Finite Operator Calculus with Applications to Linear Recursions

Home , Delta operator, Operator (mathematics), Sheffer sequence, Umbral calculus

Finite Operator Calculus With Applications to Linear Recursions

Heinrich Niederhausen Florida Atlantic University Boca Raton [email protected] www.math.fau.edu/Niederhausen 2 Contents

Contents i

1 Prerequisites from the Theory of Formal Power Series 1 1.1 Generating Functions and Linear Recursions ...... 2 1.1.1 Roots ...... 9 1.1.2 Exercises ...... 10 1.2 Composition and Inverses ...... 13 1.2.1 Exercises ...... 15 1.3 Multivariate Power Series ...... 17 1.3.1 Exercises ...... 20

2 Finite Operator Calculus in One Variable 21 2.1 Polynomials, Operators, and Functionals ...... 21 2.1.1 The Vector Space of Polynomials, and Their Bases . . . . . 22 2.1.2 Standard Bases and Linear Operators ...... 25 2.1.3 Exercises ...... 26 2.2 Finite Operators ...... 27 2.2.1 Translation Operators ...... 28 2.2.2 Basic Sequences and Delta Operators ...... 30 2.2.3 Special Cases ...... 32 2.2.4 Exercises ...... 40 2.3 Sheﬀer Sequences ...... 44 2.3.1 Initial Values Along a Line ...... 47 2.3.2 The Umbral Group ...... 51 2.3.3 Special Cases ...... 54 2.3.4 Exercises ...... 63 2.4 Transfer Theorems ...... 69 2.4.1 Umbral shifts and the Pincherle derivative ...... 73 2.4.2 Proof of the Transfer Formula ...... 74 2.4.3 Exercises ...... 76 ii CONTENTS

3 Applications 79 3.1 The Functional Expansion Theorem ...... 79 3.1.1 Some Applications of the Functional Expansion Theorem . 83 3.1.2 Exercises ...... 87 3.2 Diagonals of Riordan Matrices as Values of Sheﬀer Sequences . . . 89 3.2.1 Exercises ...... 93 3.3 Determinants of Hankel Matrices ...... 94 3.3.1 Exercises ...... 100 3.4 Classical Umbral Calculus ...... 102 3.4.1 The Cumulant Umbra ...... 109 3.4.2 Exercises ...... 111

4 Finite Operator Calculus in Several Variables 113 4.1 Polynomials and Operators in Several Variables ...... 114 4.1.1 Exercises ...... 117 4.2 The Multivariate Transfer Formulas ...... 119 4.2.1 Transfer with constant coeﬃ cients ...... 119 4.2.2 Operator based transfer ...... 121 4.2.3 The multivariate Pincherle derivative ...... 123 4.2.4 Transfer with operator coeﬃ cients ...... 128 4.2.5 Exercises ...... 131 4.3 The Multivariate Functional Expansion Theorem ...... 134 4.3.1 Exercises ...... 136

5 Special Constructions in Several Variables 137 5.1 Multi-indexed Sheffer Sequences ...... 137 5.1.1 Delta Operators for multi-indexed Sheffer sequences. . . . . 139 5.1.2 Translation invariance of diagonalization, and some examples.140 5.1.3 Abelization of Multi-Indexed Sequences ...... 142 5.1.4 Exercises ...... 147 5.2 Polynomials with all but one variable equal to 0 ...... 150 5.2.1 Exercises ...... 153 5.3 Cross-Sequences and Steffensen Sequences ...... 153 5.3.1 Exercises ...... 155

6 A General Finite Operator Calculus 157 6.1 Transforms of Operators ...... 157 6.2 Reference Frames, Sheﬀer Sequences, and Delta Operators . . . . . 162 6.2.1 Reference Frames ...... 162 6.2.2 Sheﬀer Sequences and Delta Operators ...... 165 6.2.3 Exercises ...... 169 6.3 Transfer Formulas ...... 171 6.3.1 General Umbral Shifts and the Pincherle Derivative . . . . 171 6.3.2 Equivalent Transfer Formulas ...... 173 CONTENTS iii

6.3.3 Exercises ...... 175 6.4 Functionals ...... 176 6.4.1 Augmentation ...... 179 6.4.2 Orthogonality ...... 179 6.4.3 Exercises ...... 184

7 Applications of the General Theory 187 7.1 The Binomial Reference Frame ...... 188 7.1.1 Orthogonal Binomial Reference Sequences ...... 189 7.1.2 Generalized Catalan Operators ...... 195 7.1.3 Dickson Polynomials ...... 197 7.1.4 Exercises ...... 200 7.2 Eulerian Diﬀerential Operators ...... 204 7.2.1 Exercises ...... 209

8 Solutions to Exercises 211

Bibliography 257 iv CONTENTS

Preface

The following text originated from various lecture notes for graduate courses in combinatorics. I would very much appreciate receiving your comments, additions and corrections. My apologies to all those mathematicians not mentioned in the text, their important contributions to the theory of the Finite Operator Calculus skipped over because of ignorance or by design in order to keep the material manageable. Your views can still be included - please let me know: [email protected].

Introduction

n Every linear operator T on polynomials has a representation T = n 0 M(pn) , ≥ D where is the derivative operator, and M(pn) stands for multiplication by the D P polynomial pn (x) (Pincherle [74, 1901]). When applied to polynomials, the operator T reduces to a finite sum, of course, and may therefore be called a “finite” n operator. A special case of this concept, when T can be written as T = n 0 cn , ≥ D where c0, c1,... are scalars, is called a called a finite operator in the “Finite Op- erator Calculus” [83, 1973] by G.-C. Rota and his students D. KahanerP and A. n Odlyzko. Hence T is isomorphic to the formal power series n 0 cnt , and it is of course exactly this isomorphism that made the Finite Operator≥ Calculus so widely applicable. We adopt Rota’s approach in this book, but consider,P in the last two n chapters, also linear operators of the form T = n 0 cn , where can be any linear operator reducing the degree by 1, deg ( q) =≥ deg qR 1 for all polynomialsR q of degree larger than 0, and q = 0 for polynomialsR P of degree− 0 (we identify polynomials of degree 0 with scalarsR c = 0; however, we let deg (0) = , as usual). The fundamental difference between6 Rota’s approach and the generalized−∞ version we n present, due to J. M. Freeman [35, 1985], also his student, is that T = n 0 cn is translation invariant, TEc = EcT , where Ec : f (x) f (x + c) is the operator≥ D translating by c. The generalized version does not have7→ this property.P Translation invariance, however, is an important feature in many applications. The applications we have in mind are usually the solutions to recursive equations. Suppose your analysis of a given enumerative problem resulted in a recursive expression for the numbers you are looking for; to be specific, let us assume you arrived at

Fm = Fm 1 + Fm 2, − − the recursion for the Fibonacci numbers, starting at F0 = F1 = 1. A computer will give us “special”answers in a very short time, even F10,000 makes no problem whatsoever. From this point of view, you will not need this book. However, finding n+1 n+1 n+1 out that Fn = 1 + √5 1 √5 / 2 √5 is as surprising as it is − − rewarding. In addition, it is even “practical”; a scientific calculator will show that 2089 F10,000 5.4 10 . It also tells you something about the ratio, Fn/Fn 1, and ≈ × − CONTENTS v its famous limit, the Golden Ratio. Generating functions are the standard tool for solving this type of linear recursion. We give a brief introduction in the first chapter. The reader familiar with formal power series may just want to browse through section 1.2 for the notation. Now suppose the recursion you found was

Fn (m) = Fn (m 1) + Fn 1 (m 2) , − − − with initial conditions Fn (n) = Fn 1 (n) for n 1, and F0 (m) = 1 for all m. The recursion is still easy, but the− initial values≥ are also “recursive” - we have to know Fn 1 (n) before we can say what the value of Fn (n) is. We will see − that F0 (x) ,F1(x),... is a sequence of polynomials, actually a basis, and that the operator T : Fn (x) Fn 1 (x) is a translation invariant operator, satisfying the operator equation 7→ − 1 2 I = E− + E− T. We will show in chapters 2 and 3.1 how to ﬁnd the solution with given initial values from such an equation. Suppose you found the system of recursion

sm,n (u, v) = sm,n (u 1, v) + sm 1,n (u, v + 2) and − − sm,n (u, v) = sm,n (u, v 1) + sm,n 1 (u + 1, v + 1) sm,n 2 (u, v) , − − − − with initial values sm,n (0, 0) = 0 for all m, n 0, except s0,0 (0, 0) = 1. This system of recursions is two dimensional and linear,≥ the initial values are explicit. We show how to write sn,m (u, v) as a sum of binomial coeffi cients in chapters 3.1 and 5, dedicated to the multivariate Finite Operator Calculus. Technically the problems get more diffi cult to solve, of course, when it comes to higher dimensions. The theory, however, remains quite easy. Somewhere between the univariate and the multivariate case fall the Steffensen sequences (section 5.3) and the multi- indexed sequences (section 5.1). Rota’s goal was to create a solid foundation for the “Umbral Calculus”, to purge it of the “witchcraft”, as he called it. This was one of his favorite themes, and he wrote more papers on Umbral Calculus later. Several young (at that time) mathematicians took up his work, and showed that it had applications to a va- riety of mathematical topics, including approximation theory, signal processing, probability theory, and, of course, combinatorics. After all, the “Finite Operator Calculus”was published a part VIII in a series of papers “On the Foundations of Combinatorial Theory”. A complete survey of papers relating to Umbral Calcu- lus until the year 2000 has been compiled by Di Bucchianico and Loeb [26]. An application of the Finite Operator Calculus can also be found in Taylor [96, 1998]. We assume that after 5 chapters the reader will get interested in the theory itself. J. M. Freeman explored this generalization in some depth [35], and we follow it literally in the last two chapters. Finally, the expert may wonder why Finite Operator Calculus and not Um- bral Calculus? Umbral Calculus is a highly “symbolic” language; no operators, vi CONTENTS just umbrae! As a rule of thumb, every Finite Operator statement gets shorter in Umbral Calculus; for example, the contents of section 2.2.3, Basic sequences with polynomial coeffi cients, is reduced to α˜D χ.α in umbral notation. However, it may be exactly this brevity, achieved through≡ a multitude of special definitions, that prevents Umbral Calculus from being widely known. We give an introduction to Umbral Calculus in section 3.4. Most of the examples and exercises in this book refer to combinatorial problems, with few exceptions. Yet this is not a book on enumerative combinatorics, because we begin where combinatorics ends, that is at the crucial point, where combinatorics has delivered a recursion and initial conditions. Without question, this is the hard part of combinatorics; solving the recursion is the technical part. Finite Operator Calculus can help with the technical part, when applicable. We will therefore describe the combinatorics only briefly, just enough to introduce the recursion. There is a special sequence of examples taken from [84] about the enumeration of lattice paths containing patterns; they are Example 1.2.3, 2.3.15, 2.4.5, 5.1.6, and 5.1.8. The reader of this text should have access to a computer algebra package (CAS) like Mathematica, Maple, MuPad, etc. This will allow for checking coeffi - cients of formal power series, getting conjectures on new results, and verifying the polynomial formulas in the examples and exercises. Chapter 1

Prerequisites from the Theory of Formal Power Series

The isomorphism between Finite Operator Calculus and formal power series allows us to express “everything”we can do with operators also in terms of power series. Why then do we prefer one over the other? May be the criterion should be the ease of use; how diﬃ cult it is to formulate, classify, and solve a given recursion in one way or the other. However, ease of use heavily depends on the availability of a commonly known (mathematical) language. For example, linear recursions are easily translated into functional relations between formal power series, or into equations between operators. We will see in section 2.4 how to solve such an equation directly for the polynomials involved in the recursion, without knowing the operators explicitly.

We give a brief introduction to the powerful method of solving recursions by generating functions. A more detailed discussion can be found in many text books; an excellent resource is the “generatingfunctionolgy” by H. Wilf [102]. If the generating function is rational, a quotient of polynomials, an explicit form for the coeﬃ cients can be obtained from the roots of the denominator, at least in principle. We show an example in subsection 1.1.1.

Most important for the following chapters is the Lagrange-Bürmann inversion formula. A proof in algebraic form can be found in Henrici’s Applied and Computational Complex Analysis, Vol. 1 [41]. See also Hofbauer [43]. Applying Lagrange inversion in several variables gets more tedious; for pedagogical reasons we separate the multivariate from the univariate case. It is, however, not the La- grange inversion that is the principle obstacle in solving linear recursions in several variables. See section 1.3 for more details. 2 Chapter 1. Prerequisites from the Theory of Formal Power Series

1.1 Generating Functions and Linear Recursions

Suppose you have computed the number Cn of certain structures on a set of f (n) elements for every n = 0, 1,... , and you want to “store”the results. For example, you counted the number Cn of sequences (c0, c1, c2, . . . , c2n) of length f (n) = 2n i consisting of +1 and 1 in equal numbers such that the partial sum k=0 ck are never below 0, − i 2n P ck 0 and ck = 0 ≥ k=0 k=0 X X for all i = 1 to 2n. The first few Cn’sare C1 = # (1, 1) = 1, { − } C2 = # (1, 1, 1, 1) , (1, 1, 1, 1) = 2, { − − − − } C3 = 5, etc. Let us assume that C0 = 1 (assuming that C0 = 0 would also make sense; see Example 1.2.1). The generating function of the sequence (Cn)n 0 is defined as the 0 1 2 ≥ “series”γ (t) = C0t + C1t + C2t + ... , a formal sum using a formal variable t. The generating function (or formal power series, which we use as synonyms) γ (t) is in essence the sequence C0,C1,... ; convergence of the series is not assumed. Some arithmetic with generating functions is obvious, like addition and scalar multiplication. Actually, the formal power series are a vector space over the integers Z, or the rational Q, or some other ring. If we call this ring k, the power series will be denoted by k [[t]]. So the ring k contains all the coeffi cients like C0,C1,... . It will turn out that a ring structure on the coeffi cients is not enough. We will assume in this chapter that k is an integral domain, i.e., we assume that for two coeffi cients a and b their product ab cannot be zero, if a and b are both different from 0. In the following chapters, we even will assume that k is a field, and we write F [[t]] for those power series. However, in section 2.4 on transfer theorems, we will need again power series that have an integral domain (and not a field) as their coeffi cient ring. By convention, there is only one “value” of t that can be substituted for t, and that is t = 0, giving γ (0) = C0. Again, this is nothing but a notational trick, but very helpful, as we will see below. Next we need a device to extract the n- th coeffi cient from a generating function. As a notation, we define the coeffi cient n n functional [t ] on k [[t]] such that Cn = [t ] γ (t), or we could write Cn = [γ]n, realizing that the name of the formal variable is not of interest. Analysis has given us a simple method to find the n-th coeffi cient of an analytic function γ (t), Cn = 1 dn n! dtn γ (t) t=0. We can also define the formal derivative of a formal power series: n d γ (t) has coeffi cients d γ = (n + 1) C , and therefore C = 1 d γ (t) dt dt n n+1 n n! dtn t=0 can be defined for formal power series as well. We saw that convergence it not an issue for generating functions, because we never evaluate them at a specific t, except for t = 0, a case which does not need convergence. However, having some positive radius of convergence is a great help if we want to use γ (t) as a storage device. Consider the above example, the 1.1. Generating Functions and Linear Recursions 3

balanced sequences of 1’s and 1’s. In this case the numbers Cn are the Catalan numbers, probably one of the most− studied sequences in combinatorics! And it is well-known that ∞ 2 C tn = (1.1) n √ n=0 1 + 1 4t X − (see Exercise 2.3.8 for a proof), a series that converges for all t < 1/4, viewed as a function in t. Writing 2/ 1 + √1 4t is clearly a very convenient notation! − We can get back the numbers Cnfrom 2/ 1 + √1 4t as C0 = 2/ 1 + √1 = 1, − d 2 4 C1 = = 2 = 1, dt 1+√1 4t 1+√1 4t √1 4t − t=0 ( ) t=0 − − 1 d 4 3 √1 4t+1 C2 = 2 = 4 − 3 = 2, 2 dt (1+√1 4t) √1 4t (1 4t)3 /2(1+√1 4t) − − t=0 − − t=0 4 d 3√1 4t+1 2√1 4t+3 10t C3 = − 3 = 16 − − 4 = 5 , etc. 3 dt (1 4t)3/2(1+√1 4t ) ( 1+4t)2√1 4t(1+ √1 4t) − − t=0 − − − t=0 The above calculations should convince you that finding the coeffi cients by the differentiation method is a recursive procedure; from the n -th derivative you calculate the n+1-th derivative. You want a computer algebra package to find C100 this way. We can solve that recursion by finding Cn explicitly, which means that we have to expand 2/ 1 + √1 4t in powers of t. Of course, 2/ 1 + √1 4t = − − 1 √1 4t / (2t), a function that is slightly easier to expand. Thus − − 1 √1 4t ∞ 1/2 n+1 2n 1 n 1 − − = ( 1) 2 − t − . 2t n n=1 − X Substitute n 1 1/2 ( 1) − 2n 1 = − − n 22n 1 (2n 1) n − − 1 2n and get Cn = n+1 n . Note that for large n this formula needs a computer also; however, Cn can be fairly accurately approximated by Stirling’s formula, 2n 2n n / (n + 1) 2 / ((n + 1) √nπ), with a relative error of roughly 1/1000 if n is around 100. ≈ Multiplication of convergent power series is defined with the help of the Cauchy product, n n [t ](γ (t) σ (t)) = cksn k, − k=0 X k k if t γ (t) = ck and t σ (t) = sk. This definition is carried over to the formal power series. Note that multiplication is commutative, because multiplication in k is commutative. The reciprocal of a formal series γ (t) can exist only when γ (0) = 0; otherwise we would obtain negative powers of t. We write 1/γ (t) for the reciprocal,6 and 4 Chapter 1. Prerequisites from the Theory of Formal Power Series

1 sometimes γ (t)− . We have 1 = γ (t) (1/γ (t)); this tells us all about the coeffi cients n of 1/γ (t) = n∞=0 cñt : 1 = c0c˜0, hence c˜0 = 1/c0 (showing again that c0 = 0), P 2 6 0 = c1c˜0 + c0c˜1, thus c˜1 = c1/c0, and in general n 1 − cñ = k=0− c˜kcn k /c0, for all n 1. − − ≥ There will only be a positive power of c in the denominator of the expression P 0 for cñ. We can now refine our statement about the existence of a reciprocal. Lemma 1.1.1. A formal power series γ has a reciprocal iff γ (0) has a nonzero multiplicative inverse in the coeffi cient ring for all n 0. ≥ For example, if the coeffi cient ring equals Z, a power series must start with 1 or 1 in order to have a reciprocal. If we return to the Catalan generating function− c (t) = 2/ 1 + √1 4t , we get the reciprocal 1/c (t) = 1 + √1 4t /2, but something ‘surprising’happens− in this special case: − 1/c (t) = 1 tc (t) . (1.2) − The order of a power series σ (t) is the smallest n 0 such that [tn] σ (t) = 0. We saw above that if σ has a reciprocal then ord σ =≥ 0. A power series β (t6 ) of order 1 is called a delta series in the Finite Operator Calculus, if ord β (t) = 1 and β(t)/t has a reciprocal. Remark 1.1.2. We defined scalar multiplication, addition, and multiplication of formal power series in purely algebraic terms. For the reader interested in the combinatorics behind all this we recommend the 1981 paper by Joyal [48], and the book on Combinatorial Species and Tree-like Structures by F. Bergeron, G. Labelle, and P. Leroux [10]. How do we “combinatorially”calculate the coeffi cient k 1 2 of β (t) if k is a positive integer and β (t) = b1t + b2t + ... ? If we write

k n β (t) = t bj bj 1 ··· k n k j1+ +jk=n X≥ ···jXi>0 we can sort the vectors (j1, . . . , jk) of positive integers and obtain vectors (λ1, . . . , λk) of sorted integers, each with a certain multiplicity. Of course, it still holds that λ1 + + λk = n. If we sort such that λ1 λk, then λ = (λ1, . . . , λk) is called··· a partition of n. In symbols, λ n. The≥ number · · · ≥ of parts is k, λ = k. There ` | | is an equivalent representation of a partition as a multiset 1`1 , . . . , n`n , where ì n n the term i means ì occurrences of i in λ. Hence i=1 ì = k, and i=1 iì = n. For every partition of n we think of these two equivalent representations simul- taneously! The above mentioned multiplicity is theP number of permutationsP of λ, k which is . (Choose `1 places for the ones in λ, then `2 places for the twos, `1,...,`n etc.) Hence k k n k β (t) = t bλi . (1.3) `1, . . . , `n n k λ n, λ =k i=1 X≥ ` X| | Y 1.1. Generating Functions and Linear Recursions 5

For example,

k k k k k t β (t) = bλi = b1 `1 λ k, λ =k i=1 ` X| | Y k k+1 k k k 1 t β (t) = bλi = kb2b1− (1.4) `1, `2 λ k+1, λ =k i=1 ` X| | Y k+2 k k 1 k 2 k 2 t β (t) = kb b − + b b − 3 1 2 2 1 k+3 k k 1 k k 2 k 3 k 3 t β (t) = kb b − + 2 b b b − + b b − . 4 1 2 3 2 1 3 2 1 There is a second concept of partitioning in combinatorics, the partitions of an n-set. Here a set of n-elements is written as the union of k nonempty and disjoint subsets. The number of such set partitions is S (n, k), the Stirling number of the second kind (Stanley [89]). We write S (n, k) for the set of all partitions of an n- set into k parts, thus S (n, k) = Sn,k The parts B1,...,Bk in a set partition are written in no particular order; we| think| of them sorted decreasingly by magnitude, B1 Bk . Parts with the same number of elements have to be sorted in | | ≥ · · · ≥ | | some way (lexicographically). Again, the numbers B1 Bk will make a partition λ of n, with k parts, but with a certain multiplicity.| | ≥ · · · ≥ The | multiplicity| is n! , i.e., this is the number of partitions in Sn,k such that Bi = λi. λ1! λk!`1! `n! ··· ··· | | Hence (1.3 shows that β (t)k

k n k! n! = t λi!bλi n! λ1! λk!`1! `n! n k λ n, λ =k i=1 X≥ ` X| | ··· ··· Y k n k! = t λi!bλ n! i × n k λ n, λ =k i=1 ! X≥ ` X| | Y (B1,...,Bk) Sn,k such that Bi = λi × |{ ∈ | | }| k n k! = t Bi !b Bi . (1.5) n! | | | | n k (B1,...,Bk) Sn,k i=1 X≥ X∈ Y

Example 1.1.3 (Fibonacci numbers). The Fibonacci numbers Fn can be defined by F0 = F1 = 1 and Fn = Fn 1 + Fn 2 for all n 2. (Note that Z can serve as − − ≥ the coeffi cient ring.) However, defining Fn = 0 for negative n lets us express this recursion as Fn Fn 1 Fn 2 = δ0,n for all n 0. Multiplying this formula by tn and summing− up over− − all n− 0 gives the generating≥ function of the Fibonacci ≥ 6 Chapter 1. Prerequisites from the Theory of Formal Power Series numbers,

n n 2 1 = (Fn Fn 1 Fn 2) t = Fnt 1 t t − − − −   − − n 0 n 0 X≥ X≥   n 2 thus n 0 Fnt = 1/ 1 t t . The roots of the denominator are easy to ﬁnd; hence ≥ − − P 1 F = [tn] − n t2 + t 1 − n 1/√5 1/√5 = [t ] 1 1 1 1 t + + √5 − t + √5! 2 2 2 − 2 1 1 = n n √5 1 + 1 √5 1 √5 1 − √5 1 1 √5 1 √5 1 2 2 − 2 − 2 2 − 2 2 − 2 n+1 n+1 1 + √5 1 √5 = − − n+1 2 √5

On the ﬁrst glance, this formula for Fn does not look integer, but a closer look will easily convince you. Of course we could also expand 1/ 1 t t2 in powers of t. − − We get F = n/2 n k , certainly an integer, and so we arrive at the identity n k=0 −k P n/2 n+1 n+1 n k 1 + √5 1 √5 − = − − k 2n+1√5 k=0 X n A recursion of the form σn = j=1 αjσn j + γn, where αj and γn are given − for all j 1 and n 0, is called a linear recursion for σn in one variable ( n). ≥ ≥ P The starting value in this recursion, σ0, equals γ0, and if γ1, γ2,... are diﬀerent from 0, the recursion is called inhomogeneous. However, if the sequence of inhomogeneous terms eventually becomes 0, so that the last nonzero term is γ` 1, then we usually do not say that the recursion is inhomogeneous, but has initial− values n σ0, σ1, . . . , σ` 1, and then follows the (homogeneous) recursion σn = j=1 αjσn j − − for n `. Of course, the terms γ0, . . . , γ` 1 can be recovered from the initial values, ≥ k − P as γk = σk j=1 αjσk j. − − Theorem 1.1.4.P Suppose the numbers σn solve for n 0 the (inhomogeneous) linear recursion ≥ n

σn = αjσn j + γn − j=1 X where α1, α2,... and γ0, γ1, ... are sequences of given constants. Then

n n k 1 σn = γk t − j 1 ∞ αjt k=0 j=1 X − P 1.1. Generating Functions and Linear Recursions 7 and k ∞ n k∞=0 γkt σnt = 1 ∞ α tj n=0 j=1 j X −P k k P j Proof. From ∞ γkt = ∞ σkt 1 ∞ αjt follows the Theorem. k=0 k=0 − j=1 The formP of the generatingP function for (σPn) shows us that it will be rational if both (γn) and (αn) are eventually 0. This characteristic will be important in section 1.1.1. Explicit expressions can be derived from Theorem 1.1.4 in a large number of applications. We discuss Fibonacci-like sequences σn = uσn 1 + vσn 2 + γn in Exercises 1.1.2 and 1.1.3. − − n j The sequences 1, a1, a2,... and [t ] 1/ 1 j∞=1 αjt are some- − − − n 0 ≥ times called orthogonal (because they are reciprocals), P and the sequences (σn) and (γn) are an inverse pair. For examples of inverse pairs see section 2.3.3.

Example 1.1.5 (Derangement Numbers). The derangement numbers dn denote the number of permutations π of [n] that are derangements, i.e., πi = i for all i = n 6 1, . . . , n. They follow the recursion dn = ndn 1 + ( 1) ,with initial value d0 = 1. − − n In the notation of Proposition 1.1.4, σn := dn/n!, αn = δn,1, and γn = ( 1) /n!. Thus −

n k n k ( 1) n k 1 ( 1) dn = n!σn = n! 1 + − t − = n! − k! 1 α1t k! k=1 ! k=0 X − X k k n ∞ n k∞=0 ( 1) t /k! t dnt /n! = σnt = − = e− / (1 t) 1 t − n 0 n=0 P X≥ X − The generating function is not rational.

Example 1.1.6 (Bernoulli Numbers). The Bernoulli numbers Bn solve the system n n+1 1 of equations δn,0 = k=0 Bn k for n 0. Dividing by n! shows that the k+1 n+1 − ≥ numbers Bn/n! can be calculated from the linear recursion P n 1 Bn/n! = Bn k/ (n k)!. − (k + 1)! − − k=1 X

Applying the proposition with ak = 1/ (k + 1)! for k 1 gives the (exponential) generating function − ≥

B B t n tn = 0 = n! 1 + tj/ (j + 1)! t + tj+1/ (j + 1)! n 0 j 1 j 1 X≥ ≥ ≥ t t = P = . P t + (et 1 t) et 1 − − − 8 Chapter 1. Prerequisites from the Theory of Formal Power Series

The generating function is not rational. The Bernoulli polynomials ϕn (x) := n k n Bk x − t xt k=0 k! (n k)! have generating function et 1 e by convolution. Note that − − P t (x+1)t t xt t x( t) e = e = − e− − et 1 1 e t e t 1 − − − − − n hence ϕn (x + 1) = ( 1) ϕn ( x). See Exercise 2.3.2 for more details on Bernoulli numbers. An explicit− formula− for the Bernoulli numbers is obtained in Exercise 2.3.15. A detailed discussion of Bernoulli numbers (58 pages) can be found in the “Calculus of Finite Differences” by Jordan [47]. Example 1.1.7. A polyomino is a union of a finite number of squares with vertices in Z2 such that every square shares at least one side with some other square. Translations do not change a polyomino, but reflections and rotations do. In a horizontally convex polyomino P any line segment parallel to the x-axis with both end points in P must be completely in P . For example, there are 19 polyominos with 4 squares; all 19 polyominos are horizontally convex.

Hickerson [42] found a combinatorial proof in one dimension that the number f (n) of horizontally convex polyominos made of n+1 squares follows the recursion f (n + 3) = 5f (n + 2) 7f (n + 1) + 4f (n) − for n 1, with initial values f (0) = 1, f (1) = 2, f (2) = 6, f (3) = 19. By Proposition≥ 1.1.4

` 1 n 1 − k f (n) t = j σk αjσk j t 1 ∞ αjt  − −  n 0 j=1 k=0 j 1 X≥ − X X≥ 1 +P (2 5) t + (6 10 + 7) t2 + (19 30 + 14 4) t3 = − − − − 1 5t + 7t2 4t3 − − (1 t)3 = − 1 5t + 7t2 4t3 − − This generating function begins with 1 + 2t + 6t2 + 19t3 + 61t4 + 196t5 + 629t6 + 2017t7 + 6466t8 + 20 727t9 + 66 441t10 + 212 980t11 + 682 721t12 + 2188 509t13 + 7015 418t14 + ... 1.1. Generating Functions and Linear Recursions 9

We could expand the generating function in terms of tn, but we will ﬁnd a “simple” expression for these coeﬃ cients in Example 1.1.8 Of course, deriving the recursion is the hard part. Stanley [89, p. 259] obtained the recursion from the generating function, which he found by the transfer- matrix method. His approach starts from the observation that f (n 1) = (n1 + n2 1) (n2 + n3 1) + + (ns + ns+1 1), summing over n− 1 − − ··· − all 2 − compositions of n1 + +ns+1 = n ( s = 0 contributes 1). For example, if n = 5, thenP the composition 5···(s = 0; 1 term) contributes 1, the compositions into 2 terms (s = 1) contribute 16, then 3 terms contribute also 16, 4 terms add 15, 5 terms 12, and 6 terms contributes 1. For example, f (4) = 1+16+16+15+12+1 = 61.

1.1.1 Roots

Suppose the recursion is homogeneous (γk = δ0,k and ` = 1). If there are only n ﬁnitely many factors α1, . . . , αd, the generating function n 0 σnt is rational. By Stanley’sTheorem 4.1.1 [89] ≥ P k n σn = pi (n) ri (1.6) i=1 X where the 1/ri’sare the k distinct roots of the polynomial

d k j mi 1 αjt = (1 rit) − − j=1 i=1 X Y and each pi (n) is a polynomial (in n) of degree less than the multiplicity mi of root 1/ri. The polynomials pi (n) can be determined from the the ﬁrst few values of σn. We followed Stanley’s procedure in the example of the Fibonacci numbers, disguised as a partial fraction decomposition. We will do partial fraction decomposition again in the more involved next example. Example 1.1.8. Applying the root formula (1.6) is possible even when the numer- ator of the generating function is a polynomial diﬀerent from 1. In Example 1.1.7 we found the generating function for the number f (n) of horizontally convex polyominos made of n + 1 squares,

3 2 n (1 t) (t 1) (1 t) f (n) t = − 2 3 = − − 1 5t + 7t 4t 4 (t t1)(t t2)(t t3) n 0 X≥ − − − − − With r = 3 71 + 6√177we determine the roots of the denominator as 11 r2+7r r2 11+14r r2+11 t1 = − , t2 = − i√3 , and t3 = t∗, the complex conjugate of 12rp 24r − 24r 2 t2 (the calculations were made with an algebra package; it can be cumbersome to 10 Chapter 1. Prerequisites from the Theory of Formal Power Series do this by hand!). We wrote the generating function already in a form to suggest a partial fractions decomposition,

t 1 A B C f (n) tn = − + + 4 t t1 t t2 t t n 0 2∗ X≥ − − − 2 We need A (t t2)(t t∗) +B (t t1)(t t∗) +C (t t2)(t t1) = (1 t) , thus − − 2 − − 2 − − − A + B + C = 1

A (t2 + t∗) B (t1 + t∗) C (t1 + t2) = 2 − 2 − 2 − − At2t2∗ + Bt1t2∗ + Ct1t2 = 1

This system has the solution 2 2 (5r+r2 11) ( r( 5 i√3+ 5 )+r2 11 i√3+ 11 ) A = 1 − , B = 1 − 2 2 − 2 2 , and C = B . 3 r4 11r2+121 3 (r2 11 i√3 11 )(r2+11) ∗ − − 2 − 2 If we write ξ = i√3 1 /2, ξ∗ = i√3 + 1 /2 (both are third roots of unity) 2 − 2 − (5rξ∗+r 11ξ) 1 − then B = 3 r4 11ξr2+121ξ . The generating function can be written as − ∗ (1 t) A B f (n) tn = − + 2 Re 4 t t1 t t2 n 0 X≥ − − − For n 1 we ﬁnd ≥ (1 t) A B f (n) = [tn] − + 2 Re 4 t t1 t t2 − − − 1 n A B 1 n 1 A B = [t ] + 2 Re + t − + 2 Re −4 t t1 t t2 4 t t1 t t2 − − − − A n 1 n B n 1 n = t− − t− + 2 Re t− − t− 4 1 − 1 4 2 − 2 A n 1 B n 1 = t− − (1 t1) + 2 Re t− − (1 t2) 4 1 − 4 2 −

2 3 2 3 (5r+r 11) ξ(11ξ 5rξ∗ r ) 1 − − − With A (1 t1) = 36 (r4 11r2+121)r and B (1 t2) = 36(r4 11ξr2+121ξ )r we get − − − − ∗

2 3 2 3 1 5r + r 11 n 1 ξ 11ξ 5rξ∗ r n 1 f (n) = − t− − 2 Re − − t− − . 144r r4 11r2 + 121 1 − (r2 11ξ )(r2 + 11) 2 − − ∗ !! 1.1.2 Exercises 1.1.1. Find an explicit expression for the number of horizontally convex polynomials, i.e., expand the generating function (1 t)3 / 1 5t + 7t2 4t3 in terms of t. − − − 1.1. Generating Functions and Linear Recursions 11

1.1.2. A Fibonacci-like sequence σ0, σ1,... solves a recurrence of the form

σn = uσn 1 + vσn 2 + γn − − for n 2, with initial values σ0, σ1, and given inhomogeneous terms γ2, γ3,... . Show≥ that k σ0 + (σ1 uσ0) t ∞ γkt − + k=2 . (1.7) 1 ut vt2 1 ut vt2 − − P− − is the generating function of σn. For example, the Fibonacci recursion Fn = Fn 1+ − Fn 2 has initial values F0 = 1, F1 = 1; this means that in Proposition 1.1.4 we − have the terms γn = δ0,n and α1 = 1, α2 = 1, αn = 0 for all n 2. Hence n n k 1 n 1 ≥ Fn = γk t j = [t ] 2 . Show that for any Fibonacci-like k=0 − 1 ∞ αj t 1 t t − j=1 − − sequence holds P P n n k − n k j n k 2j j σn = γk −j − u − − v (1.8) k=0 j=0 X X where γ0 = σ0 and γ1 = σ1 u. Derive sums like (1.8) for the following special cases: −

1. The Lucas recursion Ln = Ln 1 + Ln 2 for all n 2, L0 = 2 and L1 = 1. − − ≥ 2. The recursion σn = (a 1) σn 1 + aσn 2 + n 1 for all n 2, and a = 1. − − − − ≥ 6 3. The recursion Pn+1 (x) = 2xPn (x) + Pn 1 (x) for the Pell polynomials [69], − with P0 (x) = 1 and P1 (x) = x. The polynomials have the generating function

1 xt 1 1 + t2 P (x) tn = − = + (1.9) n 1 2xt t2 2 2 4xt 2t2 n 0 X≥ − − − − 1 1 + t2 1 = + 2 2 (1 t2) 1 2xt/ (1 t2) − − −

The Pell numbers pn are deﬁned as pn = Pn (1). How do diﬀerent values for p1 change the subsequent numbers?

4. The recursion Un (x) = 2xUn 1 (x) Un 2 (x) for the Chebychev polynomials − − − of the second kind, with U0 (x) = 1 and U1 (x) = 2x (see (7.17)). In Exercise 1.1.4 the root formula (1.6) is applied to the above cases. More related results on Chebychev polynomials in Exercise 7.1.13.

2 1.1.3. Show that for all Fibonacci-like sequences (Exercise 1.1.2) holds σn = n n σn 1σn+1 + ( 1) v c for all n 1. Determine the constant c, and show that − − ≥ 2 n 2 for the Pell polynomials holds Pn (x) = Pn 1 (x) Pn+1 (x) + ( 1) 1 + x , for 2 − n − the Fibonacci numbers Fn = Fn 1Fn+1 + ( 1) , and for the Chebychev polynomi- 2 − − als Un (x) = Un 1 (x) Un+1 (x) 1. − − 12 Chapter 1. Prerequisites from the Theory of Formal Power Series

1.1.4. Consider again the Fibonacci-like numbers deﬁned in Exercise 1.1.2,

σn = uσn 1 + vσn 2 + γn − − for n 2, with initial values σ0, σ1, and given inhomogeneous terms γ2, γ3,... . ≥ This time we want an expression for σn in terms of the roots of the polynomial d j 2 1 αjt = 1 ut vt . Show that − j=1 − − r+1 r+1 P 2 2 r 1 r 1 u + √4v + u u √4v + u [t ] = 2− − − − 1 ut vt2 √4v + u2 − − if the discriminant 4v + u2 = 0. If 4v + u2 = 0 then 1 ut vt2 = (ut 2)2 /4 and 6 − − − 1 u r [tr] = (r + 1) . (1 ut/2)2 2 − Derive expressions in terms of the roots for the special recursions in Exercise 1.1.2 .

1.1.5. In the Fibonacci sequence let Fn = 0 for negative n. Show that Fn = k LkFn k ( 1) Fn 2k for all n 2k 1 > 0, if the numbers Lk are the Lu- − − − − ≥ − cas sequence following the Fibonacci-like recursion Lk = Lk 1 + Lk 2 with initial − − values L0 = 2,L1 = 1, hence

k k 1 + √5 + 1 √5 L = − . k 2k 1.2. Composition and Inverses 13

1.2 Composition and Inverses

Before we go deeper into the structure of formal power series, we want to introduce a notation. Remember that k is an integral domain: k has a unit element 1, is commutative and has no zero divisors, ab = 0 when a and b are in k and both 6 different from 0. We write k [t] for the polynomials in t with coeffi cients in the ring k, and k [[t]] for the formal power series with coeffi cient ring k as before. Note that we can view k [t] as being imbedded in k [[t]]. However, any element a from k can be used to evaluate a polynomial at a, giving again an element from k. In k [[t]], evaluation is only possible at a = 0. We saw that σ (0) k must have a ∈ multiplicative inverse in k if 1/σ (t) exists. Such an element of k is called a unit. Hence 1/σ (t) exists over an integral domain iff σ (0) is a unit in k. Composition of a power series γ (t) into a power series β (t) is achieved by substituting γ (t) for t in β (t). We write β (γ (t)) for the composition. Because the composition has to be in k [[t]], the term β (γ (0)) has to be in k. But only if β is a polynomial we know β (a) for a k and a = 0. Hence we will require that γ (t) is not of order 0, so γ can be substituted∈ into6 any power series β. Underlining the operational aspect we also write C(γ)β, where C (γ): k [[t]] k [[t]] is the linear operator that does the substitution of γ (t). → In analysis we have a concept of convergence; we know that eln t = t. However, ln t is not a formal power series (it is not defined at 0, the only place we can evaluate a formal power series). We can say that eln(1+t)) 1 = t, which means that C(ln (1 + t)) (et 1) = t. In other words, ln (1 + t) is the−compositional inverse of − et 1. If a power series β is of order 1, and [t] β (t) is a unit in k [t], then β − 1 1 has a unique compositional inverse β− such that β β− (t) = t. Such power series we called delta series. Often it is more convenient to use another symbol, 1 like γ, for the inverse of β. Note that the notation β− (t) for the (compositional) 1 inverse is very similar to the notation of the reciprocal β (t)− = 1/β (t), and it becomes indistinguishable, if the argument t is omitted. This shows why we like the notation 1/β for the reciprocal. Of course, they usually do not exist both for the same β k [[t]]! ∈ Example 1.2.1. Consider the sequence 0, 1, 2, 3, 4,... by looking at the formal n+1 n − − power series β (t) = n 1 ( 1) nt (a delta series in Z [[t]]); we want to know its compositional inverse.≥ We− have Pn+1 n d 1 2 β (t) = t n 0 ( 1) (n + 1) t = t dt 1+t = t/ (1 + t) , hence − ≥2 − 1 − t = 1 + 2t + t β and t = 2β 1 √1 4β 1. Therefore, the compositional P1 − − − inverse β− (t) equals the generating function of the Catalan numbers (section 1 1 1.1) without the constant term, β− (t) = 1 √1 4t 1. 2t − − − Remember that every statement about formal power series really means a statement about an infinite sequence of coeffi cients. For a series of order 1, the sequence starts with 0, and then a coeffi cient not equal to zero follows. Suppose we have a power series β with coeffi cients (0, a, b, c, . . . ). The compositional inverse 14 Chapter 1. Prerequisites from the Theory of Formal Power Series

γ has coeffi cients 0, a,˜ ˜b, c,˜ . . . such that the composition gives (0, 1, 0,... ). For calculating the first few term of the composition we need the n-th coeffi cient of k 2 ˜ ˜2 the k-th power of the inverse, γ n. For k = 2 we get 0, 0, a˜ , 2ãb, b + 2ãc,˜ . . . k and for = 3 we have 0, 0, 0, a˜3, 3ã˜b + 3ãc˜ + 3˜bc,˜ . . . . Hence β (γ) has the coeffi cient sequence (0, aa,˜ a˜b + ba˜2, ac˜ + 2ba˜˜b, ad˜ + b ˜b2 + 2ãc˜ + ca˜3,... ), and this has to equal 1 (0, 1, 0,... ). We can now recursively find the coeffi cients of β− = γ,

0, 1/a, b/a3, 2b2 ac /a5, ad2 5abc + 5b3 /a7,... (1.10) − − − Algebraically it is important that the powers of the coeffi cient a have a nonzero multiplicative inverse in k, if we want to find the compositional inverse of a power series. This is the only coeffi cient we divide by repeatedly; the remaining terms are sums of products. Hence, a compositional inverse γ (t) will exist, if β (t) is a delta series! Lagrange-Bürmann inversion, which provides a formula for [tn] γk, has been shown for this general setting ([41, Theorem 1.9a], [43]). We only state the result, assuming that β (t) = t/φ (t), which means that φ (t) is invertible. In this way, Laurent series are not needed. For more on Lagrange-Bürmann inversion see Exercise 1.2.2. Theorem 1.2.2. If β (t) = t/φ (t) is a delta series (hence φ has a reciprocal), and γ (t) is the compositional inverse of β, then for all 0 k n holds ≤ ≤ k n n γ n = k [φ ]n k . (1.11) − Again we want to point out that the name (t) of the formal variable is of no significance. We can as well write (1.11) as

n k n k n n [s ] γ(s) = k u − φ (u) , where γ k [[s]] and φ k [[u]]. For example, in ﬁnding the inverse of β (t) = et 1 we usually∈ proceed by∈ letting u = et 1, say, and solving for t = ln (1 + u). Then− 1 − we call t = β− (u), and change the variable from u back to t. Note that the presentation (1.10) is variable free! Lagrange-Bürmann inversion is routinely applied in combinatorics. We show an example from lattice path enumeration. Example 1.2.3. We are interested in ﬁnding the number D (n; k) of , lattice path from (0, 0) to (2n, 0), staying weakly above the x-axis, and having{% &} exactly k occurrences of the pattern %& , which we also write as uddu (d = and u = ). The following path to (10&%, 0) contains the pattern uddu twice (counting& overlaps).% %& %&%& %&%& 1.2. Composition and Inverses 15

n k The generating function F (x, t) = n,k 0 D (n; k) t x is a power series in two variables, so it would be beyond the scope of≥ this chapter. However, certainly k n, so we can see F (x, t) as a power seriesP in t with coeﬃ cients that are polynomials≤ in x. If we set D (0; 0) = 1, then

tF 3 ((1 x) t + 1) F 2 + (1 + 2 (1 x) t) (1 x) t = 0 − − − − − (see [84]). The need for an inverse arises, because we can easily write t as a function of F , F (F 1) t = − F 3 (1 x)(F 1)2 − − − but we want F as a function of t. Remember that we can only invert a delta series, so we define u = F (x, t) 1. Thus − (u + 1) u u β (u) := t = 3 = (u + 1) (1 x) u2 (u + 1)3 (1 x) u2 / (u + 1) − − − − n and we need the inverse, u = γ (t). We have to check that the negative powers a− of the linear term in β (u) are different from 0., We could get a by differentiating, or by noting that (u + 1) / (u + 1)3 (1 x) u2 starts with 1, hence β (u) = − − u + ... and therefore a = 1. We can now find the compositional inverse γ of β from (1.11) as n 3 2 n 1 (u + 1) (1 x) u n [γ]n = u − − − u + 1 ! Routine calculations give

(n 1)/2 i b − c n 2n 3i i k i k n [γ] = − x ( 1) − n n i n 2i 1 k − i=0 k=0 X − − − X (n 1)/2 1 b − c n 2n 3i i i k D (n; k) = − ( 1) − n n i n 2i 1 k − i=k X − − − The sample path above is one of 4077 counted by D (5; 2). Remark 1.2.4. It is not true that a power series must be of order 1 to have an inverse. For example, the power series 1 + t has the inverse t 1. Add your own examples in Exercise 1.2.1. −

1.2.1 Exercises

1.2.1. Find power series in R [[t]] that are of order 0 but have a compositional inverse. 16 Chapter 1. Prerequisites from the Theory of Formal Power Series

1.2.2. Write the Lagrange-Bürmann inversion formula (1.11) in terms of β instead of φ (t) = t/β (t). This formulation needs negative powers of β, which only n exist in the ﬁeld of Laurent series, where series of the form σ (t) = n k σnt ≥ are deﬁned for all k Z. An important functional on Laurent series is the 1 ∈ P residue, res(σ) = t− σ (t). Note that res (σ0 (t)) = 0 for all Laurent series σ (t). k Let ρ (t) = k 0 rkt be any formal power series. Show the original Lagrange- Bürmann Theorem,≥ which says that for any delta series β (t) holds P 1 ρ (γ) = r + res ρ β n tn 0 n 0 − n 1 X≥ n 1.2.3. Show that for every power series σ (t) = n 0 σnt holds ≥ 1 t n n n σ = n 0 t k=0 σn k. P 1 t 1 t k − − − ≥ P P 1.2.4. The smallest nontrivial integral domain is Z2, the integers modulo 2. + 0 1 0 1 0 0 1 0∗ 0 0 1 1 0 1 0 1 Addition and multiplication in Z2

1 Show β (t) = β− (t) holds for more power series than just β (t) = t when k = Z2. 1.2.5. If k is an integral domain, then k [[t]] is also an integral domain. 1.2.6. Show that for the Catalan generating function c (t) = 2/ 1 + √1 4t holds n 1 − 1/c (t) = 1 tc (t). Hence Cn = k=0− CkCn 1 k for all n 1. − − − ≥ 1.2.7. [10, Chapter 3.2]Let γ(t) =Pt + G (γ(t)), where 0 = G (0) = G0 (0). Then

k k 1 j 1 k 1 j γ (t) = t + D − kt − G (t) for all k 1. j! ≥ j 1 X≥ 1.3. Multivariate Power Series 17

1.3 Multivariate Power Series

Let r be an integer larger than 1. The r-dimensional array

S = (σn ,...,n ) r 1 r (n1,...,nr ) N ∈ 0 can be represented as the formal power series

n1 nr σ (t1, . . . , tr) = σn1,...,nr t1 tr . r ··· (n1,...,nr ) N X ∈ 0

n1 nr The coeffi cient functional [t1 tr ] σ = σn1,...,nr recovers the coeffi cients of the ··· ∂ series. If they are in k, we say that σ k [t1, . . . , tr]. The partial derivatives ∈ ∂ti are defined on k [[t1, . . . , tr]] as ∂ n1 ni 1 nr σ = niσn1,...,nr t1 t − tr ∂ti r ··· ··· (n1,...,nr ) N X ∈ 0 and we have ∂n1+ +nr ··· n1 nr n1 nr σ (0,..., 0) = n1! nr![t1 tr ] σ = n1! nr!σn1,...,nr . ∂t ∂tr ··· ··· ··· 1 ··· Similar to the univariate case, evaluation of σ is only allowed by setting some or all of the ti’s equal to 0. However, there is a new concept in multivariate power series: we can equate some of the formal variables, making them equal to a new formal variable s, say. For example,

m n3 n4 [s t3 t4 ] σ (s, s, t3, t4) = σi,j,n3,n4 . i+j=m X n A univariate formal power series φ (w) = n 0 φnw , on the other hand, can be made into a multivariate series by replacing≥ the formal variable by a linear combination of new formal variables. For example,P m + n [smtn] φ (s + t) = φ . m m+n For notational simplicity, we continue the discussion of multivariate formal power series in the bivariate case. Addition of bivariate power series is defined as expected; multiplication needs the Cauchy product as in the univariate case, m n m n γ (s, t) σ (s, t) = s t γi,jσm i,n j − − m,n 0 i=0 j=0 X≥ X X m n 0 0 when [s t ] γ (t) = γm,n. We will say that γ is of order 0 iff s t γ is different from 0. If s0t0 γ is a unit in k, then γ (s, t) has a reciprocal τ (s, t) such that 1 γ (s, t) τ (s, t) = 1. We denote the reciprocal by 1/γ (s, t), or γ (s, t)− . 18 Chapter 1. Prerequisites from the Theory of Formal Power Series

Suppose the numbers σm,n solve for (m, n) N N the (inhomogeneous) linear recursion ∈ × m n γm,n = αi,jσm i,n j − − i=0 j=0 X X where (αi,j)i,j 0 and (γi,j) are double sequences of given constants, and α00 = 1. We obtain, of≥ course, the generating function identity

γ (s, t) σ (s, t) = . α (s, t)

The above linear recursion is equivalent to

σm,n = γm,n αi,jσm i,n j (1.12) − − − (0,0)<(i,j) (m,n) X≤ where we consider the partial order (i, j) (k, l) iﬀ i j and k l on Z2. Note that (k, l 1) and (k 1, l) are both less≤ than (k, l). In≤ order to≤ calculate σ (s, t) − − as the simple expression γ (s, t) /α (s, t) the numbers σm,n must depend on the 2 numbers σi,j in a cone in N ; each σm,n depends only on σi,j for (0, 0) (i, j) (m, n) and (i, j) = (m, n). For more discussions of higher dimensional recursions≤ ≤ see Bousquet-Mélou6 and Petkovšek in [16]. Example 1.3.1. The lattice walk in the ﬁrst quadrant with steps , , can cycle back to itself, so we better count the number of ways that such{% a walk← reaches↓} (m, n) in k steps, starting at (0, 0). We call the number of walks d (m, n; k). The obvious recursion

d (m, n; k) = d (m + 1, n; k 1) + d (m, n + 1; k 1) + d (m 1, n 1; k 1) − − − − − is not of the form (1.12). Bousquet-Mélou and Petkovšek show how the “kernel method” of generating functions can be applied to solve this problem. It has been solved before in other ways by Kreweras [53, 1965]. The notation for power series in r > 1 variables can be a challenge; we avoid it by considering just two variables, in some examples three, but leave the general case to the reader. A multi-series (ρ, σ) is a pair (or more general an r-tuple) of formal power 2 series in two (or r) variables, (ρ, σ) k [[s, t]] . We say that (γ1, γ2) is a delta multi- ∈ series iﬀ γ1 (s, t) = sφ (s, t) and γ2 (s, t) = tψ (s, t) where φ (0, 0) and ψ (0, 0) are units in k, and φ (s, t) and ψ (s, t) are in k [[s, t]] (both having order 0). Thus γ1 (0, t) = 0, and γ1 (s, 0) is a power series in s. Analogously, γ2 (s, 0) = 0 and γ2 (0, t) k [[t]]. We∈ need the concept of the compositional inverse of a delta multi-series. The compositional inverse of the delta multi-series (β1, β2) is the multi-series (γ1, γ2) such that β1 (γ1 (s, t) , γ2 (s, t)) = s and β2 (γ1 (s, t) , γ2 (s, t)) = t. The inverse of 1.3. Multivariate Power Series 19

a delta multi-series is also a delta multi-series. If (γ1, γ2) is inverse to (β1, β2), then (β1, β2) is also inverse to (γ1, γ2). The inverse of (β1 (s, t) , β2 (s, t)) is usually 1 1 denoted by β1− (s, t) , β2− (s, t) . The partial derivatives ∂ φ (s, t) and ∂ φ (s, t) of a bivariate power series ∂s ∂t φ (s, t) inherit their properties (for example, the product rule) from the univariate case. We remember from Calculus that the derivative of a multi-series γ = (γ1, . . . , γr) is deﬁned as γ , where γ stands for the Jacobian. In the bivariate case, |J | J

∂ ∂ ∂(γ1, γ2) ∂s γ1 (s, t) ∂s γ2 (s, t) γ = = ∂ ∂ |J | ∂ (s, t) γ1 (s, t) γ2 (s, t) ∂t ∂t

∂ ∂ ∂ ∂ = γ1 γ2 γ2 γ1 . ∂s ∂t − ∂s ∂t The following multivariate Lagrange-Good inversion formula needs the Jaco- bian determinant.

Theorem 1.3.2. If γ1 (s, t) , γ2 (s, t) is a multi-series with compositional inverse β1 (s, t) , β2 (s, t), where we can write γ1 (s, t) = s/ε1 (s, t) and γ2 (s, t) = t/ε2 (s, t) with ε1 and ε2 of order 0, then

k l m+1 n+1 β1 (s, t) β2 (s, t) = ε1 (s, t) ε2 (s, t) γ . m,n |J | m k,n l h i h i − − For an elegant proof in Finite Operator Calculus terms see Hofbauer [43]. Note that the Theorem already assumes that an inverse exists: We have written it in a form that forces γ1 and γ2 to be a pair of delta series. Hence γ = 0 |J | 6 (Exercise 1.3.1), and an inverse pair of delta series will always exist. If β1 (s, t) = s/φ1 (s, t) and β2 (s, t) = t/φ2 (s, t) with (φ1, φ2) of order 0, then the Lagrange- Good inversion formula can also be written as

k l m+1 k n+1 l φ1 (s, t) φ2 (s, t) = ε1 (s, t) − ε2 (s, t) − γ (1.13) m,n |J | m,n h i h i (Exercise 1.3.3). Example 1.3.3. A multiseries does not need to be a delta series for having an inverse. Suppose γ1 (s, t) = s and γ2 (s, t) = s+t. We ﬁnd Jγ = 1, and β1(s, t) = | | s, β2 (s, t) = t s. − Example 1.3.4. The pair γ1 (s, t) = as/ (1 bt) and γ2 (s, t) = at/ (1 bs) has the inverse pair − −

1 1 1 1 a− bt 1 1 a− bs β (s, t) = a− s − and β (s, t) = a− t − 1 1 a 2b2st 2 1 a 2b2st − − − − (Exercise 1.3.4). Clearly, a needs to be a unit. Only nonnegative powers of b occur in the expansion of β1 and β2. Hence b does not has to be a unit. 20 Chapter 1. Prerequisites from the Theory of Formal Power Series

1.3.1 Exercises

1.3.1. Show that for any delta multi-series γ1 (s, t) and γ2 (s, t) holds γ = 0. |J | 6 1 1 1.3.2. Show that the compositional inverse β1− (s, t) , β2− (s, t) of a delta series (β1 (s, t) , β2 (s, t)) is also a delta series. 1.3.3. Show the inversion formula (1.13).

1.3.4. Let γ1 (s, t) = as/ (1 bt)and γ2 (s, t) = at/ (1 bs), where a is a unit. Apply Theorem 1.3.2 to show− that −

1 1 1 1 a− s 1 a− bt a− t 1 a− bs − , − 1 a 2b2st 1 a 2b2st − − − − is the pair inverse to (γ1, γ2). Convince yourself that b does not have to be invertible in k by choosing k = Z. 1.3.5. Show that the Lagrange-Good formula in Theorem 1.3.2 is equivalent to

∂ ∂ k l m n 1 γ1 ∂s ε1 γ2 ∂s ε2 β1 (s, t) β2 (s, t) = ε1 (s, t) ε2 (s, t) − ∂ − ∂ m,n γ1 ε1 1 γ2 ε2 − ∂t − ∂t m k,n l h i − −

1.3.6. Show that the multiseries (1 + s + t, 1 + s t) of order 0 has an inverse. − 1.3.7. Let φ and ψ be bivariate power series in (s, t), and let θ and η be a delta pair. Show that ∂ (φ, ψ) ∂ (φ, ψ) ∂ (θ, η) = ∂ (s, t) ∂ (θ, η) ∂ (s, t)

(hint: ∂φ = ∂φ ∂θ + ∂φ ∂η ). ∂s ∂θ ∂s ∂η ∂s 1.3.8. The mapping β (s) (β (s) , t) is an embedding of k [[s]] into k [[s, t]]. If β (s) = s/φ (s) is a delta series7→ and γ (s) its inverse, derive the univariate inversion formula (1.11)

k n n γ n = k [φ ]n k for 0 k n − ≤ ≤ from the bivariate Lagrange-Good formula in Theorem 1.3.2 applied to (β (s) , t). Chapter 2

Finite Operator Calculus in One Variable

The isomorphism between finite operators on polynomials, defined in section 2.2, and formal power series is central to the Finite Operator Calculus. It allows for transfer theorems, enabling us to find basic polynomial solutions to certain operator equations, and for a “functional expansion theorem”, helping us to determine a specific solution under given initial conditions. All our operators act on polynomials, and it is a recursion on polynomials, that defines the operator equation. Therefore, we begin by looking at polynomials in section 2.1. Even if the set of operators we study seems to be “small”compared to all operators imaginable, this set - the delta operators - has a powerful property, the translation invariance. Rota et al. studied it in detail [83, 1973], after an earlier paper with Mullin [62, 1970], and followed by work with Roman [81, 1978].

