℤ → (0, +∞) be a mapping from the set of integers to the set of positive real numbers such that β(0) = 1. For 1 ≤ p < +∞ and f (z) = ∑n∈ℤ ̂f (n)z n in 𝕃(ℂ), the mapping ‖ ⋅ ‖β,p : 𝕃(ℂ) → ℝ, defined by 1/p
p ‖f ‖β,p = ( ∑ βp (n)̂f (n) ) n∈ℤ
= ‖ ⋅ ‖β
238  7 Formal Laurent series is a ℝvalued function if the righthand side is a real number. We call the sequence {β(n)}n∈ℤ the βsequence as we did in Section 6.3. The mapping ‖ ⋅ ‖β defined above is a norm on ℒp (β). The next propositions are similar to the Propositions 6.3.2, 6.3.3, and 6.3.6. The proof for each proposition here is similar to the proof of the analogous proposition in the Section 6.3 and, therefore, we omit those proofs. Proposition 7.7.8. Let fk (z) = ∑ ̂fk (n)z n ∈ 𝕃(ℂ) n∈ℤ
be given such that ̂fk (n) = δk (n) for k ∈ ℤ. Then {fk }k∈ℤ forms a base for ℒp (β), 1 ≤ p, and ‖fk ‖β = β(k). Proposition 7.7.9. The space ℒp (β) is a Banach space under the norm ‖ ⋅ ‖β . Definition 7.7.10. Let X be a normed space. The dual space of X or conjugate space of X is the normed space B(X) of all bounded linear functionals on X with the operator norm. This space is denoted by X ∗ . Definition 7.7.11. Let T ∈ B(X) be given where X is a Banach space. For each x ∈ X, the orbit of x under T, denoted by orb(x, T), is the set of images of x under the successive iterates of T, or orb(T, x) = {x, Tx, T 2 x, . . . }, where T 2 x = T ∘ T(x) = T(Tx). The vector x is called hypercyclic for T if orb(T, x) is dense in X. The hypercyclic operator is the one that has a hypercyclic vector. If orb(T, x) is dense in X, we also say that the vector x is Tuniversal [35]. It is clear that Mz is a linear operator on the Banach space ℒp (β) and Mz ∈ B(ℒp (β)) if it is bounded. The following proposition will help us to have a theorem late. Proposition 7.7.12 ([35]). Let X be a separable Fspace and let T ∈ B(X) be given. If T has a universal vector, then it has a dense Gδ set of universal vectors. Proof. An Fspace is a topologically complete metric space. A subset E in a topological space is called Gδ if E is the intersection of a sequence of open subsets of this space. Let {yk }k∈ℕ ⊆ X be a countable dense subset. Suppose that T has a universal vector, it follows that orb(T, x) is dense in X for each hypercyclic vector x. Then, for any N, j, k ∈ ℕ, we set F(j, N, k) = {x ∈ X : T n x − yj < 1/k
for some n ≥ N}.
7.7 Topological spaces of formal Laurent series  239
Since T ∈ B(X), the set T −n (B1/k (yj )) is open, where B1/k (yj ) = {x ∈ X : ‖Tx − yj ‖ < 1/k}. Then each of such F(j, N, k) is a union of sets T −n (B(yj , 1/k)), which are open by the continuity of T, is itself open. Let E be the set of all Tuniversal vectors of X, then E=
⋂ F(j, N, k),
j,N,k∈ℕ
which is a Gδ subset of X. If x is a universal vector, then so is every T n x in the dense orbit orb(T, x). This completes the proof. The following properties of ℒp (β) can be found in [103] and the lemma was slightly modified by the author. Theorem 7.7.13. Let Mz be bounded on ℒp (β). If Mz is hypercyclic, then for all ϵ > 0 and m ∈ ℕ we have β(j + n) < ϵβ(j)
and
β(j − n) < ϵβ(j)
for all −m < j < m and all n > 2m. α Proof. Let ϵ > 0 and m ∈ ℕ be given. Choose α > 0 such that 1−α < ϵ. Since Mz is hypercyclic on the Banach space ℒp (β), Proposition 7.7.12 yields that the set of hy1 percyclic vectors for Mz is dense in ℒp (β). Then, for ∑j≤m β(j) fj , there exists a vector
x = ∑j
̂ x(j) f β(j) j
∈ ℒp (β) for Mz such that
1 f < α, x − ∑ β(j) j β j≤m p where {fk }k∈Z is a base for ℒp (β) which can be formed as the base {fk }∞ k=0 for ℋ (β) in Proposition 6.3.2. We write
fk (z) = ∑ fk̂ (n)z n , n∈ℤ
fk̂ (n) = δk (n)
for all k ∈ ℤ. By Proposition 6.3.2 and a simple extension from ℕ to ℤ for ‖fk ‖β , we have ‖fk ‖β = β(k) for all k ∈ ℤ. Therefore, we have p fj fj p 1 ̂ ̂ − 1) fj = ∑ x(j) + ∑ (x(j) x − ∑ β(j) β j>m β(j) j≤m β(j) β j≤m p ̂ p ̂ < αp . = ∑ x(j) + ∑ 1 − x(j) j>m
j≤m
240  7 Formal Laurent series Then ̂ x(j) < α for j > m and ̂ x(j) > 1 − α for j ≤ m.
(∗)
Since orb(Mz , x) is dense in ℒp (β), there exists n > 2m such that 1 n f < α. Mz x − ∑ β(j) j β j≤m Then fj p fj p fj+n n = ∑ x(j) < αp . ̂ − ∑ Mz x − ∑ β β(j) β(j) β(j) β j j≤m j≤m If j ≤ m, then n + j > m for n > 2m. Then the above inequality becomes p x(j) ‖fj ‖pβ ̂ p p ∑ ‖fj+n ‖β + ∑ p < α , β(j) β (j) j j≤m because fj+n ∉ {f−m , f−m+1 , . . . , f0 , f1 , . . . , fm }. Then x(j) n ̂ fj < α Mz β(j)
̂ β(n + j) 2m.
f
j Since ‖ β(j) ‖β = 1, it follows that
x(j ̂ − n)β(j) − β(j − n) fj fj n ̂ − n) = x(j fj−n − ⋅ Mz β(j − n) β(j) β β(j − n) β(j) β β(j) ̂ − 1 < α. = x(j − n) β(j − n) Thus, for j ≤ m we have ̂ β(n + j) 1 − α. x(j − n) β(j − n)
̂ − n) < α by (∗). Applying (∗), we If j ≤ m, then j − n < −m for n > 2m, and hence x(j have β(n + j) < Since
α 1−α
α α β(j) and β(j − n) < β(j). 1−α 1−α
< ϵ, the proof is completed.
7.7 Topological spaces of formal Laurent series  241
Lemma 7.7.14. Let Mz be bounded on ℒp (β). Assume that for δ > 0 and g, h ∈ span{fj : j ≤ m} there exists n ∈ ℕ large enough and u ∈ span{fj : j + n ≤ m} such that and Mzn h < δ.
‖u‖ < Mzn u − g < δ Then Mz is hypercyclic.
Proof. It is clear that ‖Mz ‖ > 0 because ‖fj ‖β = β(j) > 0 and β(0) = 1. Let F = { ∑ aj fj : aj ∈ ℂ, k ∈ ℕ}. j≤k
Then F is dense in ℒp (β), and then there exists a countable subset E ⊆ F such that E is dense in ℒp (β). Write E = {gk : k ∈ ℕ}. We now construct a sequence {qk } ⊆ ℒp (β) such that lim Mznk qk − gk β = 0,
k→∞
where {nk } ⊆ ℕ ∪ {0} is a rapidly increasing sequence to be specified as follows. Let ϵ > 0 be given. Set η = max{1, ‖Mz ‖}. Let n1 = 0 and q1 = g1 . By the assumption, for δ = 2−2 ηϵ,
g = g2
and
h = q1 ,
we have n2 ∈ ℕ with n2 > n1 and then we have q2 = u ∈ span{fj : j + n2  ≤ 2} such that ‖q2 ‖β < Mzn2 q2 − g2 β < δ
and
n2 Mz q1 β < δ.
Next, put δ = 2−3 η−n2 ϵ,
g = g3 ,
and h = q1 + q2 ∈ span{fj : j ≤ 3}.
By the assumption again, there exists n3 ∈ ℕ with n3 > n2 and q3 ∈ span{fj : j+n3  ≤ 3} such that ‖q3 ‖ < Mzn3 q3 − g3 β < δ
and Mzn3 (q1 + q2 )β < δ.
Inductively, we can have an increasing sequence {nk }∞ k=1 ⊆ ℕ ∪ {0} and a sequence {qk } such that qk ∈ span{fj : j + nk  ≤ k},
242  7 Formal Laurent series ‖qk ‖β ≤ 2−k η−nk−1 ϵ, nk −k Mz qk − gk β < 2 ϵ, and nk −k Mz (q1 + q2 + ⋅ ⋅ ⋅ qk−1 )β < 2 ϵ, for all k ≥ 2. Now let q = ∑∞ j=1 qj and we show that q is a hypercyclic vector for Mz by n showing that {Mz k q}n∈ℕ is dense in ℒp (β). Applying the above inequalities, we have ∞ n k−1 nk n n Mz q − gk β = [Mz k ( ∑ qj ) + Mz k qk + ∑ Mz k qj ] − gk j=1 j=k+1 β ∞ k−1 ≤ Mznk ( ∑ qj ) + Mznk qk − gk β + ∑ Mznk qj j=1 β j=k+1 β ∞
≤ 2−k ϵ + 2−k ϵ + ‖Mz ‖nk ∑ ‖qj ‖β j=k+1
≤2
−k+1
nk −nk
ϵ + 2 ‖Mz ‖ η −k
ϵ.
Since ‖Mz ‖nk η−nk ≤ 1, it follows that nk −k Mz q − gk β < 3 ⋅ 2 ϵ. n
Since {gk }k∈ℕ is dense in ℒp (β), it follows that {Mz k q}k∈ℕ is also dense in ℒp (β). That is, q is a hypercyclic vector for Mz . Theorem 7.7.15. Let Mz be bounded on ℒp (β). If for all ϵ > 0 and m ∈ ℕ, there exists a large n ∈ ℕ such that β(j + n) < ϵβ(j)
and
β(j − n) < ϵβ(j)
for all j ≤ m, then Mz is hypercyclic. Proof. Let ϵ > 0 and m ∈ ℕ be given. Suppose that there is n > 2m such that β(j + n) < ϵβ(j) and
β(j − n) < ϵβ(j)
for all j ≤ m. For g and h in span{fj : j ≤ m}, we have p p n p n ̂ ̂ M h = M ( h(j)f ) = h(j)f ∑ ∑ z β z j j+n β β j≤m j≤m p
̂ p p β(j + n) ) = ∑ h(j) β(j) ( β(j) j≤m
≤ sup( j≤m
p
β(j + n) ) ‖h‖pβ , β(j)
7.7 Topological spaces of formal Laurent series  243
and similarly p p β(j − n) −n p ̂ ≤ sup ( g(j)f = M ) ‖g‖pβ . g ∑ z β j−n β(j) j≤m β j≤m
So we have n Mz hβ ≤ ϵ‖h‖β
and Mz−n g β ≤ ϵ‖g‖β .
By setting u = Mz−n g, we see that the conditions in the hypothesis of Lemma 7.7.14 are satisfied. Thus, Mz is hypercyclic. Let (ℒp (β))∗ = ℒq (βp/q ), where
1 p
f (z) = ∑ f ̂(n)z n ∈ ℒp (β) n∈ℤ
+
1 q
= 1. For
n ̂ and g(z) = ∑ g(n)z ∈ (ℒp (β)) , ∗
n∈ℤ
it is welldefined that p ̂ ⟨f , g⟩ = ∑ f ̂(n)g(n)β (n), n∈ℤ
̂ ̂ where g(n) is the complex conjugate of g(n). Corollary 7.7.16. There exists a sequence {β(n)}n∈ℤ such that Mz and Mz∗ are both hypercyclic on ℒp (β) and on ℒq (βp/q ), respectively. Proof. Let {β(n)}n∈ℤ be a sequence such that: (i) c1 ≤ β(n) ≤ c2 for c1 > 0, c2 > 0, (ii) for every ϵ > 0 and m ∈ ℕ, there exists n ∈ ℕ such that β(j − n) < ϵβ(j) and β(j + n) < ϵβ(j), for all j ≤ m. Actually (i) ensures that the operators Mz and Mz∗ are bounded on ℒp (β) and on ℒq (βp/q ), respectively. The condition (ii) implies that Mz is hypercyclic on ℒp (β) by Theorem 7.7.15. We may represent Mz∗ as multiplication by z on ℒq (β1p/q ) with β1 (n) = β(−n) for all n ∈ ℤ. Under the condition (ii), we have β1 (j − n) = β(n − j) < ϵβ(−j) = ϵβ1 (j),
and
β1 (j + n) = β(−j − n) < ϵβ(−j) = ϵβ1 (j),
for all j ≤ m. Then Mz is hypercyclic on ℒp (β) by Theorem 7.7.15 and this implies that Mz∗ is also hypercyclic on ℒq (βp/q ).
244  7 Formal Laurent series Corollary 7.7.17. If β(0) = 1 and β(n) = β(−n + 1)−1 for every integer n ≠ 1, then Mz is not hypercyclic on Lp (β). Proof. Assume that Mz is hypercyclic, by letting j = 0 in the relations of Theorem 7.7.13, for any ϵ > 0 we have β(n) < ϵ
and β(−n) < ϵ,
for n large enough. This implies that β(n + 1)−1 = β(−n) < ϵ. Then β(n + 1) 1 > 2 β(n) ϵ
for
n ∈ ℕ large enough
and so ‖Mz ‖ > ϵ12 . Then ‖Mz ‖ is unbounded because ϵ > 0 is arbitrary, which is a contradiction to that Mz ∈ B(ℒp (β)).
8 Iteration and iterative roots The functional equation f [f (x)] = g(x) was investigated by Pfeiffer [63] as early as 1918. Pfeiffer found that there was an analytic g for which this functional equation has no analytic solution, and every formal solution is divergent for all x except x = 0. Please review Theorem 4.2.2 in this book. If f is a solution of this functional equation, or more generally, a solution of f [n] (x) = g(x),
n ∈ ℕ,
(8.1)
where f [n] = f ∘ f [n−1] , whether g is analytic or not, say that g is a formal power series, then f is called an nth iterative root series of g or just an nth iterative root of g. If g is a unit and f is a solution of the functional equation (8.1), then ord(f ) = 0 is the only possibility for f . This case is very complicated. If g ∈ 𝕏0 (ℂ) with ord(g) > 1, then, by Theorem 1.2.5, f must satisfy the condition n
(ord(f )) = ord(g). This case is also not simple. We look for the case that g ∈ 𝔻(ℂ), an almost unit formal power series. After we investigate this case, we will return to the other two cases. Iterative conjugacy plays an important role in iterations and iterative roots. Section 8.1 provides some useful results about iterative conjugacy. The canonical form for iterative conjugacy is a very useful tool in the studies of conjugacy. The contributions of Scheinberg and Kimura are collected in this section. The author expanded on their beautiful work in order to make those theorems easier for our readers to understand. Section 8.2 introduces the iteration, mainly the iteration for formal power series in 𝔻. The recurrence formula for f [n] is not difficult, but it is very complicated to find any nonrecurrence formula, even for some particular f . Section 8.3 discusses the iterative roots for formal power series. Schröder’s problem, a wellknown problem in functional analysis, was developed from Schröder’s equation ψ(mζ ) = F(ψ(ζ )). This problem will be discussed in Section 8.4 with respect to the formal power series. A proof of Cayley’s lemma will be provided in this section with an application to formal power series. We also show the close relationship, mainly investigated by M. Cohen, between Schröder’s problem and formal power series with finite composition order.
8.1 Conjugacy of formal power series in 𝔻 We would like to indicate that all formal power series discussed in this section are almost units in 𝔻(ℂ) unless the series is clearly specified. https://doi.org/10.1515/9783110599459008
246  8 Iteration and iterative roots Iterative conjugacy or equivalency provides us some alternative way to investigate the iteration and iterative roots. The work below can be found in [88]. The technique Scheinberg used is very useful. Definition 8.1.1. Let f , ϕ ∈ 𝕏 = 𝕏(ℂ) be given. If there is a formal power series h ∈ 𝕏 such that h[−1] ∘ f ∘ h = ϕ, we say that f and ϕ are iterative conjugate or iterative equivalent, or just conjugate, and denote it by f ∼ ϕ. It is known that h ∈ 𝔻(ℂ) if h[−1] exists. If a, b ∈ 𝔻 such that a(z) = a1 z + a2 z 2 + ⋅ ⋅ ⋅ ,
b(z) = b1 z + b2 z 2 + ⋅ ⋅ ⋅ .
We say that a and b have the same canonical form at the level m if aj = bj = 0
for
1 < j < m,
and
am bm ≠ 0,
for some integer m ≥ 2. It is clear that if f and ϕ are conjugate, say that h[−1] ∘ f ∘ h = ϕ for some h ∈ 𝔻(ℂ), then ϕ[n] = h[−1] ∘ f [n] ∘ h for all n ∈ ℕ. Thus, in order to study f [n] , it is convenient to find a formal power series h such that the composite formal power series ϕ = h[−1] ∘ f ∘ h has the iterates of a simple form. Some developments about such simple forms will be introduced in the next section. The following property of the iterative conjugacy is trivial but it is worth knowing. If a, b ∈ 𝔻 such that a ∼ b and a(z) = a1 z + a2 z 2 + ⋅ ⋅ ⋅ ,
b(z) = b1 z + b2 z 2 + ⋅ ⋅ ⋅ ,
then a1 = b1 . A seminorm on a vector space X is a real valued function p on X such that the following conditions are satisfied by all members x and y of X and each scalar α: (i) p(αx) = αp(x); (ii) p(x + y) ≤ p(x) + p(y). Let 𝒜 be an algebra over real or complex field. Suppose that 𝒜 is equipped with a Hausdorff topology defined by a family {p} of seminorms and that 𝒜 is complete in the sense: If a sequence {an } ⊆ 𝒜 is Cauchy relative to each of the seminorms, then there exists a ∈ 𝒜 such that an → a in each seminorm.
8.1 Conjugacy of formal power series in 𝔻

247
Definition 8.1.2. Let 𝒜 be an algebra over the real or complex field. A derivation of 𝒜 is a linear operator D : 𝒜 → 𝒜 such that D(ab) = (Da)b + a(Db) for all
a, b ∈ 𝒜.
An automorphism is an invertible linear operator T : 𝒜 → 𝒜 such that T(ab) = (Ta)(Tb) for all a, b ∈ 𝒜. As usual, we denote by B(𝒜) the family of all linear operators on 𝒜 that are bounded under each seminorm For example, 𝕏(ℂ) is such an algebra and taking formal derivative for all elements of 𝕏(ℂ) is such a derivation. We provide another example. Example 8.1.3. Let 𝒜 be the set of all analytic functions on the unit disc with topology of uniform convergence on compact subsets. The family of seminorms can be the set {pr : 0 < r < 1}, where pr (a) = sup{a(z) : z ≤ r}, for a ∈ 𝒜. The differential operator D = d/dz is a continuous derivation of 𝒜, but does not belong to B(𝒜). Lemma 8.1.4. If D is a derivation belonging to B(𝒜), then eD is an automorphism belonging to B(𝒜) where eD = 1 +
Dn D D2 + + ⋅⋅⋅ + + ⋅⋅⋅. 1! 2! n!
n
D Proof. The operator eD = ∑∞ n=0 n! is welldefined and belongs to B(𝒜) because of completeness of 𝒜 and boundedness of each seminorm; of course, the inverse to eD is e−D . Applying Leibniz’s formula to
D(ab) = (Da)b + a(Db) with induction we have n n n! Dn (ab) = ∑ ( )(Dn−k a)(Dk b) = ∑ (Dm a)(Dk b), k m!k! k=0 k+m=n
for each n ∈ ℕ ∪ {0}. Then Dn (ab) Dm a Dk b = ∑ ( )( ). n! m! k! k+m=n
248  8 Iteration and iterative roots An algebra is also a ring [95] and, therefore, we may consider the formal power series Dm a m z m! m=0 ∞
ϕ(z) = ∑
and
Dk b k z . k! k=0 ∞
ψ(z) = ∑
Then the Cauchy product of ϕ and ψ yields that ∞ Dk b k Dm a m z )( ∑ z ) m! k! m=0 k=0 ∞
ϕ(z)ψ(z) = ( ∑ ∞
= ∑[ ∑ ( n=0 k+m=n
∞ Dm a Dk b Dn (ab) n )( )]z n = ∑ z . m! k! n! n=0
Since D ∈ B(𝒜), it follows that both ϕ and ψ are convergent for all z ∈ ℂ. Take z = 1, we have ∞ ∞ Dk b Dm a Dn (ab) )( ∑ )= ∑ = eD (ab). m! k! n! m=0 n=0 k=0 ∞
eD (a)eD (b) = ( ∑
Thus, eD is an automorphism and belongs to B(𝒜). Recall the general formal logarithm Ln defined in Section 6.4 which, of course, is not a linear operator on 𝕏. We are now introducing a transformation on 𝒜, the set of all analytic function on the unit disk, that has a similar form as Ln. The notation log is used to distinct it from Ln and ln although they are related. Lemma 8.1.5 ([88]). Let T ∈ B(𝒜) be an automorphism of 𝒜. Suppose that the series (−1)n−1 (T − I)n , n n=1 ∞
D = log T = log(I + (T − I)) = ∑
converges absolutely in each seminorm. Then D ∈ B(𝒜) and eD = T, and D is a derivation of 𝒜. Proof. By hypothesis on convergence, D ∈ B(𝒜) and eD = T, since 1 1 1 log(1 + x) = x − x 2 + x 3 − x 4 + ⋅ ⋅ ⋅ 2 3 4 whenever the series rearrangements are permissible. It remains to show that D is a derivation of 𝒜. Put S = T − I. For any a, b ∈ 𝒜 we have ab + S(ab) = T(ab) = (Ta)(Tb) = (a + Sa)(b + Sb) = ab + (Sa)b + a(Sb) + (Sa)(Sb).
