finite operator calculus with applications to linear recursions

Finite Operator CalculusWith Applications to Linear Recursions

Heinrich NiederhausenFlorida Atlantic University

Boca [email protected]/Niederhausen

Contents

Contents i

1 Prerequisites from the Theory of Formal Power Series 11.1 Generating Functions and Linear Recursions . . . . . . . . . . . . . 2

1.1.1 Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.1.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2 Composition and Inverses . . . . . . . . . . . . . . . . . . . . . . . 131.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.3 Multivariate Power Series . . . . . . . . . . . . . . . . . . . . . . . 171.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2 Finite Operator Calculus in One Variable 212.1 Polynomials, Operators, and Functionals . . . . . . . . . . . . . . . 21

2.1.1 The Vector Space of Polynomials, and Their Bases . . . . . 222.1.2 Standard Bases and Linear Operators . . . . . . . . . . . . 252.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.2 Finite Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.2.1 Translation Operators . . . . . . . . . . . . . . . . . . . . . 282.2.2 Basic Sequences and Delta Operators . . . . . . . . . . . . 302.2.3 Special Cases . . . . . . . . . . . . . . . . . . . . . . . . . . 322.2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.3 Sheffer Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.3.1 Initial Values Along a Line . . . . . . . . . . . . . . . . . . 472.3.2 The Umbral Group . . . . . . . . . . . . . . . . . . . . . . . 512.3.3 Special Cases . . . . . . . . . . . . . . . . . . . . . . . . . . 542.3.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

2.4 Transfer Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 692.4.1 Umbral shifts and the Pincherle derivative . . . . . . . . . . 732.4.2 Proof of the Transfer Formula . . . . . . . . . . . . . . . . . 742.4.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

ii CONTENTS

3 Applications 793.1 The Functional Expansion Theorem . . . . . . . . . . . . . . . . . 79

3.1.1 Some Applications of the Functional Expansion Theorem . 833.1.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

3.2 Diagonals of Riordan Matrices as Values of Sheffer Sequences . . . 893.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

3.3 Determinants of Hankel Matrices . . . . . . . . . . . . . . . . . . . 943.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

3.4 Classical Umbral Calculus . . . . . . . . . . . . . . . . . . . . . . . 1023.4.1 The Cumulant Umbra . . . . . . . . . . . . . . . . . . . . . 1093.4.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

4 Finite Operator Calculus in Several Variables 1134.1 Polynomials and Operators in Several Variables . . . . . . . . . . . 114

4.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1174.2 The Multivariate Transfer Formulas . . . . . . . . . . . . . . . . . 119

4.2.1 Transfer with constant coeffi cients . . . . . . . . . . . . . . 1194.2.2 Operator based transfer . . . . . . . . . . . . . . . . . . . . 1214.2.3 The multivariate Pincherle derivative . . . . . . . . . . . . . 1234.2.4 Transfer with operator coeffi cients . . . . . . . . . . . . . . 1284.2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

4.3 The Multivariate Functional Expansion Theorem . . . . . . . . . . 1344.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

5 Special Constructions in Several Variables 1375.1 Multi-indexed Sheffer Sequences . . . . . . . . . . . . . . . . . . . 137

5.1.1 Delta Operators for multi-indexed Sheffer sequences. . . . . 1395.1.2 Translation invariance of diagonalization, and some examples.1405.1.3 Abelization of Multi-Indexed Sequences . . . . . . . . . . . 1425.1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

5.2 Polynomials with all but one variable equal to 0 . . . . . . . . . . . 1505.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

5.3 Cross-Sequences and Steffensen Sequences . . . . . . . . . . . . . . 1535.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

6 A General Finite Operator Calculus 1576.1 Transforms of Operators . . . . . . . . . . . . . . . . . . . . . . . . 1576.2 Reference Frames, Sheffer Sequences, and Delta Operators . . . . . 162

6.2.1 Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . 1626.2.2 Sheffer Sequences and Delta Operators . . . . . . . . . . . . 1656.2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

6.3 Transfer Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . 1716.3.1 General Umbral Shifts and the Pincherle Derivative . . . . 1716.3.2 Equivalent Transfer Formulas . . . . . . . . . . . . . . . . . 173

CONTENTS iii

6.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1756.4 Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

6.4.1 Augmentation . . . . . . . . . . . . . . . . . . . . . . . . . . 1796.4.2 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . 1796.4.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

7 Applications of the General Theory 1877.1 The Binomial Reference Frame . . . . . . . . . . . . . . . . . . . . 188

7.1.1 Orthogonal Binomial Reference Sequences . . . . . . . . . . 1897.1.2 Generalized Catalan Operators . . . . . . . . . . . . . . . . 1957.1.3 Dickson Polynomials . . . . . . . . . . . . . . . . . . . . . . 1977.1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

7.2 Eulerian Differential Operators . . . . . . . . . . . . . . . . . . . . 2047.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

8 Solutions to Exercises 211

Bibliography 257

iv CONTENTS

Preface

The following text originated from various lecture notes for graduate courses incombinatorics. I would very much appreciate receiving your comments, additionsand corrections.

My apologies to all those mathematicians not mentioned in the text, theirimportant contributions to the theory of the Finite Operator Calculus skipped overbecause of ignorance or by design in order to keep the material manageable. Yourviews can still be included - please let me know: [email protected].

Introduction

Every linear operator T on polynomials has a representation T =∑n≥0M(pn)Dn,

where D is the derivative operator, and M(pn) stands for multiplication by thepolynomial pn (x) (Pincherle [74, 1901]). When applied to polynomials, the oper-ator T reduces to a finite sum, of course, and may therefore be called a “finite”operator. A special case of this concept, when T can be written as T =

∑n≥0 cnDn,

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. Kahaner and A.Odlyzko. Hence T is isomorphic to the formal power series

∑n≥0 cnt

n, and it is ofcourse exactly this isomorphism that made the Finite Operator Calculus so widelyapplicable. We adopt Rota’s approach in this book, but consider, in the last twochapters, also linear operators of the form T =

∑n≥0 cnRn, where R can be any

linear operator reducing the degree by 1, deg (Rq) = deg q−1 for all polynomials qof degree larger than 0, andRq = 0 for polynomials of degree 0 (we identify polyno-mials of degree 0 with scalars c 6= 0; however, we let deg (0) = −∞, as usual). Thefundamental difference between Rota’s approach and the generalized version wepresent, due to J. M. Freeman [35, 1985], also his student, is that T =

∑n≥0 cnDn

is translation invariant, TEc = EcT , where Ec : f (x) 7→ f (x+ c) is the operatortranslating by c. The generalized version does not have this property. Translationinvariance, however, is an important feature in many applications.

The applications we have in mind are usually the solutions to recursive equa-tions. Suppose your analysis of a given enumerative problem resulted in a recursiveexpression for the numbers you are looking for; to be specific, let us assume youarrived at

Fm = Fm−1 + Fm−2,

the recursion for the Fibonacci numbers, starting at F0 = F1 = 1. A computerwill give us “special”answers in a very short time, even F10,000 makes no problemwhatsoever. From this point of view, you will not need this book. However, finding

out that Fn =((

1 +√

5)n+1 −

(1−√

5)n+1

)/(2n+1

√5)is as surprising as it is

rewarding. In addition, it is even “practical”; a scientific calculator will show thatF10,000 ≈ 5.4× 102089. 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 toolfor solving this type of linear recursion. We give a brief introduction in the firstchapter. The reader familiar with formal power series may just want to browsethrough 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 haveto know Fn−1 (n) before we can say what the value of Fn (n) is. We will seethat F0 (x) , F1(x), . . . is a sequence of polynomials, actually a basis, and that theoperator T : Fn (x) 7→ Fn−1 (x) is a translation invariant operator, satisfying theoperator equation

I = E−1 + E−2T.

We will show in chapters 2 and 3.1 how to find the solution with given initialvalues 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. Thissystem 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 chapters3.1 and 5, dedicated to the multivariate Finite Operator Calculus. Technically theproblems get more diffi cult to solve, of course, when it comes to higher dimensions.The theory, however, remains quite easy. Somewhere between the univariate andthe 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”, topurge 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 OperatorCalculus”was published a part VIII in a series of papers “On the Foundations ofCombinatorial Theory”. A complete survey of papers relating to Umbral Calcu-lus until the year 2000 has been compiled by Di Bucchianico and Loeb [26]. Anapplication 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 theoryitself. J. M. Freeman explored this generalization in some depth [35], and we followit 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 inUmbral Calculus; for example, the contents of section 2.2.3, Basic sequences withpolynomial coeffi cients, is reduced to αD ≡ χ.α in umbral notation. However, itmay be exactly this brevity, achieved through a multitude of special definitions,that prevents Umbral Calculus from being widely known. We give an introductionto Umbral Calculus in section 3.4.

Most of the examples and exercises in this book refer to combinatorial prob-lems, with few exceptions. Yet this is not a book on enumerative combinatorics,because we begin where combinatorics ends, that is at the crucial point, wherecombinatorics 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. Wewill therefore describe the combinatorics only briefly, just enough to introducethe recursion. There is a special sequence of examples taken from [84] about theenumeration 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 thepolynomial formulas in the examples and exercises.

Chapter 1

Prerequisites from the Theory ofFormal Power Series

The isomorphism between Finite Operator Calculus and formal power series allowsus 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 theease of use; how diffi cult it is to formulate, classify, and solve a given recursionin one way or the other. However, ease of use heavily depends on the availabilityof a commonly known (mathematical) language. For example, linear recursionsare easily translated into functional relations between formal power series, or intoequations between operators. We will see in section 2.4 how to solve such anequation directly for the polynomials involved in the recursion, without knowingthe operators explicitly.

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

Most important for the following chapters is the Lagrange-Bürmann inver-sion formula. A proof in algebraic form can be found in Henrici’s Applied andComputational Complex Analysis, Vol. 1 [41]. See also Hofbauer [43]. ApplyingLagrange inversion in several variables gets more tedious; for pedagogical reasonswe 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 severalvariables. 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

consisting of +1 and −1 in equal numbers such that the partial sum∑ik=0 ck are

never below 0,i∑

k=0

ck ≥ 0 and2n∑k=0

ck = 0

for all i = 1 to 2n. The first few Cn’s areC1 = # (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; seeExample 1.2.1). The generating function of the sequence (Cn)n≥0 is defined as the“series”γ (t) = C0t

0 + C1t1 + C2t

2 + . . . , 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 scalarmultiplication. Actually, the formal power series are a vector space over the integersZ, or the rational Q, or some other ring. If we call this ring k, the power serieswill 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 willassume in this chapter that k is an integral domain, i.e., we assume that for twocoeffi cients a and b their product ab cannot be zero, if a and b are both differentfrom 0. In the following chapters, we even will assume that k is a field, and wewrite 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) astheir 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 cientfunctional [tn] on k [[t]] such that Cn = [tn] γ (t), or we could write Cn = [γ]n,realizing that the name of the formal variable is not of interest. Analysis has givenus a simple method to find the n-th coeffi cient of an analytic function γ (t), Cn =1n!

dn

dtn γ (t)∣∣t=0. We can also define the formal derivative of a formal power series:

ddtγ (t) has coeffi cients

[ddtγ]n

= (n+ 1)Cn+1, and therefore Cn = 1n!

dn

dtn γ (t)∣∣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 notneed convergence. However, having some positive radius of convergence is a greathelp 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 Catalannumbers, probably one of the most studied sequences in combinatorics! And it iswell-known that

∞∑n=0

Cntn =

2

1 +√

1− 4t(1.1)

(see Exercise 2.3.8 for a proof), a series that converges for all t < 1/4, viewedas 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,

C1 = ddt

21+√

1−4t

∣∣∣t=0

= 4

(1+√

1−4t)2√

1−4t

∣∣∣∣t=0

= 1,

C2 = 12ddt

4

(1+√

1−4t)2√

1−4t

∣∣∣∣t=0

= 4 3√

1−4t+1

(1−4t)3/2(1+√

1−4t)3

∣∣∣∣t=0

= 2,

C3 = 43ddt

3√

1−4t+1

(1−4t)3/2(1+√

1−4t)3

∣∣∣∣t=0

= 16 2√

1−4t+3−10t

(−1+4t)2√

1−4t(1+√

1−4t)4

∣∣∣∣t=0

= 5 , etc.

The above calculations should convince you that finding the coeffi cients bythe differentiation method is a recursive procedure; from the n-th derivative youcalculate 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 thatwe 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

2t=

∞∑n=1

(1/2

n

)(−1)

n+122n−1tn−1.

Substitute (1/2n

)=

(−1)n−1

22n−1 (2n− 1)

(2n−1n

)and get Cn = 1

n+1

(2nn

). Note that for large n this formula needs a computer

also; however, Cn can be fairly accurately approximated by Stirling’s formula,(2nn

)/ (n+ 1) ≈ 22n/ ((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,

[tn] (γ (t)σ (t)) =

n∑k=0

cksn−k,

if[tk]γ (t) = ck and

[tk]σ (t) = sk. This definition is carried over to the formal

power series. Note that multiplication is commutative, because multiplication ink is commutative.

The reciprocal of a formal series γ (t) can exist only when γ (0) 6= 0; otherwisewe would obtain negative powers of t. We write 1/γ (t) for the reciprocal, and


sometimes γ (t)−1. We have 1 = γ (t) (1/γ (t)); this tells us all about the coeffi cients

of 1/γ (t) =∑∞n=0 cnt

n :1 = c0c0, hence c0 = 1/c0 (showing again that c0 6= 0),0 = c1c0 + c0c1, thus c1 = −c1/c20, and in generalcn = −

(∑n−1k=0 ckcn−k

)/c0, for all n ≥ 1.

There will only be a positive power of c0 in the denominator of the expressionfor cn. 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 nonzeromultiplicative inverse in the coeffi cient ring for all n ≥ 0.

For example, if the coeffi cient ring equals Z, a power series must start with1 or −1 in order to have a reciprocal. If we return to the Catalan generatingfunction 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) 6= 0.We saw above that if σ has a reciprocal then ordσ = 0. A power series β (t) oforder 1 is called a delta series in the Finite Operator Calculus, if ordβ (t) = 1 andβ(t)/t has a reciprocal.

Remark 1.1.2. We defined scalar multiplication, addition, and multiplication offormal power series in purely algebraic terms. For the reader interested in thecombinatorics behind all this we recommend the 1981 paper by Joyal [48], andthe book on Combinatorial Species and Tree-like Structures by F. Bergeron, G.Labelle, and P. Leroux [10]. How do we “combinatorially”calculate the coeffi cientof β (t)

k if k is a positive integer and β (t) = b1t1 + b2t

2 + . . . ? If we write

β (t)k

=∑n≥k

tn∑

j1+···+jk=nji>0

bj1 · · · bjk

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 thatλ1 + · · · + λk = n. If we sort such that λ1 ≥ · · · ≥ λk, then λ = (λ1, . . . , λk) iscalled a partition of n. In symbols, λ ` n. The number of parts is k, |λ| = k. Thereis an equivalent representation of a partition as a multiset

1`1 , . . . , n`n

, where

the term iì means ì occurrences of i in λ. Hence∑ni=1 ì = k, and

∑ni=1 iì = n.

For every partition of n we think of these two equivalent representations simul-taneously! The above mentioned multiplicity is the number of permutations of λ,which is

(k

`1,...,`n

). (Choose `1 places for the ones in λ, then `2 places for the twos,

etc.) Hence

β (t)k

=∑n≥k

tn∑

λ`n,|λ|=k

(k

`1, . . . , `n

) k∏i=1

bλi . (1.3)


For example,

[tk]β (t)

k=

∑λ`k,|λ|=k

(k

`1

) k∏i=1

bλi = bk1

[tk+1

]β (t)

k=

∑λ`k+1,|λ|=k

(k

`1, `2

) k∏i=1

bλi = kb2bk−11 (1.4)

[tk+2

]β (t)

k= kb3b

k−11 +

(k

2

)b22b

k−21[

tk+3]β (t)

k= kb4b

k−11 + 2

(k

2

)b3b2b

k−21 +

(k

3

)b32b

k−31 .

There is a second concept of partitioning in combinatorics, the partitions of ann-set. Here a set of n-elements is written as the union of k nonempty and disjointsubsets. The number of such set partitions is S (n, k), the Stirling number of thesecond 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 arewritten 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 insome way (lexicographically). Again, the numbers |B1| ≥ · · · ≥ |Bk| will make apartition λ of n, with k parts, but with a certain multiplicity. The multiplicity is

n!λ1!···λk!`1!···`n! , i.e., this is the number of partitions in Sn,k such that |Bi| = λi.

Hence (1.3 shows that β (t)k

=∑n≥k

tnk!

n!

∑λ`n,|λ|=k

n!

λ1! · · ·λk!`1! · · · `n!

k∏i=1

λi!bλi

=∑n≥k

tnk!

n!