2.1 Polynomials, Operators, and Functionals

One main ingredient in the Finite Operator Calculus are bases of polynomials. Therefore our polynomials must come from a vector space where every sequence (pn) of polynomials of degree n is a basis, and therefore their coeffi cients must lie in a field F; we must allow division by any coeffi cient except 0. We will also assume that F has characteristic 0, hence the field is infinite. We could define our polynomials as having coeffi cients in an integral domain k, and define a basis such that the leading coeffi cient in a basis polynomial is a unit. For example, if our polynomials have integer coeffi cients, we would have the basis xn : n 0 . { ≥ } However, we will embed these polynomials in Q [x], and find many more bases. 22 Chapter 2. Finite Operator Calculus in One Variable

2.1.1 The Vector Space of Polynomials, and Their Bases Pascal’s Triangle recurrence can serve as a prototype of a recurrence in two variables, pn (m) = pn (m 1) + pn 1 (m) . (2.1) − − We present it in a rectangular array, shown on the left of the following tables. 1 m 1 6 21 56 126 1 1 4 1 5 15 35 70 1 2 1 3 1 4 10 20 35 1 3 3 1 2 1 3 6 10 15 1 4 6 4 1 1 1 2 3 4 5 1 5 10 10 5 1 0 1 1 1 1 1 0 1 2 3 4 n Pascal’sTriangle in its usual form pn (m) written as an array The columns on the right side of this table look like values of polynomials in n+m m, of degree n, and because we know that pn (m) = n this is easily shown. Actually, we know that the recurrence (2.1) must have a polynomial solution, as long as p0 (m) = 1 for all m 0, because of the following Theorem. This Theorem helps to decide when a given≥ diﬀerence recursion (like (2.1), written as pn (m) pn (m 1) = pn 1 (m)) has a solution in terms of polynomials. − − − Theorem 2.1.1. Let x0, x1,... be a sequence of given initial points, xi Z, and ∈ let Fn (m) be recursively deﬁned for all integers n 0 by ≥ n

Fn (m) = Fn (m 1) + aiFn i (m + bi) − − i=1 X where ak, bk F, for k = 0, 1,... , a1 = 0, and integers m > xn. Let Fn (m) = 0 ∈ 6 for all m < xn. We assume that bi xn i xn 1 for all n 1, and i = 1, . . . , n. ≥ − − − ≥ If F0 (m) is a constant different from 0 for all m x0, and if Fn (xn) is a known ≥ “initial value” at xn for all n, then there exists a sequence of polynomials fn (x) such that Fn (m) = fn (m) for all m xn and n 0 (inside the recursive domain). ≥ ≥ Proof. We can extend F0 (m) to all x F by making it the same constant every- ∈ where, f0 (x) := F0 (x0). Note that the new values (for x < x0) have no effect on the recursion F0 (m) = F0 (m 1) for all m > x0. Now suppose Fn (m) has been extended to a polynomial of− degree n for all m and for all 0 i n. Then ≤ ≤ Fn+1 (m) Fn+1 (m 1), the backwards difference, is a polynomial of degree n in − − m for all m > xn+1, as long as the right hand side of Fn+1 (m) Fn+1 (m 1) = n+1 − − i=1 aiFn+1 i (m + bi) is a sum of polynomials with highest degree n. This is − the case if all the terms Fn+1 i (m + bi) refer to x-values above the correspond- P − ing initial points, i.e., m + bi xn+1 i for all i = 1, . . . , n, and m > xn+1, thus ≥ − bi xn+1 i xn+1 1. But if the backwards difference of a function is a polynomial,≥ then− the− function− itself can be chosen as a polynomial of one degree higher 2.1. Polynomials, Operators, and Functionals 23

(see [47]), Fn+1 (m) = fn+1(m), say, for all m > xn+1. We can actually express this polynomial for m > xn as

n m

fn (m) = fn (xn) + ai fn i (k + bi) (2.2) − i=1 k=x +1 X Xn and we can ﬁnd fn (x) for other values of x by Lagrange interpolation (the La- grange interpolation formula is applied in the proof of Lemma 2.2.12).

m 1 7 17 21 21 6 1 6 12 13 13 5 1 5 8 8 8 4 1 4 5 5 5 3 1 3 3 3 3 2 1 2 2 2 2 1 1 1 1 1 1 0 1 1 1 1 1 0 1 2 3 4 n Fn (m) = Fn (m 1) + Fn 1 (m 2) and Fn (0) = 1 for all n 0 − − − ≥

The above table shows an example of a recursion that does not have a polynomial extension. The above Theorem does not apply, because the initial points are all 0, and b1 = 2 xn 1 xn 1 = 1 (but do you recognize some numbers? See Example 3.1.8).− − − − − Going back to Pascal’s Triangle we see that the Theorem above asks us for the initial points, which we can choose as xi = 1; the factors ai are 0 except for − a1 = 1, and the translations bi can all be chosen as 0. The condition b1 = 0 ≥ xn 1 xn 1 is satisﬁed for all i, and p0 (m) = 1 for m 0. We (implicitly) − − − ≥ assumed pn ( 1) = 0 for n 1. The discrete integral (2.2) tells us that pn (m) = m − ≥ k=0 pn 1 (k), which means − P n + m m n 1 + k = − n n 1 k=0 X − for n 1, if we would know the solution to the recurrence already! For example, ≥ suppose we keep Pascal’s recursion, but assume that p0 (x) = 1 for all x, and n 1 pn (1 n) = − pi (n 2i) for all n 1; what is the solution in this case? The − i=0 − ≥ Theorem tells us that pn (m) is a polynomial of degree n, and we will see that this information isP extremely valuable. How to make the most of it will be the topic of the following sections. 24 Chapter 2. Finite Operator Calculus in One Variable

m 1 4 10 20 35 56 m 1 4 12 35 107 344 2 1 3 6 10 15 21 2 1 3 8 23 72 237 1 1 2 3 4 5 6 1 1 2 5 25 49 165 0 1 1 1 1 1 1 0 1 1 3 10 24 116 1 1 0 0 0 0 0 1 1 0 2 7 24 82 −2 1 -1 0 0 0 0 −2 1 -1 2 5 17 58 −3 1 -2 1 0 0 0 −3 1 -2 3 3 12 41 −4 1 -3 3 -1 0 0 −4 1 -3 5 0 9 29 − − 0 1 2 3 4 5 n 0 1 2 3 4 n Pascal’srecursion continued to Pascal’srecursion with n 1 negative integers pn (1 n) = − pi (n 2i) − i=0 − P We will begin the study of polynomials with a discussion of bases, and then use those bases to define linear operators, from polynomials to polynomials. Be- cause all our operators are linear, we will drop the word “linear” in the future. There is a special subset of these operators that maps polynomials to coeffi cients, which can be viewed as a polynomial of degree 0. Such operators are called linear functionals. Again, we drop the word “linear”. The vector space of polynomials with coeffi cients in F (a field containing Z) will be denoted by F [x], where we think of x as a formal variable. This way it is clear how F [x] is embedded in the much larger space k [[x]], the formal power series over k F. Remember that a basis must have the properties that (1) ⊇ every element p k [x] can be written in a unique way as the finite sum p (x) = n ∈ akpk (x) with coeffi cients ak F, and (2) the elements pn of the basis are k=0 ∈ linearly independent. The dimension of F [x] is infinite; the set of polynomials P pn : n = 0, 1,... can serve as a basis as long as deg pn = n, and p0 = 0. In the{ following, we} will always assume that these last two properties define6 a basis pn of F [x]. This condition requires F to be a field and not just some integral domain.{ } The polynomials solving Pascal’srecursion above just have coeffi cients in the integral domain Z; embedding them in a vector space requires F = Q. In case of an infinite dimensional space the term Hamel basis is sometimes used; we will use “basis”for finite as well as infinite bases. Note that the power series k [[t]] do not have a basis - however, it is clear that tn : n = 0, 1,... acts like a basis: (1) Every element f k [[t]] can be written in { } k ∈ a unique way as the sum f (x) = n∞=0 akt with coeffi cients ak k, and (2) the elements tn of the basis are linearly independent. The problem∈ is that the sum k P f (x) = n∞=0 akt is not finite in general. This type of “basis” is often called a pseudobasis. If (φn) is a sequence in k [[t]], where k is an integral domain, such P n that ord (φn) = n and [t ] φn is a unit in k, then (φn) is a pseudobasis (Exercise 2.1.2). 2.1. Polynomials, Operators, and Functionals 25

2.1.2 Standard Bases and Linear Operators

A basis pn : n = 0, 1,... of F [x] can also be seen as the sequence p0, p1,... . We write{(f ) or (f ) }for sequences, and if the set p is a basis, we call the n n N n n ∈ { } sequence (pn) a basis. Remember that we required deg pn = n . Every such basis (pn) uniquely determines a linear operator Q from F [x] onto itself by deﬁning

Qpn = pn 1 − for all n 1, and Qp0 = 0. By linear extension every polynomial p gets the image ≥ n n

Qp = αkQpk = αkpk 1 − k=0 k=1 X X n if p (x) = k=0 αkpk (x). The set of such operators Q has the properties that Q reduces the degree of polynomials by 1 and Q maps constants polynomials into P 0. Thus Q has a kernel consisting of all constants (which we can identify with F), ker Q = F. We denote the set of all such operators by Ω. They are also called Gel‘fond-Leontiev operators (or generalized “diﬀerence-tial”operator in [56]).

Lemma 2.1.2. For every operator Q Ω exists a basis (pn) such that Qpn = pn 1 ∈ − and Qp0 = 0.

Proof. Let p0 (x) = 1. Because of the second property we have Qp0 = 0. From Qx = c, say, we see that c = 0 (otherwise x ker Q), and we define p1 (x) = x/c. 6 ∈ Assume that p0, p1, . . . , pn are already defined, Qpi = pi 1 for i = 1, . . . , n. The n+1 n+1 n − polynomial Qx has degree n, thus Qx = k=0 akpk, with an = 0. It follows n+1 n 1 n 1 6 that Qx /a = p + − a p /a = p + − a (Qp ) /a . Hence p = n n k=0 k k n n Pk=0 k k+1 n n+1 n+1 n 1 x − akpk+1 /an. By induction, (pn) is defined. The polynomials pn are − k=0 P P of degree n, and p0 = 0, thus (pn) is a basis. P 6 The proof of the Lemma shows a unique basis corresponding to every Q. However, there are more bases than the one constructed - every nonzero multiple, for example. It is easy to check that if (rn) and (sn) both satisfy the condition Qrn = rn 1 and Qsn = sn 1, and if deg (arn + bsn) = n for all n 0, ar0 + bs0 = − − ≥ 6 0, then (arn + bsn) is also a basis corresponding to Q. We need to “standardize” the bases to bring them into a one-to-one correspondence with Ω, by requiring specific initial values. The proof above shows that we can ask that they evaluate to 0 at 0 for all n 1, and are identically 1 for n = 0. Such a basis we call a standard basis. ≥

Lemma 2.1.3. Let Q Ω. There exists a unique standard basis (qn) such that ∈

Qqn = qn 1 for all n 1, and Qq0 = 0 − ≥ qn (0) = 0 for all n 1, and q0 (x) = 1. ≥ 26 Chapter 2. Finite Operator Calculus in One Variable

Proof. See the construction in the proof to the previous Lemma. Functionals are special operators mapping F [x] F, which can be seen as a vector space over itself, containing all polynomials of→ degree 0. For functionals L we have two notations: Lp and L p . In the special case where L stands for the n h | i n coeﬃ cient of x in p (x), we also write [x ] p (x) and [p]n, as in the case of formal power series.

Remark 2.1.4. If F F we can think of F [x] being embedded in F [[x]]; then x ⊆ must be a formal variable. The coeffi cient functional [p]n can be extended to F [[x]]. However, polynomials differ from formal power series in that they can be evaluated at any element a F. Such an evaluation functional we denote by Evala : p (x) ∈ 7→ p (a). Let p¯ be the polynomial p (a), a F. If F would be a finite finite field then ∈ degp ¯ can be smaller than deg p. We avoid this complication by assuming that F has charactericstic 0. Note that only when a = 0 the evaluation can be extended to F [[x]].

2.1.3 Exercises

2.1.1. Apply Theorem 2.1.1 to show that the recursion pn (x) = pn (x 1) + n 1 − pn 1 (x) for all x > 1 n with initial values pn (1 n) = i=0− pi (n 2i) for − − − − n 1, and p0 (1) = 1, has a solution that can be extended to a polynomial. ≥ P 2.1.2. Suppose F is an integral domain. Show that any sequence (φn) in F [[t]] is n a pseudobasis of F [[t]] if ord (φn) = n and [t ] φn is a unit in F.

2.1.3. Let dn (m) = dn (m 1) + dn 1 (m) dn 2 (m 2)for all m > n + 1, with − − − − − initial values at dn (n + 1) ,and d0 (m) = 1 for all m 1. Show that the solution to this problem can be extended to a polynomial sequence.≥ 2.1.4. Show that the following operators are in Ω: The forward diﬀerence operator ∆ : p (x) p (x + 1) p (x), the backwards diﬀerence operator : p (x) p (x) p (x 1)7→, the division− operator χ : p (x) (p (x) p (0)) /x, and∇ the deriv-7→ − − 7→ − ative operator : p (x) p0 (x). Find the standard basis (Lemma 2.1.3) for each operator. D 7→ n a 2.1.5. Prove that [x ], Evala and 0 are functionals on F [x]. R 2.2. Finite Operators 27

2.2 Finite Operators

Take any operator that is degree reducing, deg (T xn) = n k for all n k and some fixed k 1, and T xi = 0 for all i = 0, . . . , k 1.− Special examples≥ are the Gel‘fond-Leontiev≥ operators defined in section 2.1.− The powers T,T 2,T 3,... 1 n are defined (and are degree reducing), and therefore a0 + a1T + + anT p is n ··· i defined, where T p = 0 if deg p nk. Hence the infinite sum i 0 aiT is defined ≤ ≥ on any polynomial p F [x], if ai F for all i 0. Linear operators of this form, i ∈ ∈ ≥ P i 0 aiT where T is degree reducing , are called finite operators, because on ≥ n/k i anyP polynomial of degree n they act like the finite sum ib=0 c aiT . The integral domain of all such finite operators for fixed T is denoted by ΣT . Any operator Q P in ΣT is given by T and the sequence of coeffi cients (a0, a1,... ). We saw that the formal power series are a way of “storing”such a sequence of coeffi cients. Hence we i i have a bijection between i 0 aiT and i 0 ait . The additive structure on ΣT ≥ ≥ and F [[t]] are preserved under this bijection. Is there more structure preserved? To answer this we have toP look at the compositionP of two linear operators R and Q, (RQ) p := R (Qp) i j If R and Q are both in ΣT , Q = i 0 aiT and R = j 0 bjT , then ≥ ≥ i P j i P j i R (Qp) = R ai T p = bjT ai T p = bj aiT T p i 0 j 0 i 0 j 0 i 0 X≥ X≥ X≥ X≥ X≥ i i+j i i = bj aiT p = bj ai jT p = bjai j T p −  −  j 0 i 0 j 0 i j i 0 j=0 X≥ X≥ X≥ X≥ X≥ X   Therefore, the coeffi cients in the composition RQ are the same as in the Cauchy product in the corresponding power series. The integral domains ΣT and F [[t]] are i j isomorphic. We can write Q = α (T ) = i 0 aiT and R = β (T ) = j 0 bjT , and we have α (T ) a ti =: α (t),≥β (T ) b tj =: β (t≥). From i 0 i P j 0 j P α(t)β (t) RQ and 'α (t) β≥(t) = β (t) α (t) follows'RQ =≥ QR; any two opera- ' P P tors in ΣT commute! Three remarks about the isomorphism between formal power series and ΣT : Composition of operators corresponds to multiplication of power series. This is due to an unfortunate choice of the word “composition”for what should be called the product of operators. This gets even more confusing later, when we need the composition of power series. The second remark concerns the choice of letters. We say that ΣT F [[t]], ' but ΣT F [[s]], or any other formal variable, as long as F remains the same, because' the name of the formal variable does not matter. But it has to be a formal variable! For any degree reducing operator A we could write F [[A]] = ΣA, but we will not do this, because we believe that F [[t]] ΣA is a “cleaner”notation. The process of replacing the formal variable by A is' called evaluation. 28 Chapter 2. Finite Operator Calculus in One Variable

Example 2.2.1. Suppose the basis (pn) follows Pascal’srecursion pn (x) pn (x 1) = − − pn 1 (x). The operator : f (x) f (x) f (x 1) is called the backwards difference− operator. Taylor’sTheorem∇ 7→ tells us that− f (−x 1) = ( 1)n nf (x) /n!, − n 0 − D where = d/dx. Hence = I ( 1)n n/n! = I ≥ e . We find in n 0 P −D Σ . TheD backwards differeence∇ operator− ≥ is− also degreeD reducing.− ∇ D P If we choose the coeffi cients (a0, a1,... ) from some integral domain F F, i ⊇ then the finite operators i 0 aiT may map F [x] to some different space. How- ≥ i ever, if S is a degree reducing operator on F [x] and F = ΣS, then i 0 AiT , P ≥ with Ai ΣS, maps F [x] into itself. Such operator coeffi cients will not give us any new operators,∈ but they simplify the description of operators. For example,P we can see the operator as an element of Σ with coeffi cients (0, 0, 1, 1/2!, 1/3!,... ) ∇D D − in F, Σ F [[t]] as before. But we can also see as an element of Σ with coeffi cientsD '(0, , 0,... ) in Σ , Σ Σ [[t]], or as∇D an element of Σ withD coeffi - cients (0, , 0,...∇ ) in Σ , Σ ∇ ΣD '[[t]].∇ We will do this in section 2.4∇ on transfer D D ∇ ' D theorems, but it will be clearly stated. In general, the coeffi cients in ΣS will come from the same field F as in F [x], the polynomials.

2.2.1 Translation Operators Besides the degree reducing operators there is another important set of operators, the translation operators Ea : p (x) p (x + a). The translation operators are 7→ deﬁned for all a F, and they form a group, EaEb = Ea+b. They are degree ∈ preserving. Because x is a formal variable, and a F, we should say what we n ∈n n n k n k mean by the polynomial (x + a) . Of course, (x + a) := k=0 k a x − . Note that on the basis (xn) we have P n n k a n n n k n k k n E x = (x + a) = a x − = a D x (2.3) k k! k=0 k=0 X X where k = dk/dxk is the k-th power of the derivative operator. Hence D k a k a E = a D = e D k! k 0 X≥ This shows that Ea Σ , and it follows that Ea commutes with every other ∈ D operator in Σ . We call a linear operator T on F [x] translation invariant, if a aD TE p (x) = E T p (x) for all a F and p F [x]. The operators in ΣD are exactly the translation invariant operators,∈ as we∈ show in the following Lemma (the “First Expansion Theorem”in [83, p. 691]).

Lemma 2.2.2. A linear operator T on F [x] is translation invariant iﬀ T Σ . In i i ∈ D that case, T = i 0 Eval0 T x /i! ≥ | D P 2.2. Finite Operators 29

Proof. All we have left to show is that every operator T is in Σ if it satisﬁes a n a n n i D TE x = E T x for all n 0 and a F. Let T x = i 0 cn,ix , where for given ≥ ∈ ≥ n the ring elements cn,i are eventually 0 for large enough i. Thus P

i i i + k EaT xn = Ea c xi = c ai kxk = xk aic n,i n,i k − k n,i+k i 0 i 0 k=0 k 0 i 0 X≥ X≥ X X≥ X≥ and

n n n a n n n i i n n i k k n i TE x = T a − x = a − ci,kx = x a cn i,k i i i − i=0 i=0 k 0 k 0 i=0 X X X≥ X≥ X Both expressions are equal iﬀ the coeﬃ cients of xk agree,

n i + k i n i a cn,i+k = a cn i,k k i − i 0 i=0 X≥ X for all k 0 and for all a F. Both sides are a polynomial in a F. The right hand side is of≥ degree at most∈n, hence the left hand side is likewise,∈ and we conclude that 0 i n also holds on the left side. The coeﬃ cients of ai must be equal for all 0 ≤i ≤n, hence ≤ ≤

(i + k)!cn,i+k/n! = k!cn i,k/ (n i)! for all 0 i n and 0 k n i. − − ≤ ≤ ≤ ≤ −

If k = 0, then i!cn,i/n! = cn i,0/ (n i)!,thus − − n n n i ci,0 n! n i ci,0 i n T x = cn,ix = x − = x . i! (n i)! i! D i=0 i=0 i 0 X X − X≥ and therefore T Σ . ∈ D Remark 2.2.3. The degree-by-one reducing operators in Ω are of general interest. Let a (x) be a polynomial and deﬁne the multiplication operator M (a): p (x) 7→ a (x) p (x) for all p (x) F [x]. If a1 (x) and a2 (x) deﬁne M (a1 + a2) = M (a1) + ∈ M (a2), and M (a1a2) = M (a1) M (a2) = M (a2) M (a1). For any linear operator T on F [x] there exists a sequence of polynomials (an (x)) such that T = n n 0 M (an) R for some given R Ω (the polynomials an (x) are not neces- sarily≥ of deg n). See Exercise 2.2.2.∈ We saw in Lemma 2.2.2 that for R = the Ptranslation invariant operators are obtained by the choice of constant polynomialsD n cn (x) = Eval0 T x /n!. This and most of the other results in this section were already knownh at| thei end of the 19th century (see Pincherle [74]). 30 Chapter 2. Finite Operator Calculus in One Variable

2.2.2 Basic Sequences and Delta Operators Even the set of standard bases is too large for our purposes. We will focus on a “tiny” subset, often called the sequences of binomial type. These sequences (bn) follow the binomial theorem (see also section 2.3),

bn (x + y) = bi (y) bn i (x) − i=0 X x for all n 0. Examples are bn (x) = n (Vandermonde convolution) and, if Q F, ≥ n ⊂ bn (x) = x /n! (giving the original binomial theorem). Standard sequences that are of binomial type are called basic sequences ; not a great name, but from here on, the only bases in F [x] we will consider in this section, are basic sequences. The binomial theorem is a convolution identity. It says that for the generating n function b (x, t) = n 0 bn (x) t holds b (x + y, t) = b (x, t) b (y, t), where b (0, t) = b (x, 0) = 1. If b (x, t)≥is of the form f (t)x, where f (0) = 1, then the binomial P theorem holds. In this case, f (t) = eβ(t) for some delta series β. In other words, the logarithm of f (t) = 1+ higher order terms in t exists, and log f (t) = β (t). We will investigate that approach in this chapter and the next. 1 Because β (t) is a delta series, the compositional inverse β− (t) of β (t) exists and is also a delta series (section 1.2). If γ (t) is any delta series, we call the operator 1 γ ( ) a delta operator, hence β− ( ) and β ( ) are both delta operators. We will D 1 D D study the delta operator β− ( ) and and see what it does to (bn). The meaning of β ( ) has to wait until sectionD 2.3.2. D

Transforms of Operators on F [x] [[t]] The investigation follows the ideas of J. M. Freeman [35] in his “Transform of Operators”, opening the door to a unify- ing theory beyond the Finite Operator Calculus. First we note that b (x, t) is in F [x] [[t]], the formal power series with coeffi cients in the ring of polynomials F [x]. We worked already with a power series in F [x] [[t]] in Example 1.2.3. An element from F [x] [[t]] can be understood as an infinite matrix, whose rows stand for the coeffi cients of t, the n-th row containing the coeffi cients of a polynomial. The matrix is triangular, if the polynomials are a basis for F [x]. An operator A on F [x] is extended to an operator on F [x] [[t]] by defin- i i ing A i 0 pi (x) t := i 0 (Api (x)) t . Such a t-linear extension is called an ≥ ≥ x-operator on F [x] [[t]]. For example, ext Q [x] [[t]], and P P ∈ next = tnext. (2.4) D In the same way, there are t-operators on F [x] [[t]], extended by x-linearity. Substituting a delta series α (t) is a linear operator on F [[t]] (see again section 1.2), and by x-linearity C(α)f (x, t) = f (x, α (t)) for all f (x, t) F [x] [[t]]. Note that x-operators commute with t-operators! ∈ 1 xβ(t) Now we are ready to see what the x-operator β− ( ) does to b (x, t) = e , D 2.2. Finite Operators 31

1 1 xβ(t) 1 xt 1 xt β− ( ) b (x, t) = β− ( ) e = β− ( ) C(β)e = C(β)β− ( ) e D D D D 1 n because x- and t-operators commute. Suppose β− ( ) = n 1 γn . Then D ≥ D 1 xt n xt 1 xtP β− ( ) e = γnt e = β− (t) e , D n 1 X≥ 1 xt as in (2.4), showing that the action of the x-operator β− ( ) on e is the same 1 D 1 as multiplication by the formal power series β− (t). We say that β− ( ) and 1 xt D M β− are transforms of each other, with respect to e . Hence

1 1 xt 1 xβ(t) β− ( ) b (x, t) = C(β)M β− e = β− (β (t)) e D n = tb (x, t) = bn 1 (x) t − n 0 X≥ 1 We have shown that the operator β− ( ) Σ is exactly the operator B : bn D ∈ D 7→ bn 1. This is the content of the following theorem. − n Theorem 2.2.4. The basic sequence (bn) has generating function n 0 bn (x) t = xβ(t) 1 1 ≥ e , iﬀ β− ( ): bn bn 1 for all n 1, and β− ( ) b0 = 0. D 7→ − ≥ D P Example 2.2.5. We list some frequently occurring basic sequences.

n n n xt 1 1. If bn (x) = x /n!, then n 0 x t /n! = e hence β (t) = t = β− (t), and 1 n n ≥n 1 β− ( ) x /n! = x /n! = x − / (n 1)!. D D P − n 1+x n 1+x n x 2. If bn(x) = −n then n 0 −n t = (1 t)− hence β (t) = 1/ ln (1 t) 1 t ≥ − 1 n 1+x − and β− (t) = 1 e− . We obtain the delta operator = 1 E− : −n n 2+x − P ∇ − 7→ −n 1 . − x x n x 1 3. If bn(x) = n then n 0 n t = (1 + t) and β− ( ) = E 1, mapping x x ≥ D − n into n 1 . This delta operator is called ∆. Thus ∆ = E . − P ∇ 1 1 4. Suppose we interchange β and β− in the previous example, then β− ( ) = n+1 n D ln (1 + ) = n 1 ( 1) /n. How can we ﬁnd the basic sequence for this deltaD operator?≥ One− wayD will be shown in Example 2.3.11, another in section 2.4. P Remark 2.2.6. (a) Not every sequence of binomial type is a standard sequence. For n example, let b2n (x) = x /n! and b2n+1 (x) = 0 for all n 0. The sequence (bn) is of binomial type, but it is not a basis according to our deﬁnition.≥ (b) The binomial theorem involves the binom x+y, a sum of two formal variables. We could view it as a statement about polynomials in two variables, F [x, y], with F [x] and F [y] embedded in F [x, y]. A more fruitful interpretation is given in in 32 Chapter 2. Finite Operator Calculus in One Variable terms of Hopf algebras (see [61]). Closer to our approach is choosing y F, and reading the binomial theorem as ∈

n n y i E bn (x) = Evaly bi bn i (x) = Evaly bi B bn (x) (2.5) h | i − h | i i=0 i=0 X X y i where B : bn bn 1. The expansion E = i 0 Evaly bi B is just a special case of Exercise7→ 2.2.1,− hence gives an independent≥ h proof of| thei binomial theorem for basic sequences. P 1 (c) If B = β− ( ) is a delta operator, then =β (B), hence every operator in D D Σ is also expandable in B, so Σ ΣB for all B Σ . We saw already one D y D ' i y ∈ D example in (2.5), E = ∞ Evaly bi B , thus E ΣB. Every A ΣB has its i=0 h | i ∈ ∈ isomorphic image A˜ Σ such that A = A˜ as operators on F [x], but A = α (B), PD say, and A˜ =α ˜ ( )∈, hence α =α ˜ (β (t)). Even when A and A˜ are the same D operator on F [x], the power series representation of A and A˜ are diﬀerent, when B and are diﬀerent. Of course, all operators in ΣB commute with all operators in Σ . D D

2.2.3 Special Cases We present three special classes of basic sequences; the first two arise in probability theory. The relationship of Finite Operator Calculus to probability theory has been studied by DiBucchianico [24, Theorem 3.5.10], but also later by [29] in the connection with Umbral Calculus. Finally we consider basic sequences (bn) having coeffi cients that are themselves values of basic sequences. While this looks like a rather esoteric event on the first glance, it actually happens frequently; it happens, whenever the coeffi cient of x in the quadratic b2(x) is not zero!

Basic Sequences and Moment Generating Functions Suppose the basic sequence xα(t) (an) with generating function e is related to the random variable X through n E[X ] = n!an (1), hence for the moment generating function of X holds

Xt α(t) mX (t) := E e = e .

Of course, this requires a distribution where all the moments E[Xn] exist and are in F, and because a1 (1) = 0 for any basic sequence (a1 (0) = 0), it also 6 requires µ = E [X] = 0 (but see Example 2.2.9). For the basic sequence (an) this 2n6 2 means that E X = (2n)!a2n (1) 0; especially σ = V ar (X) > 0 implies ≥ that a (1) > a (1)2 /2 > 0. Hence not all basic sequences generate probability 2 1 moments, and not all moment generating functions generate basic sequences! We deﬁne the cumulant generating function

2 2 KX (t) := ln mX (t) = α (t) = µt + σ t + ... 2.2. Finite Operators 33

As for the moment generating function, it holds that KX0 (0) = E [X], but

2 K00 (0) = V ar [X] = 2a2(1) a1(1) X − = 2α2

n if α (t) = n 1 αnt . Note that KX is either a quadratic polynomial (normal distribution) or≥ a true power series, i.e., no polynomial at all (Lucas [55]). The P n k numbers κn = n!αn are called the cumulants of X. If an (x) = k=0 an,kx /k!, then the relationship between the cumulants and the moments of X can be described by Lemma 2.2.10, P

n n n k E [X ] = n!an(1) = n! an,k/k! = n! α n /k! k=0 k=0 X X n k

= Bi !α Bi = κ B | | | | | | k=1 (B1,...,Bk) S(n,k) i=1 π S(n) B π X X∈ Y ∈X Y∈ where S (n, k) is the set of all partitions of n into k parts (see Remark 1.1.2), and S (n) is the set oﬀ all partitions of n. Because of the two combinatorial methods k of representing α n, as explained in Remark 1.1.2, there is a second formula for expressing moments through cumulants, n k n n! E [X ] = αλi . `1! `n! k=0 λ n, λ =k i=1 X ` X| | ··· Y 1 We can say that KX− ( ) is the delta operator for (an). For example, if X D N k N k has the binomial distribution Pr (X = k) = k p (1 p) − with parameters N and p, then − n N t n N k N k t N mX (t) = k p (1 p) − = 1 + e 1 p n! k − − n 0 k=0 X≥ X and

n n x Nx k n t k Nx k k! an (x) = [t ] mX (t) = p [t ] e 1 = p S (n, k) . k − k n! k 0 k=0 X≥ X 1 Hence we ﬁnd A = KX− ( ) = N ln (1 + p∆). D k When x is a nonnegative integer k, say, then mX (t) = mX1+ +Xk (t), where ··· X1,...,Xk are independent random variables with the same distribution as X, kα(t) thus e = mX1+ +Xk (t), and ··· n an (k) = E [(X1 + + Xk) ] /n! ··· 34 Chapter 2. Finite Operator Calculus in One Variable

What can we say if x is not an integer? DiBucchianico [24, Theorem 3.5.10] has shown that for every weakly continuous convolution semigroup (µx)x 0 of proba- ≥ bility measures on R exist a basic sequence (an) that

n ∞ y a (x) = dµ (y) n n! x Z−∞ for x 0 iff the above equation holds for n = 1. Note that this definition excludes ≥ the semigroups for which the first moment of µ1 is zero, because a1 (1) = 0 for any basic sequence, but here are cases when this can be overcome. If we6 choose n n for µx the point mass at x, called δx, then ∞ (y /n!) dδx (y) = x /n!, the basic polynomial for . −∞ D R x k Example 2.2.7. The Poisson semigroup µx = e− k 0 x /k! δk leads to the basic polynomials ≥ P x k n ∞ e x k φ (x) = (yn/n!) dµ (y) = − , n x n! k! k 0 Z−∞ X≥ known as the exponential polynomials (see Dobinski’s formula (2.26)).

Example 2.2.8. The basic polynomials for the Gamma semigroup

yx 1 dµ (y) = e y − dy x − Γ(x) are n x 1 ∞ y y − y n 1+x g (x) = e dy = − , n n! Γ(x) − n Z0 associated to the backwards difference operator . ∇ Example 2.2.9. The normal distribution with expectation 0 is a semigroup with 2 2 moment generating function eσ t /2. Of course, we cannot find a basic sequence y2/(2x) agreeing with the moments of dµx (y) = e− /√2xπ for x > 0, because the expectation is 0, and so are all odd moments. However, we can define

2n 1 ∞ y y2 1 pn (1) = e− 2 dy = , √2π (2n)! 2nn! Z−∞ skipping the odd moments. Thus

xn p (x) = . n 2nn! The delta operator is 2 . D 2.2. Finite Operators 35

Distributions of Binomial Type Let F = R, and let X be a random variable on 0, . . . , n with probability distribution function (p.d.f.) p (i) = Pr (X = i). We {say that} the p.d.f. is of binomial type iﬀ

bi (α) bn i (β) 0 p (i) = − 1 (2.6) ≤ bn (α + β) ≤ for some basic sequence (bn) and α, β R such that bn (α + β) = 0. It can be shown that all symmetric distributions,∈p (i) = p (n i), are of binomial6 type if − p0 > 0. The expectation of any random variable X having a distribution of binomial type equals αn E [X] = α + β (Exercise 2.2.13). The second moment equals for all n 2 ≥ 2 2 αβn S bn 2 (α + β) V ar [X] = 2 αβ − , (2.7) (α + β) − bn (α + β) where Sbn (x) = (n + 1) bn+1 (x) /x, a translation invariant operator (Exercise 2.2.14 and Lemma 2.3.4). The variance is a symmetric function in α and β. For n αβn2 n(n 1) example, if bn (x) = x /n!, then S = I, and V ar [X] = 2 αβ − 2 = (α+β) − (α+β) αβn/ (α + β)2. It is easy to verify that the binomial and the hypergeometric distribution are of binomial type; however, it is interesting that both distributions are of binomial type because of a general construction based on conditioning (see Exercises 2.2.9 and 2.2.10). A subclass of the distributions of binomial type is obtained by conditioning m xβ(t) as follows. Let β (t) be the delta series such that n 0 bn (x) t = e . Suppose ≥ that for some parameter space Θ the family of random variables Xθ, θ Θ, takes P ∈ values in N0 such that for all i N0 ∈ i θ Pr (Xθ = i) = τ σ bi (θ) (2.8) β(τ) for some τ R where β (τ) converges, and σ = e− . Furthermore, assume ∈ that the process Xθ is stationary with independent increments at α and β, where α, β, α + β Θ, ∈ Pr (Xα = i and Xα+β = i + j) = Pr (Xα = i) Pr (Xβ = j) . (2.9)

Then the conditional distribution p (i) := Pr (Xα = i Xα+β = n) is of binomial type on 0, . . . , n , | { } Pr (Xα = i) Pr (Xβ = n i) p (i) = Pr (Xα = i Xα+β = n) = − | Pr (Xα+β = n) i α n i β τ σ bi (α) τ − σ bn i (β) − = n α+β . τ σ bn (a + β) 36 Chapter 2. Finite Operator Calculus in One Variable

Basic Sequences with Polynomial Coeﬃ cients We assume that F equals Q, R, or C, so that a polynomial of degree n that is known at n + 1 integers places is n xβ(t) known everywhere. Let (bn (x)) be a basic sequence, n 0 bn (x) t = e . We denote the coeﬃ cients of xk/k! in b (x) by b = xk/k!≥ b (x), n n,k P n n xk b (x) = b n n,k k! k=0 X Lemma 2.2.10. For all 0 k n holds ≤ ≤ k bn,k = β n . Proof. From

n xk xk exβ(t) = b (x) tn = tn b = b tn n n,k k! k! n,k n 0 n 0 k=0 k 0 n k X≥ X≥ X X≥ X≥ and xk exβ(t) = β (t)k k! k 0 X≥ k n follows bn,k = β n by comparing coeffi cients of x /n!. We say that (bn) has polynomial coeffi cients iff there exists a polynomial sequence ˜bn (x) such that bn,i = ˜bn i (i) for all n 0 and for all 0 i n. − ≥ ≤ ≤ Surprisingly, such coeffi cients exist under very mild conditions on (bn). We call

˜bn (x) the sequence of coeﬃ cient polynomials for (bn (x)), and we will prove

that ˜bn is the basic sequence for ln (β (t) /t).

r Theorem 2.2.11. Let (bn (x)) be a basic sequence such that bn,n = 1 for all n N0, n xβ(t) ∈ and n 0 bn (x) t = e . The following statements are equivalent: ≥ 1. P(bn) has polynomial coeﬃ cients.

2. There exists a basic sequence ˜bn (x) such that bn,i = ˜bn i (i) for all n 0 − ≥ and for all 0 i n. ≤ ≤ 3. b2,1 = 0. 6

4. There exists a basic sequence ˜bn (x) such that bn,1 = ˜bn 1 (1) for all n > 0. − ˜ 5. There exists a β˜ (t) F [[t]] such that β (t) = teβ(t). ∈

The basic sequence ˜bn occurring in (2) and (4) in the Theorem is actually xβ˜(t) the same sequence, and it has the generating function e . Note that b2,1 = β2. 2.2. Finite Operators 37

Proof. (1) (2): Let bn,k = cn k (k) where deg cn = n. The generating function ⇒ − of (cn) equals

n n k n k k n k cn (k) t = bn+k,kt = β n+k t = t− β n t = (β (t) /t) . n 0 n 0 n 0 n k X≥ X≥ X≥ X≥ 0 From t β (t) /t = b1,1 = 1 follows log (β (t) /t) exists and is a delta series, thus (cn (x))is a basic sequence. (2) (3): The polynomial ˜b1 (x) has only one root, ⇒ and that is at 0. Hence b2,1 = ˜b1 (1) = 0. (3) (4): Let b2,1 = 0. The num- 6 ⇒ 6 ˜ ˜ n bers bn (1), n 0, deﬁne the basic sequence bn (x) , because n 0 bn (x) t = ≥ x ≥ n ˜ ˜ ˜ n 0 bn (1) t . The only conditions on bn (1) are b0 (1) = 1Pand b1 (1) = 0. ≥ 6 ˜ BothP conditions are satisﬁed if b2,1 = 0 and bn,1 = bn 1 (1). (4) (5): The 6 − ⇒ generating function of ˜bn in (4) equals

x xβ˜(t) n n e = ˜bn (x) t = ˜bn (1) t   n 0 n 0 X≥ X≥   hence β˜(t) ˜ n n+1 n+1 te = t bn (1) t = bn+1,1t = [β]n+1 t = β (t) . n 0 n 0 n 0 X≥ X≥ X≥ (5) (1): The series β˜ (t) must be a delta series, because log (β (t) /t) exists and is a⇒ delta series. From

k n kβ˜(t) (β (t) /t) = bn+k,kt = e n 0 X≥ xβ˜(t) follows that for the basic sequence ˜bn with generating function e holds bn+k,k = ˜bn (k). The condition bn,n = 1 is equivalent to β (t) = t + ... . Suppose [β (t)]1 = λ = 0. Let an (x) := bn (x/λ). Theorem 2.2.11 applies to (an (x)), which has the 6 n n n coeﬃ cients an,k = [β (t) λ− ]k = λ− bn,k. The following statements are equivalent:

1. (bn (x/λ)) has polynomial coeﬃ cients.

i 2. There exists a basic sequence ˜bn (x) such that bn,i = λ ˜bn i (i) for all n 0 − ≥ and for all 0 i n. ≤ ≤ 3. b2,1 = 0. 6 4. There exists a basic sequence ˜bn (x) such that bn,1 = λ˜bn 1 (1) for all − n > 0. 38 Chapter 2. Finite Operator Calculus in One Variable

˜ 5. There exists a delta series β˜ (t) F [[t]] such that β (t) = λteβ(t). ∈ From β (t) = λteβ˜(t) follows

β (t) b ˜ = 1 + 2,1 t + = eβ(t) = 1 + β˜ (t) + = 1 + β˜ t + ... λt λ ··· ··· 1 h i n which shows that the coeffi cient ˜bn,n := [x /n!] ˜bn (x) equals n bn ˜ ˜ n b2,1 2,1 bn,n = β (t) = = n . (2.10) n λ b1,1 h i Thus ˜bn (x) would not be of degree n if b2,1 = 0. One application of basic sequences with polynomial coeffi cients will be given in connection with Riordan matrices (Corollary 3.2.2). Another application occurs in Lagrange-Bürmann inversion. If we want an explicit answer from the Lagrange- Bürmann formula for β (t), explicit in the sense of Stanley [89], with a fixed number of sums over integer intervals, then it gets diffi cult if we are not able to explicitly determine φ (t)in β (t) = t/φ (t). For example, let β (t) = eat ebt R [t], a = b, both different from 0. Of course we have φ (t) = t/ eat ebt ,− but how∈ do we6 find k n k n − n an explicit expression for n t − φ (t) , the coeffi cient of t in the k-th power of 1 β− (t)? In this situation it may be worthwhile checking the coeffi cient β2 of the 2 n k n quadratic term of β (t) = β1t + β2t + ... . Because in that case, t − φ (t) = n˜ ˜ β1− bn k ( n) for all integers n k, where bn is the sequence of “coeffi cient − − ≥ xβ˜(t) β˜(t) polynomials”with generating function e , andβ (t) = β1te . If we know the x basic sequence ˜bn explicitly, from their generating function (β (t) / (β1t)) , then we are done, k k n k n˜ γ n = [φ ]n k = β1− bn k ( n) (2.11) n − n − − for all n k. Otherwise, the following lemma may help. ≥ 2 Lemma 2.2.12. [67]If β (t) = β1t + β2t + ... with β1 = 0 and β2 = 0, and (bn) is the basic sequence with generating function exβ(t), then6 the positive6 powers of the compositional inverse γ (t) of β (t) can be expanded in positive powers of β (t) as

n j k 2n + k n+k n ( 1) n k j γ (t) = k t − β− − − b n j n + k + j 1 n+j,j n 0 j=0 X≥ X j+n j j where bn+j,j = t β (t) are also the coeﬃ cients of x /j! in bn+j (x). Proof. The Lagrange interpolation formula (E. Waring [99], 1779) applied to ˜bn k (x) says that − n k n k − − x xi ˜bn k (x) = ˜bn k (xj) − . − − xj xi j=0 j=i=0 X 6 Y − 2.2. Finite Operators 39

Suppose we set xi = i and x = n, then − n k n k − − n i ˜bn k ( n) = ˜bn k (j) − − − − − j i j=0 j=i=0 X 6 Y − n k j − n k 2n k ( 1) j = n − − − β1− bn k+j,j. (2.12) j n n + j − j=0 X

Certainly, the above Lemma should only be applied if “ordinary” Lagrange inversion gets too complicated! Such a case is discussed in Example 2.2.13. If the Lagrange inversion formula (1.11) can be applied, Lemma 2.2.12 may lead to some identity (Exercises 2.2.19 and 2.2.20). Example 2.2.13. The factorial numbers F (k, j; a, b) of the second kind are deﬁned j for k j 0 as the coeﬃ cients of j!tk/k! in eat ebt , ≥ ≥ − j j! k at bt j j j i (ai+(j i)b)t F (k, j; a, b) t = e e = ( 1) − e − k! − i − k j i=0 X≥ X where the parameters a and b may be real or complex. It easy to determine F (k, j; a, b) from this generating function via the ordinary binomial theorem,

j k j i (ia + (j i) b) F (k, j; a, b) = ( 1) − − . − i!(j i)! i=0 X − The best known example of factorial numbers of the second kind are the Stirling numbers S (k, j) of the second kind, S (k, j) = F (k, j; 1, 0). The factorial numbers of the ﬁrst kind, f (n, k; a, b), occur in the inverse γ (t) of eat ebt =: β (t), − k! γ(t)k = f (n, k; a, b) tn. n! n k X≥ We want to ﬁnd an explicit expression for f (n, k; a, b), so we try Lemma 2.2.12 in the case when a2 = b2: 6 n j 1 n k j 2n + k n ( 1) − (a b)− − − j! γ(t)k = k tn+k − − F (j + n, j; a, b) , n j n + k + j (j + n)! n 0 j=0 X≥ X hence

n k j j n k! 2n k − n k k ( 1) (a b)− − j! f (n, k; a, b) = − − − − F (j + n k, j; a, b) . n! n k j n + j (j + n k)! − j=0 − X − 40 Chapter 2. Finite Operator Calculus in One Variable

Inserting the expression we found for F (k, j; a, b) gives f (n, k; a, b) =

n k j n j j+n k − 2n k (a b)− − (ia + (j i) b) − k − − ( 1)i − n k j, n k + j, k n + j − i!(j i)! j=0 i=0 X − − − X − for a2 = b2. If a = 1 and b = 0 we obtain an explicit expression for the Stirling numbers6 of the ﬁrst kind,

n k j j+n k − 2n k k i − s (n, k) = − ( 1)i . n k j, n k + j, k n + j − i!(j i)! j=0 i=0 X − − − X − 2.2.4 Exercises 2.2.1. Show that for every translation invariant operator T holds i T = i 0 Eval0 T bi B , where B is any delta operator with basic sequence (bn) (a generalization≥ h of| Lemmai 2.2.2). P 2.2.2. Suppose that (rn) is the standard basis for Ω, : rn rn 1. Let T R ∈ R 7→ − be an arbitrary linear operator on F [x] such that T rn (x) = an (x) for all n 0 ≥ (an (x) may have any degree!). Show that

n ∞ l(c) n T = M (an k) ( 1) M rc1 rc2 rc .  − − ··· l(c)  R n=0 k=0 c C(k) X X ∈X   The set C (k) stands for all compositions (ordered partitions) of k,

k l

C (k) = (c1, . . . , cl): ci 1, ci = k ≥ l=1 ( i=1 ) [ X and l (c) is the length of c C (k). Note that c C(0) = 1. Find ∈ ∈ ··· l(c) P ( 1) M rc rc rc − 1 2 ··· l(c) c C(k) ∈X when = . R D 2.2.3. Show that the division operator χ can be written in the form

k 1 k χ = ciM x − . D k 1 X≥ 2.2.4. The binomial theorem can be applied to the forward diﬀerence operator ∆, showing that for all nonnegative integers n and polynomials p holds

n n n n k ∆ p (x) = ( 1) − p (x + k) . k − k=0 X 2.2. Finite Operators 41

This formula holds for a much larger class of functions than polynomials; we can n 1 take, for example, ∆ z− for complex z, and ﬁnd n n k n 1 n ( 1) − ∆ z− = − , k z + k k=0 X as long as z = k for k = 0, . . . , n. Show by induction for such z that 6 − n n 1 ( 1) ∆ z− = n! − . z (z + 1) (z + n) ··· You have proven the identity n n z 1 ( 1)k = k − z + k z+n k=0 n X 1 2.2.5. Suppose B = β− ( ) is a delta operator, and (bn(x)) a basic sequence for D B. Show: If a = 0, then the scaled polynomials bn(ax) are the basic polynomials 6 1 for the delta operator β− ( /a) D 2.2.6. Suppose the radius r of convergence of β (t) is larger than τ. Find the moment generating function of the distribution (2.8). Show that E[Xθ] = pβ0 (p) and 2 V ar [Xθ] = pβ0 (p) + p β00 (p).

2.2.7. Let Y,Y1,Y2 ... be i.i.d. random variables with finite moments and nonzero expectation. Show that there exists a unique basic sequence (an) such that an (m) := n E [(Y1 + + Ym) ] /n! for all m, n = 0, 1, 2,... . ··· 2.2.8. Let Y be a geometric random variable, i.e. Pr (Y = k) = p (1 p)k, the probability of k failures before the first success in independent trials with− success probability 0 < p < 1. Show that n x+k 1 k k k! bn (x) = − (1 p) p− S (n, k) k − n! k=0 X is the n-th basic polynomial obtained from this distribution. (The numbers S (n, k) are the Stirling numbers of the first kind; see Exercise 2.3.15 for their generating function). θ i 2.2.9. Suppose Xθ has the Poisson distribution Pr (Xθ = i) = e− θ /i! for all i N0 and θ R+. Show by conditioning that the binomial distribution is of binomial∈ type. ∈

2.2.10. Take a sample of size n from a population of size α + β, α, β N0, which contains α marked and β unmarked subjects. Show by conditioning on the∈ binomial distribution that the hypergeometric distribution f (i) := Pr (i marked objects in the sample) is of binomial type. 42 Chapter 2. Finite Operator Calculus in One Variable

2.2.11. Show that every symmetric distribution on 0, . . . , n with p0 > 0 is a distribution of binomial type, { }

bi (1/2) bn i(1/2) p (i) = − . bn (1)

2.2.12. Find the basic sequence in (2.6) when p0 = pi = 1/ (n + 1) for all i 1 ∈ 0, . . . , n (discrete uniform distribution). Show that V ar [X] = 12 n (n + 2) by applying{ } (2.7). αn n 2.2.13. Show that is the expected value i=0 ibi (α) bn i (β) /bn (α + β) of a α+β − random variable that has a distribution of binomial type (2.6). P 2.2.14. Let X be a random variable having a distribution of binomial type (see 2 Exercise 2.2.13). Show that the second moment µ2 = E X equals

2 2 αn S bn 2 (α + β) µ2 = αβ − α + β − bn (α + β) where Sbn (x) = (n + 1) bn+1 (x) /x. As an example, calculate the second moment α β α+β of the hypergeometric distribution p (i) = i n i / n . − 2.2.15. Let c be a nonnegative real number, α and β positive real numbers. Show that i 1 n 1 i n αβ (ci + α) − (c (n i) + β) − − p (i) = − i (α + β) (cn + α + β)n 1 − for i = 0, . . . , n is a distribution of binomial type, generalizing the binomial distribution (c = 0). For n 2 show that the variance equals ≥ 2 n k 2 αβn (k 1) c − αβn! − . 2 − k (α + β) k=2 (cn + α + β) (n k)! X − 2.2.16. Find the moment generating function of the distribution (2.8). Show that 2 E [Xθ] = pβ0 (p) and V ar [Xθ] = pβ0 (p) + p β00 (p)

2.2.17. Find the basic sequence (bn) such that bn+k,k = bn (k) for all n, k 0 ( ≥ bn (x) is its own coeﬃ cient polynomial in Theorem 2.2.11)

2.2.18. Show that for any delta series β (t) = t/φ (t) with [β]2 = 0 and β (t) = β˜(t) ˜ n k n 6 β1te holds bk (n) = β1− t φ (t)− for all integers n (β1 = [β]1). 2.2.19. The Lagrange interpolation formula, even in its simple form (2.12), can be the reason for identities that are not obvious. Show that

n n j x x + n n ( 1) − x n + j 1 = − − n n j x + j n j=0 X for all x 0 by applying (2.12) to β (t) = t/ (1 t). ≥ − 2.2. Finite Operators 43

2.2.20. Show that the compositional inverse of β (t) = tet has the same expansion when calculated from (1.11) or from (2.11). Substituting β (t) shows that

n 1 ∞ (n + 1) − n ( 1)n ett = e t. n! − n=0 − X Calculate the same inverse from Lemma 2.2.12. Comparing the answers shows that n 1 n k n x − n ( 1) − = − kn x+n k x + k n k=0 X for all n 1. This is a special case of identity (1.47) in [38]. ≥ 2.2.21. Use the factorial numbers of the ﬁrst kind to show that the coeﬃ cient of tn/n!in (sin t)m equals 0 if m + n is odd, and equals

m (m+n)/2 m m k n ( 1) 2− ( 1) (2k m) − k − − k=0 X if n + m is even (n, m N0). ∈ 2.2.22. Use Lemma 2.2.12 and Exercise 2.2.21 to show that the coeﬃ cient of tn/n! in the compositional inverse of t2 + sin t equals

(n 1)/2 2l n+l 1 j − (2n 1)! ( 1) − − F (2l + j, j; 1, 1) 1 − − − 2j (2l j)! (n 1 2l)! (2l + j)! 2 (n l) 1 + j l=0 j=0 X X − − − − − 44 Chapter 2. Finite Operator Calculus in One Variable

2.3 Sheﬀer Sequences

We have three diﬀerent objects deﬁning each other: A basic sequence (bn), a delta n 1 series β (t) = ln n 0 bn (1) t , and a delta operator β− ( ): bn (x) bn 1 (x). How can we enlarge≥ the set of sequences under consideration,D if they7→ are already− isomorphic to theP power series of order 1? We choose to investigate pairs of power series of the form σ (t) exβ(t), where σ is a power series of order 0.

Definition 2.3.1. (Sheffer [86, 1945]) A Sheffer sequence (sn) is a basis for F [x] with generating function

n xβ(t) sn (x) t = σ (t) e , (2.13) n 0 X≥ where β (t) is a delta series, and σ (t) has a reciprocal in F [[t]].

1 In the language we introduced to prove Theorem 2.2.4, β− ( ) is an x- operator, commuting with the t-operator “multiplication by σ (t)”, henceD

1 β− ( ) sn = sn 1. D − 1 On the other hand, if (sn) is a basis for F [x] such that β− ( ) sn = sn 1 holds 1 D − for all n 1, then β− ( ) is the transform of multiplication by t with respect to ≥ nD s (x, t) := n 0 sn (x) t . Hence s (x, t) = β (t) s (x, t); solving this differential ≥ xβ(tD) n equation gives s (x, t) = σ (t) e , where σ (t) = n 0 sn (0) t . P ≥ Thus (bn) and (sn) follow the same recursion, but differ in the initial values; n P sn (0) = [t ] σ (t), as the definition of (sn) shows. The additional power series σ (t) is brought in to take care of initial values! Of course, if we define σ (t) = 1, then 1 sn = bn for all n 0. We say that (sn) is a Sheffer sequence for B = β− ( ), ≥ D and it is associated to (bn). The Sheffer sequence is uniquely defined by the pair of power series (σ, β). For different solutions to the operator equation Bsn = sn 1 holds the superposition principle in the same way as for differential equations:− If (tn) is another solution, Btn = tn 1, then sn (x) + tn (x) also solves this equation. − Even sn (x) + tn k (x) solves the same equation, for fixed k 0, as long as we − ≥ interpret tn (x) = 0 for n < 0. However, for k = 0 we get a Sheffer sequence only if deg (sn + tn) = n. From the generating function (2.13) follows directly the binomial theorem (for Sheffer sequences),

sn (x + y) = si (y) bn i (x) (2.14) − i=0 X In the future, when we refer to the binomial theorem we will always mean this version for Sheﬀer sequences, which includes the “ordinary”binomial theorem. 2.3. Sheﬀer Sequences 45

In the alternative interpretation of the binomial theorem using the evaluation functional (see Remark 2.2.6 (b), with y = 0) we can write

i i sn (x) = Eval0 si B bn (x) = [σ] B bn (x) = σ (B) bn (x) . h | i i i 0 i 0 X≥ X≥

The operator S := σ (B) is called Sheffer operator for (sn), Sbn = sn. The Sheffer operator S is invertible, because σ (t) has a reciprocal 1/σ (t), and S commutes with all operators in Σ . D We need an initial value yn = sn (xn) for every sn (x) to get a unique solution to the operator equation Bsn = sn 1. The initial points x0, x1,... can be any − elements in F; as soon as the Sheffer sequences (sn) and (tn) for B agree at only one value xn for all n 0, ≥ sn (xn) = tn (xn) (2.15) the two Sheffer sequences must be equal (show by induction; the kernel of B only contains constants!). Example 2.3.2. We want to show the identity

n a + kz x kz n a + x k − = − zk (2.16) k n k n k k=0 k=0 X − X − for all a, x, z C (identity 3.144 in Gould’s list [38]). We apply ∆ to sn (x) = n a+kz x∈ kz k=0 k n− k (of degree n in x) obtaining sn 1 (x), hence (sn) is a Sheffer − − sequence for ∆. In the same way the right hand side tn (x), say, is also a Sheffer P sequence for ∆. We want to show that sn (x) = tn (x) for all n 0. Take the initial ≥n+1 points xn = n a. Use identity (2.29) to show that sn (xn) = z 1 / (z 1). − − − Of course, tn (xn) gives the same value, hence sn (xn) = tn (xn), and therefore sn (x) = tn (x) for all x, by property (2.15). Example 2.3.3. (Cartier [19]) This example from signal processing shows how the terms from Finite Operator Calculus can be interpreted in such an applied area. The first trivial change we have to make is defining our polynomials in t, the time variable, instead of x. A linear and stationary transmission device can be modeled as a linear operator V on R [t], mapping a suitable input f (t) into an output F (t) such that V f (t + τ) = F (t + τ) (stationarity = translation invariance!). The impulse response, I (t), is defined as the image of the pulse δ (t) (Dirac function), and from f (t) = ∞ f (τ) δ (t τ) dτ follows that −∞ − R ∞ F (t) = V f (t) = f (t τ) I (τ) dτ (2.17) − Z−∞

We assume that ∞ τ nI (τ) dτ exists for all n 0, and ∞ I (τ) dτ = 0. Trans- lating this setting−∞ into the language of the Finite≥ Operator−∞ Calculus6 we see that R R 46 Chapter 2. Finite Operator Calculus in One Variable the operator V is translation invariant, V Σ . Equation (2.17) shows that V acts as a convolution integral; the impulse response∈ D I (t) is deﬁned such that

∞ V = T ti i/i! = ( 1)i τ iI (τ) dτ i/i! x=0 D − D i 0 i 0 Z−∞ X≥ X≥ exists (as a formal power series; see Lemma 2.2.2), and we also assume that V is invertible (V 1 = 0). Hence V is a Sheﬀer operator, and the polynomials vn (t) := n 6 d V t /n! must be Sheﬀer polynomials for t = dt , such that tvn (t) = vn 1 (t) for all n 0. They have the generating functionD D − ≥ n tp vn (t) p = Θ (p) e n 0 X≥ (in signal processing we write p instead of t), where Θ(p) is the spectral gain, pn Θ(p) = v (0) pn = [T tn] n x=0 n! n 0 n 0 ≥ ≥ X X n n ∞ n p ∞ τp tp tp = ( 1) τ I (τ) dτ = e− I (τ) dτ = e− V e . − n! n 0 X≥ Z−∞ Z−∞ Note that the above identity says that Θ(p) etp = V etp, hence Θ(p) and V are transforms of each other (see section 2.2.2), and therefore V = Θ ( t). Now assume that the response F (t) = V f (t) has the Taylor expansionD

n n F (t) = (t t0) ( F )(t0) /n! − Dt n 0 X≥ around t0. We want to show that this implies the solution of the inverse problem, recovering the input as

n f (t) = un (t t0)( F )(t0) − Dt n 0 X≥ n tp where n 0 un (t) p = e /Θ(p). First we note that the response is no longer ≥ F (t) in C [t], but in C [t] [[p]], and to say that the Taylor expansion exist at t0 P n n is the same as saying that F (t, p) := n 0 (t p) ( t F )(p0) /n! can be evalu- n ≥ − D ated at p = t0 (think of ( t F )(p0) as a given sequence of constants). We deﬁne D 1 Pn the Sheﬀer polynomials un (t) = t /n! for t, having generating function Θ( t) n 1 tp 1 tp D D un (t) p = e = V e . Hence n 0 Θ( t) − ≥ D

P 1 t0 n n f (t) = V − F (t) = UF (t, t0) = UE− t /n! ( F )(t0) (2.18) Dt n 0 X≥ n = un (t t0)( F )(t0) . − Dt n 0 X≥ 2.3. Sheﬀer Sequences 47

n Especially f (t0) = n 0 un (0) ( t F )(t0). If F is analytic, then ≥ D

P n f (t) = un (0) ( F )(t) . Dt n 0 X≥ Suppose the response is the average input over one time unit,

t F (t) = f (τ) dτ. t 1 Z − Before we can solve the inverse problem, we have to find the spectral gain, which t means finding the impulse response I (t) first. We have F (t) = t 1 f (τ) dτ = − ∞ f (τ) I (t τ) dτ, thus I (t) = 1 for t [0, 1], and 0 else. Adding an input R that−∞ averages− to zero over every time unit∈ will not change the response. Hence Rthe input cannot uniquely be recovered from the response, but up to those zero- p 1 τp 1 e− average inputs. We find the spectral gain Θ(p) = 0 e− dτ = −p , thus (un) 1 tp p pt p p(t+1) has the generating function e = p e = p e , which also gener- Θ( t) 1 e e 1 D − − R − ates the Bernoulli polynomials ϕn (t + 1). We saw in Example 1.1.6 that un (t) = n ϕn (t + 1) = ( 1) ϕn ( t). Formula (2.18) shows that − −

n n n f (t) = ϕn (t t0 + 1) ( F )(t0) = ( 1) ϕn (t0 t)( F )(t0) . − Dt − − Dt n 0 n 0 X≥ X≥

If t0 = t we obtain the Euler —MacLaurin summation formula,

n n 1 B2n 2n f (t) = ( 1) ϕn (0) ( F )(t) = F (t) + F 0 (t) + F (t) − Dt 2 (2n)!Dt n 0 n 1 X≥ X≥ t f (t) f (t 1) B2n 2n 1 = f (τ) dτ + − − + − (f (t) f (t 1)) . 2 (2n)!Dt − − t 1 n 1 Z − X≥ There are other applications of Finite Operator Calculus on the continuous scale. For example, Ismail [45], and Ismail and May [46] investigated applications in approximation theory.