8.1 Conjugacy of formal power series in 𝔻

249
Then S(ab) = (Sa)b + a(Sb) + (Sa)(Sb). Since S is a linear operator and S2 = S ∘ S, it follows that S2 (ab) = (S2 a)b + a(S2 b) + 2(Sa)(Sb) + 2(S2 a)(Sb) + 2(Sa)(S2 b) + (S2 a)(S2 b).
Then S2 (ab) − (S2 a)b − a(S2 b) = 2(Sa)(Sb) + 2(S2 a)(Sb)
+ 2(Sa)(S2 b) + (S2 a)(S2 b),
or S2 (ab) − (S2 a)b − a(S2 b) = ∑ cm,n (Sm a)(Sn b), m,n∈ℕ
(8.2)
with 1 ≤ m, n ≤ 2 and m + n ≥ 2, where cm,n are some real numbers. For any k ∈ ℕ, k ≥ 2, inductively we have Sk (ab) − (Sk a)b − a(Sk b) = ∑ cm,n (Sm a)(Sn b), m,n∈ℕ
(8.21 )
where m, n ∈ ℕ, m, n ≤ k, and m + n ≥ k, for some real numbers cm,n . Since D=S−
S2 S3 + − ⋅⋅⋅, 2 3
it follows that
D(ab) − (Da)b − a(Db) = [S(ab) − (Sa)b − aS(b)] 1 − [s2 (ab) − (S2 a)b − a(S2 b)] + ⋅ ⋅ ⋅ 2 (−1)k−1 k + [S (ab) − (Sk a)b − a(Sk b)] + ⋅ ⋅ ⋅ k = ∑ cm,n (Sm a)(Sn b) = Σ(a, b) m,n∈ℕ
by applying (8.21 ), where cm,n are certain real numbers. The proof will be completed by showing that cm,n = 0
for all m, n.
(8.3)
By induction on q = m + n, let us assume that cm,n = 0 for all m + n < q. We start the induction with q = 2 because m, n ∈ ℕ. The terms cm,n (Sm a)(Sn b) with m + n = 2 in
Σ(a, b) can only be occurred in S −
S2 2
D=S−
inside the expression S2 S3 S4 + − + ⋅⋅⋅. 2 3 4
250  8 Iteration and iterative roots By equation (8.2) and the equation for S(ab), the only term contributes to m + n = 2 in S is (Sa)(Sb), and in S2 is 2(Sa)(Sb), then ∑ cm,n (Sm a)(Sn b) = (Sa)(Sb) −
m+n=2
2(Sa)(Sb) = 0, 2
or cm,n = 0,
if
m + n = 2.
Then (8.3) is true for q = 2. For the inductive step, we may choose an algebra, automorphism, and elements a and b so that Σ(a, b) will reduce to Σ(a, b) = ∑ cm,n (Sm a)(Sn b) m+n=q
and the terms (Sm a)(Sn b) will be linearly independent. Let 𝒜 be the algebra of all polynomials F(x, y) in two variables x and y, modulo the ideal spanned by the monomials xm+1 yn+1 for m, n ≥ −1 and m + n > q. Since 𝒜 is finite dimensional, it is superfluous to consider seminorms. Let D be the derivation x2 (𝜕/𝜕x) + y2 (𝜕/𝜕y). D is nilpotent because Dq+2 = 0. Let T be the automorphism eD , then S = eD − I is nilpotent since D is. Then the identity x = log(1 + (ex − 1)) = (ex − 1) −
(ex − 1)2 + ⋅⋅⋅ 2
holds when we put x = D and ex − 1 = T − I = S. Now Dk x = k!xk+1 and Sm = Dm + (higher powers of D). Thus, Sm x = m!xm+1 + (higher powers of x) n
S y = n!y
n+1
and
+ (higher powers of y).
Then Σ(x, y) = ∑m+n=q cm,n m!n!x m+1 yn+1 , because cm,n = 0 for m+n < q by the inductive hypothesis and x m+1 yn+1 = 0 for m + n > q due to modulo. However, Σ(x, y) = D(xy) − D(x)y − xD(y) = 0, because D is a derivation. Since the terms (Sm a)(Sn b) are set to be linearly independent, cm,n = 0 the induction is completed.
for m + n = q,
8.1 Conjugacy of formal power series in 𝔻
 251
Henceforth, let 𝒜 be the algebra of all formal power series such as a(z) = n ∑∞ n=1 an z , or 𝒜 = 𝔻(ℂ). By Definition 8.1.2 and the induction, we may obtain that Dz n = nz n−1 Dz,
n ∈ ℕ,
for every derivation D, and hence D(z n 𝒜) ⊆ z n 𝒜. For every automorphism T of 𝒜, we also have T(z n 𝒜) ⊆ z n 𝒜 since z n 𝒜 = 𝒜n+1 . Thus, D and T belong to B(𝒜) and induce operators on 𝒜/(z n 𝒜). We call an operator U : 𝒜 → 𝒜 quasinilpotent if U(z n 𝒜) ⊆ z n+1 𝒜. The next lemma indicates those observations. The details of the proof can be found in [88]. Lemma 8.1.6. An automorphism T of 𝒜 is the exponential of a quasinilpotent derivation D if and only if Tz = z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ . In this case, D is unique among quasinilpotent operators. Moreover, Tz − z and Dz have the same lowest degree term. We now start discussing the conjugacy of formal power series. Proposition 8.1.7. If a ∈ 𝔻(ℂ) with a(z) = a1 z + a2 z 2 + ⋅ ⋅ ⋅ and an1 ≠ 1 for all n ∈ ℕ, then a(z) ∼ a1 z. We may also say that a(z) and a1 z have the same canonical form at the level 1. Proof. We prove it inductively. Assume that we have a(z) ∼ a1 z + an z n + an+1 z n+1 + ⋅ ⋅ ⋅ , for some n ∈ ℕ, n ≥ 2. Let gn (z) = z + bn z n where bn will be determined. If we write gn[−1] as gn[−1] (z) = c1 z + c2 z 2 + ⋅ ⋅ ⋅ , we may easily compute that c1 = 1, cj = 0 for 2 ≤ j < n, and cn = −bn , and so on, or g [−1] (z) = z − bn z n + pn+1 (z) where pn+1 is a formal power series with ord(pn+1 ) ≥ n + 1. Then gn[−1] ∘ a ∘ gn = gn[−1] ∘ (a1 z + an z n + ⋅ ⋅ ⋅) ∘ gn
= gn[−1] ∘ [a1 z + (an + a1 bn )z n + ⋅ ⋅ ⋅]
= a1 z + (an + a1 bn − bn an1 )z n + pn+1 (z),
252  8 Iteration and iterative roots where pn+1 is a formal power series with ord(pn+1 ) ≥ n + 1. Put bn = an (an1 − 1)−1 where an1 − 1 ≠ 0, we have a(z) ∼ a1 z + pn+1 (z), and we complete the induction. Proposition 8.1.8. Let a ∈ 𝕏0 (ℂ) be given such that a(z) = ak z k + ak+1 z k+1 + ⋅ ⋅ ⋅ ,
k ≥ 2,
ak ≠ 0,
and aj = 0 for j < k, k ∈ ℕ. Then a(z) ∼ z k . Proof. Let g(z) = cz for some c ∈ ℂ. Then (g [−1] ∘ a ∘ g)(z) = ak ck−1 z k + pk+1 (z), where pk+1 ∈ 𝕏0 with ord(pk+1 ) ≥ k + 1. Put ck−1 = a−1 k we have a(z) ∼ z k + pk+1 (z). We now aim to eliminate inductively all the terms of pk+1 (z). Say that a(z) ∼ z k + pk+n (z)
and pk+n (z) = a z n+k + pn+k+1 (z),
where pn+k+1 ∈ 𝕏0 with ord(pn+k+1 ) ≥ n + k + 1, for some n ∈ ℕ. Let g(z) = z + bz n+1 ,
and then g [−1] (z) = z − bz n+1 + ⋅ ⋅ ⋅ .
Then (g [−1] ∘ a ∘ g)(z) = g [−1] ∘ (z k + a z n+k + ⋅ ⋅ ⋅) ∘ g(z)
= g [−1] ∘ [z k + (kb + a )z n+k + ⋅ ⋅ ⋅] = z k + (kb + a )z n+k + qn+k+1 (z),
where qn+k+1 is a formal power series with ord(qn+k+1 ) ≥ n + k + 1. Choose b = −a /k, we have a(z) ∼ z k + pk+n+1 (z). We complete the induction and also complete the proof. We would like to recall the Böttcher’s equation which was introduced in Section 4.2.
8.1 Conjugacy of formal power series in 𝔻
 253
Remark 8.1.9. A popular Böttcher’s equation ((3.18) of [44]) is H(z 2 ) = f (H(z)) where f is given such that f (0) = f (0) and f (0) ≠ 0. This equation is equivalent to z 2 = H [−1] ∘ f ∘ H(z), or f (z) ∼ z 2 , if H [−1] exists. Proposition 8.1.8 not only ensures the solution of this particular Böttcher’s equation but also generalized this equation to H(z k ) = f (H(z)),
k ∈ ℕ,
k ≥ 2.
Proposition 8.1.10. Let a(z) = z + an z n + an+1 z n+1 + ⋅ ⋅ ⋅, where an ≠ 0. Then a(z) ∼ z + z n + cz 2n−1 , for some constant c. Proof. Let h(z) = bz for some b ≠ 0, it follows that h[−1] = b1 z. Let α = h[−1] ∘ a ∘ h with bn−1 = a−1 n , we have 1 a(z) ∼ α(z) = ( )a(bz) = z + z n + an+1 z n+1 + pn+2 (z), b where pn+2 ∈ 𝕏 such that ord(pn+2 ) ≥ n + 2. If n > 2, let g2 (z) = z + b2 z 2 , we want to choose the value of b2 so that g2[−1] ∘ α ∘ g2 (z) = z + z n + an+2 z n+2 + pn+3 (z) = β(z), without the term of z n+1 in the expression of β, where pn+3 ∈ 𝕏 with ord(pn+3 ) ≥ n + 3. This can be done by letting α∘g2 = g2 ∘β, and then determining he value of b2 . Actually, α ∘ g2 (z) = z + b2 z 2 + z n + (an+1 + nb2 )z n+1 + pn+2 (g2 (z)), and g2 ∘ β(z) = z + z n + an+2 z n+2 + pn+3 (z) + b2 [β(z)]
2
= z + b2 z 2 + z n + 2b2 z n+1 + an+2 z n+2 + pn+3 (z).
Letting α ∘ g2 (z) = g2 ∘ β(z) and equating the term of z n+1 in the both sides, we have that an+1 + nb2 = 2b2 . Since n > 2, by letting b2 = an+1 (2 − n)−1 we have g2[−1] ∘ α ∘ g2 (z) = z + z n + an+2 z n+2 + pn+3 (z) for some pn+3 ∈ 𝕏 with ord(pn+3 ) ≥ n + 3.
254  8 Iteration and iterative roots The formula g2[−1] ∘ α ∘ g2 (z) = β(z) can also be understood by α ∘ g2 (z) = g2 ∘ (z + z n )
(modulo z n+2 ).
If n > 3, we eliminate an+2 by conjugation with g3 = z + b3 z 3 , where b3 = an+2 (3 − n) . Continuing this approach inductively until we have −1
a(z) ∼ z + z n + cz 2n−1 + ⋅ ⋅ ⋅ . Now we eliminate inductively all the power of z with exponents greater than or equal to 2n. Assume that there is a nonnegative integer k such that a(z) ∼ z + z n + cz 2n−1 + dz 2n+k + ⋅ ⋅ ⋅ , where d ≠ 0. Let g(z) = z + bz n+k+1 , we claim that b can be chosen so that g ∘ (z + z n + cz 2n−1 + dz 2n+k + ⋅ ⋅ ⋅) = (z + z n + cz 2n−1 ) ∘ g (modulo z 2n+k+1 ). On expansion this equation becomes z + z n + cz 2n−1 + dz 2n+k + bz n+k+1 + b(n + k + 1)z 2n+k = z + z n + bz n+k+1 + nbz 2n+k + cz 2n−1 .
Let d + b(n + k + 1) = nb, or b = −d/(k + 1), our clam is true. This completes the induction. We would like to inform the readers that the iteration of the formal power series a(z) = z + an z n + an+1 z n+1 + ⋅ ⋅ ⋅ with an ≠ 0 was investigated by Fatou [21] and Szekeres [97], too. This topic will be revisited after the completion of the canonical forms. Proposition 8.1.11. If z + z n + cz 2n−1 ∼ z + z m + dz 2m−1 , then n = m and c = d. Proof. Let α(z) = z + z n + cz 2n−1
and
β(z) = z + z m + dz 2m−1 .
Modulo z n , α(z) = z. Since α ∼ β, it follows that there exists some g ∈ 𝔻(ℂ) such that β(z) = g ∘ α ∘ g [−1] (z) = z
(modulo z n ).
Then m ≥ n. Similarly, we may have that n ≥ m. So α(z) = z + z n + cz 2n−1
and β(z) = z + z n + dz 2n−1 .
 255
8.1 Conjugacy of formal power series in 𝔻
Write g(z) = b1 z + b2 z 2 + ⋅ ⋅ ⋅. Expanding g ∘ α = β ∘ g and comparing the coefficients of z n , we obtain bn + b1 = bn + bn1 ,
bn−1 = 1. 1
or
Now b1 z commutes with both α and β, so we may replace g by g1 (z) = g(
1 z) = z + b2 z 2 + b3 z 3 + ⋅ ⋅ ⋅ . b1
We claim: b2 = b3 = ⋅ ⋅ ⋅ = bn−1 = 0 if n > 2. Assume the claim is not true. Let bk ≠ 0 for some k ≤ n − 1, but bj = 0 for 2 ≤ j < k. Expanding g1 ∘ α = β ∘ g1 and comparing coefficients of the term of z n+k−1 , we have k
g1 ∘ α(z) = z + z n + cz 2n−1 + bk (z + z n + cz 2n−1 ) + ⋅ ⋅ ⋅ + bn+k−1 (z + z n + cz 2n−1 )
n+k−1
+ ⋅⋅⋅.
The coefficient of the term of z n+k−1 of g1 ∘ α is bk (k1 ) + bn+k−1 because the term cz 2n−1 in (z + z n + cz 2n−1 )k does not contribute anything to the coefficient of z n+k−1 . Similarly, n
β ∘ g1 (z) = (z + bk z k + ⋅ ⋅ ⋅) + (z + bk z k + ⋅ ⋅ ⋅) + d(z + bk z k + ⋅ ⋅ ⋅)
2n−1
,
and the coefficients of z n+k−1 rely only on the term bn+k−1 z n+k−1 and the term (z + bk z k + ⋅ ⋅ ⋅)n that produce the coefficient n bn+k−1 + bk ( ) = bn+k−1 + nbk . 1 Then we have bn+k−1 + kbk = bn+k−1 + nbk
or
(n − k)bk = 0,
which is a contradiction because k ≠ n and bk ≠ 0. Thus, the claim is true, or g1 (z) = z + bn z n + bn+1 z n+1 + ⋅ ⋅ ⋅ + b2n−1 z 2n−1 + ⋅ ⋅ ⋅ . Then g1 ∘ α(z) = α(z) + bn αn (z) + ⋅ ⋅ ⋅ + b2n−1 α2n−1 (z) + ⋅ ⋅ ⋅ n
= z + z n + cz 2n−1 + bn (z + z n + cz 2n−1 ) + ⋅ ⋅ ⋅ + b2n−1 (z + z n + cz 2n−1 )
2n−1
+ ⋅⋅⋅.
The coefficient of the term z 2n−1 inside b2n−1 αn (z) is b2n−1 , the coefficient of the term z 2n−1 inside the binomial (z + z n + cz 2n−1 )n is determined by the term (z + z n )n (cz 2n−1 )0 = (z + z n )n . By binomial formula again, the term of z 2n−1 inside the bn (z + z n + cz 2n−1 )n is n 1 )nn−1 (z n ) = bn nz 2n−1 . n−1
bn (
256  8 Iteration and iterative roots Thus, the coefficient of the term z 2n−1 in g1 ∘ α(z) is c + nbn + b2n−1 . Similarly, we have the coefficient of the term z 2n−1 in β ∘ g1 (z) is b2n−1 + nbn + d. Finally, comparing the coefficients of the term z 2n−1 in the equality g1 ∘ α = β ∘ g1 , we have c + nbn + b2n−1 = b2n−1 + nbn + d, which yields that c = d. Proposition 8.1.12. Let k ∈ ℕ be the smallest integer such that ak1 = 1. Then a1 z + a2 z 2 + ⋅ ⋅ ⋅ = a(z) ∼ a1 z if and only if a[k] (z) = z where a[k] = a ∘ a[k−1] . Proof. Suppose that a(z) ∼ a1 z = b(z). Then a[k] = b[k] and it is obvious that b[k] (z) = ak1 z = z. Conversely, we suppose that a[k] = I𝔻 or a[k] (z) = z. Take gn (z) = z + bn z n . As what we did in the proof of Proposition 8.1.7, gn[−1] (z) = z − bn z n + ⋅ ⋅ ⋅, and we have gn (z) = gn[−1] (z) = z
(modulo z n ).
Then gn[−1] ∘ a ∘ gn (z) = gn[−1] ∘ [a1 (z + bn z n ) + ⋅ ⋅ ⋅ + an gnn + ⋅ ⋅ ⋅] 2
= [a1 (z + bn z n ) + a2 (z + bn z n ) + ⋅ ⋅ ⋅ + an gnn + ⋅ ⋅ ⋅] 2
n
− bn [a1 (z + bn z n ) + a2 (z + bn z n ) + ⋅ ⋅ ⋅] + ⋅ ⋅ ⋅
= a1 z + a2 z 2 + ⋅ ⋅ ⋅ an−1 z n−1 + Az n + ⋅ ⋅ ⋅ ,
where A is the coefficient of z n . in the first bracket above, only a1 bn in a1 (z + bn z n ) and an in an gzn contribute to A, and in the second bracket a1 (z + bn z n ) is the only term whose nth power may contribute to A. Thus A = a1 bn + an − bn an1 . Then we have n−1
gn[−1] ∘ a ∘ gn (z) = ∑ aj z j + (an + a1 bn − bn an1 )z n + ⋅ ⋅ ⋅ . j=1
The first n − 1 coefficients remain unchanged after modulo z n . If (n − 1) is not a multiple of k, then a1 bn − bn an1 = bn a1 (1 − an−1 1 ) ≠ 0, and hence we can choose bn , n − 1 ≠ lk, l ∈ ℕ ∪ {0}, such that bn =
an1
an − a1
and make A = a1 bn + an − bn an1 = 0.