∑λ`n,|λ|=k

(k∏i=1

λi!bλi

)×

× |(B1, . . . , Bk) ∈ Sn,k such that |Bi| = λi|

=∑n≥k

tnk!

n!

∑(B1,...,Bk)∈Sn,k

k∏i=1

|Bi|!b|Bi|. (1.5)

Example 1.1.3 (Fibonacci numbers). The Fibonacci numbers Fn can be defined byF0 = F1 = 1 and Fn = Fn−1 + Fn−2 for all n ≥ 2. (Note that Z can serve asthe coeffi cient ring.) However, defining Fn = 0 for negative n lets us express thisrecursion as Fn − Fn−1 − Fn−2 = δ0,n for all n ≥ 0. Multiplying this formula bytn and summing up over all n ≥ 0 gives the generating function of the Fibonacci


numbers,

1 =∑n≥0

(Fn − Fn−1 − Fn−2) tn =

∑n≥0

Fntn

(1− t− t2)thus

∑n≥0 Fnt

n = 1/(1− t− t2

). The roots of the denominator are easy to find;

hence

Fn = [tn]−1

t2 + t− 1

= [tn]

(1/√

5

t+ 12 + 1

2

√5− 1/

√5

t+ 12 −

12

√5

)

=1√

5(

12 + 1

2

√5) (− 1

2

√5− 1

2

)n − 1√5(

12 −

12

√5) (

12

√5− 1

2

)n=

(1 +√

5)n+1 −

(1−√

5)n+1

2n+1√

5

On the first glance, this formula for Fn does not look integer, but a closer look willeasily convince you. Of course we could also expand 1/

(1− t− t2

)in powers of t.

We get Fn =∑n/2k=0

(n−kk

), certainly an integer, and so we arrive at the identity

n/2∑k=0

(n− kk

)=

(1 +√

5)n+1 −

(1−√

5)n+1

2n+1√

5

A recursion of the form σn =∑nj=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).The starting value in this recursion, σ0, equals γ0, and if γ1, γ2, . . . are differentfrom 0, the recursion is called inhomogeneous. However, if the sequence of inho-mogeneous terms eventually becomes 0, so that the last nonzero term is γ`−1, thenwe usually do not say that the recursion is inhomogeneous, but has initial valuesσ0, σ1, . . . , σ`−1, and then follows the (homogeneous) recursion σn =

∑nj=1 αjσn−j

for n ≥ `. Of course, the terms γ0, . . . , γ`−1 can be recovered from the initial values,as γk = σk −

∑kj=1 αjσk−j .

Theorem 1.1.4. Suppose the numbers σn solve for n ≥ 0 the (inhomogeneous)linear recursion

σn =

n∑j=1

αjσn−j + γn

where α1, α2, . . . and γ0, γ1, . . . are sequences of given constants. Then

σn =n∑k=0

γk[tn−k

] 1

1−∑∞j=1 αjt

j


and∞∑n=0

σntn =

∑∞k=0 γkt

k

1−∑∞j=1 αjt

j

Proof. From∑∞k=0 γkt

k =(∑∞

k=0 σktk) (

1−∑∞j=1 αjt

j)follows the Theorem.

The form of the generating function for (σn) shows us that it will be rational ifboth (γn) and (αn) are eventually 0. This characteristic will be important in section1.1.1. Explicit expressions can be derived from Theorem 1.1.4 in a large numberof applications. We discuss Fibonacci-like sequences σn = uσn−1 + vσn−2 + γn inExercises 1.1.2 and 1.1.3.

The sequences 1,−a1,−a2, . . . and(

[tn] 1/(

1−∑∞j=1 αjt

j))

n≥0are some-

times called orthogonal (because they are reciprocals), 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 thenumber of permutations π of [n] that are derangements, i.e., πi 6= i for all i =1, . . . , n. They follow the recursion dn = ndn−1 + (−1)

n,with initial value d0 = 1.In the notation of Proposition 1.1.4, σn := dn/n!, αn = δn,1, and γn = (−1)

n/n!.

Thus

dn = n!σn = n!

(1 +

n∑k=1

(−1)k

k!

)([tn−k

] 1

1− α1t

)= n!

n∑k=0

(−1)k

k!∑n≥0

dntn/n! =

∞∑n=0

σntn =

∑∞k=0 (−1)

ktk/k!

1− t = e−t/ (1− t)

The generating function is not rational.

Example 1.1.6 (Bernoulli Numbers). The Bernoulli numbers Bn solve the systemof equations δn,0 =

∑nk=0

(n+1k+1

)1

n+1Bn−k for n ≥ 0. Dividing by n! shows that thenumbers Bn/n! can be calculated from the linear recursion

Bn/n! = −n∑k=1

1

(k + 1)!Bn−k/ (n− k)!.

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

∑n≥0

Bnn!tn =

B0

1 +∑j≥1 t

j/ (j + 1)!=

t

t+∑j≥1 t

j+1/ (j + 1)!

=t

t+ (et − 1− t) =t

et − 1.


The generating function is not rational. The Bernoulli polynomials ϕn (x) :=∑nk=0

Bkk!

xn−k

(n−k)! have generating functiont

et−1ext by convolution. Note that

t

et − 1e(x+1)t =

t

1− e−t ext =

−te−t − 1

e−x(−t)

hence ϕn (x+ 1) = (−1)nϕn (−x). See Exercise 2.3.2 for more details on Bernoulli

numbers. An explicit formula for the Bernoulli numbers is obtained in Exercise2.3.15. 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 verticesin 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 ahorizontally convex polyomino P any line segment parallel to the x-axis with bothend points in P must be completely in P . For example, there are 19 polyominoswith 4 squares; all 19 polyominos are horizontally convex.

Hickerson [42] found a combinatorial proof in one dimension that the numberf (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. ByProposition 1.1.4

∑n≥0

f (n) tn =1

1−∑∞j=1 αjt

j

`−1∑k=0

σk −∑j≥1

αjσk−j

tk

=

(1 + (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 with1 + 2t+ 6t2 + 19t3 + 61t4 + 196t5 + 629t6 + 2017t7 + 6466t8 + 20 727t9 + 66 441t10 +212 980t11 + 682 721t12 + 2188 509t13 + 7015 418t14 + . . .


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] ob-tained the recursion from the generating function, which he found by the transfer-matrix method. His approach starts from the observation thatf (n− 1) =

∑(n1 + n2 − 1) (n2 + n3 − 1) + · · ·+ (ns + ns+1 − 1), summing over

all 2n−1 compositions of n1 + · · ·+ns+1 = n ( s = 0 contributes 1). For example, ifn = 5, then the composition 5 (s = 0; 1 term) contributes 1, the compositions into2 terms (s = 1) contribute 16, then 3 terms contribute also 16, 4 terms add 15, 5terms 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 onlyfinitely many factors α1, . . . , αd, the generating function

∑n≥0 σnt

n is rational.By Stanley’s Theorem 4.1.1 [89]

σn =

k∑i=1

pi (n) rni (1.6)

where the 1/ri’s are the k distinct roots of the polynomial

1−d∑j=1

αjtj =

k∏i=1

(1− rit)mi

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 fewvalues of σn. We followed Stanley’s procedure in the example of the Fibonaccinumbers, disguised as a partial fraction decomposition. We will do partial fractiondecomposition 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 1.1.7we found the generating function for the number f (n) of horizontally convex poly-ominos made of n+ 1 squares,

∑n≥0

f (n) tn =(1− t)3

1− 5t+ 7t2 − 4t3=

(t− 1)

4

(1− t)2

(t− t1) (t− t2) (t− t3)

With r =3√

71 + 6√

177we determine the roots of the denominator ast1 = 11−r2+7r

12r , t2 = r2−11+14r24r − i

√3 r

2+1124r , and t3 = t∗2, the complex conjugate of

t2 (the calculations were made with an algebra package; it can be cumbersome to


do this by hand!). We wrote the generating function already in a form to suggesta partial fractions decomposition,∑

n≥0

f (n) tn =t− 1

4

(A

t− t1+

B

t− t2+

C

t− t∗2

)

We need A (t− t2) (t− t∗2)+B (t− t1) (t− t∗2)+C (t− t2) (t− t1) = (1− t)2, thus

A+B + C = 1

−A (t2 + t∗2)−B (t1 + t∗2)− C (t1 + t2) = −2

At2t∗2 +Bt1t

∗2 + Ct1t2 = 1

This system has the solution

A = 13

(5r+r2−11)2

r4−11r2+121 , B = 13

(−r( 52 i√

3+ 52 )+r2− 11

2 i√

3+ 112 )

2

(r2− 112 i√

3− 112 )(r2+11)

, and C = B∗.

If we write ξ =(i√

3− 1)/2, ξ∗ = −

(i√

3 + 1)/2 (both are third roots of unity)

then B = 13

(5rξ∗+r2−11ξ)2

r4−11ξr2+121ξ∗ . The generating function can be written as∑n≥0

f (n) tn =(1− t)−4

(A

t− t1+ 2 Re

B

t− t2

)For n ≥ 1 we find

f (n) = [tn](1− t)−4

(A

t− t1+ 2 Re

B

t− t2

)= −1

4[tn]

(A

t− t1+ 2 Re

B

t− t2

)+

1

4

[tn−1

]( A

t− t1+ 2 Re

B

t− t2

)=A

4

(t−n−11 − t−n1

)+ 2 Re

B

4

(t−n−12 − t−n2

)=A

4t−n−11 (1− t1) + 2 Re

B

4t−n−12 (1− t2)

With A (1− t1) = 136

(5r+r2−11)3

(r4−11r2+121)r and B (1− t2) =ξ(11ξ−5rξ∗−r2)

3

36(r4−11ξr2+121ξ∗)r we get

f (n) =1

144r

((5r + r2 − 11

)3r4 − 11r2 + 121

t−n−11 − 2 Re

(ξ(11ξ − 5rξ∗ − r2

)3(r2 − 11ξ∗) (r2 + 11)

t−n−12

)).

1.1.2 Exercises

1.1.1. Find an explicit expression for the number of horizontally convex polynomi-als, i.e., expand the generating function (1− t)3

/(1− 5t+ 7t2 − 4t3

)in terms of

t.


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

σ0 + (σ1 − uσ0) t

1− ut− vt2 +

∑∞k=2 γkt

k

1− ut− vt2 . (1.7)

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 wehave the terms γn = δ0,n and α1 = 1, α2 = 1, αn = 0 for all n ≥ 2. HenceFn =

∑nk=0 γk

[tn−k

]1

1−∑∞j=1 αjt

j = [tn] 11−t−t2 . Show that for any Fibonacci-like

sequence holds

σn =

n∑k=0

γk

n−k∑j=0

(n−k−j

j

)un−k−2jvj (1.8)

where γ0 = σ0 and γ1 = σ1 − u. Derive sums like (1.8) for the following specialcases:

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 6= 1.

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

∑n≥0

Pn (x) tn =1− xt

1− 2xt− t2 =1

2+

1 + t2

2− 4xt− 2t2(1.9)

=1

2+

1 + t2

2 (1− t2)

1

1− 2xt/ (1− t2)

The Pell numbers pn are defined as pn = Pn (1). How do different values forp1 change the subsequent numbers?

4. The recursion Un (x) = 2xUn−1 (x)−Un−2 (x) for the Chebychev polynomialsof 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. Morerelated results on Chebychev polynomials in Exercise 7.1.13.

1.1.3. Show that for all Fibonacci-like sequences (Exercise 1.1.2) holds σ2n =

σn−1σn+1 + (−1)nvnc for all n ≥ 1. Determine the constant c, and show that

for the Pell polynomials holds Pn (x)2

= Pn−1 (x)Pn+1 (x) + (−1)n (

1 + x2), for

the Fibonacci numbers F 2n = Fn−1Fn+1 + (−1)

n, and for the Chebychev polynomi-als U2

n (x) = Un−1 (x)Un+1 (x)− 1.


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 polynomial1−

∑dj=1 αjt

j = 1− ut− vt2. Show that

[tr]1

1− ut− vt2 = 2−r−1

(u+√

4v + u2)r+1 −

(u−√

4v + u2)r+1

√4v + u2

if the discriminant 4v + u2 6= 0. If 4v + u2 = 0 then 1 − ut − vt2 = (ut− 2)2/4

and[tr]

1

(1− ut/2)2 = (r + 1)

(u2

)r.

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 =LkFn−k − (−1)

kFn−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 initialvalues L0 = 2, L1 = 1, hence

Lk =

(1 +√

5)k

+(1−√

5)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 introducea notation. Remember that k is an integral domain: k has a unit element 1, iscommutative and has no zero divisors, ab 6= 0 when a and b are in k and bothdifferent from 0. We write k [t] for the polynomials in t with coeffi cients in thering k, and k [[t]] for the formal power series with coeffi cient ring k as before. Notethat we can view k [t] as being imbedded in k [[t]]. However, any element a fromk can be used to evaluate a polynomial at a, giving again an element from k. Ink [[t]], evaluation is only possible at a = 0. We saw that σ (0) ∈ k must have amultiplicative 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 bysubstituting γ (t) for t in β (t). We write β (γ (t)) for the composition. Because thecomposition has to be in k [[t]], the term β (γ (0)) has to be in k. But only if β isa polynomial we know β (a) for a ∈ k and a 6= 0. Hence we will require that γ (t)is not of order 0, so γ can be substituted into any power series β. Underlining theoperational aspect we also write C(γ)β, where C (γ) : k [[t]] → k [[t]] is the linearoperator 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 canevaluate a formal power series). We can say that eln(1+t))−1 = t, which means thatC(ln (1 + t)) (et − 1) = t. In other words, ln (1 + t) is the compositional inverse ofet − 1. If a power series β is of order 1, and [t]β (t) is a unit in k [t], then βhas a unique compositional inverse β−1 such that β

(β−1 (t)

)= t. Such power

series we called delta series. Often it is more convenient to use another symbol,like γ, for the inverse of β. Note that the notation β−1 (t) for the (compositional)inverse is very similar to the notation of the reciprocal β (t)

−1= 1/β (t), and it

becomes indistinguishable, if the argument t is omitted. This shows why we likethe notation 1/β for the reciprocal. Of course, they usually do not exist both forthe same β ∈ k [[t]]!

Example 1.2.1. Consider the sequence 0, 1,−2, 3,−4, . . . by looking at the formalpower series β (t) =

∑n≥1 (−1)

n+1ntn (a delta series in Z [[t]]); we want to know

its compositional inverse. We haveβ (t) = −t

∑n≥0 (−1)

n+1(n+ 1) tn = −t ddt

11+t = t/ (1 + t)

2, hencet =

(1 + 2t+ t2

)β and t = 1

2β

(1−√

1− 4β)− 1. Therefore, the compositional

inverse β−1 (t) equals the generating function of the Catalan numbers (section1.1) without the constant term, β−1 (t) = 1

2t

(1−√

1− 4t)− 1.

Remember that every statement about formal power series really means astatement about an infinite sequence of coeffi cients. For a series of order 1, thesequence starts with 0, and then a coeffi cient not equal to zero follows. Suppose wehave a power series β with coeffi cients (0, a, b, c, . . . ). The compositional inverse


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

the k-th power of the inverse,[γk]n. For k = 2 we get

(0, 0, a2, 2ab, b2 + 2ac, . . .

)and for k = 3 we have(

0, 0, 0, a3, 3ab+ 3ac+ 3bc, . . .). Hence β (γ) has the coeffi cient sequence

(0, aa, ab + ba2, ac + 2bab, ad + b(b2 + 2ac

)+ ca3, . . . ), and this has to equal

(0, 1, 0, . . . ). We can now recursively find the coeffi cients of β−1 = 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 nonzeromultiplicative inverse in k, if we want to find the compositional inverse of a powerseries. This is the only coeffi cient we divide by repeatedly; the remaining termsare sums of products. Hence, a compositional inverse γ (t) will exist, if β (t) isa 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 statethe result, assuming that β (t) = t/φ (t), which means that φ (t) is invertible. Inthis way, Laurent series are not needed. For more on Lagrange-Bürmann inversionsee Exercise 1.2.2.

Theorem 1.2.2. If β (t) = t/φ (t) is a delta series (hence φ has a reciprocal), andγ (t) is the compositional inverse of β, then for all 0 ≤ k ≤ n holds

n[γk]n

= k [φn]n−k . (1.11)

Again we want to point out that the name (t) of the formal variable is of nosignificance. We can as well write (1.11) as

n [sn] γ(s)k = k[un−k

]φ (u)

n,

where γ ∈ k [[s]] and φ ∈ k [[u]]. For example, in finding the inverse of β (t) = et−1we usually proceed by letting u = et − 1, say, and solving for t = ln (1 + u). Thenwe call t = β−1 (u), and change the variable from u back to t. Note that thepresentation (1.10) is variable free!

Lagrange-Bürmann inversion is routinely applied in combinatorics. We showan example from lattice path enumeration.