2.3.1 Initial Values Along a Line

Sheffer sequences (sn) must carry the initial condition information that gives us a 1 specific solution to the recursion formula Bsn (x) = sn 1 (x), where B = β− ( ). If the initial condition means fixed (i.e. a priori known)− initial values at initialD points xn, n 0, then the easiest of such problems occurs when xn = c, a constant, ≥ for all n 0. Suppose sn (c) = yn is known. By the binomial theorem for Sheffer ≥ 48 Chapter 2. Finite Operator Calculus in One Variable

sequences we can expand (sn) in terms of the associated basic sequence (bn) as

n n

sn (x) = si (c) bn i (x c) = yibn i (x c) , or (2.19) − − − − i=0 i=0 X X

n n (x c)β(t) sn (x) t = ynt e − .   n 0 n 0 X≥ X≥   Fortunately there is a larger class of initial value problems that can be solved: Initial points xn along any line, xn = an + c for some a, c F. To expand such a ∈ Sheﬀer sequence in terms of (bn) we need the following fact about basic sequences.

Lemma 2.3.4. [83] Let (bn) be a basic sequence, and define rn (x) = (n + 1) bn+1 (x) /x for n 1, and r0 (x) = 1. Then (rn) is a Sheffer ≥ sequence for the same delta operator as (bn). n xβ(t) Proof. There exists a delta series β (t) such that n 0 bn (x) t = e . Take ≥ the t-derivative on both sides, (n + 1) b (x) tn = xβ (t) exβ(t). Note that n 0 n+1 P 0 xβ(≥t) β0 (t) has order 0, hence β0 (t) e is the generating function of the Sheffer se- n+1 P 1 quence bn+1 (x) . The delta operator remains β− ( ) in both cases, (bn) x n 0 D and n+1 b (x) . ≥ x n+1 We can only say that rn (x) = (n + 1) bn+1 (x) /x has the initial values n rn (0) = [t ] β0 (t); however, the superposition tn (x) := bn (x c) arn 1 (x c) = − − − − (x an c) bn (x c) / (x c) is also Sheffer sequence for B and satisfies the ini- − − − − tial condition tn (an + c) = δ0,n.

Corollary 2.3.5. If (bn (x)) is the basic sequence for B, and a and c are constants in F, then bn (x c) tn (x) = (x an c) − − − x c − is the Sheﬀer polynomial for B with initial values tn (an + c) = δ0,n. n n 1 Suppose c = 0 and bn (x) = x /n!, thus tn (x) = (x an) x − /n!. The binomial theorem for Sheﬀer sequences tells us that −

n n 1 n i 1 n i (x + y an)(x + y) − = (y ai) y − x − . − i − i=0 X

This formula is called Abel’sidentity. In general, the process of moving from (bn) to ((x an c) bn (x c) / (x c))n 0 is called Abelzation. By Abelization, the binomial− theorem− (2.14)− becomes− ≥

n bn (x + y) bi (y) (x + y an) = (y ai) bn i (x) . − x + y − y − i=0 X 2.3. Sheﬀer Sequences 49

Example 2.3.6 (Catalan Numbers). The numbers Cn of , lattice paths from (0, 0) to (n, n) staying weakly above the diagonal y = x are{→ the↑} Catalan numbers. The Catalan numbers also count the , lattice paths from (0, 0) to (2n, 0) staying weakly above the x-axis. There{% are&} more than hundred diﬀerent combinatorial problems where Catalan numbers occur; R. Stanley has compiled a list in his book [90], and maintains a website with additional examples [88]. Returning to the ﬁrst interpretation of Cn above, we see that Cn equals the values dn (n) when dn (m) dn (m 1) = dn 1 (m) (Pascal’srecursion), and dn (m) has initial values − − − dn (n 1) = δ0,n. − m 1 8 35 110 275 572 1001 1430 1430 7 1 7 27 75 165 297 429 429 0 6 1 6 20 48 90 132 132 0 -429 5 1 5 14 28 42 42 0 -132 -429 4 1 4 9 14 14 0 -42 -132 -297 3 1 3 5 5 0 -14 -42 -90 -165 2 1 2 2 0 -5 -14 -28 -48 -75 1 1 1 0 -2 -5 -9 -14 -20 -27 0 1 0 -1 -2 -3 -4 -5 -6 -7 0 1 2 3 4 5 6 7 n The Catalan numbers (bold) and the polynomials dn (m)

n 1+x We saw that the basic polynomials −n solve Pascal’s recursion, hence the x+1 n n+x Abelization dn (x) = x+1− n solves the same recursion with the right initial 2n values dn (n 1) = δ0,n. Therefore, Cn = dn (n) = / (n + 1). The numbers − n dn (m) for m n are often called the ballot numbers, giving the number of ways a two candidate≥ ballot can be realized such that candidate B is never ahead of candidate A during the whole ballot procedure. The table suggest that dn (m) = dm+1 (n 1) for all m 1; prove this conjecture. − For classical− lattice≥ path − counting we refer to Mohanty [60] and Narayana [63]. Modern approaches can be found in Krattenthaler [52], [50], and Kratten- thaler, Guttman and Viennot [51], to name just a few.

It is easy to check that every Sheffer sequence (sn) for B transforms into a a Sheffer sequence (sn (an + x)) for E− B, which is again a delta operator. There- fore, taking for sn the special Sheffer polynomials (x an) bn (x) /x for B, we − a see that (xbn (x + an) / (x + an))n 0 must be a Sheffer sequence for E− B, and ≥ because of the initial values, (xbn (an + x) / (an + x)) is the basic sequence for a E− B. We denote these basic polynomials by b (an + x) b(a) (x) := x n . (2.20) n an + x

If we are searching for Sheﬀer polynomials for B with initial values tn (an + c) = a yn, y0 = 0, we are searching for Sheﬀer polynomials for E− B with initial values 6 50 Chapter 2. Finite Operator Calculus in One Variable

sn (c) = yn. Hence

n n (a) bi(ai + x c) tn (an + x) = sn (x) = yn ibi (x c) = (x c) yn i − − − − − ai + x c i=0 i=0 − X X (2.21) according to (2.19).

Example 2.3.7. We noticed in the Catalan example above that

dn (n 2) = dn 1 (n 1) = Cn 1, − − − − − − and d0 ( 2) = 1. This is an example of recursive initial values; we may not know − explicitly the initial value of dn (n 2), but we will be able to calculate it from − the previous initial value(s), because we can calculate dn 1 (x). We observe a = 1 2n+x+1 x+2 n−1 2i+x+1 x+2 and c = 2, so that dn (n + x) = i=0− Cn i 1 . − n n+x+2 − − − i i+x+2 For the Catalan numbers we get the recursion C = d (n) = 2n+1 2 nP n n n+2 n 1 2i+1 2 2n+2 1 n 1 2i+2 1 − i=0− Cn i 1 = i=0− Cn i 1 . Knowing that − − i i+2 n+1 (n+2) − − − i+1 i+2 C = 2n 1 gives the even more beautiful recursion C = n C C , Pn n n+1 P n+1 i=0 n i i which has a nice combinatorial interpretation (ﬁrst return decomposition; see− also Exercise 1.2.6). P

We now would like to get an idea about the relationship between β (t) = n (a) n (a) ln n 0 bn (1) t and βa (t) = ln n 0 bn (1) t . Because bn is the basic se- ≥ ≥ a quenceP for E− B we get immediatelyP that

at 1 1 e− β− (t) = βa− (t) . (2.22)

aβ(t) 1 1 aβa(t) β(t) Hence e− t = βa− (β (t)), or β− (βa (t)) = te− . If φ (t) := e is given, βa(t) and we want to know φa (t) := e , we can also try to solve the equation

a φa (t) = φ (tφa (t) ) . (2.23)

If we want to ﬁnd the generating function of the Abelization

sn (x) = (x an) bn (x) /x − of (bn), it is often easier to calculate

bn (x + an) s (x + an) tn = x tn = b(a) (x) tn = exβa(t). n x + an n n 0 n 0 n 0 X≥ X≥ X≥ Example 2.3.8 (Catalan numbers continued). In the Catalan example we have 1 t 1 t t β− (t) = 1 e− , hence β− = e− (1 e− ), which can be inverted (quadratic − 1 − 2.3. Sheﬀer Sequences 51 equation; pick the root that is 0 at t = 0) to 1 1 β1 (t) = ln + √1 4t . Hence − 2 2 − b (x + 1 + n) d (x + n) tn = (x + 1) n tn = b(1) (x + 1) tn n x + 1 + n n n 0 n 0 n 0 X≥ X≥ X≥ 2 x+1 = e(x+1)β1(t) = 1 + √1 4t − and n n 2 1 √1 4t Cnt = dn (n) t = = − − 1 + √1 4t 2t n 0 n 0 X≥ X≥ − Example 2.3.9. Let U1,...,UM be identically and independently distributed random variables with common density f (y) = 1 for 0 y 1, and 0 else (uniform distribution). Denote the order statistic by U ,U≤ ,...,U≤ , where U U (1) (2) (M) (1) ≤ (2) ≤ U(M). Select constants a and c such that 0 < a+c < c+Ma < 1 Elementary probability· · · ≤ theory tells us that for c + na x 1 ≤ ≤

Pr U(1) c + a, . . . , U(n 1) c + (n 1) a, x U(n) c + na /n! ≥ − ≥ − ≥ ≥ x un un 1 u3 u2 − = 1du1du2 dun 2dun 1dun c+na c+(n 1)a c+(n 2)a ··· c+2a c+a ··· − − Z Z − Z − Z Z

(see Wald and Wolfowitz [98]). Call the above probability pn (x), extendable to a polynomial in x of degree n. Of special interest is the value pM (1). The sequence p0, p1, . . . , pM satisfies the system of differential equations pn (x) = pn 1 (x), D − and the initial values pn (c + na) = 0 for all n 1. Furthermore, p0 (x) = x ≥ 1du = 1. Hence p0, p1, . . . , pM is the beginning piece of a Sheffer se- D a+b (1) quence for , and pn (a + cn) = δ0,n. From Corollary 2.3.5 we obtain pn (x) = R D n 1 (x c an)(x c) − /n!, a close relative to the original Abel polynomial. − − − 2.3.2 The Umbral Group

On the operator side, a Sheﬀer sequence (sn) can be represented by the pair (S, B), 1 where S = σ (B) is the Sheﬀer operator, and B = β− ( ) is the delta operator D for (sn). On the formal power series side, we consider the pair (σ, β). Suppose we have another pair of formal power series, (ρ, α), where ρ has a reciprocal and α is a delta series. The pair (σ (α) ρ, β (α)) has the same properties, σ (α (t)) ρ (t) has a reciprocal and β (α (t)) is a delta series. We say that the set of such pairs form the umbral group with respect to the operation

(σ, β) (ρ, α) := (σ (α) ρ, β (α)) . ◦ The identity in this group is the pair (1, t) corresponding to the basic sequence (xn/n!) for . How does the Sheﬀer sequence belonging to (σ (α) ρ, β (α)) look like? D 52 Chapter 2. Finite Operator Calculus in One Variable

n We ﬁrst ﬁnd the basic sequence (pn), say, with generating function n 0 pn (x) t = ≥ exβ(α(t)). Note that the delta operator P for (p ) must be written as n P 1 1 1 P = α− β− ( ) = α− (B) . D n xβ(α(t)) Lemma 2.3.10. If n 0 pn (x) t = e , where α and β are delta series with basic polynomials a (≥x) = n a xi/i! and b (x), respectively, then Pn i=0 n,i n P n pn (x) = an,ibi (x) i=0 X for all n 0. ≥ i Proof. We know from Lemma 2.2.10 that an,i = α n. Hence

n xβ(t) i n pn (x) t = C (α) e = bi (x) α (t) = bi (x) an,it n 0 i 0 i 0 n i X≥ X≥ X≥ X≥ n n = t an,ibi (x) . n 0 i=0 X≥ X

We have already seen that multiplication of power series corresponds to com- 1 1 position of operators; now we have composition of power series α− β− (t) corresponding to the umbral operator UB, mapping (an) to the basic sequence n (an (b (x))), where an (b (x)) := i=0 an,ibi (x). We have the following equivalent statements, occurring in the literature. P 1. Delta series β (t): The delta series α (t) is mapped onto the delta series β (α (t)). 1 1 2. Delta series β− (t): The delta series α− (t) is mapped onto the delta series 1 1 α− β− (t) . 1 3. Delta operator B: The delta operator A = α− ( ) is mapped onto the delta 1 D operator α− (B).

4. Umbral operator UB: The basic sequence (an (x)) is mapped onto the basic sequence (an (b (x))) 5. Umbral subgroup: (1, β) (1, α) = (1, β (α)) . (2.24) ◦

If we apply UB to the Sheffer polynomials rn (x) := ρ ( ) an (x), we obtain D xβ(α(t)) rn (b (x)) for this composition, having generating function ρ (t) e . Instead n of inserting (bn) for (x /n!), we could also insert the Sheffer sequence (sn) with xβ(t) generating function σ (t) e , giving the composition (rn (s (x))), with generating 2.3. Sheffer Sequences 53 function ρ (t) σ (α (t)) exβ(α(t)) (see Exercise 2.3.12). This is the meaning of the operation (σ, β) (ρ, α) = (σ (α) ρ, β (α)) in terms of Sheffer sequences, that is ◦ n i xk n xk n r (s (x)) = r s = r s . n n,i i,k k! k! n,i i,k i=0 k=0 k=0 i=k X X X X n If rn (s (x)) = x /n!, then the two Sheffer sequences (or the corresponding group elements) are inverse to each other. If

n tn (x) = rn (s (x)) = rn,isi (x) i=0 X then the coeffi cients rn,i are called connection coeffi cients. The problem of connection coeffi cients refers to finding these coeffi cients when (tn) and (sn) are given. If (tn) is represented by (τ, γ) in the umbral group, then (τ, γ) = (σ (α) ρ, β (α)). i Hence the connection coeffi cients rn,i = x /i! rn (x) can be obtained from

1 1 1 1 (ρ, α) = τ, β− (γ) = 1/σ β− , β− (τ, γ) (2.25) σ (β 1 (γ)) ◦ − Example 2.3.11. [83, p. 747] Denote by S (n, k) the Stirling numbers of the second kind; S (n, k) is the number of set partitions of an n-set into k nonempty blocks (re- n k gardless of order). Deﬁne the polynomial sequence Sn (x) = k=0 S (n, k) x /n!. t It is well-known that (S ) has the generating function S (x) tn = ex(e 1). n n 0 Pn − 1 ≥ Hence (Sn) is the basic sequence with delta operator β− ( ) = ln (1 + ). As a t P D D member of the umbral group we write (Sn) as (1, e 1). Now we change our view 1 − 1 t point and consider the delta operator α− ( ) for which α− (t) = e 1. This delta operator is the forward diﬀerence operatorD −

1 1 ∆ = α− ( ) = E 1 : f (x) f (x + 1) f (x) . D − 7→ − x The binomial coeﬃ cients bn (x) = n are the basic polynomials; they are (1, ln (1 + t)) in the umbral group. From α (β (t)) = t follows that the two group elements are inverse, hence by umbral composition follows xn S (x) k! = = S (b (x)) = S (n, k) x . n! n n k n! k 0 X≥ x The exponential polynomials Sn (x) are inverse to n (and vice versa). From S (x) xk xk k tn = ext = e x (1 + t)k = e x tn n − k! − k! n n 0 k 0 n 0 k 0 X≥ X≥ X≥ X≥ follows k k S (x) x k x x x x = e− = e− Evalk n n k! | n k! k 0 k 0 X≥ X≥ 54 Chapter 2. Finite Operator Calculus in One Variable

x k and by linearity p (S(x)) = e− k 0 p (k) x /k! for all polynomials p. The special case p (x) = xn/n! is known as Dobinski’s≥ formula, P xk knxk S (x) = S (n, k) = e x . (2.26) n n! − n!k! k 0 k 0 X≥ X≥ We can also ask which umbral element (σ, β) will bring the pair (ρ, α) to the pair (1, ln (1 + t)), which corresponds to the binomial coeﬃ cients. In other words, we are asking for (σ, β) such that

(σ, β) (ρ, α) = (σ (α) ρ, β (α)) = (1, ln (1 + t)) , ◦ 1 1 hence β (t) = ln 1 + α− (t) , and σ (t) = 1/ρ α− (t) . If we define (rn) and (sn) as Sheffer sequences with generating function xα(t) 1 x 1 ρ (t) e and 1 + α− (t) /ρ α− (t) , x respectively, then rn (s (x)) = n . Remark 2.3.12. The umbral group occurs in different forms in the literature; in our description we focus on power series. The similarity to the Riordan group has been pointed out by He, Hsu, and Shiue [40]. As defined above, the umbral group and umbral composition are terms used in the Finite Operator Calculus, introduced already by Rota, Kahaner and Odlyzko [83]. In section 3.4 on classical Umbral Calculus we also use the word umbra, but in a different meaning.

2.3.3 Special Cases Both cases we study in this subsection are connected to the name of J. Riordan, who wrote in 1968 the (at that time) popular book Combinatorial Identities [72]. Riordan Matrices are closely related to Sheffer sequences, but the coeffi cients - not the values - of the polynomials are seen as counting some combinatorial objects. For the cases when coeffi cients become values see Corollary 3.2.2. The second topic is about Inverse Pairs of number sequences. We present several example from Riordan’sbook and interpret them through Sheffer Sequences.

Riordan Matrices Because of our focus on recursions we will define a Riordan matrix in a somewhat unusual way. An infinite triangular array S = (s ) , n,k n,k N ∈ sn,k = 0 if n < k, is a Riordan matrix if and only if it satisfies the recursions

n i − sn+1,i+1 = sn,i+kak for all 0 i n, and ≤ ≤ k=0 Xn

sn+1,0 = sn,klk. k=0 X 2.3. Sheﬀer Sequences 55

where (an)n 0 and (ln)n 0 are given numerical sequences with a0 = 0 ([75], [58]). The existence≥ of an a-sequence≥ is useful in determining whether a6 given matrix is a Riordan matrix or not. Often, trying to determine the a-sequence from a matrix that is not Riordan will come to an end after a few values.

0 1 2 3 4 k 0 1 2 3 4 k 0 1 0 1 1 1 1 1 2 1 2 3 1 1 2 4 2 1 3 6 4 1 1 3 9 5 2 1 4 16 9 5 1 1 4 22 13 6 2 1 5 41 26 12 6 1 1 5 57 35 17 7 2 1 6 113 71 37 15 7 1 6 154 97 49 21 8 2 7 316 204 106 49 18 8 7 429 275 143 64 25 9 n 907 590 316 146 62 21 n 1223 794 422 195 80 29

Both the above matrices are Riordan matrices; the matrix on the left has l-sequence l0 = 1, l1 = 2, lk = 1 for k 2. For finding the a-sequence on the left, we begin ≥ with s0,0 = 1, and s1,0 = 1. Hence a0 = 1, which not only makes s1,1 = 1, but produces all 1’s along the main diagonal. Next, the l-sequence gives s1,0 = 1; assuming that a1 = 0 gives all the 1’s on the first subdiagonal. In the next step we get s2,0 = 3, and a2 = 1 generates the numbers 4, 5, 6, 7,... . This way the table can be filled, if we assume that a3 = 2, a4 = 3, a5 = 3, a6 = 0, a7 = 8, − etc. Why this curious a-sequence? Because we only have to know sn,0 and sn,1 if the matrix has the Riordan property (see Theorem 2.3.13 below); the sequences (and thus sn,k for k 2) follow. The table on the left has been constructed from ≥ n s0,0 = 1 and sn,1 + sn 1,0 = sn,0 + ( 1) for n 1. − − ≥ The table on the right is constructed from s0,0 = 1 and sn,1 + sn 1,0 = sn,0. This leads to the same a-sequence as on the left, and on the first glance− it looks like the same l-sequence too. However, this simple l-sequence does not work for finding s1,0. The true sequence on the right is l0 = 2, l1 = 0, l2 = 1, l3 = 1, l4 = 5, l5 = 6,... . We will return to this problem later (Example 2.3.15). We state as a Theorem what we just saw, the definition of a Riordan matrix through its first two columns.

Theorem 2.3.13. The matrix S = (sn,k) is a Riordan matrix such that sn+1,i+1 = n i n k=0− sn,i+kak for all 0 i n, and sn+1,0 = k=0 sn,klk where (an)n 0 and ≤ ≤ ≥ n k P(ln)n 0 are given numerical sequences, a0 = 0, if andP only if sn,k = [t ] σ (t) β (t) ≥ 6 for all n 0, 0 k n, where β (t) F [[t]] is the delta series ≥ ≤ ≤ ∈

1 n β− (t) = t/ ant n 0 X≥ 56 Chapter 2. Finite Operator Calculus in One Variable and

n σ (t) = s0,0/ 1 t lnβ (t)  −  n 0 X≥   is of order 0.

n i Proof. Suppose sn+1,i+1 = k=0− sn,i+kak for all 0 i n. From a0 = 0 follows n k ≤ ≤ 6 sn,n = 0, hence sn (x) := sn,kx /k! is a polynomial of degree n. We ﬁnd 6 Pk=0

n+1P n+1 n k+1 xk − xk sn+1 (x) = sn+1,k = sn,k 1+jaj D D k! D − k! k=0 k=0 j=0 X X X n+1 n k j+1 x − j = aj sn,k = aj sn (x) D (k j + 1)!  D  j=0 k=j 1 j 0 X X− − X≥   j j The power series t/ j 0 ajt is a delta series, thus / j 0 aj a delta oper- ≥ D ≥ D xβ(t) ator, and (sn) a Sheﬀer sequence for this delta operator. If we write σ (t) e P 1 P j for the generating function of (sn), then β− ( ) = / aj , and of course D D j 0 D s = σβk . Hence ≥ n,k n P n 1 ∞ n ∞ − n σ (t) = sn,0t = s0,0 + sn 1,klkt − n=0 n=1 k=0 X X X ∞ ∞ n ∞ k = s0,0 + t lk sn,kt = s0,0 + t lkσ (t) β (t) k=0 n=k k=0 X X X and therefore

k σ (t) 1 t lkβ (t) = s0,0.  −  k 0 X≥   We leave showing the other direction as Exercise 2.3.17.

The above Theorem identifies a Riordan matrix S = (sn,k) as the coeffi cient n k matrix sn,k = [t ] σ (t) β (t) . This is the usual definition of Riordan matrices. Therefore, solving recurrence relations with the Riordan matrix approach utilizes a different aspect of Sheffer sequences, where we up to now only looked at the n kβ(t) values of the polynomials sn (k) = [t ] σ (t) e . However, we will investigate a special case in section 3.2, where both concepts come together. Example 2.3.14. We first consider the left of the two Riordan matrices above. The 2.3. Sheffer Sequences 57

l-sequence is l1 = 2, and li = 1 for the other li’s. We ﬁnd

k 1 1 = σ (t) 1 t lkβ (t) = σ (t) 1 tβ (t) t  −  − − 1 β (t) k 0 X≥ −   2 1 β (t) t 1 + β (t) β (t) = σ (t) − − − . 1 β (t) − n The recurrence s0,0 = 1 and sn,1 + sn 1,0 = sn,0 ( 1) for n 1, gives the generating function equation − − − ≥

t σ (t) 1 + = σ (t) β (t) + tσ (t) . − 1 + t Now we have two equations for σ and β; thus

1 β (t) = 1 t2 (1 + t2)2 4t and 2 − − − q 2 σ (t) = . (1 + t) (t 1)2 + (1 + t2)2 4t − − q Example 2.3.15. There is an interpretation of the matrix on the right hand side as a ballot problem: Consider the number D (n, m) of , -paths that reach (n, m), staying weakly above y = x, and avoiding the pattern{→ ↑}uruu. Note that the same number of such paths reach (n, n) and (n 1, n) for positive n. Let −

sn,k = D (n k, n + 1) . −

m 1 2 8 21 49 97 154 6 1 2 7 17 35 57 57 5 1 2 6 13 22 22 4 1 2 5 9 9 3 1 2 4 4 2 1 2 2 1 1 1 0 1 0 1 2 3 4 5 n Table of D (n, m) From the lattice path problem follows the recursion

sn,k = sn 1,k 1 + sn,k+1 sn 2,k 1 + sn 2,k − − − − − − 58 Chapter 2. Finite Operator Calculus in One Variable for all n > 0 and 0 k n. This implies the existence of an a-sequence (Exercise 2.3.18) such that ≤ ≤ k k j 1 2j 1 2j 1 ( 1) − − a = − − − , k j k j 2j 1 j=0 − − X the same sequence as in Example 2.3.14. Hence Theorem 2.3.13 implies that

1 t 1 2 β− (t) = n = 1 1 4t (t 1) , n 0 ant 2 (1 t) − − − ≥ − q and by inversion P 2 2 1 2 2 2 1 t 4t 4t β (t) = 1 t (1 + t ) 4t = − 1 1 − 2 . 2 − − − 2 − s − (1 t2) ! q − Now sn,0 = sn,1 + sn 1,0 − for n 1 and s0,0 = 1, thus ≥ σ (t) 1 tσ (t) = σ (t) β (t) − − and therefore 2 σ (t) = (1 t)2 + (1 + t2)2 4t − − q 2 2t2 2t + 1 t2 1 1 4t 4t (1 −t2)2 1 β (t) = − − − − − = 1 . 2 (t2 t3)q t t t2 − − − The numbers m 1 m 1 n D (n, m) = sm 1,m 1 n = t − σ (t) β (t) − − − − − m n m β (t) − m 1 n = [t ] β (t) − − t t2 − − ! can be expanded for m > n 0 as ≥ n+1 m 2j + n + 1 D (n, m) = − ( 1)j (2.27) m j − × j=0 X − j/2 m j i 2j + m + n + 1 m n m 1 n − − − − − × j 2i i m j + 1 − i 2j + m + n + 1 i=0 X − − − n j/2 m n 1 n j 2i j 1 1 − − ( 1) − − − n j − i j + 1 2i× j=0 i=0 X − X − 2j + m n 3i 2j + m n 3i (m n) − − + − − . × − j 2i 1 j 2i − − − 2.3. Sheﬀer Sequences 59

(see Exercise 2.3.19). Theorem 2.3.13 shows that Riordan matrices offer another look at the umbral group, from the coeffi cient level, because we can take any basic sequence (an) xα(t) with generating function e , and construct the Sheffer polynomial sn(a (x)) = n n k k=0 sn,kak (x), where sn,k = [t ] σ (t) β (t) . This Sheffer polynomial has co- n effi cients sn,k = [ak (x)] sn (a(x)) and generating function n 0 sn (a (x)) t = P ≥ σ (t) exα(β(t)), hence (s (a (x))) belongs to the group element (1, α) (σ, β) (Exer- n P cise 2.3.20). In matrix notation, ~s = S ~a, where ◦

T T ~s = (s0 (a (x)) , s1 (a (x)) ,... ) and ~a = (a0 (x) , a1 (x) ,... ) .

Viewing the matrix elements as coeﬃ cients of Sheﬀer polynomials can lead to elegant proofs. For example, if we want the generating function of the row sums, k + x 1 n s , we can apply the basic polynomials a (x) = − with gener- k=0 n,k k k x n Pating function (1 t)− , so that sn,k = sn(a (1)). Hence − k=0 n P n 1 t sn,k = σ (t) (1 β(t))− . − n 0 k=0 X≥ X k Choosing ak (x) = x /k! gives

n n β(t) t sn,k/k! = σ (t) e . n 0 k=0 X≥ X

More general, take two Riordan matrices S = (sn,k) and R = (rn,k), where n k n k sn,k = [t ] σ (t) β (t) and rn,k = [t ] ρ (t) α (t) . The product

SR = sn,jrj,k   j=k X   has matrix elements with generating function

n n ∞ j j k t sn,jrj,k = σ (t) β (t) t ρ (t) α (t) n 0 j=k j=k X≥ X X = σ (t) ρ (β (t)) α (β (t))k .

We have shown the group (anti-) isomorphism S (σ, β) where SR (ρ, α) (σ, β) = (ρ (β) σ, β (α)), when R (ρ, α) (see also [40]).' ' ◦ ' We denote by bn,k the coeﬃ cients of the basic polynomials n k 1 bn(x) = k=0 bn,kx /k! for β− ( ). The coeﬃ cients follow recursions reminiscent of the binomial theorem. D P 60 Chapter 2. Finite Operator Calculus in One Variable

Theorem 2.3.16. Let S = (sn,k) be the Riordan matrix deﬁned by n k 1 sn,k = [t ] σ (t) β (t) , and let (bn) be the basic sequence for β− ( ). Then D n j − sn,i+j = sk,ibn k,j − k=i X for all 0 i + j n. ≤ ≤ The proof follows from σ (t) β (t)i+j = σ (t) β (t)i β (t)j.

Corollary 2.3.17. The lower triangular matrix S is a Riordan matrix iﬀ sn,n = 0 6 and there exists a sequence c1, c2,...,with ci in F, such that

n 1 − sn,i+1 = sk,icn k − k=0 X for all 0 i < n. ≤ k We leave the proof as Exercise 2.3.22. Note that β (t) = k 1 ckt if sn,k = ≥ n k [t ] σ (t) β (t) . P Inverse Pairs In its simplest form, an inverse pair is nothing but a pair of power series (σ (t) , γ (t)) related to each other through a third known power series α (t), say, whose reciprocal is also “known”,

1 γ (t) = α (t) σ (t) and σ (t) = γ (t) α (t)

If an inverse pair of relations is given, only one of the two must be shown; the other will hold automatically. The linear recursion in one variable is an inverse j pair (Proposition 1.1.4), with α (t) = 1 ∞ αjt . − j=1 t t t Example 2.3.18. If α (t) = e then 1/α (tP) = e− , and if we write γ (t) = e σ (t) t then σ (t) = e− γ (t). This elementary relationship looks more interesting on the n n coeﬃ cient level. Let γ (t) = n 0 γnt /n! and σ (t) = n 0 σnt /n!, then ≥ ≥ n P n P n n n k γn = σk σn = ( 1) − γk k ⇐⇒ k − k=0 k=0 X X n (Riordan [72, Chpt. 2]). For the power series σ˜ (t) = n 0 σnt and γ˜ (t) = n ≥ n 0 γnt holds ≥ P 1 t P σ˜ =γ ˜ (t) (2.28) 1 t 1 t − − 2.3. Sheﬀer Sequences 61

(see Exercise 1.2.3). The relationship between the numbers Cn and the Motzkin numbers Mn is an application of this inverse pair, namely n n Cn+1 = Mn k k − k=0 X (the Motzkin numbers Mn count the number of , , path to (n, 0) stay- {% & →} ing weakly above the x-axis, while the Catalan numbers Cn count the number of , path to (2n, 0) also staying weakly above the x-axis [92]). The argument {% &} for this identity goes as follows. Strip the Catalan paths counted by Cn+1 of their ﬁrst (up) and last (down) step. Make the remaining 2n steps into n double steps, always grouping two consecutive steps together. Then map the double steps onto single steps as in the following table.

double step: % & single step: &% % & %&= −→ % & ⇒ Now we have a , , , - path to (n, 0) staying weakly above the x-axis. It {% & → ⇒} is called a bicolored Motzkin path, thus Cn+1 equals the number of such paths to n (n, 0). This path can have any number k = 0, 1, . . . , n of steps at k places; taking them out leaves a Motzkin path to (n k, 0) behind.⇒ Hence − n n Cn+1 = Mn k. k − k=0 X The above inverse relationship tells us that

n n n k Mn = ( 1) − Ck+1. k − k=0 X It also tells us that 1 t n 1 t C tn = M =: M n+1 1 t n 1 t 1 t 1 t n 0 n 0 X≥ − X≥ − − − (see (2.28)). We saw in Example 2.3.8 that c (t) = C tn = 2 , and n 0 n 1+√1 4t ≥ − t t P c (t) 1 = M . − 1 t 1 t − − The compositional inverse of t/ (1 t) is t/ (1 + t), hence − t 1 t √1 2t 3t2 M (t) = c 1 /t = − − − − 1 + t − 2t2 is the generating function for the Motzkin numbers. 62 Chapter 2. Finite Operator Calculus in One Variable

The general inverse pairs are not that trivially constructed; they originate in matrix algebra rather than generating functions. Suppose we have an inﬁnite matrix A = (αi,j)i,j 0 that is triangular, αi,j = 0 for all i < j. Further more, ≥ assume that A is i; there exists an inﬁnite matrix B = (βi,j)i,j 0 such that BA = i ≥ I = (δi,j)i,j 0. Hence k=j βi,kαk,j = δi,j for all i, j 0. For such a matrix and its inverse holds≥ ≥ P n n γn = αn,kσk σn = βn,kγk ⇐⇒ k=0 k=0 X X for all n 0, as can be easily checked (Gould class of inverse relations). In matrix ≥ notation, let γ = (γ0, γ1,... )>, a column vector, and let σ = (σ0, σ1,... )>. Then

1 γ = Aσ σ = A− γ. ⇐⇒ The problem is finding such pairs of inverse matrices explicitly! We started with diagonal matrices, but we now want to show a larger class that can be found from basic sequences by Abelization. Let (bn (x))n 0 be a basic sequence for some delta operator B. Let u and v be two arbitrary but≥ nonzero constants, and define the Sheffer sequence (sn) for B by the initial values sk (0) = σk for all k 0. This is a ≥ Sheffer sequence only if σ0 = 0, so we omit leading zeroes. The binomial theorem 6 n (2.14) shows that sn (un + v) = k=0 σkbn k (un + v). We let γn = sn (un + v), hence − Pn γn = σkbn k (un + v) . − k=0 X To get the inverse direction, let tn (x) = sn (un + v + x), a Sheffer polynomial for u u E− B (see 2.3.1). The delta operator E− B has the basic sequence (xb (x + un) / (x + un)). By the binomial theorem

n un v tn ( un v) = tk (0) − − bn k((n k) u un v), − − (n k) u un v − − − − k=0 X − − − thus n un + v σn = γk bn k( ku v). uk + v − − − k=0 X ui+v In matrix form, A = (bi j (ui + v))i,j 0 and B = uj+v bi j( ju v) are − ≥ − − − i,j 0 inverses (see Exercise 2.3.25). ≥ x un + v The special case b (x) = gives γ = n σ n n n k=0 k n k − n n k v+un n + v P+ (u 1) k σn = γk ( 1) − − . ⇐⇒ k=0 − v+n+(u 1)k n k − − P 2.3. Sheﬀer Sequences 63

2.3.4 Exercises 2.3.1. Are the Pell polynomials (see Exercise 1.1.2) Sheﬀer polynomials?

2.3.2. Show that the Bernoulli polynomials (of the ﬁrst kind) ϕn (x) with generating function text/ (et 1) (Example 1.1.6) are a Sheﬀer sequence for . Show that the binomial theorem− implies D

n i n 1 n i n!ϕn (x) = − − Bi (x 1) − . i − i=0 X n n n i n n i With the same tools show that 0 = Bi x − ( 1) (1 x) − . This i=0 i − − − n n n n n i implies Bn = ( 1) i=0 i Bi, andP0 = i=0 i 2 Bi for odd n. Get B2k+1 = 0 − 1 for k 1 from the generating function, and B = 1/2 (show that t (et 1)− + P P 1 t/2 is≥ symmetric). − − 2.3.3. One way of defining the Bernoulli polynomials of the second kind is as x Sheffer sequence (ϕ ˆn (x)) for ∆ that satisfies the initial condition ϕˆn (x) = n 1 D − for all n 0. Show that these two conditions uniquely define (ϕ ˆn (x)), and that ≥ n bn :=ϕ ˆn (0) has the generating function n 0 bnt = t/ ln (1 + t). Show that n n j ≥ ϕˆn (x) = bn + s (n 1, j 1) x , where the Stirling numbers s (n, m) of j=1 j − − P the first kind are defined in Exercise 2.3.16. Use ϕˆ (x) = x and ϕˆ (x) = P n n 1 n n x 1 x Dx y − i=0 bi n i to show that 0 n dx = bn. Write t 0 (1 + t) dx as a power series in Stirling− numbers of the first kind to show that P R R n 1 1 − xm ϕˆ (x) = b + s (n 1, m 1) . n n (n 1)! m m=1 − − − X

The polynomials ϕˆ2n (n 1 + x) are an even function of x (check that their deriv- − ative is odd!). Apply ∆ to show that odd degrees are odd, ϕˆ2n 1 (n 1 + x) = − n − ϕˆ2n 1 (n 1 x). We can write both statements together as ϕˆn 2 1 + x = − n− −n − − ( 1) ϕˆn 1 x . − 2 − − 2.3.4. Show that for the Bernoulli polynomials (ϕn) of the ﬁrst kind holds ∆ϕn (x) = n 1 x − / (n 1)! [3, Exc. 1-44]. − 2.3.5. The Stirling numbers of the second kind S (n, k) can be deﬁned as the number of partitions of an n-set into k nonempty blocks. Show that

n! S (n, k) = r1 r2 rn (1!) r1! (2!) r2! (n!) rn! 1r1+2r2+ +nrn=n ··· r1+r2+X···+rn=k ···

2.3.6. Let x0, x1,... be a given sequence in F. Suppose (tm) is a polynomial se- r quence and B a delta operator such that Evalx B tm = δm,r for all 0 r m h r | i ≤ ≤ 64 Chapter 2. Finite Operator Calculus in One Variable

(we do not assume that (tm) is a Sheﬀer sequence for B). Show that for the B-basic sequence (bm) holds

bm (x) = bm j (xj) tj (x) . − j 0 X≥ 2.3.7. Use (2.22) to show that the generating function for the (translated) Abel polynomials equals

n 1 n ∞ n 1 n x (x + an) − t /n! = exp x (an) − t /n! . n 0 n=1 ! X≥ X 2.3.8. Show that

n n x + kz x kz zn+1 1 − − = − (2.29) k n k z 1 k=0 X − − for all x, z C (identity 3.145 in Gould’s list [38]). Begin by showing that ∈ n n x + kz x nz x kz 1 = − − − . k x kz n k k=0 X − −

Then let fn (x) stand for the left hand side in (2.29), and show that fn (x) − zfn 1 (x 1) = 1. − − 2.3.9. The power series of order 0 are a group (G, ) with respect to multiplication, and the power series of order 1 are a group (H,· ) with respect to composition. Show that the identity in the direct product G ◦H is also the identity in the umbral group. Is G H isomorphic to the umbral× group? Show that the umbral subgroup (2.24) is isomorphic× to H. 2.3.10. The set G¯ = (ρ, t): order(ρ) = 0 is a subgroup of the umbral group. Show that G in Exercise{ (2.3.9) is isomorphic} to G¯. Show that for any (σ, β) in the umbral group the right coset (σ, β) G¯ represents all Sheﬀer sequence for the 1 ◦ same delta operator β− ( ). Characterize the left cosets. Is G¯ normal? D 2.3.11. Find the inverse of (σ, β) in the umbral group.

2.3.12. We use the same notation as in subsection 2.3.2. Write US,B for the umbral n n (x-) operator that replaces x /n! by sn (x) for all n 0, US,Bx /n! = sn(x). Note xt xt ≥ that US,Be = M (σ) C (β) e ; the umbral operator is the transform of M (σ) C (β). Use the transform approach to show that (rn (s (x))) has the generating function ρ (t) bσ (α (t)) exβ(α(t)).

2.3.13. Use the notation of Exercise 2.3.12. Show that the umbral operator US,B is not in Σ (except when B = ). D D 2.3. Sheﬀer Sequences 65

t a+t x 2.3.14. The Poisson-Charlier polynomials pn (x; a) have generating function e− a , hence n n j x ( 1) − p (x; a) = − n j aj (n j)! j=0 X − for a = 0. If 6 n pn (x; a) = rn,ipi (x; b) i=0 X find the connection coeffi cients rn,i using (2.25). 2.3.15. The Stirling number S (n, k) is the number of set partitions of an n-set into k nonempty blocks (regardless of order; see Exercise 2.3.5). Show that k!S (n, k) is also the number of mappings from an n-set to a k-set. Use either definition k k k j n to prove that k!S (n, k) = ( 1) − j . Derive this result also from the j=0 j − generating function φ (x) in Example 2.3.11. The Stirling numbers S (n, k) (of n P the second kind) have the generating function

tn (et 1)k S (n, k) = − n! k! n k X≥ which has been already applied in Examples 2.3.11 and 2.2.13 . Show that

tn n ( 1)k k! t − S (n, k) = , n! k + 1 et 1 n 0 k=0 X≥ X − which implies an explicit formula for the Bernoulli numbers Bn,

n n j n 1 k Bn = ( 1) j . − k + 1 j j=0 k=j X X (Hint: Use Exercise 2.3.4.) x 2.3.16. In Example 2.3.11 we found that n and (φn) are inverse to each other, thus n φ x /n! = φn (b (x)) = n , x x n xi n n where bn (x) = n . Let n = i=0 an,i i! . Hence x /n! = i=0 an,iφi (x). Show n that this implies k!an,iS (i, k) /i! = δn,k for all 0 k n. The numbers i=k P ≤ P≤ s (n, i) = n!an,i/i! are the (signed) Stirling numbers of the ﬁrst kind, P ∞ ∞ tn s (n, i) xi = (1 + t)x . n! i=0 n=i X X n n We have i=k s (n, i) S (i, k) = δn,k. Show also that k=j S (n, k) s (k, j) = δn,j. P P 66 Chapter 2. Finite Operator Calculus in One Variable

k 2.3.17. Show: If S = (sn,k)n,k 0 satisﬁes sn,k = σβ n, where σ admits a recip- ≥ rocal and β is a delta series, then there exist two sequences (an)n 0 and (ln)n 0 n i ≥ ≥ in F, a0 = 0, such that sn+1,i+1 = k=0− sn,i+kak for all 0 i n, and 6 n ≤ ≤ sn+1,0 = sn,klk. k=0 P 2.3.18. ShowP that for a Riordan matrix (sn,k) holds

n i l − sn+1,i+1 = sn+1,i sn,i 1a0 + sn 1,i 1+l (ak ak+1) al k (2.30) − − − − − − l=0 k=0 X X for all 0 < i n, where (an) is the a-sequence (a0 = 0). Vice versa, this condition ≤ 6 together with a suitable sn,0 implies that (sn,k) is a Riordan matrix. Show that in Example 2.3.15 the recursion

sn,k = sn 1,k 1 + sn,k+1 sn 2,k 1 + sn 2,k − − − − − − for all n > 0 and 0 k n, implies the a-sequence ≤ ≤ k k j 1 2j 1 2j 1 ( 1) − − a = − − − . k j k j 2j 1 j=0 − − X 2.3.19. Show expansion (2.27). It is also true that

n j/2 i m n 1 n j j i ( 1) D (n, m) = − − ( 1) − − − n j − i j + 1 2i× j=0 i=0 X − X − 2j + m n 3i 2j + m n 3i (m n) − − + − − × − j 2i 1 j 2i − − − for m > n. k 2.3.20. If S is a Riordan matrix, sn,k = σβ n, and (an) a basic sequence with xα(t) n generating function e , show that sn(a (x)) = k=0 sn,kak (x) is a Sheffer n xα(β(t)) polynomial with generating function n 0 sn (a (x)) t = σ (t) e . ≥ P 2.3.21. Use a given Riordan matrixPS = (sn,k) to find the corresponding basic n k sequence for the Sheffer sequence sn (x) = k=0 sn,kx /k!. 2.3.22. Prove Corollary 2.3.17. P n n 2.3.23. Let α (t) = n 0 αnt and 1/α (t) = n 0 βnt Show: If ≥ ≥ P n 1 P n 1 n αn = c (c + un) − /n!, then βn = c (c un) − ( 1) /n! − − This α (t) is defined with the help of the Abel polynomials n 1 α bn(x) = x (x + αn) − /n!, the basic polynomials for E− . It occurs with u = 1 as No. 1 in Riordan’s [72, Table 2.2] Table 3.1 of Abel inverseD relations. 2.3. Sheffer Sequences 67

2.3.24. The reciprocals No. 2

n 2 2 n 2 n αn = (c + un) /n! and βn = c u n (c un) − ( 1) /n! − − − in the same table use the same basic polynomials as in Exercise 2.3.23, but we can n u view (αk) as an evaluation of the Sheﬀer sequence ((x + un) /n!) for E− at c. D Show that βk = bk ( c) ubk 1 (u c). − − − − 2.3.25. Let (bn) be a basic sequence. Show that A = (bi j (ui + v))i,j 0 and B = − ≥ ui+v uj+v bi j( ju v) are inverse matrices. − − − i,j 0 ≥ n 2.3.26. Suppose that γn = k=0 σkbn k (un + v) for all n 0. Let a F. Show that this equivalent to − ≥ ∈ P m+n n+m am u (n + m) bn+m k (am + v) σk = γk − bn+m k(am uk) − am uk − − k=0 k=0 X X − for all m, n N0. A special case is ∈ n m n n m n n k γn = σk σn+k = ( 1) − γm+k. k ⇐⇒ k k − k=0 k=0 k=0 X X X 2.3.27. Class 1 and Class 2 in Riordan’s Table [72, Table 2.2] of Gould classes of inverse relations is the pair p (q 1) n 1 α = − − − − and n,k k p + (q 1) (n k) p + (q 1) (n k) β = − − + q − − n,k k k 1 − n k (In Riordan’s notation βn,n k = An,k and βn,n k = ( 1) − Bn,k). Show: − − − (αi,j)i,j 0 is the inverse matrix to (βi,j)i,j 0. ≥ ≥ αk αk 1 The next 3 exercises are based on the special properties k = α k −1 and k k 1 − (αk) (αk) − k! = α (k 1)! . − 2.3.28. Show: c 2αn + αk α = − and n,k k 2αn c α (2n k) c β = − − − n,k α (2n k) c k − − 2αn α 1 c α (2n k) 1 c α − − − − − − − α (2n k) 1 c k 1 − − − − deﬁne a pair of inverse matrices. 68 Chapter 2. Finite Operator Calculus in One Variable

2.3.29. Show:

k k 1 (c + u (n k)) k (c + un) − αn,k = − and βn,k = ( 1) (c + u (n k)) k! − − k! deﬁne a pair of inverse matrices. For u = 1 this is No. 3 in Riordan’s [72, Table 3.1] table of Abel inverse relations (In Riordan’s notation αn,n k = An,k/ (n k)! n k − − and βn,n k = ( 1) − Bn,k/ (n k)!). − − − 2.3.30. Show:

(c uk + 2un)k α = − and n,k k! k 2 k (c + 2un uk) − 2 βn,k = ( 1) − (2un + c) + uk (u 4un 2c) − k! − − deﬁne a pair of inverse matrices. For u = 1 and c = 0 this is No. 5 in Riordan’s [72, Table 3.1] table of Abel inverse relations. 2.4. Transfer Theorems 69

2.4 Transfer Theorems

Suppose, we have two delta operators A and B connected by an operator equation. We consider A as a “known”operator, with known basic sequence (an). Our goal is to ﬁnd an expression for the basic polynomials bn (x) for B. This could be the n generating function b (x, t) = n 0 bn (x) t , or an explicit expansion of bn (x) in ≥ terms of the basis (an). Of course, the form of theP operator equation is the key point in such expansions. If the operator equation can be written as

1 1 1 1 B = β− ( ) = τ − (A) = τ − α− ( ) D D where α (t), β (t), and τ (t) are delta series, α and τ known, then we ﬁnd β (t) from β (t) = α (τ (t)), and b (x, t) = exα(τ(t))

(see equation (2.32) below). More interesting is the case when τ is not in F [[t]], but has coeﬃ cients that are operators themselves. We have to restrict this settings to translation invariant operators, but the result, Theorem 2.4.2, is still fairly general. We begin with an example showing how natural such a generalization is. Example 2.4.1. Very often, a linear recursion will lead us straight to an operator equation. Consider the number D (n, m) of , lattice paths from (0, 0) to (n, m), staying weakly above the diagonal y ={→x,↑} and avoiding the pattern

• →• , which we also write as rrur (r = and u = ). The following ↑ → ↑ path→• → to (5•, 5) contains the pattern rrur twice (counting overlaps).

m • → • × 4 ↑ • → • → • × 3 ↑ • → • → • × 2 ↑ • × 1 ↑ • × 0 ↑ 0• × 1 2 3 4 5 n

The recursion for avoiding this pattern is

∞ D (n, m) D (n, m 1) = D (n 1, m) ( 1)i D (n 3 2i, m 1 i) − − − − − − − − − i=0 X with initial values D (n, n 1) = δ0,n. Does this recursion have a polynomial solu- − tion? In the notation of Theorem 2.1.1 we have the initial points xn = n 1, the i − factors a1+2i = ( 1) for all i 0, and the translations b1 = 0 xk xk+1 = 1, − ≥ ≥ − − b3+2i = 1 i k 1 (k + 3 + 2i 1) = 3 2i for all i 0. Hence the − − ≥ − − − − − ≥ 70 Chapter 2. Finite Operator Calculus in One Variable

solution to the recursion can be extended to a polynomial dn (x), say, of degree n. In operator notation, the recursion reads as

k k 2k+1 B = ( 1) E− B = (2.31) ∇ − 1 + E 1B2 k 0 − X≥ where Bdn = dn 1. Note that the known delta operator ΣB is written as a − 1 3 2 5 ∇ ∈ power series in B, = B E− B + E− B ... , with coeﬃ cients ∇ − − n (n 1)/2 (n 1)/2 [B ] = ( 1) − E− − Σ ∇ − ∈ D if n is odd, and 0 else.

In the example above we expended A (= ) in ΣB. We saw in Remark 2.2.6 ∇ that every operator in Σ has also a representation in ΣB. In other words, if A is a delta operator in Σ , andD D 2 3 3 A = T1B + T2B + T B + ... where Ti Σ for all i 1, and T1 is invertible, then B is a delta operator in Σ , ∈ D ≥ i D and A ΣB. Certainly A ΣB in the example, because E− ΣB, but we do not ∈ i ∈ ∈ know how E− is expanded in ΣB, only in Σ . We will show how to ﬁnd the basic sequence for B expanded in terms of the basicD sequence for A without knowing i the expansion of E− in ΣB (but we know that it exists!). To do this, we need a technical device, the Pincherle derivative, introduced in the next subsection. For now we state the main Transfer Formula, and prove it later.

Theorem 2.4.2. Let A be a delta operator whose basic sequence (an) is known to us. Suppose A can be expanded as

j A = τ (B) = TjB j 1 X≥ where Tj Σ , T1 invertible. It follows that B is a delta operator with basic ∈ D sequence b0 (x) = 1 and

n 1 b (x) = x τ i a (x) n n x i i=1 X for all n 1. ≥ Remark 2.4.3. The power series τ in the above theorem is an element of Σ [[t]], the power series in t with coeﬃ cients that are translation invariant operators,D an integral domain. For example, τ (t) = t + Bt2 is such a power series. Now “evaluate” τ at B, giving A = τ (B) = B + B3, and

n n/2 i 1 n i i 1 bn (x) = x (1 + Bt) ai (x) = x − B an i (x) n i x i x − i=1 i=0 X h i − X 2.4. Transfer Theorems 71 by the theorem above. Of course, B + B3 would also be the evaluation at B of η (t) = t + t3 F [[t]], ∈

n/3 n 2i bn (x) = − an 2i (x) . i − i=0 X Finally, φ (t) = B3 + t also evaluates to B + B3, but φ (t) is not a delta series. We want to emphasize again, that τ (B) means τ (t) evaluated at t = B.

Consider the special case τ (t) F [[t]], which means that each operator Tj ∈ n i just multiplies by some scalar tj F, j 1. Now bn (x) = τ ai (x), and ∈ ≥ i=1 n n xα(τ(t)) P bn (x) t = e (2.32) n 0 X≥ n 1 where α (t) = ln n 0 an (1) t . If A = ∆ = E I we get the following noteworthy special case. ≥ − P Corollary 2.4.4. Suppose E1 = I + τ (B) where τ (t) F [[t]] is a delta series. Then ∈

n x bn (x) t = (1 + τ (t)) . (2.33) n 0 X≥ Proof. The forward diﬀerence operator ∆ = E1 I is a delta operator with ex- 1 − pansion ∆ = α− ( ) = eD 1, hence α(t) = ln (1 + t) and D − exα(τ(t)) = (1 + τ (t))x .