8.1 Conjugacy of formal power series in 𝔻
 257
Inductively, we may use this method to eliminate all the terms of z j such that j − 1 ≠ lk, l = 0, 1, 2, . . . . Then we have a(z) ∼ a1 z + ak+1 z k+1 + a2k+1 z 2k+1 + ⋅ ⋅ ⋅ . However, all the coefficients other than a1 must be zero by the assumption a[k] (z) = z. Assume not, say that amk+1 is the first nonzero one, then an easy induction shows that j
j−1
a[j] (z) = a1 z + ja1 amk+1 z mk+1 + ⋅ ⋅ ⋅ , for j ∈ ℕ. Then mk+1 a[k] (z) = z + kak−1 + ⋅ ⋅ ⋅ ≠ z, 1 amk+1 z
a contradiction to that a[k] (z) = z. Thus, a(z) ∼ a1 z. Theorem 8.1.13. Let a(z) = a1 z + a2 z 2 + ⋅ ⋅ ⋅
and
b(z) = b1 z + b2 z 2 + ⋅ ⋅ ⋅ ,
j
with ak1 = 1 and a1 ≠ 1, 2 ≤ j < k. Then g [−1] ∘ a[k] ∘ g = b[k]
if and only if
g [−1] ∘ a ∘ g = b.
Proof. It is obvious that g [−1] ∘ a[k] ∘ g = b[k]
if g [−1] ∘ a ∘ g = b.
Put c = g [−1] ∘ a ∘ g and let c[k] = b[k] . We show that c = b. Suppose that an ≠ 0 for some n ∈ ℕ, and aj = 0 for 2 ≤ j < n. Then a[k] (z) = z + rn z n + rn+1 z n+1 + ⋅ ⋅ ⋅ . Applying Proposition 8.1.10, there exists h ∈ 𝔻 such that (h[−1] ∘ c ∘ h) (z) = h[−1] ∘ c[k] ∘ h(z) = z + z n + dz 2n−1 . [k]
Then (h[−1] ∘ c ∘ h)
[k]
= h[−1] ∘ c[k] ∘ h(z) = (h[−1] ∘ b ∘ h) . [k]
Thus, it is enough to show that if f (z) = a1 z + an z n + ⋅ ⋅ ⋅ ,
F(z) = a1 z + cn z n + ⋅ ⋅ ⋅ ,
258  8 Iteration and iterative roots where cn ≠ 0, and f [k] (z) = F [k] (z) = z + z n + dz 2n−1 + ⋅ ⋅ ⋅ , then f = F. Since f ∈ 𝔻(ℂ), it follows that f commutes with f [k] with composition by (1.10). Comparing the coefficient of z n in the equality f ∘ f [k] = f [k] ∘ f , or in the expression a1 z + (a1 + an )z n + ⋅ ⋅ ⋅ = a1 z + (an + an1 )z n + ⋅ ⋅ ⋅ , we have a1 = an1 , or an−1 = 1, and hence n − 1 is a multiple of k. Then 1 f [k] ∘ (a1 z) = a1 z + (a1 z)n + d(a1 z)2n−1 = a1 f [k] (z) = (a1 z) ∘ f [k] . a
Similarly a1 f (z) = z + an z n + ⋅ ⋅ ⋅ commutes with z + z n + dz 2n−1 or commutes with f [k] . 1 1 By Lemma 8.1.5 and Lemma 8.1.6, z + z n + dz 2n−1 = eD z
and
1 f (z) = eD1 z, a1
for some quasinilpotent derivations D and D1 . Actually we have that D = (z n +⋅ ⋅ ⋅)d/dz. It is seen by Lemma 8.1.5 that D consists of all powers of (eD − 1) and D1 consists of all powers of (eD1 − 1). Since eD1 commutes with eD , it follows that D1 commutes with D. Now it is easy to see that this means D1 = λD for some scalar λ. Indeed, if λ is the coefficient of z n in D1 z, then D = D1 − λD commutes with D and has the form D = (qz m + ⋅ ⋅ ⋅)d/dz
with m ≠ n.
Then D D = D ∘ D = (qz m + ⋅ ⋅ ⋅)d/dz((z n + ⋅ ⋅ ⋅)d/dz)
= (qz m + ⋅ ⋅ ⋅)[(nz n−1 + ⋅ ⋅ ⋅)d/dz + (z n + ⋅ ⋅ ⋅)d2 /(dz)2 ]
= ((qn)z m+n−1 + ⋅ ⋅ ⋅)d/dz + (qz m + ⋅ ⋅ ⋅)(z n + ⋅ ⋅ ⋅)d2 /(dz)2 . Similarly, we have DD = ((qm)z m+n−1 + ⋅ ⋅ ⋅)d/dz + (qz m + ⋅ ⋅ ⋅)(z n + ⋅ ⋅ ⋅)d2 /(dz)2 . Then D D − DD = ((n − m)qz m+n−1 + ⋅ ⋅ ⋅)d/dz. However, D D − DD = 0, we have ((n − m)qz m+n−1 + ⋅ ⋅ ⋅)d/dz = 0.
8.1 Conjugacy of formal power series in 𝔻
 259
Since n ≠ m, it follows that q = 0. But q was the first nonzero coefficient, if any existed. Therefore, D = 0. We have now shown that f (z) = a1 eλD z, where the automorphism Ta1 z : a → a ∘ (a1 z) commutes with D. Then ak1 ekλD z = f [k] (z) = z + z n + dz 2n−1 = eD z, which implies that λk = 1. Thus, f (z) = a1 e(1/k)D z. Of course, the same analysis holds for F(z); so F(z) = a1 e(1/k)D z, and hence f = F. Proposition 8.1.14. Let a(z) be set as in Theorem 8.1.13. Let k be the smallest positive integer such that ak1 = 1 and suppose that a[k] ∼ z + z n + cz 2n−1 . Then a(z) ∼ a1 z + b1 z n + c1 z 2n−1 , where b1 =
1 ka1
and
c1 =
ca1 (k − 1)n . − k 2k 2 a31
Proof. Let α(z) = a1 z + b1 z n + c1 z 2n−1 . We know that an−1 = 1 by the proof of Theo1 rem 8.1.13. An induction then shows that m−1 n α[m] = am 1 z + ma1 b1 z
+ (mam−1 1 c1 +
m(m − 1) m−2 2 2n−1 na1 b1 )z + ⋅⋅⋅. 2
Put b1 =
1 ka1
and c1 = (
ca1 (k − 1)n − ); k 2k 2 a31
we see that α[k] = z + z n + cz 2n−1 + ⋅ ⋅ ⋅ . By Proposition 8.1.10, we have α[k] ∼ z + z n + cz 2n−1 ∼ a[k] . By Theorem 8.1.13, a ∼ α.
260  8 Iteration and iterative roots n Table 8.1: Canonical form ∼ a(z) = ∑∞ n=1 an z .
zk
When a1 = 0 and k ≥ 2 is the smallest such that ak ≠ 0.
a1 z
When a1 ≠ 0 and
z + z n + cz 2n−1
When a1 = 1 and an ≠ 0, but aj = 0 for 1 < j < n (c is computable from a(z) mod z 2n as in Proposition 8.1.10).
a1 z + b1 z n + c1 z 2n−1
When ak1 = 1 and a[k] ∼ z + z n + cz 2n−1 , b1 = 1/(ka1 ), c1 = c
an ≠ 1 all n ≠ 0 { { { 1 k {or a1 = 1 and { { [k] {a ≡ z where k ∈ ℕ is the smallest.
(n − 1) is a multiple of k.
a1 k
−
n(k−1) ; 2k 2 a31
As a conclusion, Scheinberg introduced a table of canonical forms. Table 8.1 summan rizes the canonical form, where the formal power series a(z) = ∑∞ n=1 an z is an arbitrary member of 𝒜 or 𝔻. Table 8.1 consists of 4 categories, from the top to the bottom. In Table 8.1, k = min{n ∈ ℕ : an ≠ 0} in the first category for z k , and k is the minimal integer greater than 1 with ak1 = 1 in the category 2 and 4. In category 4, we see the iteration a[k] which will be further discussed in the next section. By the end of Section 8.3, we will see a remark and find an improvement of category 4. Almost at the same time of the establishment of the Table 8.1, the conjugacy of the formal power series f (z) = z + an z n + an+1 z n+1 + ⋅ ⋅ ⋅ was investigated by Kimura [51]. It was proved that for f (z) = z + am z m+1 + am+1 z m+2 + ⋅ ⋅ ⋅ there exists a g ∈ 𝔻, g ∼ f and g(z) = z(1 + λz m + μz 2m ). This result is very close to that of Proposition 8.1.10. The approach used by Kimura was similar to the method used in Proposition 8.1.10 but developed in different way. Most important point, we believe, is that both the works were independently completed. Let us introduce this achievement. Let f ∈ 𝔻 = 𝔻(ℂ) be given such that ∞
f (z) = z(1 + ∑ aj z j ), j=m
am = λ ≠ 0.
(8.4)
If ϕ ∈ 𝔻 such that ∞
ϕ(z) = z(1 + ∑ pj z j ), j=1
(8.5)
8.1 Conjugacy of formal power series in 𝔻
261

then it is easy to see that ϕ[−1] has the similar form, that is, ∞
ϕ[−1] (z) = z(1 + ∑ qj z j ). j=1
Then the conjugate g = ϕ ∘ f ∘ ϕ[−1] has the same form as that of f : ∞
g(z) = z(1 + ∑ bj z j ), j=m
which can be verified by letting g ∘ ϕ = f ∘ ϕ. We want to determine ϕ in such a way that the expression of g becomes as simple as possible. Let {ϕk } ⊆ 𝔻 be a sequence such that ϕk (z) = z(1 + pk z k ),
k ∈ ℕ.
(8.6)
Such ϕk is called a change of variable, and the sequence {ϕk } is called a sequence of changes of variable, and the conjugate g = ϕ ∘ f ∘ ϕ[−1] is called the transform of f by ϕ. We prefer calling g a conjugate of f with respect to ϕ. By the sequence {ϕk }, the composition ϕk ∘ ϕk−1 ∘ ⋅ ⋅ ⋅ ∘ ϕ1 has the form k
ψk = ϕk ∘ ϕk−1 ∘ ⋅ ⋅ ⋅ ∘ ϕ1 = z(1 + ∑ pj z j + Pk+1 (z)), j=1
where Pi ∈ 𝕏 with ord(Pi ) ≥ i for all i ∈ ℕ. Actually such a ϕk plays a role like gk+1 did in the proof of Proposition 8.1.10. It is clear that the sequence {ψk } converges to the formal power series (8.5) ∞
ϕ(z) = z(1 + ∑ pj z j ) j=1
in the ultrametric topology defined in Definition 2.4.7. of For any ϕk (z) = z(1 + pk z k ), k ∈ ℕ, applying Corollary 1.5.13, the inverse ϕ[−1] k ϕk can be expressed as ϕ[−1] (z) = z(1 − pk z k + P2k (z)), k and the relation ϕk ∘ ϕ[−1] = I𝔻 can be expressed as k k
(1 − pk z k + P2k )(1 + pk z k (1 − pk z k + P2k ) ) = 1.
(8.7)
Using this formula, we obtain ∞
j
f ∘ ϕ[−1] (z) = z(1 − pk z k + P2k )(1 + ∑ aj z j (1 − pk z k + P2k ) ) k j=m
∞
j
= z(1 − pk z k + P2k + (1 − pk z k + p2k ) ∑ aj z j (1 − pk z k + P2k ) ). j=m
262  8 Iteration and iterative roots In this formula, the only thing we want to verify is the coefficient of the term z m+k+1 or the term z m+k inside the big bracket. This term can only occur in the expressions ∞
j
∑ aj z j (1 − pk z k + P2k )
j=m
and
∞
j
(−pk z k ) ∑ aj z j (1 − pk z k + P2k ) . j=m
This term is m (−pk z k )am z m + am+k z m+k + am z m (−pk ( )z k ), 1 and the coefficient is −pk am + am+k − am mpk = am+k − (m + 1)λpk , where λ = am . Then (x) f ∘ ϕ[−1] k ∞
j
= z(1 − pk z k + P2k + (1 − pk z k + p2k ) ∑ aj z j (1 − pk z k + P2k ) ) j=m
m+k−1
= z(1 − pk z k + P2k + ∑ aj z j j=m
+ (am+k − (m + 1)λpk )z m+k + Pm+k+1 ). Then we have ϕk ∘ f ∘ ϕ[−1] k
m+k−1
= z(1 − pk z k + P2k + ∑ aj z j + (am+k − (m − 1)λpk )z m+k j=m
m+k−1
+ Pm+k+1 ) × (1 + pk z k [1 − pk z k + P2k + ∑ aj z j j=m
k
+ (am+k − (m + 1)λpk )z m+k + pm+k+1 ] ) m+k−1
= z(1 − pk z k + P2k + ∑ aj z j + (am+k − (m − 1)λpk )z m+k j=m
k
+ Pm+k+1 ) ⋅ (1 + pk z k [1 − pk z k + P2k ] + pk z k (kλz m + pm+1 )) m+k−1
= z(1 + ∑ aj z j + (am+k + (k − m)λpk )z m+k + ⋅ ⋅ ⋅), j=m
8.1 Conjugacy of formal power series in 𝔻

263
by applying (8.7) in the last step. Therefore, if we set ∞
ϕk ∘ f ∘ ϕ[−1] (z) = z(1 + ∑ bj z j ), k j=m
we have aj
bj = {
aj + (k − m)λpk
for m ≤ j ≤ m + k − 1, for j = m + k.
This means that the first k coefficients am , am+1 , . . . , am+k−1 of f are not affected, but the (k + 1)th coefficient am+k can be changed to an arbitrary value bm+k if k ≠ m. For the sequence of changes of variable {ϕk } where ϕk (z) = z(1 + pk z k ),
k ∈ ℕ,
define a sequence {gk } recursively by g1 = ϕ1 ∘ f ∘ ϕ[−1] 1 , , gk = ϕk ∘ gk−1 ∘ ϕ[−1] k
k = 2, 3, . . . .
Then gk = (ϕk ∘ ⋅ ⋅ ⋅ ∘ ϕ1 ) ∘ f ∘ (ϕk ∘ ⋅ ⋅ ⋅ ∘ ϕ1 )[−1] = ψk ∘ f ∘ ψk[−1] . j Since f (z) = z(1 + ∑∞ j=m aj z ), we may set ∞
gk (z) = z(1 + ∑ bk,j z j ), j=m
and have from the preceding that bk,m = λ,
k = 1, 2, . . . ,
bk,m+k = bk+1,m+k = bk+2,m+k = ⋅ ⋅ ⋅ ,
k = 1, 2, . . . .
Then the sequence {gk } converges to a formal power series ∞
g(z) = z(1 + ∑ bj z j ), j=m
bm = λ.
On the other hand, the sequence {ψk } converges to ∞
ϕ(z) = z(1 + ∑ pj z j ). j=1
264  8 Iteration and iterative roots So we have g = ϕ ∘ f ∘ ϕ[−1] . In the preceding, we have seen that we can so choose ϕk , or pk , that each bj assumes arbitrary value given in advance (for k ≠ m) possibly with the exception of b2m (for k = m). It should be noted that ϕm , or pm , can be arbitrarily given. Therefore, we can put bj = 0 except for b2m which is uniquely determined. We obtain the following. Theorem 8.1.15. Let f ∈ 𝔻 be such that ∞
f (z) = z(1 + ∑ aj z j ), j=m
am = λ ≠ 0.
Then there exists a conjugate ϕ ∈ 𝔻 ∞
ϕ(z) = z(1 + ∑ pj z j ), j=m
such that ϕ ∘ f ∘ ϕ[−1] = g = z(1 + λz m + μz 2m ). The coefficient pm can be arbitrary but μ is uniquely determined. If we review Proposition 8.1.10, we may find the beauty of each approach.
8.2 The iteration of formal power series The lefthand side of equation (8.1), or f [n] , is called the nth iteration of the function f . If f is a formal power series, f [n] is called an nth iterative power of the formal power series f . In Section 1.5, particularly by equations (1.10) and (1.11), we see some interesting properties for iteration of an almost unit formal power series. With the establishment of the general composition for formal power series, it is possible to extend the investigation for the iteration of formal power series beyond the space 𝔻. Example 8.2.1. Considering E1 ∈ 𝕏(ℝ) E1 (x) = 1 + x +
1 2 1 3 x + x + ⋅⋅⋅. 2! 3!
8.2 The iteration of formal power series  265
By Theorem 5.4.1, we know that E1 ∘ E1 = E1[2] ∈ 𝕏(ℝ). By Theorem 5.7.2, CE [2] = (CE1 )2 1 where 1 [1 [ [ [ [1 =[ [. . . [ [ [1 [. . .
CE1
0 1
0
2 ... k ...
0
0
1 2! 22 2!
1 3! 23 3!
1 4! 24 4!
...
...
...
...
...
...
k2 2!
k3 3!
... . . .] ] ] ] . . .] ]. . . .] ] ] . . .] . . .]
k4 4!
If we write E1[2] (x) = E1 (E1 (x)) = c0 + c1 x + c2 x2 + ⋅ ⋅ ⋅ + ck x k + ⋅ ⋅ ⋅ , then c0 = ∑∞ n=0
1 n!
= e. For k ∈ ℕ, 1 ∞ nk−1 1 nk = . ∑ k! n=1 (n − 1)! n=1 n! k! ∞
ck = ∑ For example, c1 = ∑∞ n=1
1 (n−1)!
= ∑∞ n=0
c2 =
1 n!
= e.
1 ∞ 1 ∞ m+1 n = ∑ . ∑ 2 n=1 (n − 1)! 2 m=0 m!
By Abel’s limit theorem, f (r) =
1 ∞ n+1 n r ∑ 2 n=0 n!
converges uniformly on [0, 1] with f (1) = c2 . But ∫ f (r) dr =
1 ∞ 1 n+1 1 ∑ r + C = rer + C. 2 n=0 n! 2
Then f (r) =
d 1 (∫ f (r) dr) = er (1 + r). dr 2
Then, c2 = f (1) = e. Thus, x
ee = e + ex + ex2 + ⋅ ⋅ ⋅ . x
The Maclaurin series of E1[2] (x) = f (x) = ee = ∑∞ n=0 x
ee = e + ex + ex 2 +
f (n) n!
is
5e 3 15e 4 x + x + ⋅⋅⋅. 3! 4!
(8.8)
266  8 Iteration and iterative roots Then we have ∞ (n + 1)2 n2 = ∑ , n! n=1 (n − 1)! n=0 ∞
5e = c3 = ∑
∞ n3 (n + 1)3 = ∑ . n! n=1 (n − 1)! n=0 ∞
15e = c4 = ∑
We obtain two different series expressions for e that, of course, converge to e slower than the convergence of e=1+1+
1 1 + + ⋅⋅⋅. 2! 3! x
It is obvious that the higher order derivative of ee is very complicated. Practically the formula (8.8) is useful. Generally, the iteration of formal power series is determined by the existence of the composition. The general composition theorem, or Theorem 5.4.1, is a criterion n for the iteration. For example, f (x) = ∑∞ n=1 n!x is an almost unit formal power series which diverges everywhere except at x = 0, but f [n] ∈ 𝔻(ℝ) for every n ∈ ℕ. Theorem 8.2.2. Let f ∈ 𝕏 = 𝕏(F) be given such that deg(f ) ≠ 0, where F = ℝ or F = ℂ. Suppose that there exists n ∈ ℕ such that f [n−1] = ⏟⏟ f ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ∘ f ∘ ⏟f⏟⏟∘⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ⋅ ⋅ ⋅ ∘ ⏟f⏟ ∈ 𝕏. (n−1) compositions
Then f [n] ∈ 𝕏 if and only if f (k) (f [n−1] (0)) ∈ F
for every
k ∈ ℕ,
where f (k) is the kth formal derivative of f . Proof. It is a general case for Proposition 5.4.13. 𝔻(ℝ) is a group under the composition of formal power series. We now further investigate the iteration of almost unit formal power series. The proof of the proposition below can be seen in Proposition 1.5.19. Proposition 8.2.3. Let f ∈ 𝔻 = 𝔻(ℝ) be given such that f (x) = a0 + a1 x + a2 x 2 + ⋅ ⋅ ⋅ with a0 = 0, a1 ≠ 0. Let n ∈ ℕ be given. Write (n) (n) 2 (n) k f n (x) = a(n) 0 + a1 x + a2 x + ⋅ ⋅ ⋅ + ak x + ⋅ ⋅ ⋅ ,
f [n] (x) = f0[n] + f1[n] x + f2[n] x 2 + ⋅ ⋅ ⋅ + fk[n] x k + ⋅ ⋅ ⋅ . Then for every k ∈ ℕ, [n−1] (2) fk[n] = an−1 ak + ⋅ ⋅ ⋅ + fk[n−1] ak1 . 1 ak + f2
(8.9)
8.2 The iteration of formal power series  267
Example 8.2.4. For any n ∈ ℕ, applying formula (8.9) we have n f1[n] = an−1 1 a1 = a1 ,
[n−1] 2 f2[n] = an−1 a1 1 a2 + f2
2 n−2 [n−2] 2 = an−1 a1 ) 1 a2 + a1 (a1 a2 + f2 2
2 [n−2] = an−1 1 a2 (1 + a1 ) + (a1 ) f2 2
2 n−3 2 [n−3] = an−1 ) 1 a2 (1 + a1 ) + (a1 ) (a1 a2 + a1 f2 3
2 2 [n−3] = an−1 1 a2 (1 + a1 + a1 ) + (a1 ) f2
⋅⋅⋅ = =
⋅⋅⋅
⋅⋅⋅
an−1 1 a2 (1 an−1 1 a2 (1
n−1 [1] f2
2 + a1 + a21 + ⋅ ⋅ ⋅ + an−2 1 ) + (a1 )
+ a1 + a21 + ⋅ ⋅ ⋅ + an−2 + an−1 1 1 ).