Example 1.2.3. We are interested in finding the number D (n; k) of , latticepath from (0, 0) to (2n, 0), staying weakly above the x-axis, and having exactly koccurrences of the pattern , which we also write as uddu (d = andu =). The following path to (10, 0) contains the pattern uddu twice (countingoverlaps).

1.2. Composition and Inverses 15

The generating function F (x, t) =∑n,k≥0D (n; k) tnxk 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 series in t with coeffi cients that are polynomialsin 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 afunction of F ,

t =F (F − 1)

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

β (u) := t =(u+ 1)u

(u+ 1)3 − (1− x)u2

=u(

(u+ 1)3 − (1− x)u2

)/ (u+ 1)

and we need the inverse, u = γ (t). We have to check that the negative powers a−n

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 βfrom (1.11) as

n [γ]n =[un−1

]( (u+ 1)3 − (1− x)u2

u+ 1

)nRoutine calculations give

n [γ]n =

b(n−1)/2c∑i=0

(n

n− i

)(2n− 3i

n− 2i− 1

) i∑k=0

(i

k

)xk (−1)

i−k

D (n; k) =1

n

b(n−1)/2c∑i=k

(n

n− i

)(2n− 3i

n− 2i− 1

)(i

k

)(−1)

i−k

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 aninverse. For example, the power series 1 + t has the inverse t− 1. Add your ownexamples 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 compositionalinverse.


1.2.2. Write the Lagrange-Bürmann inversion formula (1.11) in terms of β in-stead of φ (t) = t/β (t). This formulation needs negative powers of β, which onlyexist in the field of Laurent series, where series of the form σ (t) =

∑n≥k σnt

n

are defined for all k ∈ Z. An important functional on Laurent series is theresidue, res(σ) =

[t−1]σ (t). Note that res (σ′ (t)) = 0 for all Laurent series σ (t).

Let ρ (t) =∑k≥0 rkt

k be any formal power series. Show the original Lagrange-Bürmann Theorem, which says that for any delta series β (t) holds

ρ (γ) = r0 +∑n≥1

1

nres(ρ′β−n

)tn

1.2.3. Show that for every power series σ (t) =∑n≥0 σnt

n holds

11−tσ

(t

1−t

)=∑n≥0 t

n∑nk=0

(n

k

)σn−k.

1.2.4. The smallest nontrivial integral domain is Z2, the integers modulo 2.

+ 0 1 ∗ 0 10 0 1 0 0 01 1 0 1 0 1Addition and multiplication in Z2

Show β (t) = β−1 (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

1/c (t) = 1− tc (t). Hence Cn =∑n−1k=0 CkCn−1−k for all n ≥ 1.

1.2.7. [10, Chapter 3.2]Let γ(t) = t+G (γ(t)), where 0 = G (0) = G′ (0). Then

γ (t)k

= tk +∑j≥1

1

j!Dj−1

(ktk−1G (t)

j)for all k ≥ 1.

1.3. Multivariate Power Series 17

1.3 Multivariate Power Series

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

S = (σn1,...,nr )(n1,...,nr)∈Nr0

can be represented as the formal power series

σ (t1, . . . , tr) =∑

(n1,...,nr)∈Nr0

σn1,...,nr tn11 · · · tnrr .

The coeffi cient functional [tn11 · · · tnrr ]σ = σn1,...,nr recovers the coeffi cients of theseries. If they are in k, we say that σ ∈ k [t1, . . . , tr]. The partial derivatives ∂

∂tiare defined on k [[t1, . . . , tr]] as

∂

∂tiσ =

∑(n1,...,nr)∈Nr0

niσn1,...,nr tn11 · · · tni−1 · · · tnrr

and we have(∂n1+···+nr

∂tn11 · · · ∂tnrrσ

)(0, . . . , 0) = n1! · · ·nr! [tn11 · · · tnrr ]σ = n1! · · ·nr!σn1,...,nr .

Similar to the univariate case, evaluation of σ is only allowed by setting some orall of the ti’s equal to 0. However, there is a new concept in multivariate powerseries: we can equate some of the formal variables, making them equal to a newformal variable s, say. For example,

[smtn33 tn44 ]σ (s, s, t3, t4) =∑

i+j=m

σi,j,n3,n4 .

A univariate formal power series φ (w) =∑n≥0 φnw

n, on the other hand, canbe made into a multivariate series by replacing the formal variable by a linearcombination of new formal variables. For example,

[smtn]φ (s+ t) =

(m+ n

m

)φm+n.

For notational simplicity, we continue the discussion of multivariate formalpower series in the bivariate case. Addition of bivariate power series is defined asexpected; multiplication needs the Cauchy product as in the univariate case,

γ (s, t)σ (s, t) =∑m,n≥0

smtnm∑i=0

n∑j=0

γi,jσm−i,n−j

when [smtn] γ (t) = γm,n. We will say that γ is of order 0 iff[s0t0

]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

γ (s, t) τ (s, t) = 1. We denote the reciprocal by 1/γ (s, t), or γ (s, t)−1.


Suppose the numbers σm,n solve for (m,n) ∈ N × N the (inhomogeneous)linear recursion

γm,n =

m∑i=0

n∑j=0

αi,jσm−i,n−j

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

σ (s, t) =γ (s, t)

α (s, t).

The above linear recursion is equivalent to

σm,n = γm,n −∑

(0,0)<(i,j)≤(m,n)

αi,jσm−i,n−j (1.12)

where we consider the partial order (i, j) ≤ (k, l) iff i ≤ j and k ≤ l on Z2. Notethat (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 thenumbers σi,j in a cone in N2; each σm,n depends only on σi,j for (0, 0) ≤ (i, j) ≤(m,n) and (i, j) 6= (m,n). For more discussions of higher dimensional recursionssee Bousquet-Mélou and Petkovšek in [16].

Example 1.3.1. The lattice walk in the first quadrant with steps ,←, ↓ cancycle 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). Theobvious 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 “kernelmethod” of generating functions can be applied to solve this problem. It has beensolved before in other ways by Kreweras [53, 1965].

The notation for power series in r > 1 variables can be a challenge; we avoidit by considering just two variables, in some examples three, but leave the generalcase to the reader.

A multi-series (ρ, σ) is a pair (or more general an r-tuple) of formal powerseries in two (or r) variables, (ρ, σ) ∈ k [[s, t]]

2. 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) areunits in k, and φ (s, t) and ψ (s, t) are in k [[s, t]] (both having order 0). Thusγ1 (0, t) = 0, and γ1 (s, 0) is a power series in s. Analogously, γ2 (s, 0) = 0 andγ2 (0, t) ∈ k [[t]].

We need the concept of the compositional inverse of a delta multi-series. Thecompositional 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 usuallydenoted by

(β−1

1 (s, t) , β−12 (s, t)

).

The partial derivatives ∂∂sφ (s, t) and ∂

∂tφ (s, t) of a bivariate power seriesφ (s, t) inherit their properties (for example, the product rule) from the uni-variate case. We remember from Calculus that the derivative of a multi-seriesγ = (γ1, . . . , γr) is defined as |J γ|, where J γ stands for the Jacobian. In thebivariate case,

|J γ| =∣∣∣∣∂(γ1, γ2)

∂ (s, t)

∣∣∣∣ =

∣∣∣∣ ∂∂sγ1 (s, t) ∂

∂sγ2 (s, t)∂∂tγ1 (s, t) ∂

∂tγ2 (s, t)

∣∣∣∣=

(∂

∂sγ1

)(∂

∂tγ2

)−(∂

∂sγ2

)(∂

∂tγ1

).

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 inverseβ1 (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[

β1 (s, t)kβ2 (s, t)

l]m,n

=[ε1 (s, t)

m+1ε2 (s, t)

n+1 |J γ|]m−k,n−l

.

For an elegant proof in Finite Operator Calculus terms see Hofbauer [43].Note that the Theorem already assumes that an inverse exists: We have writtenit in a form that forces γ1 and γ2 to be a pair of delta series. Hence |J γ| 6= 0(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[

φ1 (s, t)kφ2 (s, t)

l]m,n

=[ε1 (s, t)

m+1−kε2 (s, t)

n+1−l |J γ|]m,n

(1.13)

(Exercise 1.3.3).

Example 1.3.3. A multiseries does not need to be a delta series for having aninverse. 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 theinverse pair

β1 (s, t) = a−1s1− a−1bt

1− a−2b2stand β2 (s, t) = a−1t

1− a−1bs

1− a−2b2st

(Exercise 1.3.4). Clearly, a needs to be a unit. Only nonnegative powers of b occurin the expansion of β1 and β2. Hence b does not has to be a unit.


1.3.1 Exercises

1.3.1. Show that for any delta multi-series γ1 (s, t) and γ2 (s, t) holds |J γ| 6= 0.

1.3.2. Show that the compositional inverse(β−1

1 (s, t) , β−12 (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

a−1s(1− a−1bt

)1− a−2b2st

,a−1t

(1− a−1bs

)1− a−2b2st

is the pair inverse to (γ1, γ2). Convince yourself that b does not have to be invertiblein k by choosing k = Z.

1.3.5. Show that the Lagrange-Good formula in Theorem 1.3.2 is equivalent to[β1 (s, t)

kβ2 (s, t)

l]m,n

=

[ε1 (s, t)

mε2 (s, t)

n

∣∣∣∣ 1− γ1∂∂sε1 −γ2

∂∂sε2

−γ1∂∂tε1 1− γ2

∂∂tε2

∣∣∣∣]m−k,n−l

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 deltapair. Show that ∣∣∣∣∂ (φ, ψ)

∂ (s, t)

∣∣∣∣ =

∣∣∣∣∂ (φ, ψ)

∂ (θ, η)

∣∣∣∣ ∣∣∣∣∂ (θ, η)

∂ (s, t)

∣∣∣∣(hint: ∂φ∂s = ∂φ

∂θ∂θ∂s + ∂φ

∂η∂η∂s ).

1.3.8. The mapping β (s) 7→ (β (s) , t) is an embedding of k [[s]] into k [[s, t]]. Ifβ (s) = s/φ (s) is a delta series and γ (s) its inverse, derive the univariate inver-sion formula (1.11)

n[γk]n

= k [φn]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 OneVariable

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 fortransfer theorems, enabling us to find basic polynomial solutions to certain opera-tor equations, and for a “functional expansion theorem”, helping us to determinea 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 polynomialsin section 2.1. Even if the set of operators we study seems to be “small”comparedto 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 earlierpaper 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 mustlie in a field F; we must allow division by any coeffi cient except 0. We will alsoassume that F has characteristic 0, hence the field is infinite. We could defineour polynomials as having coeffi cients in an integral domain k, and define a basissuch that the leading coeffi cient in a basis polynomial is a unit. For example, ifour 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 vari-ables,

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 1261 1 4 1 5 15 35 70

1 2 1 3 1 4 10 20 351 3 3 1 2 1 3 6 10 15

1 4 6 4 1 1 1 2 3 4 51 5 10 10 5 1 0 1 1 1 1 1

0 1 2 3 4 nPascal’s Triangle in its usual form pn (m) written as an array

The columns on the right side of this table look like values of polynomials inm, of degree n, and because we know that pn (m) =

(n+mn

)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. ThisTheorem helps to decide when a given difference recursion (like (2.1), written aspn (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, andlet Fn (m) be recursively defined for all integers n ≥ 0 by

Fn (m) = Fn (m− 1) +

n∑i=1

aiFn−i (m+ bi)

where ak, bk ∈ F, for k = 0, 1, . . . , a1 6= 0, and integers m > xn. Let Fn (m) = 0for 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 effecton the recursion F0 (m) = F0 (m− 1) for all m > x0. Now suppose Fn (m) hasbeen extended to a polynomial of degree n for all m and for all 0 ≤ i ≤ n. ThenFn+1 (m)−Fn+1 (m− 1), the backwards difference, is a polynomial of degree n inm for all m > xn+1, as long as the right hand side of Fn+1 (m)− Fn+1 (m− 1) =∑n+1i=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-ing initial points, i.e., m + bi ≥ xn+1−i for all i = 1, . . . , n, and m > xn+1, thusbi ≥ xn+1−i − xn+1 − 1. But if the backwards difference of a function is a polyno-mial, 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 expressthis polynomial for m > xn as

fn (m) = fn (xn) +

n∑i=1

ai

m∑k=xn+1

fn−i (k + bi) (2.2)

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 21 216 1 6 12 13 135 1 5 8 8 84 1 4 5 5 53 1 3 3 3 32 1 2 2 2 21 1 1 1 1 10 1 1 1 1 1

0 1 2 3 4 nFn (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 poly-nomial extension. The above Theorem does not apply, because the initial pointsare 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 forthe initial points, which we can choose as xi = −1; the factors ai are 0 except fora1 = 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) =∑mk=0 pn−1 (k), which means

(n+m

n

)=

m∑k=0

(n− 1 + k

n− 1

)

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, andpn (1− n) =

∑n−1i=0 pi (n− 2i) for all n ≥ 1; what is the solution in this case? The

Theorem tells us that pn (m) is a polynomial of degree n, and we will see that thisinformation is extremely valuable. How to make the most of it will be the topic ofthe following sections.


m 1 4 10 20 35 56 m 1 4 12 35 107 3442 1 3 6 10 15 21 2 1 3 8 23 72 2371 1 2 3 4 5 6 1 1 2 5 25 49 1650 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 nPascal’s recursion continued to Pascal’s recursion withnegative integers pn (1− n) =

∑n−1i=0 pi (n− 2i)

We will begin the study of polynomials with a discussion of bases, and thenuse 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 linearfunctionals. 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 itis clear how F [x] is embedded in the much larger space k [[x]], the formal powerseries 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) =∑nk=0 akpk (x) with coeffi cients ak ∈ F, and (2) the elements pn of the basis are

linearly independent. The dimension of F [x] is infinite; the set of polynomialspn : n = 0, 1, . . . can serve as a basis as long as deg pn = n, and p0 6= 0. Inthe following, we will always assume that these last two properties define a basispn of F [x]. This condition requires F to be a field and not just some integraldomain. The polynomials solving Pascal’s recursion above just have coeffi cients inthe integral domain Z; embedding them in a vector space requires F = Q.

In case of an infinite dimensional space the term Hamel basis is sometimesused; 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 thattn : n = 0, 1, . . . acts like a basis: (1) Every element f ∈ k [[t]] can be written ina unique way as the sum f (x) =

∑∞n=0 akt

k with coeffi cients ak ∈ k, and (2) theelements tn of the basis are linearly independent. The problem is that the sumf (x) =

∑∞n=0 akt

k is not finite in general. This type of “basis” is often called apseudobasis. If (φn) is a sequence in k [[t]], where k is an integral domain, suchthat ord (φn) = n and [tn]φn is a unit in k, then (φn) is a pseudobasis (Exercise2.1.2).

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 (fn)n∈N or (fn) for sequences, and if the set pn is a basis, we call thesequence (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

Qp =

n∑k=0

αkQpk =

n∑k=1

αkpk−1

if p (x) =∑nk=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 into0. Thus Q has a kernel consisting of all constants (which we can identify with F),kerQ = F. We denote the set of all such operators by Ω. They are also calledGel‘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. FromQx = c, say, we see that c 6= 0 (otherwise x ∈ kerQ), and we define p1 (x) = x/c.Assume that p0, p1, . . . , pn are already defined, Qpi = pi−1 for i = 1, . . . , n. Thepolynomial Qxn+1 has degree n, thus Qxn+1 =

∑nk=0 akpk, with an 6= 0. It follows

that Qxn+1/an = pn+∑n−1k=0 akpk/an = pn+

∑n−1k=0 ak (Qpk+1) /an. Hence pn+1 =(

xn+1 −∑n−1k=0 akpk+1

)/an. By induction, (pn) is defined. The polynomials pn are

of degree n, and p0 6= 0, thus (pn) is a basis.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 conditionQrn = 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 requiringspecific initial values. The proof above shows that we can ask that they evaluateto 0 at 0 for all n ≥ 1, and are identically 1 for n = 0. Such a basis we call astandard 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.


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 Lwe have two notations: Lp and 〈L | p〉. In the special case where L stands for thecoeffi cient of xn in p (x), we also write [xn] p (x) and [p]n, as in the case of formalpower series.

Remark 2.1.4. If F ⊆ F we can think of F [x] being embedded in F [[x]]; then xmust 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 evaluatedat 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 thendeg p can be smaller than deg p. We avoid this complication by assuming that Fhas charactericstic 0. Note that only when a = 0 the evaluation can be extendedto F [[x]].

2.1.3 Exercises

2.1.1. Apply Theorem 2.1.1 to show that the recursion pn (x) = pn (x− 1) +

pn−1 (x) for all x > 1 − n with initial values pn (1− n) =∑n−1i=0 pi (n− 2i) for

n ≥ 1, and p0 (1) = 1, has a solution that can be extended to a polynomial.