Example 2.4.5. In our operator equation (2.31) for the avoidance of the pattern rrur we can use that

B = 1 2 ∇ 1 + E− B i n i i B n i i k 2k i − τ = = B − − E− B = − E− 2 n 1 + E 1B2 k (n i) /2 − n k 0 − X≥ if n i is even, and 0 otherwise. The coeﬃ cients τ i are not in F but in Σ . − n D 72 Chapter 2. Finite Operator Calculus in One Variable

Hence n 1 i 1 + x b (x) = x τ i − n n x i i=1 X (n 1) /2 − n + 2i i 1 n 2i 1 + x = x − E− − − i n 2i n 2i 1 i=0 X − − − (n 1)/2 i − n 1 i n 3i 1 + x ( 1) x = − − − − − i n 2i 1 n 2i i=0 X − − − for n 1. This deals only with the problem of ﬁnding the basic sequence! However, ≥ we saw that dn (n 1) = δ0,n, hence we have ‘initial values along a line’(section 2.3.1), − x n + 1 d (x) = − b (x + 1) n x + 1 n (n 1)/2 i − n 1 i n 3i + x ( 1) = (x n + 1) − − − − . − i n 2i 1 n 2i i=0 X − − − The generating function for (bn) we can obtain from (2.33), because we can 1 1 1 2 solve for E in 1 E− = B/ 1 + E− B . We obtain − (B2 + 1)2 4B3 + 1 B2 E1 = − − q 2 (1 B) − (take the root which is not 0 at 0). Therefore, x n 1 2 2 3 bn (x) t = (t + 1) + (t + 1) 4t / (2 (1 t)) . 2 − − n 0 X≥ q We ﬁnd the generating function of xbn (n + x) / (n + x) from (2.23),

2 2 3 (tφ1 (t)) + 1 4 (tφ1 (t)) 1 − φ1 (t) = (tφ1 (t) + 1) + r 2 2 (1 tφ1 (t)) − which has the power series solution 1 √t4 + 2t2 + 1 4t φ (t) = + − . 1 2 2 (t2 t) − Hence x+1 1 √t4 + 2t2 + 1 4t d (n + x) tn = + − . 2 2 (t2 t) n 0 ! X≥ − 2.4. Transfer Theorems 73

In Theorem 2.4.2 we start with the operator equation

j A = TjB j 1 X≥ and then proceed expanding the basic sequence of B. If this equation can be rewritten in the form 0 = F (E,B) for some function F , then a generating function for (bn) can be found explictly when we can solve 0 = F (E,B) for E explicitly 1 1 2 (Corollary 2.4.4). In the above example we had I E− = B/ 1 + E− B , which is easily solved. − 2.4.1 Umbral shifts and the Pincherle derivative The umbral shift (associated to ) is a linear operator denoted by θ such that D θxn = xn+1 for n 0. ≥ Note that θn1 = xn. From

θEax = x2 + ax = (x + a)2 = Eaθx 6 follows that the umbral shift is not a member of Σ . We will write just x for θ in this section. Note that the umbral shift is not translationD invariant. With the help of the umbral shift we define the Pincherle derivative T 0 of any operator T , T 0 = T x xT. − In which sense is T 0 a derivative? The product rule of differentiation holds, (ST )0 = S0T + ST 0 (see Exercise 2.4.5), but there is more. If T Σ , T = τ ( ), say, then d ∈ D d D d dt τ (t) exists, and we can define the (ordinary) derivative of T as d T = d τ ( ), d D D D again in Σ . The two concepts agree if T Σ : d T = T x xT . In other words, the PincherleD derivative equals the ordinary∈ derivativeD D if T is− translation invariant (see Exercise 2.4.4 for a proof). For example,

n n n n n n (E )0 = E x xE = E x E (x n) = nE . − − − The way we defined the umbral shift and the Pincherle derivative above is focused on the derivative operator . We can switch to any other delta operator D A, say, with basic sequence (an), and define the umbral shift θA associated to A as θAan = (n + 1) an+1 for n 0. ≥ Note that θAAan = nan for all n 1. ≥ The corresponding Pincherle derivative of an operator T is defined as the commu- tator of θA and T , T 0 = T θA θAT, A − 74 Chapter 2. Finite Operator Calculus in One Variable

d and we can easily prove that dA T = TA0 if T is translation invariant. We will need this type of derivative (or derivation) in section 4.2.4. Proposition 2.4.6. If A is a delta operator, then

θA = θ 0 . DA Proof. We have

d θ a (x, t) = a (x, t) = M (x) M (α ) exα(t) A dt 0

From Aˆ = M (t) with respect to a (x, t) follows

d θ a (x, t) = M (x) α (A) exα(t) = M (x) α α 1 (A) exα(t) A 0 dA − d = M (x) exα(t). dAD

Hence θA = x 0 . DA n 1+x n+x 1 n 1+x Example 2.4.7. If A = , then θ − = (n + 1) = xE − , thus ∇ ∇ n n+1 n θ = xE1. Note that E = . For T we get (T x xT ) E1 = T E1 if T is 0 0 0 translation∇ invariant. We canD check∇ this∇ by calculating− D

dT dT d dT d ln (1 ) dT 1 dT = D = − ∇ = = E1. d d d −d d d 1 d ∇ D ∇ D ∇ D − ∇ D As a concrete example we ﬁnd

Ek 0 = EkxE1 xE1Ek = Ekx Ek (x k) E1 = kEk+1. ∇ − − − It follows that Ekθ = (x + k) Ek+1. In general, ∇ d d k k k 0 k 0 k E θA θAE = E A = E D = kE D − D dA dA hence d kEk Ekθ = θ Ek + kEk D = θ Ek + . A A dA A Ax xA − 2.4.2 Proof of the Transfer Formula

1 Because B is of order 1, we can write B = P − , where P Σ is invertible. n+1 n D ∈ D Thus sn (x) = B0P x /n! is a Sheﬀer polynomial for B, because

n n 1 n x n x − Bsn (x) = B0P = B0P = sn 1 (x) . D n! (n 1)! − − 2.4. Transfer Theorems 75

Here we only need commutativity in Σ . What are the initial values of (sn (x))? n+1 D We calculate B0P in more details: For n 1 holds ≥ n+1 1 n+1 1 1 n+1 n 2 n+1 B0P = P − 0 P = P − + P − 0 P = P P − P 0P D D − D n 1 n = P (P )0 . − n D All this can be done with the ordinary derivative, it does not need the Pincherle derivative. However, using this concept, we see that

n n n n n+1 x n x 1 n n x x n x B0P = P (P x xP ) = P n! n! − n − D n! n D n!

1 n n n n because n P x x /n!=P x /n!. Note that this cancelling out eﬀect only occurs n+1 D n n+1 when B0P is applied to x /n! (the n in B0P must agree with the n in xn/n!). Thus xn xn 1 xn 1 s (x) = B P n+1 = xP n − = θP n − n 0 n! n! n! for all n 1. Now Eval0 θp = 0 for any polynomial p (x) = 0, thus sn (0) = 0 for all n ≥ 1. For nh= 0 we| geti 6 ≥ 1 0 1 s0 (x) = B0P 1 = P 1 P 0P − 1 = 1. − D

Therefore, the Sheﬀer sequence (sn) for B actually is the basic sequence (bn) for B. We have proven the Transfer Theorem [83],

n+1 n bn (x) = B0P x /n!. (2.34)

If the Pincherle derivative is applied, we call the formula, given in the following Theorem, the Transfer Formula with Pincherle derivative.

Theorem 2.4.8. If B = P is a delta operator with basic sequence (bn), then D xn 1 b (x) = xP n − (2.35) n n! The ﬁrst Transfer Theorem also shows how to transfer from the basic sequence (an) of some delta operator A to the basic sequence (bn) for B. Corollary 2.4.9. If A = VB, where V is invertible, then for all n 1 holds ≥ a (x) b (x) = xV n n (2.36) n x n an 1 (x) = θ V − A n 76 Chapter 2. Finite Operator Calculus in One Variable

1 1 1 Proof. Let B = P − and A = S− , hence PS− = V . By the previous Theo- n 1 Dn x − D rem, an (x) = xS . For n 1, the polynomials an (x) are 0 at 0, hence they n! ≥ are divisible by x (we have seen that (n + 1) an+1 (x) /x are Sheﬀer polynomials for n n 1 n n 1 A). Therefore an (x) /x = S x − /n!, or equivalently S− (an (x) /x) = x − /n!. Substituting this into (2.35) gives the Corollary. We saw in Proposition 2.4.6 that θA = x 0 , hence the second expression for DA bn (x) follows,

n an 1 (x) n an 1 (x) n 1 an 1 (x) n an (x) θAV − = xV 0 − = xV θA − = xV . n DA n x n x

Finally, the proof of the Transfer Theorem 2.4.2 follows by Lagrange Inversion 1 n of (2.36). We have to apply PS− to the Sheﬀer polynomial nan (x) /x for A, thus we need an expansion of this operator in terms of powers of A. We only know that A = τ (B) in the Transfer Formula. We want to ﬁnd

1 n i 1 n i i n i PS− = A PS− A = A (A/B) A . | | i 0 i 0 X≥ D E X≥ In the last expression we think of B as a power series in A, B = A/φ (A), where φ is of order 0. Hence

1 n i n i n i PS− = A φ (A) A = [φ ] A . | i i 0 i 0 X≥ X≥ k n Now apply the inversion formula (1.11), n γ n = k [φ ]n k, from right to left; we want to know [φn] , where t/φ (t) is the compositional inverse− of τ (t). Hence i n [φn] = τ n i i n i − n − and

n n PS 1 = τ n i Ai and − n i − n i 0 − X≥ n n i i n i bn (x) = x τ − A an (x) /x = x τ − an i (x) /x. n i n n − i 0 − i 0 X≥ X≥ 2.4.3 Exercises d 2.4.1. Use the Transfer Formula to show that the basic sequence for E− A equals (xan (x + dn) / (x + dn)), if (an) is the basic sequence for A. 2.4.2. Let B be the delta operator that satisﬁes the recursion

2 2 = B E− B . ∇ − 2.4. Transfer Theorems 77

Use the Transfer Formula (Theorem 2.4.2) to show that

n i x n i 3i 2n 1 + x bn (x) = ( 1) − − − , n i i − i 1 i=1 X − − and use formula (2.33) to ﬁnd the generating function

x 1 + 1 4t2 + 4t3 / (2 (1 t)) − − p for (bn). 1 2.4.3. Show: θ = xE− ∇ k k 1 2.4.4. Show: If T = k 0 τk then T 0 = k 0 τkk − . ≥ D ≥ D 2.4.5. Prove the productP rule of diﬀerentiationP for the Pincherle derivative, (ST )0 = n n 1 n 1 n 1 n 2 S0T +ST 0. This implies (P )0 = P 0P − +P P − 0 = P 0P − +P (n 1) P 0P − . − k d 2.4.6. Show: If A is a delta operator and T = k 0 τkA ΣA then TA0 = dA T . ≥ ∈ k 0 k 1 P 2.4.7. Show: E ∆ = kE − . 2.4.8. Show directly, without the Pincherle derivative, that

n k bn (x) = τ n ak (x) k=0 X if A = τ (B) and τ F [[t]] (see (2.32)). ∈ 2.4.9. A Schröder path is a random walk talking steps , , 2, 0 weakly above {% & h i} the x-axis. Give the horizontal step the weight ω C. The number of paths S (n; ω) from the origin to (2n, 0) are the ω-weighted Large∈ Schröder numbers [87, 1870][95]. Rotate the paths by 45◦. The equivalent paths with steps , , are {↑ → %} counted by dn (m), say, where dn+m (m n) counts the corresponding Schröder − paths. We have dn (n) = S (n; ω). The path counts dn (m) can be extended to a Sheﬀer sequence (dn). Use (2.33) to show that

1 + ωt x b (x) tn = n 1 t n 0 X≥ − is the generating function of the basic sequence for the same delta operator as (dn). Apply (2.23) to get the generating function for the large Schröder numbers,

n n 2 S (n; ω) t = bn (n + 1) t = . 2 2 n 0 n 0 1 ωt + 1 2t (ω + 2) + t ω X≥ X≥ − − p 78 Chapter 2. Finite Operator Calculus in One Variable Chapter 3

Applications

The binomial theorem (2.14) together with Abelization is at this time our only tool for ﬁnding Sheﬀer sequences that satisfy certain initial conditions. With the help of the Functional Expansion Theorem 3.1.4 we will enlarge the pool of problems that can be solved; the theorem contains the binomial theorem as a corollary. To demonstrate the scope of this theorem, we show three examples in section 3.1.1. We revisit Riordan matrices in section 3.2. In many cases, a Riordan matrix

S = (si,j)i,j 0 has elements which are not only coeffi cients of formal power series generating Sheffer≥ sequences, they are also values of Sheffer sequences. More precisely, these are the matrices where s2,1 = s1,0s1,1/s0,0. This is a consequence of Theorem 2.2.11. 6 In the subsection on determinants of Hankel matrices, Finite Operator Cal- culus allows for a systematic description of a rather simple case. However, the general case is diffi cult, and only some special cases have been considered. Finally, we look at the relationship between Umbral Calculus and Finite Operator Calculus. Both theories are similar; one can as well view Finite Operator Calculus as an application of Umbral Calculus. Only an abbreviated introduction into Umbral Calculus is given. More details can be found in [30].

3.1 The Functional Expansion Theorem

We denote by F [x]∗ the vector space of all functionals from F [x] to F. We saw two main examples, the coeffi cient functionals and the evaluation functionals. By linear extension, functionals are defined on all of F [x] if we define them just on a n basis of F [x]. For example, if we know L x /n! =: an for all n 0, then L is h | i ≥ defined as a functional on F [x],

deg p deg p k L p = πk L x /k! = πkak h | i | k=0 k=0 X X 80 Chapter 3. Applications

deg p k if p (x) = k=0 πkx /k!. The sequence of coeﬃ cients (in F) a0, a1,... we will store n xt as λ(t) = n 0 ant = Le , and we see immediately that F [x]∗ F [[t]] with respect toP addition.≥ For convenience we will make this also into an' isomorphism respectingP products by introducing the “right”product of functionals,

n n k n k L N xn := L x N x − , h ∗ | i k | | k=0 X i.e., (L N) ext = Lext Next = λ(t)ν(t). ∗ This mapping from F [x]∗ to F [[t]]is a ring isomorphism [35, p. 129]. We write k L∗ for the k-th power under this product. The multiplicative unit in F [[t]] is 0t 1 = e , hence the evaluation at 0, Eval0, is the multiplicative unit in F [x]∗. We call Eval0 also the identity functional. A linear functional L has a reciprocal (w.r.t. -multiplication), iff L 1 has a reciprocal in F. ∗ h | i Thus the functionals F [x]∗ and the translation invariant operators Σ are D both isomorphic to F [[t]]. This means that everything we did with operators in the preceding sections we could as well have achieved with functionals. Indeed, Roman’s Umbral Calculus [76] is based on functionals. However, we found the interpretation of a recursion as an operator equation more natural, and chose the operator approach. On the other hand, we like to view initial conditions on polynomial solutions as functionals. We took the functional L, made it into a formal power series λ (t) = Lext, and can now make a translation invariant operator (in Σ ) out of it by considering λ ( ), the operator associated to L. Note that λ (t) ext = Dλ ( ) ext, because tnext = nDxt D e . If L has a reciprocal, then λ ( ) is invertible. For example, let L = Evala D D xt at for some a F. We first get λ (t) = Evala e = e , and then the associated ∈ a a operator λ ( ) = e D = E , the translation by a. D The notation λ ( ) for the associated operator of L stresses the dependents D k on more than necessary. It holds that λ ( ) = k 0 L bk B for any delta D D ≥ h | i operator B with basic sequence (bn) (Exercise 3.1.1). We will also use the notation op (L) for the operator associated to L. For example,P if we want λ ( ) = op (L) = x D ∆, we must define L k = δ1,k for all k 0. Alternatively, from ∆ = ED 1 follows L xn = 1 for| all n 1, and L 1≥= 0. − h | i ≥ h | i We now prove a more technical lemma, that sheds some light on the “purpose”of the products of functionals.

Lemma 3.1.1. For every p F [x] and L, N F [x]∗ holds ∈ ∈ L N p = N op (L) p = L op (N) p h ∗ | i h | i h | i Proof. We show only the ﬁrst statement; the second follows from commutativity. 3.1. The Functional Expansion Theorem 81

We can expand p (x) in terms of the basic (xn/n!). Hence it suﬃ ces to show that

(L N) ext = L N xn tn/n! = N λ ( ) xn tn/n! ∗ h ∗ | i h | D i n 0 n 0 X≥ X≥ = Nλ ( ) ext, D which follows from λ (t) ext = λ ( ) ext. D Corollary 3.1.2. Let T Σ , and J, L F [x]∗. If J p = L T p for all ∈ D ∈ h | i h | i p F[x], then op (J) = op (L) T . ∈ Proof. Let N be the functional such that T = op (N). The above Lemma shows that op (J) = op (L N) = op (L) op (N) = op (L) T . We have seen a∗ special case of this Corollary in (2.5). Remark 3.1.3. We defined Lext = λ (t), but that means λ (t) must equal Lˆ, the transform of L. Looking closer at xn xn x0 Leˆ xt = L tn = Ltˆ n = λ (t) | n! n! 0! n 0 n 0 X≥ X≥ shows that the t-operator Lˆ maps 1 to λ (t), and Ltˆ n = 0 for all n > 0. However, we are using λ (t) as the multiplication operator M (λ). The transform of M (λ) is the x-operator λ ( ) = op (L). The transform of Lˆ is the functional L, of course. D We have seen how to find Sheffer sequences with initial values sn (an + c) = yn for all n 0. We now want to find the Sheffer sequence (for B) with initial ≥ conditions L sn = yn, say, for some functional L F [x]∗. Let ush begin| i with the functional L and its isomorphic∈ power series λ (t) = xt Le . Suppose λ (t) has a reciprocal 1/λ (t). Define the Sheffer sequence (ln (x)) 1 n for the delta operator B = β− ( ) by the generating function n 0 ln (x) t = D ≥ exβ(t)/λ (β (t)). Thus l (0) tn = 1/λ (β (t)). Applying L to the generating n 0 n P function gives ≥ P 1 1 L exβ(t) = C (β) Lext = C (β) 1 = 1 λ (β (t)) λ (t) and therefore L ln = δ0,n. The identity h | i k ln (x) = L ln k lk (x) = L B ln lk (x) h | − i | k 0 k 0 X≥ X≥ is trivial, but it shows that every polynomials p F [x] can be written as p (x) = k ∈ L B p lk (x), because (ln) is a basis. Using the Sheffer operator S : bk k 0 | 7→ l for≥ all k 0 we can write this as p (x) = L Bkp Sb (x). However, k k 0 k P 1 ≥ ≥ | λ ( )− = S, as can be seen from D P 1 1 1 λ ( ) exβ(t) = λ ( ) C (β) ext = C (β) λ ( ) ext = exβ(t), D λ (β (t)) D λ (t) D λ ( ) D 82 Chapter 3. Applications

hence λ ( ) ln = bn. This means that D k 1 p (x) = L B p λ ( )− bk (x) . | D k 0 X≥ The special sequence (ln) is no longer needed! We apply this identity to the Sheﬀer sequence (sn) for B,where the initial conditions are given in terms of L, and obtain the functional expansion theorem, writing op (L) for λ (D):

Theorem 3.1.4. If L is a functional such that L 1 = 0, and (sn) is a Sheﬀer h | i 6 sequence and (bn) the basic sequence for the same delta operator, then n 1 sn (x) = L sn k op (L)− bk (x) , h | − i k=0 X where op (L) is the invertible operator

n x n k op (L) = L = L ak A | n! D h | i n 0 k 0 X≥ X≥ for any delta operator A with basic sequence (an). The ﬁrst thing we want to do with this Theorem is showing that it implies the binomial theorem. Let L sn = sn (y) for some y F. Thus L = Evaly, and we have seen in section 3.1h that| opi (L) = Ey. Hence ∈

y sn (x) = sn k (y) E− bk (x) = sn k (y) bk (x y) . − − − k 0 k 0 X≥ X≥ Using the same notation as in the functional expansion Theorem, we ﬁnd the generating function of (sn),

xβ(t) k n e k 0 L sk t ≥ h | i sn (x) t = n (3.1) L bn t n 0 nP0 X≥ ≥ h | i Remark 3.1.5. In a typical application ofP the functional expansion theorem we have some recursive initial values for sn (x), like sn (b) = γn +a0sn (x0)+a1sn 1 (x1)+ − a2sn 2 (x2) + ... . However, we have to watch the highest degree n that occurs in − such a recursion; for example, sn (b) = γn + a0sn (x0) + a1sn 1 (x1) becomes the functional − L = Evalb a0 Evalx a1 Evalx B − 0 − 1 where L sn = γn, but L 1 = 1 a0 has to be a unit so that L has a reciprocal. h | i h | i − Hence the “deﬁning recursion” for (sn) cannot be used if it is of the type we discussed in the Transfer Theorem 2.4.2,

j Asn = TjB sn j 1 X≥ 3.1. The Functional Expansion Theorem 83 thus j L = Evalb A Evalb TjB − j 1 X≥ for some b F, because ∈ L 1 = Evalb A1 Evalb T1B1 = 0. h | i h | i − h | i 3.1.1 Some Applications of the Functional Expansion Theorem We will look at some problems that require initial conditions diﬀerent from the initial value problems we have seen so far. Example 3.1.6. Suppose we are asked to solve the system of diﬀerential equations 1 sn (x) = sn0 (x) under the condition 0 sn (x) dx = 1 for all n 0. The integral 1 xt et 1 ≥ 1 is a functional on [x], we have e dx = =: λ (t), hence λ ( )− = F 0R −t 1 Bn n D op (L)− = / eD 1 = n 0 n! , where B0 = 1, B1 = 1/2, B2 = 1/6, D − ≥ R D − B3 = 0, etc., are the Bernoulli numbers. Note that B2n+1 = 0 for all n 1 1 P ≥ (because t (et 1)− + t/2 is an even function). The delta operator has basic sequence (xn/n−!), hence D

n k n k k n k k j 1 x Bj j x Bj x − sn (x) = λ ( )− = = D k! j! D k! j! (k j)! k=0 k=0 j=0 k=0 j=0 X X X X X − n k n n j xk k B − xk = B x j = j . k! j j − j! k! k=0 j=0 j=0 k=0 X X X X 2 For the first few n we get s0 = 1, s1 (x) = x + 1/2, s2 (x) = x + x + 7/6 /2, x3 x2 1 x2 1 n n s3(x) = 6 + 2 +x+1 2 2 + x + 1 + 12 (x + 1), etc. Using that k=0 k Bk = n − ( 1) Bn we find − n P ( 1)k s (1) = − B . n k! k k=0 X If we do the same for the forward difference operator, i.e., looking for a 1 solution (rn (x)) such that ∆rn = rn 1, and rn (x) dx = 1 for all n 0, then − 0 ≥ 1 R n n λ ( )− = ln (1 + ∆) /∆ = ( 1) / (n + 1) · − · n 0 X≥ and

n n k j n j k 1 x ( 1) x x ( 1) r (x) = λ (∆)− = − = − . n k j + 1 k j n j k + 1 k=0 k=0 j=0 j=0 k=0 X X X − X − X Hence limn rn (0) = ln 2, and limn rn (1) = 2 ln 2. →∞ →∞ 84 Chapter 3. Applications

Example 3.1.7. In chapter 2 we posed the problem of finding the Sheffer sequence n 1 (pn) that satisfies the initial condition pn (1 n) = − pi (n 2i) for all n 1, − i=0 − ≥ p0 ( 1) = 1, and follows Pascal’s recursion pn (x) = pn (x 1) + pn 1 (x), which implies− (see also Exercise 2.1.1) that we are lookingP for a Sheffer− sequence− for n 1+x ∇ with basic polynomials bn (x) = −n . m 1 4 12 35 107 344 2 1 3 8 23 72 237 1 1 2 5 15 49 165 0 1 1 3 10 34 116 1 1 0 2 7 24 82 −2 1 -1 2 5 17 58 −3 1 -2 3 3 12 41 −4 1 -3 5 0 9 29 − 0 1 2 3 4 n n 1 Pascal’s recursion with pn (1 n) = − pi (n 2i) − i=0 − n 1 We have for n 1 the information that 0 = pn (1P n) i=0− pi (n 2i), which implies that ≥ − − − P n 2i i 0 = Eval1 pn (x n) Eval0 E pn (x n) . h | − i − | ∇ − * i=1 + X Hence we define a functional L such that E2 L pn (x n) = Eval1 pn (x n) Eval0 ∇ pn (x n) h | − i h | − i − | 1 E2 − − ∇ = 0 for n 1, and L p0 = 1. The polynomials (pn (x n))n 0 are a Sheffer se- ≥ h | i 1 − 2 ≥1 1 E (E +E )∆ 1 1 1 E ∆ − quence for E = ∆, with op (L) = E 1 E1∆ = 1 E1∆ , and op (L)− = 1 1 ∇ − − − E− (1 E ∆) − 1 (E1+1)∆ . Hence we derive from the Functional Expansion Theorem that − 1 x 1 1 1 j j x pn (x n) = op (L)− = E− 1 E ∆ E + 1 ∆ − n − n j 0 X≥ n j x 1 = 1 E1∆ E1 + 1 − − n j j=0 − X n j n j 1 j i + x 1 − j 1 i + x = − − i n j − i n j j=0 i=0 j=1 i=0 X X − X X − n 1 j x 1 − j i + x 1 = − + − n i n 1 j j=0 i=0 X X − − 3.1. The Functional Expansion Theorem 85

n 1+x n 1 j j n 1+i+x thus pn (x) = −n + j=0− i=0 i n− 1 j . − − Example 3.1.8. Also inP chapterP 2 we saw the example Fn (m) = Fn (m 1) + − Fn 1 (m 2), Fn (0) = 1 for all n 0, which does not have a polynomial extension.− − ≥ m 1 9 30 50 55 55 55 m 1 9 30 50 55 55 55 8 1 8 23 33 34 34 34 8 1 8 23 33 34 34 34 7 1 7 17 21 21 21 21 7 1 7 17 21 21 21 21 6 1 6 12 13 13 13 13 6 1 6 12 13 13 13 13 5 1 5 8 8 8 8 8 5 1 5 8 8 8 8 9 4 1 4 5 5 5 5 5 4 1 4 5 5 5 4 11 3 1 3 3 3 3 3 3 3 1 3 3 3 4 -2 26 2 1 2 2 2 2 2 2 2 1 2 2 1 6 -15 69 1 1 1 1 1 1 1 1 1 1 1 2 -2 13 -43 167 0 1 1 1 1 1 1 1 0 1 0 3 -7 28 -98 364 0 1 2 3 4 5 n 0 1 2 3 4 5 n Fn (0) = 1 The polynomial extension

However, if we take the same recursion but with recursive initial values Fn (n) = Fn 1 (n) for n 1, F0 (0) = 1, then the initial points are xn = n, and Theorem − ≥ 2.1.1 shows that there is a sequence of polynomials such that fn (m) = Fn (m) for all m n, because b1 = 2 n 1 n 1 for all n 0 in that Theorem. It is easy ≥ − ≥ − − − ≥ to show that for the remaining 0 m < n also holds Fn(m) = Fn 1 (m). Therefore ≤ − 1 we deﬁne the functional L on fn (x + n), a Sheﬀer polynomial for E− B, as

L fn (x + n) = fn (n) fn 1 (n) h | i − − = Eval0 fn (x + n) Eval0 Bfn (x + n) , h | i − h | i 2 1 1 x where = E− B. The basic polynomials for E− B = E = ∆ equal n . We ﬁnd op∇(L) = 1 B, and from the Functional Expansion Theorem∇ follows − 1 x n x n x + k f (x + n) = = Ek∆k = n 1 E1∆ n n n k k=0 k=0 − X X − The generating function is also easy to check: For x 0 holds ≥ n x 2 Fn (n + x) t = (1 + t) / 1 t t . − − n 0 X≥ We get the same result if we work with the initial values Fn (n) = Fn 1 (n 1) + − − Fn 2 (n 2), hence the numbers on the diagonal are the Fibonacci numbers. − − Example 3.1.9. The number d (n, m) of , - lattice path avoiding the pattern rruu follow the recursion {→ ↑} d (n, m) = d (n, m 1) + d (n 1, m) d (n 2, m 2) − − − − − 86 Chapter 3. Applications but only for m n+2. They take their initial values from d (n, n + 1) (unknown!), a value that is≥ calculated from the Pascal recursion

d (n, m) = d (n, m 1) + d (n 1, m) − − for all m n + 1. Therefore, dn (n + 1) = dn 1 (n) + dn 1 (n + 1), the right hand side belonging≤ to the polynomial of degree n −1. − − m 1 8 30 77 163 306 519 794 794 7 1 7 23 53 103 178 275 275 0 6 1 6 17 35 62 97 97 0 5 1 5 12 22 35 35 0 4 1 4 8 13 13 0 3 1 3 5 5 0 2 1 2 2 0 Avoiding the pattern 1 1 1 0 rruu 0 1 0 0 1 2 3 4 5 6 7 n

We take the functional

L dn (x + n) = dn (n + 1) dn 1 (n) dn 1 (n + 1) h | i − − − − = Eval1 dn (x + n) Eval0 + Eval1 Bdn (x + n) , h | i − h | i where Bdn (x) = dn 1(x) for all n 0. We know that L dn (x + n) = δ0,n, and op (L) = E1 1 + −E1 B, thus ≥ h | i − 1 k 1 E− k j 1 k k op (L)− = = E E B . 1 (1 + E1) E 1B j − − − k 0 j=0 − X≥ X

Apply the Functional Expansion Theorem 3.1.4 to (dn (n + x))n 0, a Sheﬀer se- ≥ 1 (1) quence for E− B, with (still unknown) basic polynomials bn (x)

k n k k j 1 k k (1) k j 1 (1) dn (n + x) = E − E− B bn (x) = E − bn k (x) . j j − k 0 j=0 k=0 j=0 X≥ X X X The basic polynomials for B we found in Exercise 2.4.2 ,

n i n i 3i 2n 1 + x bn (x) = x ( 1) − − − /i n i − i 1 i=1 X − − n k k j 1 (1) hence dn (x) = k=0 j=0 j E − bn k (x) − n k i n k k j 1 n k i x( 1) − − 3i n+k 1+x = k=0 j=0 Pj E −P i=1− n k i − i − i 1− − − − P P P 3.1. The Functional Expansion Theorem 87

n k i n n 1 n 1 k n k i ( 1) − − 3i 2n+k+j 2+x = 2 + j=0− (x n + j 1) k=−j j i=1− n k i − i − i 1 − − − − − − for all x n + 1. We ﬁnd d (n) from dn 1 (n). The generating function of P ≥ P P − dn (x + n) can be calculated from (3.1),

xβ1(t) k n e k 0 L dk (x + k) t dn (x + n) t = ≥ h | i (1) n n 0 Pn 0 L bn t X≥ ≥ | P D Eexβ1(t) = xβ (t) 1 1 xβ (t) Eval1 e 1 (Eval0 + Eval1) E (E B) e 1 − − exβ1(t) e(x 1)β1(t) − . = β (t) (x+1)β (t) = β (t) e 1 (Eval0 + Eval1) te 1 1 t te 1 − − − t 1 1 Now we have to determine β1 (t) from e− β− (t) = β1− (t). We have β (t) = 1+√1 4t2+4t3 ln 2(1− t) (see Exercise 2.4.2) and ﬁnd (by taking the root that gives us a power series− of order 1)

1 1 2t 1 4t t 2t β− (t) = e e + 4e 4e . 2 − 2 − p 1 1 t 1 √ 2t t Hence β1− (t) = 2 e 2 e + 4e− 4, and solving again a quadratic equation we get − − 1 eβ1(t) = 1 + t2 (1 + t2)2 4t . 2t − − q Finally,

x 1 − 1 2 2 2 2t 1 + t (1 + t ) 4t n − − dn (x + n) t = q 1 1 1 2 n 0 t t2 + (1 + t2) 4t X≥ 2 − − 2 2 − q for all x 1. We ﬁnd the generating function for dn (n) as ≥

n n 1 2t dn (n) t = 1 + t dn 1 (n) t − = 1 + − 2 n 0 n 1 1 2t t2 + (1 + t2) 4t X≥ X≥ − − − q (see [84]).

3.1.2 Exercises

3.1.1. Let B be any delta operator and (sn) any Sheﬀer sequence for B. Show that

k k 0 L sk B ≥ h | i λ( ) = n . D n 0 Eval0 sn B P≥ h | i P 88 Chapter 3. Applications

Especially for the basic sequence (bn) of B holds

k λ( ) = L bk B . D h | i k 0 X≥ This also shows that the Sheﬀer sequence (sn) is uniquely deﬁned by L, if λ ( ) is invertible , because D k n k 0 L sk t s (0) t = ≥ h | i . n λ (β (t)) n 0 P X≥ 3.1.2. Prove the generating function (3.1). 3.1.3. Prove that n n

L bk sn k (x) = L sk bn k (x) , h | i − h | i − k=0 k=0 X X in the notation of the Functional Expansion Theorem 3.1.4 . 3.1.4. Use the notation from Example 2.3.9. Let k be a fixed index such that 0 c + ka 1. Find ≤ ≤ qn (x) = Pr(U 0 for i = 1, . . . , k 1,..., and (i) ≥ − U(k) c + ka, . . . , U(n 1) c + (n 1) a, x U(n) c + na)/n! ≥ − ≥ − ≥ ≥ for 0 n M. ≤ ≤ 3.1.5. Show that n ( 1)j n x s (x) = − n j + 1 k j j=0 k=j X X − is the solution to the difference equation sn (x + 1) sn (x) = sn 1 (x) under the 1 − − condition sn (x) dx = 1 for all n 0. 0 ≥ 3.1.6. UseR the Functional Expansion Theorem to show that x + 1 n n + x s (x) = − n x + 1 n is the solution to the difference equation sn (x) sn (x 1) = sn 1 (x) under the n − − − condition sn (i) = 1 for all n 0. See also Example 2.3.6. i=0 ≥ 3.1.7. UseP the Functional Expansion Theorem to show that x x + 1 sn (x) = n − n 1 − is the solution to the difference equation sn (x + 1) sn (x) = sn 1 (x) under the n − − condition i=0 sn i (n + i) = 1 for all n 0. − ≥ P 3.2. Diagonals of Riordan Matrices as Values of Sheffer Sequences 89

3.2 Diagonals of Riordan Matrices as Values of Sheﬀer Sequences

The following Riordan matrix B has been constructed with a ﬁrst column bn,0 = n 1 δn,0, and the k + 1-th column according to bn,k+1 = j=0− bj,kCn j (Corollary − 2.3.17), where Ck stands for the n-th Catalan number. We know (Theorem 2.3.13) Pk n k that the rows as coeﬃ cients of polynomials, bn,k = x /k! bn (x) = [t ] β (t) , where b (x) tn = exβ(t). However, it is striking that the numbers on the n 0 n diagonal and≥ subdiagonals seem to be values of polynomials too! P bn,k 0 1 2 3 4 5 6 7 k 0 1 1 0 1 2 0 1 1 3 0 2 2 1 4 0 5 5 3 1 5 0 14 14 9 4 1 6 0 42 42 28 14 5 1 7 0 132 132 90 48 20 6 1 n 0 429 429 297 165 75 27 7 1 n 1 bn,k+1 = j=0− bj,kCn j by Corollary 2.3.17 − Theorem 2.2.11 tells usP that this must happen because b2,1 = 0. In that case there 6 exists a basic sequence ˜bn (x) with the property bn,k = ˜bn k (k) − In other word, the entries along the subdiagonal lines in the Riordan matrix B are values of the polynomials ˜bn (x) with generating function β (t) β˜ (t) = ln . t

Example 3.2.1. The Riordan matrix B = (bn,k) above is generated by σ (t) = 1 n 1 n and by bn,i+1 = k=0− sk,iCn k, hence β (t) = n 1 Cnt (Corollary 2.3.17), and − ≥ b = [tn] β (t)k. Here C = 2n / (n + 1) is the n-th Catalan number. We saw in n,k P n n P k Example 2.3.8 that β (t) = 1 √1 4t /2 . The matrix shows that b2,1 = 1. − − We ﬁnd β˜ (t) = ln 1 √1 4t / (2t) , and − − n x ˜bn (x) t /n ! = 1 √1 4t / (2t) . − − n 0 X≥ In Example 2.3.7 we saw that n 1 2n + x x + 1 − 2i + x x + 1 ˜bn (x) = Cn i 1 . n n + x + 1 − − − i i + x + 1 i=0 X 90 Chapter 3. Applications

Therefore, bn,k = ˜bn k (k) = − n k 1 2n k k + 1 − − 2 (n k 1 i) 1 2i + k k + 1 − − − − . n k n + 1 − n k 1 i n k i i i + k + 1 i=0 − X − − − − − In addition to the condition b2,1 = 0, Theorem 2.2.11 requires that bn,n = 1 6 for all n 0, so that a given Riordan matrix B = [tn] β (t)k has diagonals that ≥ are values of a basic sequence. Suppose bn,n is not the constant 1. It follows from n the deﬁnition of a Riordan matrix B that bn,n = b0,0a0 for n > 0. If b0,0 = c = 0, then divide every element of the matrix B by c. Now all main diagonal elements6 of n ˜ the new matrix are a0 for some a0. If a0 = 1, we are done; we ﬁnd bn,k = cbn k (k), − where the basic sequence ˜bn (x) has generating function

x ˜ 1 β (t) exβ(t) = (3.2) c t n (see Exercise 3.2.1). Next we assume that bn,n = a0 for all n, and that a0 = 1. Depending on the matrix (see Example 3.2.4) we may choose one of the6 two methods. k 1. (Column standardization) Divide the elements of the k-th column by a0 .The k result is a new matrix with elements a0− bn,k, satisfying the condition that k˜ the main diagonal equals one everywhere. We ﬁnd bn,k = a0 bn k (k), where − the basic sequence ˜bn (x) has generating function

x ˜ β (t) exβ(t) = (3.3) a t 0 (see Exercise 3.2.2). n 2. (Row standardization) Divide the elements of the n-th row by a0 .The result is n a new matrix with elements a0− bn,k, satisfying the condition that the main nˆ diagonal is all ones. We ﬁnd bn,k = a0 bn k (x), where the basic sequence − ˆbn (x) has generating function

x ˆ β (t/a ) exβ(t) = 0 (3.4) t (see Exercise 3.2.3).

Note that in both cases a0 = b1,1 = β1. If a basic sequence through the diagonals of a Riordan matrix exist,

bn,k = ˜bn k (k) , − 3.2. Diagonals of Riordan Matrices as Values of Sheﬀer Sequences 91

xβ˜(t) x how does it help us ﬁnding bn,k? The relationship e = (β (t) /t) is not very n k useful, as the example above shows, because expanding bn,k = [t ] β (t) or ˜bn (k) = n k [t ](β (t) /t) is basically the same work. This will happen if we know c1, c2,... n 1 in Corollary 2.3.17 such that bn,i+1 = k=0− bk,icn k. If the matrix is given by the n i − 1 a-sequence, b = − b a for all 0 i n, then we only get β (t) n+1,i+1 k=0 n,i+k kP − (Theorem 2.3.13), and β (t) must be obtained through≤ ≤ an inversion process, which may not be possible or ugly.P However, it can be easily checked that in this situation

∞ k k ∆ = akE B˜ (3.5) k=1 X (Exercise 3.2.4), where B˜ is the delta operator mapping ˜bn to ˜bn 1. We can use − the Transfer Formula (Theorem 2.4.2) to construct ˜bn this way.

Finally we discuss the case when S = (sn,k) is a Riordan matrix of the general n k kind sn,k = [t ] σ (t) β (t) , but can be brought into a form where the diagonal are values of a Sheﬀer sequence. Because 2 σ (t) β (t) = (σ0 + σ1t + ... ) β1t + β2t + ... 2 2 = σ0β1t + (σ1β1 + σ0β2) t + = s1,1t + s2,1t + ... ··· we see that β2 = 0 is equivalent to s2,1 = σ1β1 = s1,0s1,1/s0,0 = s1,0a0 = l0a0. 6 6 Corollary 3.2.2. Let S = (sn,k) be a Riordan matrix such that

n k sn,k = [t ] σ (t) β (t) for a delta series β and a power series σ that admits a reciprocal. There exists a n s1,1 Sheffer sequence (ˆsn (x)) such that sn,k = sˆn k (k) with generating func- s0,0 − n xβˆ(t) tion n 0 sˆn (x) t =σ ˆ (t) e iff s2,1 = s1,0s1,1/s0,0. In this case, ≥ 6 x P ˆ s β (s t/s ) σˆ (t) exβ(t) = σ 0,0 t 0,0 1,1 . s t 1,1 k s1,1 There also exists a Sheffer sequence (˜sn (x)) such that sn,k = sñ k (k) s0,0 − n xβ˜(t) with generating function n 0 sñ (x) t =σ ˜ (t) e iff s2,1 = s1,0s1,1/s0,0. In that case, ≥ 6 x P ˜ s β (t) σ˜(t)exβ(t) = σ (t) 0,0 . s t 1,1 Remark 3.2.3. We obtain the same operator equation for (˜sn (x)) as for ˜bn (x) , ∞ k 1 k ˜k ∆ = aka0− E B . (3.6) k=1 X 92 Chapter 3. Applications

(Exercise 3.2.4). To determine the Sheffer sequence (˜sn) we need initial values. There may be some special combinatorial properties that can supply initial values, n 1 k or we can use the l-sequence to determine sñ (0) = k=0− a0 lksñ 1 k (k); in this case we define the functional − − P

∞ k k+1 L sñ = Eval0 sñ a lk Evalk B˜ sñ = 0 h | i h | i − 0 | k=0 X D E for all n > 0, and L s˜0 = s0,0 = 0. The Functional Expansion Theorem tells us that h | i 6 s0,0 sñ (x) = ˜bn (x) . (3.7) k k k+1 1 ∞ a lkE B˜ − k=0 0 Example 3.2.4. Consider the followingP lattice path problem: Instead of taking steps from , , say, allow any step with coordinates (i, j) in N0 N0, except (0, 0). We want{→ ↑} to count the number D (n, m) of such paths from the× origin to (n, m). m 1 For example, D (0, 0) = 1, D (0, m) = 2 − for m 1, D (1, 1) = 3, D (2, 2) = 26, and D (n, m) = D (m, n). We have ≥

n m 1 n 1 − − D (n, m) = D (i, j) + D (i, m) i=0 j=0 i=0 X X X directly from the deﬁnition of the step set. We let D (n, m) = 0 if n < 0 or m < 0. It follows that

n D (n, m) = 2D (n, m 1) + 2iD (n i, m 1) (3.8) − − − i=1 X for m n 0, m 2. We make a Riordan matrix S = (sn,k) out of D (n, m) by ≥ ≥ ≥ defining sn,k = D (n k, n) for all 0 k n, with the exception of s0,0 which we have to define as 1/2−for making S Riordan≤ ≤ (this fixes the problem that (3.8) does not hold for m = n = 1).

sn,k 0 1 2 3 4 5 k 0 1/2 1 3 1 2 26 8 2 3 252 76 20 4 4 2568 768 208 48 8 5 26 928 8016 2208 544 112 16 n 287 648 85376 23776 6080 1376 256 32 n k j sn,k = 2sn 1,k 1 + j=1− 2 sn 1,k 1+j − − − −

Obviously, the condition of Corollary 3.2.2P is satisﬁed, 8 = s2,1 = s1,0s1,1/s0,0 = 6. n i k 6 Equation (3.8) implies that sn+1,i+1 = 2sn,i + − 2 sn,i+k for all 0 i n, k=1 ≤ ≤ P 3.2. Diagonals of Riordan Matrices as Values of Sheﬀer Sequences 93

k hence S has the a-sequence a0 = 2, ak = 2 for all k 1. This means that we 1 1 ≥ know β− (t), and it is easy to ﬁnd β (t) from β− (t) (see Exercise 3.2.6), but we want to follow the alternative route outlined in Remark 3.2.3. From (3.8) follows n k+2 that sn+1,0 = 6sn,0 + k=1 2 sn,k (see Exercise 3.2.5), and therefore S has the k+2 l-sequence l0 = 6, lk = 2 for k 1. We divide the columns by powers of 2, because we get all integersP this way,≥ except on the diagonal.

0 1 2 3 4 5 k 0 1/2 1 3 1/2 2 26 4 1/2 3 252 38 5 1/2 4 2568 384 52 6 1/2 5 26928 4008 552 68 7 1/2 n 287648 42688 5944 760 86 8 1/2 k sn,k/2 =s ˜n k (k) −

k 1 k k Equation (3.6) shows that ∆ = ∞ aka − E B˜ = 2EB/˜ 1 4EB˜ , hence k=1 0 − P n i i n ( 1) 2 t x n 1 2n i x ˜b − (x) = = − 2 − n i i n i i i=1 "(1 4t) # i=0 X − n X − is for n 0 the basic sequence for EB˜. We will use the l-sequence in Exercise 3.2.5; now we≥ exploit a fact much closer to the given problem: D (n, m) = D (m, n) for all 2k m, n 0. This means for the Sheffer sequence (˜sn) that sñ k (k) = 2− sñ ( k) ≥ 1 2n − − for all k n. Hence 2 =s ˜0 (n) = 2− sñ ( n), and by the binomial theorem for Sheffer sequences≤ (2.14) follows −

n n n i 2k 1˜( 1) 2n i 1 x − n k 1 s˜n (x n) = 2 − bn− k (x) = 2 − − − − − − i n k i k=0 i=0 k=0 X X X − − n x n = 22n i 1 . − − i i i=0 X Therefore, n k k − 2n k i 1 n n k sn,k = 2 s˜n k (k) = 2 − − − − . − i i i=0 X 3.2.1 Exercises 3.2.1. Prove formula (3.2). 3.2.2. Prove formula (3.3). 94 Chapter 3. Applications

3.2.3. Prove formula (3.4). 3.2.4. Prove formula (3.5). k 3.2.5. In Example 3.2.4 show that the a-sequence equals a0 = 2, ak = 2 for all k+2 k 1, and the l-sequence equals l0 = 6, lk = 2 for k 1. ≥ ≥ 1 3.2.6. In Example 3.2.4 show that the a-sequence implies β− (t) = t (1 2t) / (2 2t), thus β (t) = 1 + 2t √1 12t + 4t2 /4. Use Theorem 2.3.13 to show− that σ (t)− = − − 1/ 2√1 12t + 4t2 . −

1 t t 1 t (1 2t) β− (t) = n = n n = − . n 0 ant 2 + n 1 2 t 2 1 t ≥ ≥ − By Theorem 2.3.13 P P

1 s0,0 s0,0 σ β− (t) = 1 n = 1 1 1 β− (t) n 0 lnt 1 6β− (t) 8tβ− (t) / (1 2t) − ≥ − − − 1 1 t = − P. 2 1 4t + 2t2 − Now replace t by β (t).

3.2.7. Let S = (sn,k) be a Riordan matrix such that

n k sn,k = [t ] σ (t) β (t) . Eliminate the ﬁrst m rows and columns of S and call the new matrix R, rn,k = n sn+m,k+m. Show that R has the same a-sequence as S, and that n 0 rn,0t = σ (t)(β (t) /t)m. ≥ P 3.3 Determinants of Hankel Matrices

A Hankel matrix is constant along parallels to the second diagonal; more precisely h is a Hankel matrix iﬀ h = h (i + j), i.e., h depends only on i,j0 i,j=0,...,n 1 i,j0 i,j0 − T i + j. If A is any square matrix, then AA is a Hankel matrix. C. Radoux [71], [70], and Martin Aigner [1] considered matrices A = (am,k)m,k 0 ≥ such that am,m = 1, am,k = 0 for m > k, and

am+n,0 = am,kan,k (3.9) k X for all m, n 0, i.e., the matrix is lower triangular, with ones on the diagonal, and for all choices≥ of m and n the inner product of the m-th and n-th row is the same as long as m+n remains the same. We call the sequence an,0, n 0, an unweighted Radoux sequence. Not every sequence can be an unweighted Radoux≥ sequence; we 3.3. Determinants of Hankel Matrices 95

m 2 have for even indices a2m,0 = k=0 am,k, which implies a dependence on previous 2 terms. For example, a1,0 can be freely selected, but then a2,0 = a1,0 + 1. Next, 3 P a3,0 = a1,0 + a1,0 + a2,1 is arbitrary, because a2,1 can be chosen accordingly, but 2 2 2 a4,0 = a1,0 + 1 + a2,1 + 1 is determined. Let An be the restriction of A to the (n + 1) (n + 1) matrix (am,k)0 m,k n, × ≤ ≤ thus det An = 1. Because of the inner product property (3.9), the Hankel matrix T AnAn = (ai+j,0)0 i,j n has the determinant det (ai+j,0)0 i,j n = 1. More interesting is the determinant≤ ≤ of the “second”Hankel matrix, ≤ ≤

dn := det (ai+j+1,0)0 i,j n 1 . ≤ ≤ −

Elementary matrix manipulations (Exercise 3.3.1) show that dn =

a10 1 0 0 ··· 1 a21 a10 0 0  0− 1 ··· 0 0  ··· det  ...   ············   0 0 1 0   ···   0 0 an 1,n 2 an 2,n 3 1   ··· − − − − −   0 0 1 an,n 1 an 1,n 2   ··· − − − −    and hence dn = sn 1dn 1 dn 2 (3.10) − − − − where d0 = 1 and sn 1 = an,n 1 an 1,n 2 for all n 1, s0 = a1,0. As Aigner − − −T − − ≥ points out, the fact that det AnAn = 1 together with the numbers dn ( n 1) uniquely determine the matrix A. Therefore, we can ask for the matrix A, given≥ the sequence of determinants d1, d2,... Let cn (k) = an+k,k; in this notation cn (k) = skcn 1 (k) + cn 2 (k + 1) for ∇ − − all k 1 (see (3.12) below), thus sk = c1 (k) = ak+1,k ak,k 1. For k = 0 we ≥ ∇ − − have the initial values cn (0) = c1(0)cn 1(0) + cn 2 (1). We assume that cn (k) = 0 − − for all n < 0. The numbers c1 (k), k = 0, 1,...,completely determine the matrix A. A variation of the condition (3.9) appears in Zhang and Feng [103], introducing a nonzero sequence fk such that f0 = 1. In an abuse of standard notation we k will deﬁne (only in this section!) fk! := Πi=0fi. Deﬁne the matrix A by am,k = 0 for k > m, amm = 1, and

ai+j,0 = ai,kaj,kfk! (3.11) k 0 X≥ ˜ for all i, j 0 (a weighted inner product). If we deﬁne An := (ai,kfk!)0 i,k n, then ≥ ≤ ≤ ˜T det (ai+j,0)0 i,j n = det AnAn = fn!!, ≤ ≤ 96 Chapter 3. Applications

n k n n+1 k where fn!! = Πk=0Πi=0fi = Πk=1fk − . We will call (an,0) a weighted Radoux sequence in this case; the sequence (fn) is determined by det (ai+j,0)0 i,j n = fn!!. ≤ ≤ A sequence (an) is a weighted Radoux sequence, if

1. a0 = 1,

2. there exists a lower triangular matrix A = (ai,j)i,j 0 and a sequence (fn), ≥ f0 = 1 and fk = 0 for all k 0, such that an = an,0 = k 0 ai,kaj,kfk! for all i + j = n. 6 ≥ ≥ P By deﬁning the weights to be nonzero, we are still excluding some sequences m 1 2 from being weighted Radoux sequences, because a2m − a fk! = fm! = 0. − k=0 m,k 6 If we would allow f = 0, we would reduce the nonzero columns of the matrix A˜ k P to the ﬁrst k columns. For example, the Fibonacci sequence can be obtained this way, letting f0 = f1 = 1, and f2 = 0. Then an,0 = Fn, and an,1 = Fn 1. Still, the set of weighted Radoux sequences is much larger then in the case when− all the weights are 1.

Lemma 3.3.1. A lower triangular matrix A = (ai,j)i,j 0 with diagonal elements equal to 1 satisﬁes the weighted inner product condition≥ (3.11) iﬀ

an,k = an 1,k 1 + skan 1,k + fk+1an 1,k+1 (3.12) − − − − where sk = ak+1,k ak,k 1. − − Proof. The cases n = 1, 2 (and all k 0) can be veriﬁed directly; assume al,k = ≥ fk+1al 1,k+1 + al 1,k 1 + skal 1,k holds for all integers l n. Denote by ri the − − − − ˜ ≤ i-th row of An, and r˜i the i-th row of An := (ai,kfk!)0 i,k n. Then ≤ ≤

rn r˜l = an,kal,kfk! = (fk+1an 1,k+1 + an 1,k 1 + skan 1,k) al,kfk! ◦ − − − − k 0 k 0 X≥ X≥ = (fk+1!an 1,k+1 + fk!an 1,k 1 + fk!skan 1,k) al,k − − − − k 0 X≥ = (al,k 1fk! + al,k+1fk+1! + fk!skal,k) an 1,k − − k 0 X≥ = (al,k 1 + al,k+1fk+1 + skal,k) an 1,kfk! − − k 0 X≥

Hence the condition rn r˜l = rn 1 r˜l+1 is equivalent to al+1,k = al,k 1 + ◦ − ◦ − al,k+1fk+1 + skal,k, ﬁnishing the induction proof. In terms of the functions cn (k) = an+k,k we have

cn (k) = skcn 1 (k) + fk+1cn 2 (k + 1) for k 1 (3.13) ∇ − − ≥ cn (0) = s0cn 1 (0) + f1cn 2 (1) for all n 1 − − ≥ c0 (k) = 1 for all k 0 ≥ 3.3. Determinants of Hankel Matrices 97

The second Hankel determinant det (ai+j+1,0)0 i,j n 1 equals ≤ ≤ −

s0 1 0 0 ··· f1 s1 1 0 0  ···  fn 1!! det 0 f2 s2 1 0 = fn 1!!dn (3.14) − ··· −    ···   fn 1 sn 1   − −    (Exercise 3.3.1) where

dn = sn 1dn 1 fn 1dn 2 for n 2 (3.15) − − − − − ≥ d1 = s0, and d0 = 1.

Zhang and Feng consider further Hankel determinants of the form

det (ai+j+k,0)0 i,j m ≤ ≤ for k = 2, 3,... , but the results get more unpleasant. We will ask the following questions: Given both determinants, what is the Radoux sequence? In terms of cn (k) we can ask the same question: Given (fn) and (dn), what is cn k (k) for all n k? We can also prescribe the sequence (sn) − ≥ instead of dn, because of (3.15). And ﬁnally, how do we know that the sequences (fn) and (dn) generate a Radoux sequence? For example, if f1 = f2 = 2 and s0 = 1, s1 = 1, and sk = 0 for k 2, then − ≥ 1 0 0 1 1 0 0 A = 3  2 0 1 0   2 2 + f2 0 1      2 Hence a4,0 = a2,0 + 0f1 + 1f2 = 4 + f2, but also a4,0 = a1,0a3,0 + 1a3,1f1 = 2 + f1 (2 + f2). Therefore, f1 and f2 determine each other in this case. We cannot give an answer to those questions in the stated generality. How- ever, we can answer it completely when (cn) is a Sheﬀer sequence. Equation (3.13) tells us that this will happen when sk is a constant s, say, for all k 1, and ≥ fk+1 equals a constant u = 0 for all k > 1. Hence c1(k) = sk + a, where we set 6 s0 = a1,0 =: a. The weight f1 is another constant free to choose; for reasons that will appear later we set f1 = 2vu, for v = 0. We obtain the recursion 6

cn (k) = scn 1 (k) + ucn 2 (k + 1) ∇ − − with initials conditions

cn (0) = acn 1(0) + 2uvcn 2 (1) − − 98 Chapter 3. Applications

for n 1, and c0 (k) = 1. This setup implies for the sequence (fn 1!!dn) of determinants≥ −

dn = sdn 1 udn 2 for n 3 − − − ≥ d1 = a, d2 = sa 2uv. −

Let B be the delta operator for (cn); it follows from the recursion for cn (x) that 1 1 2 1 E− = sB + uE B , thus − 1 E = 1 sB (1 sB)2 4uB2 2uB2 − − − − q and therefore, by (2.33),

x 1 st (1 st)2 4ut2 n b (x, t) = bn (x) t = − − − −  q2ut2  n 0 X≥ x  −  = 2x 1 st + (1 st)2 4ut2 − − − q where (bn) is the basic sequence for B. The initial conditions are described by the functional 2 L = Eval0 a Eval0 B 2uv Eval1 B − − giving L bn = δ0,n. It follows from the Functional Expansion Theorem 3.1.4 that h | i b (x, t) c (x) tn = . n 1 at 2uvt2b (1, t) n 0 X≥ − −

We obtain the generating function for the Radoux sequence an,0 = cn(0)

n 1 1 an,0t = = 1 at 2uvt2b (1, t) 2 n 0 − − 1 v + t (vs a) + v (1 st) 4ut2 X≥ − − − − q (3.16)

1 v + t (vs a) v (1 st)2 4ut2 = − − − − − 1 2v + 2 (vs a + va) t q(2vsa a2 4v2u) t2 − − − − −

The generating function of the numbers dn can be calculated using Proposition 1.1.4, a 2uvt d tn = − . (3.17) n+1 1 st + ut2 n 0 X≥ − 3.3. Determinants of Hankel Matrices 99

This generating function has a quadratic in the denominator, and therefore it is of Fibonacci type. As in Exercise 1.1.4,

n 1 n+1 n+1 n n 2− − dn+1 = a s + √δ s √δ 4uν s + √δ s √δ √δ − − − − − 2 n 1 n n = − − a s + √δ 4uν s + √δ a s √δ 4uv s √δ √δ − − − − − if the discriminant δ := s2 4u = 0. If s2 = 4u then 1 st + 4ut2 = (st 2)2 /4 and − 6 − − n n 1 dn+1 = (as (n + 1) 4uνn) 2− s − . − So we can identify the Radoux sequence (an,0) as values of a Sheﬀer sequence with generating function (3.16) if

n n n(n+1)/2 det (ai+j,0)0 i,j n = fn!! = 2 v u ≤ ≤ and (dn) has a generating function of the above form. Example 3.3.2 (Motzkin Numbers). We want to determine the Radoux sequence n+1 ( 2 ) (an,0) such that det (ai+j,0)0 i,j n = (m + 1) for some nonnegative inte- ≤ ≤ ger m, and dn = det (ai+j+1,0)0 i,j n 1 / det (ai+j,0)0 i,j n 1 has the generating function ≤ ≤ − ≤ ≤ − 1 (m + 1) t d tn = − . n+1 1 + (m + 1) t (t 1) n 0 X≥ − (n+1)n/2 We ﬁnd immediately from fn!! = (m + 1) that fk = m + 1 for all k 1. a 2≥uvt Hence u = m+1 and 2v = 1. The generating function for dn is of the form 1 −st+ut2 − if we choose a = 1, and s = m + 1. The Radoux sequence (an,0) = (cn (0)) has the generating function

1 − n 2 2 an,0t = 2 1 + t (m 1) + (1 (m + 1) t) 4 (m + 1) t − − − n 0 X≥ q 1 + t (m 1) (1 (m + 1) t)2 4 (m + 1) t2 = − − − − q 2t (t + m)

The case m = 0 gives the generating function of Motzkin path, i.e., , , - paths weakly above the x-axis and ending on the x-axis. For larger values{% & of→}m, the corresponding lattice path problem is described in [57]. Besides the Sheﬀer sequences, there are other cases known of Radoux sequences. One such application, still closely related to Sheﬀer sequences, is given in Exercises 3.3.6. 100 Chapter 3. Applications

3.3.1 Exercises

3.3.1. Show that det (ai+j+1,0)0 i,j n 1 equals the expression in (3.14). (Hint: Use (3.12)). ≤ ≤ −

3.3.2. [1], [103] Determine the Radoux sequence (an,0) such that

n det (ai+j,0)0 i,j n = (1 + µ) = det (ai+j+1,0)0 i,j n 1 . ≤ ≤ ≤ ≤ − If both determinants are 1, we obtain the Catalan numbers.