If a1 = 1, then f2[n] = na2 . If a1 ≠ 1, then f2[n] = an−1 1 a2
1 − an1 . 1 − a1
For k = 3, we have [n−1] (2) f3[n] = an−1 a3 + f3[n−1] a31 . 1 a3 + f2
Importing the formula of f2[n−1] and using the formula a(2) 3 = 2a1 a2 we have n−2 n−2 [n−1] 3 f3[n] = an−1 a1 1 a3 + a1 a2 (1 + a1 + ⋅ ⋅ ⋅ + a1 )2a1 a2 + f3 n−1 2 n−2 [n−1] 3 = an−1 a1 1 a3 + 2a1 a2 (1 + a1 + ⋅ ⋅ ⋅ + a1 ) + f3 n−1 2 n−2 = an−1 1 a3 + 2a1 a2 (1 + a1 + ⋅ ⋅ ⋅ + a1 )
n−2 2 n−3 [n−2] 3 + a31 [an−2 a1 ] 1 a3 + 2a1 a2 (1 + a1 + ⋅ ⋅ ⋅ + a1 ) + f3
2 n−1 2 n−2 = an−1 1 a3 (1 + a1 ) + 2a1 a2 [1 + a1 + ⋅ ⋅ ⋅ + a1 2
3 [n−2] + a21 (1 + a1 + ⋅ ⋅ ⋅ + an−3 , 1 )] + (a1 ) f3
continuing the precess we have n−2
n−2
j
j
n−3
j
2 n−1 2 2 f3[n] = an−1 1 a3 ∑ (a1 ) + 2a1 a2 ( ∑ a1 + a1 ∑ a1 + ⋅ ⋅ ⋅ j=0
n−2
+ (a21 )
j=0
0
j
∑ a1 ) + (a31 )
j=0
j=0
n−1 [1] f3 .
Then n−1
2j
n−2
2j
n−2−j
n−1 2 f3[n] = an−1 1 a3 ∑ a1 + 2a1 a2 ∑ a1 (1 + a1 + ⋅ ⋅ ⋅ + a1 j=0
j=0
).
268  8 Iteration and iterative roots The reader may compare this example with the formulas established in Proposition 1.5.19. The iteration of sine was introduce by Towse several years ago. Example 8.2.5 ([99]). Let us begin with the familiar power series 1 3 1 5 1 7 x + x − x + ⋅⋅⋅ 3! 5! 7!
sin x = x −
which can be considered as a formal power series although it converges on ℝ. Then 1 1 8 7 sin[2] (x) = sin(sin(x)) = x − x 3 + x 5 − x + ⋅⋅⋅, 3 10 315 1 11 5 731 7 sin[3] (x) = x − x3 + x − x + ⋅⋅⋅. 2 40 5040 Of course, Towse showed us how he obtained those formal power series. Proposition 8.2.6. Let f ∈ 𝔻 = 𝔻(ℝ) be given such that ∞
f (x) = ∑ an xn = x + a3 x 3 + a4 x 4 + a5 x 5 + ⋅ ⋅ ⋅ , n=0
where a0 = 0, a1 = 1, k ∈ ℕ ∪ {0}, and write [n] [n] 2 [n] 3 [n] 4 f [n] (x) = a[n] 0 + a1 x + a2 x + a3 x + a4 x + ⋅ ⋅ ⋅ .
Then a[n] 0 = 0,
a[n] 1 = 1,
a[n] 2 = 0,
a[n] 3 = na3 ,
a[n] 4 = na4 ,
and 2 a[n] 5 = na5 + 3a3 n(n − 1)/2.
Proof. Since f [n] (x) = f (f [n−1] (x)), we have f (f (x)) = x + a[n−1] x3 + a[n−1] x4 + ⋅ ⋅ ⋅ 4 3
3
+ a3 (x + a[n−1] x 3 + a[n−1] x 4 + ⋅ ⋅ ⋅) 4 3
4
+ a4 (x + a[n−1] x 3 + a[n−1] x 4 + ⋅ ⋅ ⋅) + ⋅ ⋅ ⋅ . 4 3
[n] [n−1] [n] + a3 . Summing We can easily see that a[n] 0 = 0, a1 = 1, a2 = 0, and a3 [n] = a3 [i] [i−1] z3 − a3 from i = 2 to n yields a telescope sum and we get n
[i] [i−1] a[n] ) = (n − 1)a3 , 3 − a3 = ∑(a3 − a3 i=2
a[n] 3
so = na3 . Similarly, we have a[4] = na4 . The formula for a[n] get complicated 4 k quickly. The formula for a[n] can be found in [99]. 5
8.3 The iterative roots of formal series in 𝔻
269

Now we have sin[n] (x) = x −
n 3 1 1 5 3 1 41 7 x + ( n2 − n)x5 − ( n − n2 + x ) + ⋅⋅⋅. 6 24 30 432 45 3780
We would like to introduce a more general formula for a[n] without proof to close k this section. Proposition 8.2.7 ([99]). Let f ∈ 𝔻 = 𝔻(ℝ) be given such that ∞
f (x) = ∑ an xn = x + a3 x3 + a4 x 4 + a5 x 5 + ⋅ ⋅ ⋅ , n=0
where a0 = 0, a1 = 1, k ∈ ℕ ∪ {0}, and write [n] [n] 2 [n] 3 [n] 4 f [n] (x) = a[n] 0 + a1 x + a2 x + a3 x + a4 x + ⋅ ⋅ ⋅ .
is a polynomial in n with leading term of the form For k odd, the coefficient a[n] k γk (a3 n)(k−1)/2 , where γk ∈ ℚ. For k even, the coefficient a[n] is a polynomial in n of degree at most (k − 2)/2. k
8.3 The iterative roots of formal series in 𝔻 Let g ∈ 𝕏(ℂ) be given and let n ∈ ℕ be given with n > 1 (it is trivial if n = 1). If there exists a formal power series f ∈ 𝕏(ℂ) such that f [n] = f ∘ f [n−1] = g,
(8.10)
f is called an nth iterative root or iterated root of g. We immediately face the following questions: 1. What kind of g has iterative roots? 2. What kinds of iterative roots could be if g has some or all nth iterative roots? 3. How to find the iterative roots for g if g has such roots? The problem of finding iterative roots of functions dates back at least to 1881 by Abel [2]. Since then it has attracted the attention of many mathematicians. In 1918, Pfeiffer investigated the functional equation f [f (x)] = g(x) (see Theorem 4.2.2), where the solution f , no matter it is analytic or formal, is actually an iterative root of g with the order 2. This topic has been developing deeply and widely so far by many mathematicians such as J. Schwaiger, H. Fripertinger, L. Reich, etc. [22, 73, 92]. The most recent such work by W. Jabtoński can be found in [45] and [46].
270  8 Iteration and iterative roots Proposition 8.3.1. Let g ∈ 𝕏 = 𝕏(ℂ) be given and let n ∈ ℕ be given with n > 1. If there is f ∈ 𝕏 such that f [n] = g, then: (i) (ord(f ))n ≤ ord(g); (ii) In particular, ord(f ) = 0 if ord(g) = 0. Proof. Denote ord(f ) = m and ord(g) = M, both of them are nonnegative integers. By Proposition 5.4.12, n
M = ord(g) = ord(f [n] ) ≥ [ord(f )] = mn . (i) and (ii) follow from this inequality. Please be noticed that the other direction of (ii) above is false. For example, if f (x) = 1 − x, then f [2] (x) = 1 − (1 − x) = x. The General Composition Theorem 5.4.1 allows the existence of the iterative roots of g either ord(g) = 0 or ord(g) > 1. We will discuss it in some other place if we can. This problem is open. In this section, we only discuss the case ord(g) = 1, or g ∈ 𝔻 = 𝔻(ℂ). We would like to indicate that the iterative roots or iterative power can work on any function if the composition exists, not only work for the formal power series. The proposition below claims that there is no function can be a square iterative root of a quadratic polynomial on ℂ. The proof of this proposition can be found in [80]. If we restrict the functions to be formal power series, the claim can be easily obtained by applying Proposition 8.3.1. Proposition 8.3.2. Let P be a polynomial of degree 2 define on the entire complex plane ℂ. Then P has no iterative roots of any order r with r ≥ 2, that is, there is no function f whatever mapping ℂ into itself such that f [r] = P. The following lemma comes from number theory. An element ϵ is called a primitive nth root of unity if ϵn = 1 and ϵm ≠ 1 for 1 ≤ m < n. We choose a version from Hardy’s book (p. 67, [36]) without proof. Lemma 8.3.3. Any qth root of unity is a primitive rth root, for some divisor r of q. It is clear that g ∈ 𝔻(ℂ) has an mth iterative root if and only if its conjugate f does, where g = h ∘ f ∘ h[−1] . Then, given a set of representatives of formal power series with respect to the equivalence relation “∼”, we only need to investigate those formal power series being representatives. In addition to Lemma 8.3.3, Schwaiger provided the following lemma and theorem [92]. i m Lemma 8.3.4. Let f (z) = ∑∞ i=1 ai z ∈ 𝔻(ℂ) be given and let b1 = a1 , then there is an mth iterative root g of f with b1 as coefficient of z if and only if there is a conjugate f ∗ of
8.3 The iterative roots of formal series in 𝔻
 271
f of the form ∞
f ∗ (z) = ∑ a∗i z i
where
i=1
a∗i = 0
whenever
bi−1 1 ≠ 1.
Such an f ∗ is called semicanonical with respect to b1 . By the canonical form Table 8.1 in Section 8.1, we have the theorem below. Theorem 8.3.5. Let 𝔻 = 𝔻(ℂ), then: (a) If an almost unit formal power series f ∈ 𝔻 is conjugate to a canonical form of type 0,
a1 z,
or
z + z n + cz 2n−1 ,
then f has iterative roots of any order m. (b) If f ∈ 𝔻 is conjugate to some canonical form a1 z + b1 z n + c1 z 2n−1 , where a1 is a primitive kth root of unity, then f has an mth iterative root if and only if (m, n − 1) divides
n−1 . k
(c) If f ∈ 𝔻 is conjugate to z k for some integer k ≥ 2, then f has an mth iterative root if and only if k is a perfect mth power (k = lm for some integer l). Furthermore, in this case all mth roots of f are given by g = bz l , where lm = k
k−1
and
b l−1 = 1.
Proof. It is clear that we may suppose that f itself is canonical. (a) By Lemma 8.3.4, we need only check whether f is semicanonical. In fact, f (z) = a1 z is semicanonical with respect to any mth root b1 of a1 , 0 has 0 as a root of mth order, and f (z) = z + z n + cx2n−1 is semicanonical with respect to the mth root 1 of 1. (b) Let f (z) = a1 z + b1 z n + c1 z 2n−1 , where a1 is a primitive kth root of unity, k divides (n−1), and f [k] ∼ z +z n +cz 2n−1 . It is the canonical form category 4 in Table 8.1. Suppose that g is an mth root of f , we show that (m, n − 1) divides (n − 1)/k. Write g(z) = dz + d2 z 2 + d3 z 3 + ⋅ ⋅ ⋅ and f1 (z) = z + z n + cz 2n−1 . Then dm = a1 because g [m] = f . Since g [mk] = (g [m] )
[k]
= f [k] ∼ f1 ,
it follows that the canonical form f1 has an (mk)th iterative root with d as the coefficient of z. Then, by Lemma 8.3.4, ∞
f1 ∼ f2 = z + ∑ a∗i z i i=2
with a∗i = 0 if di−1 ≠ 1. But f1 (z) = z + z n + cz 2n−1 is canonical and conjugate to f2 which implies, by Table 8.1 (category 3) of Section 8.1 for z + z n + cz 2n−1 , that a∗n ≠ 0
272  8 Iteration and iterative roots and consequently dn−1 = 1 (see the proof of Theorem 8.1.13). Now let d0 be a primitive (n − 1)th root of unity such that d0(n−1)/k = a1 . Since dn−1 = 1, it follows that there exists some integer u with d0u = d. Now dm = a1 implies that d0um = d0(n−1)/k or um ≡
n−1 k
(mod(n − 1)).
. But this is possible for such an u if and only if (m, n − 1) divides n−1 k Conversely, suppose that (m, n − 1) divides n−1 , we need show that f has mth iterk ative root. For this purpose, let u be a solution of um ≡
n−1 k
(mod(n − 1)),
and put d = d0u where d0 is defined as above. Then dm = d0um = d0 = a1 and 1 = dn−1 = d2(n−1) = d(2n−1)−1 . So f (z) = z+a1 z n−1 +b1 z 2n−1 is semicanonical with respect to d implying by Lemma 8.3.4 the existence of an mth root of f . (c) Let f be conjugate to z k for some k ≥ 2. We consider the structure of g with g ∘ f = f ∘ g and we may simply let f (z) = z k . Let us claim that g is of the form g(z) = bz l . For otherwise, we would have g(z) = bz l + cz r ⋅ h(z),
c ≠ 0 ≠ b,
l ≥ 0,
r>l
and
h(0) = 1,
for some h ∈ 𝕏. The conjugacy g ∘ f = f ∘ g and the fact g ∘ f (z) = g(z k ) = bz lk + cz rk h(z k ) imply that f ∘ g(z) = g k (z), but k k k g k (z) = (bz l + cz r ⋅ h(z)) = ∑ ( )bk−j cj z lk+(r−l)j ⋅ hj j j=0 k k = bk z kl + ∑ ( )bk−j cj z lk+(r−l)j ⋅ hj . j j=1
Equating the coefficients of z lk+(r−l) , g ∘ f only has the terms whose exponents are multiples of k, by take j = 1 in the righthand side of f ∘ g we have that kbk−1 ch as the coefficient of z lk+(r−l) . Then 0 = kbk−1 ch, which is a contradiction because h(0) = 1 and bc ≠ 0. Thus, g = bz l for some b and l. Let g be an mth iterative root of f . Then g commutes with f because of g [m] = f . By induction on j, we see that 2
j−1
g [j] (z) = b1+l+l +⋅⋅⋅+l z l
j
implying that
8.4 Schröder’s problem
 273
g [m] = f (or g [m] (z) = z k ) if and only if 2
b1+l+l +⋅⋅⋅+l
m−1
=1
and lm = k,
or, k−1
b l−1 = 1. This is the desired result. Remark 8.3.6. Part (b) above shows that f (z) = a1 z + b1 z n + c1 z 2n−1 has infinitely many iterative roots but no root of order (n − 1). This corrects Proposition 8.1.14 saying that f has only finitely many roots [92].
8.4 Schröder’s problem Ernst Schröder’s foundational 1871 paper [91] set the stage for much to come. This paper formally introduced the concept of the rth iterate of a function F(z), F [r] = ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ F ∘ F ∘ F ⋅ ⋅ ⋅ ∘ F , including the investigation of the coefficients of F [r] and a functional r times
equation which late was called the Schröder’s equation. The conjugacy of linear functional transformations was another bright point of the paper. Marshall Cohen recently extended the Schröder’s work and provided some interesting results. Some of those results are introduced in this section [18]. Let us start with the Cayley’s lemma [10], which had been used by Schröder in his 1871 paper. It is interesting that both Cayley and Schröder did not prove this lemma. Cayley actually even did not state it as a lemma but just used it in an example, which will be provided here soon. Schröder stated this Lemma as Cayley’s and then used it as a jumping point for his discussion of iterated composition of formal power series. It is Marshall Cohen who recently investigated Schröder’s problem and provided a proof of this lemma. His work is listed below which can be found in [15] and [16]. Lemma 8.4.1. Let S be a commutative ring and let f : {0, 1, 2, . . . } → S be a mapping, then r r a a f (r) = ∑ [( ) ∑ (−1)b ( )f (a − b)], a b a=0 b=0
for all r ∈ ℕ ∪ {0}. Proof. Let Mr = (ma,b )0≤a,b≤r be the lower triangular matrix given by m0,0 [m [ 1,0 [ Mr = [ m2,0 [ [ ... [ mr,0
0 m1,1 m2,1 ... mr,1
0 0 m2,2 ... mr,2
... ... ... ... ...
0 0 ] ] ] 0 ], ] ... ] mr,r ]
274  8 Iteration and iterative roots where r a ma,b = (−1)b ( )( )f (a − b), a b
ma,b = 0
for b > a.
Cayley’s lemma claims that f (r) =
∑
ma,b ∈Mr
ma,b .
Please also be noticed that we set up (00 ) = 1 and then r r a a f (r) = ∑ [( ) ∑ (−1)b ( )f (a − b)] a b a=0 b=0
r = f (0) + [( )(f (1) − f (0))] 1 r r + ( )[( )(f (2) − 2f (1) − f (0))] + ⋅ ⋅ ⋅ 2 1 r r + [(f (r) − ( )f (r − 1) + ( )f (r − 2) + ⋅ ⋅ ⋅ + (−1)r f (0))]. 1 2
The value of f (r) is the sum of the horizontal rowsums of Mr . But we may also express this sum as the sum of the rowsums of diagonal rows parallel to the main diagonal, along which f (a − b) = f (0), a constant. For 0 ≤ b ≤ r, the bth diagonal row Db runs from (r − b, 0) to (r, b). The diagonal row D0 represents the main diagonal D0 = {mr,0 }. Then r
r
b
f (r) = ∑ ( ∑ mc,d ) = ∑ ( ∑ mr−b+k,k ). b=0 mc,d ∈Db
b=0 k=0
Recalling the combinatorial identity r x r r−y ( )⋅( )=( )⋅( ), x y y x−y
(8.11)
we can see that the sum along each nondegenerate diagonal Db (b > 0) is zero. We have b
∑ mc,d = ∑ mr−b+k,k
mc,d ∈Db
k=0 b
r r−b+k )( ) ⋅ f ((r − b + k) − k) r−b+k k
= ∑ (−1)k ( k=0 b
r r−b+k )( ) ⋅ f (r − b) r−b+k r−b
= ∑ (−1)k ( k=0
8.4 Schröder’s problem
b
= ∑ (−1)k ( k=0
 275
r b )( ) ⋅ f (r − b) (by (8.11)) r−b k
b r b = f (r − b) ⋅ ( ) ⋅ ∑ (−1)k ( ) b k=0 k
r = f (r − b) ⋅ ( ) ⋅ (1 − 1)b = 0. b Thus, r
∑ ma,b = ∑ ( ∑ mc,d ) =
ma,b ∈Mr
b=0 mc,d ∈Db
∑ mc,d
mc,d ∈D0
= mr,0 = f (r), as Cayley’s lemma states. Remark 8.4.2. 1. If S is a commutative ring, then 𝕏(S) is also a commutative ring. 2. The mapping f is very flexible. The next example will show us how flexible the mapping f is. Example 8.4.3. Let G ∈ 𝔻(ℂ) be an almost unit formal power series such that G(z) = z. Define f : ℕ ∪ {0} → 𝕏(ℂ) be such that f (r) = Gr = ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ G ⋅ G ⋅ G⋅⋅⋅G r times
for each r ∈ ℕ ∪ {0}.
For r = 3, we know that G3 = z 3 and also we have 3 3 a a f (3) = ∑ [( ) ∑ (−1)b ( )f (a − b)] a b a=0 b=0
= 1 + 3(z − 1) − 3(z 2 − 2z + 1) + (z 3 − 3z 2 + 3z − 1)
= 1 + 3(z − 1) + 3(z − 1)2 + (z − 1)3 3
= (1 + (z − 1)) = z 3 = G3 (z). For any f ∈ 𝔻, if f [n] = I𝔻 for some n ∈ ℕ, we say that f has finite composition order, otherwise we say that f has infinite composition order. If f has finite composition order, then the smallest n such that f [n] (z) = z is called the composition order of f . We always use composition order to distinguish it from the order of a formal power series.