2.1.2. Suppose F is an integral domain. Show that any sequence (φn) in F [[t]] isa pseudobasis of F [[t]] if ord (φn) = n and [tn]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, withinitial values at dn (n+ 1) ,and d0 (m) = 1 for all m ≥ 1. Show that the solutionto this problem can be extended to a polynomial sequence.

2.1.4. Show that the following operators are in Ω: The forward difference opera-tor ∆ : p (x) 7→ p (x+ 1) − p (x), the backwards difference operator ∇ : p (x) 7→p (x)−p (x− 1), the division operator χ : p (x) 7→ (p (x)− p (0)) /x, and the deriv-ative operator D : p (x) 7→ p′ (x). Find the standard basis (Lemma 2.1.3) for eachoperator.

2.1.5. Prove that [xn], Evala and∫ a

0are functionals on F [x].

2.2. Finite Operators 27

2.2 Finite Operators

Take any operator that is degree reducing, deg (Txn) = n − k for all n ≥ k andsome fixed k ≥ 1, and Txi = 0 for all i = 0, . . . , k − 1. Special examples arethe Gel‘fond-Leontiev operators defined in section 2.1. The powers T, T 2, T 3, . . .are defined (and are degree reducing), and therefore

(a0 + a1T

1 + · · ·+ anTn)p is

defined, where Tnp = 0 if deg p ≤ nk. Hence the infinite sum∑i≥0 aiT

i is definedon any polynomial p ∈ F [x], if ai ∈ F for all i ≥ 0. Linear operators of this form,∑i≥0 aiT

i where T is degree reducing , are called finite operators, because on

any polynomial of degree n they act like the finite sum∑bn/kci=0 aiT

i. The integraldomain of all such finite operators for fixed T is denoted by ΣT . Any operator Qin ΣT is given by T and the sequence of coeffi cients (a0, a1, . . . ). We saw that theformal power series are a way of “storing”such a sequence of coeffi cients. Hence wehave a bijection between

∑i≥0 aiT

i and∑i≥0 ait

i. The additive structure on ΣTand F [[t]] are preserved under this bijection. Is there more structure preserved?To answer this we have to look at the composition of two linear operators R andQ,

(RQ) p := R (Qp)

If R and Q are both in ΣT , Q =∑i≥0 aiT

i and R =∑j≥0 bjT

j , then

R (Qp) = R∑i≥0

ai(T ip)

=∑j≥0

bjTj∑i≥0

ai(T ip)

=∑j≥0

bj∑i≥0

aiTjT ip

=∑j≥0

bj∑i≥0

aiTi+jp =

∑j≥0

bj∑i≥j

ai−jTip =

∑i≥0

i∑j=0

bjai−j

T ip

Therefore, the coeffi cients in the composition RQ are the same as in the Cauchyproduct in the corresponding power series. The integral domains ΣT and F [[t]] areisomorphic. We can write Q = α (T ) =

∑i≥0 aiT

i and R = β (T ) =∑j≥0 bjT

j ,and we have α (T ) '

∑i≥0 ait

i =: α (t), β (T ) '∑j≥0 bjt

j =: β (t). Fromα(t)β (t) ' RQ and α (t)β (t) = β (t)α (t) follows RQ = QR; any two opera-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 isdue to an unfortunate choice of the word “composition”for what should be calledthe product of operators. This gets even more confusing later, when we need thecomposition 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 aformal 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.


Example 2.2.1. Suppose the basis (pn) follows Pascal’s recursion pn (x)−pn (x− 1) =pn−1 (x). The operator ∇ : f (x) 7→ f (x)−f (x− 1) is called the backwards differ-ence operator. Taylor’s Theorem tells us that f (x− 1) =

∑n≥0 (−1)

nDnf (x) /n!,where D = d/dx. Hence ∇ = I −

∑n≥0 (−1)

nDn/n! = I − e−D. We find ∇ inΣD. The backwards differeence operator is also degree reducing.

If we choose the coeffi cients (a0, a1, . . . ) from some integral domain F ⊇ F,then the finite operators

∑i≥0 aiT

i may map F [x] to some different space. How-ever, if S is a degree reducing operator on F [x] and F = ΣS , then

∑i≥0AiT

i,with Ai ∈ ΣS , maps F [x] into itself. Such operator coeffi cients will not give us anynew operators, but they simplify the description of operators. For example, we cansee the operator ∇D as an element of ΣD with coeffi cients (0, 0, 1,−1/2!, 1/3!, . . . )in F, ΣD ' F [[t]] as before. But we can also see ∇D as an element of ΣD withcoeffi cients (0,∇, 0, . . . ) in Σ∇, ΣD ' Σ∇ [[t]], or as an element of Σ∇ with coeffi -cients (0,D, 0, . . . ) in ΣD, Σ∇ ' ΣD [[t]]. We will do this in section 2.4 on transfertheorems, but it will be clearly stated. In general, the coeffi cients in ΣS will comefrom 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) 7→ p (x+ a). The translation operators aredefined for all a ∈ F, and they form a group, EaEb = Ea+b. They are degreepreserving. Because x is a formal variable, and a ∈ F, we should say what wemean by the polynomial (x+ a)

n. Of course, (x+ a)n

:=∑nk=0

(nk

)akxn−k. Note

that on the basis (xn) we have

Eaxn = (x+ a)n

=

n∑k=0

(n

k

)akxn−k =

n∑k=0

akDkk!xn (2.3)

where Dk = dk/dxk is the k-th power of the derivative operator. Hence

Ea =∑k≥0

akDkk!

= eaD

This shows that Ea ∈ ΣD, and it follows that Ea commutes with every otheroperator in ΣD. We call a linear operator T on F [x] translation invariant, ifTEap (x) = EaTp (x) for all a ∈ F and p ∈ F [x]. The operators in ΣD are exactlythe translation invariant operators, as we show in the following Lemma (the “FirstExpansion Theorem”in [83, p. 691]).

Lemma 2.2.2. A linear operator T on F [x] is translation invariant iff T ∈ ΣD. Inthat case, T =

∑i≥0

⟨Eval0 | Txi

⟩Di/i!


Proof. All we have left to show is that every operator T is in ΣD if it satisfiesTEaxn = EaTxn for all n ≥ 0 and a ∈ F. Let Txn =

∑i≥0 cn,ix

i, where for givenn the ring elements cn,i are eventually 0 for large enough i. Thus

EaTxn = Ea∑i≥0

cn,ixi =

∑i≥0

cn,i

i∑k=0

(i

k

)ai−kxk =

∑k≥0

xk∑i≥0

(i+ k

k

)aicn,i+k

and

TEaxn = T

n∑i=0

(n

i

)an−ixi =

n∑i=0

(n

i

)an−i

∑k≥0

ci,kxk =

∑k≥0

xkn∑i=0

(n

i

)aicn−i,k

Both expressions are equal iff the coeffi cients of xk agree,

∑i≥0

(i+ k

k

)aicn,i+k =

n∑i=0

(n

i

)aicn−i,k

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

Txn =

n∑i=0

cn,ixi =

n∑i=0

ci,0i!

n!

(n− i)!xn−i =

∑i≥0

ci,0i!Dixn.

and therefore T ∈ ΣD.

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 oper-ator T on F [x] there exists a sequence of polynomials (an (x)) such that T =∑n≥0M (an)Rn 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 = D thetranslation invariant operators are obtained by the choice of constant polynomialscn (x) = 〈Eval0 | Txn〉 /n!. This and most of the other results in this section werealready known at the end of the 19th century (see Pincherle [74]).


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) =

n∑i=0

bi (y) bn−i (x)

for all n ≥ 0. Examples are bn (x) =(xn

)(Vandermonde convolution) and, if Q ⊂ F,

bn (x) = xn/n! (giving the original binomial theorem). Standard sequences thatare of binomial type are called basic sequences; not a great name, but from hereon, 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 generatingfunction b (x, t) =

∑n≥0 bn (x) tn 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

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.

Because β (t) is a delta series, the compositional inverse β−1 (t) of β (t) existsand is also a delta series (section 1.2). If γ (t) is any delta series, we call the operatorγ (D) a delta operator, hence β−1 (D) and β (D) are both delta operators. We willstudy the delta operator β−1 (D) and and see what it does to (bn). The meaningof β (D) has to wait until section 2.3.2.

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 inF [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 elementfrom F [x] [[t]] can be understood as an infinite matrix, whose rows stand for thecoeffi cients of t, the n-th row containing the coeffi cients of a polynomial. Thematrix 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-ing A

∑i≥0 pi (x) ti :=

∑i≥0 (Api (x)) ti. Such a t-linear extension is called an

x-operator on F [x] [[t]]. For example, ext ∈ Q [x] [[t]], and

Dnext = tnext. (2.4)

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 section1.2), and by x-linearity C(α)f (x, t) = f (x, α (t)) for all f (x, t) ∈ F [x] [[t]]. Notethat x-operators commute with t-operators!

Now we are ready to see what the x-operator β−1 (D) does to b (x, t) = exβ(t),


β−1 (D) b (x, t) = β−1 (D) exβ(t) = β−1 (D)C(β)ext = C(β)β−1 (D) ext

because x- and t-operators commute. Suppose β−1 (D) =∑n≥1 γnDn. Then

β−1 (D) ext =∑n≥1

γntnext = β−1 (t) ext,

as in (2.4), showing that the action of the x-operator β−1 (D) on ext is the sameas multiplication by the formal power series β−1 (t). We say that β−1 (D) andM(β−1

)are transforms of each other, with respect to ext. Hence

β−1 (D) b (x, t) = C(β)M(β−1

)ext = β−1 (β (t)) exβ(t)

= tb (x, t) =∑n≥0

bn−1 (x) tn

We have shown that the operator β−1 (D) ∈ ΣD is exactly the operator B : bn 7→bn−1. This is the content of the following theorem.

Theorem 2.2.4. The basic sequence (bn) has generating function∑n≥0 bn (x) tn =

exβ(t), iff β−1 (D) : bn 7→ bn−1 for all n ≥ 1, and β−1 (D) b0 = 0.

Example 2.2.5. We list some frequently occurring basic sequences.

1. If bn (x) = xn/n!, then∑n≥0 x

ntn/n! = ext hence β (t) = t = β−1 (t), andβ−1 (D)xn/n! = Dxn/n! = xn−1/ (n− 1)!.

2. If bn(x) =(n−1+xn

)then

∑n≥0

(n−1+xn

)tn = (1− t)−x hence β (t) = 1/ ln (1− t)

and β−1 (t) = 1−e−t. We obtain the delta operator ∇ = 1−E−1 :(n−1+xn

)7→(

n−2+xn−1

).

3. If bn(x) =(xn

)then

∑n≥0

(xn

)tn = (1 + t)

x and β−1 (D) = E − 1, mapping(xn

)into

(x

n−1

). This delta operator is called ∆. Thus ∆ = E∇.

4. Suppose we interchange β and β−1 in the previous example, then β−1 (D) =

ln (1 +D) =∑n≥1 (−1)

n+1Dn/n. How can we find the basic sequence forthis delta operator? One way will be shown in Example 2.3.11, another insection 2.4.

Remark 2.2.6. (a) Not every sequence of binomial type is a standard sequence. Forexample, let b2n (x) = xn/n! and b2n+1 (x) = 0 for all n ≥ 0. The sequence (bn) isof 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], withF [x] and F [y] embedded in F [x, y]. A more fruitful interpretation is given in in


terms of Hopf algebras (see [61]). Closer to our approach is choosing y ∈ F, andreading the binomial theorem as

Eybn (x) =

n∑i=0

〈Evaly | bi〉 bn−i (x) =

n∑i=0

〈Evaly | bi〉Bibn (x) (2.5)

where B : bn 7→ bn−1. The expansion Ey =∑i≥0 〈Evaly | bi〉Bi is just a special

case of Exercise 2.2.1, hence gives an independent proof of the binomial theoremfor basic sequences.

(c) If B = β−1 (D) is a delta operator, then D =β (B), hence every operator inΣD is also expandable in B, so ΣD ' ΣB for all B ∈ ΣD. We saw already oneexample in (2.5), Ey =

∑∞i=0 〈Evaly | bi〉Bi, thus Ey ∈ ΣB. Every A ∈ ΣB has its

isomorphic image A ∈ ΣD such that A = A as operators on F [x], but A = α (B),say, and A = α (D), hence α = α (β (t)). Even when A and A are the sameoperator on F [x], the power series representation of A and A are different, whenB and D are different. Of course, all operators in ΣB commute with all operatorsin ΣD.

2.2.3 Special Cases

We present three special classes of basic sequences; the first two arise in probabilitytheory. The relationship of Finite Operator Calculus to probability theory hasbeen studied by DiBucchianico [24, Theorem 3.5.10], but also later by [29] in theconnection with Umbral Calculus. Finally we consider basic sequences (bn) havingcoeffi cients that are themselves values of basic sequences. While this looks like arather 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(an) with generating function exα(t) is related to the random variable X throughE [Xn] = n!an (1), hence for the moment generating function of X holds

mX (t) := E[eXt]

= eα(t).

Of course, this requires a distribution where all the moments E [Xn] exist andare in F, and because a1 (1) 6= 0 for any basic sequence (a1 (0) = 0), it alsorequires µ = E [X] 6= 0 (but see Example 2.2.9). For the basic sequence (an) thismeans that E

[X2n

]= (2n)!a2n (1) ≥ 0; especially σ2 = V ar (X) > 0 implies

that a2 (1) > a1 (1)2/2 > 0. Hence not all basic sequences generate probability

moments, and not all moment generating functions generate basic sequences!We define the cumulant generating function

KX (t) := lnmX (t) = α (t) = µt+ σ2t2 + . . .


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

K ′′X (0) = V ar [X] = 2a2(1)− a1(1)2

= 2α2

if α (t) =∑n≥1 αnt

n. Note that KX is either a quadratic polynomial (normaldistribution) or a true power series, i.e., no polynomial at all (Lucas [55]). Thenumbers κn = n!αn are called the cumulants of X. If an (x) =

∑nk=0 an,kx

k/k!,then the relationship between the cumulants and the moments of X can be de-scribed by Lemma 2.2.10,

E [Xn] = n!an(1) = n!

n∑k=0

an,k/k! = n!

n∑k=0

[αk]n/k!

=

n∑k=1

∑(B1,...,Bk)∈S(n,k)

k∏i=1

|Bi|!α|Bi| =∑

π∈S(n)

∏B∈π

κ|B|

where S (n, k) is the set of all partitions of n into k parts (see Remark 1.1.2), andS (n) is the set off all partitions of n. Because of the two combinatorial methodsof representing

[αk]n, as explained in Remark 1.1.2, there is a second formula for

expressing moments through cumulants,

E [Xn] =

n∑k=0

∑λ`n,|λ|=k

n!

`1! · · · `n!

k∏i=1

αλi .

We can say that K−1X (D) is the delta operator for (an). For example, if X

has the binomial distribution Pr (X = k) =(Nk

)pk (1− p)N−k with parameters N

and p, then

mX (t) =∑n≥0

tn

n!

N∑k=0

kn(Nk

)pk (1− p)N−k =

(1 +

(et − 1

)p)N

and

an (x) = [tn]mX (t)x

=∑k≥0

(Nxk

)pk [tn]

(et − 1

)k=

n∑k=0

(Nxk

)pkk!

n!S (n, k) .

Hence we find A = K−1X (D) = N ln (1 + p∆).

When x is a nonnegative integer k, say, thenmX (t)k

= mX1+···+Xk (t), whereX1, . . . , Xk are independent random variables with the same distribution as X,thus ekα(t) = mX1+···+Xk (t), and

an (k) = E [(X1 + · · ·+Xk)n] /n!


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

an (x) =

∫ ∞−∞

yn

n!dµx (y)

for x ≥ 0 iff the above equation holds for n = 1. Note that this definition excludesthe semigroups for which the first moment of µ1 is zero, because a1 (1) 6= 0 forany basic sequence, but here are cases when this can be overcome. If we choosefor µx the point mass at x, called δx, then

∫∞−∞ (yn/n!) dδx (y) = xn/n!, the basic

polynomial for D.

Example 2.2.7. The Poisson semigroup µx = e−x∑k≥0

(xk/k!

)δk leads to the

basic polynomials

φn (x) =

∫ ∞−∞

(yn/n!) dµx (y) =e−x

n!

∑k≥0

xkkn

k!,

known as the exponential polynomials (see Dobinski’s formula (2.26)).

Example 2.2.8. The basic polynomials for the Gamma semigroup

dµx (y) = e−yyx−1

Γ (x)dy

are

gn (x) =

∫ ∞0

yn

n!

yx−1

Γ (x)e−ydy =

(n−1+xn

),

associated to the backwards difference operator ∇.

Example 2.2.9. The normal distribution with expectation 0 is a semigroup withmoment generating function eσ

2t2/2. Of course, we cannot find a basic sequenceagreeing with the moments of dµx (y) = e−y

2/(2x)/√

2xπ for x > 0, because theexpectation is 0, and so are all odd moments. However, we can define

pn (1) =1√2π

∫ ∞−∞

y2n

(2n)!e−

y2

2 dy =1

2nn!,

skipping the odd moments. Thus

pn (x) =xn

2nn!.