3.3.3. [103] Determine the Radoux sequence (an,0) such that

n n(n+1)/2 det (ai+j,0)0 i,j n = 2 µ ≤ ≤ and n 1 n(n 1)/2 n det (ai+j+1,0)0 i,j n 1 = 2 − µ − (1 + µ ) . ≤ ≤ − 3.3.4. Determine the Radoux sequence (an,0) such that

n+1 n+1 ( 2 ) ( 2 ) det (ai+j,0)0 i,j n = ( 1) (ω + 1) ≤ ≤ − for some complex parameter ω = 1. Suppose 6 −

dn = det (ai+j+1,0)0 i,j n 1 / det (ai+j,0)0 i,j n 1 ≤ ≤ − ≤ ≤ − has the generating function

(ω + 1) (1 + t) d tn = . n+1 1 (ω + 2) t (ω + 1) t2 n 0 X≥ − −

Show that an,0 = S (n; ω), the Large Schröder numbers in Exercise 2.4.9.

3.3.5. Let u be a nonzero constant. Show: If (an) is a Radoux sequence with given n sequence sn = an+1,n an,n 1 and weights (fn), then añ := anu is a Radoux − − 2 sequence such that a˜i+j,0 = k 0 a˜i,ka˜j,k u fk ! and sñ =a ñ+1,n añ,n 1 = usn. ≥ − − n 3.3.6. [71], [103] Show thatP the Bell numbers n = S (n, k), the number of B k=0 partitions of an n-set , are a Radoux sequence. Let an,k = n!rn k (k) /k!, where P − (rn) is the Sheffer sequence with generating function

t x n et 1 e 1 r (x) t = e − − n t n 0 X≥ et 1 t x Show that n!rn (0) = n. Apply the operator x+tDt to e − ((e 1) /t) to show that B − (n + x) rn (x) = rn 2 (x + 1) + xrn (x 1) + (x + 1) rn 1 (x) . − − − 3.3. Determinants of Hankel Matrices 101

Hence fk = k and sk = k + 1. The ﬁrst Hankel determinant equals

det (ai+j,0)0 i,j n = n!!, ≤ ≤ and the second

det (ai+j,0)1 i n,0 j n 1 = (n 1)!!. ≤ ≤ ≤ ≤ − − In [103] the slightly more general case fk = kµ and sk = k+µ is considered, giving the Hankel determinants

n(n+1)/2 det (ai+j,0)0 i,j n = n!!µ ≤ ≤ and n(n+1)/2 det (ai+j+1,0)0 i,j n 1 = (n 1)!!µ . ≤ ≤ − − n k Show that an,0 = S (n, k) µ for n 0 is the Radoux sequence in this case. k=0 ≥ 3.3.7. [21]DenoteP by D (n, m; ω) the weighted number of lattice paths from (0, 0) to (n, m) with step vectors (1, 1), (1, 1), and ω-weighted step vector (1, 0). The weight of a path is the product− of the weight of its steps (the diagonal steps have weight 1). The weighted paths enumerated by D (n, 0; ω) are often called weighted grand Motzkin paths. They have the generating function 1 D (n, 0; ω) tn = 2 n,m 0 (1 ωt) 4t2 X≥ − − q If ω = 0, the path does not take horizontal steps; the counts are the central binomial n coeﬃ cients n/2 for even n. Show that the generating function is of the type (3.16), and calculate the Hankel determinant D (i + j, 0; ω) 0 i,j n 1. Noting that | | ≤ ≤ −

[n/2] n+1 n+1 k n k k n 2k x y ( 1) − (xy) (x + y) − = − − k x y k=0 X − we let ω = (x + y) /√xy. Show that

n 1 n/2 n n D (i + j + 1, 0; ω) 0 i,j n 1 = 2 − (xy)− (x + y ) . | | ≤ ≤ − 3.3.8. [21]The weighted paths in Exercise 3.3.7 become the Motzkin paths, if we make the additional assumption that they stay weakly above the x-axis. Denote the weighted Motzkin path to (n, m) by M (n, m; ω). TheMotzkin numbers have the generating function

1 ωt (1 ωt)2 4t2 M (n, 0; ω) tn = − − − − .  q2t2  n 0 X≥   102 Chapter 3. Applications

Show that this generating function is of the type (3.16), and calculate the Hankel determinant M (i + j, 0; ω) 0 i,j n 1. Show also that | | ≤ ≤ − n+1 n+1 n/2 x y M (i + j + 1, 0; ω) 0 i,j n 1 = (xy)− − | | ≤ ≤ − x y − if ω = (x + y) /√xy.

3.4 Classical Umbral Calculus

G.-C. Rota’s original intentions, when writing the “Finite Operator Calculus”, was creating a solid foundation for the Blissard Calculus [13, 1861], or Umbral Calculus, as Sylvester called it (see E.T. Bell’s “History of Blissard’s Symbolic Calculus”[11]). In Rota’snotation, an umbra α is a formal variable such that the functional E F [α]∗ is defined by linear extension of ∈ n E [α ] = an for some sequence of scalars a0 = 1, a1, a2,... in F. Here we follow Rota’sconven- tion of requiring that E α0 = 1; a point can be made for allowing E α0 = 0 [37], but we will not do so. The functional E is called the evaluation. A different n umbra β evaluates differently, E [β ] = bn (we do not distinguish between different evaluation functionals in our notation). We think of an umbral alphabet A = α, β, . . . , and require that distinct umbrae are uncorrelated, { } E cαiβjγk ... = cE αi E βj E γk ... where c F, and only a finite number of exponents i, j, k, . . . is different form ∈ 0 (c F). Two umbral polynomials are equivalent ( ), if their evaluation (first moment)∈ agrees. ' The word “uncorrelated” is a statistical term, and there a more similarities between umbrae and random variables. Suppose X is a random variable such that n all moments of X exist, E[X ] = xn, say, for all n 0. Then X “behaves” like n ≥ the umbra ξ with evaluation E [ξ ] = xn. The converse is not true, of course; for 2 example, if x2 < x1, then X would have negative variance. However, we will keep the language of random variables and their moments in this section, keeping in mind that random variables and umbrae just share a common intersection. We talk about the moment generating function mα (t) of an umbra α,

αt n n n mα = mα (t) = E e = E [α ] t /n! = ant /n!, n 0 n 0 X≥ X≥ even when α can not be interpreted as a random variable. We will reserve the symbol E for the random variable case. A cursive E means evaluation in this section. 3.4. Classical Umbral Calculus 103

We will follow the work of DiNardo and Senato [29] in this presentation. A large number of applications can be found in Gessel [37]. The connection between Finite Operator Calculus and Umbral Calculus was also investigated in Rota, Shen, and Taylor [82, 1998]. In the same way as there are sequences of i.i.d. random variables, each with the same moments, there are sequences of (uncorrelated) umbrae α, α0, α00,... all with the same moment generating function. Such umbrae are called similar, α α0 in symbols, ≡ n n α α0 α (α0) for all n 0. ≡ ⇐⇒ ' ≥ All diﬀerent and similar umbrae together make the (saturated) umbral alphabet, i j which we call again A. For two similar umbrae α and α0 holds E α (α0) = aiaj, i j i+j but E α α = E α = ai+j. Hence h i n n n n i n i n i n i E (α + α ) = E α (α ) − = E α E (α ) − 0 i 0 i 0 "i=0 # i=0 X X h i Both have the same moment sequence (an), hence n n n E (α + α0) = aian i. i − i=0 X n The subscript n is the shadow (umbra) of the superscript ! Evaluating and distin- guishing between similar umbrae are new ideas Rota brought to Umbral Calculus. n n n Blissard would have written (α + α) when he meant i=0 i aian i. We write k.α for the sum of k similar umbrae. This is diﬀerent from kα, of course,− because αkt P mkα = E e , and

(α+ +α0)t αt α0t k k mk.α (t) = E e ··· = E e E e = mα (t) = m ··· α h i h i Note the analogy to the sum of k independent copies of a random variable! It is no surprise, that k.α has the same properties as an i.i.d. sequence X1,...,Xk; we only mention j. (k.α) = (jk) .α. Of course, X1 Xk is also a random variable, n n ···n with moments E [(X1 Xk) ] = E [X1 ] E[Xk ], and we deﬁne the (auxiliary) umbrae α.k to have moments··· ···

.k n k E α = an h i for all n 0. Umbrae≥ are a strong symbolic language, but they are just in one-to-one correspondence with formal power series. The strength is the result of the many deﬁnitions made for umbrae. We will give here only the most important ones. For all n 0 we assume the existence of a unity umbra u, ≥ n t E [u ] = 1 hence mu = e , 104 Chapter 3. Applications and an augmentation ε,

n E [ε ] = δ0,n hence m = 1. We note that for any umbra γ holds

n n E [(ε + γ)n] = E εn i E γi = E [γn] . i − i=0 X Finally we need the singleton umbra χ,

n n E [χ ] = 1 for n = 0, 1 and E [χ ] = 0 else; hence ,mχ = 1 + t. We saw that

k 1 n i1 ik mk.α = mα (t) = E (tα1) E (tαk) n! i1, . . . , ik ··· n 0 i1+ +ik=n X≥ ···X h i h i = E [(k.α)n] tn/n! n 0 X≥ (see also Remark 1.1.2). Hence for n 1 ≥

n n i1 ik E [(k.α) ] = E α1 E αk i1, . . . , ik ··· i1+ +ik=n ···X k n = al1 alj (3.18) j l1, . . . , lj ··· j 1 l1+ +lj =n; lν >0 X≥ ··· X Note that in the above sum for any j the largest possible lν such that l1+ +lj = n is n j + 1. Therefore, ··· − n n E [(k.α) ] = (k j + 1)j Bn,j (a1, . . . , an j+1) − − j=1 X where 1 n Bn,j (a1, . . . , an j+1) = al1 alj . − j! l1, . . . , lj ··· l1+ +lj =n; lν >0 ··· X The functions Bn,j are called the partial Bell exponential polynomials in [73]. We see that E [(k.α)n] is a polynomial in k. Therefore, we can substitute the real number x for the positive integer k and get the deﬁnition of x.α through x mx.α (t) = mα (t) .

The moments of x.α are in F [x]. If the ﬁrst moment of α is diﬀerent from 0 then E [(x.α)n] is a polynomial of degree n. More details are discussed in Exercises 3.4.3 - 3.4.5. 3.4. Classical Umbral Calculus 105

Example 3.4.1. (In this and the following examples we write terms from the Finite Operator Calculus with a hat, like aˆn (x), to distinguish them from umbral terms.) 1 Let (ˆan) be the basic sequence for the delta operator αˆ− ( ), with corresponding n D umbra α such that E [α ] = n!ˆan (1), thus αˆ(t) mα (t) = e . n It follows that E [(x.α) /n!] =a ˆn(x) because

(x.α)t x xaˆ(t) E e = mα (t) = e . h i j j n From x = E (x.u) follows aˆn (x.u) (x.α) /n!. The polynomial aˆn (x.u) is a ' simple exampleh of ani umbral polynomial. We will do more with this type of umbral polynomials later.

Let γ be an umbra, and deﬁne the new factorial umbra γ(k) by its falling factorial moments E γ = E [γ (γ 1) (γ k + 1)] . (k) − ··· − We deﬁned

n x n E [(x.α) ] = E αi1 E αij j i1, . . . , ij ··· j 1 i1+ +ij =n; il>0 X≥ ··· X for all n > 0 as a consequence of substituting x for k in (3.18). We can take an umbra γ and substitute it for x, creating a new umbra γ.α with moment generating function tn m (t) = E [(γ.α)n] γ.α n! n 0 X≥ tn γ n = E 1 + αi1 αij  n! j i1, . . . , ij ···  n 1 j 1 i1+ +ik=n; il>0 X≥ X≥ ··· X  k  · γ m m γ k = 1 + E α t /m! = E (mα 1)  k     k −  k 1 m 1 k 0 X≥ X≥  X≥      (3.19)

γ log(1+mα(t) 1) = E e − = mγ (log mα (t)) . (3.20) h i The logarithm of mα (t) is deﬁned because mα (t) = 1+ higher order terms in t. Immediately we get γ.u γ, and from E u = 1 = δ1,n for n 1 we obtain ≡ (n) (n) ≥ n tn αt E n 0 (u.α) n! = E [e ], i.e., u acts a right and left identity in this product, ≥ h i P u.γ = γ = γ.u (3.21) 106 Chapter 3. Applications for all umbrae γ. If γ χ, the singleton umbra, then ≡ k 1 E χ = (k 1)! ( 1) − , and (3.22) (k) − − E χ tk/k! = ln (1 + t)  (k)  k 1 X≥   for k 1, thus by (3.19) ≥

n k 1 t ( 1) − .k E e(χ.α)t = E [(χ.α)n] = 1 + E − eαt 1 n!  k −  n 0 k 1 h i X≥ X≥ = 1 + log E 1 + eαt 1 = 1 + log E eαt .  − Of course we can get this much quicker from (3.20): mχ.α = mχ (log mα) = 1 + log mα. Now let α χ in γ.α, then ≡ .γ E eχt = E [(1 + t).γ ] = E [(1 + t)γ ] or

γ log(1+t) γ mγ.χ = mγ (log mχ) = mγ (log (1 + t)) = E e = E [(1 + t) ] , h i the γ-factorial umbra, (γ.χ)n γ (3.23) ' (n) n for all n 0. Especially (χ.χ) χ(n) (see (3.22)). For≥ any three umbrae α, γ,' and η the product is left-distributive, (α + η) .γ α.γ + η.γ, but in general not right-distributive, ≡

γ. (α + η) γ.α + γ.η 6≡ This follows from (3.20), because for general γ

m (t) = mγ (log mα (t) + log mη (t)) = mγ (log mα (t)) mγ (log mη (t)) γ.(α+η) 6 = mγ.α+γ.η (t) .

ct For mγ = e equality will hold, i.e., γ cu (c F). The Bell umbra β is the important≡ umbra∈ where all factorial moments are 1, E β(n) = 1 for all n 0. Hence β.χ u by (3.23). Note that the moment generating function of the≥ Bell umbra equals≡

n n n n βt t t et 1 mβ = E e = E S (n, k) β = S (n, k) = e −  n! (k) n! n 0 k=0 n 0 k=0 X≥ X X≥ X   3.4. Classical Umbral Calculus 107

(see Exercise 2.3.15 on Stirling numbers of the second kind). This shows that

t mχ.β = 1+ log mβ = e thus χ.β u β.χ (3.24) ≡ ≡ We also have

mα 1 mβ.α = mβ (log mα) = e − t mα.β = mα (log mβ) = mα e 1 . − The Bell umbra can be used to define the composition umbra of α and γ as γ.β.α. The associativity of the product needs a proof, of course (Exercise 3.4.8). Be aware of the different definition of umbral composition in section 2.3.2. We saw in (3.24) and (3.21) that

χ.β.α u.α α = α.u α.β.χ ≡ ≡ ≡ hence χ acts as left and right identity with respect to the composition. The inverse 1 1 composition umbra of α is denoted by αh− i; it satisﬁes the relation αh− i.β.α χ. ≡ 1 Lemma 3.4.2. If the inverse composition umbra of α is αh− i, then

1 1 αh− i.β.α χ α.β.αh− i χ ≡ ⇐⇒ ≡ 1 Proof. Let αh− i.β.α. Then

1 1 1 1 1 1 αh− i.β.α.β.αh− i αh− i.β.α .β.αh− i χ.β.αh− i (χ.β) .αh− i ≡ ≡ ≡ 1 1 1 u.αh− i αh− i αh− i.u, ≡ ≡ ≡ 1 by (3.24) and (3.21), hence β.α.β.αh− i u (see Exercise 3.4.9) and ≡ 1 1 α.β.αh− i χ.β.α.β.αh− i χ.u χ. ≡ ≡ ≡ 1 1 So we have α.β.αh− i χ and αh− i.β.α χ at the same time. ≡ ≡ We have 1 uh− i χ.χ (3.25) ≡ for the inverse composition umbra of u, because

1 1 1 1 χ.χ = uh− i.β.u .χ = uh− i.β.χ = uh− i.u = uh− i by (3.21) and (3.24). 108 Chapter 3. Applications

Example 3.4.3. We saw in Example 3.4.1 that the umbra x.α with moments

n E [(x.α) ] = n!ˆan(x) satisﬁes the generating function identity

x xαˆ(t) mα (t) = e .

n In other words, the umbral polynomial aˆn (x.u) is equivalent to (x.α) /n!,

n aˆn (x.u) (x.α) /n!. ' This is only possible if E [α] = 0, because aˆ1 (1) = 0 for every basic sequence (ˆan). 6 6 Because of mα (t) = 1 + E [α] t + ... we know that log mα(t) exists, and there is a formal power series f (t) of order 0 such that 1 + log mα (t) = f (t). The form of f (t) tells us that f (t) = mη (t) for some umbra η, and

E [η] = [t] log mα (t) = [t] mα (t) = 0. 6 Hence η undergoes the same restrictions as α. From

mη (t) 1 mα (t) = e −

1 n follows α β.η, the adjoint [30] of ηh− i, and therefore E [(x.β.η) ] = n!ân(x). ≡ 1 We have 1+ˆα (t) = mη (t). Note that ηh− i.α = χ. Let c = 0. If ρ is another umbra, 6 x(mη (t) 1) then cρ + x.β.η has moment generating function cmρ (t) e − . This is the 1 generating function of the Sheffer sequence (ˆsn) for αˆ− ( ), with initial values n D sˆn (0) = cE [ρ /n!]. The choice cρ ε brings us back to the umbral polynomial ≡ with moments n!ân(x). We write σx for the umbral polynomial cρ + x.α. The binomial theorem for Sheffer sequences becomes

σx+y sσx + y.α, ≡ because

n (cσ + x.α)k y.αn k sˆ (x + y) = E [σ /n!] = E E − n x+y k! (n k)! k=0 " # − n X

= sˆk(x)ˆan k (y) . − k=0 X Example 3.4.4. Remember that the composition of two basic sequences gives another basic sequence. For example, we can ﬁnd a basic sequence (ˆgn) such that x aˆn (ˆg(x)) = n (see section 2.3.2). This means that γˆ (ˆα (t)) = log (1 + t), or 1 γˆ(t) αˆ(t) γˆ (t) = log 1 +α ˆ− (t) In umbral terms, mγ = e and mα = e , where n n gˆn (x.u) (x.γ) /n! and aˆn (x.u) (x.α) /n!. We get ' ' 1 + t = mχ = mγ (log mα) = mγ.α 3.4. Classical Umbral Calculus 109

(see 3.20), hence χ γ.α. We solve this for γ by observing that ' 1 χ = αh− i.β.α = γ.α

1 t implies αh− i.β = γ. Hence mγ = m 1 (log mβ) = m 1 (e 1), or αh− i αh− i − n 1 k γ α k E = E h− i [tn] et 1 , n!  k! −  k 0 X≥   and therefore k n 1 γ αh− i S (n, k) . ' k 0 X≥ 3.4.1 The Cumulant Umbra Again we think of E [eαt] at the moment generating function of α. The moment generating function minus 1 of χ.α, mχ.α 1, is the cumulant generating function − log mα of α. Therefore, χ.α is called the α-cumulant umbra, also denoted as κα. Observe that n (χ.α) χ( λ )dλαλ ' | | λ n X` `1 `n (see Exercise 3.4.5), where λ = (λ1, λ2 ) 1 , . . . , n , and ··· ∼ n! dλ = . `1 `2 `n (1!) `1! (2!) `2! (n!) `n! ··· Furthermore, 1 .`1 2 .`1 n .`n αλ α α (a ) ≡ 1 2 ··· n where E[αi] E [α], hence ≡ `1 `n E [αλ] = a a . 1 ··· n The moments of the α-cumulant umbra κα are called the cumulants of α, n n λ 1 `j E [κ ] = ( λ 1)! ( 1)| |− dλ a , (3.26) α | | − − j λ n j=1 X` Y written in terms of the moments of α. Hence

n κα ( 1)( λ 1) dλαλ. ' − | |− λ n X` In the last expression, replace the 1 by any umbra δ, −

δ( λ 1)dλαλ. | |− λ n X` 110 Chapter 3. Applications

Lemma 3.4.5. For all umbrae α, δ and all positive integers n holds n 1 δ( λ 1)dλαλ α (α + δ.α) − . | |− ' λ n X` Proof. We have n n 1 n 1 i n i α (α + δ.α) − − α (δ.α) − ' i 1 i=1 − Xn n 1 i − α δ( λ )dλαλ. ' i 1 | | i=1 λ n i X − X` − To show why this last equivalence equals the equivalence in the Lemma, we have to write the partitions λ n i in the multiset form 1`1 , 2`2 ,... . Hence n 1 ` − α (α + δ.α) − ' n 1 .`1 i .ì+1 n 1 δ(` +` +... ) (n i)! α α − αi 1 2 − 1 ··· i ··· `1 `2 i 1 (1!) `1! (2!) `2! i=1 1`1+2`2+ =n i X X··· − − ··· .` .` +1 n i (` + 1) δ n! α1 1 αi i i (`1+`2+... ) 1 ··· i ··· `1 ì+1 ' n (1!) `1! (i!) (ì + 1)! i=1 1`1+...i(ì+1)+ =n X X ··· ··· ··· ¯ ¯ 1 .`1 i .ì δ ¯ ¯ n! α α ¯ ( 1+`1+`2+... ) 1 i i ` − ··· ··· k ik ¯ ¯ . ' (1!)`1 `¯ ! (i!)ì `¯ ! n 1`¯1+2`¯2+ =n and `¯i >0 1 i k ···X k ··· ··· X ¯ ¯ From k ikìk = i iì = n follows the Lemma. As a first application of the above Lemma we see that P P n n 1 κ α (α 1.α) − . (3.27) α ' − From u β.χ (see (3.24)) follows ≡ β.κa β.χ.α u.α α. ≡ ≡ ≡ Therefore, the moments of α can be expressed in terms of the cumulants by evaluating n n α (β.κa) β( λ )dλ (κα)λ dλ (κα)λ . ' ' | | ' λ n λ n X` X` This is the inverse relation to (3.26). Applying the same Lemma as before,

n n 1 α κα (κa + β.κα) − . (3.28) ' Remark 3.4.6. (a) The pair (3.27) and (3.28) of relationships between cumulants and moments can be extended to boolean and to free cumulants ([28]). (b) We did not dwell on the connections between Umbral Calculus and probability theory, except for motivating some of the deﬁnitions. Most of the material in section 2.2.3 also has a natural formulation in umbral terms. 3.4. Classical Umbral Calculus 111

3.4.2 Exercises n n n x 3.4.1. Verify that E [(x.u) ] = x and E [(x.χ) /n!] = n . (x.α)t x 3.4.2. Substitute x for k in (3.18) and show that E e = mα (t) . 3.4.3. The partial Bell polynomials are deﬁned as 1 n Bn,j (α1, α2, . . . , αn j+1) = ai1 ai2 aij − j! i1, . . . , ij ··· i1+ +ij =n; il>0 ··· X

We say that λ = (λ1, λ2,... ) is a partition of n, λ n, if λ1 λ2 ... , λi N, ` ≥ ≥ ∈ and i 1 λi = n. We denote by λ the number of nonzero parts of λ. If `j is the ≥ | | n n frequency of every integer occurring in λ, then j`j = n and `j = λ . P j=1 j=1 | | We also write λ as a multiset, λ = 1`1 , 2`2 ,... . Show that P P Bn,j (α1, α2, . . . , αn j+1) = − n! a`1 a`2 a`n `1 `2 `n 1 2 n (1!) `1! (2!) `2! (n!) `n! ··· 1`1+2`2+ +n`n=n ··· `1+`2+X···+`n=j ··· = dλaλ λ n, λ =j ` X| |

`1 `1 `n n! where aλ = a a a , and dλ = ` ` `n . 1 2 n (1!) 1 `1!(2!) 2 `2! (n!) `n! ··· ··· n 3.4.4. Let α be an umbra with moments E [α ] = an for all n 1. Show that ≥ n ` n (k) λ ai i E [(k.α) /n!] = | | `1! `n! i! λ n i=1 X` ··· Y n 3.4.5. Let α be an umbra with moments E [α ] = an for all n 1. The (auxiliary) umbra x.α has moments ≥

n E [(x.α) ] = (x)( λ ) dλaλ | | λ n X` (Exercise 3.4.3).Show that E [(x.α)n] is obtained by substituting x for k in E [(k.α)n]. n Verify that the degree of E [(x.α) ] is n iﬀ a1 = 0. Show that for the unity umbra u holds E [(x.u)n] = xn. 6 3.4.6. Show the following properties of x.α. 1. x.α x.γ α γ ≡ ⇐⇒ ≡ 2. x. (cα) c (x.α) for c F ≡ ∈ In F [x, y] holds 3. (x + y) .α x.α + y.α (analogue of the binomial theorem!) ≡ 112 Chapter 3. Applications

4. x. (α + γ) x.α + x.γ ≡ 5. x. (y.α) y. (x.α). ≡ 3.4.7. If E[α] = a, then E [(α a.u)n] is the n-th central moment of α. Find an expression for E [(α a.u)n] in− terms of the moments of α. − 3.4.8. Show that for three umbrae α, γ, and η holds

(α.γ) .η α. (γ.η) ≡ (associativity). 3.4.9. Let α, γ, and η be umbrae . If γ.α γ.η, then α η, and if γ.α η.α, then γ η. ≡ ≡ ≡ ≡ 3.4.10. Suppose the umbra ξ corresponds to the random variable X with moment n n generating function mξ (t), i.e., E [ξ ] = E [X ] or all n 0. Furthermore, let N be a random variable on the natural numbers, with moment≥ generating function n n mν (t), where ν is an umbra such that E [ν ] = E [N ] for all n. Show that the random sum X1 + + XN has moment generating function ν.ξ. ··· n 3.4.11. Prove Theorem 2.2.11 with Umbral Calculus. Let n!ân (1) = E [α ], the moments of an umbra α. We assume that E [α] = 1. The basic polynomials aˆn (x) have polynomial coeffi cients, aˆn,i =a ñ i(i) for some basis (ãn). Let n!ãn (1) = n n 1 n − E [˜α ]. Show that nα˜ − (χ.α) for all n 1. The umbra with n-th moment n 1 ' ≥ equal to nα˜ − is also written as α˜D. Hence α˜D χ.α. ≡ Chapter 4

Finite Operator Calculus in Several Variables

We will present the higher dimensional Finite Operator Calculus in two variables only, with a few exceptions; generalizations to a larger number of variables require a more streamlined notation, but no additional insight. An exception to this rule is the bivariate transfer formula (Lemma 4.2.12). More than two variables would give substantially more complicated results. Multivariate Finite Operator Calculus has been considered from many angles (Roman [80], Verde-Star [97], Watanabe [100],[101], and others). We restrict the discussion in this section to the minimum needed for solving recursions. In this chapter we will make use of the results on multivariate power series (section 1.3). Interesting for applications are the special constructions in several variables (chapter 5). All we can do is giving a ﬂavor of the many possibilities; only a few have been explored! 114 Chapter 4. Finite Operator Calculus in Several Variables

4.1 Polynomials and Operators in Several Variables

Let F be a field of characteristic 0. A polynomial p (u, v) F [u, v] is an expression of the form ∈ i j p (u, v) = a0,0 + a1,0u + a0,1v + + ai,ju v . ··· We define for all k, l 0 the (linear) coeffi cient functional ukvl on F [u, v] as ≥ ukvl p = a . We will also use the shorter notation [p] , especially when we do k,l k,l not really care about the variables u and v. Hence, any matrix with finitely many nonzero entries from F can be interpreted as a bivariate polynomial. The evaluation functional Evala,b is defined as Evala,b p = p (a, b), where the scalars a, b F are substituted for the formal variables u and v. A polynomial p has degree m in∈u if umvl p (u, v) = 0 for some l 0, and ukvl p (u, v) = 0 for all k > m and all l. The degree in v6 is defined correspondingly.≥ Finally, a polynomial p (u, v) has degree deg p = (m, n) if it has degree m in u and degree n in v. For instance, u5v3 + u2v7 has degree 5 in u, degree 7 in v, hence deg u5v3 + u2v7 = (5, 7). We call (bm,n) a polynomial sequence, iff deg bm,n = (m, n). Note m,n N0 that the degree of m in u∈may change if we substitute a specific value (scalar) for v. For instance, the polynomial sequence

m n + m j 1 n + m 2j m j n j bm,n (u, v) = − − − u − v − / (n + m 2j)! (4.1) j m j − j=0 X − for m, n 0 has degree m n in u if v = 0, ≥ −

deg bm,n (u, 0) = m n for m n − ≥

(more about this sequence in Example 5.2.1). A polynomial sequence (bm,n) is a m n basis of F [u, v], if bm,n (u, v) contains the term u v (Exercise 4.1.2). The sequence in (4.1) is a basis. ∂ The partial derivative operators uand v on F [u, v] are deﬁned as ∂u p (u, v) ∂ D D and ∂v p (u, v), respectively. Hence we obtain the ring of (linear) ﬁnite operators Σ u, v on F [u, v] by letting D D

G Σ u, v iﬀ G = γ ( u, v) for some γ F [[s, t]] . ∈ D D D D ∈

The isomorphism Σ u, v F [s, t] we get from D D '

GS γσ if G = γ ( u, v) and S = σ ( u, v) . ' D D D D

The operators in Σ u, v commute. We call an operator G partial, if G has a D D a power series expansion only in u, or v. The (partial) translation operators Eu b a u D b v D and Ev are deﬁned as e D and e D , respectively. A linear operator T on F [u, v] a b a b is translation invariant iﬀ TEuEv = EuE2T for all a, b F. As expected, we have the following important characterization of translation∈ invariant operators. 4.1. Polynomials and Operators in Several Variables 115

Lemma 4.1.1. A linear operator T on F [u, v] is translation invariant iff T ∈ Σ u, v . In that case, D D i j u v i j T = Eval0,0 T | i!j! DuDv i,j 0 X≥ The proof is left as Exercise 4.1.3. Similar to multi-series, we have multi-operators (T1,T2); a delta pair (B1,B2) Σ2 is multi-operator such that B = γ ( , ), and B = γ ( , ), where∈ u, v 1 1 u v 2 2 u v D D D D D D (γ1, γ2) is a delta multi-series (see section 1.3). Note that we denote pairs of operators by (T1,T2) or T1,T2, but write Fu and Fv for the partial operators, acting on u and on v only. We will make that distinction throughout this chapter, except under the (rare) circumstances when we are dealing with more than two variables. For example, when our polynomials are in F [x1, x2, x3], we are writing ρ for the partial derivatives (ρ = 1, 2, 3). D By Lemma 4.1.1 delta pairs consist of translation invariant operators. The following Lemma is useful and easy to prove. m n Lemma 4.1.2. If (β1, β2) is a delta multi-series and m, n 0 then [β1 β2 ]k,l = 0 ≥ m n 6 m n implies k m and l n. Furthermore, [β β ] = [β1] [β2] = 0. ≥ ≥ 1 2 m,n 1,0 0,1 6 Finally, we define a (bivariate) Sheffer sequence (sm,n) as a basis, i.e., deg sm,n = (m, n) and [sm,n] = 0, solving the system m,n 6

B1sn,m (u, v) = sn 1,m (u, v) and B2sn,m (u, v) = sn,m 1 (u, v) − − for all m, n 0, where (B1,B2) is a delta pair (sn,m (u, v) = 0 for all m < 0 or ≥ n < 0). The Sheffer sequence (bn,m) with initial values bn,m (0, 0) = δm,0δn,0 is the basic sequence for (B1,B2). The standard example is of course the basic se- m n m n quence (u v / (m!n!)) for ( u, v). Note that the coeffi cient of u v in a Sheffer D D m n polynomial sm,n (u, v) is never 0. This excludes lots of ‘easy’cases like u + v from the Sheffer family! The quest for solutions to initial value problems needs the observation that in the intersection of the kernel of B1 and the kernel of B2 only the constant polynomials survive; thus we can prescribe a value sm,n (um,n, vm,n) for all m, n 0. In Exercise 4.3.3 you are asked to find the Sheffer sequence for ≥ ( u, v) with initial values sm,n (m + n, m n) = 1/ (m + n)!. D D − In general, the operators B1 and B2 in a delta pair (B1,B2) act both on both variables u and v; they are usually not the partial operators acting only on one variable, as it is the case for u and v in the (special) delta pair ( u, v). In u D D u D 1D addition, we write ∆u = eD I = Eu I, and u = I e−D = I Eu− (same for v) for the partial difference− operators.− Furthermore,∇ taking− any two− univariate delta operators A and B on F [x], we can make them into the delta pair (A, B), and A will be the partial operator with respect to u, B for v. The basic sequence factors in this case, (am (u) bn (v)). Of course, this is the case we are not really interested in, but it is often the starting point for getting to more useful constructions. 116 Chapter 4. Finite Operator Calculus in Several Variables

The transforms we introduced in section 2.2.2 can be generalized to the multivariate case, and the multivariate analogue of Theorem 2.2.4 follows in the same way. This proves the following theorem.

Theorem 4.1.3. The basic sequence (bm,n) has generating function m n uβ1(s,t)+vβ2(s,t) 1 m,n 0 bm,n (u, v) s t = e iff β1− ( u, v): bm,n bm 1,n and 1 ≥ D D 7→ − β2− ( u, v): bm,n bm,n 1 for all m, n 0. P D D 7→ − ≥ In Exercise 4.1.6 we verify that the basic sequence as defined in the Theorem above really is a basis of the vector space of polynomials in two variables. A bivariate Sheffer sequence (sm,n) has generating function

m n uβ1(s,t)+vβ2(s,t) sm,n (u, v) s t = σ (s, t) e m,n 0 X≥ for some unit σ (s, t) F [[s, t]] (note that F [[s, t]] is not a ﬁeld; if σ (0, 0) = 0 then ∈ 6 σ is a unit in F [[s, t]].). The generating function shows that m n sm,n (u + u0, v + v0) = si,j (u, v) bm i,n j (u0, v0) , (4.2) − − i=0 j=0 X X the multivariate version of the binomial theorem. Example 4.1.4. To show the general structure of this example, we need at least three variables u, v, and w, say. The reader should have no problem generalizing the above setting to the trivariate case. Suppose that for ρ = 1, 2, 3 we have a n univariate basic sequence (aρ,n) with generating function aρ,n (x) t = n 0 n N0 x ≥ ∈ γρ (t) where γρ (t) is a univariate power series such that γρ (0) = 1, i.e., γρ (t) = αρ(t) P e for some delta series αρ. Deﬁne

bl,m,n (u, v, w) = a1,l (u) a2,m (v + l) a3,n (w + l + m) and

ω3 (r, s, t) = γ3 (t)

ω2 (r, s, t) = γ2 (sω3 (r, s, t)) = γ2 (sγ3 (t))

ω1 (r, s, t) = γ1 (rω2 (r, s, t) ω3 (r, s, t)) = γ1 (rγ2 (sγ3 (t)) γ3 (t)) . u v w The polynomial sequence (bl,m,n) has the generating function ω1 ω2 ω3 , because l m n l,m,n 0 bl,m,n (u, v, w) r s t = ≥ l m n P a1,l (u) a2,m (v + l) a3,n (w + l + m) r s t l 0 m 0 n 0 X≥ X≥ X≥ w l m = ω3 a1,l (u) a2,m (v + l)(rω3) (sω3) l 0 m 0 X≥ X≥ w l v+l w v l = ω3 a1,l (u)(rω3) γ2 (sω3) = ω3 ω2 a1,l (u)(rω3ω2) l 0 l 0 X≥ X≥ w v u uβ1(r,s,t)+vβ2(r,s,t)+wβ3(r,s,t) = ω3 ω2 ω1 = e 4.1. Polynomials and Operators in Several Variables 117

where βρ (r, s, t) = ln ωρ (r, s, t). We verify (without deriving it!) that 1 1 s t 1 1 t 1 1 β1− (r, s, t) = α1− (r) e− − , β2− (r, s, t) = α2− (s) e− , and β3− (r, s, t) = α3− (t). We get

1 1 1 1 1 1 1 β3 β1− , β2− , β3− = ln ω3 β1− , β2− , β3− = α3 α3− (t) = t 1 1 1 1 1 γ3 α3− (t) 1 β β− , β− , β− = α α− (s) = α α− (s) = s 2 1 2 3 2 2 et 2 2 ! 1 t 1 1 1 1 1 1 γ2 α2− (s) e− γ3 α3− (t) γ3 α3− (t) β β− , β− , β− = α α− (r) 1 1 2 3 1 1 es et ! 1 1 γ2 α2− (s) 1 = α α− (r) = α α− (r) = r. 1 1 es 3 1 ! 1 We also check that βρ− ( 1, 2, 3), ρ = 1, 2, 3, really is the corresponding delta 1 D D D triple; β− ( 1, 2, 3) bl,m,n (u, v, w) ρ D D D

1 ρ+1 3 = α− ( ρ) e−D −···−D bl,m,n (u, v, w) ρ D 1 1 1 = α− ( ρ) E− E− bl,m,n (u, v, w) ρ D ρ+1 ··· 3 hence we get for ρ = 1

1 1 1 α− ( u) E− E− bl,m,n (u, v, w) 1 D 2 3 = a1,l 1 (u) a2,m (v + l 1) a3,n (w + l 1 + m) − − − = bl 1,m,n (u, v, w) , − and similar for the other values of ρ. We will discuss this setting again in Example 4.2.10, in a slightly more general way. See Exercise 4.1.1 for an application to u 1 + l v 1 + l + m w 1 + l + m + n b (u, v, w) = − − − . (4.3) l,m,n l m n 4.1.1 Exercises

4.1.1. Apply Example 4.1.4 to ﬁnd the polynomials (bl,m,n (u, v, w)) generated by x+n 1 aρ,n (x) = n− for ρ = 1, 2, 3. Find the delta triple (B1,B2,B3) for (bl,m,n). 4.1.2. Show that every p (u, v) F [u, v] can be uniquely written in terms of a basis. ∈ 4.1.3. Prove Lemma 4.1.1.

4.1.4. Show that [sm,n] = 0 (m, n 0) for any Sheﬀer sequence (sm,n). m,n 6 ≥ 4.1.5. Show that the basic sequence (bm,n (u, v)) can be reconstructed from the two univariate sequences (bm,n (u, 0))m,n 0 and (bm,n (0, v))m,n 0. Is (bm,n (u, 0))m 0 for ﬁxed n a univariate basic sequence?≥ ≥ ≥ 118 Chapter 4. Finite Operator Calculus in Several Variables

4.1.6. Show that the coeﬃ cient of smtn in euβ1(s,t)+vβ2(s,t) equals

m n m n k ul vk − βl βk l! k! 1 i,j 2 m i,n j l=0 k=0 i=l j=0 − − X X X X and that this coeﬃ cient has degree (m, n). Hence the sequence (bm,n) in Theorem 4.1.3 is a basis in F [u, v]. 4.2. The Multivariate Transfer Formulas 119

4.2 The Multivariate Transfer Formulas

The transfer formulas show us how to express a basic sequence in terms of another basic sequence, when a relation (the transfer) between the corresponding delta pairs of operators is known. This is not so straightforward in the multivariate setting as it was in the one dimensional case. Suppose the delta pair (A1,A2) with known basic sequence (am,n) can be written as a multi-series in (B1,B2), Aρ = τρ (B1,B2). We want a way to express (bm,n) in terms of (am,n) using only the coeffi cients of (τ1, τ2). Now all depends on the coeffi cients: If they are in the same ring F as the coeffi cients of the polynomials, the result follows from Lagrange-Good 2 inversion, and is stated in Lemma 4.2.1. In this case we write (τ1, τ2) F [[s, t]] . This simple case can be seen as a Corollary of the more general transfer∈ formula; however, we want to emphasize its importance and show it independently in the first subsection. An elementary example is A1 = Ev∆u, A2 = ∆v and τ1 (s, t) = s, α τ2 (s, t) = (t t ) / (1 t) (Example 4.2.2). − − If the coeffi cients of (τ1, τ2) are translation invariant operators themselves, as it is so often the case in applications, a lot more work has to be done. In 2 this case we write (τ1, τ2) Σ u, v [[s, t]] . For example, τ1 (s, t) = Evs and 2 ∈ D D τ2 (s, t) = EuEvt t (Example 4.2.14). We consider such transfer functions in the last subsection.− In between, there is the operator based transfer formula, when

Aρ = VρBρ, where Vρ is invertible in Σ u, v . This can be seen as the “extreme D D case”of having operators as coeﬃ cients, τ1 (s, t) = V1s and τ2 (s, t) = V2t.

4.2.1 Transfer with constant coeﬃ cients

Suppose Aρ = τρ (B1,B2) and τρ F [[s, t]]; power series and polynomials have ∈ 1 coeffi cients from the infinite field F. We can write Bρ = τρ− (A1,A2), hence Bρ 1 1 is also a delta operator. From Bρ = βρ− ( u, v) and Aρ = αρ− ( u, v), say, 1 1 1 1 D D D D follows τρ− α1− , α2− = βρ− . Therefore

m n uβ1(s,t)+vβ2(s,t) uα1(τ1,τ2)+vα2(τ1,τ2) bm,n (u, v) s t = e = e m,n 0 X≥ k l = ak,l (u, v) τ1 (s, t) τ2 (s, t) . k,l 0 X≥ We have proven the following lemma, the bivariate version of Exercise 2.4.8.

Lemma 4.2.1. If the delta pair (A1,A2) with basic sequence (am,n) can be written as 2 a delta multi-series (τ1, τ2) F [[s, t]] in the linear operators B1,B2, so that Aρ = ∈ τρ (B1,B2) ΣB1,B2 , then (B1,B2) is also a delta pair, and its basic sequence has the generating∈ function

m n uα1(τ1,τ2)+vα2(τ1,τ2) bm,n (u, v) s t = e (4.4) m,n 0 X≥ 120 Chapter 4. Finite Operator Calculus in Several Variables and the expansion m n k l bm,n (u, v) = τ1 τ2 m,n ak,l (u, v) . k=0 l=0 X X Note that the above Transfer Formula only holds if the multi-series τ1, τ2 connecting (A1,A2) and (B1,B2) has coeffi cients free of operators; we must have 2 (τ1, τ2) F [[s, t]] . Similar to Corollary 2.4.4 in the univariate case, we have ∈ d special cases where the generating function (4.4) simplifies. For example, if EuEv = d Ev + τ1 (B1,B2) for some d F, and Ev = 1 + τ2 (B1,B2), then ∈ u m n τ1 (s, t) v bm,n (u, v) s t = 1 + d (1 + τ2 (s, t)) (4.5) m,n 0 (1 + τ2 (s, t)) ! X≥ (Exercise 4.2.2). Such cases can arise when α1 and α2 are of a very simple form. Example 4.2.2. Suppose a B2 B2 (Ev∆u, ∆v) = B1, − , 1 B2 − α hence τ1 (s, t) = s and τ2 (s, t) = (t t ) / (1 t) in Lemma 4.2.1. The delta − u− v m operator (E2∆u, ∆v) has the basic sequence m −n m,n 0; this can either be checked directly or has to wait until Example 4.2.10. We obtain≥ from Lemma 4.2.1 m n u v k b (u, v) = τ kτ l − m,n 1 2 m,n k l k=0 l=0 X X u n v m t ta l = − [tn] − . m l 1 t l=0 X − The coeffi cient of tn in ((t tα) / (1 t))l has been written as l by Euler − − n l α 1 [32], − − (n l)/(a 1) l b − − c l n (a 1) j 1 = ( 1)j − − − n l − j l 1 α 1 j=0 − − X − for l 1. In this notation, ≥ u n v m l b (u, v) = − m,n m l n l l=0 α 1 X − − We can also arrive at the generating function of this basic sequence via formula (4.5), letting d = 1: 1 t u 1 tα v b (u, v) smtn = 1 + s − − . m,n 1 tα 1 t m,n 0 X≥ − − 4.2. The Multivariate Transfer Formulas 121

4.2.2 Operator based transfer

Suppose T = φ ( u, v) is an operator in Σ u, v . We define the partial derivatives ∂ D D ∂ D D ∂ of T as T = φs ( u, v) and T = φt ( u, v), where φs = φ (s, t) and ∂ u ∂ v ∂s ∂ D D D D D D φt = ∂t φ (s, t), the partial derivatives defined in section 1.3. Similar, the Jacobian determinant of T1 = φ1 ( u, v) and T2 = φ2 ( u, v) is defined as D D D D ∂ ∂ ∂(φ , φ ) φ1 ( u, v) φ2 ( u, v) 1 2 ∂ u ∂ u (φ1, φ2) = = D∂ D D D∂ D D |J | ∂ ( u, v) φ1 ( u, t) φ2 ( u, v) ∂ v D ∂ v D D D D D D ∂ ∂ ∂ ∂ = φ1 φ2 φ2 φ1 . ∂ u ∂ v − ∂ v ∂ v D D D D

If (T1,T2) is in Σ u, v , then its derivative (Jacobian determinant) will be in D D ∂ ∂ Σ , . Note that the s- and t-derivative are now denoted by and , re- u v ∂s ∂t spectively.D D This slightly cumbersome notation has the advantage that it is clear ∂ what happens under the above isomorphism: If Tρ = τρ ( u, v) then T1 is ∂ v ∂ D D D isomorphic to ∂t τ1 (s, t), for example. The choice of coeﬃ cient ring is very important: For a simple case consider

∂( u, v ) u v ∇ ∇ . If u = τ1 ( u, v) = 1 e , and v = τ2 ( u, v) = 1 e , ∂( u, v ) −D −D D D ∇ D D − ∇ D D − thus τρ Q [[s, t]], then ∈

u v v u ∂ 1 e−D ∂ 1 e−D ∂ 1 e−D ∂ 1 e−D ( u, v) = − − − − |J ∇ ∇ | ∂ u ∂ v − ∂ u ∂ v D D D D u v 1 1 = e−D e−D 0 = E− E− . − u v 0 0 If however u = ι1 ( u, v) = u , and v = ι2 ( u, v) = v , hence ∇ D D ∇ Du ∇ D D ∇ Dv τρ Σ u, v [[s, t]], then ( u, v) = 0. Note that in the ﬁrst version (τ1, τ2) is ∈ D D |J ∇ ∇ | a delta pair. In the second case, (ι1, ι2) is of order 0.

Example 4.2.3. We consider another example from Σ u, v [[s, t]]. Let γ1 (s, t) = D D as/ (1 bt) and γ2 (s, t) = at/ (1 bs) be the pair from Example 1.3.4. This pair has the− inverses −

1 1 1 1 a− bt 1 1 a− bs β (s, t) = a− s − and β (s, t) = a− t − . 1 1 a 2b2st 2 1 a 2b2st − − − −

We saw that b does not have to be invertible; if we take Σ u, v as the coeﬃ cient D D ring, we can choose a = I and b = u, obtaining the inverses D

1 ut β1 (s, t) = s − D Σ u, v [[s, t]] and 1 2 st ∈ D D − Du 1 us β2 (s, t) = t − D Σ u, v [[s, t]] . 1 2 st ∈ D D − Du 122 Chapter 4. Finite Operator Calculus in Several Variables

Hence the delta pair γ1 ( u, v) = u/ (1 u v) Σ u, v [[s, t]] and γ2 ( u, v) = 2 D D D − D D ∈ D D D D v/ 1 u Σ u, v [[s, t]] has the inverses D − D ∈ D D 2 2 u u v v u v β1 ( u, v) = D − D3 D and β2 ( u, v) = D − D3 D . D D 1 v D D 1 v − DuD − DuD

As written above, we do not distinguish between u as “formal variable” and u as coeﬃ cient. This will be a mistake when we try toD verify the inverses, because notD every occurrence u in γ1 ( u, v) can be replaced by β1.Let us write γ1 ( u, v) = D D D D D s/ (1 t)| and γ ( , ) = t/ (1 s)| . Thus we u (s,t)=( u, v ) 2 u v u (s,t)=( u, v ) − D | D D D D − D | D D must calculate γ1 (β1, β2) = β1/ (1 uβ2) = u and γ2 (β1, β2) = β2/ (1 uβ1) = − D D − D v. D The following Lemma shows the easiest case of an operator equation Aρ =

τρ (B1,B2) where τρ Σ 1, 2 [[s, t]] is a delta multiseries with operator coeﬃ - ∈ D D cients, namely τρ (B1,B2) = VρBρ. Of course, Vρ is invertible.

Lemma 4.2.4. Let (A1,A2) and (B1,B2) be delta pairs, with basic sequences (am,n) and (bm,n), respectively. If Aρ = VρBρ for ρ = 1, 2, then

∂ (B ,B ) b (u, v) = V m+1V n+1 1 2 a (u, v) . (4.6) m,n 1 2 ∂ (A ,A ) m,n 1 2

Proof. Because (A1,A2) and (B1,B2) are both delta pairs, there must exist a delta 2 pair of multiseries (τ1, τ2) F [[s, t]] such that Aρ = τρ (B1,B2). Therefore ∈ m n m n m k n l k l bm,n (u, v) s t = τ1 − τ2 − m,n A1 A2am,n (u, v) (4.7) m,n 0 k=0 l=0 X≥ X X 1 1 as in Lemma 4.2.1. From Bρ = τρ− (A1,A2) = Vρ− Aρ we see that Vρ = υρ (Au,Av) 1 1 if we deﬁne τ1− (s, t) = s/υ1 (s, t) and τ2− (s, t) = t/υ2 (s, t). By Lagrange-Good inversion 1 1 m k n l m+1 n+1 ∂ τ1− , τ2− τ1 − τ2 − m,n = υ1 υ2 . " ∂ (s, t) # k,l

Hence the operator on the right hand side of (4.7) equals

1 1 m+1 n+1 ∂ τ1− , τ2− m+1 n+1 ∂ (B1,B2) υ1 (A1,A2) υ2 (A1,A2) = V1 V2 . ∂ (A1,A2) ∂ (A1,A2)

We will rephrase this lemma with the help of the Pincherle derivative, albeit the bivariate case only. Examples are postponed until section 4.2.3. 4.2. The Multivariate Transfer Formulas 123

4.2.3 The multivariate Pincherle derivative The umbral shift is the key to the Pincherle derivative. We deﬁne the bivariate umbral shifts for the delta pair A1,A2 as

θA : am,n (u, v) (m + 1) am+1,n (u, v) 1 7→ θA : am,n (u, v) (n + 1) am,n+1 (u, v) . 2 7→

Note that θA1 A2 = A2θA1 , and

j j 1 θA A am,n (u, v) = (m j + 1) A − am,n 1 1 − 1 for j > 0. The umbral shifts are not translation invariant! From ∂ ∂ θ a (u, v) smtn = a (u, v) smtn = euα1(s,t)+vα2(s,t) A1 m,n ∂s m,n ∂s m,n 0 m,n 0 X≥ X≥ ˆ ∂ m n (which means that θA1 = ∂s ) follows θA1 m,n 0 am,n (u, v) s t = ≥ ∂ ∂ P u α (s, t) + v α (s, t) a (u, v) smtn ∂s 1 ∂s 2 m,n m,n 0 X≥ ∂ ∂ m n = u α1(A1,A2) + v α2 (A1,A2) am,n (u, v) s t ∂A1 ∂A1 m,n 0 X≥ ∂ u ∂ v m n = u D + v D am,n (u, v) s t . ∂A1 ∂A1 m,n 0 X≥ Hence

∂ u ∂ v θA1 = u D + v D and (4.8) ∂A1 ∂A1 ∂ u ∂ v θA2 = u D + v D . ∂A2 ∂A2

∂ ∂ ∂ ∂ This is the multivariate form of Proposition 2.4.6. It follows from ∂s ∂t = ∂t ∂s that

θA1 θA2 = θA2 θA1 . (4.9)

This result follows also from the observation that θA1 θA2 am,n (u, v) = θA2 θA1 am,n (u, v) for all natural numbers m and n. Note that when A1 is a function of u only, then D u is a function of A1 only, but v can be a power series in A1 and A2, thus D D ∂ v θA2 = v D (4.10) ∂A2 in this special case. We demonstrate this behavior in the following example. 124 Chapter 4. Finite Operator Calculus in Several Variables

1 Example 4.2.5. Let A1 = u, and A2 = E− v. For ﬁnding the umbral shifts with ∇ u ∇ respect to A1and A2 we need to express u and v in terms of A1 and A2. We u D D v have A1 = I e−D , hence u = ln (I A1). From A2 = (I A1) I e−D − D − − − − follows v = ln (I A2/ (I A1)). Hence D − − − 2 ∂ u ∂ v I A2/ (I A1) θA1 = u D + v D = u + v − = uEu + vEuEv v ∂A1 ∂A1 1 A1 I A2/ (I A1) ∇ − − − ∂ (I A1) ∂ ln (I A2/ (I A1)) θA2 = u − v − − = vEuEv − ∂A2 − ∂A2

We check the commutation rule (4.9): θA1 θA2

2 = ((u v) Eu + vEuEv) vEuEv = v (u v) E Ev + vEuEvvEuEv − − u = vEuEv (u + v) Eu vEuEvvEuEv = vEuEv ((u + v) Eu vEuEv) − − = θA2 θA1 .

m+n 1+u n 1+v The basic sequence for this delta pair A1,A2 is am,n (u, v) = m− −n , and we check that θA am,n (u, v) 1 m + n 1 + u n 1 + v = (uEu + vEvEu v) − − ∇ m n m + n + u n 1 + v n 1 + v m + n + u n 1 + v = u − + v − = (u + n) − m n n 1 m n 1 − − m + n + u n 1 + v = (m + 1) − = (m + 1) a (u, v) m + 1 n m+1,n and θA2 am,n (u, v)

m + n 1 + u n 1 + v m + n + u n + v = vE E − − = v v u m n m n = (n + 1) am,n+1 (u, v) .

As in the univariate case, the Pincherle derivative with respect to Aρ is deﬁned with the help of the umbral shifts.

Deﬁnition 4.2.6. Let T be a linear operator on F [u, v]. The Pincherle derivatives of T with respect to the delta pair A1,A2 are the operators

T 0 = T θA θA T and A1 1 − 1 T 0 = T θA θA T. A2 2 − 2 Of course we want to know when T = ∂T . This is answered in the following A0 ρ ∂Aρ Lemma. 4.2. The Multivariate Transfer Formulas 125

Lemma 4.2.7. Let T be a translation invariant operator on F [u, v], and A1,A2 a delta pair. Then ∂T TA0 ρ = . ∂Aρ and T is translation invariant. A0 ρ Proof.

∂ u ∂ v ∂ u ∂ v T 0 = T u D + v D u D + v D T Aρ ∂A ∂A − ∂A ∂A ρ ρ ρ ρ ∂ u ∂ v ∂ u ∂ v = (T u uT ) D + (T v vT ) D = Tθ0u D + Tθ0v D − ∂Aρ − ∂Aρ ∂Aρ ∂Aρ ∂T ∂ u ∂T ∂ v ∂T = D + D = ∂ u ∂Aρ ∂ v ∂Aρ ∂Aρ D D where θu and θv are the univariate umbral shifts deﬁned in section 2.4.1,

umvn umvn θ = u u m!n! m!n!

(θu is a partial operator on u; the other variables are treated a constants). If the delta pair A1,A2 is expressed in terms of the delta pair B1,B2, then

∂A1 ∂A2 TB0 ρ = TA0 1 + TA0 2 . ∂Bρ ∂Bρ

Special results for the bivariate case We can rephrase the Jacobian determinant of translation invariant operators in terms of umbral shifts. In

∂ (R,S) = (RθA θA R)(SθA θA S) (RθA θA R)(SθA θA S) ∂ (A ,A ) 1 − 1 2 − 2 − 2 − 2 1 − 1 1 2

the commuting factors on the right hand side can be written in four diﬀerent orders, resulting in four diﬀerent expansions. We give one version in the following lemma.

Lemma 4.2.8. For every delta pair A1,A2 and every pair of translation invariant operators R and S holds ∂(R,S) = ∂(A1,A2)

RθA1 SθA2 θA1 RSθA2 + θA1 Rθ A2 S RθA2 SθA1 + θA2 RSθA1 θA2 RθA1 S. − − − This can be further simpliﬁed with the help of the Pincherle derivatives. 126 Chapter 4. Finite Operator Calculus in Several Variables

Proof. ∂(R,S) = ∂(A1,A2)

(RθA θA R)( SθA θA S) (RθA θA R)(SθA θA S) 1 − 1 2 − 2 − 2 − 2 1 − 1 = RθA SθA θA RSθA + θA RθA S RθA SθA + θA RSθA θA RθA S 1 2 − 1 2 1 2 − 2 1 2 1 − 2 1 because θA1 θA2 = θA2 θA1 .

If we exchange RθA1 θA1 R with SθA2 θA2 S, and RθA2 θA2 R with − ∂(R,S)− − SθA θA S in the above proof, we get = 1 − 1 ∂(A1,A2)

SθA RθA θA SRθA +θA SθA R Sθ A RθA +θA SRθA θA SθA R (4.11) 2 1 − 2 1 2 1 − 1 2 1 2 − 1 2 We can now rephrase Lemma 4.2.4.