276  8 Iteration and iterative roots Lemma 8.4.4. Let f ∈ 𝔻(ℂ) be given such that f (z) = z + ak z k + ak+1 z k+1 + ⋅ ⋅ ⋅ , where ak ≠ 0. Then for every n ∈ ℕ f [n] (z) = z + nak z k + Pk+1 (z),
(∗)
where Pk+1 ∈ 𝕏(ℂ) with ord(Pk+1 ) ≥ k + 1. Also f has infinite composition order in 𝕏(ℂ). Proof. Equation (∗) can be obtained by Induction. Since ak ≠ 0, it follows that nak ≠ 0 for all n ∈ ℕ and then f has infinite composition order. The theorem below with a interesting approach was recently published [16]. Of course, the necessary and sufficient condition for f [n] (z) ∼ a1 z for any f ∈ 𝔻(ℂ) such that f (z) = a1 z + a2 z 2 + ⋅ ⋅ ⋅, an1 = 1 has already been introduced in Proposition 8.1.12, by Scheinberg in 1970 [88]. We recommend the readers to read the comprehensive investigation for such topic by Reich and Tomaschek in 2014 [78]. Theorem 8.4.5. Let f ∈ 𝔻(ℂ) be given such that f (z) = ωz + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ , and f [n] = I𝔻 , or f [n] (z) = z, where ω is the nth primitive root of unity. Define f ∗ by f∗ =
1 n n−j [j] ∑ω f . n j=1
Then f ∗ ∈ 𝔻(ℂ), and f∗ ∘ f ∘ f∗
[−1]
(z) = ωz.
Two elements of composition order n in 𝔻(ℂ) are conjugate if and only if their leading coefficients are the same primitive nth root of unity. Proof. We first show that f has nth composition order implies that ω is the nth primitive root of unity. By Corollary 1.5.4, we have n
(D(f )) = D(f [n] ). Then f [n] = I𝔻 implies that (D(f ))n = unit matrix. and hence ωn = 1. Assume that ωa = 1 for some a ∈ ℕ and a < n, then ab = n for some b ∈ ℕ. Then f [a] (z) = ωa z + P2 (z) = z + P2 (z),
8.4 Schröder’s problem
 277
where P2 ∈ 𝕏(ℂ) and ord(P2 ) ≥ 2, saying that dz 2 ∈ P2 , d ≠ 0. By Lemma 8.4.4, we have (f [a] ) (z) = z + bdz 2 + P3 (z). [b]
However, (f [a] ) (z) = f [ab] (z) = f [n] (z) = z. [b]
This contradiction shows that ω is the nth primitive root of unity. It is obvious that for all g, h, k ∈ 𝔻(ℂ) we have (g + h) ∘ k = g ∘ k + h ∘ k
and ℓω ∘ (g + h) = ℓω ∘ g + ℓω ∘ h,
where ℓω ∈ 𝔻 such that ℓω (z) = ωz. Then we have ℓω ∘ f ∗ = f∗ ∘ f =
1 n n−j+1 [j] f , ∑ω n j=1
and
1 n n−j [j+1] . ∑ω f n j=1
Since ωn = 1 and f [n] = I𝔻 , it follows that 1 n [1] [ω f + ωn−1 f [2] + ⋅ ⋅ ⋅ + ωf [n] ] n 1 = [f [1] + ωn−1 f [2] + ⋅ ⋅ ⋅ + ωf [n] ] n 1 = [ωn−1 f [2] + ⋅ ⋅ ⋅ + ωf [n] + ω0 f [1] ] n = f∗ ∘ f.
ℓω ∘ f ∗ =
Then f∗ ∘ f ∘ f∗
[−1]
= ℓω .
Finally, for every g ∈ 𝔻(ℂ), g ∘ ℓω ∘ g [−1] (z) = ωz + P2 (z),
where
ord(P2 ) ≥ 2.
Then g ∘ ℓω ∘ g [−1] = ℓβ We complete the proof.
if and only if ω = β.
278  8 Iteration and iterative roots Definition 8.4.6. Let ℓω ∈ 𝔻(ℂ) be given where ω ∈ ℂ, then the centralizer of ℓω in 𝔻(ℂ) is the set Zω = {h ∈ 𝔻(ℂ)  h ∘ ℓω = ℓω ∘ h}. Lemma 8.4.7. Let ω be a primitive nth root of unity and let h ∈ 𝔻(ℂ). Then h ∈ Zω if and only if ∞
h(z) = ∑ hnj+1 z nj+1 , j=0
where h(z) = h1 z + z2 z 2 + h3 z 3 + ⋅ ⋅ ⋅. Proof. By definitions of ℓω and h, we have ∞
∞
k=1 ∞
k=1
h ∘ ℓω (z) = ∑ hk (ωz)k = ∑ ωk hk z k ,
and
ℓω ∘ h(z) = ∑ ωhk z k . k=1
∞ k k k Then h ∈ Zω if and only if ∑∞ k=1 ω hk z = ∑k=1 ωhk z , and hence if and only if
ωhk = ωk hk
for all k ∈ ℕ,
which is equivalent to hk = 0
for all k ≠ 1 (mod n).
Corollary 8.4.8. Let f ∈ 𝔻(ℂ) be given such that f (z) = ωz + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ , and f [n] = I𝔻 , or f [n] (z) = z, where n is the composition order of f . Let g ∈ 𝔻(ℂ) be given, then g ∘ f ∘ g [−1] = ℓω if and only if there exists a sequence {hnj+1 }∞ j=0 ⊆ ℂ such that ∞
h(z) = ∑ hnj+1 z nj+1 ∈ 𝔻(ℂ) and j=0
g = h ∘ f ∗,
where f is the formal power series set as in Theorem 8.4.5. ∗
Proof. By Theorem 8.4.5, f = f ∗
[−1]
g ∘ (f ∗
∘ ℓω ∘ f ∗ . Then g ∘ f ∘ g [−1] = ℓω is equivalent to
[−1]
∘ ℓω ∘ f ∗ ) ∘ g [−1] = ℓω
which is equivalent to that g ∘ f ∗ ∈ Zω . Applying Lemma 8.4.7, we have an almost nj+1 unit formal power series h(z) = ∑∞ h in 𝕏(ℂ) such that j=0 nj+1 z [−1]
g ∘ f∗
[−1]
= h,
or
g = h ∘ f ∗.
8.4 Schröder’s problem
 279
Corollary 8.4.9. Let f ∈ 𝔻(ℂ) be given such that f (z) = ωz + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ , and f [n] = I𝔻 , or f [n] (z) = z, where n is the composition order of f . For every sequence {gnj+1 }∞ j=0 ⊆ ℂ with g1 ≠ 0, there exists a unique sequence k {gk }1 1, fk[n−1] = ak ωn−2 (1 + ωk−1 + ⋅ ⋅ ⋅ + ω(k−1)(n−2) ) + Pn−1,k .
280  8 Iteration and iterative roots For simplicity we introduce the further notation bk = fk[n−1]
(including b1 = ωn−1 )
[z k ]Q(z) = coefficient of z k in the polynomial Q(z). Note that ak [z k ](b1 z)k = ak ω(n−1)k but that, by induction, bi = fi[n−1]
does not contain ak
if
1 ≤ i < k.
Thus we have ∞
∞
f [n] (z) = f (f [n−1] (z)) = f ( ∑ fk[n−1] z k ) = f ( ∑ bk z k ) k=1
∞
∞
k
k
2
k=1
∞
k
3
= ω( ∑ bk z ) + a2 ( ∑ bk z ) + +a3 ( ∑ bk z ) + ⋅ ⋅ ⋅ . k=1
k=1
k=1
The term of z k can only occur in the expression (b1 z + b2 z 2 + ⋅ ⋅ ⋅)j , 1 ≤ j ≤ k, and relies on (b1 z + ⋅ ⋅ ⋅ + bk−j+1 z k−j+1 )j for such j. Applying the inductive hypothesis, we have j
k k−j+1 fk[n] = ωbk + ⋅ ⋅ ⋅ + a ) +⋅⋅⋅ j [z ](b1 z + ⋅ ⋅ ⋅ + bk−j+1 z ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ k
k
no ak in it
+ ak [z ](b1 z)
= ω[ak ωn−2 (1 + ωk−1 + ⋅ ⋅ ⋅ + ω(k−1)(n−2) ) + Pn−1,k ] + ⋅ ⋅ ⋅ + aj (polynomial with no ak s) + ⋅ ⋅ ⋅ + ak ω(n−1)k
= ak ωn−1 [ak ωn−2 (1 + ωk−1 + ⋅ ⋅ ⋅ + ω(k−1)(n−2) )] + ωPn−1,k + ak ω(n−1)k + Pn,k (ω, . . . , ak−1 )
= ak ωn−1 (1 + ωk−1 + ⋅ ⋅ ⋅ + ω(k−1)(n−1) ) + Pn,k (ω, . . . , ak−1 ), where Pn,k (a1 , a2 , . . . , ak−1 ) is the result of summing all the terms which do not contain ak . By induction on Pn−1,k and the definition of Pn,k , it follows that if aj = 0 for all j with 1 < j < k, then Pn,k (a1 , 0, . . . , 0) = 0. Moreover, since all computations involve taking integral powers, the coefficients of Pn,k are integers. Theorem 8.4.11. If n ∈ ℕ and ω is a primitive nth root of unity in ℂ, then for every infinite sequence {ak }1 1. Let B = Bϵ (0) be an open ball with a sufficiently small radius ϵ in which f (x) ≠ 0 if x ≠ 0. Let f [p] be pth iterate of f such that fp approached to zero uniformly as p → ∞. Inside B every f [p] exists, vanished only for x = 0 at which it has a zero of order mp . The mp th roots of f (x) will be uniform within B, with simple zeros at x = 0. The first (mp −1)/(m−1) coefficient in the development of f [p] is am . We can then select a sequence of functions 1/m
[f (x)]
1/m2
, [f [2] (x)]
, . . . , [f [p] (x)]
1/mp
,...,
(8.13)
such that the coefficients of x in the terms of the sequence approach any chosen (m − 1)th root of am . Now we present the Böttcher’s theorem. The proof is made by Ritt and can be found in [82].
8.4 Schröder’s problem
 283
Theorem 8.4.12 (Böttcher’s theorem). In the circle B described above, the functions in (8.13) converges uniformly to an analytic function ϕ(x), which satisfies the functional equation m
ϕ(f (x)) = [ϕ(x)] .
9 Formal series and general exponents Since the space of formal power series 𝕏 was systematically established, the completeness of this space has become a task for mathematicians to work out. By means of the completeness axiom, the system ℝ gives us a lot of important properties, including the exponents bs where b > 0, s ∈ ℝ. We realize that some parts of the power (under multiplication) of the formal power series have been missing from 𝕏 except the power of formal series of the integer exponents. This chapter will add those missed parts to 𝕏, or to the formal analysis. The main work of this chapter was done by Dariusz Bugajewski and the author. This chapter will answer the following questions: 1. For a formal power series f ∈ 𝕏(ℂ) such that f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ , is r
∞
n
r
f (z) = ( ∑ an z ) , n=0
2. 3.
or
or
f = (a0 , a1 , a2 , . . . ),
(a0 , a1 , a2 , . . . )r ,
where r ∈ ℝ, still a formal power series in 𝕏(ℂ)? If (a0 , a1 , a2 , . . . )r = (c0 , c1 , c2 , . . . ) ∈ 𝕏(ℂ), what is the formal power series (c0 , c1 , c2 , . . . )? Is there any algorithm for computing all ck in #2 above?
We introduce the power of formal power series with general exponents via three steps. Section 9.2 introduces the formal root series g 1/n , n ∈ ℕ and the algorithm of computing g 1/n . Section 9.3 introduces the formal power series with rational exponents, or g r , r ∈ ℚ. We would like to indicate that in both Section 9.2 and Section 9.3 the approach of our investigation is the algebraic approach. The third step is introducing the power of the formal power series with real exponent. The power of formal power series with real exponents are introduced by the analytical approach in Section 9.4. Of course, radicals and rationals are also real numbers, and so there are two ways to understand and study the formal root series and the power of formal power series with rational exponents. We will see that there is no conflict no matter we treat rational exponents independently or as a subset of the power of formal power series with real exponents. The integral exponent of a formal power series, or the nth power of a formal power series f , has been investigated for many years, especially after 2002. Some studies were introduced in Chapter 5. The sets ℤ and ℚ are subsets of ℝ; therefore, the algorithms for the integer exponents and for the rational exponents should work well with the algorithms for real exponents. We will see that there is no conflict between the algorithms for the formal power series. At the end of this chapter, the powerful Exponent Algorithm will be introduced. https://doi.org/10.1515/9783110599459009
286  9 Formal series and general exponents
9.1 Introduction Different multiplication generates different power and, therefore, we may anticipate that they should generate different roots. The iterative roots have been developed for years and obtained many interesting results, which we have seen in Chapter 8. It is clear that 𝕏(S) itself is a ring and, if the S is commutative, 𝕏(S) is an ideal under the addition, scalar multiplication and the multiplication or the Cauchy product. It is then natural and necessary to investigate the radical of 𝕏(S) if S is ℝ or ℂ, or investigate the possible formal root power series in 𝕏(S). Before the formal root series was established systematically, certain kinds of root formal power series had been investigated already. For example, Niven introduced the formal logarithm of a rational power of a formal power series and obtained that L(αr ) = rL(α) where r ∈ ℚ [61]. Also, Proposition 7 in [47] described the properties of Riordan matrices involving the real exponents of formal power series. All these indicate the necessity of the establishment of the algebraic roots in the ring of formal power series, including the existence and uniqueness of such exponents n1 of a formal power series g, or g 1/n . Given a real number b > 0, the value of bs , s ∈ ℝ, the real exponent of the base b, could be an example for us to try to set up f s , s ∈ ℝ, f ∈ 𝕏(ℝ), the formal exponent series. We immediately realize that it is not helpful because we do not assign any value to the variable in the formal power series and, therefore, we cannot use the limit or convergence as we did for bx . Let f (x) = 1 + x + 2!x2 + 3!x3 + ⋅ ⋅ ⋅, the following expressions: f 3 (x),
f 1/3 (x),
f 3/2 (x),
f π (x),
may represent the powers of f with possible integer exponent, rational exponent and real exponent of f if we can define them mathematically. We want to know what formal power series they may represent and to know, if they exist, the existence, and the uniqueness of them. We answered some questions about f n , n ∈ ℕ in Chapter 5. The investigations of the multiplicative root f 1/n in 𝕏(S) for S = ℝ and S = ℂ, or the formal root power series with respect to the Cauchy product, have been developed by Reich, Smítal, and Štefánková since 2005 [76, 77]. It is known that f n exists for a formal power series f over any commutative ring S. Unfortunately, it is not clear whether systematically there exists a formal power series f such that f n = g for a given formal power series g and a given positive integer n, although the constant term of such f has been investigated by Henrici in [42]. It is also not clear whether such f , if exists, is unique in the sense of the first nonzero coefficient. More importantly, we need some algorithm to compute he formal root series if it exists. We should remind the readers that we have seen some series such as Puiseux series whose terms may have rational exponents, and we also have some sequence of
9.2 Formal root series and its algorithm
 287
functions with the power 1/mp , m ∈ ℕ in (8.13). Comparing them with what we will see in this chapter may help us to understand and study the formal root series. Computation of formal power series, including the computation of inverse formal power series, no matter computing by recurrent formula or by the explicit Wrowski formula, is a importance subject for mathematics. The computation of certain composition of formal power series such as J. C. P. Miller formula has special interest in computational mathematics and combinatorics. In addition to establish certain sufficient and necessary conditions for the existence of the formal root power series, we provide algorithms to compute those formal root power series if they exist. The uniqueness of such existing formal root series will be investigated, too. Before we go to the next section, we provide the definition of a formal root (power) series. Definition 9.1.1. Let S be a ring. Let k ∈ ℕ be given and let g ∈ 𝕏(S) be given. If there exists a formal power series f ∈ 𝕏(S) such that f k = g, then f is called a kth root formal root series of g, and f is denoted by g 1/k . The rational exponents of formal power series will be defined on 𝕏+ (ℝ), very similar to the definition of the rational exponents br , b > 0. The real exponents of formal power series is a big challenge. It is a new accomplishment of the formal analysis by the author, which will be discussed in the final section.
9.2 Formal root series and its algorithm At the beginning, we consider some examples which show that some formal power series have root series and some formal power series do not have any nontrivial root series. Example 9.2.1. Let S = ℝ and let ∞
−2 n )z , n
g(z) = ∑ ( n=0
a binomial formal power series Ba with a = −2. Then g(z) =
1 . (1+z)2
We have
g 1/2 (z) = 1 − z + z 2 − z 3 + ⋅ ⋅ ⋅ , because 2
2
(1 + z)2 (1 − z + z 2 − z 3 + ⋅ ⋅ ⋅) = [(1 + z)(1 − z + z 2 − z 3 + ⋅ ⋅ ⋅)] = 1. Let us notice that we did not have to consider the convergence of the above series.
288  9 Formal series and general exponents k Example 9.2.2. Let S = ℝ and let g(z) = ∑∞ k=1 z . Assume that there exists a formal n power series f such that f = g for some integer n ≥ 2. Then ord(f ) = 0 yields that ord(f n ) = 0 ≠ ord(g), and ord(f ) > 0 would provide that
ord(f n ) = n ord(f ) ≥ 2 ≠ 1 = ord(g), Thus, there is no formal power series f such that f n (z) = g(z) if n ≥ 2. Example 9.2.1 and Example 9.2.2 show that some formal power series has certain root series and some formal power series does not have any nontrivial root series. Moreover, it might happen that some formal power series has all nth formal root series. n Remark 9.2.3. Let S = ℝ and let g(z) = ∑∞ n=0 n! z . It is clear that the regular power ∞ series ∑n=0 n! xn converges nowhere except x = 0. However, we will prove that g has nth formal root series for every n ∈ ℕ.
Now we prove some auxiliary results. Lemma 9.2.4. Let f (z) = as z s +as+1 z s+1 +⋅ ⋅ ⋅ be a formal power series with s = ord(f ) > 0 and let (n) (n) 2 (n) k f n (z) = a(n) 0 + a1 z + a2 z + ⋅ ⋅ ⋅ + ak z + ⋅ ⋅ ⋅ .
= 0, 0 ≤ k < sn. Then a(n) k For k ≥ sn, we have
n − rs − rs+1 − ⋅ ⋅ ⋅ − rk−1 rs rs+1 n n − rs r )as as+1 ⋅ ⋅ ⋅ akk , )⋅⋅⋅( a(n) = ∑ ( )( k rk rs+1 rs
(9.1)
where the sum is extended over all possible tuples (rs , rs+1 , . . . , rk ) of nonnegative integers such that rs + rs+1 + ⋅ ⋅ ⋅ + rk = n
and
srs + (s + 1)rs+1 + ⋅ ⋅ ⋅ + krk = k.
(9.2)
Proof. Since ord(f ) = s > 0, we have ai = 0, 0 ≤ i < s and, therefore, it is clear that r
r
r
a00 a11 ⋅ ⋅ ⋅ akk = 0 if any ri ≠ 0, 0 ≤ i < s. Since (m ) = 1 for any m ∈ ℕ, by (5.4) we have 0 n − r0 − r1 − ⋅ ⋅ ⋅ − rk−1 r0 r1 n n − r0 n − r0 − r1 r )a0 a1 ⋅ ⋅ ⋅ akk )( )⋅⋅⋅( a(n) = ∑ ( )( k rk r2 r1 r0 n − rs − rs+1 − ⋅ ⋅ ⋅ − rk−1 rs rs+1 n n − rs n − rs − rs+1 r )as as+1 ⋅ ⋅ ⋅ akk , )⋅⋅⋅( = ∑ ( )( )( rs+1 rs+2 rk rs where the sum is extended over all possible (rs , rs+1 , . . . , rk ) of nonnegative integers such that rs + rs+1 + ⋅ ⋅ ⋅ + rk = n
and srs + (s + 1)rs+1 + ⋅ ⋅ ⋅ + krk = k.