The delta operator is 2D.


Distributions of Binomial Type Let F = R, and let X be a random variable on0, . . . , n with probability distribution function (p.d.f.) p (i) = Pr (X = i). Wesay that the p.d.f. is of binomial type iff

0 ≤ p (i) =bi (α) bn−i (β)

bn (α+ β)≤ 1 (2.6)

for some basic sequence (bn) and α, β ∈ R such that bn (α+ β) 6= 0. It can beshown that all symmetric distributions, p (i) = p (n− i), are of binomial type ifp0 > 0.

The expectation of any random variable X having a distribution of binomialtype equals

E [X] =αn

α+ β

(Exercise 2.2.13). The second moment equals for all n ≥ 2

V ar [X] =αβn2

(α+ β)2 − αβ

S2bn−2 (α+ β)

bn (α+ β), (2.7)

where Sbn (x) = (n+ 1) bn+1 (x) /x, a translation invariant operator (Exercise2.2.14 and Lemma 2.3.4). The variance is a symmetric function in α and β. Forexample, if bn (x) = xn/n!, then S = I, and V ar [X] = αβn2

(α+β)2− αβ n(n−1)

(α+β)2=

αβn/ (α+ β)2.

It is easy to verify that the binomial and the hypergeometric distribution areof binomial type; however, it is interesting that both distributions are of binomialtype because of a general construction based on conditioning (see Exercises 2.2.9and 2.2.10).

A subclass of the distributions of binomial type is obtained by conditioningas follows. Let β (t) be the delta series such that

∑n≥0 bn (x) tm = exβ(t). Suppose

that for some parameter space Θ the family of random variables Xθ, θ ∈ Θ, takesvalues in N0 such that for all i ∈ N0

Pr (Xθ = i) = τ iσθbi (θ) (2.8)

for some τ ∈ R where β (τ) converges, and σ = e−β(τ). Furthermore, assumethat the process Xθ is stationary with independent increments at α and β, whereα, 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 binomialtype on 0, . . . , n,

p (i) = Pr (Xα = i |Xα+β = n) =Pr (Xα = i) Pr (Xβ = n− i)

Pr (Xα+β = n)

=τ iσαbi (α) τn−iσβbn−i (β)

τnσα+βbn (a+ β).


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 isknown everywhere. Let (bn (x)) be a basic sequence,

∑n≥0 bn (x) tn = exβ(t). We

denote the coeffi cients of xk/k! in bn (x) by bn,k =[xk/k!

]bn (x),

bn (x) =

n∑k=0

bn,kxk

k!

Lemma 2.2.10. For all 0 ≤ k ≤ n holds

bn,k =[βk]n.

Proof. From

exβ(t) =∑n≥0

bn (x) tn =∑n≥0

tnn∑k=0

bn,kxk

k!=∑k≥0

xk

k!

∑n≥k

bn,ktn

and

exβ(t) =∑k≥0

xk

k!β (t)

k

follows bn,k =[βk]nby comparing coeffi cients of xn/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).

Theorem 2.2.11. Let (bn (x)) be a basic sequence such that bn,n = 1 for all n ∈ Nr0,and

∑n≥0 bn (x) tn = exβ(t). The following statements are equivalent:

1. (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 6= 0.


)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

the same sequence, and it has the generating function exβ(t). Note that b2,1 = β2.


Proof. (1) ⇒ (2): Let bn,k = cn−k (k) where deg cn = n. The generating functionof (cn) equals∑n≥0

cn (k) tn =∑n≥0

bn+k,ktn =

∑n≥0

[βk]n+k

tn = t−k∑n≥k

[βk]ntn = (β (t) /t)

k.

From[t0]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) 6= 0. (3) ⇒ (4): Let b2,1 6= 0. The num-

bers bn (1), n ≥ 0, define the basic sequence(bn (x)

), because

∑n≥0 bn (x) tn =(∑

n≥0 bn (1) tn)x. The only conditions on bn (1) are b0 (1) = 1 and b1 (1) 6= 0.

Both conditions are satisfied if b2,1 6= 0 and bn,1 = bn−1 (1). (4) ⇒ (5): The

generating function of(bn

)in (4) equals

exβ(t) =∑n≥0

bn (x) tn =

∑n≥0

bn (1) tn

x

hence

teβ(t) = t∑n≥0

bn (1) tn =∑n≥0

bn+1,1tn+1 =

∑n≥0

[β]n+1 tn+1 = β (t) .

(5) ⇒ (1): The series β (t) must be a delta series, because log (β (t) /t) exists andis a delta series. From

(β (t) /t)k

=∑n≥0

bn+k,ktn = ekβ(t)

follows that for the basic sequence(bn

)with generating function exβ(t) holds

bn+k,k = bn (k).The condition bn,n = 1 is equivalent to β (t) = t + . . . . Suppose [β (t)]1 =

λ 6= 0. Let an (x) := bn (x/λ). Theorem 2.2.11 applies to (an (x)), which has thecoeffi cients an,k = [β (t)

nλ−n]k = λ−nbn,k. The following statements are equiva-

lent:

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


)such that bn,i = λibn−i (i) for all n ≥ 0

and for all 0 ≤ i ≤ n.3. b2,1 6= 0.


)such that bn,1 = λbn−1 (1) for all

n > 0.


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

From β (t) = λteβ(t) follows

β (t)

λt= 1 +

b2,1λt+ · · · = eβ(t) = 1 + β (t) + · · · = 1 +

[β]

1t+ . . .

which shows that the coeffi cient bn,n := [xn/n!] bn (x) equals

bn,n =[β (t)

n]n

=

(b2,1λ

)n=bn2,1bn1,1

. (2.10)

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 occursin 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 numberof sums over integer intervals, then it gets diffi cult if we are not able to explicitlydetermine φ (t)in β (t) = t/φ (t). For example, let β (t) = eat − ebt ∈ R [t], a 6= b,both different from 0. Of course we have φ (t) = t/

(eat − ebt

), but how do we find

an explicit expression for kn

[tn−k

]φ (t)

n, the coeffi cient of tn in the k-th power ofβ−1 (t)? In this situation it may be worthwhile checking the coeffi cient β2 of thequadratic term of β (t) = β1t + β2t

2 + . . . . Because in that case,[tn−k

]φ (t)

n=

β−n1 bn−k (−n) for all integers n ≥ k, where(bn

)is the sequence of “coeffi cient

polynomials”with generating function exβ(t), and β (t) = β1teβ(t). If we know the

basic sequence(bn

)explicitly, from their generating function (β (t) / (β1t))

x, then

we are done, [γk]n

=k

n[φn]n−k =

k

nβ−n1 bn−k (−n) (2.11)

for all n ≥ k. Otherwise, the following lemma may help.Lemma 2.2.12. [67]If β (t) = β1t+ β2t

2 + . . . with β1 6= 0 and β2 6= 0, and (bn) isthe basic sequence with generating function exβ(t), then the positive powers of thecompositional inverse γ (t) of β (t) can be expanded in positive powers of β (t) as

γ (t)k

= k∑n≥0

(2n+ k

n

)tn+k

n∑j=0

(n

j

)(−1)

j

n+ k + jβ−n−k−j1 bn+j,j

where bn+j,j =[tj+n

]β (t)

j are also the coeffi cients of xj/j! in bn+j (x).

Proof. The Lagrange interpolation formula (E. Waring [99], 1779) applied tobn−k (x) says that

bn−k (x) =

n−k∑j=0

bn−k (xj)

n−k∏j 6=i=0

x− xixj − xi

.


Suppose we set xi = i and x = −n, then

bn−k (−n) =

n−k∑j=0

bn−k (j)

n−k∏j 6=i=0

−n− ij − i

= n

n−k∑j=0

(n− kj

)(2n− kn

)(−1)

j

n+ jβ−j1 bn−k+j,j . (2.12)

Certainly, the above Lemma should only be applied if “ordinary”Lagrangeinversion gets too complicated! Such a case is discussed in Example 2.2.13. If theLagrange inversion formula (1.11) can be applied, Lemma 2.2.12 may lead to someidentity (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 definedfor k ≥ j ≥ 0 as the coeffi cients of j!tk/k! in

(eat − ebt

)j,

∑k≥j

j!

k!F (k, j; a, b) tk =

(eat − ebt

)j=

j∑i=0

(j

i

)(−1)

j−ie(ai+(j−i)b)t

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,

F (k, j; a, b) =

j∑i=0

(−1)j−i (ia+ (j − i) b)k

i! (j − i)! .

The best known example of factorial numbers of the second kind are the Stirlingnumbers S (k, j) of the second kind, S (k, j) = F (k, j; 1, 0). The factorial numbersof the first kind, f (n, k; a, b), occur in the inverse γ (t) of eat − ebt =: β (t),

γ(t)k =∑n≥k

k!

n!f (n, k; a, b) tn.

We want to find an explicit expression for f (n, k; a, b), so we try Lemma 2.2.12in the case when a2 6= b2:

γ(t)k = k∑n≥0

(2n+ k

n

)tn+k

n∑j=0

(n

j

)(−1)

j−1

n+ k + j

(a− b)−n−k−j j!(j + n)!

F (j + n, j; a, b) ,

hence

k!

n!f (n, k; a, b) =

(2n− kn− k

) n−k∑j=0

(n− kj

)k (−1)

j

n+ j

(a− b)−j−n j!(j + n− k)!

F (j + n− k, j; a, b) .


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

k

n−k∑j=0

(2n− k

n− k − j, n− k + j, k

)(a− b)−j−n

n+ j

j∑i=0

(−1)i (ia+ (j − i) b)j+n−k

i! (j − i)!

for a2 6= b2. If a = 1 and b = 0 we obtain an explicit expression for the Stirlingnumbers of the first kind,

s (n, k) =

n−k∑j=0

(2n− k

n− k − j, n− k + j, k

)k

n+ j

j∑i=0

(−1)i ij+n−k

i! (j − i)! .

2.2.4 Exercises

2.2.1. Show that for every translation invariant operator T holdsT =

∑i≥0 〈Eval0 | Tbi〉Bi, where B is any delta operator with basic sequence (bn)

(a generalization of Lemma 2.2.2).

2.2.2. Suppose that (rn) is the standard basis for R ∈ Ω, R : rn 7→ rn−1. Let Tbe an arbitrary linear operator on F [x] such that Trn (x) = an (x) for all n ≥ 0(an (x) may have any degree!). Show that

T =

∞∑n=0

n∑k=0

M (an−k)∑

c∈C(k)

(−1)l(c)M(rc1rc2 · · · rcl(c)

)Rn.The set C (k) stands for all compositions (ordered partitions) of k,

C (k) =

k⋃l=1

(c1, . . . , cl) : ci ≥ 1,

l∑i=1

ci = k

and l (c) is the length of c ∈ C (k). Note that

∑c∈C(0) · · · = 1. Find∑

c∈C(k)

(−1)l(c)M(rc1rc2 · · · rcl(c)

)when R = D.2.2.3. Show that the division operator χ can be written in the form

χ =∑k≥1

ciM(xk−1

)Dk.

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

∆np (x) =

n∑k=0

(n

k

)(−1)

n−kp (x+ k) .


This formula holds for a much larger class of functions than polynomials; we cantake, for example, ∆n

(z−1)for complex z, and find

∆n(z−1)

=

n∑k=0

(n

k

)(−1)

n−k

z + k,

as long as z 6= −k for k = 0, . . . , n. Show by induction for such z that

∆n(z−1)

= n!(−1)

n

z (z + 1) · · · (z + n).

You have proven the identityn∑k=0

(n

k

)(−1)

k z

z + k=

1(z+nn

)2.2.5. Suppose B = β−1 (D) is a delta operator, and (bn(x)) a basic sequence forB. Show: If a 6= 0, then the scaled polynomials bn(ax) are the basic polynomialsfor the delta operator β−1 (D/a)

2.2.6. Suppose the radius r of convergence of β (t) is larger than τ . Find the mo-ment generating function of the distribution (2.8). Show that E [Xθ] = pβ′ (p) andV ar [Xθ] = pβ′ (p) + p2β′′ (p).

2.2.7. Let Y, Y1, Y2 . . . be i.i.d. random variables with finite moments and nonzeroexpectation. Show that there exists a unique basic sequence (an) such that an (m) :=E [(Y1 + · · ·+ Ym)

n] /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, theprobability of k failures before the first success in independent trials with successprobability 0 < p < 1. Show that

bn (x) =

n∑k=0

(x+k−1k

)(1− p)k p−k k!

n!S (n, k)

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 generatingfunction).

2.2.9. Suppose Xθ has the Poisson distribution Pr (Xθ = i) = e−θθi/i! for alli ∈ N0 and θ ∈ R+. Show by conditioning that the binomial distribution is ofbinomial type.

2.2.10. Take a sample of size n from a population of size α+ β, α, β ∈ N0, whichcontains α marked and β unmarked subjects. Show by conditioning on the binomialdistribution that the hypergeometric distribution

f (i) := Pr (i marked objects in the sample)

is of binomial type.


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

p (i) =bi (1/2) bn−i(1/2)

bn (1).

2.2.12. Find the basic sequence in (2.6) when p0 = pi = 1/ (n+ 1) for all i ∈0, . . . , n (discrete uniform distribution). Show that V ar [X] = 1

12n (n+ 2) byapplying (2.7).

2.2.13. Show that αnα+β is the expected value

∑ni=0 ibi (α) bn−i (β) /bn (α+ β) of a

random variable that has a distribution of binomial type (2.6).

2.2.14. Let X be a random variable having a distribution of binomial type (seeExercise 2.2.13). Show that the second moment µ2 = E

[X2]equals

µ2 =αn2

α+ β− αβS

2bn−2 (α+ β)

bn (α+ β)

where Sbn (x) = (n+ 1) bn+1 (x) /x. As an example, calculate the second momentof the hypergeometric distribution p (i) =

(αi

)(βn−i)/(α+βn

).

2.2.15. Let c be a nonnegative real number, α and β positive real numbers. Showthat

p (i) =

(n

i

)αβ

(α+ β)

(ci+ α)i−1(c (n− i) + β)n−1−i

(cn+ α+ β)n−1

for i = 0, . . . , n is a distribution of binomial type, generalizing the binomial distri-bution (c = 0). For n ≥ 2 show that the variance equals

αβn2

(α+ β)2 − αβn!

n∑k=2

(k − 1) ck−2

(cn+ α+ β)k

(n− k)!.

2.2.16. Find the moment generating function of the distribution (2.8). Show thatE [Xθ] = pβ′ (p) and V ar [Xθ] = pβ′ (p) + p2β′′ (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 6= 0 and β (t) =

β1teβ(t) holds bk (n) = β−n1

[tk]φ (t)

−n for all integers n (β1 = [β]1).

2.2.19. The Lagrange interpolation formula, even in its simple form (2.12), canbe the reason for identities that are not obvious. Show that(

x

n

)=

(x+ n

n

) n∑j=0

(n

j

)(−1)

n−jx

x+ j

(n+ j − 1

n

)for all x ≥ 0 by applying (2.12) to β (t) = t/ (1− t).


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

∞∑n=0

(−1)n (n+ 1)

n−1

n!

(ett)n

= e−t.

Calculate the same inverse from Lemma 2.2.12. Comparing the answers showsthat

xn−1(x+nn

) =

n∑k=0

(n

k

)(−1)

k−n

x+ kkn

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 oftn/n!in (sin t)

m equals 0 if m+ n is odd, and equals

(−1)(m+n)/2

2−mm∑k=0

(m

k

)(−1)

k(2k −m)

n

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∑l=0

2l∑j=0

(2n− 1)! (−1)n+l−1−j

F (2l + j, j; 1,−1)

2j (2l − j)! (n− 1− 2l)! (2l + j)!

1

2 (n− l)− 1 + j


2.3 Sheffer Sequences

We have three different objects defining each other: A basic sequence (bn), a deltaseries β (t) = ln

∑n≥0 bn (1) tn, and a delta operator β−1 (D) : bn (x) 7→ bn−1 (x).

How can we enlarge the set of sequences under consideration, if they are alreadyisomorphic to the power series of order 1? We choose to investigate pairs of powerseries 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≥0

sn (x) tn = σ (t) exβ(t), (2.13)

where β (t) is a delta series, and σ (t) has a reciprocal in F [[t]].

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

β−1 (D) sn = sn−1.

On the other hand, if (sn) is a basis for F [x] such that β−1 (D) sn = sn−1 holdsfor all n ≥ 1, then β−1 (D) is the transform of multiplication by t with respect tos (x, t) :=

∑n≥0 sn (x) tn. Hence Ds (x, t) = β (t) s (x, t); solving this differential

equation gives s (x, t) = σ (t) exβ(t), where σ (t) =∑n≥0 sn (0) tn.