Theorem 4.2.9. Let (A1,A2) and (B1,B2) be delta pairs, with basic sequences (am,n) and (bm,n), respectively. If Aρ = VρBρ for ρ = 1, 2, then for m, n 1 then bivariate transfer formula holds, ≥

1 m n 1 m m n bm,n (u, v) = θA2 V1 V2 am,n 1 + (θA1 V1 θA2 θA2 V1 θA1 ) V2 am 1,n 1. n − mn − − −

Proof. We know from Lemma 4.2.4 and Exercise 4.2.3 that bm,n (u, v) =

∂ (B ,B ) V m+1V n+1 1 2 1 2 ∂ (A ,A ) 1 2 n m m n m n 1 m ∂V 2 1 n ∂V1 1 ∂ (V1 ,V2 ) = V V V A2 V A1 + A1A2 . 1 2 − n 1 ∂A − m 2 ∂A mn ∂ (A ,A ) 2 1 1 2

m n In this proof we will write R for V1 and S for V2 , slightly shortening the long expressions. Replacing the partial derivatives by the Pincherle derivatives we get

∂ (B ,B ) V m+1V n+1 1 2 a 1 2 ∂ (A ,A ) m,n 1 2

1 1 = RS S (RθA θA R) A1 R (SθA θA S) A2 am,n − m 1 − 1 − n 2 − 2 1 ∂ (R,S) + A A a mn ∂ (A ,A ) 1 2 m,n 1 2 1 1

= SθA1 Ram 1,n RSam,n + RθA2 Sam,n 1 m − − n − 1 ∂ (R,S) + am 1,n 1. mn ∂ (A ,A ) − − 1 2

4.2. The Multivariate Transfer Formulas 127

Now apply Lemma 4.2.8 to expand ∂ (R,S) /∂ (A1,A2) . Thus bm,n (u, v) = | | 1 1 SθA1 Ram 1,n (u, v) + RθA2 Sam,n 1 (u, v) RSam,n (u, v) m − n − − 1 + ( θA1 RSθA2 RθA2 SθA1 + RθA1 SθA2 ) am 1,n 1 (u, v) mn − − − − 1 + (θA2 RSθA1 + θA1 RθA2 S θA2 RθA1 S) am 1,n 1 (u, v) mn − − − 1 1 = (SθA1 θA1 S) Ram 1,n + RθA1 Sam 1,n RSam,n m − − m − − 1 1 + θA2 RSam,n 1 + (θA1 RθA2 θA2 RθA1 ) Sam 1,n 1. n − mn − − − The ﬁrst line of the last two-line equation equals 0, because of translation invariance of the Pincherle derivatives. Note that formula (4.6) contains 2 terms (from the 2 2 determinant), each containing 2 partial derivatives. Replacing them by Pincherle× derivatives gives us 8 terms. Theorem 4.2.9 reduces the 8 terms to 3. In the case of three dimensions, the 3 3 determinant will give us 6 terms, and the Pincherle derivatives will expand them× by a factor of 23, giving 48 terms. They can be reduced to 13 terms.

c f Example 4.2.10. Let B1 = EvA1, and B2 = Eu A2, where the delta pair (A1,A2) c f has the basic polynomials am,n (u, v). We ﬁnd V1 = Ev− and V2 = Eu− . We are looking for the basic sequence for (B1,B2). If we apply Theorem 4.2.9 we get

1 cm fn 1 cm cm fn bm,n (u, v) = θA2 Ev− Eu− am,n 1 + θA1 Ev− θA2 θA2 Ev− θA1 Eu− am 1,n 1 n − mn − − − 1 cm fn cu ∂ ( u, v) cm fn = θA2 Ev− Eu− am,n 1 D D Ev− Eu− am 1,n 1 n − − n ∂ (A ,A ) − − 1 2

∂ v ∂ u If A2 has an expansion in v only, then D = 0, and θA = u D (see (4.10)), ∂A1 1 ∂A1 hence D 1 ∂ v 1 cm fn bm,n (u, v) = θA2 cmu D Ev− Eu− am,n 1 n − ∂A2 u fn − − If, in addition, A1 is power series in u only, meaning that A1 and A2 are partial D operators, then am,n (u, v) = pm (u) qn (v), where A1pm = pm 1 and A2qn = qn 1. In this case, − −

pm (u fn) qn (v cm) bm,n (u, v) = (vu vfn cum) − − . − − u fn v cm − −

For example, if A1 = u and A2 = v then D D m 1 n 1 (u fn) − (v m) − bm,n (u, v) = (uv vfn ucm) − − − − m!n! 128 Chapter 4. Finite Operator Calculus in Several Variables

If A1 = ∆u and A2 = ∆v, then

vu fnv cmu u fn v dm b (u, v) = − − − − . m,n (u fn)(v cm) m n − − fn cm The sequence sm,n := Eu Ev bm,n is a Sheﬀer sequence for (A1,A2), because

fn cm fn c(m 1) c fn c(m 1) A1Eu Ev bm,n = Eu Ev − EvA1bm,n = Eu Ev − B1bm,n = sm 1,n, − and the same holds for A2. If we choose Aρ = ∆ρ then

uv fndm u v s (u, v) = − m,n uv m n with generating function

m n u v u 1 v 1 sm,n (u, v) s t = (1 + s) (1 + t) fdst (1 + s) − (1 + t) − − m,n 0 X≥ u 1 v 1 = (1 + s) − (1 + t) − (1 + t + s + st (1 fd)) . −

4.2.4 Transfer with operator coeﬃ cients

In Theorem 4.2.9 we found a way to express the basic sequence (bm,n) in terms of (am,n), but the operators Aρ and Bρ were connected by an invertible operator Vρ, not by a power series τρ. The following Corollary holds for any number of variables. However, the conclusion we draw from if, the second Transfer Formula, is based on special properties of a 2 2 determinant, and therefore only holds for the bivariate case. ×

Corollary 4.2.11. Let (A1,A2) and (B1,B2) be delta pairs, with basic sequences (an) and (bn), respectively, and Aρ = τρ (B1,B2) for ρ = 1, 2, where (τ1, τ2) 2 ∈ Σ u, v [[s, t]] is a delta multi-series with translation invariant operator coeﬃ - D D cients. If Aρ = Bρυρ (A1,A2), then

m n m 1 i n 1 j ∂ (τ1, τ2) i j υ (A ,A ) υ (A ,A ) = τ − − τ − − A A 1 1 2 2 1 2 1 2 ∂ (s, t) 1 2 i 0 j 0 m 1,n 1 X≥ X≥ − − i+1 m j+1 n ∂ (τ1, τ2) i j = ε − ε − A A 1 2 ∂ (s, t) 1 2 i 0 j 0 i,j X≥ X≥

where ε1 (s, t) = s/τ1 (s, t) and ε2 (s, t) = t/τ2 (s, t).

Proof. From Bρ = Aρ/υρ (A1,A2) we see that (s/υ1 (s, t) , t/υ2 (s, t)) equals the (delta multi-series) inverse of (τ1, τ2). Therefore we can express the coeﬃ cients of 4.2. The Multivariate Transfer Formulas 129

m n υ1 υ2 in terms of the coeﬃ cient of the inverse (τ1, τ2) by applying the Lagrange- Good inversion formula (1.13),

∂τ1 ∂τ2 m n i+1 m j+1 n − ∂s ∂s [υ1 υ2 ]i,j = ε1 − ε2 ∂τ1 ∂τ2 ∂t ∂t i,j

letting ε1 (s, t) = s/τ1 (s, t) and ε2 (s, t) = t/τ2 (s, t).

Special results for the bivariate case The second transfer formula applies to the situation of Lemma 4.2.1, but with connecting bivariate series (τ1, τ2) Σ u, v [[t]] that may contain translation invariant operators among its coeﬃ cients.∈ D D

Lemma 4.2.12. Let (A1,A2) and (B1,B2) be delta pairs, with basic sequences (am,n) and (bm,n), respectively. If Aρ = τρ (B1,B2) for ρ = 1, 2, where

2 (τ1, τ2) Σ u, v [[s, t]] ∈ D D is a bivariate delta series with translation invariant coeﬃ cients, then bm,n (u, v)

1 m n 1 m m n = θA2 υ1 υ2 am,n 1 + (θA1 υ1 θA2 θA2 υ1 θA1 ) υ2 am 1,n 1. n − mn − − − m n where υ1 υ2 is the power series in A1 and A2 given in Corollary 4.2.11, Aρ = Bρυρ (A1,A2), and

m n m 1 i n 1 j ∂ (τ1, τ2) i j υ υ (A ,A ) = τ − − τ − − A A 1 2 1 2 1 2 ∂ (s, t) 1 2 i,j 0 m 1,n 1 X≥ − − i+1 m j+1 n ∂ (τ1, τ2) i j = ε − ε − A A 1 2 ∂ (s, t) 1 2 i,j 0 i,j X≥

The proof is a combination of Theorem 4.2.9 (note that Vρ = υρ (A1,A2)) and Corollary 4.2.11. Symmetry demands an alternate expression for bm.n,

1 m n 1 n n m θA1 υ1 υ2 an,m 1 + (θA2 υ2 θA1 θA1 υ2 θA2 ) υ1 am 1,n 1 m − mn − − −

(Exercise 4.2.4). The Lemma simpliﬁes, of course, if A1 = B1, say. In this case υ1 = 1, and we state the result as a Corollary.

Corollary 4.2.13. Let (A1,A2) and (B1,B2) be delta pairs, with basic sequences (am,n) and (bm,n), respectively. If A1 = B1and A2 = τ2 (B1,B2), where (s, τ2) 2 ∈ Σ u, v [[s, t]] is a delta multi-series with translation invariant coeﬃ cients, then D D

1 n bm,n (u, v) = θA2 υ2 (A1,A2) am,n 1 (u, v) n − 130 Chapter 4. Finite Operator Calculus in Several Variables where

n m 1 i n 1 j ∂τ2 i j υ (A ,A ) = s τ − − A A 2 1 2 − − 2 ∂t 1 2 i,j 0 m 1,n 1 X≥ − − 1+j n ∂τ2 i j = ε − A A 2 ∂t 1 2 i,j 0 i,j X≥

(ε2 (s, t) = t/τ2 (s, t)). Proof. m m θA υ θA θA υ θA = θA θA θA θA = 0 1 1 2 − 2 1 1 1 2 − 2 1 (see (4.9)). Of course there must be a connection between Lemma 4.2.12 and the transfer formula m n k l bm,n (u, v) = τ1 τ2 m,n ak,l (u, v) k=0 l=0 X X shown in Lemma 4.2.1, when τρ F [[s, t]], i.e., τρ has no operators as coeﬃ cients. We refer to Exercise 4.2.6 for more∈ details. 1 Example 4.2.14. Let A1 = E− u, and A2 = v, hence v ∇ ∇ m 1 + u n + m 1 + v a (u, v) = − − . m,n m n If A1 = B1 and A2 = (EuEv B2) B2, then by Corollary 4.2.13 − 1 n bm,n (u, v) = θA2 υ2 am,n 1. n −

We calculated the Pincherle derivative θA2 in Example 4.2.5,

θA = uEuEv uEv + vEv. 2 −

Note that τ2 is a power series in t only; therefore we get in Corollary 4.2.13

n 1 j ∂τ2 n j 1 τ2 − − = (EuEv t) − − (EuEv 2t) ∂t n 1 − − j − h i n n j n 2j n 2j j = − E − E − ( 1) n j j u v − − for 0 j < n. Hence ≤ n ∞ n n j j n 2j j υ (A2) = − ( 1) (EuEv) − A 2 n j j − 2 j=0 − X n n n 2 2 = (EuEv) 2− 1 + 1 4Eu− Ev− A2 . − q 4.2. The Multivariate Transfer Formulas 131

n j n n j 1 n n n n n n From − = − − follows [A ] υ = ( 1) E− E− if n > 0. Thus j n j j 1 j 2 2 − u v b (u, v) = − − m,n

1 n (uEuEv + (v u) EvEv) υ2 am,n 1 n − − n 1 j − ( 1) n j m + n 1 2j + u 2n + m 1 3j + v = (uEu u + v) − − − − − − − n j j m n 1 j j=0 X − − − n 1 j − ( 1) n j m + n 1 2j + u = − − − − n j j m j=0 X − um 2n + m 1 3j + v v + − − × n 2j + u n 1 j − − − This formula for bm,n also holds if n = 0; of course, b0,0 (u, v) = 1. A gnerating 1 function for (bm,n) can be obtained as follows. The equations E− u = B1, and v ∇ v = (EuEv B2) B2 can be solved as Eu = 1/ (1 EvB1) and ∇ − −

2 2 2 2 1 + B + B1 + 2 (B2 1) 2B1 1 + (B1 B ) 2 − − − − 2 Ev = 1 + τ2 (B1,B2) = 2 q2 (B2 + B1 + B1B2 ) hence m n u v bm,n (u, v) s t = (1 s (1 + τ2 (s, t))) (1 + τ2 (s, t)) − m,n 0 X≥ 4.2.5 Exercises

4.2.1. [39] Suppose (bm,n) is the basic sequence for the delta pair 1 1 β1− ( u, v) , β2− ( u, v) . Show that for any multi-series η F [[s, t]] holds D D D D ∈

[η (β1 (s, t) , β2 (s, t))] = Eval0,0 η ( u, v) bm,n . m,n h | D D i 2 4.2.2. Show that the generating function (4.5) holds, if (τ1, τ2) F [[s, t]] and d ∈ A1 = Ev ∆ = τ1 (B1,B2), A2 = ∆ = τ2 (B1,B2). 4.2.3. Show that the operator in Lemma ?? can be written as

n m m+1 n+1 ∂ (B1,B2) m n 1 m ∂V2 1 n ∂V1 V V = V V V A2 V A1 1 2 ∂ (A ,A ) 1 2 − n 1 ∂A − m 2 ∂A 1 2 2 1 1 ∂ (V m,V n) + A A 1 2 . mn 1 2 ∂ (A ,A ) 1 2

4.2.4. Show (4.11), and prove the following variation of Theorem 4.2.9:

1 m n 1 n n m bm,n (u, v) = θA1 V1 V2 an,m 1 + (θA2 V2 θA1 θA1 V2 θA2 ) V1 am 1,n 1. m − mn − − − 132 Chapter 4. Finite Operator Calculus in Several Variables

4.2.5. Suppose τρ and υρ are given by Aρ = τρ (B1,B2) = Bρυρ (A1,A2). Show k l that τ1 (s, t) τ2 (s, t) m,n h i n m m n m n t m ∂υ2 s n ∂υ1 1 ∂ (υ1 , υ2 ) = [υ1 (s, t) υ2 (s, t) υ1 (s, t) υ2 (s, t) + st ]m k,n l. −n ∂t −m ∂s mn ∂ (s, t) − −

4.2.6. Suppose υρ and τρ are deﬁned as in Exercise 4.2.5, but now assume that both have coeﬃ cients in F only (no operators). Show that in this case

1 m n 1 m m n θA2 υ1 υ2 am,n 1 + (θA1 υ1 θA2 θA2 υ1 θA1 ) υ2 am 1,n 1 n − mn − − − m n m n 1 m ∂ n 1 n ∂ m A1A2 ∂ (υ1 , υ2 ) = υ υ am,n A2υ υ am,n A1υ υ am,n + am,n 1 2 − n 1 ∂A 2 − m 2 ∂A 1 mn ∂ (A ,A ) 2 1 1 2

Together with Exercise 4.2.5 this implies the equivalence of Lemma 4.2.1 and

Lemma 4.2.12 in the bivariate case when υρ and τρ have no operators as coef- ﬁcients. ∂En n 1 ∂En n+1 4.2.7. Show that ∂∆ = nE − and ∂ = nE . ∇ c d f g 4.2.8. Find the basic sequence for the delta pair EuEv ∆u,Eu Ev ∆v 1 4.2.9. Suppose we have the operator equations ∆u = Ev− B1 (1 + αB2) and ∆v = B2. Use Lemma 4.2.1 to show that the basic sequence for (B1,B2) has the generating function s (1 + αt) u b (u, v) smtn = 1 + (1 + t)v . m,n 1 + t m,n 0 X≥ Derive from the generating function or otherwise that n n u m v m k u m k v k bm,n (u, v) = − α = (α 1) − . m k n k m k − n k k=0 k=0 X − X − In terms of hypergeometric functions this means x m n, m; z x n, m; 1 z − F − − = F − − − n 2 1 x m n + 1 n 2 1 x − − − for all positive integers n and m, where a, b; z (a) (b) F = n b zn 2 1 c n!(c) n 0 n X≥ and (a) = a (a + 1) (a + n 1). This is part of the general identity n ··· − a, b; z Γ(c)Γ(c a b) a, b; 1 z F = − − F − 2 1 c Γ(c a)Γ(c b) 2 1 a + b + 1 c − − − c a b Γ(c)Γ(a + b c) c a, c b; z + (1 z) − − − 2F1 − − − Γ(a)Γ(b) 1 + c a b − − 4.2. The Multivariate Transfer Formulas 133 for all complex numbers a, b, c and z, which converges when arg (1 z) < π. The reader interested in hypergeometric functions may consult the| “Special− Functions”| by Andrews, Askey, and Roy [3]. A classic reference to this topic is the small book by Bailey [8, 1935].

us t 1 t +v 1 s 4.2.10. The bivariate basic sequence (am,n) with generating function e − − can be expanded as

n m ui vj n + j i 1 m + i j 1 a (u, v) = − − − − m,n i! j! n i m j i=0 j=0 X X − −

Suppose for the delta pair A1,A2 for (am,n) holds the recursion A1 = B1 (1 B2) / (1 B1B2) − − and A2 = B2 (1 B1) / (1 B1B2). Find the basic sequence for (bm,n). − − 134 Chapter 4. Finite Operator Calculus in Several Variables

4.3 The Multivariate Functional Expansion Theorem

The Functional Expansion Theorem 3.1.4 is easily adopted to the multivariate case. A functional on F [u, v] is a special x-operator mapping F [u, v] to F.A functional L is invertible if L 1 = 0. The bivariate power series λ (s, t) = us+vt h | i 6 Le becomes a translation invariant operator λ ( u, v) (in Σ u, v ) by deﬁn- us+vt us+vt D D D D ing λ ( u, v) e = λ (s, t) e . We call λ ( u, v), or op (L), the operator associatedD D to L. D D Theorem 4.3.1 (Bivariate Functional Expansion Theorem). Suppose L is an invertible linear functional on F[x, y], and (bm,n) is the basic sequence for the delta pair (B1,B2) = 1 1 β1− ( u, v) , β2− ( u, v) . Any polynomial in F [u, v] can be expanded in the form D D D D k l 1 p(u, v) = L B B p op(L)− bk,l(u, v). | 1 2 k,l 0 X≥ If (sm,n) is a Sheﬀer sequence for (B1,B2), then

m n 1 sm,n(u, v) = L sk,l op(L)− bm k,n l(u, v) h | i − − k=0 l=0 X X uβ1(s,t)+vβ2(s,t) m n k l e sm,n (u, v) s t = L sk,l s t and  h | i  Leuβ1(s,t)+vβ2(s,t) m,n 0 k,l 0 X≥ X≥ m n  m n 

L bk,l sm k,n l (u, v) = L sk,l bm k,n l (u, v) . h | i − − h | i − − k=0 l=0 k=0 l=0 X X X X Proof. Exercise 4.3.1 Example 4.3.2. Suppose we join the “pause step” 0, 0 to the steps , , and . Consider a random walk from the origin to theh pointi (n, m) in k→(discrete↑ time)% steps, under the following restriction: To reach any point (i, j) on the path the random walker needs l > ai + bj steps, where a and b are given nonnegative integers. Because we require that the number of steps is large at each point, we are talking about a “slow”walker, respecting a speed limit (see Exercise 5.1.8). Suppose the slow walker gets a ticket whenever her speed exceeds the limit ai+bj+1. Denote by D (n, m; k; l) the number of such walks from (0, 0) to (n, m) in k steps and l tickets. Let us take the generating function of those tickets,

∞ D (n, m; k) := D (n, m; k; l) tl. l=0 X The recurrence for D (n, m; k) is

D (n, m; k + 1) = D (n, m; k)+D (n 1, m; k)+D (n, m 1; k)+D (n 1, m 1; k) , − − − − 4.3. The Multivariate Functional Expansion Theorem 135 and because tickets are picked up at time k = an + bm + 1, the initial conditions are

D (n, m; an + bm + 1) = τ(D (n 1, m; an + bm) + D (n, m 1; an + bm) − − + D (n 1, m 1; an + bm)) − − or equivalently

(τ 1) D (n, m; an + bm + 1) = τD (n, m; an + bm) − for all (n, m) = (0, 0). 6 D (n, m; k) and its polynomial extension when a = 2 and b = 1 (counting results in bold face; initial values D (n, m; an + bm + 1) are in boxes) . . . m . . . τ 4 τ 3 . . . − . . . 3 3 . (τ 1)3 . . τ 3 τ 2 τ − . − . . . 2 . . 2 τ 2 3τ+3 (τ 1)2 . τ(τ 1) 3τ 3 5τ 2 τ . τ 2+τ+1 . − − − − 2 1 τ 2 2(τ 1)(τ 2) τ 1 2τ 2 4τ τ 2τ 2 2τ 2 τ+1 2τ 2 2 τ+2 2τ +2τ − − − − − − − − 0 1 τ 3 1 τ 2 1 τ 1 1 τ 1 τ + 1 − − − 0 1 n 0 1 n 0 1 n 0 1 n 0 1 n k = 0→ k = 1→ k = 2→ k = 3→ k = 4→ At each k, the values D (n, m; k) can be extended to polynomials of degree m+n in k, with coeffi cients in Z [τ]. For instance, if n = m = 1, then D (1, 1; k) = 2 (τ 1) (τ 2) + (2τ 5) k + k2. − − − Let D (n, m; ξ) = δsm,n (u, y). In Exercise 4.2.9 we found the basic sequence (bm,n (u, v)) for (sm,n) as u v b (u, v) = . m,n m n This basic sequence does not contain the variable τ at all. The initial values will bring τ to the solution, such that D (n, m; k) is of degree n+m. In view of Theorem 4.3.1 we let rm,n (u, v) = sm,n (u + an + bm, v + an + bm), a Sheffer sequence for b b a a Eu− Ev− ∆u,Eu− Ev− ∆v . This delta pair has the basic polynomials uv+ anu + bmv u + an + bm v + an + bm b(b,a) (u, v) = . m,n (u + an + bm)(v + an + bm) m n (see Exercise 4.2.8). In terms of (rm,n) we can write the (recursive) initial values as (τ 1) rm,n (1, 1) = τrm,n(0, 0), so we define the functional − L rm,n = τrm,n (0, 0) (τ 1) rm,n (1, 1) h | i − − = (τ Eval0,0 (τ 1) Eval1,1) rm,n = δm,0δn,0. − − 136 Chapter 4. Finite Operator Calculus in Several Variables

The operator associated to L is

1 1 1 1 1 1 µL = τ (τ 1) E E = E E τE− E− τ + 1 − − u v u v u v − with inverse 1 1 1 Eu− Ev− µL− = 1 1 . 1 τ 1 Eu− Ev− − − The bivariate Functional Expansion Theorem 4.3.1 implies

1 (b,a) 1 1 l 1 1 l (b,a) rm,n (u, v) = µ− b (u, v) = E− E− τ 1 E− E− b (u, v) . L m,n u v − u v m,n l 0 X≥ We ﬁnd

an bm an bm D (n, m; ξ) = δsm,n (u, v) = δEu− − Ev− − rm,n (u, v) l an bm 1 l l i i i (b,a) = δ (EuEv)− − − τ ( 1) E− E− b (u, v) i − u v m,n l 0 i=0 X≥ X l l ξ i 1 an bm ξ i 1 ξ i 1 = τ l ( 1)i − − − − − − − − , i − ξ i 1 m n l 0 i=0 X≥ X − − hence

l l k i 1 an bm k i 1 k i 1 D (n, m; k; l) = ( 1)i − − − − − − − − i − k i 1 m n i=0 X − − 4.3.1 Exercises 4.3.1. Prove the bivariate Functional Expansion Theorem.

4.3.2. Show that in Example 4.3.2 sm,n (u, v) =

1 l l 1 l uv bmu anv u v = E− τ E− − (∆u + v) − − . v u ∇ uv m n l 0 X≥ Expand this expression to show that D (n, m; k; l) =

l l k i 1 an bm k i 1 k i 1 ( 1)i − − − − − − − − i − k i 1 m n i=0 X − − 4.3.3. Find the solution to the system of diﬀerential equations ∂ ∂ sm,n (u, v) = sm 1,n (u, v) and sm,n (u, v) = sm,n 1 (u, v) ∂u − ∂v − such that sm,n (m + n, m n) = 1/ (m + n)! − Chapter 5

Special Constructions in Several Variables

Three “degenerated” types of multivariate polynomial sequences will be considered in this chapter. First we study sequences (bn1,...,nr (ξ)) which can be thought as “diagonalizations”of multivariate Sheffer sequences (bn1,...,nr (x1, . . . , xr)), setting xρ = ξ for all ρ = 1, . . . , r. We call the sequences (bn1,...,nr (ξ)) multi-indexed. An important example for a multi-indexed polynomial in combinatorics is the ξ+n1+ +nr multinomial coeffi cient ··· . A different kind of multi-indexed polynomi- n1,...,nr als is considered in section 5.2, where we set all but one of the variables equal to 0. The third type reduces to one the number of indices but keeps the variables, b (x , . . . , x ) = b (x , . . . , x ). The sequence (b (x , . . . , x )) n 1 r i1+ +ir =n i1,...,ir 1 r n 1 r is called a Steffensen sequence··· [83]. P

5.1 Multi-indexed Sheﬀer Sequences

As before, we consider the bivariate case as a model for r variables. We take any bivariate Sheﬀer polynomial and reduce it to a univariate polynomial by setting u = v. To distinguish the new polynomial clearly from the univariate Sheﬀer polynomials we have seen before, we call the new variable ξ. On the generating function level, we begin with the multivariate formal power series σ (s, t) euβ1(s,t)+vβ2(s,t), and transform it into

ξ(β1(s,t)+β2(s,t)) m n σ (s, t) e = sm,n (ξ, ξ) s t . m,n 0 X≥ 138 Chapter 5. Special Constructions in Several Variables

If we write β1 (s, t) = sφ1 (s, t) and β2 (s, t) = tφ2 (s, t), where φρ (s, t) is of order 0, then the coeﬃ cient of ξm+n in [smtn] eξ(β1(s,t)+β2(s,t)) equals

m+n m + n m i i m i m+n i s t φ (s, t) φ (s, t) − i − − 1 2 i=0 X m + n m + n = s0t0 φ (s, t)m φ (s, t)n = φ (0, 0)m φ (0, 0)n m 1 2 m 1 2 which is always different from 0. Hence sm,n (ξ, ξ) has degree m+n in ξ. We call the operator that replaces u and v by ξ diagonalization, and we will write sn1,...,nr (ξ) instead of sn1,...,nr (ξ, . . . , ξ), for any number of variables. For the diagonalization operator we write δ : sn ,...,n (x1, . . . , xr) sn ,...,n (ξ); there should be no 1 r → 1 r confusion with the Kronecker delta δi,j. The polynomials sn1,...,nr (ξ) are called a multi-indexed Sheffer sequence. From the generating function it is clear that multi-indexed Sheffer sequences and their basic sequence (i.e., σ = 1) satisfy the binomial theorem. For any delta pair (β1, β2) holds that β (s, t) := β1 (s, t)+β2 (s, t) is a bivariate power series such that β (0, 0) = 0, and [β] = 0, [β] = 0. Vice versa, any β 1,0 6 0,1 6 with these properties can be decomposed as a sum of two delta series β1 and β2, but β1 and β2 are not uniquely defined! This ambiguity should be seen as a strength, not a weakness of the theory! Example 5.1.1. In Example 4.1.4 we found the trivariate basic sequence

u 1 + l v 1 + l + m w 1 + l + m + n b (u, v, w) = − − − . l,m,n l m n Diagonalization gives

ξ 1 + l + m + n b (ξ) = − . l,m,n l, m, n Note that ξ 1 + l + m + n ξ 2 + l + m + n ξbl,m,n (ξ) := − − ∇ l, m, n − l, m, n = bl 1,m,n (ξ) + bl,m 1,n (ξ) + bl,m,n 1 (ξ) , − − − thus 1 1 1 ξδ = δ (B1 + B2 + B3) = δ E− E− 1 + E− 2 + 3 . ∇ 2 3 ∇ 3 ∇ ∇ The polynomials bl,m,n (ξ) have the generating function

ξ 1 + l + m + n l m n ξ − r s t = (1 r s t)− . l, m, n − − − l,m,n 0 X≥ 5.1. Multi-indexed Sheﬀer Sequences 139

The same polynomials could be obtained by diagonalizing the trivariate basic sequence u 1 + l + m + n v 1 + m w 1 + m + n − − − . l m n Any delta triple (β1, β2, β3) adding up to ln (1 r s t) will generate ξ 1+l+m+n − − − − − l,m,n . 5.1.1 Delta Operators for multi-indexed Sheﬀer sequences.

In the Example above we found that ξ is the diagonalization of the sum of a delta triple of basic operators. In general,∇ we call the (univariate) delta operator Bξ on F [ξ] that satisfies Bξsm,n (ξ) = δ (B1 + B2) sm,n (u, v) the delta operator of the multi-indexed Sheffer sequence (sm,n (ξ)), if deg Bξsm,n = m + n 1. This − last condition is not automatically satisfied, because (sm,n (ξ)) is not a basis.

Example 5.1.2. A multi-indexed sequence (bm,n (ξ)) may not have a delta operator (but still is of binomial type), even if we know a multivariate basic sequence (bm,n (u, v)) such that bm,n (ξ) = δbm,n (u, v). In this case, all we can say is that δ (B1 + B2) bm,n (u, v) = bm 1,n (ξ) + bm,n 1 (ξ). For example, let −um ( v)n − (B1,B2) = ( u, v), thus bm,n (u, v) = − . Then D −D m! n! m 1 n m n 1 u − ( v) u ( v) − δ (B + B ) b (u, v) = δ − + − 1 2 m,n (m 1)! n! m! (n 1)! − − ! m+n 1 n 1 ξ − ( 1) − = − (n m) . m!n! −

The mapping bm,n (ξ) bm 1,n (ξ) + bm,n 1 (ξ) has bn,n (ξ) in its kernel, for all 7→ − − n > 0. This contradicts the fact that deg (Bξbn,n (ξ)) = 2n 1 for a basic sequence. − Even if (bm,n (u, v)) is a basic sequence, (bm,n (ξ)) is not a basis of F [ξ]; we need the following theorem about the existence of a delta operator Bξ on F [x] such that Bξbm,n = bm 1,n + bn,m 1. − − Theorem 5.1.3. Let (bn,m (ξ)) be a multi-indexed sequence with generating function

m n ξβ(s,t) bm,n (ξ) s t = e , m,n 0 X≥ where β (0, 0) = 0, and [β] = 0, [β] = 0. There exists a delta operator Bξ on 1,0 6 0,1 6 F [ξ] such that Bξbm,n (ξ) = bm 1,n (ξ) + bn,m 1 (ξ) − − ˆ ˆ 1 for all m, n 0 iﬀ there is a univariate delta series β, such that Bξ = β− ( ) ≥ D and βˆ (s + t) = β (s, t). 140 Chapter 5. Special Constructions in Several Variables

Proof. We want

ξβ(s,t) 1 ξβ(s,t) 1 ξw 1 ξw (s + t) e = βˆ− ( ) e = C(β)βˆ− ( ) e = C(β)βˆ− (w) e D D 1 ξβ(s,t) = βˆ− (β (s, t)) e

ˆ 1 hence (s + t) = β− (β (s, t)). Because a delta operator does not always exist, we will say that (bm,n (ξ)) is the basic sequence for δ (B1 + B2), if (B1,B2) is a delta pair such that Bρ = 1 β− ( u, v), and β1 (s, t) + β2 (s, t) = β (s, t). ρ D D In Example 5.1.2 we have β (s, t) = s t, and we would need a delta series βˆ − such that βˆ (s t) = s + t, which does not exist. Note that βˆ (s + t) = β (s, t) iﬀ ˆ 1 − s + t = β− (β (s, t)). In Example 5.1.1 we found β (r, s, t) = ln (1 r s t), ˆ 1 ˆ 1 − − − − and β− ( ) = ξ = 1 e−D, hence β− ( ln (1 r s t)) = r + s + t. D ∇ − − − − − 5.1.2 Translation invariance of diagonalization, and some examples. The diagonalization δ is translation invariant in the following sense. a Lemma 5.1.4. Let Eξ be the translation by a F on F [ξ]. The diagonalization δ is translation invariant, ∈ a a a Eξ δ = δEuEv . Proof. Let p (u, v) F [u, v]. Then ∈ a a a a δEuEv p (u, v) = δp (u + a, v + a) = p (ξ + a) = Eξ p (ξ) = Eξ δp (u, v) .

Of course, δ is not invertible, hence there can be infinitely many bivariate Sheffer sequences and delta pairs giving the same multi-indexed Sheffer sequence and delta operator. us(1+t)+vt(1 s) Example 5.1.5. Let α1 (s, t) = s + st and α2 (s, t) = t st, thus e − = − k n+m k n k m n k+m (us + vt + st (u v)) ∞ u − v − (u v) − − = smtn − . k! (k n)! (k m)! (n + m k)! k 0 n,m=0 k=0 X≥ X X − − − Hence

m+n k n k m m+n k am,n (u, v) = u − v − (u v) − / ((k n)! (k m)! (n + m k)!) . − − − − k=0 X

If we proceed to the multi-indexed polynomials am,n (ξ) = δam,n (u, v), we are left m+n with am,n (ξ) = ξ / (m!n!). This should be clear from the generating function, because β (s, t) := a1 (s, t) + a2 (s, t) = s + t. Hence, we can also choose Bρ = ρ um vn D and get bm,n (ξ) = δ m! n! = am,n (ξ). The delta operator Bξ certainly exists in 5.1. Multi-indexed Sheﬀer Sequences 141

this case, because β (s, t) = s + t, thus Bξ = ξ,and therefore ξδ = δ ( u + v). In other words, D D D D

m+n m+n 1 m+n 1 ξξ / (m!n!) = ξ − / ((m 1)!n!) + ξ − / (m!(n 1)!) . D − −

If we go back to the ﬁrst interpretation, α1 (s, t) = s + st and α2 (s, t) = t st, we ﬁnd −

1 1 2 α− (s, t) = s + t + 1 (s + t + 1) 4s 1 2 − − q 1 1 2 α− (s, t) = s + t 1 + (1 + s + t) 4s 2 2 − − q 1 1 hence α1− (s, t) + α2− (s, t) = s + t, and therefore A1 + A2 = B1 + B2. This will ˆ 1 ˆ 1 happen only if ξ = β− ( ξ), because the condition β− (α1 + α2) = s + t is also D ˆ D1 1 equivalent to s + t = β α1− + α2− . In general, only δ (A1 + A2) = δ (B1 + B2) will hold. More applied is the following example, counting paths with k occurrences of a certain pattern. Example 5.1.6. Let D (n, m; k) enumerate the number of , -paths that stay above the diagonal, reach the point (n, m), and contain the{→ pattern↑} u2r2 exactly k times. They satisfy the recursion

D (n, m; k) = D (n, m 1, k) + D (n 1, m; k) D (n 2, m 2; k) − − − − − + D (n 2, m 2, k 1) − − − with initial values are D (2k + n, 2k + n 1; k) = δn+k,0 for n, k 0. − ≥

m 1 8 28 62 110 176 7 48 150 285 15 84 222 7 1 7 21 40 63 90 6 35 92 162 10 45 93 6 1 6 15 24 33 40 5 24 51 72 6 20 27 5 1 5 10 13 15 13 4 15 24 23 3 6 0 4 1 4 6 6 5 0 3 8 8 0 1 0 3 1 3 3 2 0 2 3 0 0 2 1 2 1 0 1 0 1 1 1 0 0 0 1 0 n 0 1 2 3 4 5 2 3 4 5 4 5 6 → k = 0 k = 1 k = 2 D (n, m; k) for k = 0, 1, 2 142 Chapter 5. Special Constructions in Several Variables

Let tk,n (m) = D (n + 2k, m + 2k + n; k), hence tk,n( 1) = δn+k,0 and − deg tk,n = n + k. We have the recursion

tk,n (ξ) = tk,n (ξ 1) + tk,n 1 (ξ + 1) tk,n 2 (ξ) + tk 1,n (ξ) − − − − − and we view (tk,n (ξ)) as the diagonalization of tk,n (u, v) = tk,n (u 1, v 1) + tk,n 1 (u + 1, v + 1) tk,n 2 (u, v) + tk 1,n (u, v) . − − − − − − In terms of operators,

1 1 2 I = E− E− + EuEvB2 B + B1 u v − 2 hence

1 1 1 2 I E− + E− 1 E− = EuEvB2 B + B1, or − v v − u − 2 1 2 v + E− u = EuEvB2 B + B1. ∇ v ∇ − 2 2 We let v = EuEvB2 B2 , and u = EvB1, the operators in Example 4.2.14. Because∇ of the initial values,− ∇

tk,n (ξ) = bk,n (ξ + 1) , thus D (n, m; k)

= tk,n 2k (m n) = bk,n 2k (m n + 1) − − − − (n/2) k j − n 2k j (m n + 1) ( 1) m k 2j + 1 = − − − − − − j m k 2j + 1 k × j=0 X − − m + n 3k 3j − − . × n 2k j − − 5.1.3 Abelization of Multi-Indexed Sequences

Let β (s, t) be a bivariate series such that β (0, 0) = 0, and [β]1,0 = 0, [β]0,1 = ∂ 6 ∂ 6 0. The bivariate power series ∂s β (s, t) is of order zero 0, and so is ∂t β (s, t). Let (sm,n (ξ)) be the multi-indexed Sheﬀer sequence with generating function ∂ ξβ(s,t) ∂s β (s, t) e , hence ∂ ξs (ξ) smtn = ξ β (s, t) eξβ(s,t) m,n ∂s m,n 0 X≥ ∂ ∂ = eξβ(s,t) = b (ξ) smtn ∂s ∂s m,n m,n 0 X≥ m n = (m + 1) bm+1,n (ξ) s t m,n 0 X≥ 5.1. Multi-indexed Sheﬀer Sequences 143 and therefore b (ξ) s (ξ) = (m + 1) m+1,n . m,n ξ

In the same way we find tm,n (ξ) = (n + 1) bn+1 (ξ) /ξ if we define the Sheffer ∂ ξβ(s,t) sequence (tm,n (u, v)) with generating function ∂t β (s, t) e . We combine both results in the following theorem. Theorem 5.1.7. Let (bm,n (ξ)) be a multi-indexed basic sequence with generating function eξβ(s,t), and let c, d F such that c [β] +d [β] = 1. The multi-indexed ∈ 1,0 0,1 6 Sheffer sequence (rm,n (ξ)) with generating function

m n ∂ ∂ (ξ η)β(s,t) rm,n (ξ) s t = 1 c β (s, t) d β (s, t) e − − ∂s − ∂t m,n 0 X≥ has roots in ξ = η + cm + dn (η F), ∈ ξ η cm dn rm,n (ξ) = − − − bm,n (ξ η) . ξ η − − Proof. By superposition, any linear combination of multi-indexed Sheﬀer sequences of total degree m+n is again a multi-indexed Sheﬀer sequence; especially, bm,n (ξ η), − sm 1,n (ξ η), and tm,n 1 (ξ η) can be combined to − − − −

bm,n (ξ η) rm,n (ξ) = bn,m (ξ η) (cm + dn) − . − − ξ η − (c,d) c d (c,d) The polynomials bm,n (ξ) are basic for E− − Bξ because bm,n (0) = δ0,m+n and

c d E− − Bξrm,n (ξ + cm + dn)

= rm 1,n (ξ + c (m 1) + d (n 1)) + rm,n 1 (ξ + c (m 1) + d(n 1)) − − − − − − = δrm 1,n (u + c (m 1) + d (n 1)) , v + c (m 1) + d (n 1)) − − − − − + rm,n 1 (u + c (m 1) + d (n 1) , v + c (m 1) + d (n 1)) − − − − − c d c d = δ Eu− Ev− B1 + Eu− Ev− B2 rm,n (u + cm + dm, v + cm + dn) ,

c d c d c d thus one way of arriving at E− − Bξ is by diagonalizing Eu− Ev− B1,Eu− Ev− B2 . We call (rm,n (ξ)) the (multi-indexed) Abelization of (bm,n (ξ)). If (sm,n (u, v)) is a Sheﬀer sequence for the delta pair (B1,B2), then (sm,n (u + cm + dn, v + cm + dn)) is a Sheﬀer sequence for

c c d d Eu− Ev− B1,Eu− Ev− B2 , thus (sm,n (ξ + cm + dn)) is a multi-indexed Sheﬀer for

c c d d δ Eu− Ev− B1 + Eu− Ev− B2 . 144 Chapter 5. Special Constructions in Several Variables

Choosing for (sm,n (ξ)) the special Sheﬀer sequence (rm,n (ξ)) from Theorem 5.1.7, with η = 0, we obtain ξ r (ξ + cm + dn) = b (ξ + cm + dn) , m,n ξ + cm + dn m,n

c c d d the basic sequence for δ Eu− Ev− B1 + Eu− Ev− B2 . Example 5.1.8. Suppose D(n, m; k) is the number of , -paths reaching (n, m) staying weekly above the diagonal, and having exactly{→k↑}occurrences of the pattern rα, where α is some integer greater than 1. If k = 0 then D(n, m; 0) = D (n, m 1; 0) + D (n 1, m; 0) D (n α, m 1; 0), because the paths we subtract are− those where the− pattern occurs− at− the end,− and is preceded by an -step. If k > 0 then we may have subtracted the k-th occurrence of the pattern, so↑ we have to add it back in. However, the pattern rα may already occur once at the end of the path counted by D (n 1, m; k), and we can not have the pattern rα+1 occurring, so we have to subtract− from D (n 1, m; k) those cases that end in rα. They come from − D (n α 1, m 1; k 1), hence D (n, m; k) = − − − − D (n, m 1; k) + D (n 1, m; k) D (n α, m 1; k) (5.1) − − − − − + D (n α, m 1; k 1) D (n α 1, m 1; k 1) . − − − − − − − −

m 1 10 54 208 630 1572 7 60 275 15 90 225 9 1 9 44 154 423 951 6 45 180 9 36 0 8 1 8 35 110 270 536 5 32 108 4 0 7 1 7 27 75 161 273 4 21 56 0 6 1 6 20 48 87 118 3 12 21 5 1 5 14 28 40 36 2 5 0 4 1 4 9 14 13 0 1 0 3 1 3 5 5 0 0 2 1 2 2 0 1 1 1 0 k = 0 k = 1 0 1 0 n 0 1 2 3 4 5 4 5 6 8 9 10 → k = 0 k = 1 k = 2 D (n, m; k) for α = 4 and k = 0, 1, 2

The initial values are D (n, n 1; k) = δn+k,0 for n, k 0. − ≥ Let sk,n (m) = D (αk + n, m; k), then sk,n(αk + n 1) = δk+n,0, and deg sk,n = n + k. The recursion (5.1) becomes −

sk,n (ξ + 1) sk,n (ξ) = sk,n 1 (ξ + 1) sk,n α (ξ) + sk 1,n (ξ) sk 1,n 1 (ξ) . − − − − − − − − 5.1. Multi-indexed Sheﬀer Sequences 145

We can think of sk,n (ξ) as the diagonalization of the polynomial sk,n (u, v) where

sk,n (u + 1, v + 1) sk,n (u, v) − = sk,n 1 (u + 1, v + 1) sk,n α (u, v) + sk 1,n (u, v) sk 1,n 1 (u, v) . − − − − − − −

In operator notation, where B1sk,n = sk 1,n and B2sk,n = sk,n 1, we get − − α B2 B2 Ev (Eu I) + Ev I = Ev∆u + ∆v = − + B1. − − 1 B2 − α B2 B2 Hence we can assume that we have the relation (Ev∆u, ∆v) = B1, − be- 1 B2 − tween delta pairs. The delta operator (Ev∆u, ∆v) has the basic sequence u v k k −n (Example 4.2.10). We found the basic sequence (bk,n) for (B1B2) k,n 0 ≥ in Example 4.2.2, u n v k l b (u, v) = − . k,n k l n l l=0 α 1 X − − For sk,n (ξ) we need the Abelization of bk,n (ξ) where sk,n(αk + n 1) = δk+n,0, hence − ξ + 1 n αk s (ξ) = − − b (ξ + 1) k,n ξ + 1 k,n by Theorem 5.1.7. Finally, D (n, m; k) m + 1 n = sk,n αk (m) = − bk,n αk (m + 1) − m + 1 − n αk m + 1 n m + 1 − m + 1 k l = − − m + 1 k l n αk l l=0 α 1 X − − − n αk m + 1 n m + 1 − m + 1 k = − − m + 1 k l × l=0 X (n αk l)/(α 1) b − − − c l n αk (α 1) j 1 ( 1)j − − − − × − j l 1 j=0 X − c d In Example 4.2.10 we found the basic sequence (bm,n) for B1 = EuEv A1, f g B2 = Eu Ev A2, but only when the basic polynomials for (A1,A2) factor, am,n (u, v) = I II a1 (u) a2 (v). Hence sm,n (u, v) = bm,n (u + cm + fn, v + dm + gn) is the Sheﬀer polynomial for (A1,A2) with roots in um,n = cm fn and vm,n = dm gn, − − − −

sm,n ( cm fn, dm gn) = δm,0δn,0. − − − − We will show in the following example how this type of Sheﬀer sequence can be applied after diagonalization. 146 Chapter 5. Special Constructions in Several Variables

3 Example 5.1.9. We consider lattice paths in N0 that take the step vectors (1, 0, 0), (0, 1, 0), and (0, 0, 1). The number of paths from the origin to (m, n, ξ) is the trino- ξ+m+n mial coeﬃ cient m,n , a Sheﬀer sequence for ξ. We can interpret each of the three steps as a vote given to candidate A, B, or∇C, respectively, and we want at any time candidate C to get at least as many votes as candidate A. For the number of paths dm,n (ξ) this means that dm,n (m 1) = 0 for all m > 0. In addition, we want candidate C to have at least as many− votes as candidate B has more votes than A; for example, if B gets n = 3 votes and A gets m = 1, than C must get at least n m = 2 votes. Thus a coalition of A and C can balance or defeat B at any − time. This condition holds when dm,n (n m 1) = 0. Of course, d0,0 (ξ) = 1. − − (Why does dm,0 (ξ) agree for all ξ 2 and 0 m ξ with dm (ξ) in Example 2.3.6?) ≥ ≤ ≤ m ξ = 1 ξ = 2 ξ = 3 4 0 0 0 0 0 0 0 3 0 0 0 0 0 0 5 30 100 240 450 660 660 2 0 0 0 0 0 2 8 18 28 28 0 5 25 70 140 210 210 0 1 1 2 2 0 2 6 10 10 0 3 12 27 42 42 0 0 1 1 0 1 2 2 0 1 3 5 5 0 0 1 2 3 n 0 1 2 3 4 n 0 1 2 3 4 5 n

Lattice paths to (m, n, ξ) when dm,n (m 1) = 0 − and dm,n (n m 1) = 0 (m + n > 0) − − Remember that

1 1 1 1 ξ = δ E− u + v = δ E− E− ∆u + E− ∆v , ∇ v ∇ ∇ u v v 1 u+m 1 v+m+n 1 where E− u, v has basic polynomials bm,n (u, v) = − − , which v ∇ ∇ m n satisfy the condition bm,n (u, v m) = δm,0. We now need the basic polynomials 1 1 −1 rm,n (u, v) for E− E− u,E− v , which equal u v ∇ v ∇ u v + m u + 2m v + m + 2n r (u, v) = m,n u + 2m v + m + 2n m n by Example 4.2.10. They allow us to ﬁnd the Sheﬀer polynomials

m+1 n+1 dm,n (u, v) = Eu− Ev− rm,n (u, v) 1 for E− u, v , with roots in u = m 1 and v = n 1 m. Hence v ∇ ∇ − − − m+1 n+1 dm,n(ξ) = δEu− Ev− rm,n (u, v) (ξ m + 1) (ξ + m n + 1) ξ + m + n + 1 = − − (ξ + m + 1) (ξ + m + n + 1) m, n is the Sheﬀer polynomial for ξ with the requested roots in ξ = m 1 and ξ = n m 1, for m + n > 0. ∇ − − − 5.1. Multi-indexed Sheﬀer Sequences 147

5.1.4 Exercises

5.1.1. We begin with a univariate basic sequence (an (ξ)) for A with generating n N0 n ξα(t) ∈ (l+m+n)! function n 0 an (ξ) t = e , and deﬁne pl,m,n (ξ) := l!m!n! al+m+n (ξ) for ≥ r ξα(t1+t2+t3) all n N0. Check that the generating function of (pl,m,n (ξ)) equals e , ∈ P and that Aξ := A is the delta operator for (pl,m,n (ξ)),

Aξpl,m,n (ξ) = pl 1,m,m (ξ) + pl,m 1,n (ξ) + pl,m,n 1 (ξ) . − − −

However, for Aξ to be the delta operator for (pl,m,n (ξ)) we need a trivariate basic sequence (bl,m,n (x1, x2, x3)) such that pl,m,n (ξ) = δbl.m,n (x1, x2, x3). Let 3 3 βρ (t1, t2, t3) = α tσ α tσ . Show that σ=ρ − σ=ρ+1 1. (β1, β2, β3) isP a delta multiseries,P 1 1 1 2. the basic sequence (bl,m,n (x1, x2, x3)) for β1− , β2− , β3− has the generating function l m n ξα(t1+t2+t3) δ bl,m,n (x1, x2, x3) t1t2 t3 = e , l,m,n 0 X≥ 1 1 1 1 3. α− (t1 + t2 + t3) = β1− (t1, t2, t3) + β2− (t1, t2, t3) + β3− (t1, t2, t3),

1 1 3 1 3 4. β− (t) = α− tσ α− tσ , ρ σ=ρ − σ=ρ+1 5. the delta operatorP Bξ for (bl,m,n (ξP)) satisﬁes the condition

1 Bξδ = α− ( ) δ D and therefore Bξ = Aξ.

l+m+n+ξ 1 As an example, show that l,m,n − is the multi-indexed basic sequence for ξ. ∇ 5.1.2. Characterize the polynomials that are in the kernel of δ. 5.1.3. A special case of Example 4.1.4 is the multivariate basic polynomial 3 bn1,n2,n3 (x1, x2, x3) = ρ=1 anρ (xρ + n1 + + nρ 1) where anρ (xρ) is − nρ N0 ··· ∈nρ the same univariate basic sequence for A for all ρ, hence an (xρ) tρ = nρ 0 ρ Q ≥ xρ 1 γ (tρ) and A = (ln γ)− ( ). D P

1. Let pn1,n2,n3 (ξ) := δbn1,n2,n3 (x1, x2, x3) for all nρ 0. Find the generating function p (ξ) tn1 tn2 tn3 . Show≥ that the projections n1,n2,n3 0 n1,n2,n3 1 2 3 ≥ pn ,0,0 (ξ) , p0,n ,0 (ξ) and p0,0,n (ξ) are all equal. What does that imply for 1 2P 3 the delta operator for (pn1,n2,n3 (ξ)), if such an operator exists? 2. Consider the generating function p (ξ) tn1 tn2 tn3 in the n1,n2,n3 0 n1,n2,n3 1 2 3 ≥ special case A = , and show that there is no delta operator for (pn1,n2,n3 (ξ)) in this case. D P 148 Chapter 5. Special Constructions in Several Variables

3. Let A = . Check the generating function of (pn ,n ,n (ξ)) to show that the ∇ 1 l 3 delta operator Pξ for (pn1,n2,n3 (ξ)) is equal to ξ. In this case, pn1,n2,n3 (ξ) ξ 1+n1+n2+∇n3 is the “right” factorization to give − n1,n2,n3 n n 5.1.4. Show that δ = δ ( u + + r) for all n 0. Dξ D ··· D ≥ 1 1 1 5.1.5. Show that the components A1 = 1E− E− , A2 = 2E− , and A3 = 3 ∇ 2 3 ∇ 3 ∇ are a possible solution of the operator equation ξδ = δ (A1 + A2 + A3). ∇ 5.1.6. Show that in Theorem 5.1.3 the condition βˆ (s + t) = β (s, t) is equivalent ˆ 1 1 to s + t = β β1− + β2− , where β (s, t) = β1 (s, t) + β2 (s, t).

5.1.7. In Example 5.1.9 show that dn 1,n+1 (0) = (n 2) Cn for all n 1. Is there a combinatorial interpretation for this− relationship to− the Catalan numbers?≥ 5.1.8. Suppose we join the “pause step” 0, 0 to the steps , , and . Consider h i → ↑ % a random walk taking these steps by introducing the probabilities pij = Pr ( i, j ) where i, j 0, 1 corresponds to the four step vectors, 1, 0 = , etc. Weh wanti to ﬁnd the∈ probability { } Pr (n, m; k) that this random walk startsh i at (0→, 0) and reaches the point (n, m) in k (discrete time) steps, under the following restriction: To reach any point (i, j) on the path the random walker needs l > ai + bj steps, where a and b are given nonnegative integers. Because we require that the number of steps is large at each point, we are talking about a “slow” walker, respecting a speed limit Pr (i, j; ai + bj) = 0 when (i, j) = (0, 0). 6 Pr (n, m; k) when a = 2 and b = 1 m 0 3 3 0 p00p01 2 2 2 2 0 p00p01 3p00p01 0 2 3 2 1 0 p00p01 2p00p01 0 3p00p01 p00 (3p10p01 + p00p11) 2 3 2 4 3 0 1 p00 p00 p00 p00p10 p00 2p00p10 n 0 0 0 0 1 0 1 → k = 0 k = 1 k = 2 k = 3 k = 4

m+n k n m Let sm,n (k) := p00 − p10− p01− Pr (n, m; k) (note the switch in n and m); show that the numbers sm,n (k) follow the recursion

sm,n (k + 1) = sm,n (k) + sm,n 1 (k) + sm 1,n (k) + αsm 1,n 1 (k) − − − − where α = p00p11/ (p10p01), sn,m (k) = 0 if n or m are negative, s0,0 (k) = k p00− Pr (0, 0; k) = 1 for k N0, and sm,n (k) = 0 if k = an + bm when (m, n) 2 ∈ ∈ N0 (0, 0). The condition s0,0 (k) = 1 can be extended to negative k without chang- \ ing the values of sm,n (k) for k > am + bn. Extend sm,n (k) to a multi-indexed polynomial sequence sm,n (ξ) = δsm,n (u, v) solving the diﬀerence equation sm,n (u + 1, v + 1) sm,n (u, v) = sm,n 1 (u, v) + sm 1,n (u, v) + αsn 1,m 1 (u, v) , − − − − − 5.1. Multi-indexed Sheﬀer Sequences 149

Show that

ξ an bm ξ m m ξ m s (ξ) = − − − αj m,n ξ m j n j j=0 X − ξ m m ξ m (an + bm) ξ 1 m m ξ m = − αj − − αj m j n j − m m j n j j=0 j=0 X − X − by Theorem 5.1.7. In terms of probabilities,

k m n n m Pr (n, m; k) = p00− − p10p01sm,n (k) m k an bm k m k m j+k m n j n j m j = − − − p − − p p − p − k m j n j 00 11 10 01 j=0 X − whenever k an + bm, and Pr (n, m; k) = 0 otherwise. Of course, Pr (0, 0; 0) = 1. ≥ If there are no pauses, p00 = 0, then we can have between max (m, n) and m + n steps. In addition, assume that a = b = 0. Then

k m+n k k m k n Pr (n, m; k) = p − p − p − . m + n k, k m, k n 11 10 01 − − − 150 Chapter 5. Special Constructions in Several Variables

5.2 Polynomials with all but one variable equal to 0

If we take a bivariate Sheﬀer sequence (sm,n (u, v)), say, and set v = 0, we obtain a bi-indexed sequence of polynomials sm,n (u) := sm,n (u, v), say. The binomial theorem (4.2) tells us that

m n

sm,n (x + y) = si,j (x) bm i,n j (y, 0) − − i=0 j=0 X X where (bm,n (u, v)) is the basic sequence corresponding to (sm,n (u, v)). The generating function of (sm,n (u)) is of the form

m n uβ1(s,t) sm,n (u) s t = σ (s, t) e . m,n 0 X≥

m n uβ1(s,t) If we write bm,n (u) for bm,n (u, 0), then m,n 0 bm,n (u) s t = e , where ≥

m Pn uβ1(s,t)+vβ2(s,t) bm,n (u, v) s t = e m,n 0 X≥ for some bivariate delta series β2 (s, t). As far as (bm,n (u)) is concerned, β2 (s, t) can be just any delta series, so we choose a new β¯2 (s, t) = t. The selected delta 1 operator pair β− ( u, v), v has the basic sequence ¯bm,n (u, v) , where 1 D D D n vn j vn ¯b (u, v) = b (u) − = B¯ and m,n m,j (n j)! m n! j=0 X − vn b (u) = ¯b (u, 0) = B¯ m,n m,n m n! v=0 (m) vn (m) (m) j We see that ¯bm,n (u, v) = B¯ where B¯ is the operator B¯ = ∞ bm,j (u) : n! j=0 Dv F [v] F [u, v] for all m 0, translation invariant in v. Therefore, → ≥ P (m) (m 1) B1B¯ = B¯ − .