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 289
Corollary 9.2.5. Let g ∈ 𝕏(ℂ) be a nonunit formal power series with ord (g) = k > 0. If the root formal power series g 1/m exists, then ord(g 1/m ) = l, where ml = k for some l ∈ ℕ. Corollary 9.2.6. Let g be a nonzero formal power series over ℂ. If g 1/n exists for all n ∈ ℕ, then ord(g) = 0. Recalling the formulas (5.3) and (5.4), we have the following. Lemma 9.2.7. Let f (x) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ be a formal power series in 𝕏(ℂ). Let a(n) k be defined as in (5.3). Then a(n) = B0 (f ) + B1 (f ), k where
n − r0 − r1 − ⋅ ⋅ ⋅ − rk−2 r0 r1 n n − r0 rk−1 )a0 a1 ⋅ ⋅ ⋅ ak−1 )⋅⋅⋅( , B0 (f ) = ∑ ( )( rk−1 r1 r0 B1 (f ) = nan−1 0 ak
and the sum is extended over all possible sequences (r0 , r1 , . . . , rk−1 ) of nonnegative integers satisfying the formulae r0 + r1 + ⋅ ⋅ ⋅ + rk−1 = n,
r1 + 2r2 + ⋅ ⋅ ⋅ + (k − 1)rk−1 = k.
Proof. By (5.3), we have n − r0 − r1 − ⋅ ⋅ ⋅ − rk−1 r0 r1 n n − r0 n − r0 − r1 r )a0 a1 ⋅ ⋅ ⋅ akk , )⋅⋅⋅( )( a(n) = ∑ ( )( k rk r2 r1 r0 where the sum is extended over all sequences (r0 , r1 , . . . , rk ) of nonnegative integers that satisfy (5.4). By the second equation in (5.4), it is clear that rk can admit only two values, that is, rk = 0
and rk = 1.
If rk = 0, then the summation above becomes
n n − r0 n − r0 − r1 − ⋅ ⋅ ⋅ − rk−2 r0 r1 rk−1 )⋅⋅⋅( ) a0 a1 ⋅ ⋅ ⋅ ak−1 , ∑ ( )( r0 r1 rk−1
where the sum is extended over all possible sequences (r0 , r1 , . . . , rk−1 ) of nonnegative integers such that r0 + r1 + ⋅ ⋅ ⋅ + rk−1 = n, however, this is B0 (f ).
r1 + 2r2 + ⋅ ⋅ ⋅ + (k − 1)rk−1 = k;
290  9 Formal series and general exponents If rk = 1, then r1 = r2 = ⋅ ⋅ ⋅ = rk−1 = 0
and then r0 = n − 1,
by the first equation in (5.4). Then the summation in this case reduces to (
n )a = nan−1 0 ak ; n−1 k
however, this is B1 (f ). We provide some necessary and sufficient conditions for a formal power series g to have formal root series g 1/n for some n ∈ ℕ. Some algorithms of computing the nth formal root series will be presented, too [29]. Let 𝕂 ∈ {ℝ, ℂ}. j Theorem 9.2.8. Fix g(z) = ∑∞ j=0 bj z ∈ 𝕏(𝕂) with b0 ≠ 0 and let n be a positive integer. j n If g 1/n (z) = ∑∞ j=0 aj z exists, then a0 = b0 . Conversely, for any fixed solution a0 ∈ 𝕂 of the equation an0 = b0 , the unique nth n formal root g 1/n (z) = ∑∞ n=0 an z can be computed by the following algorithm:
(i) a1 = b1 /(nan−1 0 ); (ii)
n − r0 − r1 − ⋅ ⋅ ⋅ − rk−2 r0 r1 n n − r0 rk−1 )a0 a1 ⋅ ⋅ ⋅ ak−1 ] )⋅⋅⋅( ak = [bk − ∑ ( )( rk−1 r1 r0 /(nan−1 0 ) for
= (bk − ∑
k≥2
n! rk−1 r r a 0 a 1 ⋅ ⋅ ⋅ ak−1 )/(nan−1 0 ), r0 !r1 ! ⋅ ⋅ ⋅ rk−1 ! 0 1
where by (9.3) below n! n − r0 − r1 − ⋅ ⋅ ⋅ − rk−2 n n − r0 )= , )⋅⋅⋅( ( )( rk−1 r0 !r1 ! ⋅ ⋅ ⋅ rk−1 ! r0 r1 and the sum is extended over all the sequences (r0 , r1 , . . . , rk−1 ) of nonnegative integers such that r0 + r1 + ⋅ ⋅ ⋅ + rk−1 = n,
r1 + 2r2 + ⋅ ⋅ ⋅ + (k − 1)rk−1 = k.
Proof. Suppose that f (z) = g 1/n is an nth root of g. Then f n (z) = g(z). Let us write (n) (n) k f n (z) = a(n) 0 + a1 z + ⋅ ⋅ ⋅ + ak z + ⋅ ⋅ ⋅ . (n) (n) n n Then a(n) 0 = b0 , so a0 = b0 , because a0 = a0 . Also, bk = ak for all n ∈ ℕ. Now we are going to prove the uniqueness. For that, let ∞
h(z) = ∑ cj z j = g 1/n . j=0
(9.3)
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 291
Then c0 = b1/n 0 = a0 . Suppose that ai = ci
for
0≤i 0, then: (a) there exists a unique root g 1/n (z) for arbitrary odd positive integer n; (b) there exist exactly two different nth roots, or ±g 1/n , for even positive integer n. We always pick the notation g 1/n to represent the root with g 1/n (0) > 0 unless we specify otherwise. The formal power series g 1/n with the positive first nonzero coefficient is called the principal nth root of g. 2. If 𝕂 = ℝ and b0 < 0, then there exists a unique root g 1/n (z) if and only if n is arbitrary odd positive integer. 3. If 𝕂 = ℂ, then for arbitrary positive integer n there exist exactly n different roots g 1/n (z).
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 293
We should understand that the uniqueness of the nth root of g is in the sense of the first nonzero coefficient rule that is, the first nonzero coefficient of the formal root series determines the formal root series. This rule includes (b) of Corollary 9.2.9 in which the uniqueness of the notation g 1/n is determined when n is even. j Example 9.2.10. Let g(x) = ∑∞ j=0 j! x ∈ 𝕏(ℝ) and let n ∈ ℕ be given. We are going to
compute the nth formal root series of g. Writing g 1/n (x) = a0 + a1 x + a2 x 2 + ⋅ ⋅ ⋅, and applying the algorithm of Theorem 9.2.8 for bj = j!, j = 0, 1, 2, . . ., we have a0 = 1,
a1 = 1/n,
1 3 1 a2 = ( + 2 ), 2 n n
....
Then g 1/n (x) = 1 +
1 1 3 1 x + ( + 2 )x 2 + ⋅ ⋅ ⋅ . n 2 n n
Taking n = 2, we have 1 1 3 1 g 1/2 (x) = 1 + x + ( + 2 )x 2 + ⋅ ⋅ ⋅ , 2 2 2 2 and then 2
(g 1/2 (x)) = 1 + 1!x + 2!x 2 + ⋅ ⋅ ⋅ . Example 9.2.11. Let ∞
−2 n )z = 1 − 2z + 3z 2 − 4z 3 + ⋅ ⋅ ⋅ , n
g(z) = ∑ ( n=0
be given. By Example 9.2.1, we know that g 1/2 (z) = 1 − z + z 2 − z 3 + ⋅ ⋅ ⋅ . Now, we get the same outcome applying the algorithm given in Theorem 9.2.8 and we provide the first four coefficients of g 1/2 and the formula for kth coefficient with k ≥ 4. For our convenience, we would like to use the equality n n − r0 n − r0 − r1 − ⋅ ⋅ ⋅ − rk−2 n! ( )( )⋅⋅⋅( )= r0 r1 rk−1 r0 !r1 ! ⋅ ⋅ ⋅ rk−1 ! where all rj satisfy (9.3). Now suppose that g 1/2 (z) = a0 + a1 z + a2 z 2 + a3 z 3 + ⋅ ⋅ ⋅ .
294  9 Formal series and general exponents It is obvious that a0 = 1 and a1 = −1. For k = 2, by (9.3), we get r0 + r1 = 2, r1 = 2, and then r0 = 0,
r1 = 2
is the only possibility for the summation, and hence 2 2 − r0 r0 r1 1 )a0 a1 ] a2 = [b2 − ( )( 2 r0 r1 1 = [3 − 1 ⋅ 1 ⋅ 1 ⋅ 1] = 1. 2 For k = 3, by (9.3) and the fact that all rj ’s are nonnegative integers, r0 = 0,
r1 = 1,
r2 = 1
is the only selection for those rj ’s and, therefore, 1 2 2 1 a3 = [−4 − ( )( )( ) a00 a11 a12 ] 2 0 1 1 1 = [−4 − 1 ⋅ 2 ⋅ 1 ⋅ 1 ⋅ (−1) ⋅ 1] = −1. 2 We claim: ak = (−1)k ,
k∈ℕ
by using (ii) of Theorem 9.2.8.
Now let k ∈ ℕ be such that k ≥ 4 and assume that ai = (−1)i for 0 ≤ i ≤ k − 1. Then bk = (−1)k (k + 1), and let the nonnegative integers r0 , r1 , . . . , rk−1 satisfy (9.3). First, we investigate r0 . If r0 = 2, then rj = 0 for all other rj and hence the second equation in (9.3) is not true. If r0 = 1, then there is only one rj = 1, 1 ≤ j ≤ k − 1, and all other rj are zeros. Then the second equation in (9.3) would be j = k for j ≤ k − 1 which is a contradiction. Therefore, we infer that r0 = 0. Then (9.3) becomes (a) r1 + r2 + ⋅ ⋅ ⋅ + rk−1 = 2,
(b) r1 + 2r2 + ⋅ ⋅ ⋅ + (k − 1)rk−1 = k.
We show that for the tuples (r1 , r2 , . . . , rk−1 ) satisfying (a) and (b) above, 2 2 − r0 2 − r0 − r1 − ⋅ ⋅ ⋅ − rk−2 r0 r1 rk−1 )⋅⋅⋅( )a0 a1 ⋅ ⋅ ⋅ ak−1 ∑ ( )( r0 r1 rk−1 2! rk−1 r r =∑ a 0 a 1 ⋅ ⋅ ⋅ ak−1 = (−1)k (k − 1). r1 !r2 ! ⋅ ⋅ ⋅ rk−1 ! 0 1
(∗)
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 295
Suppose that we have obtained that ai = (−1)i
for
0 ≤ i ≤ k − 1.
Case 1: rj = 2 for some 1 ≤ j ≤ k − 1. It occurs if and only if ri = 0 for all the other i ≠ j. Then k must be even because 2j = k by (b). That is, in this situation j = k/2 is the only such j. By inductive hypothesis, 2! 2! rk−1 r r a 0 a 1 ⋅ ⋅ ⋅ ak−1 = (−1)rj = 1. r1 !r2 ! ⋅ ⋅ ⋅ rk−1 ! 0 1 rj ! Case 2: 0 ≤ ri ≤ 1, 0 ≤ i ≤ k − 1. In this case, there are exactly only two such ri for which ri = 1. Let us assume that ri = 1 and rj = 1, and 1 ≤ i < j ≤ k − 1 and all the other rm are zeros. Then, by (b) above, j = k − i. Since each rm , 1 ≤ m ≤ k − 1, can only take the value 1 once, it follows that the total number of such pairs is (k−1)/2 if k is odd, and the total number of such pairs is (k−2)/2 2! if k is even because there is no such a pair for rk/2 . In this case, r !r !⋅⋅⋅r k−1 = 2. Then if k 1 2 is odd, ∑
2! rk−1 r r a 0 a 1 ⋅ ⋅ ⋅ ak−1 r1 !r2 ! ⋅ ⋅ ⋅ rk−1 ! 0 1 k−1 (−1)1r1 (−1)2r2 ⋅ ⋅ ⋅ (−1)(k−1)rk−1 =2⋅ 2 = (−1)k (k − 1) = −(k − 1) = (−1)k (k − 1),
and if k is even, ∑
k−2 2! rk−1 r r =2⋅ a 0 a 1 ⋅ ⋅ ⋅ ak−1 (−1)k = k − 2. r1 !r2 ! ⋅ ⋅ ⋅ rk−1 ! 0 1 2
For k even, combining Case 1 and Case 2, we have ∑
2! rk−1 r r a 0 a 1 ⋅ ⋅ ⋅ ak−1 = (Case 1) + (Case 2) r1 !r2 ! ⋅ ⋅ ⋅ rk−1 ! 0 1 k−2 =1+2⋅ (−1)k = 1 + (k − 2) = k − 1 = (−1)k (k − 1). 2
The sum is extended over all such rj , 1 ≤ j ≤ k − 1 satisfying (a) and (b). Thus, (∗) is proved. Applying (ii) of Theorem 9.2.8, we get ak = [bk −
2! rk−1 r r a 0 a 1 ⋅ ⋅ ⋅ ak−1 ]/(2an−1 0 ) r1 !r2 ! ⋅ ⋅ ⋅ r k−1 0 1
= [(−1)k (k + 1) − (−1)k (k − 1)]/2 = (−1)k . Thus ak = (−1)k for every nonnegative integer k.
296  9 Formal series and general exponents The next example shows that a polynomial p(x) on ℝ usually does not have a rootpolynomial (a polynomial f such that f k = p) but may have a formal root series. Example 9.2.12. Let p(x) = 1 + x be a polynomial on ℝ. Then p does not have a rootpolynomial f such that f 2 = p. However, there is a formal power series f ∈ 𝕏(ℝ) such that f 2 = p and 1 1 1 f (x) = 1 + x − x 2 + x 3 + ⋅ ⋅ ⋅ . 2 8 16 Denote f (x) = a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅, applying Theorem 9.2.8, we easily obtain that a1 = b1 /(2a2−1 0 ) = 1/2.
a0 = 1,
For all integer k ≥ 2, formula (9.3) becomes r0 + r1 + ⋅ ⋅ ⋅ + rk−1 = 2,
r1 + 2r2 + ⋅ ⋅ ⋅ + (k − 1)rk−1 = k,
applying (ii) of Theorem 9.2.8 with bk = 0, we have n − r0 − ⋅ ⋅ ⋅ − rk−2 r0 r1 1 2 2 − r0 rk−1 )a0 a1 ⋅ ⋅ ⋅ ak−1 . )⋅⋅⋅( ak = − ∑ ( )( rk−1 r1 2 r0 For k = 2, (9.3) becomes r0 + rk−1 = r0 + r1 = 2,
r1 = k = 2.
1
Then a2 = − 21 a21 = − 21 ( 21 )2 = − 81 = −( 22 ). For k = 3, (9.3) becomes r0 + r1 + r2 = 2,
r1 + 2r2 = 3.
Since r1 ≤ 2, the only solution for r1 + 2r2 = 3 is r1 = r2 = 1 which yields that r0 = 0. Then 1 1 2 2 1 1 1 1 1 a3 = − ( )( )( )a1 a2 = − ⋅ 2 ⋅ ⋅ (− ) = = ( 2 ). 2 0 1 1 2 2 8 16 3
In fact, 1
1
1
f (x) = B1/2 (x) = 1 + ( 2 )x + ( 2 )x 2 + ( 2 )x 3 + ⋅ ⋅ ⋅ , 1 3 2 a formal binomial series defined in Proposition 2.2.4 whose function representation is (1 + x)1/2 . Now, we discuss the case of nonunit formal power series, which was investigated in [77], too.
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 297
Theorem 9.2.13. Let G ∈ 𝕏0 (ℂ) be given such that ∞
∞
j=0
j=0
G(z) = ∑ bj z s+j = z s ( ∑ bj z j ) = z s g(z) and mn = s, where m, n and s are positive integers. Then ∞
∞
j=0
j=0
F(z) = ∑ aj z m+j = z m ( ∑ aj z j ) = z m f (z) is an nth root series of G(z) if and only if f (z) is an nth root series of g(z). Proof. Let f n (z) = g(z). Then obviously n
F n (z) = (z m f (z)) = z mn f n (z) = z s g(z) = G(z), so F(z) is an nth root of G(z). Conversely, let F n (z) = G(z). Then n
(z m f (z)) = z mn g(z),
z mn f n (z) = z mn g(z).
Since 𝕏(ℂ) is an integral domain, we have that f n (z) = g(z) and end the proof. By Theorem 9.2.8 and Theorem 9.2.13, we get the following. Corollary 9.2.14. Let ∞
∞
j=0
j=0
g(z) = ∑ bj z s+j = z s ∑ bj z j be a formal power series in 𝕏(𝕂) such that ord(g) = s > 0 and let n ∈ ℕ. If a formal j power series f (z) = ∑∞ j=0 aj z is an nth formal root series of g, then s = mn with suitable positive integer m, aj = 0
for all
0≤j ≤m−1
and
anm = b0 .
Conversely, if mn = s, then for every fixed solution a0 of the equation an0 = b0 , the unique nth formal root series ∞
∞
j=0
j=0
f (z) = ∑ aj z m+j = z m ∑ aj z j can be computed by the algorithm presented in Theorem 9.2.8. The last property of formal root series we would like to present in this section is given in the following.
298  9 Formal series and general exponents Proposition 9.2.15. Let f (z) = a0 + a1 z + ⋅ ⋅ ⋅ ,
g(z) = b0 + b1 z + ⋅ ⋅ ⋅
be formal power series in 𝕏(𝕂) such that f k = g, where k is a positive integer. If f1 (z) = a0 − a1 z + ⋅ ⋅ ⋅ + (−1)n an z n + ⋅ ⋅ ⋅ ,
g1 (z) = b0 − b1 z + ⋅ ⋅ ⋅ + (−1)n bn z n + ⋅ ⋅ ⋅ , then f1k = g1 . Proof. Let ℓ(z) = −z. We have f1 (z) = (f ∘ ℓ)(z) and g1 (z) = (g ∘ ℓ)(z). Since ℓ(z) is a nonunit, applying the right distributive law we obtain f1k = (f ∘ ℓ)k = (f k ∘ ℓ) = (g ∘ ℓ) = g1 , which completes the proof.
9.3 Power of formal power series with rational exponents There are two approaches to construct the rational exponents for formal power series: algebraical approach and analytical approach. In this section, we use the algebraical approach. The analytical approach will be discussed in the next section where the real exponents, including rational exponents in it, will be introduced. Remark 9.3.1. For r = m/n ∈ ℚ, three questions must be answered before setting up g r for g ∈ 𝕏(ℝ): 1. Is it true that (g 1/n )m = (g m )1/n ? 2. For positive integers m, n, p, q with m/n = p/q, is g m/n = g p/q ? 3.
What is g r if r ∈ ℚ and r < 0? Let us recall the subset 𝕏+ (ℝ) ⊆ 𝕏(ℝ) 𝕏+ (ℝ) = {f ∈ 𝕏(ℝ)  f (0) > 0},
defined in Definition 6.4.1. We would like to remind the readers that the notation g 1/n represents the principal nth root of g if g 1/n ∈ 𝕏(ℝ) and n is even. Lemma 9.3.2. Let f , g ∈ 𝕏+ = 𝕏+ (ℝ) be given such that f (0) > 0 and g(0) > 0. Then, for m, n ∈ ℕ, (i) (fg)1/n = f 1/n g 1/n ,
9.3 Power of formal power series with rational exponents  299
(ii) (f m )1/n = (f 1/n )m , 1 (iii) (f 1/n )1/m = f mn , 1 1 m+n (iv) f 1/n f 1/m = f n + m = f nm . Proof. Let n ∈ ℕ be given. By Corollary 9.2.9, all f 1/n , g 1/n and (fg)1/n are welldefined formal power series in 𝕏+ (ℝ). Let a = f 1/n , b = g 1/n , then an = f , bn = g and then fg = an bn = (ab)n . Then (fg)1/n = ab = f 1/n g 1/n . (i) is true. Taking f = g in (i), we have (f 2 )
1/n
2
= (f 1/n ) .
Inductively, we can obtain (ii). For (iii), let c = (g 1/n )1/m , d = g 1/(nm) , then g 1/n = cm
and dm = g 1/n ,
and then cm = dm . By the uniqueness of the formal root series and Corollary 9.2.9, we have (g 1/n )
1/m
= c = d = g 1/(nm) .
Applying (i) and (iii), we have 1
1
f n+m = f
m+n nm
1
1
= (f m+n ) nm = (f m ⋅ f n ) nm 1
1
= (f m ) nm (f n ) nm = ((f m ) =f
1/n 1/m
f
1/m 1/n
)
1/m 1/n
((f m )
)
.