Thus (bn) and (sn) follow the same recursion, but differ in the initial values;sn (0) = [tn]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, thensn = bn for all n ≥ 0. We say that (sn) is a Sheffer sequence for B = β−1 (D),and it is associated to (bn). The Sheffer sequence is uniquely defined by the pairof 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 weinterpret tn (x) = 0 for n < 0. However, for k = 0 we get a Sheffer sequence onlyif deg (sn + tn) = n.

From the generating function (2.13) follows directly the binomial theorem(for Sheffer sequences),

sn (x+ y) =

n∑i=0

si (y) bn−i (x) (2.14)

In the future, when we refer to the binomial theorem we will always mean thisversion for Sheffer sequences, which includes the “ordinary”binomial theorem.

2.3. Sheffer Sequences 45

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

sn (x) =∑i≥0

〈Eval0 | si〉Bibn (x) =∑i≥0

[σ]iBibn (x) = σ (B) bn (x) .

The operator S := σ (B) is called Sheffer operator for (sn), Sbn = sn. The Shefferoperator S is invertible, because σ (t) has a reciprocal 1/σ (t), and S commuteswith all operators in ΣD.

We need an initial value yn = sn (xn) for every sn (x) to get a unique solutionto the operator equation Bsn = sn−1. The initial points x0, x1, . . . can be anyelements in F; as soon as the Sheffer sequences (sn) and (tn) for B agree at onlyone 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 onlycontains constants!).

Example 2.3.2. We want to show the identity

n∑k=0

(a+ kz

k

)(x− kzn− k

)=

n∑k=0

(a+ x− kn− k

)zk (2.16)

for all a, x, z ∈ C (identity 3.144 in Gould’s list [38]). We apply ∆ to sn (x) =∑nk=0

(a+kzk

)(x−kzn−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 Sheffersequence for ∆. We want to show that sn (x) = tn (x) for all n ≥ 0. Take the initialpoints xn = n−a. Use identity (2.29) to show that sn (xn) =

(zn+1 − 1

)/ (z − 1).

Of course, tn (xn) gives the same value, hence sn (xn) = tn (xn), and thereforesn (x) = tn (x) for all x, by property (2.15).

Example 2.3.3. (Cartier [19]) This example from signal processing shows how theterms 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 timevariable, instead of x. A linear and stationary transmission device can be modeledas a linear operator V on R [t], mapping a suitable input f (t) into an outputF (t) such that V f (t+ τ) = F (t+ τ) (stationarity = translation invariance!). Theimpulse 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

F (t) = V f (t) =

∫ ∞−∞

f (t− τ) I (τ) dτ (2.17)

We assume that∫∞−∞ τnI (τ) dτ exists for all n ≥ 0, and

∫∞−∞ I (τ) dτ 6= 0. Trans-

lating this setting into the language of the Finite Operator Calculus we see that


the operator V is translation invariant, V ∈ ΣD. Equation (2.17) shows that Vacts as a convolution integral; the impulse response I (t) is defined such that

V =∑i≥0

[Tti]x=0Di/i! =

∑i≥0

(−1)i

(∫ ∞−∞

τ iI (τ) dτ

)Di/i!

exists (as a formal power series; see Lemma 2.2.2), and we also assume that V isinvertible (V 1 6= 0). Hence V is a Sheffer operator, and the polynomials vn (t) :=V tn/n! must be Sheffer polynomials for Dt = d

dt , such that Dtvn (t) = vn−1 (t) forall n ≥ 0. They have the generating function∑

n≥0

vn (t) pn = Θ (p) etp

(in signal processing we write p instead of t), where Θ (p) is the spectral gain,

Θ (p) =∑n≥0

vn (0) pn =∑n≥0

[Ttn]x=0

pn

n!

=∑n≥0

(−1)n∫ ∞−∞

τnI (τ) dτpn

n!=

∫ ∞−∞

e−τpI (τ) dτ = e−tpV etp.

Note that the above identity says that Θ (p) etp = V etp, hence Θ (p) and V aretransforms of each other (see section 2.2.2), and therefore V = Θ (Dt).Now assume that the response F (t) = V f (t) has the Taylor expansion

F (t) =∑n≥0

(t− t0)n

(Dnt F ) (t0) /n!

around t0. We want to show that this implies the solution of the inverse problem,recovering the input as

f (t) =∑n≥0

un (t− t0) (Dnt F ) (t0)

where∑n≥0 un (t) pn = etp/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 t0is the same as saying that F (t, p) :=

∑n≥0 (t− p)n (Dnt F ) (p0) /n! can be evalu-

ated at p = t0 (think of (Dnt F ) (p0) as a given sequence of constants). We definethe Sheffer polynomials un (t) = 1

Θ(Dt) tn/n! for Dt, having generating function∑

n≥0 un (t) pn = 1Θ(Dt)e

tp = V −1etp. Hence

f (t) = V −1F (t) = UF (t, t0) =∑n≥0

(UE−t0tn/n!

)(Dnt F ) (t0) (2.18)

=∑n≥0

un (t− t0) (Dnt F ) (t0) .


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

f (t) =∑n≥0

un (0) (Dnt F ) (t) .

Suppose the response is the average input over one time unit,

F (t) =

∫ t

t−1

f (τ) dτ.

Before we can solve the inverse problem, we have to find the spectral gain, whichmeans finding the impulse response I (t) first. We have F (t) =

∫ tt−1

f (τ) dτ =∫∞−∞ f (τ) I (t− τ) dτ , thus I (t) = 1 for t ∈ [0, 1], and 0 else. Adding an inputthat averages to zero over every time unit will not change the response. Hencethe input cannot uniquely be recovered from the response, but up to those zero-average inputs. We find the spectral gain Θ (p) =

∫ 1

0e−τpdτ = 1−e−p

p , thus (un)

has the generating function 1Θ(Dt)e

tp = p1−e−p e

pt = pep−1e

p(t+1), which also gener-ates the Bernoulli polynomials ϕn (t+ 1). We saw in Example 1.1.6 that un (t) =ϕn (t+ 1) = (−1)

nϕn (−t). Formula (2.18) shows that

f (t) =∑n≥0

ϕn (t− t0 + 1) (Dnt F ) (t0) =∑n≥0

(−1)nϕn (t0 − t) (Dnt F ) (t0) .

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

f (t) =∑n≥0

(−1)nϕn (0) (Dnt F ) (t) = F (t) +

1

2F ′ (t) +

∑n≥1

B2n

(2n)!D2nt F (t)

=

∫ t

t−1

f (τ) dτ +f (t)− f (t− 1)

2+∑n≥1

B2n

(2n)!D2n−1t (f (t)− f (t− 1)) .

There are other applications of Finite Operator Calculus on the continuousscale. For example, Ismail [45], and Ismail and May [46] investigated applicationsin approximation theory.

2.3.1 Initial Values Along a Line

Sheffer sequences (sn) must carry the initial condition information that gives us aspecific solution to the recursion formula Bsn (x) = sn−1 (x), where B = β−1 (D).If the initial condition means fixed (i.e. a priori known) initial values at initialpoints 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


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

sn (x) =

n∑i=0

si (c) bn−i (x− c) =

n∑i=0

yibn−i (x− c) , or (2.19)

∑n≥0

sn (x) tn =

∑n≥0

yntn

e(x−c)β(t).

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 aSheffer 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 definern (x) = (n+ 1) bn+1 (x) /x for n ≥ 1, and r0 (x) = 1. Then (rn) is a Sheffersequence for the same delta operator as (bn).

Proof. There exists a delta series β (t) such that∑n≥0 bn (x) tn = exβ(t). Take

the t-derivative on both sides,∑n≥0 (n+ 1) bn+1 (x) tn = xβ′ (t) exβ(t). Note that

β′ (t) has order 0, hence β′ (t) exβ(t) is the generating function of the Sheffer se-quence

(n+1x bn+1 (x)

)n≥0

. The delta operator remains β−1 (D) in both cases, (bn)

and(n+1x bn+1 (x)

).

We can only say that rn (x) = (n+ 1) bn+1 (x) /x has the initial valuesrn (0) = [tn]β′ (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 constantsin F, then

tn (x) = (x− an− c) bn (x− c)x− c

is the Sheffer polynomial for B with initial values tn (an+ c) = δ0,n.

Suppose c = 0 and bn (x) = xn/n!, thus tn (x) = (x− an)xn−1/n!. Thebinomial theorem for Sheffer sequences tells us that

(x+ y − an) (x+ y)n−1

=

n∑i=0

(n

i

)(y − ai) yi−1xn−i.

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

(x+ y − an)bn (x+ y)

x+ y=

n∑i=0

(y − ai) bi (y)

ybn−i (x) .


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 combina-torial problems where Catalan numbers occur; R. Stanley has compiled a list inhis book [90], and maintains a website with additional examples [88]. Returning tothe first interpretation of Cn above, we see that Cn equals the values dn (n) whendn (m)−dn (m− 1) = dn−1 (m) (Pascal’s recursion), and dn (m) has initial valuesdn (n− 1) = δ0,n.

m 1 8 35 110 275 572 1001 1430 14307 1 7 27 75 165 297 429 429 06 1 6 20 48 90 132 132 0 -4295 1 5 14 28 42 42 0 -132 -4294 1 4 9 14 14 0 -42 -132 -2973 1 3 5 5 0 -14 -42 -90 -1652 1 2 2 0 -5 -14 -28 -48 -751 1 1 0 -2 -5 -9 -14 -20 -270 1 0 -1 -2 -3 -4 -5 -6 -7

0 1 2 3 4 5 6 7 nThe Catalan numbers (bold) and the polynomials dn (m)

We saw that the basic polynomials(n−1+xn

)solve Pascal’s recursion, hence the

Abelization dn (x) = x+1−nx+1

(n+xn

)solves the same recursion with the right initial

values dn (n− 1) = δ0,n. Therefore, Cn = dn (n) =(

2nn

)/ (n+ 1). The numbers

dn (m) for m ≥ n are often called the ballot numbers, giving the number of waysa two candidate ballot can be realized such that candidate B is never ahead ofcandidate 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 aSheffer sequence (sn (an+ x)) for E−aB, which is again a delta operator. There-fore, taking for sn the special Sheffer polynomials (x− an) bn (x) /x for B, wesee that (xbn (x+ an) / (x+ an))n≥0 must be a Sheffer sequence for E

−aB, andbecause of the initial values, (xbn (an+ x) / (an+ x)) is the basic sequence forE−aB. We denote these basic polynomials by

b(a)n (x) := x

bn(an+ x)

an+ x. (2.20)

If we are searching for Sheffer polynomials for B with initial values tn (an+ c) =yn, y0 6= 0, we are searching for Sheffer polynomials for E−aB with initial values


sn (c) = yn. Hence

tn (an+ x) = sn (x) =

n∑i=0

yn−ib(a)i (x− c) = (x− c)

n∑i=0

yn−ibi(ai+ x− c)ai+ x− c

(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 knowexplicitly the initial value of dn (n− 2), but we will be able to calculate it fromthe previous initial value(s), because we can calculate dn−1 (x). We observe a = 1

and c = −2, so that dn (n+ x) =(

2n+x+1n

)x+2

n+x+2 −∑n−1i=0 Cn−i−1

(2i+x+1

i

)x+2i+x+2 .

For the Catalan numbers we get the recursion Cn = dn (n) =(

2n+1n

)2

n+2 −∑n−1i=0 Cn−i−1

(2i+1i

)2i+2 =

(2n+2n+1

)1

(n+2) −∑n−1i=0 Cn−i−1

(2i+2i+1

)1i+2 . Knowing that

Cn =(

2nn

)1

n+1 gives the even more beautiful recursion Cn+1 =∑ni=0 Cn−iCi,

which has a nice combinatorial interpretation (first return decomposition; see alsoExercise 1.2.6).

We now would like to get an idea about the relationship between β (t) =

ln∑n≥0 bn (1) tn and βa (t) = ln

∑n≥0 b

(a)n (1) tn. Because

(b(a)n

)is the basic se-

quence for E−aB we get immediately that

e−atβ−1 (t) = β−1a (t) . (2.22)

Hence e−aβ(t)t = β−1a (β (t)), or β−1 (βa (t)) = te−aβa(t). If φ (t) := eβ(t) is given,

and we want to know φa (t) := eβa(t), we can also try to solve the equation

φa (t) = φ (tφa (t)a) . (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

∑n≥0

sn (x+ an) tn =∑n≥0

xbn (x+ an)

x+ antn =

∑n≥0

b(a)n (x) tn = exβa(t).

Example 2.3.8 (Catalan numbers continued). In the Catalan example we haveβ−1 (t) = 1 − e−t, hence β−1

1 = e−t (1− e−t), which can be inverted (quadratic


equation; pick the root that is 0 at t = 0) toβ1 (t) = − ln

(12 + 1

2

√1− 4t

). Hence

∑n≥0

dn (x+ n) tn =∑n≥0

(x+ 1)bn (x+ 1 + n)

x+ 1 + ntn =

∑n≥0

b(1)n (x+ 1) tn

= e(x+1)β1(t) =

(2

1 +√

1− 4t

)x+1

and ∑n≥0

Cntn =

∑n≥0

dn (n) tn =2

1 +√

1− 4t=

1−√

1− 4t

2t

Example 2.3.9. Let U1, . . . , UM be identically and independently distributed randomvariables with common density f (y) = 1 for 0 ≤ y ≤ 1, and 0 else (uniform dis-tribution). Denote the order statistic by U(1), U(2), . . . , U(M), where U(1) ≤ U(2) ≤· · · ≤ U(M). Select constants a and c such that 0 < a+c < c+Ma < 1 Elementaryprobability 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

c+na

∫ un

c+(n−1)a

∫ un−1

c+(n−2)a

· · ·∫ u3

c+2a

∫ u2

c+a

1du1du2 · · · dun−2dun−1dun

(see Wald and Wolfowitz [98]). Call the above probability pn (x), extendable to apolynomial in x of degree n. Of special interest is the value pM (1). The sequencep0, p1, . . . , pM satisfies the system of differential equations Dpn (x) = pn−1 (x),and the initial values pn (c+ na) = 0 for all n ≥ 1. Furthermore, p0 (x) =D∫ xa+b

1du(1) = 1. Hence p0, p1, . . . , pM is the beginning piece of a Sheffer se-quence for D, and pn (a+ cn) = δ0,n. From Corollary 2.3.5 we obtain pn (x) =

(x− c− an) (x− c)n−1/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),where S = σ (B) is the Sheffer operator, and B = β−1 (D) is the delta operatorfor (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 areciprocal 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 ofsuch 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!) forD. How does the Sheffer sequence belonging to (σ (α) ρ, β (α)) look like?


We first find the basic sequence (pn), say, with generating function∑n≥0 pn (x) tn =

exβ(α(t)). Note that the delta operator P for (pn) must be written as

P = α−1(β−1 (D)

)= α−1 (B) .

Lemma 2.3.10. If∑n≥0 pn (x) tn = exβ(α(t)), where α and β are delta series with

basic polynomials an (x) =∑ni=0 an,ix

i/i! and bn (x), respectively, then

pn (x) =

n∑i=0

an,ibi (x)

for all n ≥ 0.

Proof. We know from Lemma 2.2.10 that an,i =[αi]n. Hence∑

n≥0

pn (x) tn = C (α) exβ(t) =∑i≥0

bi (x)α (t)i

=∑i≥0

bi (x)∑n≥i

an,itn

=∑n≥0

tnn∑i=0

an,ibi (x) .

We have already seen that multiplication of power series corresponds to com-position of operators; now we have composition of power series α−1

(β−1 (t)

)corresponding to the umbral operator UB , mapping (an) to the basic sequence(an (b (x))), where an (b (x)) :=

∑ni=0 an,ibi (x). We have the following equivalent

statements, occurring in the literature.

1. Delta series β (t): The delta series α (t) is mapped onto the delta seriesβ (α (t)).

2. Delta series β−1 (t): The delta series α−1 (t) is mapped onto the delta seriesα−1

(β−1 (t)

).

3. Delta operator B: The delta operator A = α−1 (D) is mapped onto the deltaoperator α−1 (B).

4. Umbral operator UB : The basic sequence (an (x)) is mapped onto the basicsequence (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 (D) an (x), we obtainrn (b (x)) for this composition, having generating function ρ (t) exβ(α(t)). Insteadof inserting (bn) for (xn/n!), we could also insert the Sheffer sequence (sn) withgenerating function σ (t) exβ(t), giving the composition (rn (s (x))), with generating


function ρ (t)σ (α (t)) exβ(α(t)) (see Exercise 2.3.12). This is the meaning of theoperation (σ, β) (ρ, α) = (σ (α) ρ, β (α)) in terms of Sheffer sequences, that is

rn (s (x)) =

n∑i=0

rn,i

i∑k=0

si,kxk

k!=

n∑k=0

xk

k!

n∑i=k

rn,isi,k.