1 1 Note that B1 = β− ( u, v), where β− (β1 (s, t) , t) = s, which shows that B1 1 D D 1 depends on our choice of B2 = v (but β1 (s, t) is independent of that choice). D Example 5.2.1. Let

m m j n j n + m 2j n + m j 1 u − v − b (u, v) = − − − m,n m j j (n + m 2j)! j=0 X − − min(m,n) m j n j n + m j 1 u − v − = − − , j (m j)! (n j)! j=0 X − − 5.2. Polynomials with all but one variable equal to 0 151 the polynomial sequence in the introduction to this chapter. We obtain the generating function

n + m + j 1 (us)m (tv)n b (u, v) smtn = sjtt − m,n j m!n! m,n 0 j 0 m,n j ≥ ≥ ≥ X X Xm n (us) (tv) m n us+vt uβ (s,t)+vβ (s,t) = (1 st)− − = e 1 st = e 1 2 , m!n! − − m,n 0 X≥ m 1 m n and we see that bm,n (u) = n− u − / (m n)! if m n, and 0 otherwise. We ﬁnd − ≥ n n j m m j n v − m 1 u − j v ¯bm,n (u, v) = bm,j (u) = − (n j)! j (m j)!Dv n! j=0 j=0 X − X − m n m n n m 1 j j u v m 1 u v (m) v = − = (1 + u v) − = B¯ , j DuDv m! n! D D m! n! n! j 0 X≥ (m) m 1 m n thus B¯ = (1 + u v) − u /m! for m 1. If m = 0 then ¯b0,n (u, v) = v /n!, (0)D D ≥ and therefore B¯ = I. From β1 (s, t) = s/ (1 st) and β¯2 (s, t) = t follows B1 = 1 1 − β− ( u, v), where β− (s, t) = s/ (1 + st), hence 1 D D 1 u B1 = D . 1 + u v D D (m) We check our calculations by applying B1 to B¯ for m 1, ≥ m m 1 (m) u m 1 u m 2 u − (m 1) B1B¯ = D (1 + u v) − = (1 + u v) − = B¯ − . 1 + u v D D m! D D (m 1)! D D − Suppose we are given B1 as the solution of the operator equations A1 = τ1 (B1, v), A2 = v, and we would like to expand bm,n (u) in terms of the basic D D sequence (am,n) for (A1, v), D n vn l vn a (u, v) = a (u, 0) − = A¯(m) m,n m,l (n l)! n! l=0 X − From Corollary 4.2.13 we know that

1 m bm,n (u, v) = θA1 υ1 (A1, v) am 1,n (u, v) m D − m 1 n 1 − 1+i m ∂τ1 = θA1 ε1 − am 1 i,n j (u, v) m ∂s − − − i=0 j=0 i,j X X where ε1 (s, t) = s/τ1 (s, t)). 152 Chapter 5. Special Constructions in Several Variables

Example 5.2.2. Suppose we want to know the number of ballot paths (taking steps r = and u = , and staying weakly above the diagonal y = x) containing the pattern→ rur exactly↑ k times. The number D (n, m; k) of such paths from (0, 0) to (m, n) follows the recursion D (n, m; k) =

k i l l + 1 D(n 1, m; k)+D(n, m 1; k) ( 1) − D(n 2 l, m 1 l; k i) . − − − − − − − − − i i=0 l 0 X X≥ m 1 10 45 128 273 0 9 72 273 0 8 84 9 1 9 36 91 174 0 8 86 186 0 7 63 8 1 8 28 62 105 0 7 42 120 0 6 45 7 1 7 21 40 59 0 6 30 72 0 5 30 6 1 6 15 24 30 0 5 20 39 0 4 18 5 1 5 10 13 13 0 4 12 18 0 3 9 4 1 4 6 6 4 0 3 6 6 0 2 3 3 1 3 3 2 0 0 2 2 0 0 1 0 2 1 2 1 0 0 1 0 0 0 1 1 1 0 k = 0 0 0 k = 1 0 0 1 0 0 n 0 1 2 3 4 1 2 3 4 2 3 4 → k = 0 k = 1 k = 2 D (n, m; k) for k = 0, 1, 2

If we deﬁne bk,n (v) = D (n + k, v 1 + k + n; k), then bk,n (v) can be extended − to a polynomial of degree n for all k 0, and bk,n (0) = δkδn. The recursion for ≥ bk,n (v) equals

k i l l + 1 bk,n (v) = bk,n 1 (v + 1)+bk,n (v 1) ( 1) − bk i,n+i 2 l (v + 1) , − − − − − − − i i=0 l 0 X X≥

k i ¯ k u − ¯ and for bk,n (u, v) = i=0 bi,n (v) (k i)! it remains the same, bk,n (u, v) = −

P k l + 1 i l ¯bk,n 1 (u, v + 1) + ¯bk,n (u, v 1) ( 1) − ¯bk i,n+i 2 l (u, v + 1) . − − − i − − − − i=0 l 0 X X≥ In terms of operators,

EvB¯2 B¯2 u v = B¯2Ev − D , hence ∇ − 1 + B2 u − D t s 1 τ2 (s, t) = Evt 1 − = Evt . − 1 + t s 1 + t s − − 5.3. Cross-Sequences and Steﬀensen Sequences 153

We ﬁnd for n > 0

¯ 1 n bk,n (u, v) = θ v υ2 ( u, v) ak,n 1 (u, v) n ∇ D ∇ − k n 1 1 − 1+j n ∂τ2 = θ v ε2 − ak i,n 1 j (u, v) n ∇ ∂t − − − i=0 j=0 i,j X X k n 1 1 − j n i + n i + j n j = θ v ( 1) Ev − ak i,n 1 j (u, v) n ∇ − i + n i + j j − − − i=0 j=0 X X uk n 1+v for n > 0, where θ 2 = (uEu u + v) Ev and ak,n (u, v) = − . Hence ∇ − k! n n 1 j − ( 1) k + n k + j 2 (n j) + v 1 b (v) = v − − − , k,n i + n k + j j n 1 j j=0 X − − and ﬁnally D (n, m; k) = bk,n k (m + 1 n) for m n k. We have bk,0 (v) = − − ≥ ≥ δk,0.

5.2.1 Exercises 5.3 Cross-Sequences and Steﬀensen Sequences

Suppose the Sheﬀer polynomials pm,n (u, v) have the generating function

m n uβ(s,t)+vα(s,t) pm,n (u, v) s t = ρ (s, t) e . m,n 0 X≥ The polynomials m [λ] sm (x) := pm n,n (x, λ) − n=0 X m 0, are a Steffensen sequence [83, Section 8] (note that in [83, Section 8] a Steffensen≥ sequence may be of degree less than m in λ). Our Steffensen sequence has the generating function

m ∞ [λ] m ∞ m n n xβ(r,r)+λα(r,r) sm (x) r := pm n,n (x, λ) r − r = ρ (r, r) e − m=0 m=0 n=0 X X X Of course, ρ (r, r) is a power series of order 0 in r, and β (r, r) and α (r, r) are both [λ] univariate delta series in r. Hence sm (x) is for m 0 ≥ 1 fixed λ a Sheffer sequence for β− ( x) in the variable x, and for • D 1 fixed x a Sheffer sequence for α− ( λ) in the variable λ. • D 154 Chapter 5. Special Constructions in Several Variables

[λ] If ρ (s, t) = 1, the resulting Steﬀensen sequences cm (x) is called a cross- sequence. 1 If (bm (x)) is the basic sequence for β− ( x), and (am (λ)) the basic sequence 1 D for α− ( λ), then D m [λ] [λ] 1 k [λ] sm (x + y) = sk (x) bm k (y) = bk (y) α− ( λ) sm (x) − D k=0 k 0 X X≥ 1 yβ(α− ( λ)) [λ] = e D sm (x) m [λ+µ] [λ] 1 k [λ] sm (x) = sk (x) am k (µ) = ak (µ) β− ( x) sm (x) − D k=0 k 0 X X≥ 1 µα(β− ( x)) [λ] = e D sm (x) m [λ+µ] [λ] [µ] sm (x + y) = sk (x) cm k (y) . − k=0 X

1 µα(β− ( x)) Thus the operator e D on F [x] acts as the translation by µ in the para- 1 yβ(α− ( λ)) meter λ, and the operator e D on F [λ] acts as the translation by y in the parameter x. Seen as operators on F [x, λ] we get

1 µα(β− ( x)) µ λ µ e D = e D = Eλ and 1 yβ(α− ( λ)) y x y e D = e D = Ex,

1 1 or β− ( x) = α− ( λ). Hence D D [λ] λ [0] x [λ] sm (x) = T sm (x) = P sm (0)

1 1 where T := eα(β− (Dx)), and P := eβ(α− (Dλ)). Because T (and P ) are invertible, 1 1 the operators T β− ( x) and P α− ( λ) are delta operators (on F [x] and F [y], D D 1 respectively). According to (2.36) the delta operator T β− ( x) has for m > 0 m 1 D the basic polynomials xT bm (x) /x. Similar, P α− ( λ) has basic polynomials m 1 D λP λ am (λ). [λ] Suppose pm (x) is the Steﬀensen sequence with generating function m 0 ≥ xβ(r)+λα(r) [0] [0] β0 (r) e , thus pm (x) = (m + 1) bm+1 (x) /x for m > 0, and p0 (x) x+y [m+1] ≡ β0 (0). For this Abelization type we ﬁnd (Exercise 5.3.1) m+1 pm (x + y) =

m y [m+1] x [m+1] x [k] y [m+1 k] pm (y) + pm (x) + pk 1 (x) pm k − (y) (5.2) m + 1 m + 1 k − m + 1 k − k=1 X − 5.3. Cross-Sequences and Steffensen Sequences 155 for all m 0 [83, Proposition 8.3]. In the same way we can define the Steffensen ≥ [λ] xβ(r)+λα(r) [λ] sequence tm (x) with generating function α0 (r) e , thus tm (0) = m 0 ≥ [λ] λ+µ [λ+µ] (m + 1) am+1 (λ) /λ for m > 0, and t (0) 1. Now tm (m + 1) = 0 ≡ m+1 m λ [λ] µ [µ] λ [λ] µ [µ] tm (m + 1)+ tm (m + 1)+ tk 1 (k) tm k (m + 1 k) m + 1 m + 1 k − m + 1 k − − k=1 X − for all m 0 . ≥ Example 5.3.1. Suppose the basic polynomials lm,n (x, y) have the generating function m n us/(s 1) v lm,n (u, v) s t = e − (1 t)− , − m,n 0 X≥ thus n + v 1 m ( u)i m 1 l (u, v) = − − − . m,n n i! m i i=0 X − xr/(r 1) α 1 The cross-sequence with generating function e − (1 r)− − is the sequence of Laguerre polynomials −

m m i ( x) m + α (α) lm n,n (x, α + 1) = − = Lm (x) . (5.3) − i! m i n=0 i=0 X X − 5.3.1 Exercises 5.3.1. [83, Proposition 8.3] Show identity (5.2). 5.3.2. Show that for the Laguerre polynomials holds

m i m i (α) y − (α) k + i 1 Lm (x y) = Lm i k (x) − − i! − − k i=0 k=0 X X 5.3.3. Show that for the Laguerre polynomials holds

m (α+β) (α) (β 1) Lm (x + y) = Lk (x) Lm −k (y) − k=0 X 5.3.4. Show

m (α+1) (α) Lm (x) = Lk (x) and k=0 X (α+1) (α+1) [α] Lm (x) Lm 1 (x) = Lm (x) − − 156 Chapter 5. Special Constructions in Several Variables

5.3.5.

(m+2) (m+2) (m+2) yLm (y) + xLm (x) (x + y) Lm (x + y) m − m + 1 (k+1) (m+2 k) = xy Lk 1 (x) Lm k − (y) k (m + 1 k) − − k=1 X − 5.3.6. The Hermite polynomials of variance σ2 are deﬁned as

2 [1/σ ] 2/(2σ2) n Hn (x) = e−D x /n!

[1] for positive σ. The special case Hn is called “Hermite polynomial” later (see [λ] (6.36)). Show that Hn is not a cross sequence as we deﬁned it, but has similar properties. For example,

m/2 [λ+µ] k k k [λ] Hm (x) = ( 1) 2− µ Hm 2k (x) − − k=0 X and m [λ+µ] [λ] [µ] Hm (x + y) = Hk (x) Hm k (y) − k=0 X Chapter 6

A General Finite Operator Calculus

The generalized Finite Operator Calculus in this chapter follows J. M. Freeman’s [35] Transforms of Operators on k[x][[t]], which we already introduced in section 2.2.2. Other approaches are closely related [97], or more general [9]; however, we believe that this setting is exactly at the right level for our purpose, providing a better understanding of the Finite Operator Calculus. In the following we give up translation invariance, but retain the commutativity of the operators under study. Of course, no translation invariance means no binomial theorem! From now on we will call Rota’s Finite Operator Calculus exponential, because it is based on ext. The coeﬃ cient ring k will be an integral domain containing Z, as before.

6.1 Transforms of Operators

An element f (x, t) in k[x][[t]] is a formal power series (in t) that has polynomials n (in x) as coeffi cients. The polynomial pn (x) = [t ] f (x, t) can be of any degree. The power series f k[x][[t]] can also be seen as an element of k [[t]] [[x]], ∈ n n n k k n f (x, t) = pn (x) t = t pn,kx = x pn,kt , (6.1) n 0 n 0 k=0 k 0 n 0 X≥ X≥ X X≥ X≥ k but only finitely many of the coeffi cients pn,k = x pn (x) are different from 1 1 zero for every n. For example, the series (1 x)− (1 t)− k [[t]] [[x]] is not − − ∈ in k[x][[t]]. More can be said if we consider the special case when deg pn (x) = n for all n 0. The following two closely related results of Freeman [35] give some insight into≥ the structure of this kind of series. 158 Chapter 6. A General Finite Operator Calculus

n Lemma 6.1.1. Suppose s(x, t) = n 0 sn (x) t where (sn (x)) is a basis of k[x] ≥ (i.e., deg sn = n and s0 = 0). If for some sequence (pn) of polynomials and (βn) of power series holds 6 P

s(x, t) = pn(x)βn (t) (6.2) n 0 X≥ then (pn) is a basis of k[x], if and only if βn is of order n.

Proof. Denote by (sn,k)n,k 0, (pn,k)n,k 0, and (βn,k)n,k 0 the coeﬃ cient matrices ≥ ≥ ≥ k of (sn (x)), (pn (x)), and (βn (t)), respectively, thus sn,k = x sn (x), etc. The ﬁrst statement says that if in the matrix product (s ) = (p )(β )T the matrix n,k n,k n,k (sn,k) and one of the matrices on the right hand side are lower triangular and invertible, then the remaining one is of the same type. n Example 6.1.2. The standard basis r0(x) := 1, rn(x) := x (1 x ) /(1 x) for xt − − n > 0, generates the series 1 + (1 t)(1 xt) in k[x][[t]]. From − − xt tn 1 + = 1 + xn (1 t) (1 xt) 1 t n 1 − − X≥ − n follows that the basis (x ) corresponds to the pseudobasis β0 (t) = 1, βn (t) = n t / (1 t) for n 1, in k [[t]], where deg βn = n. In matrix notation, − ≥ T 1 1 1 0 0 0 0 0 1 0 1 1 1 1 1       0 1 1 = 0 0 1 1 1 1  0 1 1 1   0 0 0 1   1 1         . . . . .   . . . . .   .   . . . . .   . . . . .   .              Lemma 6.1.3. Suppose (pn) is a given basis of k[x]. If for some sequence (sn) of polynomials and (βn) of power series holds

n sn(x)t = pn(x)βn(t) (6.3) n 0 n 0 X≥ X≥ then (sn) is a basis of k[x], if and only if βn is of order n.

Proof. In the notation of the previous proof, consider the matrix product (sn,k) = T (pn,k)(βn,k) , where (pn,k) is lower triangular and invertible. The matrix (sn,k) is T of the same kind iﬀ (βn,k) is lower triangular and invertible. In developing the ﬁnite operator calculus we now make the important deci- sion that we want to discuss operators that are isomorphic to the additive and multiplicative structure of formal power series. Therefore, we restrict ourselves to n the case βn (t) = β (t) , where β (t) is of order 1. Other choices are possible, and indeed have been pursued [36]. 6.1. Transforms of Operators 159

Remark 6.1.4. If we choose β (t) to be of order w 1, then ord(β (t)n) = wn, ≥ hence βn,k = 0 for all k = 0, . . . , wn 1, n 1, and βn,wn = 0. If the matrix − ≥ 6 T (pn,k) is again lower triangular and invertible, then (sn,k) = (pn,k)(βn,k) , where 2 deg (sn) = n/w in Lemma 6.1.3. For example, when w = 2, β (t) = t / (1 t), b n c m/2 1 m k 2− k and pn (x) = x (compare to example 6.3.3), then sn (x) = x bk=0 c− −k − x .

An operator X on k[x][[t]] is called k[[t]]-linear iﬀ X isP linear on k[x][[t]], and

n n X pn (x) t = (Xpn (x)) t n 0 n 0 X≥ X≥ n for all pn (x) t k[x][[t]]. The k[[t]]-linear operators on k[x][[t]] are called x- ∈ operators; their ring is denoted by x. For an x-operator X we only have to P L describe its action on some basis of k [x]. The linear operators on k [x] can be identiﬁed with x, if we remember the k [[t]]-linearity. L Example 6.1.5. (1) , (2) M (x) ( multiplication by x); (3) η := M (x) : xn nxn (note that η +D 1 is invertible); (4) χ : p (x) (p (x) p (0)) /x D( division7→ 7→ 1− operator), with χc = 0 for all c k. We have χ = (η + 1)− , but χ is also the left inverse of M (x), χM (x) = I∈. D

n Let s (x, t) = n 0 sn (x, t) t be in k[x][[t]] such that (sn) is a basis of k[x], ≥ and let X be an x-operator mapping sn to qn k[x] for all n,(qn may be of any degree). Hence P ∈

n n n Xs (x, t) = qn (x) t = t qn,ksk (x) = sk (x) t qn,k n 0 n 0 k 0 k 0 n 0 X≥ X≥ X≥ X≥ X≥ where qn,k = [sk (x)] qn (x) is diﬀerent from 0 for only ﬁnitely many k for very n. n Therefore, the sequence n 0 qn,kt is a sequence of elements from k[x][[t]] ≥ n 0 ≥ that is exactly of the type occurring in the presentation of any f(x, t) k[x][[t]] P ∈ as an element of k[[t]][x] (see (6.1)). We call the mapping

k n Xˆs sk (x, t) t = sk (x) t qn,k k 0 k 0 n 0 X≥ X≥ X≥ k n Xˆst = t qn,k n 0 X≥ ˆ the s-transform of X (with respect to s (x, t)). The operator Xs is a k[x]- linear, and is called a t-operator. The set t = Xˆs : X t is the set of all L ∈ L t-operators on k[x][[t]]. It is easy to see that t doesn not dependo on s (x, t). n L Let f (x, t) = n 0 pn (x) t be an arbitrary element from k[x][[t]]. Deﬁne ≥ P 160 Chapter 6. A General Finite Operator Calculus

n the x-operator X : x pn (x). Then 7→ n 1 ˆ 1 k ˆ k f (x, t) = pn (x) t = X (1 xt)− = X1/(1 xt) (1 xt)− = x X1/(1 xt)t − − − − n 0 k 0 X≥ X≥ k n = x pn,kt k 0 n 0 X≥ X≥ 1 n ˆ k n hence the (1 xt)− -transform of X : x pn (x) is X1/(1 xt) : t n 0 pn,kt . − 7→ − 7→ ≥ ˘ Vice versa, we can take any T t and deﬁne the s-transform ofPT , Ts, say, as the x-operator ∈ L

T˘ss (x, t) = T s (x, t) .

Of course, T˘s = T . The existence and uniqueness proof for T˘s is simple: We now that T s(x, t) k[x][[t]], hence there exists a unique sequence of polynomials (qn) (not necessarilyb ∈ of degree n) such that

n qn (x) t = T s(x, t). n 0 X≥

Hence T˘ssn (x) = qn (x) for all n 0. ≥ We only know t-operators as transforms of x-operators. In general, it may not be easy to decide wether a linear operator on k [[t]] is a t-operator or not. For k n example, if T t = n 0 pn,kt is of order 0 for all k 0, then for any basis (sn) we will get ≥ ≥ P k n n T sk (x) t = sk (x) pn,kt = t pn,ksk (x) k 0 k 0 n 0 n 0 k 0 X≥ X≥ X≥ X≥ X≥ 0 where the coeffi cient of t equals k 0 p0,ksk (x), which is an infinite sum of basis ≥ polynomials, and therefore not in k[x]. Operators T where ord T tn n k for a P ≥ − fixed integer k, and for all n 0, are in t. ≥ L Example 6.1.6. The t-operator Dt, and multiplication by t, are examples for operators in t. We denote differentiation with respect to the power series variable t L by Dt, while we write for the (more commonly used) derivative with respect to the variable of the polynomials.D We now give two important classes of t-operators where the image of tn is a formal power series of order larger than or equal to n. Let σ, β k [[t]] such that ∈ ord (β) > 0. The k[x]-linear operators composition C(β) and multiplication M(σ) are defined on k[x][[t]] such that

n M(σ)C(β)s(x, t) = M(σ)s(x, β(t)) = sn(x)σ(t)β(t) . (6.4) n 0 X≥ 6.1. Transforms of Operators 161

Any two multiplication operators commute. Note that

C(β)M(σ) = M(σ (β))C(β). (6.5)

We often talk about the transform, omitting s from the notation when it is clear from the context, and whenever possible, we write just T˘ and Xˆ, without subscripts. Transforming is an anti-isomorphism,

` (T1T2) = T˘2T˘1 and X\1X2 = Xˆ2Xˆ1. (6.6)

Check that transforms of multiplication operators commute,

M(ρ)`M(σ)` = M(σ)`M(ρ)`. (6.7)

Example 6.1.7. We calculate some frequently used transforms.

xt ` 1. The t-operators Dt and M (t) have the e -transforms D˘ t = M (x), M (t) = ` , (M (t) Dt) = M (x) = η. D D n n xt 2. If ηtis the t-analog to η, i.e., ηt : t nt , then η˘t = η is the e -transform. 7→ 3. The division operator χ has the ext-transform

1 ∧ 1 ∧ 1 χˆ = (η + 1)− = ˆ (η + 1)− = M (t)(ηt + 1)− . D D n 4. If the basic sequence (bn) is the r-image of β (t) (the basic sequence with respect to ), then R ` C(β) rn (x) = bn (x) (6.8) is an example of an umbral operator.

1 ` 5. Examples for (1 xt)− -transforms are D˘ t = ηM (x), M (t) = χ, and ηˆ = − M (t) Dt. 162 Chapter 6. A General Finite Operator Calculus

6.2 Reference Frames, Sheﬀer Sequences, and Delta Op- erators

The standard basis (xn/n!), the exponential series ext, and the derivative operator are all interrelated through formal power series, and we are now looking for D n similar triples (rn(x)), r(x, t) := n 0 rn(x)t , and : rn(x) rn 1(x). We ≥ R 7→ − will call any standard basis (rn(x)) a reference sequence, i.e., deg rn(x) = n, and P rn(0) = δ0,n. The operator will be called the reference operator, r(x, t) the reference series, and all threeR ∈ together ⊗ make the reference frame. We begin with a closer look at reference frames, and continue with investi- gating Sheﬀer sequences and delta operators in general reference frames.

6.2.1 Reference Frames

The reference operator defines the sequence (rn) uniquely, because rn (0) = δ0,n. For the reference operatorR holds that is the r-transform of M(t), = M(t)`. Every linear operator thatR commutes withR can be written as a powerR series in (see Exercise 6.2.4 for details). R R n xt Example 6.2.1. 1. rn(x) = x /n! with reference series e gives the Finite Op- erator Calculus we discussed in the previous chapters. We call this reference frame exponential. 2. The sequence xn/ (n!)2 is standard; if we take it as our reference sequence, we will get the reference operator M (x) , D D xn xn xn 1 M (x) = = − . D D n!n! D n!(n 1)! (n 1)! (n 1)! − − − Note that M (x) = + M (x) 2, and D D D D k k 2 ( M (x) )k = (k i)!M xi k+i. (6.9) D D i − D i=0 X k 2 (0) (the coeffi cients i (k i)! occur in the Laquerre polynomials, m!Lm (x), Example 5.3.1). See Exercises− 6.2.2 and 6.2.3 for details. Identy (6.9) shows the “normal ordering” of the k-th power of the operator. The coeffi cients are called “Generalized Stirling Numbers” in [12]. For the reference series we obtain π ∞ xntn 1 = e2√xt cos u du = I 2√xt , n!n! π 0 n=0 0 X Z the (modified, real valued) Bessel function of the first kind. Hence tI0 2√xt = 1 M (x) π e2√xt cos u du. If we change the reference operator to π D D 0 R (m + 1) + M (x) 2, D D 6.2. Reference Frames, Sheffer Sequences, and Delta Operators 163

m/2 where m N0 is ﬁxed, then the reference series equals Im 2√xt / (xt) ∈ k n (Exercise 6.2.5). More on operators of the form n 0 k 0 cn,k M (x ) ≥ ≥ D can be found in [25]. P P 3. The operator M (x) 3 is in Ω, and therefore a reference operator. The D D −D xt t2/2 Hermite polynomials Hn (x) with generating function e − are a basis that 3 follows the recurrence M (x) Hn (x) /n! = Hn 1 (x) / (n 1)!, but D D − D − − (Hn (x) /n!) is not the reference sequence, because Hn (0) = δ0,n (see (6.36)). 6 4. In Example 6.1.2 we saw the reference sequence r0(x) := 1,

n rn(x) := x (1 x ) /(1 x) − − for n > 0, with reference series

x t xt xt r(x, t) = 1 + = 1 + . (6.10) 1 x 1 t − 1 xt (1 t) (1 xt) − − − − − The reference operator is R n n 1 x = (rn(x) rn 1(x)) = rn 1(x) rn 2(x) = x − R R − − − − − 2 0 for all n 3, x = x 1, and x = r1(x) = r0(x) = 1 = x . Hence ≥ R − R R n 1 x − if n 1, n = 2 xn = ≥ 6 R x 1 if n = 2. − We denoted the operator M (x) by η, thus ηxn = nxn. If d (n) is any D k-valued function on N0 (see [65], [35]), then we deﬁne d(η)(xn) := d(n)xn. (6.11)

n If d (n) = 0 for all n 0, and d (0) = 1, the polynomial sequence (d (n) x )n 0 is an example6 of a reference≥ sequence. Such sequences and their reference frames≥ n n are called diagonal. If (rn) is diagonal we write r (xt) = n 0 d (n) x t instead of r (x, t). With the help of the division operator χ, ≥ P p (x) p (0) χp (x) = − for all p (x) k [x] , x ∈ we can explicitly write down the reference operator , R

n n 1 d (η) n d(η)x = d (η) x − = χd(η)x R d (η + 1)

d(η) hence = χ = d (η) χ/d (η). (In Kwa´sniewski’s notation [54], = ∂ψ, R d(η+1) R when d (n) = ψn.) 164 Chapter 6. A General Finite Operator Calculus

Example 6.2.2. 1. If d(n) = n + 1, we obtain d (n) xn = (η + 1) xn. Hence

n n 2 d (n) x t = x (x/ (1 xt)) = (1 xt)− D − − n 0 X≥ n n 1 η n (η + 1) x = ηx − = χ (η + 1) x . R η + 1 This is an example from the family of 2-binomial reference frames. More on binomial reference frames in section 7.1.

1 n n 2. We obtain the exponential reference sequence from η!− x = x /n! for all n = 0, 1,... Hence

1 η!− = 1 χ = (η + 1) χ = χη = R (η + 1)!− D

n 1 k n 3. Let (z; q)n := k=0− 1 zq , and rn (x) = x / (q; q)n. The reference oper- − η ator equals = Dq := χ (1 q ), because R Q − n n n n 1 x n x χx x − Dqrn (x) = Dq = χ (1 q ) = = . (q; q)n − (q; q)n (q; q)n 1 (q; q)n 1 − − This reference frame is called q-diﬀerential by Andrews [5]. The reference series xn 1 tn = (q; q) (xt; q) n 0 n X≥ ∞ is derived in Exercise 6.2.1. The q-diﬀerential reference frame follows from the exponential reference frame by substituting 1 qη for η. − Diagonal reference sequences have been studied, among others, by Roman [78] and Kwa´sniewski [54].

Three diagonal reference series and their reference frames Binomial ( α / N0) Exponential q-diﬀerential − ∈ η+α 1 d(η) = η− 1/η! 1/ (q; q)η α xt (6.12) r(xt) = (1 xt)− e 1/ (xt; q) − ∞ n+α 1 n n n rn(x) = n− x x /n! x / (q; q)n 1 η+1 η = (η + α)− = χ = χη D := χ (1 q ) R D η+α D (q) − (continued in Example 6.3.2). 6.2. Reference Frames, Sheﬀer Sequences, and Delta Operators 165

6.2.2 Sheﬀer Sequences and Delta Operators

In Lemma 6.1.3 choose for (pn) the reference sequence (rn(x)). Every pseudo-basis (βn) of k [[t]] uniquely defines a basis (sn) of k[x], and vice versa, because n sn(x)t = rn(x)βn(t). n 0 n 0 X≥ X≥ We call (sn) the r-image of (βn). One could wish for more structure on (βn); for example we could ask for βn (t) = σ (t) γn (t), where σ (t) is of order 0, and f(i+j) γn (t) satisfies some multiplication rule, like γi (t) γj (t) = q γi+j (t), where f is some function on N0, and q is some suitable constant or formal variable. For the following, we choose f identical 0. n Definition 6.2.3. A (generalized) Sheffer sequence (sn) is the r - image of (σ(t)β(t) ) where β is a delta series and σ is invertible,

n n sn(x)t = σ(t) rn(x)β(t) . n 0 n 0 X≥ X≥ The pair (σ, β) is in the umbral group; only the reference sequence has changed. If σ(t) = 1 we call the resulting Sheffer sequence (bn) a (generalized) basic sequence. Comparing this definition to equation 6.4 shows that for any pair of delta series β (t) and invertible series σ (t) the series M(σ)C(β)r(x, t) is the generating function of a Sheffer sequence for the corresponding reference operator . We usually omit the qualifier “generalized”. A Sheffer sequence with respectR to a diagonal reference series is a Boas-Buck sequence [14]; its generating function is n n therefore of the form σ (t) n 0 d (n) x β (t) . ≥ n 2 Example 6.2.4. Let rn(x)P = (n + 1) x , ρ (t) = (1 t)− , and β (t) = t/ (1 t). The reference frame is therefore binomial with α =− 2 (2-binomial). The series−

n n 2 (n + 1) x t 1 t − = 1 x (1 t)2 1 t (1 t2) − 1 t n 0 X≥ − − − − 2 ∞ n n = (1 t (1 + x))− = (n + 1) t (1 + x) − n=0 X is the generating function of the 2-binomial Sheﬀer polynomial (n + 1) (1 + x)n = 1 E rn(x). Example 6.2.5. We saw in Example 6.1.2 that r(x, t) = 1 +xt/ ((1 t) (1 xt)) − − ∈ k[x][[t]]. With σ (t) = 1 t and β (t) = t/ (1 t) we obtain the generating function − − t t t 1 M (2t 1) C r(x, t) = C M − r(x, t) − 1 t 1 t t + 1 − − xt (1 t) = 2t 1 + − k[x][[t]]. − (1 t xt) ∈ − − 166 Chapter 6. A General Finite Operator Calculus

2 n 2 hence s0 (x) = 1, s1 (x) = 2 + x, and sn (x) = x (1 + x) − for n 2 is a generalized Sheﬀer− sequence in this frame work. ≥ There exists a simple, but useful “special case” of the binomial theorem (2.14).

Lemma 6.2.6. For every Sheﬀer sequence (sn) for a delta operator with basic sequence (bn) holds n

sn (x) = si (0) bn i (x) − i=0 X in any reference frame for all n 0. ≥ n Proof. By Deﬁnition 6.2.3, n 0 sn (0) t = σ (t) and ≥ P n n bn (x) t = rn(x)β(t) . n 0 n 0 X≥ X≥

As in the exponential reference frame, a delta operator is isomorphic to a delta series, but with respect to a general reference operator . R Definition 6.2.7. Suppose B is an x-operator. We call B the -delta operator associated to the delta series β provided that B is the C (β) r-transformR of M (t), BC (β) r(x, t) = M(t)C (β) r(x, t). (6.13) Hence is also a delta operator, when we choose β (t) = t. By Exercise 6.2.8, R 1 1 B = β− (M(t)`) = β− ( ) (6.14) R 1 1 where β− (t) is the compositional inverse of β, i.e., β− (β(t)) = t. This shows that B Σ , the set of operators that have a power series expansion in . Any - ∈ R 1 R R delta operator depends on the delta series β− (t) and the reference operator . If the frame of reference is understood from the context, we call B a delta operator.R Note that every delta operator is also a reference operator, but for the frame r(x, β (t)). In other words, if B Σ and B is a delta operator, then ΣB. This means that the set Ω of operators∈ R reducing the degree by one, is partitionedR ∈ into equivalence classes Ω Σ , where and B are equivalent iff there exists a delta series β such that β (∩B)R = .By ExerciseR 6.2.4, dividing Ω into equivalence classes of commuting operators (centralizers)R gives the same partition. n n n k Example 6.2.8. The Catalan operator : x k=1 Ck 1x − , is a χ-delta operator associated to β (t) = t t2, becauseC 7→ − − P n 2 2 1 2 n n k 1 1 1 4 (t t ) C t t = t t Ck 1x − = − − − C − 1 xt − − 2 1 x (t t2) − n 1 k=1 p− − X≥ X 1 1 = t 1 x t t2 − = M (t) C t t2 . − − − 1 xt − 6.2. Reference Frames, Sheffer Sequences, and Delta Operators 167

1 Hence β− (t) = 1 √1 4t /2, and − − ∞ k 1 1 1 = Ck 1χ = β− (χ) = 1 4χ. C − 2 − 2 − k=1 X p We saw in Deﬁnition 6.2.3 that for a Sheﬀer sequence (sn) holds

n s (x, t) = sn (x) t = M (σ) C (β) r (x, t) , n 0 X≥ where β is a delta series, and σ has a reciprocal in k [[t]]. Note that

n s (0, t) = sn (0) t = σ(t). (6.15) X Theorem 6.2.9. A polynomial sequence (sn) is a Sheﬀer sequence for B iﬀ Bsn = sn 1 for all integers n 0. − ≥ Proof. Exercise 6.2.9.

Corollary 6.2.10. Two Sheﬀer sequences (sn) and (tn) for B that agree at one argument for every n must be identical, i.e., if there is a sequence x0, x1,... such that sn (xn) = tn (xn), then sn (x) = tn (x) for all n 0. ≥ Proof. Exercise 6.2.11. The generalized Sheﬀer sequence (bn) is a basic sequence, if σ (t) = 1, i.e.,

n b (x, t) = bn (x) t = C (β) r (x, t) n 0 X≥ 1 for some delta series β and the reference series r. Again, B = β− ( ) is the associated -delta operator. Of course, we can make B into the new referenceR operator R with reference series b (x, t) and reference sequence bn (x), because Bˆ = M (t) w.r.t. C (β) r (x, t). This shows that bn (0) = δ0,n. However, not every reference frame can be obtained by substitution of a delta series. For example, the two reference series ext and 1/ (1 xt) are related by tn tn/n!, an operation that is not a substitution of a delta− series. Every reference↔ frame defines a “universe”of other reference frames, but there are infinitely many “parallel universes”! We call two reference frames r (x, t) and r0 (x, t) equivalent, if there exists delta series β (t) such that r (x, β (t)) = r0 (x, t). If both frames are diagonal, r (x, t) = n n n 0 d (n) x t and r0 (x, t) = r0 (x, t), then we also say that d (η) and d0 (η) are equivalent.≥ We saw above that 1/η! and 1 are not equivalent. The operators d (η) P η and d0 (η) are equivalent iff d (η) = a d0 (η) for some a k, because ∈ n n n n d (n) x β (t) = d0 (n) x t n 0 n 0 X≥ X≥ n n iff d (n) /d0 (n) = t /β (t) . 168 Chapter 6. A General Finite Operator Calculus

1 Sheffer operator Let B = β− ( ) be a delta operator, and (sn) a Sheffer sequence for B with generating functionR s (x, t) = σ(t)r (x, β (t)). The Sheffer opera- 1 tor of (sn) is the invertible operator S := σ(β− ( )) = σ(B). The Sheffer operator ` R is the C (β) r-transform of M (σ), S = M(σ)C(β)r. Note that S and B commute. The (generalized) B-basic sequence (bn) has the generating function C (β) r, and therefore

1 s (x, t) = M (σ) C (β) r (x, t) = C (β) M σ(β− )r (x, t) = C (β) Sr (x, t) = SC (β) r (x, t) hence

sn(x) = Sbn (x) . (6.16) Sheffer operators commute with all other operators in Σ . Every invertible operator τ ( ) in Σ is a Sheffer operator, R R R 1 τ ( ) = τ β β− ( ) (6.17) R R thus τ ( ) is the Sheffer operator for the Sheffer sequences with generating function τ(β (t))Rr (x, β (t)).

Superposition of Sheffer sequences If (sn) and (tn) are Sheffer sequences for the same -delta operator B, then sn (x) + tn k (x) is a Sheffer polynomial for B, as R − long as k > 0 remains fix. If k = 0 then sn (x) + tn (x) has to be of degree n. Let ν0, ν1,... be a sequence of scalars. Assume that t0 (x) is a non-zero constant polynomial. It can be shown by straightforward verification that (tn) defined by n

tn(x) = tj(νj)sj,n j(x), (6.18) − j=0 X is a Sheﬀer sequence for B, if (sj,n)n N0 denotes for each j N0 the generalized ∈ ∈ Sheﬀer sequence for B with roots in νn+j,

sj,n (vn+j) = δn,0.

(see also Exercise 6.2.12). Hence, if the initial values tn(νn) are given, we can expand tn (x) provided the Sheﬀer sequences (sj,n)n N0 can be found. ∈

1 1 The umbral group Suppose B = β− ( ) and A = α− ( ) are both -delta R R R operators, with basic sequences (bn) and (an), respectively, where

an (x) = an,iri (x) . i=0 X 6.2. Reference Frames, Sheﬀer Sequences, and Delta Operators 169

1 1 The operator B (A) = α− β− ( ) is also an -delta operator. The basic se- R R quence of this operator is called the umbral composition of (bn) with (an), with n associated basic polynomials an (b (x)) := i=0 an,ibi (x), because

` ` ` n C (α) C (β) r = C (α) C (β) r = C (β) CP(α) r = C (β) an (x) t (6.19) n 0 X≥ n n ` n n = an,iC (β) ri (x) t = an,ibi (x) t n 0 i=0 n 0 i=0 X≥ X X≥ X ` (see (6.8)). The sequence (an (b (x)))n 0 is basic for B (A). The operator C (β) : ≥ an (x) an (b (x)) is called an umbral operator. In exponential ﬁnite operator cal- 7→ 1 culus we denoted umbral operators by Uβ− ( ). Umbral operators do not commute with except when β (t) = t (Exercise 2.3.13).D UmbralR operators form the umbral group,

C (β)` C (α)` = C (β α)` = C (β (α))` ◦ written in terms of generating function (Exercise 6.2.14). If an (b (x)) = rn (x) for all n, we say that (bn) is inverse to (an) (see section 2.3.2). The details of the umbral group are exactly the same as in the exponential reference frame (section 2.3.2). The composition of the two Sheﬀer sequences (σ, β) and (ρ, α), representing (sn) and (tn), respectively, remains (σ, β) (τ, α) = (σ (α) τ, β (α)), but this element has the representation ◦

n n

tn (s (x)) = rk (x) tn,isi,n. k=0 i=k X X Hence (σ, β) is inverse to (τ, α) if tn (s (x)) = rn (x), i.e.,

1 β− (t) = α (t) and 1/σ (α (t)) = τ (t) . (6.20)

6.2.3 Exercises 1 6.2.1. Let 0 < q < 1. Show that limn n xtqk converges for small xt. Show →∞ k=0(1 ) that − xn Q 1 tn = (q; q) (xt; q) n 0 n X≥ ∞ (Euler). 6.2.2. Show identity (6.9) by applying ( M (x) )k to the reference sequence. D D a M(x) 6.2.3. For any 0 = a F, the operator Ua := e D D can be seen as an analog a 6 ∈ n n (0) to e D, the translation by a. Show that Ua : x /n! a Ln ( x/a) (see 5.3). As xt 1 xt/(1 at) 7→ − an x-operator, Uae = 1 at e − . More on operators of the form Ua can be found in [31]. − 170 Chapter 6. A General Finite Operator Calculus

6.2.4. Suppose T is a linear operator that commutes with . Show: R k T = Eval0 T rk h | i R k 0 X≥ The case = is equivalent to Lemma 2.2.2. R D 6.2.5. Show that for any given integer m N0 the reference sequence for the 2 ∈ m/2 operator (m + 1) + M (x) is the Bessel function Im 2√xt / (xt) D D 6.2.6. Let (sn) be a Sheffer sequence for a diagonal reference frame, s (x, t) = σ (t) r (xβ (t)). Show that sn (1) = δ0,n iff s (x, t) = r (xβ (t)) /r (β (t)). 6.2.7. Show that the q-differential reference frame follows from the exponential reference frame by substituting 1 qη for η. − 6.2.8. Let B be the C (β) r-transform of M (t), where β is a delta series. Show that 1 B = β− ( ). R 6.2.9. Let (sn) be a polynomial sequence with generating function

n s (x, t) = sn (x) t . n 0 X≥ Show: (sn) is a Sheffer sequence with generating function s (x, t) = M (σ) C (β) r (x, t) iff s (x, t) = M (β) s (x, t). Note that the latter condition is equivalent to Bsn (x) = R sn 1(x), i.e., − 1 β− ( ) s (x, t) = M (t) s (x, t) . R 1 6.2.10. Let B = β− ( ) be a delta operator, and (sn) a Sheffer sequence for B with R Sheffer operator S. Show: If pn := sn+1 for all n N0, then (pn) is a Sheffer sequence for B. Let φ (t) := β (t) /t.R Show that Sφ (B∈) is the Sheffer operator of (pn).

6.2.11. Let x0, x1,... be a given sequence of scalars, and suppose (sn) and (tn) are Sheffer sequences for B such that sn (xn) = tn (xn) for all n. Show that sn (x) = tn (x) for all n. 6.2.12. Show that (6.18) holds. 6.2.13. Let r (xt) be a diagonal reference series, with reference operator . Suppose 1 R B = β− ( ) is a delta operator, and (bn(x)) the basic sequence for B. Show: If R a is a scalar different from 0, then (bn(ax)) is the basic sequence for the delta 1 operator β− ( /a). R 6.2.14. Show that for any two umbral operators C (β)` and C (η)` holds C (β)` C (η)` = C (β η)`. ◦ 6.2.15. Let (tn) be a Sheffer sequence corresponding to the umbral element (τ, α), and (bn) the basic sequence for B. Show that (tn (b (x))) has the Sheffer operator τ (B (A)). 6.3. Transfer Formulas 171

6.3 Transfer Formulas

If two delta operators A and B in the same reference frame are connected by a formal power series τ k [[t]], we have in analogy to section 2.4 ∈ 1 1 1 1 B = β− ( ) = τ − (A) = τ − α− ( ) R R and therefore

b (x, t) = C (β) r(x, t) = C (α (τ)) r(x, t) = C (τ) a(x, t) = a (x, τ (t)) (6.21) hence n i bn (x) = τ n ai (x) (6.22) i=0 X As before, we are mainly interested in the case where τ has coeffi cients in Σ . In this case, the transfer formulas are expressed in terms of the Pincherle derivative,R and for the Pincherle derivative we have to define the general umbral shift first.

6.3.1 General Umbral Shifts and the Pincherle Derivative Deﬁnition 6.3.1. The umbral shift associated to is the linear operator denoted by θ such that R R θ rn(x) = (n + 1)rn+1(x) for n 0. R ≥ The degree reducing operator θ− is the left inverse of θ , R R 1 1 θ−rn(x) = rn 1(x) = rn(x) for n 1, and R n − nR ≥ θ−1 = 0. R Note that i i+1 θ−rn(x) = rn(x)/n for all n 1. R R R ≥ In transform notation,

˘ 1 ^ θ = Dt and θ− = Dt− R R 1 n n+1 where D− : t t / (n + 1). We obtain t 7→ n θ 1 = n!rn(x) for all n 0. R ≥ Example 6.3.2. Let be a diagonal reference operator, = d (η) χ/d (η), and n R R rn (x) = d (η) x (see (6.11)). We calculate the diagonal umbral shift as

θ d (η) xn = ηd (η) xn+1 = ηd (η) M(x)xn = ηM(x)d (η + 1) xn R 172 Chapter 6. A General Finite Operator Calculus hence d (η + 1) θ = ηM(x) R d (η) We calculate the umbral shifts for the three basic examples (6.12) of diagonal reference frames. Three important reference frames and their umbral shifts Binomial Exponential q-diﬀerential η+α 1 d (η) = η− 1/η! 1/ (q; q)η α xt r(xt) = (1 xt)− e 1/ (xt; q) − ∞ n+α 1 n n n rn(x) = n− x x /n! x / (q; q)n 1 η+1 η = a = (η+ α)− = χ = χη D(q) = χ (1 q ) R B D η+α D − η θ = M(x)(η + α) M(x) 1 qη M(x) R − 1 1 qη θ− = η+α χ χ χ −η R

In all three example we expressed θ and θ− in terms of operators applicable to any polynomial p, not just the referenceR sequence.R

Example 6.3.3. Deﬁne the reference sequence (rn) by r0(x) := 1,

n rn(x) := x (1 x ) /(1 x) − − for n > 0. We saw in Example 6.2.1 that the reference operator equals R n 1 x − if n 1, n = 2 xn = ≥ 6 R x 1 if n = 2. − We calculate the umbral shift as

n θ x = θ (rn(x) rn 1(x)) = (n + 1) rn+1(x) nrn(x) R R − − − n+1 n+1 = rn(x) + (n + 1) x = rn+1(x) + nx for n 2, and θ x = θ r1(x) = 2r2(x), θ 1 = x. Hence ≥ R R R n+1 n rn+1(x) + nx if n 0, n = 1 θ x = 2 ≥ 6 R 2x + 2x if n = 1. The inverse shift can be represented as

n 1 1 θ−x = θ− (rn(x) rn 1(x)) = rn 1(x) rn 2(x) R R − − n − − n 1 − n 1 − (n 1) rn 1(x) nrn 2(x) nx − rn 1(x) = − − − − = − − n(n 1) n(n 1) − − 2 1 for n 3, θ−x = 2 x 1, and θ−x = 1. ≥ R − R 6.3. Transfer Formulas 173

If T is an operator then the Pincherle derivative T 0 of T is deﬁned as R

T 0 = T θ θ T. (6.23) R R − R d If k is a ring of scalars, then T Σ implies T 0 = d T (Exercise 6.3.2); if ∈ R R k contains operators they must commute with θ for thisR result to hold true. Diﬀerent umbral shifts produce diﬀerent PincherleR derivatives. Note that we mark d Pincherle derivatives by a prime, and write d T if T has a power series series expansion in . R R 6.3.2 Equivalent Transfer Formulas Let B be a delta operator. As a power series in , the operator B is of order 1 R 1, hence the linear operator P − := B/ is invertible. In terms of P , PB = . R R Suppose (bn) is the basic sequence for B. The transfer formulas in [83, p. 695] carry over to the Freeman approach: If (rn) is the reference sequence, then for all positive integers n holds

d b (x) = B P n+1r (x) (6.24) n d n R n 1 d n bn(x) = P rn(x) P rn 1(x) (6.25) − n d − R n 1 n bn(x) = θ P θ−rn(x) = θ P rn 1(x) (6.26) R R n R − (see Exercise 6.3.6). Combining the ﬁrst and the last of the three formulas proves the Rodrigues’type formula

1 1 d − 1 d bn(x) = θ B bn 1(x) = θ Rbn 1(x) n R d − n R dB − R 1 d 1 n = θ β (B) = (θ β0 (B)) 1. n R dB n! R

The transfer formulas can also transfer from any basic sequence (an) to (bn).

Lemma 6.3.4. Let A, B Σ be delta operators with basic sequences (an) and (bn), respectively. Deﬁne the∈ invertibleR operator T Σ by TB = A. For all positive integers n holds ∈ R

1 d d − b (x) = B A T n+1a (x) (6.27) n d d n Rn R = θ T θ−an(x). R R Proof. In the same way as we deﬁned P through PB = we deﬁne S through d n+1 R SA = . We saw above that an(x) = d A S rn(x), and therefore rn(x) = R R 174 Chapter 6. A General Finite Operator Calculus

1 d − n 1 1 d A S− − an(x). Note that T = PS− . Substituting for rn into (6.24) gives R n the ﬁrst result. The third formula above shows that θ S− θ−an(x) = rn(x) for R all n 1. Hence R ≥ n n n bn(x) = θ P θ−rn(x) = θ P S− θ−an(x) R R R R shows the second equation. i n n i Remember that t φ is long-hand for [φ ]i, the coeﬃ cient of t in the power series φ (t)n. Combining transfer with Lagrange-Bürmann inversion gives the following useful expansion in situations where T cannot be explicitly calculated. Corollary 6.3.5. If the delta operator A Σ can be written as A = τ(B) = i ∈ R i 1 TiB Σ [[B]] for some linear operator B, such that Ti Σ for all i 1, ≥ ∈ R ∈ R ≥ and T1 is invertible, then B is also a delta operator in Σ , and the basic sequence P R (bn) of B can be expressed in terms of the basic sequence (an) of A as

n 1 − n n i i bn (x) = θ τ − n A θ−an (x) . R n i R i=0 − X for all n > 0. 1 n Proof. We saw in (6.27) that bn = θ PS− θ−an and we obtain R R n n PS 1 = τ n i Ai − n i − n i 0 − X≥ exactly in the same way as in section 2.4.2. When A and B are delta operators in Σ , then they are also delta operators R in ΣA. Hence we can think of (an) as the reference sequence, with θA−an(x) = an 1 (x) /n for positive n, and expand bn as − n 1 − 1 n i bn (x) = θA τ − an 1 i (x) (6.28) n i n − − i=0 X − n 1 1 − 1 n i d − = θ τ − A an 1 i (x) R n i n d − − i=0 − R X (see Exercise 6.3.1). If τ k [[t]], we arrive at the simple expansion (6.22). ∈ k j If A is a polynomial in B, A = τ (B) = j=1 TjB , then the coeﬃ cient of Bn in τ n i (B) equals − P k n i lj − Tj . l1, . . . , lk l1+ +lk=n i j=1 l1+2l···2+X+klk−=n Y ··· This proves the following Corollary. 6.3. Transfer Formulas 175

k j Corollary 6.3.6. If the delta operator A Σ can be written as τ(B) = j=1 TjB ∈ R such that Tj Σ , T1invertible, then B is also a delta operator, and the basic ∈ R P sequence (bn) of B can be expressed in terms of the basic sequence (an) of A as follows

n 1 k − n n i lj i bn = θ − Tj A θ−an R n i l1, . . . , lk   R i=0 − l1+ +lk=n i j=1 X l1+2l···2+X+klk−=n Y ···   n l + + l k 1 k lj n (l1+ +lk) = θ ··· Tj A − ··· θ−an. R l1 + + lk l1, . . . , lk   R l1+2l2+ +klk=n j=1 X··· ··· Y   6.3.3 Exercises 6.3.1. Suppose we make a (minor) change to the reference frame by substituting a delta series β (t) for t, i.e., our new reference series is b (x, t) := r(x, β (t)), and 1 our new reference operator is the -delta operator B = β− ( ). Find the umbral R R 1 0 shift θB associated to B, and show that θB− = β− ( ) θ−. R R k 6.3.2. Show: If T = k 0 τk Σ then ≥ R ∈ R P d k k 1 T 0 = τk = τkk − . R d R R k 0 k 0 R X≥ X≥ This will not hold in general if T has operators from Σ as coeffi cients. R d 1 d 1 6.3.3. Show that d E is different from d∆ E . D 6.3.4. Show the product rule of differentiation for the general Pincherle derivative, n n 1 (ST )0 = S0 T + ST 0 , for S, T Σ . This implies (T )0 = nT − T 0 . R R R ∈ R R R 6.3.5. In Example 6.3.3 express the umbral shifts θ and θ− in terms of χ, η, and M (x). R R 6.3.6. Show that the equalities (6.24) - (6.26) hold by verifying that the right hand sides are the same, and generate a Sheffer sequence for B, which has the correct initial values. 176 Chapter 6. A General Finite Operator Calculus

6.4 Functionals

The transform of a functional is deﬁned with respect to the reference series r(x, t),

n Lˆrr(x, t) = Lr(x, t) = L rn t . h | i n 0 X≥ n ˆ n ˆ For any t-operator T holds T r(x, t) = n 0 rn (x) T t , hence Lrt = δ0,n, and Lr maps t0 = 1 onto the power series λ (t) :=≥ L r tn. More important for P n 0 n us will be the application of λ (t) as multiplication≥ h operator| i M (λ (t)). We deﬁne the product of two functionals L and N, P n

L r N rn := L rk N rn k , h ∗ | i h | i h | − i k=0 X i.e.,

(L r N) r(x, t) = (Lr(x, t)) (Nr(x, t)) = Lˆr1 Nˆr1 = λ (t) ν (t) . (6.29) ∗ The evaluation at 0, Eval0, is the multiplicative unit. A linear functional L is invertible (w.r.t. r-multiplication), iﬀ L 1 is a unit in k, the same as in section ∗ h | i 3.1. The associated operator to the functional L K [x]∗ is the transform with respect to r (x, t) of the multiplication operator M(∈λ), i.e. λ( )r (x, t) = M(λ)r (x, t) R for all L K[x]∗. We also write op (L) for λ( ). We have the isomorphisms ∈ R K [x]∗ K [[t]] Σ . Note that λ( ) commutes with delta operators, because they are←→ all in Σ ←→. In ExerciseR 6.4.7 weR will prove that R k n k 0 L sk B ≥ h | i op (L) = L rn = n (6.30) h | i R Eval0 sn B n 0 Pn 0 X≥ ≥ h | i P for any Sheﬀer sequence (sn) with delta operator B in the reference frame r. Example 6.4.1. Let c k [x]. We get ∈ ` n op (Evalc) = M rn (c) t   n 0 X≥  n ` n = rn (c) M (t ) = rn (c) = r (c, ) . R R n 0 n 0 X≥ X≥ We can check this result applying (6.30),

k k 0 sk (c) B ≥ op (Evalc) = n = r (c, β (B)) = r (c, ) . n 0 sn(0)B R P ≥ P 6.4. Functionals 177

Lemma 3.1.1 and Corollary 3.1.2 also hold for general reference frames,

` L r N = (Lr(x, t)) (Nr(x, t)) = (Lr(x, t)) M(ν) = Lν ( ) , ∗ R where ν (t) = Nr(x, t). By defining T = op (N) andJ = L r N above, we obtain ∗ op (J) = op (L) T if J p = L T p for all p k[x]. h |Nowi weh have| i all the ingredients∈ together for the general functional expansion theorem. Theorem 6.4.2. If L is a functional such that L 1 has a reciprocal (in k), and h | i (sn) is a Sheffer sequence and (bn) the basic sequence for the same r-delta operator, then n 1 sn (x) = L sn k op (L)− bk (x) (6.31) h | − i k=0 X and k n k 0 L sk t ≥ h | i sn (x) t = n r (x, β (t)) L bn t n 0 Pn 0 X≥ ≥ h | i Proof. The proof is very much theP same as the proof of Theorem 3.1.4, and we only give a sketch. First, for every Sheffer sequence (ln) for the r-delta operator B with basic sequence (bn) holds that

k p (x) = L B p lk (x) | k 0 X≥ for all p (x) k [x]. With the help of the Sheﬀer operator S for (ln) this can be written as ∈ k p (x) = L B p Sbk (x) . | k 0 X≥ 1 In Exercise 6.4.1 it is shown that S = λ ( )− . R Example 6.4.3. [65] Suppose L is an invertible linear functional, and ` a positive integer. If B is an r-delta operator with basic sequence (bn), and (pn) a (known) Sheﬀer sequence for B, we want to solve to the initial value problem

Bsn (x) = sn 1 (x) for all n = 1, 2,... − sn (x) = pn (x) for all n = 0, . . . , ` 1 − L sn = 0 for all n = `, ` + 1,... h | i According to (6.31) we have

n ` 1 1 − 1 sn(x) = L sn k op (L)− bk (x) = L pi op(L)− bn i(x). h | − i h | i − k=0 i=0 X X 178 Chapter 6. A General Finite Operator Calculus

We can also apply (6.31) to pn (x), n 1 pn (x) = L pn k op (L)− bk (x) , h | − i k=0 X hence n 1 sn(x) = pn(x) L pi(x) op(L)− bn i. − h | i − i=` X For the generating function we get

` 1 k n k−=0 L pk t j sn (x) t = h | i k bj (x) t . L bk t n 0 Pk 0 j 0 X≥ ≥ h | i X≥ 1P 1 Suppose we take the functional pn = pn(x)dx, obtaining ∫0 | 0 n 1 R bn i(x) sn(x) = pn(x) pi(x)dx − . − 1 i=` 0 r(x, )dx X Z 0 R We discussed this functional in Example 3.1.6 for theR exponential reference frame. α If r (x, t) = (1 xt)− , the binomial framework, we must distinguish two cases. If α = 1 then =− χ : f(x) (f(x) f(0)) /x, and R 7→ − 1 1 1 1 r (x, t) = (1 xt)− dx = ln (1 t) . ∫0 − − t − Z0 Hence 1 1 − n χ n n j op x = − x = bjx − (6.32) ∫0 ln (1 χ) − j 0 − X≥ where b0, b1,... are the Bernoulli numbers of the second kind [47, (9) § 97] (Ex- ercise 2.3.3). If α = 1, 0, 1, 2,... then 6 − − 1 1 1 α α 1 (1 t) − r (x, t) = (1 xt)− dx = − − . ∫0 − t(1 α) Z0 − 1 For example, if α = 1/2, then 0 r (x, t) = 2 1 √1 t /t, which equals the Cata- ∫ − − 1 1 − t lan generating function c (t) in (1.2) at t/4. Hence 0r (x, t) = 1 4 c (t/4). In this case ∫ − 1 1 k+1 − ∞ k 1 η + 1 op = 1 4− − Ck χ ∫0 − η + 1/2 k=0 X 1 1 1 2 , η + 1; χ = + 2F1 − . 2 2 η + 1 2 6.4. Functionals 179

6.4.1 Augmentation The functional I is the identity functional with respect to the reference series ∗ r (x, t) iﬀ I rn = δ0,n, i.e., I is the identity in the r -multiplication. Because h ∗ | i ∗ ∗ we require that rn (0) = δ0,n we get I = Eval0, independent of the reference series. Another name for the identity functional∗ is augmentation. The isomorphic power series is the multiplicative identity in the ring of formal power series, and the associated operator op (I ) is the identity operator I. ∗ Proposition 6.4.4. (Freeman [35, Prop. 6.2]) Let β(t) be a delta series and σ (t) an invertible series. If L is invertible, i.e., L 1 = 0, then h | i 6 LM (σ) C (β) r (x, t) = 1 iﬀ 1 1 L = I σ β− ( ) − . ∗ R Proof. LM (σ) C (β) r (x, t) =

1 1 I M (σ) C (β) σ β− ( ) − r (x, t) ∗ R 1 1 = I M (σ) C (β) M σ β− (t) − r (x, t) = I C (β) r (x, t) = 1 ∗ ∗ 1 For any delta operator B = β− ( ) we can deﬁne the functional IB := 1 R I β− ( ) ∗ R IB rn = I Brn = I rn 1 = δ1,n h | i h ∗ | i h ∗ | − i m such that B = op (IB) (Exercise 6.4.8). Writing IB∗ for IB r r IB (m factors) we get ∗ · · · ∗

m m op (Evala I∗ ) = r (a, ) B . ∗ B R 6.4.2 Orthogonality

The definitions in this section follow T. S. Chihara’s book [20]. Choose k = C. The connection between Finite Operator Calculus and orthogonal polynomials was first seen by Kholodov [49]. We follow Freeman’sapproach because of its simplicity. We will show two applications, the orthogonal polynomials that are exponential Sheffer sequences at the end of this section, and those that are binomial Sheffer sequences later in section 7.1.1.