This is (iv). We have answered the first question of Remark 9.3.1 by (ii) above. Corollary 9.3.3. Let g ∈ 𝕏+ = 𝕏+ (ℝ) and let n ∈ ℕ be given. Then g 1/n , g −1 ∈ 𝕏+ and (g −1 )
1/n
= (g 1/n ) , −1
where g 1/n is the principal nth root of g. We define g −1/n = (g −1 )
1/n
= (g 1/n ) . −1
300  9 Formal series and general exponents Proof. Since g(0) > 0, it follows that both g 1/n and g −1 are unit formal power series and 1/n
g 1/n (0) = (g(0))
−1
and g −1 (0) = (g(0))
>0
> 0,
and hence g 1/n , g −1 ∈ 𝕏+ . Let (g −1 )1/n = a, g 1/n = b. Then g −1 = an
g = bn .
and
Then I = g −1 g = an bn = (ab)n . Then ab = I 1/n = I, or (g −1 )1/n = (g 1/n )−1 . Define 1/n
g −1/n = (g −1 )
,
we extend Lemma 9.3.2 to work for the negative root series g −1/n , and of course, 1
1
1
1
g − n ⋅ g n = g − n + n = g 0 = I. Multiplying by itself each side of (g 1/n )−1 = (g −1 )1/n with (m − 1) times, we have (g −1 )
m/n
= (g m/n ) . −1
We define g −m/n = (g −1 )
m/n
= (g m/n ) , −1
which answers the third question of Remark 9.3.1. By this, we have Corollary 9.3.4. Lemma 9.3.2 is true for all nonzero integers m and n, where g 1/n represents the principal root if n is even. Lemma 9.3.5. Let g ∈ 𝕏+ = 𝕏+ (ℝ) be given. Let m, n, p, q be positive integers such that m/n = p/q, then 1/n
(g m )
= (g p )
1/q
.
Proof. Denote A = (g m )1/n , B = (g p )1/q . Then An = g m and Bq = g p , and hence n
q
q
n
(Aq ) = Aqn = (An ) = (g m ) = g mq = g pn = (g p )
because mq = pn. The uniqueness of formal root series, or Corollary 9.2.9 provides that Aq = g p . Thus, A = (g p )
1/q
= B.
9.3 Power of formal power series with rational exponents  301
We have answered the second question of Remark 9.3.1, and hence we have answered all three questions. We are ready to define the rational exponent of formal power series. Definition 9.3.6. Let g ∈ 𝕏+ = 𝕏+ (ℝ) be given. For any rational number r = m/n, m, n ∈ ℕ, n ≠ 0 in lowest terms, the power of formal power series g with rational exponent or formal rational exponent, g r , is defined to be 1/n
g r = g m/n = (g m )
.
If r = −s < 0, then g r is defined to be (g −1 )s where s > 0. If r = 0, we define that g 0 = I𝕏 . The uniqueness of the nth root in the sense of the first nonzero coefficient rule helped us to set up this definition, similar to what we have done for rational exponents of the real numbers. Proposition 9.3.7. Let f , g ∈ 𝕏 = 𝕏(ℝ) be given such that f (0) > 0 and g(0) > 0. For r, s ∈ ℚ, we have: (i) (fg)r = f r g r ; (ii) (f r )s = f rs ; (iii) f r ⋅ f s = f r+s . Proof. Write r = m/n, s = p/q where m, n, p, q ∈ ℕ, nq ≠ 0, and write f r = f m/n = a, g r = g m/n = b. Then f m = an , g m = bn , and then we have (fg)m = f m g m = an bn = (ab)n . Then (fg)r = (fg)m/n = ((fg)m )
1/n
= ab = f r ⋅ g r .
For (ii), applying (ii) of Lemma 9.3.2, s
p 1/q
(f r ) = ((f r ) ) = ((f 1/n )
mp n
= (f
1/q mp
)
1/q
)
1
= ((f 1/n )
mp 1/q
=f
= f rs .
mp
= (f nq )
mp nq
)
mp/q
= (f 1/n )
For (iii), denote A = f r = f m/n , B = f p/q . Then An = f m , Bq = f p , and hence (A ) = (f m )q , (Bq )n = (f p )n . By (ii) above, n q
q
n
(An ) ⋅ (Bq ) = f mq ⋅ f pn = f mq+pn . By (i) and then by (ii) above, we have q
n
1
[(An ) ⋅ (Bq ) ] nq = (An )
1 q⋅ nq
(Bq )
1 n⋅ nq
= A ⋅ B.
302  9 Formal series and general exponents Then 1
f r ⋅ f s = A ⋅ B = (f mq+pn ) nq = f
mq+pn nq
= f r+s .
If r or s are negative rational, applying Definition 9.3.6. We provide an example below for computing a rational exponent of a formal power series g using the existing algorithms for computing the formal root series g 1/n and the integer exponent f m . We would like to indicate that the more efficient algorithm will be introduced in the next section. The following is a good practice for what we have learned so far. Example 9.3.8. Let g(x) = 1 + x + x2 + x3 + ⋅ ⋅ ⋅ be a formal power series in 𝕏(ℝ). We want to find the first three terms of g 2/3 by using the formulas in Theorem 9.2.8. Suppose that f (x) = g 1/3 (x) = a0 + a1 x + a2 x 2 + ⋅ ⋅ ⋅, then g 2/3 = f 2 . It is obvious that a0 = 1 and a1 = 1/(3a3−1 0 ) = 1/3 by (i) of Theorem 9.2.8. Applying (ii) of Theorem 9.2.8, we have 3 3−1 1 2 a2 = [b2 − ( )( )a0 a1 ]/(3a3−1 0 ) 1 2 2
=
1 1 [1 − 3 ⋅ 1 ⋅ 1 ⋅ ( ) ] 3 3
= 2/9,
because the only possible values for the nonnegative integers r0 and r1 are r0 = 1,
r1 = 2,
by the formula (9.3) with n = 3, k = 2, that is, r0 + r1 = 3,
r1 = 2.
For a3 , the nonnegative integers r0 , r1 , r2 must satisfy equations r0 + r1 + r2 = 3,
r1 + 2r2 = 3.
Then the solutions of (9.3) can only be (r0 , r1 , r2 ) = (0, 3, 0) and (r0 , r1 , r2 ) = (1, 1, 1). Then 3 3−0 0 3 0 3 2 1 a3 = [b3 − ( )( )a0 a1 a2 − ( )( )( )a10 a11 a12 ]/3 0 3 1 1 1 3
1 1 1 2 [1 − ( ) − 3 ⋅ 2 ⋅ 1 ⋅ 1 ⋅ ( )( )] 3 3 3 9 14 = . 81
=
9.3 Power of formal power series with rational exponents  303
Then 1 2 14 g 2/3 (x) = f 2 (x) = (1 + x + x2 + x3 + ⋅ ⋅ ⋅) 3 9 81
2
2
1 1 2 1 14 12 3 = 1 + ( + )x + (2 ⋅ + ( ) )x2 + (2 ⋅ +2⋅ )x + ⋅ ⋅ ⋅ 3 3 9 3 81 39 5 40 3 2 x + ⋅⋅⋅. = 1 + x + x2 + 3 9 81 We may verify that g 2/3 (x) ⋅ g 1/3 (x) = 1 + x + x2 + x 3 + ⋅ ⋅ ⋅ = g(x). The formal power series g in the above example is very special because g can be expressed as a formal binomial series, that is, g(x) = 1 + x + x2 + ⋅ ⋅ ⋅ = (1 − x)−1 . Then g 2/3 (x) = 1 + (
−2/3 −2/3 −2/3 )(−x) + ( )(−x)2 + ( )(−x)3 + ⋅ ⋅ ⋅ 1 2 3 5 40 3 2 x + ⋅⋅⋅. = 1 + x + x2 + 3 9 81
This method, of course, is simpler than what we have done in Example 9.3.8. The next example is a perfect formal rational exponent because we don’t have any existing formula to treat it. j Example 9.3.9. Let g(x) = ∑∞ j=0 j! x ∈ 𝕏(ℝ) and let n ∈ ℕ be given. It is clear that
there is no meaning for g 1/n in classical analysis. However, as a formal power series g 1/n is a formal power series and can be computed by the algorithm of Theorem 9.2.8 with bj = j!, j = 0, 1, 2, . . .. We have a0 = 1,
1 3 1 a2 = ( + 2 ), 2 n n
a1 = 1/n,
....
Then 1 3 1 1 x + ( + 2 )x 2 + ⋅ ⋅ ⋅ , n 2 n n 1 5 g 1/3 (x) = 1 + x + x2 + ⋅ ⋅ ⋅ , 3 9 2 11 2/3 g (x) = 1 + x + x2 + ⋅ ⋅ ⋅ . 3 9
g 1/n (x) = 1 +
Theorem 9.3.10. Let g ∈ 𝕏 = 𝕏(ℝ) be given with g(0) > 0. Then for every r ∈ ℚ, the formal derivative of g r exists and (g r ) = rg r−1 g .
(9.4)
304  9 Formal series and general exponents Proof. Since g(0) > 0, it follows that g −1 ∈ 𝕏+ , and hence g s ∈ 𝕏+ for all s ∈ ℚ including g r−1 ∈ 𝕏. Equation (9.4) is true obviously if r = 0. We suppose that r = mn for some integers m and n with mn ≠ 0. Write g m/n = F. Then F(0) > 0 and gm = Fn. Applying the Power Rule and Chain Rule for the formal derivatives, mg m−1 g = nF n−1 F . Then F =
m g m−1 g. n F n−1
But m g m−1 1 g m−1 g m−1 = g n −1 = g r−1 . = m/n n−1 = m = m n−1 +1 m− − F (g ) g n g n
We obtain (9.4). If r = −s < 0, then s > 0 and s
(g r ) = ((g −1 ) ) = s(g −1 )
= sg −s+1
s−1
(g −1 )
−g = (−s)g −s−1 g g2
= rg r−1 g .
9.4 The real exponents of formal power series The development of the exponent of a real number bs , s ∈ ℝ, in real analysis went through the path b → bn → b−n → b1/n → bm/n → bs ,
b > 0,
(∗)
where m, n ∈ ℕ. That is, b must be positive, the development started with the integer exponent of b, then the negative integer exponent, then the exponent of 1/n, then with the rational exponent, and finally arrived at the real exponent. The establishment of the integral and rational exponent of a positive real number can be made algebraically. The real exponent of a positive real number is set up analytically by importing the least upper bound property or the completeness axiom of the real number system. A beautiful procedure of setting up the value of bx can be found on page 22 of [85], which at the same time sets up the value of hs (x) for x, s ∈ ℝ, and h(x) > 0.
9.4 The real exponents of formal power series  305
The power of formal power series with real exponent, hs , s ∈ ℝ, h ∈ 𝕏(ℝ) may run through a similar path h → hn → h−n → h1/n → hm/n → hs .
(∗∗)
The formal root series and formal rational exponent series have been algebraically established in Section 9.2 and Section 9.3 for some h ∈ 𝕏. We have completed the path in (∗∗) except the last step. We begin to work on the power of the formal power series with real exponent now. Once the real exponents of formal power series is established, the formal root series and the formal rational exponent series will become the special cases of the real exponents although they are built up via the different approaches. If h is a formal power series, we do not assign any value to the indeterminate or variable in the series, we expect the outcome of any operation, including the exponent operation, is still a formal power series. The rational exponents of formal power series have been defined, including the uniqueness, for every h ∈ 𝕏+ (ℝ) with h(0) > 0. We try to keep this character for formal real exponent series, that is, hs is a formal power series as well as that bs is a real value in calculus for b, s ∈ ℝ, b > 0. Can we establish hs (x), s ∈ ℝ as a formal power series without the completeness axiom? Theorem 9.4.1. Let s ∈ ℝ be given. For c ∈ ℝ with c ≠ 0, the solution g of the formal differential equation (c + x)g = sg,
(9.5)
1 s 1 s 1 s g(x) = g(0)[1 + ( )x + 2 ( )x2 + 3 ( )x 3 + ⋅ ⋅ ⋅]. c 1 c 3 c 2
(9.6)
is
Proof. If s = 0, then the solution g of the equation (9.5) is a constant or a constant formal power series, then g = g(0). Also (0k ) = 0 for all k ∈ ℕ, and hence (9.6) becomes g = g(0). Of course, g = g(0) is a solution of (9.5). Now let us suppose that s ≠ 0. If g(x) = b0 + b1 x + b2 x2 + ⋅ ⋅ ⋅ is a solution of (9.5), then (c + x)g (x) = cb1 + (2b2 c + b1 )x + ⋅ ⋅ ⋅ + ((n + 1)bn+1 c + nbn )xn + ⋅ ⋅ ⋅ . Let (c + x)g = sg, we have that (n + 1)bn+1 c + nbn = sbn for all n ∈ ℕ ∪ {0}. Then bn+1 =
s−n b , c(n + 1) n
n = 0, 1, 2, . . . .
Then s s 1 b = ( ) b0 , and 1c 0 1 c s−1 s 1 s(s − 1) b2 = b = b = ( ) 2 b0 . 2c 1 2 c 1.2c2 0 b1 =
306  9 Formal series and general exponents Suppose that bn = (ns ) c1n b0 for some n ∈ ℕ, then bn+1 =
s−n s−n s 1 s 1 b = ( ) b =( ) b . c(n + 1) n c(n + 1) n cn 0 n + 1 cn+1 0
Thus, s 1 bn = ( ) n b0 n c
for all n ∈ ℕ ∪ {0}.
So, 1 s 1 s 1 s g(x) = b0 [1 + ( )x + 2 ( )x 2 + 3 ( )x 3 + ⋅ ⋅ ⋅], c 1 c 3 c 2 where b0 ∈ ℝ is arbitrary but it is obvious that b0 = g(0). We know that 1
2
3
x s x s x s x Bs ( ) = 1 + ( )( ) + ( )( ) + ( )( ) + ⋅ ⋅ ⋅ . c 1 c 2 c 3 c Then the solution of equation (9.5) can be denoted by x g(x) = g(0)Bs ( ). c For x < c or  xc  < 1, applying the Maclaurin series of (c + x)s and letting g(0) = cs , we obtain 1
2
3
s x s x s x g(x) = cs [1 + ( )( ) + ( )( ) + ( )( ) + ⋅ ⋅ ⋅] 1 c 2 c 3 c s
x x = cs Bs ( ) = cs (1 + ) c c = (c + x)s .
While we are considering that g is a formal power series, not the Maclaurin series of (c + x)s , we know that for any nonunit f ∈ 𝕏0 (ℝ), the composition g ∘ f ∈ 𝕏(ℝ), or 1 s 1 s 1 s cs [1 + ( )f (x) + 2 ( )f 2 (x) + 3 ( )f 3 (x) + ⋅ ⋅ ⋅] c 1 c 3 c 2 is a formal power series on ℝ. We have already had the rational exponent series (c + f )r for r ∈ ℚ. We would like to show that there is no conflict between r ∈ ℚ and r ∈ ℝ. Let us see the corollary below.
9.4 The real exponents of formal power series  307
Corollary 9.4.2. Let g ∈ 𝕏+ (ℝ) be given with g(0) = c > 0 and f (x) = g(x) − g(0) ∈ 𝕏0 . Then for any s = n1 , n ∈ ℕ, 1
2
n
1/n f (x) 1/n f (x) )( ) + ( )( ) + ⋅ ⋅ ⋅]) = g(x). 1 c 2 c
(c1/n [1 + ( Or,
1
2
1/n f (x) 1/n f (x) ) + ( )( ) + ⋅ ⋅ ⋅]. )( c c 1 2
g 1/n (x) = c1/n [1 + ( Proof.
1
2
1/n f (x) 1/n f (x) )( ) + ( )( ) + ⋅ ⋅ ⋅]) 1 c 2 c
(c1/n [1 + (
n
n
f (x) f (x) )) = cBn(1/n) ( ) c c f (x) f (x) = cB1 ( ) = c(1 + ) c c = (c1/n B1/n (
= g(x).
Or 1
2
1/n f (x) 1/n f (x) )( ) + ( )( ) + ⋅ ⋅ ⋅]. 1 c 2 c
g 1/n (x) = c1/n [1 + (
From this point, we can easily prove that there is no conflict for the case s ∈ ℚ. We are now able to define the power of formal power series with real exponent. Definition 9.4.3. For every s ∈ ℝ, the mapping E s : 𝕏+ (ℝ) → 𝕏+ (ℝ) is defined by E s (h) = g ∘ h for all h = h(0) + h ∈ 𝕏+ (ℝ), where h ∈ 𝕏0 and g is the solution of the formal differential equation (9.5) with g(0) = (h(0))s . Letting c = h(0), we have 1 s 1 s 2 1 s 3 E s (h) = cs [1 + ( )h + 2 ( )h + 3 ( )h + ⋅ ⋅ ⋅], c 1 c 3 c 2 or E s (h)(x) = cs Bs (
s
h(x) h(x) ) = cs (1 + ) = hs (x). c c
(9.7)
308  9 Formal series and general exponents The formal power series E s (h) is called the real exponent to formal power series h or just formal real exponent series of h with respect to s. By Theorem 9.4.1, E s is welldefined on 𝕏+ (ℝ) and is called the formal exponent mapping on 𝕏+ (ℝ). The formal power series g is called the generating series of formal real exponent. We usually write E s (h) = hs for h ∈ 𝕏+ . Proposition 9.4.4. Let h ∈ 𝕏+ (ℝ), s ∈ ℝ and c > 0 be given. Then: (i) E s (h) ∈ 𝕏+ (ℝ) for all h ∈ 𝕏+ (ℝ), and E 0 (h) = I𝕏 . (ii) E s (cI𝕏 ) = cs I𝕏 . (iii) E s (c + x) = (c + x)s . If x < c, (c + x)s = cs (1 + xc )s can be expressed by the Maclaurin series for (1 + xc )s , which is the same as E s (c + x) by taking h = x in (9.7). (iv) E s (h) = Bs (β) for every β ∈ 𝕏0 and h = 1 + β. (v) E s (h) = h(0)s Bs (1 + h−h(0) ). h(0) (vi) E s (h) ∈ 𝕏+ (ℝ). (vii)If we write E = E 1 , then E 1 (h) = E(h) = h. E plays a role of a composition identity for all formal power series in 𝕏+ (generalization of I𝔻 ). Proof. It is obvious that E s (h) ∈ 𝕏+ (ℝ) for all h ∈ 𝕏+ (ℝ). If s = 0, then (0n ) = 0 for all n ∈ ℕ, and then (i) is followed by E 0 (h) = c0 (1 + 0 + 0 + 0 + ⋅ ⋅ ⋅) = I𝕏 . In formula (9.7), letting h = 0 we have (ii). (iii) is derived by (9.7) with h(x) = x. (iv), (v), and (vi) are obvious. If s = 1, then (k1 ) = 0 for all k ≥ 2. Then (v) yields that E 1 (h) = h(0)B1 (1 +
h − h(0) h − h(0) ) = h(0)(1 + ) = h. h(0) h(0)
This proves (vii). Theorem 9.4.5. Let h ∈ 𝕏+ (ℝ) be given. Then (E s (h)) = sE s−1 (h)h
for every real number s. Proof. Write h(x) = h(0) + h(x), where h = h − h(0) = h − c ∈ 𝕏0 (ℝ), then 2
3
s h(x) s h(x) s h(x) ) + ( )( ) + ( )( ) + ⋅ ⋅ ⋅]) (E s (h)) = (cs [1 + ( )( 1 c 2 c 3 c
2
s h (x) s h(x) h (x) s h(x) h (x) = cs [( ) + 2( ) + 3( )( ) + ⋅ ⋅ ⋅] 1 c 2 c c 3 c c 2
s − 1 h(x) s − 1 h(x) )( )+( )( ) + ⋅ ⋅ ⋅]h (x) c 2 c 1
= scs−1 [1 + (
9.4 The real exponents of formal power series  309
= sE s−1 (h)h (x).
Since h = h , it follows the conclusion. Theorem 9.4.6. Let h ∈ 𝕏+ (ℝ) be given. Then E s+r (h) = E s (h)E r (h) for all real numbers r and s. Proof. Using (v) of Proposition 9.4.4 and Proposition 2.2.7, h − h(0) h − h(0) ))(cr Br (1 + )) h(0) h(0) h − h(0) = cs+r Bs+r (1 + ) h(0) = E s+r (h).