If rn (s (x)) = xn/n!, then the two Sheffer sequences (or the corresponding groupelements) are inverse to each other. If

tn (x) = rn (s (x)) =

n∑i=0

rn,isi (x)

then the coeffi cients rn,i are called connection coeffi cients. The problem of connec-tion 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)).Hence the connection coeffi cients rn,i =

[xi/i!

]rn (x) can be obtained from

(ρ, α) =

(1

σ (β−1 (γ))τ, β−1 (γ)

)=(1/σ

(β−1

), β−1

) (τ, γ) (2.25)

Example 2.3.11. [83, p. 747] Denote by S (n, k) the Stirling numbers of the secondkind; S (n, k) is the number of set partitions of an n-set into k nonempty blocks (re-gardless of order). Define the polynomial sequence Sn (x) =

∑nk=0 S (n, k)xk/n!.

It is well-known that (Sn) has the generating function∑n≥0 Sn (x) tn = ex(e

t−1).Hence (Sn) is the basic sequence with delta operator β−1 (D) = ln (1 +D). As amember of the umbral group we write (Sn) as (1, et − 1). Now we change our viewpoint and consider the delta operator α−1 (D) for which α−1 (t) = et − 1. Thisdelta operator is the forward difference operator

∆ = α−1 (D) = E1 − 1 : f (x) 7→ f (x+ 1)− f (x) .

The binomial coeffi cients bn (x) =(xn

)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 areinverse, hence by umbral composition follows

xn

n!=

(S (x)

n

)= Sn (b (x)) =

∑k≥0

S (n, k)(xk

)k!

n!.

The exponential polynomials Sn (x) are inverse to(xn

)(and vice versa). From∑

n≥0

(S (x)

n

)tn = ext = e−x

∑k≥0

xk

k!(1 + t)

k= e−x

∑n≥0

tn∑k≥0

xk

k!

(k

n

)follows (

S (x)

n

)= e−x

∑k≥0

(k

n

)xk

k!= e−x

∑k≥0

⟨Evalk |

(x

n

)⟩xk

k!


and by linearity p (S(x)) = e−x∑k≥0 p (k)xk/k! for all polynomials p. The special

case p (x) = xn/n! is known as Dobinski’s formula,

Sn (x) =∑k≥0

S (n, k)xk

n!= e−x

∑k≥0

knxk

n!k!. (2.26)

We can also ask which umbral element (σ, β) will bring the pair (ρ, α) to thepair (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)) ,

hence β (t) = ln(1 + α−1 (t)

), and σ (t) = 1/ρ

(α−1 (t)

). If we define (rn) and (sn)

as Sheffer sequences with generating function

ρ (t) exα(t) and(1 + α−1 (t)

)x/ρ(α−1 (t)

),

respectively, then rn (s (x)) =(xn

).

Remark 2.3.12. The umbral group occurs in different forms in the literature; inour description we focus on power series. The similarity to the Riordan grouphas been pointed out by He, Hsu, and Shiue [40]. As defined above, the umbralgroup and umbral composition are terms used in the Finite Operator Calculus,introduced already by Rota, Kahaner and Odlyzko [83]. In section 3.4 on classicalUmbral 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 - notthe 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 secondtopic is about Inverse Pairs of number sequences. We present several examplefrom Riordan’s book and interpret them through Sheffer Sequences.

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

sn+1,i+1 =

n−i∑k=0

sn,i+kak for all 0 ≤ i ≤ n, and

sn+1,0 =

n∑k=0

sn,klk.


where (an)n≥0 and (ln)n≥0 are given numerical sequences with a0 6= 0 ([75], [58]).The existence of an a-sequence is useful in determining whether a given matrix isa Riordan matrix or not. Often, trying to determine the a-sequence from a matrixthat is not Riordan will come to an end after a few values.

0 1 2 3 4 k

0 11 1 12 3 1 13 6 4 1 14 16 9 5 1 15 41 26 12 6 1 16 113 71 37 15 7 17 316 204 106 49 18 8n 907 590 316 146 62 21

0 1 2 3 4 k

0 11 2 12 4 2 13 9 5 2 14 22 13 6 2 15 57 35 17 7 2 16 154 97 49 21 8 27 429 275 143 64 25 9n 1223 794 422 195 80 29

Both the above matrices are Riordan matrices; the matrix on the left has l-sequencel0 = 1, l1 = 2, lk = 1 for k ≥ 2. For finding the a-sequence on the left, we beginwith s0,0 = 1, and s1,0 = 1. Hence a0 = 1, which not only makes s1,1 = 1, butproduces 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 stepwe get s2,0 = 3, and a2 = 1 generates the numbers 4, 5, 6, 7, . . . . This way thetable 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,1if 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 froms0,0 = 1 and sn,1 + sn−1,0 = sn,0 + (−1)

n 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 lookslike the same l-sequence too. However, this simple l-sequence does not work forfinding 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 aTheorem what we just saw, the definition of a Riordan matrix through its firsttwo columns.

Theorem 2.3.13. The matrix S = (sn,k) is a Riordan matrix such that sn+1,i+1 =∑n−ik=0 sn,i+kak for all 0 ≤ i ≤ n, and sn+1,0 =

∑nk=0 sn,klk where (an)n≥0 and

(ln)n≥0 are given numerical sequences, a0 6= 0, if and only if sn,k = [tn](σ (t)β (t)

k)

for all n ≥ 0, 0 ≤ k ≤ n, where β (t) ∈ F [[t]] is the delta series

β−1 (t) = t/∑n≥0

antn


and

σ (t) = s0,0/

1− t∑n≥0

lnβ (t)n

is of order 0.

Proof. Suppose sn+1,i+1 =∑n−ik=0 sn,i+kak for all 0 ≤ i ≤ n. From a0 6= 0 follows

sn,n 6= 0, hence sn (x) :=∑nk=0 sn,kx

k/k! is a polynomial of degree n. We find

Dsn+1 (x) = Dn+1∑k=0

sn+1,kxk

k!= D

n+1∑k=0

n−k+1∑j=0

sn,k−1+jajxk

k!

= Dn+1∑j=0

aj

n∑k=j−1

sn,kxk−j+1

(k − j + 1)!=

∑j≥0

ajDj sn (x)

The power series t/∑j≥0 ajt

j is a delta series, thus D/∑j≥0 ajDj a delta oper-

ator, and (sn) a Sheffer sequence for this delta operator. If we write σ (t) exβ(t)

for the generating function of (sn), then β−1 (D) = D/∑j≥0 ajDj , and of course

sn,k =[σβk

]n. Hence

σ (t) =

∞∑n=0

sn,0tn = s0,0 +

∞∑n=1

n−1∑k=0

sn−1,klktn

= s0,0 + t

∞∑k=0

lk

∞∑n=k

sn,ktn = s0,0 + t

∞∑k=0

lkσ (t)β (t)k

and therefore

σ (t)

1− t∑k≥0

lkβ (t)k

= s0,0.

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

matrix sn,k = [tn](σ (t)β (t)

k). This is the usual definition of Riordan matrices.

Therefore, solving recurrence relations with the Riordan matrix approach utilizesa different aspect of Sheffer sequences, where we up to now only looked at thevalues of the polynomials sn (k) = [tn]

(σ (t) ekβ(t)

). 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


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

1 = σ (t)

1− t∑k≥0

lkβ (t)k

= σ (t)

(1− tβ (t)− t 1

1− β (t)

)

= σ (t)1− β (t)− t

(1 + β (t)− β (t)

2)

1− β (t).

The recurrence s0,0 = 1 and sn,1 + sn−1,0 = sn,0 − (−1)n for n ≥ 1, gives the

generating function equation

σ (t)− 1 +t

1 + t= σ (t)β (t) + tσ (t) .

Now we have two equations for σ and β; thus

β (t) =1

2

(1− t2 −

√(1 + t2)

2 − 4t

)and

σ (t) =2

(1 + t)

((t− 1)

2+

√(1 + t2)

2 − 4t

) .Example 2.3.15. There is an interpretation of the matrix on the right hand side asa 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 samenumber 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 1546 1 2 7 17 35 57 575 1 2 6 13 22 224 1 2 5 9 93 1 2 4 42 1 2 21 1 10 1

0 1 2 3 4 5 nTable 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


for all n > 0 and 0 ≤ k ≤ n. This implies the existence of an a-sequence (Exercise2.3.18) such that

ak =

k∑j=0

(2j−1j

)(2j − 1

k − j

)(−1)

k−j−1

2j − 1,

the same sequence as in Example 2.3.14. Hence Theorem 2.3.13 implies that

β−1 (t) =t∑

n≥0 antn

=1

2 (1− t)

(1−

√1− 4t (t− 1)

2

),

and by inversion

β (t) =1

2

(1− t2 −

√(1 + t2)

2 − 4t

)=

1− t22

(1−

√1− 4t− 4t2

(1− t2)2

).

Nowsn,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

σ (t) =2

(1− t)2+

√(1 + t2)

2 − 4t

=2t2 − 2t+

(1− t2

) (1−

√1− 4t−4t2

(1−t2)2

)2 (t2 − t3)

=1

t

(β (t)

t− t2 − 1

).

The numbers

D (n,m) = sm−1,m−1−n =[tm−1

] (σ (t)β (t)

m−1−n)

= [tm]

(β (t)

m−n

t− t2 − β (t)m−1−n

)can be expanded for m > n ≥ 0 as

D (n,m) =

n+1∑j=0

(m− 2j + n+ 1

m− j

)(−1)

j × (2.27)

×j/2∑i=0

(m− jj − 2i

)(i− 2j +m+ n+ 1

i

)(m− n

m− j + 1− m− 1− ni− 2j +m+ n+ 1

)n∑j=0

(m− n− 1

n− j

)(−1)

n−jj/2∑i=0

(2i− j − 1

i

)1

j + 1− 2i×

×(

(m− n)

(2j +m− n− 3i

j − 2i− 1

)+

(2j +m− n− 3i

j − 2i

)).


(see Exercise 2.3.19).Theorem 2.3.13 shows that Riordan matrices offer another look at the um-

bral group, from the coeffi cient level, because we can take any basic sequence (an)with generating function exα(t), and construct the Sheffer polynomial sn(a (x)) =∑nk=0 sn,kak (x), where sn,k = [tn]

(σ (t)β (t)

k). This Sheffer polynomial has co-

effi cients sn,k = [ak (x)] sn (a(x)) and generating function∑n≥0 sn (a (x)) tn =

σ (t) exα(β(t)), hence (sn(a (x))) belongs to the group element (1, α) (σ, β) (Exer-cise 2.3.20). In matrix notation, ~s = S ~a, where

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

T.

Viewing the matrix elements as coeffi cients of Sheffer polynomials can lead to el-egant proofs. For example, if we want the generating function of the row sums,∑nk=0 sn,k, we can apply the basic polynomials ak (x) =

(k + x− 1

k

)with gener-

ating function (1− t)−x, so that∑nk=0 sn,k = sn(a (1)). Hence

∑n≥0

tnn∑k=0

sn,k = σ (t) (1− β(t))−1.

Choosing ak (x) = xk/k! gives

∑n≥0

tnn∑k=0

sn,k/k! = σ (t) eβ(t).

More general, take two Riordan matrices S = (sn,k) and R = (rn,k), wheresn,k = [tn]σ (t)β (t)

k and rn,k = [tn] ρ (t)α (t)k. The product

SR =

n∑j=k

sn,jrj,k

has matrix elements with generating function

∑n≥0

tnn∑j=k

sn,jrj,k =

∞∑j=k

σ (t)β (t)j([tj]ρ (t)α (t)

k)

= 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 polynomialsbn(x) =

∑nk=0 bn,kx

k/k! for β−1 (D). The coeffi cients follow recursions reminiscentof the binomial theorem.


Theorem 2.3.16. Let S = (sn,k) be the Riordan matrix defined by

sn,k = [tn](σ (t)β (t)

k), and let (bn) be the basic sequence for β−1 (D). Then

sn,i+j =

n−j∑k=i

sk,ibn−k,j

for all 0 ≤ i+ j ≤ n.

The proof follows from σ (t)β (t)i+j

= σ (t)β (t)iβ (t)

j .

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

sn,i+1 =

n−1∑k=0

sk,icn−k

for all 0 ≤ i < n.

We leave the proof as Exercise 2.3.22. Note that β (t) =∑k≥1 ckt

k if sn,k =

[tn](σ (t)β (t)

k).

Inverse Pairs In its simplest form, an inverse pair is nothing but a pair of powerseries (σ (t) , γ (t)) related to each other through a third known power series α (t),say, whose reciprocal is also “known”,

γ (t) = α (t)σ (t) and σ (t) =1

α (t)γ (t)

If an inverse pair of relations is given, only one of the two must be shown; theother will hold automatically. The linear recursion in one variable is an inversepair (Proposition 1.1.4), with α (t) = 1−

∑∞j=1 αjt

j .

Example 2.3.18. If α (t) = et then 1/α (t) = e−t, and if we write γ (t) = etσ (t)then σ (t) = e−tγ (t). This elementary relationship looks more interesting on thecoeffi cient level. Let γ (t) =

∑n≥0 γnt

n/n! and σ (t) =∑n≥0 σnt

n/n!, then

γn =

n∑k=0

(n

k

)σk ⇐⇒ σn =

n∑k=0

(n

k

)(−1)

n−kγk

(Riordan [72, Chpt. 2]). For the power series σ (t) =∑n≥0 σnt

n and γ (t) =∑n≥0 γnt

n holds

1

1− t σ(

t

1− t

)= γ (t) (2.28)


(see Exercise 1.2.3). The relationship between the numbers Cn and the Motzkinnumbers Mn is an application of this inverse pair, namely

Cn+1 =

n∑k=0

(n

k

)Mn−k

(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 argumentfor this identity goes as follows. Strip the Catalan paths counted by Cn+1 of theirfirst (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 ontosingle steps as in the following table.

double step: single step: −→ =⇒

Now we have a ,,→,⇒ - path to (n, 0) staying weakly above the x-axis. Itis called a bicolored Motzkin path, thus Cn+1 equals the number of such paths to(n, 0). This path can have any number k = 0, 1, . . . , n of ⇒ steps at

(nk

)places;

taking them out leaves a Motzkin path to (n− k, 0) behind. Hence

Cn+1 =

n∑k=0

(n

k

)Mn−k.

The above inverse relationship tells us that

Mn =

n∑k=0

(n

k

)(−1)

n−kCk+1.

It also tells us that∑n≥0

Cn+1tn =

1

1− t∑n≥0

Mn

(t

1− t

)n=:

1

1− tM(

t

1− t

)(see (2.28)). We saw in Example 2.3.8 that c (t) =

∑n≥0 Cnt

n = 21+√

1−4t, and

c (t)− 1 =t

1− tM(

t

1− t

).

The compositional inverse of t/ (1− t) is t/ (1 + t), hence

M (t) =

(c

(t

1 + t

)− 1

)/t =

1− t−√

1− 2t− 3t2

2t2

is the generating function for the Motzkin numbers.


The general inverse pairs are not that trivially constructed; they originatein matrix algebra rather than generating functions. Suppose we have an infinitematrix 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 = (δi,j)i,j≥0. Hence∑ik=j βi,kαk,j = δi,j for all i, j ≥ 0. For such a matrix and

its inverse holds

γn =

n∑k=0

αn,kσk ⇐⇒ σn =

n∑k=0

βn,kγk

for all n ≥ 0, as can be easily checked (Gould class of inverse relations). In matrixnotation, let γ = (γ0, γ1, . . . )

>, a column vector, and let σ = (σ0, σ1, . . . )>. Then

γ = Aσ ⇐⇒ σ = A−1γ.

The problem is finding such pairs of inverse matrices explicitly! We started withdiagonal matrices, but we now want to show a larger class that can be found frombasic sequences by Abelization. Let (bn (x))n≥0 be a basic sequence for some deltaoperator B. Let u and v be two arbitrary but nonzero constants, and define theSheffer sequence (sn) for B by the initial values sk (0) = σk for all k ≥ 0. This is aSheffer sequence only if σ0 6= 0, so we omit leading zeroes. The binomial theorem(2.14) shows that sn (un+ v) =

∑nk=0 σkbn−k (un+ v). We let γn = sn (un+ v),

hence

γn =

n∑k=0

σkbn−k (un+ v) .

To get the inverse direction, let tn (x) = sn (un+ v + x), a Sheffer polynomial forE−uB (see 2.3.1). The delta operator E−uB has the basic sequence(xb (x+ un) / (x+ un)). By the binomial theorem

tn (−un− v) =

n∑k=0

tk (0)−un− v

(n− k)u− un− v bn−k((n− k)u− un− v),

thus

σn =

n∑k=0

γkun+ v

uk + vbn−k(−ku− v).

In matrix form, A = (bi−j (ui+ v))i,j≥0 and B =(ui+vuj+v bi−j(−ju− v)

)i,j≥0

are

inverses (see Exercise 2.3.25).