Let (µn)n 0 be a sequence of (complex) numbers and Λ the functional defined n ≥ by Λ x = µn for all n 0. The sequence is called the (formal) moment h | i ≥ sequence, µn is the moment of order n, and Λ is the moment functional. A functional Λ is positive definite iff Λ p > 0 for all polynomials p = 0 that are nonnegative for all real x. For all positiveh | i definite functionals Λ exists6 a n n bounded nondecreasing function ψΛ such that Λ x = ∞ x dψΛ (x) (Stieltjes integral). h | i −∞ R 180 Chapter 6. A General Finite Operator Calculus

Deﬁnition 6.4.5. A polynomial sequence (pn) is orthogonal if a functional Λ on C [x] and a sequence λ (n) = 0, n 0, exist such that 6 ≥

Λ pnpm = λ(n)δn,m (6.33) h | i for all non-negative integers n, m. In that case (pn) is called the orthogonal polynomial system (OPS) corresponding to Λ. We will assume that Λ p0 = 1. h | i Note that λ (n) > 0 if Λ is positive deﬁnite. Because an OPS (pn) is also a basis of the vector space of polynomials, the functional Λ and the sequence λ (n) provide us with the coeﬃ cient functional [pn (x)] : C [x] C, such that →

pjq q (x) = Λ pj (x) | λ(j) j 0 X≥ for all polynomials q. This property explains the importance of ﬁnding an OPS k and its corresponding functional. The condition (6.33) implies Λ x pn = 0 for n | all 0 k < n, and Λ x pn = 0. ≤ h | i 6 n Let p (x, t) = n 0 pn (x) t , where (pn) is an OPS. We have ≥

P k k k n Λx p (x, t) = Λ x pn t =: fk (t) , n=0 | X where (fk)k 0 is a sequence of polynomials in C [t] such that deg fk = k, and ≥ f0 1 = λ (0). Taking transforms with respect to p (x, t) gives ≡ k k k fk (t) = ΛM x p (x, t) = M\(x )Λp (x, t) = M\(x) 1.

k If we write xˆp for M\(x) we have the simple condition xˆ 1 = fk (t) for all k 1, p ≥ which says that xˆp (a linear operator on C [[t]]) increases the degree by one when restricted to C [t], degx ˆpg (t) = 1 + deg g (t), for all g (t) C [t]. Freeman [34] pointed out the recipe for identifying an OPS with generating∈ function p (x, t): Check if n degx ˆpt = n + 1 (6.34) for all n 0. ≥ Lemma 6.4.6. If p (x, t) is the generating function for an OPS, then

n n 1 n n+1 xˆpt = αn 1t − + βnt + γn+1t − for some sequences (αn)n 0, (βn)n 0, (γn)n 1, γn = 0 for all n 1. ≥ ≥ ≥ 6 ≥ 6.4. Functionals 181

n n n+1 i Proof. From degx ˆpt = n + 1 follows that xˆpt = i=0 an,it for some scalars an,i. Hence P n+1 n i ∞ i ∞ xˆpp (x, t) = pn (x)x ˆpt = pn (x) an,it = t an,ipn (x) . n 0 n 0 i=0 i=0 n=i 1 X≥ X≥ X X X− Of course, the last inner sum has to terminate, and because

i xˆpp (x, t) = xp (x, t) = xpi (x) t , i 0 X≥ it has to terminate already at i + 1. The Lemma shows that

xpn (x) = αnpn+1 (x) + βnpn (x) + γnpn 1 (x) , (6.35) − or 1 1 pn+1 (x) = αn− (x βn) pn (x) αn− γnpn 1 (x) . − − − The latter is called the three term recurrence. Substituting for pn (x) in Λ pnpm = h | i λ(n)δn,m each of the three terms on the right hand side of (6.35) shows that Λ xpn 1pn = αn 1λ(n), Λ xpnpn = βnλ(n), and Λ xpn+1pn = γn+1λ(n). h | − i − h | i h | i The first and the third of the three relations give αnλ (n + 1) = γn+1λ(n). From n deg (xpn (x) αnpn+1 (x)) = n we find αn = kn/kn+1, where kn = [x ] pn (x). − Remark 6.4.7. (a) Let Evalt=0 be the t-operator on C [[x]] [t] that evaluates at 0. n From fn(t) = Λ x p (x, t) follows h | i n n fn (0) = ΛM (x ) Evalt=0 p (x, t) = Λ x = µn, h | i the n-th moment of Λ. It can be shown [20] that in an OPS the determinant of the Hankel matrix Mn := (µi+j)i,j=0,...,n is different from 0 for all n 0. More precisely, ≥

n kn Mn Λ x pn (x) = | | h | i Mn 1 | − | n where kn = [x ] pn (x) as before. (b) It is usually assumed that λ (n) is real and positive for all n. In this case αnγn+1 > 0.

The Meixner classification In 1934, Meixner classified all orthogonal polynomial systems that are also Sheffer sequences [59]. Freeman showed how to do this easily with the transform approach [33]. The following is based on his ideas. We will consider another application, to binomial reference frames, in section 7.1.1, 182 Chapter 6. A General Finite Operator Calculus

Suppose (pn) is a Sheﬀer sequence in the exponential reference frame, hence the generating function is

n xβ(t) α(t)+xβ(t) p (x, t) = pn (x) t = σ (t) e = e n 0 X≥ where we assume that σ (0) = 1, hence α (t) = ln σ (t). We will also assume that β0 (0) = 1. From

α(t)+xβ(t) α(t)+xβ(t) Dte = M (α0 (t) + xβ0 (t)) e follows

M (x) p (x, t) = (M (1/β0 (t)) Dt M (α0 (t) /β0 (t))) p (x, t) , − hence xˆp = M (1/β0 (t)) Dt M (α0 (t) /β0 (t)). From −

1 = deg (ˆxp1) = deg ( M (α0 (t) /β0 (t))) − we get that α0 (t) /β0 (t) must be linear,

α0 (t) /β0 (t) = ct + d, and c = 0. From 6 1 2 = deg (ˆxpt) = deg (ct + d) t β (t) − 0 n it follows that (pn) is a OPS, degx ˆpt = n + 1, iﬀ 1/β0 (t) is at most a quadratic polynomial, 1/β0 (t) = (1 at) (1 bt) . − − In this case

n n 1 n xˆpt = (1 at) (1 bt) nt − (ct + d) t − − − n+1 n n 1 = (nab c) t (n (a + b) + d) t + nt − − − must be of degree n+1 for all n 1, thus c/ (ab) is not a positive integer if ab = 0. From ≥ 6

t t x + α0 (u) /β0 (u) α (t) + xβ (t) = (xβ0 (u) + α0 (u)) du = 1 du β (u)− Z0 Z0 0 follows that for an OPS we can write t x + cu + d p (x) tn = p (x, t) = exp du , n (1 au) (1 bu) n 0 0 X≥ Z − − 6.4. Functionals 183 where abn = c for all n 1. The term d in the above generating function only shifts x, and6 can be set to≥0. If we consider real polynomials only, we must choose a and b as real or complex conjugate. The general three term recursion specializes to

xpn (x) = (n + 1) pn+1 (x) (a + b) npn (x) + (ab (n 1) c) pn 1 (x) − − − − because

n n 1 n+1 n+1 n n 1 xtˆ = (1 at) (1 bt) nt − ct = (abn c) t (a + b) nt + nt − . − − − − − Note that we are interested in any OPS up to linear transformations in x and scaling in t.It can be shown that the functional is positive deﬁnite if ab 0 and c < 0. The following types occur: ≥ 1. a = b = 0, and c = 1 : − t xt 1 t2 exp (x u) du = e − 2 . − Z0

This is the generating function of the Hermite polynomials Hn (x). They are orthogonal with respect to the normal probability measure

n 1 ∞ n x2/2 µn = Λ x = x e− dx, h | i √2π Z−∞ (see Example 2.2.9), because

1 ∞ x2/2 xt 1 t2 xs 1 s2 1 st ∞ ( (s+t)2+2x(s+t) x2)/2 e− e − 2 e − 2 dx = e e − − dx √2π √2π Z−∞ Z−∞ = est.

The three term recursion becomes

(n + 1) Hn+1 (x) = xHn (x) Hn 1 (x) − − 2 = x Hn+1(x) Hn+1 (x) . D − D Therefore,

3 (n + 1) Hn (x) = x Hn+1(x) Hn+1 (x) . (6.36) D D − D 3 Hence the polynomials H¯n (x) = Hn (x) /n! follow the recursion x H¯n = D Dn − Dn H¯n 1 for all n 0, with initial values H¯2n (0) = H2n (0) / (2n)! = ( 1) 2− / (n! (2n)!), − ≥ − and H¯2n+1 (0) = 0. The more common variation for the Hermite polynomials [3] is obtained from choosing c = 2 and replacing x by 2x, − t 2xt t2 exp 2 (x u) du = e − . − Z0 184 Chapter 6. A General Finite Operator Calculus

2. a = b = 0, and c = 1 : 6 − t x u (ax 1) t 1/α2 exp − 2 du = exp − (1 ta)− . 0 (1 au) ! a (1 ta) − Z − −

2 2 1/α xt/(t 1) Transform (ax 1) /a to x, and rescale ta as t to get (1 t)− e − . Writing a + 1 for− 1/α2 gives the generating function of the− Laguerre polynomials (Example 5.3.1). 3. a = 0, b = 0, and c < 0 : 6 t x + cu ct/a (c+ax)/a2 exp du = e− (1 at)− . 1 au − Z0 − Transform (c + ax) /a2 to x, rescale t as t/a2and let c = a3 to get the − − − x generating function of the Poisson-Charlier polynomials e t a+t (see also − a Exercise 2.3.14). 4. ab > 0, a = b, a and b real, c = 1, gives Meixner type I, and a and b conjugate complex6 gives Meixner type− II.

6.4.3 Exercises

1 6.4.1. Let (ln (x)) be the Sheﬀer sequence for the delta operator B = β− ( ) with n D generating function n 0 ln (x) t = r (x, β (t)) /λ (β (t)), where Lr (x, t) = λ (t) ≥ 1 for some invertible functional L. Show that λ ( )− is the Sheﬀer operator of (ln). P R 6.4.2. Let L K∗ [x]. For every delta series β(t), and every series φ (t) of order ∈ 0 holds that LC (β)` φ ( ) is also a functional, and therefore R κ (t) := LC (β)` φ ( ) r (x, t) R a multiplication t-operator. Show that

κ (t) = φ (t) λ (β (t)) .

6.4.3. Prove Proposition 6.4.4.

6.4.4. [65, Lemma 8]Show that the mapping op : K[x]∗ Σ , where op (L) = λ ( ), is a ring isomorphism . → R R 6.4.5. Show that for all p K [x] and L, N K[x]∗, ∈ ∈

L r N p = N λ ( ) p h ∗ | i h | R i (see [35, Proposition 6.1] and [65, Lemma 8]). 6.4. Functionals 185

1 6.4.6. Let (bn) be the basic sequence for the delta operator B = β− ( ). Suppose, R (sn) is the Sheffer sequence with generating function M(σ(t))C (β) r(x, t). Define 1 the invertible linear functional L by λ(t) = σ(β− (t)), and show that λ( ) is the 1 R Sheffer operator for (sn), sn(x) = λ( )bn(x). Let L∗− be the reciprocal of L in R 1 the -multiplication of functionals, and show that L∗− sn = δ0,n. ∗ | 6.4.7. Let B be any delta operator and (sn) any B - Sheffer sequence. Show that

k k 0 L sk B ≥ h | i λ ( ) = n . R n 0 Eval0 sn B P≥ h | i P Especially for the B - basic sequence (bn) holds

k λ ( ) = L bk B R h | i k 0 X≥

6.4.8. Suppose (bn)is the basic sequence belonging to the delta operator B = ( ). 1 R Show that for the functional IB = I β− ( ) holds ∗ R

IB bn = δn,1. h | i

6.4.9. Suppose (bn) is the basic sequence for the r-delta operator B. Show that for all m N0 ∈ m m op (Evala I∗ ) = r (a, ) B . ∗ B R 186 Chapter 6. A General Finite Operator Calculus Chapter 7

Applications of the General Theory

We pursue only two applications of the general theory. In the ﬁrst case we select the “binomial reference”,i.e., we look at delta operators obtained from delta series α β (t) substituted into (1 xt)− . We show the Freeman classiﬁcation of orthogonal polynomials for this− case, and we give some other examples, like Dickson polynomials.

The second application goes back to George Andrews in 1971 [5]. It applies to all diagonal reference frames, but it were the Eulerian diﬀerential operators Andrews had in mind. The delta operators in diagonal reference frames are in general not translation invariant. A new invariant is introduced, referring to scaling. The binomial theorem is replaced by a multiplicative version (7.22), holding for Eulerian sequences.

We only mention here another view of q-polynomials and operators, introduced by the Askey-Wilson divided difference operator q,x [7, 1985]. This op- D erator is a q-analog of the differentiation operator . The operator q,x maps n n(n 1)/4 n 1 D (n 1)(n D2)/4 (1 q) Tn (x) / q − (q; q)n into (1 q) − Un 1 (x) / q − − (q; q)n 1 , − − − − where Tn and Un are the Chebychev polynomials of the first and second kind (section 7.1.1). Ismail [44, 2001] defines a q-translation operator such that q,x commutes with this translation. He then defines q-Delta operators and ShefferD sequences in the q,x reference frame. D 188 Chapter 7. Applications of the General Theory

7.1 The Binomial Reference Frame

n+α 1 n The elements of a binomial reference frame are the reference sequence n− x , the reference series α n + α 1 n n α 1 α 1 α 1 1 (1 xt)− = − x t = χ − − M x − (1 xt)− − n D − n 0 X≥ ( α / N0), and the reference operator − ∈ η + 1 1 α = χ = (η + α)− B η + α D

n + α 1 n n + α 2 n 1 α − x = − x − B n n 1 − for all n 1. To emphasize α, we also say α-binomial reference frame. Most notable is≥ the fact that a binomial reference series also becomes an exponential reference series if we view x as fixed, x C, say, and α as the polynomial variable. This explains immediately the “binomial∈ theorem” n sn(x; ν + µ) = sk(x; ν)bn k(x; µ). (7.1) − k=0 X that holds for every Sheffer sequence (sn (x; ν)) and basic sequence (bn (x; µ)) such n µ n ν that n 0 bn (x; µ) t = (1 xβ (t)) and n 0 sn (x; v) t = σ (t) (1 xβ (t)) (delta series≥ β, and invertible− σ). Of course, this≥ is not the binomial theorem− we P P want. If we construct (tn (x)) according to convolution identity n tn(x) = tk(y)bn k(x y; α) − − k=0 X 1 we obtain a binomial Sheffer sequence for the delta operator β− ( α/ (1 + y α)) B 1 B (Exercise 7.1.1). At least, we can expand a binomial Sheffer sequence for β− ( α) when we know the initial values at 0, B n

tn(x; α) = tk(0; α)bn k(x; α). (7.2) − k=0 X We can still find the Sheffer sequence with initial values sn (c) = δ0,n in the geometric case, i.e., the binomial case with α = 1 (see Exercise 7.1.4). Note that bn (x c; α) is not a binomial Sheffer polynomial in general, because we do not have− translation invariance. An operator “similar” to the translation operator c k k E = k 0 c /k! is the invertible ≥ D P k k 1 Hα,c := c α = Σ α , B 1 c α ∈ B k 0 X≥ − B which we explore in section 7.1.1. 7.1. The Binomial Reference Frame 189

7.1.1 Orthogonal Binomial Reference Sequences We can write the generating function of λ-binomial Sheﬀer sequences in the form λ p (x, t) = (φ (t) xφ (t) β (t))− for λ / N0. We assume in this subsection that − − ∈ φ (0) = 1, β (0) = 0, and β0 (0) = 1. Let xˆp be the p (x, t)-transform of M (x). For xˆp the following general result holds. Lemma 7.1.1. Let φ (t) be of order 0 with φ (0) = 1, β (t) a delta series such that n k β0 (t) = 1, and deﬁne η (t) = φ (t) 1. Then xˆpη (t) (φ (t) β (t)) −

2n + k 2n + k + 2λ n k 1 = + η (t) η (t) (φ (t) β (t)) − (7.3) 2n + k + λ 1 2n + k + λ + 1 − λ for all n, k 0, where p (x, t) = (φ (t) xφ (t) β (t))− and λ / N0. ≥ − − ∈ Proof. Expanding p (x, t) shows that

k + λ 1 n + k + λ 1 p (x, t) = − xk ( 1)n − ηn (φβ)k . k − n k 0 n 0 X≥ X≥ Substituting the right-hand side of (7.3) for ηn (φβ)k above gives

k + λ 1 k n n + k + λ 1 n k 1 (2n + k) − x ( 1) − η (φβ) − k − n (2n + k + λ 1) k 0 n 0 X≥ X≥ − k + λ 1 k n n + k + λ 2 n k 1 (2n + k + 2λ 2) − x ( 1) − η (φβ) − − − k − n 1 (2n + k + λ 1) k 0 n 1 X≥ X≥ − − which simpliﬁes to xp (x, t). k k 1 k If p (x, t) is the generating function of an OPS, then xˆpt = αk 1t − +βkt + k+1 − γk+1t for all k 0 must hold, which restricts the possible choices for φ (t) and β (t) in Lemma 7.1.1.≥ Actually, we will show in the remainder of this section that φ (t) β (t) = t is the only possibility! We begin with an obvious simpliﬁcation: 0 Lemma 7.1.1 directly gives us xˆpt ,

2λ η (t) xˆ 1 = = b + ct = β + γ t (7.4) p λ + 1 φ (t) β (t) 0 1

λ+1 hence η (t) = 2λ (β0 + γ1t) φ (t) β (t). Note that γ1 = 0. If n = 0 in Lemma 7.1.1, then 6

k k k + 2λ λ + 1 k 1 xˆ (φβ) = + (β + γ t) φ (t) β (t) (φ (t) β (t)) − p k + λ 1 k + λ + 1 2λ 0 1 − k k 1 k + 2λ λ + 1 k = (φ (t) β (t)) − + (β + γ t)(φ (t) β (t)) k + λ 1 k + λ + 1 2λ 0 1 − 190 Chapter 7. Applications of the General Theory

k k m k Suppose, (φ (t) β (t)) = t + m>k Ck,mt , hence xˆp (φβ) =

k 1 P k k+1 αk 1t − + (βk + Ck,k+1αk) t + (γk+1 + Ck,k+1βk+1 + Ck,k+2αk+2) t − m + (Ck,m+1αm + Ck,mβm + Ck,m 1γm) t − m k+2 X≥

Comparing coeﬃ cients in both expansions shows that αk, βk, and γk are dependent on Ck,k+1 and Ck,k+2; however, these terms only depend on C1,2 and C1,3 (see (1.4)). It can be shown that

k + 1 α = (7.5) k k + λ (k + 2λ)(λ + 1) k (k + 2λ 1) βk = β0 − C1,2 2λ (k + λ + 1) − (k + λ 1) (k + λ) − and γk+1 =

(k + 2λ)(λ + 1) γ1 k (λ + 1) (λ 1) k (k + 4λ 1) + − C1,2 − C1,3 2λ (k + λ + 1) 2λ (k + λ + 1) (k + λ + 2) − (k + λ 1) (k + λ + 2) − k (k 1) (k 2) k2 (k + 1) k (k + 1) (k + 2) (k 1) k (k + 3) + − − + − C2 2 (k 1 + λ) − k + λ k + 1 + λ − 2 (k + 2 + λ) 1,2 − m Remember we want to show that m>k Ck,mt = 0. Substituting for αk, βk, and γ and comparing terms for m = k + 2 in the two representations for xˆ (φβ)k k P p gives us inﬁnitely many equations in the unknowns C1,2,C1,3, and C1,4. This system only has the solutions C1,2 = C1,3 = C1,4 = 0, which in turn implies Ck,k+1 = Ck,k+2 = Ck,k+3 = 0. If m > k 2, we obtain a recurrence for Ck,m+1 − based on the initial values Ck,k+1 = Ck,k+2 = Ck,k+3 = 0, giving us also 0. Hence m m>k Ck,mt = 0, and therefor φ (t) β (t) = t. More details can be found in Exercise 7.1.6. P

Gegenbauer polynomials We have shown that φ (t) β (t) = t, thus φ (t) 1 = λ+1 − η (t) = 2λ t (b + ct), where we write b for β0 and c for γ1. From (7.5) we obtain k+1 (λ+1)(k+2λ) (λ+1)(k+2λ 1) αk = k+λ , βk = b 2λ(k+λ+1) , and γk = c 2λ(k+λ)− . We ﬁnd

t β (t) = λ+1 1 + 2λ t (b + ct) and by rescaling t and replacing x by 2x we get the generating function of the ( λ) Gegenbauer polynomials Pn− (x),

( λ) n 2 λ P − (x)t = 1 + t 2xt − . (7.6) n − n 0 X≥ 7.1. The Binomial Reference Frame 191

The three term recursion for the Gegenbauer OPS equals

( λ) ( λ) ( λ) 2x (n + λ) Pn− (x) = (n + 2λ 1) Pn−1 + (n + 1) Pn−+1 (x) (7.7) − − From the generating function it is easy to expand the Gegenbauer polynomials as

n/2 ( λ) n k + λ 1 n k k n 2k P − (x) = − − − ( 1) (2x) − . n n k n 2k − k=0 X − − λ A diﬀerent recursion can be derived by taking the x-derivative of 1 + t2 2xt − ( λ) − (writing Pn (x) for Pn− (x)),

Pn0 (x) + Pn0 2 (x) 2xPn0 1 (x) = 2λPn 1 (x) . (7.8) − − − − Both recursions need the initial values

( λ) λ ( λ) P − (0) = − and P − = 0, (7.9) 2n n 2n+1 ( 1) following from (7.6). Note that Pn− = Un (x) is called the Chebychev polynomial ( 1/2) of the second kind, and Pn− (x) is the Legendre polynomial. Remark 7.1.2. We have found just one important example for a binomial OPS, by letting φ (t) β (t) = t. We will do an exhaustive search for the case λ = 1.

Chebychev polynomials The case λ = 1 in (??) gives us for k 1 the simple condition ≥

k k 1 k k+1 j 1 t αk 1t − + βkt + γk+1t = cj (1 + η (t)) (φ (t) β (t)) − = . − β (t) j k X≥ Hence β (t) is of the form β (t) = t/ 1 + rt + st2 . Again, we rescale t and get β (t) = t/ 1 + t2 . We also replace x by 2x. From (7.4) we obtain φ (t) = 1 + t2 / 1 bt + (1 c) t2 , hence − − 1 bt + (1 c) t2 p (x, t) = − − , 1 + t2 2xt − describing all orthogonal polynomial systems of this kind. Start by letting b = 0 and c = 1. We obtain the Chebychev polynomials of the second kind, n/2 k n k n 2k Un (x) = ( 1) − (2x) − , − k k=0 X 192 Chapter 7. Applications of the General Theory because

n/2 2 k n k n k n 2k 1 ∞ t 2 1 t ( 1) − (2x) − = − = 1 2xt + t − . − k 1 2xt 1 2xt − n 0 k=0 − k=0 − X≥ X X 1 2 The delta operator β− (χ) = 1 1 χ /χ maps Un to Un 1. The 3-term − − − recursion (??) specializes to p

Un (x) = 2xUn 1 (x) Un 2 (x) (7.10) − − − (Chihara [20]). For x = cos θ, 0 < θ < π, the Chebychev polynomials of the second kind evaluate as n/2 k n k n 2k sin ((n + 1) θ) Un (cos θ) = ( 1) − (2 cos θ) − = . − k sin θ k=0 X for all n 0. In Exercise 7.1.8 we show that ≥ n n kπ Un (x) = 2 x cos − n + 1 k=1 Y 2x 1 0 ... 0 1 2x 1 ... 0

0 1 2x . . . 0 = . (7.11) ......

0 0 0 1 2x n n × The Chebychev polynomials of the ﬁrst kind are deﬁned as Tn (x) = Un (x) − xUn 1 (x). Hence that − n 2 1 2 1 1 + 2 Tn (x) t = 2 1 2xt + t − 1 2xt 1 2xt + t − − − − − n 1 X≥ 2 1 1 t 2xt − = − 1 1 + t2 − 1 + t2 and sin ((n + 1) θ) cos θ sin nθ T (cos θ) = − = cos nθ. n sin θ 1 bt+(1 c)t2 Now consider the general case, p (x, t) = −1+t2 −2xt (D. Stanton [91, 1983]). In this case − pn (x) = Un (x) bUn 1 (x) + (1 c) Un 2 (x) − − − − for all n 0. Even a sum of the form Un (x) + + wUn 4 (x) can be orthogonal [7, (4.29)],≥ but the generating function would not··· be of the− desired type, because of initial conditions. 7.1. The Binomial Reference Frame 193

Remark 7.1.3. Anshelevich [6] studies the orthogonal Sheﬀer sequences for the case λ = 1, but in several non-commuting variables. The recursion relation is described in terms of the free cumulant generating function.

The Sheﬀer Operator Hα,y

In the special case α = 1 the binomial reference series reduces to the geometric series 1/ (1 xt) with reference operator χ and reference sequence (xn). For given − 1 y C, say, the operator H1,y = (1 yχ)− is a Sheﬀer operator for χ (see (6.17)), ∈ n − hence sn (x) := H1,yx is a 1-binomial Sheﬀer polynomial for χ and has the initial n 2 values generated by n 0 sn (y) t = (1 yt)− , thus ≥ − P n sn (y) = (n + 1) y . (7.12)

In detail,

n n+1 n+1 n n k k x y s (x) = H x = x − y = − . n 1,y x y k=0 X − By induction we can show that

n k n k k k n 2k sn (x) = ( 1) − x y (x + y) − . − k k=0 X Hence the identity

n+1 n+1 n x y k n k k k n 2k − = ( 1) − x y (x + y) − (7.13) x y − k k=0 − X follows (identity 1.60 in Gould’slist [38]). This is a special case of the identity

n/(m+1) xn+1 yn+1 b c n mk x k x n (m+1)k − = ( 1)k − ymk + y − x y − k m m k=0 − X n n k j+k k n mk − x n k j n (m + 1) k + ( 1) − y − − − − k m j k= n/m j=0 Xd e X 194 Chapter 7. Applications of the General Theory

x n (m+1)k which holds for all m N1 . We show this identity by expanding + y − = ∈ m

n/(m+1) n k b c k+j k n mk − x n k j n (m + 1) k ( 1) − y − − − − k m j k=0 j=0 X X n n k k+j k n mk − x n k j n (m + 1) k + ( 1) − y − − − − k m j k= n/m j=0 Xd e X n n k k+j k n mk − x n k j n (m + 1) k = ( 1) − y − − − − k m j k=0 j=0 X X n l l n l l n mk l n 1 + mk = x y − ( m)− − − − . − k l k l=0 k=0 X X − By identity (2.16),

l c km l 1 c + km ( m)l = − − − − k n k k=0 X − for all c. If α is any positive integer, then

α α 1 α 1 1 (1 xt)− = − M x − (1 xt)− . − D − Therefore, α α 1 α 1 1 Hα,y (1 xt)− = Hα,y − M x − (1 xt)− , − D − and from

α 1 1 α 1 α 1 1 Hα,y − = 1 − = − 1 (7.14) D 1 y (η + α)− D D 1 y (η + 1)− − D − D follows

n+α n+α n + α 1 n α 1 n+α 1 α 1 x y Hα,y − x = − H1,yx − = − − . n D D x y − α 1 α 1 The relationship Hα,y − = − H1,y in (7.14) implies D D k α 1 α 1 k H − = − H α,yD D 1,y n+α 1 n for all natural numbers k. Note that the Sheﬀer polynomial Hα,y n− x taken at x = y gives n!yn n+α , a generalization of (7.12). n 7.1. The Binomial Reference Frame 195

The evaluation functional Let c k and α be a positive integer. The functional n n ∈ Evalc x = c has the associated operator h | i α α op (Evalc) = (1 c α)− = H − B α,c in the α-binomial reference frame, according to Example 6.4.1. Hence

α 1 α α i i i op (Evalc)− = (1 c α) = ( 1) c . − B i − Bα i=0 X Suppose we want to ﬁnd the Sheﬀer sequence (sn (x))for a satisfying sn (c) = B yn k, where y0 is a unit in k. The general functional expansion theorem 6.4.2 tells∈ us that n 1 k + α 1 k sn (x) = L sn k op (L)− − x h | − i k k=0 X n k α i i k i + α 1 k i = yn k ( 1) c − − x − − i − k i k=0 i=0 − Xn X n α i i k i + α 1 k i = ( 1) c yn k − − x − i − − k i i=0 k=i X X − n n i α i i − k + α 1 k = ( 1) c yn i k − x i − − − k i=0 k=0 X X In terms of generating function,

k n k 0 ykt s (x) t = ≥ n (1 ct)α (1 xt)α n 0 P X≥ − − If c = 0, then n k + α 1 k sn (x) = yn k − x − k k=0 X n in agreement with Lemma 6.2.6. If α = 1 and yn = y , then n n n y x n n 1 sn (x) = y + (x c) − = y + (x c) H1,yx − . − y x − − 7.1.2 Generalized Catalan Operators

Let α be the generalized Catalan operator C n n n n k n 1 + α αx := Ck 1x − / − C k − k k=1 X 196 Chapter 7. Applications of the General Theory

( α / N0), where Ck is the k-th Catalan number. From − ∈ n k n 1 n αx = Ck 1 x (7.15) C − η + αD k=1 X 1 1 follows that α = 2 2 √1 4 α (see Example 6.2.8), a delta operator for the α- C − − B 1 1 binomial reference operator α = (η + α)− . Therefore we find α = β− ( α), 1 1 1 B D 2 C B where β− (t) = √1 4t, and by inversion, β (t) = t t . The basic polyno- 2 − 2 − − mials (cn) for α therefore have generating function C n 2 α cn (x; α) t = 1 x t t − . − − n 0 X≥ We expand cn (x; α) as n k + α 1 k n k k cn (x; α) = − ( 1) − x . (7.16) k n k − k=0 X − 1 2 From α = β− ( α) follows α = β ( α) = Cα C . Any Sheffer sequence (sn) C B B C − α for α must satisfy this recursion, C 1 (η + α)− sn (x) = sn 1 (x) sn 2 (x) . D − − − i In terms of the coeffi cient sn,i = x sn (x) this means

(i + 1) sn+1,i+1= (i + α)(sn,i sn 1,i) − − for all 0 i n, and where the numbers sn,0 are given initial values. If α≤= 1≤we obtain n k n k k cn (x; 1) = ( 1) − x , n k − k=0 X − an interesting sequence indeed. On one side we observe that the identity (7.13) implies n+1 n+1 n a b ( z/4) cn ( 4/z; 1) = − − − 2n (a b) − n where a = 1 + √z + 1 and b = 1 √z + 1. For z = 4 we get ( 1) cn ( 1; 1) = n n k − − − k=0 −k = Fn, the n-th Fibonacci number. On the other hand, let P n 2 Un (x) := (2x)− cn (2x) ; 1

It it easy to verify that

Un (x) = 2xUn 1 (x) Un 2 (x) (7.17) − − − and U0 (x) = 1. This recursion is solved by the Chebychev polynomials of the 2 n n kπ second kind (Exercise 7.1.13), hence cn (2x) ; 1 = (4x) x cos . k=1 − n+1 Q 7.1. The Binomial Reference Frame 197

7.1.3 Dickson Polynomials Let a be a complex number. The Dickson polynomials (of the ﬁrst kind) are the unique polynomial sequence deﬁned by

n n Dn (y + a/y; a) = y + (a/y) (7.18) for all n 0. ≥ In Exercise 7.1.16 we show that (Dn (x; a)) is a basis; it holds

(n 1)/2 − n n n/2 n k x = a + a Dn 2k (x; a) , n/2 k − k=0 X n 2 where = 0 for odd n. Obviously D0 (x) 2, D1 (x) = x,and D2 (x) = x 2a. n/2 ≡ − We get the generating function in terms of y as 1 1 D (y + a/y; a) tn = + n 1 yt 1 at/y n 0 X≥ − − In terms of x = y + a/y (it does not matter which solution we choose!) we get

1 1 2 xt D (x; a) tn = + = − n 1 yt 1 (x y) t 1 xt + t2a n 0 X≥ − − − −

This generating function becomes a Sheﬀer sequence in the 1 = χ reference frame, B if we change D0 from 2 to 1,

2 1 n 1 t a xt − Dn (x; a) t 1 = − 1 . − 1 + t2a − 1 + t2a n 0 X≥ 2 1 1 t 2xt − = − 1 1 + t2 − 1 + t2 Note that

2 1 n/2 n 1 + t xt/√ a − Dn (x; a)( a)− t 1 = 1 − − − 1 t2 − 1 + t2 n 0 X≥ − showing that we can restrict the discussion to the Sheﬀer polynomials

dn (x) := Dn (x; 1) δ0,n − − because n/2 Dn (x; a) = ( a) Dn x/√ a; 1 . − − − 198 Chapter 7. Applications of the General Theory

The Dickson polynomials with a = 1 are closely related to the Pell polynomials − Pn (x), 2Pn (x) = Dn (2x; 1) − (see (1.9)). On the other hand, the Chebychev polynomials of the ﬁrst kind, Tn (x), have the generating function

2 1 n 1 t 2xt − 2 Tn (x) t 1 = − 1 − 1 + t2 − 1 + t2 n 0 X≥ hence 2Tn (x) = Dn (2x; 1) . The deﬁnition (7.18)

n n n Dn (y 1/y; 1) = y + ( 1) /y − − − shows that for x = y 1/y holds − n n x + √x2 + 4 + x √x2 + 4 D (x; 1) = − . n 2n − 1 What properties does the basic sequence (bn) for the delta operator β− (χ) 1 2 1 have, where β− (t) is the compositional inverse of β (t) = t/ 1 t , thus β− (t) = − √1 + 4t2 1 / (2t)? In Exercise 7.1.19 it is shown that − bn (x) = xbn 1 (x) + bn 2 (x) (7.19) − − 2 for all n > 2, with initial values b0 (x) 1, b1 (x) = x, and b2 (x) = x . This ≡ recursion also holds for dn (x) and Dn (x; 1), but with diﬀerent initial values. We − get an explicit expression for bn (x) directly from the generating function,

n/2 n j 1 n 2j b (x) = − − x − . n j j=0 X From 1 + t2 d (0) = [tn] = 2 n 1 t2 − for even n 0, and 0 else, we get ≥ n/2 n j n n 2j Dn (x; 1) = − x − − j n j j=0 X − and x η bn (x) = Dn (x; 1) = (Dn (x; 1) D (0; 1)) nD − n − − − 7.1. The Binomial Reference Frame 199

for n 1. The Sheﬀer operator for (dn) equals ≥ 1 2 1 1 + β− (χ) 1 β− (χ) 1 1 = 1 + 2β− (χ) = 1 + 2β− (χ) β β− (χ) 1 2 1 2 1 β− (χ) 1 β− (χ) − 1 − = 1 + 2β− (χ) χ, and therefore Dn (x; 1) = bn (x) + 2bn 1 (x) /x − − for all n 2.Applying (7.19) we can write ≥

bn+1 (x) Dn (x; 1) = 2 bn (x) (7.20) − x − for all n 1. We≥ can also apply the factoring method (section 1.1.1) to the recursion (7.19), and obtain n n x + √x2 + 4 x √x2 + 4 bn (x) = x − − n 2 2 √x + 4 for n 1 (Exercise 7.1.20). Combining this formula with (7.20) shows again that ≥ n n x + √x2 + 4 + x √x2 + 4 D (x; 1) = − n 2n − for all n 0. If k is a positive integer, then ≥ nk nk x + √x2 + 4 x √x2 + 4 bnk (x) = x − − n 2 2 √x + 4 x + √4 + x2 x √4 + x2 = bn (x) pn,k , − , (7.21) 2 2 ! where nk nk k 1 u v − ni n(k 1 i) p (u, v) = − = u v − − n,k un vn i=0 − X is a symmetric bivariate polynomial of degree n (k 1) in u and also in v. If we let y = x + √4 + x2 /2, and y¯ = x √4 + x2 /2, the− two solutions of x = y 1/y (see (7.18)), then yy¯ = 1 and −y +y ¯ = x. By the Fundamental Theorem− of − Symmetric Functions the symmetric function pn,k (y, y¯) can be expressed as a polynomial in yy¯ and y +y ¯; hence pn,k (y, y¯) is a polynomial in x. Equation (7.21) is a factorization of bnk (x) into two polynomials in x, bn (x) and pn,k (y, y¯), where pn,k (y, y¯) must be of degree n (k 1). For k = 2 we obtain the doubling formula − n n n 2 2 b2n (x) = 2− x + 4 + x + x 4 + x bn (x) = Dn (x; 1) bn (x) . − − p p 200 Chapter 7. Applications of the General Theory

n bn (x) Dn (x; 1) n bn (x) − 0 1 2 1 x x 7 x x6 + 5x4 + 6x2 + 1 2 2 2 2 4 2 2 x x + 2 8 x x + 2 x + 4x + 2 2 2 2 6 4 2 3 x x + 1 x x + 3 9 x x + 1 x + 6x + 9x + 1 2 2 4 2 2 4 2 4 2 4 x x + 2 x + 4x + 2 10 x x + 3x + 1 x + 5x + 5 4 2 4 2 10 8 6 4 2 5 x x + 3x + 1 x x + 5x + 5 11 x x + 9x + 28x + 35x + 15x + 1 6 x2 x2 + 1 x2 + 2 12 x2 x2 + 1 x2 + 3 x2 + 2 × × × x2 + 3 x4 + 4x2 + 1 x4 + 4x2 + 1 × × × 7.1.4 Exercises 7.1.1. The Binomial Finite Operator Calculus is based on the reference series α r(xt) = (1 xt)− . −

1. Find the reference polynomials rn(x) and the reference operator α. B

2. Suppose (bn(x; α))n 0 is the basic sequence for the delta operator 1 ≥ α B = β− ( α) where α R, but α / N0, hence b (x, t) = (1 xβ (t))− . Show (7.1)B directly, without∈ using− the∈ exponential reference frame.−

3. Let y R and ∈ n sn(x) := sk(y; α)bn k(x y; α). − − k=0 X Show that (sn) is a Sheﬀer sequence for the α-binomial delta operator 1 β− ( α/ (1 + y α)). B B c η+1 7.1.2. Expand E in terms of η and α = χ, the α-binomial reference operator, B η+α c n η+α+n 1 n c k+α 1 k k+α 1 k as E = n 0 c n − α, and verify that E k− x = k− (x + c) ≥ c B c c for all c 0. Of course, E α = αE , because E / Σ α . ≥P B 6 B ∈ B 7.1.3. Let n + 1 t (x; α + 1) := b (x; α), n x n+1

n α where n 0 bn (x) t = (1 xβ (t))− . Show that (tn) is a Sheﬀer sequence for ≥ − 1 the α + 1-binomial delta operator β− ( α+1). P B 7.1. The Binomial Reference Frame 201

n 7.1.4. Let α = 1 and gn (x) = bn+1 (x; 1) /x, n = 0, 1,... , where n 0 bn (x) t = 1 1 ≥ (1 xβ (t))− . Show that (g ) is a Sheﬀer sequence for B = β ( ) with gener- n − P 1 ating− function B β (t) /t g (x)tn = . n 1 xβ (t) n 0 X≥ − c Therefore, sn (x) = 1 bn (x; 1) is a Sheﬀer sequence for B with initial values x − sn (c) = δ0,n. 7.1.5. Show that

1 α 1 α 1 1 1 − = − 1 1 y (η + α)− D D 1 y (η + 1)− − D − D (Equation (7.14)). 7.1.6. Show that in the orthogonal binomial case the three equations (7.5) hold for αk, βk, and γk. Show that φ (t) β (t) = t. ( λ) ( λ) 7.1.7. Show that the delta operator G : Pn− Pn−1 solves the operator equation → − 2G λ = B (1 + G2)

1 where λ = (η + λ)− . Find the basic sequence (gn) for G. Using the initial valuesB (7.9), we can expandD

n/2 ( λ) λ P2n− (x) = − g2n k (x) k − k=0 X with the help of (7.2). 7.1.8. Show that the determinant in (7.11) follows the Chebychev recursion (7.10). sin((n+1)θ) 7.1.9. Show that sin θ follows the Chebychev recursion (7.10).

2 7.1.10. Show that (Un) is a Sheﬀer sequence for 1 1 χ /χ. − − 7.1.11. Find the basic sequence (bn (x; α)) (formula (7.16))p for the general Catalan delta operator with the help of Corollary 6.3.6. 2 2 7.1.12. Let 1 = 1 + 1 B + B . Find the basic sequence (bn) of B in the geometric frameworkC (Cα = 1). 7.1.13. Show that for the Chebychev polynomials Un (x) holds

n+1 n+1 x + √x2 1 x √x2 1 Un (x) = − − − − 2√x2 1 − 202 Chapter 7. Applications of the General Theory

n k 7.1.14. Show that the polynomials sn (x) = k=0 sn,kx follow the recursion i+α αsn = sn 1 sn 2 if the coeﬃ cients satisfy sn+1,i+1 = (sn,i sn 1,i) for all B − − − P i+1 − − 0 i n. Verify that the basic polynomials cn (x; α)in (7.16) have this property. ≤ ≤ n i 7.1.15. Suppose we have the following recursion for sn (x) = i=0 sn,ix

n i n 1 i P n 2 i k − i i k − i sn,i Ck 1x − = sn 1,i x + Ck 1x − + sn 2,ix − − − − i=1 k=1 i=1 k=1 ! i=0 X X X X X where sn,0 = sn (0), n 0, are given initial values (Cn is the n-th Catalan num- ≥ ber). We want to ﬁnd the basic sequence (bn (x)) that satisﬁes this recursion and has the initial values bn (0) = δn,0, hence the recursion gives

n n i k bn 2 (x) + bn 1 (x) = Ck 1 (bn,i bn 1,i) x − − − − − − k=1 i=k ! X X = 1 (bn (x) bn 1 (x)) C − − if = 1 is the Catalan operator in (7.15). Show that C C 1 (1 + t) 2t + t2 1 − 1 xt 2 − − (1 t) ! − is the generating function of (bn (x)).

7.1.16. Show directly from the deﬁnition (7.18) that Dn (y + a/y; a) is a polynomial in y + a/y of degree n, and that

(n 1)/2 − n n n/2 n k x = a + a Dn 2k (x; a) . n/2 k − k=0 X n where n/2 = 0 for odd n.

n 7.1.17.Show that 2 Dn (cos θ; 1/4) = 2 cos nθ. This implies

(n 1)/2 − n n n 1 n n cos θ = 2− + 2 − cos (n 2k) θ n/2 k − k=0 X

7.1.18. For the Sheﬀer polynomials Dn (x; a) δ0,n ﬁnd the inverse in the umbral group (6.20). Compare to Exercise 7.1.16. − 7.1.19. Show that the recursion (7.19) leads to the generating function 2 1 1 xt/ 1 t − for bn (x). − − 7.1. The Binomial Reference Frame 203

7.1.20. For given x we can solve the recursion bn (x) = xbn 1 (x) + bn 2 (x) for − − n > 2 with the factoring method (Exercise 1.1.4), using the initial values σ0 = 2 b1 (x) = x and σ1 = b2 (x) = x . Show that bn (x) =

n n (n 1)/2 2 2 − x + √x + 4 x √x + 4 n+1 n n 2k 2 k x − − = 2− x − x + 4 . 2n√x2 + 4 2k+1 k=0 X 7.1.21. Prove the existence of a unique polynomial sequence (En (x; a)) such that

n+1 n+1 En (y + a/y; a) = y (a/y) / (y a/y) . − − Show that

En (x; a) = Dn (x; a) + aEn 2 (x; a) − (n 1)/2 − n/2 k = (n mod 2) a + a Dn 2k (x; a) . − k=0 X 204 Chapter 7. Applications of the General Theory

7.2 Eulerian Diﬀerential Operators

In this section we assume that R k. In 1971, G. E. Andrews presented⊂ the theory of Eulerian Differential Op- erators [5] modeled after Rota’s work on Finite Operator Calculus. As Andrews pointed out, there has been earlier work on this topic by Sharma and Chak [85], and also by Al-Salam [2]. We follow Andrews presentation, but change the order according to our theme. We give only a glimpse into the vast subject of Eulerian Differential Operators, showing how it ties in with our presentation of diagonal operators. We omit completely the combinatorial interpretation, the number of transformations on finite vector spaces. Suppose we have a diagonal reference frame. In Exercise 6.2.6 we men- 1 tioned that a Sheffer sequence (sn) for β− ( ) has initial values sn (1) = δ0,n n R iff n 0 sn (x) t = r (xβ (t)) /r (β (t)). We will see that such Sheffer sequences for ≥ the case β (t) = at (0 = a k) play an important role in the theory of Eulerian P 6 ∈ differential operators. The fact that we concentrate on sn (1) instead of sn (0) is x 0 usually expressed by changing the variable to q for q = 1, so that sn (1) = sn q . 6 x We will follow Andrews in writing sn (X), keeping in mind that X = q . Instead of the translation operator Ey we will introduce the scaling operator mapping Xn to (XY )n, and in view of Y n = eyn we denote this operator by qyη. Hence qyηp (X) = p (XY ) for any polynomial p k [X]. Writing qx for X shows why qyη : qxn q(x+y)n is the analogue to Ey.∈ Note that Andrews writes η instead of 7→ Freeman’s qη. If for an operator T on k [X] holds

T qyη = qyηT then T is called scaling invariant.

Lemma 7.2.1. An operator T on k [X] is scaling invariant, iﬀ T = f (η) for some function f : N0 k, i.e., → TXn = f (η) Xn = f (n) Xn for all n 0. ≥ n Proof. Let pn (X) = TX , where pn is of any degree. Then

yη n y n yn T q X = T (Xq ) = q pn (X) and aη n yη y q TX = q pn (X) = pn (q X) for all y k and n 0. If we think of Y = qy as the variable and X as a parameter, n∈ ≥ then Y pn (X) = pn (YX) implies that

n n TX = pn (X) = f (n) X

for some function f (n) = p (1). 7.2. Eulerian Diﬀerential Operators 205

Remark 7.2.2. Andrews calls the scaling invariant operators on k [X] Eulerian shift invariant operators. Because we reserved the word shift for a change of degree, and used scaling instead, we deviate from Andrews’terminology in this point. For brevity, we also omit the word Eulerian in this connection. Definition 7.2.3. The operator U on k [X] is an Eulerian differential operator iff and Uqyη = qy(η+1)U for all y k, and UXn 0 for all n > 0. ∈ 6≡ The operator χ : p (X) (p (X) p (0)) /X is the standard example of an Eulerian differential operator7→ − χqyηXn = χ (XY )n = Y nχXn = qy(n+1)χXn.

Deﬁnition 7.2.4. The sequence of polynomials (pn) is called an Eulerian sequence, if p0 (x) = 1, deg pn = n and n k pn (XY ) = pk (X) Y pn k (Y ) (7.22) − k=0 X for all n 0 and Y k. ≥ ∈ It follows by induction that pn (1) = δ0,n. An Eulerian sequence is a basis of k [X], and therefore there exists a linear operator P : pn pn 1. 7→ − Lemma 7.2.5. If (pn) is an Eulerian sequence, and P : pn pn 1, then P is an Eulerian diﬀerential operator. 7→ − Proof.

n n 1 yη n k − n 1 k P q pn (X) = P pn k (X) Y − pk (Y ) = Y pn 1 k (X) Y − − pk (Y ) − − − k=0 k=0 X X y(η+1) = Y pn 1 (XY ) = q P pn (X) . − n Because pn is of degree n, we have that PX 0 for all n > 0. The Eulerian differential operators and6≡ the scaling invariant operators are closely related, as stated in the following lemma. Lemma 7.2.6. If d (η) is any scaling invariant operator, and d (n) = 0 for all n > 0, then χd (η) is an Eulerian differential operator. Vice versa,6 if U is an Eulerian differential operator, then M (X) U is scaling invariant, M (X) U = d˜(η), satisfying d˜(n) = 0 for all n > 1, and d˜(0) = 0. 6 Proof. That χ is an Eulerian differential operator shows the first direction of the Lemma. Next suppose U is an Eulerian differential operator. We show for n > 0 that M (X) UXn is translation invariant: M (X) Uqyη = M (X) qy(η+1)U = qyηM (X) U. 206 Chapter 7. Applications of the General Theory

Therefore, all Eulerian differential operators are of the form χd (η), where d is a nonzero function on N1, and d (0) can be chosen as 0. Remember that we defined a diagonal reference operator as = d0 (η) χ/d0 (η), where d0 is any R R n nonzero function on N0 (section 6.2.1). Letting d0 (n) = 1/ k=1 d (k) shows that the Eulerian differential operators are exactly the diagonal reference operators. They lower the degree by one, and map constants into 0. WeQ will call the function d also diagonal, i.e., a diagonal function is 0 at 0, and d (n) = 0 for all n > 0. We summarize the above in the following Corollary. 6

Corollary 7.2.7. The polynomial sequences (pn) is Eulerian iff (pn) is the Sheffer sequences with pn (1) = δ0,n in some diagonal reference frame. The diagonal reference operators are called Eulerian differential operators by Andrews, and the corresponding Sheffer sequences are called Eulerian basic polynomials. Finding Eulerian basic polynomials is easy in principle, because we n know there generating function, n 0 pn (X) t = r (Xt) /r (t), where ≥ P n r (Xt) = Xntn/ d (k) , n 0 k=1 X≥ Y if χd (η) is the reference operator, but it can be very diffi cult in applications. Example 7.2.8. (a) The exponential reference frame is diagonal, hence n n ((X 1) /n!)n 0 must be Eulerian. We check this by defining pn (X) = (X 1) /n! and− calculating ≥ − n n n k k n (XY 1) (XY Y ) − (Y 1) n k pn (XY ) = − = − − = pn k (X) Y − pk (Y ) . n! (n k)! k! − k=0 k=0 X − X (b) The q-differentiation operator η Dq = χ (1 q ) − was defined in section 6.2.1 . Its Eulerian basic sequence has the generating function (t; q) / (Xt; q) (Exercise 6.2.1). The identity ∞ ∞ n 1 (X 1)(X q) (X q − ) n (t; q) − − ··· − t = ∞ (q; q) (Xt; q) n 0 n X≥ ∞ is due to Heine, and shown by many authors [3, Theorem 10.2.1]. We can write n 1 the Eulerian basic sequence for Dq as pn (X) = X X− ; q n / (q; q)n. Phrasing equation (7.22) for these polynomials gives n (XY ; q)n (Y ; q)k (X; q)n k n k = − Y , (7.23) (q; q) (q; q) (q; q) − n k=0 k n k X − the q-binomial theorem (Andrews [4, (3.3.10)]). 7.2. Eulerian Differential Operators 207

If d is diagonal, where do the values d (n) occur in the Eulerian basic polynomials pn (X) for χd (η)? We show in Exercise 7.2.2 that the leading coeﬃ cient n of pn (x) equals d0 (n) = 1/ k=1 d (k), where d0 (η) χ/d0 (η) = χd (η). Hence Q n n n n r (Xt) n 0 X t [X ] pn (X) ≥ pn (X) t = = n n . (7.24) r (t) t [X ] pn (X) n 0 P n 0 X≥ ≥ P We will now derive a second generating function for (pn), involving the derivative of the polynomials.

Theorem 7.2.9. If (pn) is an Eulerian basic sequence, then

n n n n 1 pn0 (1)t (X 1)/n pn (X) t = e ≥ − n 0 P X≥ (Andrews [5, Theorem 8]). n Proof. We have n 0 pn (X) t = r (Xt) /r (t), thus ≥

P n r0 (Y Xt) p0 (Y )(Xt) = Xt . n r (Xt) n 1 X≥ Letting Y = 1 we get

n Xtr0 (Xt) = r (Xt) pn0 (1) (Xt) . n 1 X≥ Seeing this as a diﬀerential equation in y = r with independent variable Xt we obtain the solution

n n 1 pn0 (1)(Xt) /n y = C (t) e ≥ = r (Xt) P where C (t) = r (0) = 1. Example 7.2.10. (a) We saw in Example 7.2.8 that ((X 1)n /n!) is the Eulerian n 1 − basic sequence for . From p0 (X) = (X 1) − / (n 1)! we see that p0 (1) = D n − − n δ1,n, and therefore we conﬁrm that

n (X 1)t (X 1) /n! = e − − n 0 X≥ by Theorem 7.2.9. We get the same generating function, eXt/et, from (7.24). η (b) The q-diﬀerentiation operator Dq = χ (1 q ) has Eulerian basic sequence n 1 − pn (X) = X X− ; q n / (q; q)n. We saw that in Example 7.2.8(b) that (t; q) p (X) tn = . n (Xt; q∞) n 0 X≥ ∞ 208 Chapter 7. Applications of the General Theory

We give a proof for this result by applying Theorem 7.2.9. First we have to ﬁnd

n 1 n 1 n 1 (X 1) (X q) X q − − (X 1) (X q) X q − p0 (X) = − − ··· − = − − ··· − n D (q; q) (X qk)(q; q) n k=0 n X − n hence pn0 (1) = (q; q)n 1 / (q; q)n = 1/ (1 q ) for all n 1. Thus − − ≥ tn ∞ ∞ tn exp p (1) tn/n = exp = exp qnk n0 n (1 qn) n n 1 n 1 (k=0 n=1 ) X≥ X≥ − X X ∞ ∞ = exp ln 1 tqk = 1/ 1 tqk = 1/ (t; q) . − − − ∞ ( k=0 ) k=0 X Y Andrews [5] gives more results in the flavor of part (b) of the above. Theorem 7.2.9 shows how an infinite sum becomes an infinite product,

n n n p0 (1)t (X 1)/n pn (X) t = e n − . n 0 n 1 X≥ Y≥ We will ﬁnish this short introduction with a result (Corollary 7.2.12) that shows how all diagonal operators can be obtained from each other. Let d be diagonal. The powers of χd (η) can be written as

(χd (η))k = χkd (η k + 1) d (η) = d (η + 1) d (η + k) χk − ··· ··· and therefore

M Xk (χd (η))k Xn = d (η k + 1) d (η) Xn − ··· if n 0. If 0 n < k then M Xk (χd (η))k Xn = 0 (remember that d (0) = 0). ≥ ≤ Lemma 7.2.11. Let d be diagonal and (pn) be the Eulerian basic sequence for χd (η). The operator T is scaling invariant iﬀ

T = Eval1 T pk d (η k + 1) d (η) h | i − ··· k 0 X≥ Proof. Let x¯ be any element of k [X], and write X¯ for qx¯. An Eulerian basic sequence satisﬁes condition (7.22), hence

n k pn XX¯ = pk X¯ X pn k (X) − k=0 X Keeping X ﬁx, this is a polynomial in X,¯ and we obtain

n k T pn XX¯ = T pk X¯ X pn k (X) − k=0 X 7.2. Eulerian Diﬀerential Operators 209 for any operator T . Only if T is scaling invariant, the left hand side equals xη q T pn X¯ . Letting X¯ = 1 we have

n k T pn (X) = Eval1 T pk X pn k (X) . h | i − k=0 X Considering this equation for all X gives the result. Corollary 7.2.12. The Eulerian operator χc (η) can be expressed as a power series in the Eulerian operator χd (η) by

k k χc (η) = χ Eval1 c (η) pk M X (χd (η)) h | i k 0 X≥ = χ (c (η) pk) (1) d (η k + 1) d (η) , − ··· k 0 X≥ where (pn) is the Eulerian basic sequence for χd (η).

7.2.1 Exercises n 1 7.2.1. Show that pn (X) = (X 1) X − for n > 0, p0 (X) = 1, satisﬁes the − condition (7.22) of an Eulerian sequence. In which reference frame is (pn) a Sheﬀer sequence? n n 7.2.2. Let χd (η) be a diagonal operator. Show that [X ] pn (X) = 1/ k=1 d (k), if n 1 (pn) is the Eulerian basic sequence for χd (η). Find an expression for X − pn (X). Q 7.2.3. Let c (η) = 1 qη and d (η) = η. Show that −

c (η) = (c (η) πk) (1) d (η k + 1) d (η) , − ··· k 0 X≥ where πk (X) is the Eulerian basic sequence for . Show also the other direction, D

η = (ηpk) (1) c (η k + 1) c (η) − ··· k 0 X≥ where pk (X) is the Eulerian basic sequence for Dq. Bibliography

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