E s (h)E r (h) = (cs Bs (1 +
Theorem 9.4.7. For every h ∈ 𝕏+ (ℝ) and s ∈ ℝ, Ln(E s (h) = s Ln(E(h)) = s Ln(h). Proof. It is obvious that E s (h) ∈ 𝕏+ (ℝ) and then Ln(E s (h)) ∈ 𝕏(ℝ). Let P = Ln(E s (h)) − s Ln(h). Then (E s (h)) h sE s−1 (h)h h −s = −s s s E (h) h E (h) h h sh − s = 0, = 1 h E (h)
P =
because E 1 (h) = h by (vii) of Proposition 9.4.4. Since P(0) = Ln(E s (h(0))) − s Ln(h(0)) = s ln(h(0)) − s ln(h(0)) = 0, It follows that P = 0. Theorem 9.4.8. Let f , g ∈ 𝕏+ (ℝ) be given. For every s ∈ ℝ, E s (fg) = E s (f )E s (g). Proof. It is obvious that fg ∈ 𝕏+ , then Theorem 9.4.7, (6.6), and (vii) of Proposition 4.4.3 yield that Ln(E s (fg)) = s Ln(fg) = s(Ln(f ) + Ln(g)), s
s
s
s
and
Ln(E (f )E (g)) = Ln(E (f )) + Ln(E (g)) = s Ln(f ) + s Ln(g). Since Ln is injective on 𝕏+ by Theorem 6.4.17, the conclusion is true.
310  9 Formal series and general exponents We now start to introduce the Exponent Algorithm for computing the real exponents of a formal power series if it exists. This algorithm consists of three equivalent formulas. Theorem 9.4.9. Let h ∈ 𝕏+ (ℝ) be given such that h(x) = h0 + h1 x + h2 x 2 + h3 x 3 + ⋅ ⋅ ⋅ . Let h(0) = h0 = c, h = h − c ∈ 𝕏0 and denote n
(n)
(n)
(n)
(n)
h (x) = h0 + h1 x + h2 x 2 + h3 x 3 + ⋅ ⋅ ⋅ , (n)
(1)
it is known that hk = 0, k < n and hk = hk for k ∈ ℕ. For every s ∈ ℝ, write (s) (s) 2 (s) 3 hs (x) = E s (h) = h(s) 0 + h1 x + h2 x + h3 x + ⋅ ⋅ ⋅ . s s Then h(s) 0 = h0 == c , and for all k ∈ ℕ,
h(s) k
(1)
(2)
(3)
(k)
s h s h s h s h = c [( ) k + ( ) k2 + ( ) k3 + ⋅ ⋅ ⋅ + ( ) kk ]. 1 c 2 c 3 c k c s
Proof. By (9.7) 2
3
s h(x) s h (x) s h (x) h (x) = E (h)(x) = c [1 + ( ) + ( ) 2 + ( ) 3 + ⋅ ⋅ ⋅]. 1 c 3 c 2 c s
s
s
s Then h(s) 0 =c . (n)
Since hk = 0 for k < n, it follows that for all k ∈ ℕ, h(s) k
(1)
(2)
(3)
(k)
s h s h s h s h = c [( ) k + ( ) k2 + ( ) k3 + ⋅ ⋅ ⋅ + ( ) kk ]. 1 c 2 c 3 c k c s
However, we have to say that the next approach of computing hs is simpler. Theorem 9.4.10. Let h ∈ 𝕏+ (ℝ) be given such that h(x) = h0 + h1 x + h2 x 2 + h3 x 3 + ⋅ ⋅ ⋅ . Let h(0) = c, h = h − c ∈ 𝕏0 and write h(x) = h0 + h1 x + h2 x 2 + h3 x 3 + ⋅ ⋅ ⋅ = 0 + h1 x + h2 x 2 + h3 x 3 + ⋅ ⋅ ⋅ .
For any s ∈ ℝ, we write (s) (s) 2 (s) 3 hs (x) = h(s) 0 + h1 x + h2 x + h3 x + ⋅ ⋅ ⋅ .
(9.8)
9.4 The real exponents of formal power series 
311
Let {an }∞ n=0 ⊆ ℝ be a sequence recurrently defined with a0 = 1, and for all n ∈ ℕ, an =
1 n−1 ∑ [s(n − k) − k]ak hn−k , nc k=0
(9.9)
an =
1 n−1 ∑ [(s + 1)k − n]an−k hk . nc k=0
(9.10)
or,
s Then h(s) 0 = c and for all n ∈ ℕ, s h(s) n = h0 an .
(9.11)
) where Proof. It is obvious that hs (x) = E s (h)(x) = cs Bs ( h(x) c h h h h h h 1 h(x) = 1 x + 2 x2 + 3 x3 + ⋅ ⋅ ⋅ = 1 x + 2 x 2 + 3 x 3 + ⋅ ⋅ ⋅ . c c c c c c c Applying the J. C. P. Miller formula (see Theorem 2.2.8) to Bs ( h(x) ) and write c Bs (
h(x) ) = a0 + a1 x + a2 x2 + a3 x 3 + ⋅ ⋅ ⋅ , c
we have that a0 = 1, and for all n ∈ ℕ, an = = =
h 1 n−1 ∑ [s(n − k) − k]ak n−k n k=0 c
1 n−1 ∑ [s(n − k) − k]ak hn−k nc k=0
1 n−1 ∑ [(s + 1)k − n]an−k hk . nc k=0
Since E s (h)(x) = cs Bs ( h(x) ), we have (9.11). c Definition 9.4.11. Formula (9.8) is equivalent to the formulas (9.9) and (9.10) with (9.11). The algorithm consists of these formulas is called the Exponent Algorithm. It is clear that formula (9.9) or formula (9.10) with formula (9.11) is simpler than formula (9.8). j + Example 9.4.12. Let h(x) = ∑∞ j=0 j!x ∈ 𝕏 (ℝ) be given. For n ∈ ℕ, write
h1/n (x) = a0 + a1 x + a2 x 2 + ⋅ ⋅ ⋅ .
312  9 Formal series and general exponents Applying (9.8) with c = 1, hj = j! for j ∈ ℕ ∪ {0}, we have 1 1 1 a1 = [ (1 − 0) − 0]a0 h1 = , 1 n n 1 1 1 1 3 1 a2 = [[ (2 − 0) − 0]a0 h2 + [ (2 − 1) − 1]a1 h1 ] = ( + 2 ). 2 n n 2 n n
c1/n = 1,
a0 = 1,
Then ∞
( ∑ j!xj )
1/n
j=0
=1+
1 1 3 1 x + ( + 2 )x 2 + ⋅ ⋅ ⋅ . n 2 n n
This is the same result of Example 9.2.10. Some properties of the formal power series f n and f 1/n and some algorithms for their coefficients have been existing for tens of years [25, 26, 42]. As we said at the beginning of this chapter, all properties of E s and the Exponent Algorithm must not conflict with those existing results when s = n or s = 1/n, n ∈ ℕ. Theoretically, there should not have any conflict, but we still like to confirm such equivalence by providing some cases. The following proposition shows the equivalence between the existing algorithms for the integral exponents and the algorithm for real exponent while the real exponent s is taking the integer value. Proposition 9.4.13. Let h ∈ 𝕏+ (ℝ) be such that h(x) = h0 + h1 x + h2 x 2 + h3 x 3 + ⋅ ⋅ ⋅ . Let s ∈ ℝ be given and write (s) (s) 2 (s) 3 hs (x) = h(s) 0 + h1 x + h2 x + h3 x + ⋅ ⋅ ⋅ .
If taking s = n ∈ ℕ, then for every k ∈ ℤ, the coefficient h(n) , obtained by (9.8) k (1)
(2)
(3)
(k)
s h s h s h s h = cs [( ) k + ( ) k2 + ( ) k3 + ⋅ ⋅ ⋅ + ( ) kk ], h(s) k 1 c 2 c 3 c k c of formula (5.12). where c = h0 , is the same as the a(n) k Proof. The conclusion is true for s = 0 obviously. We suppose that s = n ∈ ℕ. Taking f = h in (5.12), for k ∈ ℕ ∪ {0}, n ∈ ℕ, ak = hk ,
a(n) = h(n) , k k
(n)
a(n) k = hk .
Then (5.12) leads out k k n n ak (m) n a(n) = ∑ ( )an−m 0 ak = a0 ∑ ( ) m k m a0 m m=1 m=1
(m)
9.4 The real exponents of formal power series 
313
(m)
k n h = hn0 ∑ ( ) km = h(n) . k h0 m m=1
Equation (9.8) ensures the last step. Another existing algorithm we would like to confirm is the formula for f −1 in The∞ n −1 n orem 1.1.8. That is, if f (x) = ∑∞ n=0 fn x , f0 ≠ 0 and f (x) = ∑n=0 gn x , then g0 =
1 , f0
gn =
−1 n−1 ( ∑ g f ), f0 j=0 j n−j
n ∈ ℕ.
(9.12)
Corollary 9.4.14. Let h ∈ 𝕏+ (ℝ) be given such that h(x) = h0 + h1 x + h2 x2 + h3 x 3 + ⋅ ⋅ ⋅ . Write (−1) 2 (−1) 3 h−1 (x) = h(−1) + h(−1) 1 x + h2 x + h3 x + ⋅ ⋅ ⋅ , 0 ∞ where the sequence {h(−1) n }n=0 ⊆ ℝ is computed by the Exponent Algorithm (9.9). (−1) Then hn is exactly the same as gn in (9.12) when f = h.
Proof. Applying (9.9) with c = h0 , s = −1, we have that a0 = 1, and for all n ∈ ℕ, an =
1 n−1 ∑ [(−1)(n − k) − k]ak hn−k nc k=0
=
1 n−1 ∑ (−n)ak hn−k nc k=0
=
−1 n−1 ( ∑ a h ). h0 k=0 k n−k
Applying (9.11), we have h(−1) = c−1 ak and then k = c−1 an = h(−1) n =
−1 n−1 −1 ( ∑ c ak hn−k ) h0 k=0
−1 n−1 (−1) ( ∑ h hn−k ). h0 k=0 k
Recurrently, we have h(−1) = gn where gn is in (9.12) with f = h. n Before closing this chapter, we would like to remind the readers the principle of formal analysis once more. As a formal exponent of a formal power series, h(x) = (1+x)s means that s s s (1, 1, 0, 0, . . . )s = (1, ( ), ( ), ( ), . . . ). 1 2 3
314  9 Formal series and general exponents If x < 1, the power series h may take the value h(x), for example, π
π 1 3 π 1 π 1 ( ) = 1 + ( ) + ( ) 2 + ( ) 3 + ⋅⋅⋅, 3 2 2 1 2 2 2 which is a real number. We can claim now that the chain (∗∗) has been completed with the condition that h ∈ 𝕏+ (ℝ). A big challenge comes up immediately with the completeness of this chain (∗∗): what can we say for hr if r ∈ ℝ and h ∉ 𝕏+ (ℝ)? For example, what can we say for hr if h ∈ ℒs ? How many steps in (∗∗) can we make if h ∈ ℒs ? We leave this open question to the readers. This volume began with the classical binomial series s s s 1 + ( )x + ( )x2 + ⋅ ⋅ ⋅ + ( )x n + ⋅ ⋅ ⋅ , 1 2 n the earliest power series invented by Issac Newton. Fittingly, it ends with the formal real exponent series 2
3
s h(x) s h(x) s h(x) hs (x) = cs [1 + ( )( ) + ( )( ) + ( )( ) + ⋅ ⋅ ⋅] 1 c 2 c 3 c h = cs Bs ( ), c
c = h(0) > 0,
which has a strong relationship with the formal binomial series.
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Index 0 4 1/f 5 βdigits 204 βexpansion 204 βsequence 173, 238 βtransformation 204 ğ 206 δn,k 78 ℓω 277, 278 ℓp 212 ϵn (x) 204 ℂ⟦X1 , X2 , . . . , Xp ⟧ 38 𝔻 22 𝔻(S) 22 𝕂⟦z 1/n ⟧ 198 𝕃 206 𝕃p 212 𝕃p ≅ 𝕄p 225 𝕄(ψ) 231 𝕏 3 (𝕏, d) 71 𝕏g 131, 166 𝕏+ (ℝ) 180, 298 𝕏0 3 𝕏1 59, 179 𝕏(S) 3 ℒr 199 ℒp (β) 176, 237 𝒩k 37 μ(x) 203 𝜕E 172 ϕ ∗ ψ 227 ϕ ⊗ ψ 230 ℂ̃ ∗ 103 ̃f 224 ξI 224 A 17 Asequence 91 af 175
a(n) 20 k An 17 Abel’s limit theorem 146, 151 adjoint matrix 96 adjugate matrix 96 admissible sum 8, 38, 39 admitting addition 8, 38, 52, 129
admitting multiplication 45, 60 almost unit 22 annuli 189 annulus 207 Appell sequence 83, 95 Appell subgroup 93 Archimedean Property 69 associated semicirculant matrix 17 associated subgroup 94 asymptotic qadvanced Gevrey expansion f ̂(z) 104 automorphism 247 Ba 55, 64 Br (f ) 70 balanced 168 ball 70 Banach algebra 197 bases 30 basis 78 Bell polynomials 96 binomial series 1 Böttcher function 113 Böttcher’s equation 113, 252 Böttcher’s theorem 115, 283 boundary behavior test 172 B(X ) 238 Cf 133 canonical form 246, 251 canonical mapping 224 Catalan generating function 90, 122 Catalan number 122 Cauchy integral formula 207 Cauchy product 4 Cauchy–Kovalevskaya theorem 46 Cauchy’s method of majorants 46 Cayley–Hamilton theorem 96 Cayley’s lemma 273 centralizer of ℓω 278 Chain Rule 52, 150, 222 change of variable 261 characteristic function 224 characteristic polynomial 123 clopen 67 closed 66 completeness axiom 304
320  Index
composed matrix 132 composition domain 166 composition identity 308 composition of formal power series 8 composition order 275 composition order n 276 conjugate space 238 constant formal power series 4 contractive 76 convergent power series 166 convex 168 convolution 227 Dnorm 203 deg 4 degree of a formal power series 4 delta associated 132 delta associated matrix 22, 133, 158, 194 delta series 22 derivation 247 determinant 96 D(f ) 22 d(f ) 69 D(g) 166 Dhombres functional equation 110 diameter 66 digital 161 Diophantine approximation 202 Dirichlet’s theorem 203 discrete 75 discrete metric 68 distance 66 distribution rule 130 distributive laws of composition 41 division ring 5 dot product 208 DP(g) 208 dual space 238 Ea 54, 62, 64, 183 E s 307 equal formal power series 3 Exponent Algorithm 310, 311 extended step functions 224 f 49 f ∼ ϕ 246 F space 238 fk(n) 8
[f ]n 10, 52 Fq ((z −1 )) 203 f −1 5 f [−1] 26, 148 f 0 30 fn 8 f [n] 31 Faà di Bruno’s Formula 95 f (c) 3 Fermat’s little theorem 97 Fibonacci matrix 90, 94 Fibonacci numbers 95 Fibonacci sequence 126 first nonzero coefficient rule 293, 301 fixed point 72, 237 formal analysis 41, 131 formal analytic 161, 165 formal analytic domain 166 formal analytic point 147 formal binomial series 28, 55, 89, 125, 296, 314 formal derivative 49 formal differential equation 54 formal exponent mapping 308 formal exponential series 54 formal iteration group of type II 112 formal Laurent series 118, 190, 206 – complex conjugate 206 – composition 218 – formal derivative 221 – identity 209 – kth shifting 208 – order 233 – principal part 206, 219 – product 209 – regular part 206, 219 – reverse 206 – reverse shifting set 208 – ultrametric 234 formal Leibniz’s Formula 52, 154 formal logarithm 58 formal logarithmic function 58, 130, 179 formal logarithmic series 180 formal mathematics 40 formal polynomial 10 formal power series 2, 3 – general composition 132 – inverse of 18 – matrix representation associated with multiplication 132
Index  321
– order 148 – product rule 50 formal power series solution 102 formal rational exponent 301 formal real exponent series of h 308 formal root series 287 formal series 42 formal solution 105 formal sum 42 formal translation equation of type II 112 formally analytic 147 formally summable 42 function representation 37, 43 functional multiplication 230 g ⋅ f 208 gcomposition subset 166, 170 Gδ 238 g+ 206 g− 206 g−1/n 299 g1/n 287 gr 301 general Chain Rule 150, 152 general composition theorem 141 general formal Laurent series 206 general formal logarithm 180 generalized exponential functions 112 generalized logarithmic functions 112 generalized Sheffertype polynomial sequence 202 generating function 85, 125, 201 generating series for Ln 180 generating series of formal real exponent 308 greatest integer function 136 group 25 ℋp (β) 173 holomorphic 189 homogeneous 123 hypercyclic 238 I 4 I𝔻 22 IR 87 identity formal power series 4 identity of composition 22 I(g) 166 index of degree k 37
inhomogeneous 123 integer part 203 integral domain 7, 35, 215 inverse of formal power series 5 involution 92 irrationality exponent 203 isometric 225 isomorphism 225 isosceles triangle principle 66, 204 iterated functions 1 iterated inverse 26 iteration 264 iteration group 111 iterative conjugate 246 iterative equivalent 246 iterative fold 31 iterative inverse 26, 87, 148 iterative power 12, 149 iterative root 34, 245 iterative root series 245 iterative roots 269 I(x) 4, 87 J. C. P. Miller formula 57, 311 Kronecker delta function 78 kth coefficient of f 3 kth coefficient of f n 8 L 59 L3 209 Lagrange inversion 115, 158 Lagrange–Bürmann expansion 196 Lagrange–Bürmann formula 122 least upper bound property 304 Lebesgue measurable 224 lexicographic order 37 limit of difference quotient 50, 164 linear difference equation 123 linear functional 78 liner difference operator 123 Liouville number 203 ⟦⟧ 136 Ln 180 long operations 20, 21, 24, 34 M0 224 𝕄p 224 Mz 175
322  Index
Maclaurin series 58 majorant 46 Majorant lemma 47 matrix approach 158 matrix representation associated with multiplication 17 metric space 65 Minkowski sum 166 multinomial cofactor 137 multinomial theorem 136 multiplication 4 multiplication operator 178 natural boundary 162 Newton–Puiseux theorem 199 nilpotent 250 nonArchimedean metric 77 nonArchimedean norm 204 nonArchimedean valuation 69 nonexpansive mapping 75 nonformal Laurent series 207 nonunit 3 nonunit associated matrix 22 nth partial sum 10, 52 ⊙ 201 OPn 35 open 66 orbit 238 orb(T , x) 238 order 3 𝕏g 166 𝕏(S) 166 P 88 pdot product 208 Pi 26, 261 [j] Pk 32 partition function 99 Pascal matrix 88 p(n) 100 power of formal power series g with rational exponent 301 power rule 52 primitive nth root of unity 270 principal nth root 292, 298 principle of formal analysis 41, 130, 161, 197, 210, 313 product rule 50, 153
pseudo basis 30, 82 pseudoinvolution 92 pseudoorder 92 Puiseux series 198, 286 qadvanced Gevrey series 104 quasinilpotent 251 quotient field 194 quotients 193 R 87 (R, ∗) 87 Ramanujan’s partition function 100 real exponent to formal power series h 308 rearrangement of a formal power series 44 rearrangement of a series 42 reciprocal 5 region of holomorphy 162 regular Laurent series 207 residue 118, 191 reverse 26 reversed semi formal Laurent series 199 reversion 64 r(g) 166 right distributive law 13, 15, 154, 157, 222 ring isomorphism 17 Riordan array 200 Riordan group 87 Riordan group associated with Laurent series 202 Riordan identity 87 Riordan matrix 86 Riordan order 92 RNA matrix 90 Sθ (ϵ) 104 Sk 208 scalar multiplication 4 Schröder’s equation 273 Schröder’s functional equation 106, 113 Schur–Jabotinski theorem 194 section 17, 18 semi formal Laurent series 190 semicanonical 271 seminorm 246 sequence 2 sequence of changes of variable 261 series 3 Sheffer sequence 83, 95
Index  323
Sheffertype polynomial sequence 202 shifting mapping 208 shifting operator 123, 175 similar 90 singular point of f 162 space of qadvanced Gevrey series 104 spherically complete 75 standard pseudo basis 31 stochastic Riordan array 94 strong triangular inequality 65, 70 sum 4 superattracting fixed point 113 Tuniversal 238 Tβ 175 Tg 132, 172 Tg (f ) 132 term of degree k 38 termbyterm differentiation 52, 64 total degree 37 transform 261 translation equation 111 transposed matrix 21 trivial valuation 69
two rules 130, 141 ultrametric 65, 234 ultrametric space 65 ultrametric topological space 71 ultrametric topology 71, 261 umbral algebra 80 unit 3 valuation 69 valued field 69 Vandermonde’s theorem 29, 57 very well approximable 202 v(f ) 69 Wronski formula 20 (X , d) 65 X ∗ 238 [x n ] 119, 280 Zsequence 91 zero formal Laurent series 207 zero formal power series 3, 4
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