The special case bn (x) =

(x

n

)gives γn =

∑nk=0 σk

(un+ v

n− k

)⇐⇒ σn =

∑nk=0 γk (−1)

n−k v+unv+n+(u−1)k

(n+ v + (u− 1) k

n− k

).


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 generatingfunction text/ (et − 1) (Example 1.1.6) are a Sheffer sequence for D. Show that thebinomial theorem implies

n!ϕn (x) =

n∑i=0

(i− n− 1

i

)Bi (x− 1)

n−i .

With the same tools show that 0 =∑ni=0

(ni

)Bi

(xn−i − (−1)

n(1− x)

n−i). This

implies Bn = (−1)n∑n

i=0

(ni

)Bi, and 0 =

∑ni=0

(ni

)2iBi for odd n. Get B2k+1 = 0

for k ≥ 1 from the generating function, and B1 = −1/2 (show that t (et − 1)−1

+t/2 is symmetric).

2.3.3. One way of defining the Bernoulli polynomials of the second kind is asSheffer sequence (ϕn (x)) for ∆ that satisfies the initial condition Dϕn (x) =

(x

n−1

)for all n ≥ 0. Show that these two conditions uniquely define (ϕn (x)), and thatbn := ϕn (0) has the generating function

∑n≥0 bnt

n = t/ ln (1 + t). Show thatϕn (x) = bn +

∑nj=1

nj s (n− 1, j − 1)xj, where the Stirling numbers s (n,m) of

the first kind are defined in Exercise 2.3.16. Use Dϕn (x) =(x

n−1

)and ϕn (x) =∑n

i=0 bi(xn−i)to show that

∫ 1

0

(xn

)dx = bn. Write t

∫ x0

(1 + t)ydx as a power series

in Stirling numbers of the first kind to show that

ϕn (x) = bn +1

(n− 1)!

n−1∑m=1

s (n− 1,m− 1)xm

m.

The polynomials ϕ2n (n− 1 + x) are an even function of x (check that their deriv-ative is odd!). Apply ∆ to show that odd degrees are odd, ϕ2n−1 (n− 1 + x) =−ϕ2n−1 (n− 1− x). We can write both statements together as ϕn

(⌈n2

⌉− 1 + x

)=

(−1)nϕn(⌈n2

⌉− 1− x

).

2.3.4. Show that for the Bernoulli polynomials (ϕn) of the first kind holds ∆ϕn (x) =xn−1/ (n− 1)! [3, Exc. 1-44].

2.3.5. The Stirling numbers of the second kind S (n, k) can be defined as the numberof partitions of an n-set into k nonempty blocks. Show that

S (n, k) =∑

1r1+2r2+···+nrn=nr1+r2+···+rn=k

n!

(1!)r1 r1! (2!)

r2 r2! · · · (n!)rn rn!

2.3.6. Let x0, x1, . . . be a given sequence in F. Suppose (tm) is a polynomial se-quence and B a delta operator such that 〈Evalxr | Brtm〉 = δm,r for all 0 ≤ r ≤ m


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

bm (x) =∑j≥0

bm−j (xj) tj (x) .

2.3.7. Use (2.22) to show that the generating function for the (translated) Abelpolynomials equals

∑n≥0

x (x+ an)n−1

tn/n! = exp

(x

∞∑n=1

(an)n−1

tn/n!

).

2.3.8. Show that

n∑k=0

(n− x+ kz

k

)(x− kzn− k

)=zn+1 − 1

z − 1(2.29)

for all x, z ∈ C (identity 3.145 in Gould’s list [38]). Begin by showing that

1 =

n∑k=0

(n− x+ kz

k

)x− nzx− kz

(x− kzn− k

).

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 theumbral group. Is G × H isomorphic to the umbral group? Show that the umbralsubgroup (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) inthe umbral group the right coset (σ, β) G represents all Sheffer sequence for thesame delta operator β−1 (D). Characterize the left cosets. Is G normal?

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(x-) operator that replaces xn/n! by sn (x) for all n ≥ 0, US,Bxn/n! = sn(x). Notethat US,Bext = M (σ)C (β) ext; the umbral operator is the transform ofM (σ)C (β).Use the transform approach to show that (rn (s (x))) has the generating functionρ (t)σ (α (t)) exβ(α(t)).

2.3.13. Use the notation of Exercise 2.3.12. Show that the umbral operator US,Bis not in ΣD (except when B = D).


2.3.14. The Poisson-Charlier polynomials pn (x; a) have generating function e−t(a+ta

)x,

hence

pn (x; a) =

n∑j=0

(x

j

)(−1)

n−j

aj (n− j)!

for a 6= 0. If

pn (x; a) =

n∑i=0

rn,ipi (x; b)

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 intok 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 definitionto prove that k!S (n, k) =

∑kj=0

(kj

)(−1)

k−jjn. Derive this result also from the

generating function φn (x) in Example 2.3.11. The Stirling numbers S (n, k) (ofthe second kind) have the generating function

∑n≥k

S (n, k)tn

n!=

(et − 1)k

k!

which has been already applied in Examples 2.3.11 and 2.2.13 . Show that

∑n≥0

tn

n!

n∑k=0

(−1)kk!

k + 1S (n, k) =

t

et − 1,

which implies an explicit formula for the Bernoulli numbers Bn,

Bn =

n∑j=0

(−1)jjn

n∑k=j

1

k + 1

(k

j

).

(Hint: Use Exercise 2.3.4.)

2.3.16. In Example 2.3.11 we found that((xn

))and (φn) are inverse to each other,

thusxn/n! = φn (b (x)) =

(φn

),

where bn (x) =(xn

). Let

(xn

)=∑ni=0 an,i

xi

i! . Hence xn/n! =

∑ni=0 an,iφi (x). Show

that this implies∑ni=k k!an,iS (i, k) /i! = δn,k for all 0 ≤ k ≤ n. The numbers

s (n, i) = n!an,i/i! are the (signed) Stirling numbers of the first kind,

∞∑i=0

∞∑n=i

s (n, i)xitn

n!= (1 + t)

x.

We have∑ni=k s (n, i)S (i, k) = δn,k. Show also that

∑nk=j S (n, k) s (k, j) = δn,j.


2.3.17. Show: If S = (sn,k)n,k≥0 satisfies sn,k =(σβk

)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

in F, a0 6= 0, such that sn+1,i+1 =∑n−ik=0 sn,i+kak for all 0 ≤ i ≤ n, and

sn+1,0 =∑nk=0 sn,klk.

2.3.18. Show that for a Riordan matrix (sn,k) holds

sn+1,i+1 = sn+1,i − sn,i−1a0 +

n−i∑l=0

sn−1,i−1+l

l∑k=0

(ak − ak+1) al−k (2.30)

for all 0 < i ≤ n, where (an) is the a-sequence (a0 6= 0). Vice versa, this conditiontogether with a suitable sn,0 implies that (sn,k) is a Riordan matrix. Show that inExample 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

ak =

k∑j=0

(2j−1j

)(2j − 1

k − j

)(−1)

k−j−1

2j − 1.

2.3.19. Show expansion (2.27). It is also true that

D (n,m) =

n∑j=0

(m− n− 1

n− j

)(−1)

n−jj/2∑i=0

(j − ii

)(−1)

i

j + 1− 2i×

×(

(m− n)

(2j +m− n− 3i

j − 2i− 1

)+

(2j +m− n− 3i

j − 2i

))for m > n.

2.3.20. If S is a Riordan matrix, sn,k =[σβk

]n, and (an) a basic sequence with

generating function exα(t), show that sn(a (x)) =∑nk=0 sn,kak (x) is a Sheffer

polynomial with generating function∑n≥0 sn (a (x)) tn = σ (t) exα(β(t)).

2.3.21. Use a given Riordan matrix S = (sn,k) to find the corresponding basicsequence for the Sheffer sequence sn (x) =

∑nk=0 sn,kx

k/k!.

2.3.22. Prove Corollary 2.3.17.

2.3.23. Let α (t) =∑n≥0 αnt

n and 1/α (t) =∑n≥0 βnt

n Show: If

αn = c (c+ un)n−1

/n!, then βn = c (c− un)n−1

(−1)n/n!

This α (t) is defined with the help of the Abel polynomialsbn(x) = x (x+ αn)

n−1/n!, the basic polynomials for E−αD. It occurs with u = 1

as No. 1 in Riordan’s [72, Table 2.2] Table 3.1 of Abel inverse relations.


2.3.24. The reciprocals No. 2

αn = (c+ un)n/n! and βn =(c2 − u2n

)(c− un)

n−2(−1)

n/n!

in the same table use the same basic polynomials as in Exercise 2.3.23, but we canview (αk) as an evaluation of the Sheffer sequence ((x+ un)n/n!) for E−uD at c.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+vuj+v bi−j(−ju− v)

)i,j≥0

are inverse matrices.

2.3.26. Suppose that γn =∑nk=0 σkbn−k (un+ v) for all n ≥ 0. Let a ∈ F. Show

that this equivalent to

m+n∑k=0

bn+m−k (am+ v)σk =

n+m∑k=0

γkam− u (n+m)

am− uk bn+m−k(am− uk)

for all m,n ∈ N0. A special case is

γn =

n∑k=0

(n

k

)σk ⇐⇒

m∑k=0

(m

k

)σn+k =

n∑k=0

(n

k

)(−1)

n−kγm+k.

2.3.27. Class 1 and Class 2 in Riordan’s Table [72, Table 2.2] of Gould classes ofinverse relations is the pair

αn,k =

(−p− (q − 1)n− 1

k

)and

βn,k =

(p+ (q − 1) (n− k)

k

)+ q

(p+ (q − 1) (n− k)

k − 1

)(In Riordan’s notation βn,n−k = An,k and βn,n−k = (−1)

n−kBn,k). Show:

(αi,j)i,j≥0 is the inverse matrix to (βi,j)i,j≥0.

The next 3 exercises are based on the special properties(αkk

)= α

(αk−1k−1

)and

(αk)k

k! = α (αk)k−1

(k−1)! .

2.3.28. Show:

αn,k =

(c− 2αn+ αk

k

)and

βn,k =2αn− c

α (2n− k)− c

(α (2n− k)− c

k

)− α 2αn− α− 1− c

α (2n− k)− 1− c

(α (2n− k)− 1− c

k − 1

)define a pair of inverse matrices.


2.3.29. Show:

αn,k =(c+ u (n− k))

k

k!and βn,k = (−1)

k(c+ u (n− k))

(c+ un)k−1

k!

define a pair of inverse matrices. For u = 1 this is No. 3 in Riordan’s [72, Table3.1] table of Abel inverse relations (In Riordan’s notation αn,n−k = An,k/ (n− k)!

and βn,n−k = (−1)n−k

Bn,k/ (n− k)!).

2.3.30. Show:

αn,k =(c− uk + 2un)

k

k!and

βn,k = (−1)k (c+ 2un− uk)

k−2

k!

((2un+ c)

2+ uk (u− 4un− 2c)

)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 goalis to find an expression for the basic polynomials bn (x) for B. This could be thegenerating function b (x, t) =

∑n≥0 bn (x) tn, or an explicit expansion of bn (x) in

terms of the basis (an).Of course, the form of the operator equation is the key point in such expan-

sions. If the operator equation can be written as

B = β−1 (D) = τ−1 (A) = τ−1(α−1 (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]], buthas coeffi cients that are operators themselves. We have to restrict this settings totranslation 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 oper-ator 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)−∞∑i=0

(−1)iD (n− 3− 2i,m− 1− i)

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, thefactors a1+2i = (−1)

i 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


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

∇ =∑k≥0

(−1)kE−kB2k+1 =

B

1 + E−1B2(2.31)

where Bdn = dn−1. Note that the known delta operator ∇ ∈ ΣB is written as apower series in B, ∇ = B − E−1B3 + E−2B5 − . . . , with coeffi cients

[Bn]∇ = (−1)(n−1)/2

E−(n−1)/2 ∈ ΣD

if n is odd, and 0 else.

In the example above we expended A (= ∇) in ΣB . We saw in Remark 2.2.6that every operator in ΣD has also a representation in ΣB . In other words, if A isa delta operator in ΣD, and

A = T1B + T2B2 + T 3B3 + . . .

where Ti ∈ ΣD for all i ≥ 1, and T1 is invertible, then B is a delta operator in ΣD,and A ∈ ΣB . Certainly A ∈ ΣB in the example, because E−i ∈ ΣB , but we do notknow how E−i is expanded in ΣB , only in ΣD. We will show how to find the basicsequence for B expanded in terms of the basic sequence for A without knowingthe expansion of E−i in ΣB (but we know that it exists!). To do this, we need atechnical device, the Pincherle derivative, introduced in the next subsection. Fornow 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 tous. Suppose A can be expanded as

A = τ (B) =∑j≥1

TjBj

where Tj ∈ ΣD, T1 invertible. It follows that B is a delta operator with basicsequence b0 (x) = 1 and

bn (x) = x

n∑i=1

[τ i]n

1

xai (x)

for all n ≥ 1.

Remark 2.4.3. The power series τ in the above theorem is an element of ΣD [[t]],the power series in t with coeffi cients that are translation invariant operators,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

bn (x) = x

n∑i=1

[(1 +Bt)

i]n−i

1

xai (x) = x

n/2∑i=0

(n− ii

)Bi

1

xan−i (x)


by the theorem above. Of course, B + B3 would also be the evaluation at B ofη (t) = t+ t3 ∈ F [[t]],

bn (x) =

n/3∑i=0

(n− 2i

i

)an−2i (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 Tjjust multiplies by some scalar tj ∈ F, j ≥ 1. Now bn (x) =

∑ni=1

[τ i]nai (x), and∑

n≥0

bn (x) tn = exα(τ(t)) (2.32)

where α (t) = ln∑n≥0 an (1) tn. If A = ∆ = E1−I we get the following noteworthy

special case.

Corollary 2.4.4. Suppose

E1 = I + τ (B)

where τ (t) ∈ F [[t]] is a delta series. Then∑n≥0

bn (x) tn = (1 + τ (t))x. (2.33)

Proof. The forward difference operator ∆ = E1 − I is a delta operator with ex-pansion ∆ = α−1 (D) = eD − 1, hence α(t) = ln (1 + t) and

exα(τ(t)) = (1 + τ (t))x.

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

∇ =B

1 + E−1B2[τ i]n

=

[B

1 + E−1B2

]in

=[Bn−i

]∑k≥0

(−ik

)E−kB2k =

(−i

(n− i) /2

)E−

n−i2

if n − i is even, and 0 otherwise. The coeffi cients[τ i]nare not in F but in ΣD.


Hence

bn (x) = x

n∑i=1

[τ i]n

1

x

(i− 1 + x

i

)

= x

(n−1)/2∑i=0

(−n+ 2i

i

)E−i

1

n− 2i

(n− 2i− 1 + x

n− 2i− 1

)

=

(n−1)/2∑i=0

(n− 1− i

i

)(n− 3i− 1 + x

n− 2i− 1

)(−1)

ix

n− 2i

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’(section2.3.1),

dn (x) =x− n+ 1

x+ 1bn (x+ 1)

= (x− n+ 1)

(n−1)/2∑i=0

(n− 1− i

i

)(n− 3i+ x

n− 2i− 1

)(−1)

i

n− 2i.

The generating function for (bn) we can obtain from (2.33), because we cansolve for E1 in 1− E−1 = B/

(1 + E−1B2

). We obtain

E1 =

√(B2 + 1)

2 − 4B3 + 1−B2

2 (1−B)

(take the root which is not 0 at 0). Therefore,∑n≥0

bn (x) tn =

(1

2(t+ 1) +

(√(t2 + 1)

2 − 4t3)/ (2 (1− t))

)x.

We find the generating function of xbn (n+ x) / (n+ x) from (2.23),

φ1 (t) =1

2(tφ1 (t) + 1) +

√((tφ1 (t))

2+ 1)2

− 4 (tφ1 (t))3

2 (1− tφ1 (t))

which has the power series solution

φ1 (t) =1

2+

√t4 + 2t2 + 1− 4t

2 (t2 − t) .

Hence ∑n≥0

d (n+ x) tn =

(1

2+

√t4 + 2t2 + 1− 4t

2 (t2 − t)

)x+1

.


In Theorem 2.4.2 we start with the operator equation

A =∑j≥1

TjBj

and then proceed expanding the basic sequence of B. If this equation can berewritten in the form 0 = F (E,B) for some function F , then a generating functionfor (bn) can be found explictly when we can solve 0 = F (E,B) for E explicitly(Corollary 2.4.4). In the above example we had I−E−1 = B/

(1 + E−1B2

), which

is easily solved.

2.4.1 Umbral shifts and the Pincherle derivative

The umbral shift (associated to D) is a linear operator denoted by θ such that

θxn = xn+1 for n ≥ 0.

Note that θn1 = xn. From

θEax = x2 + ax 6= (x+ a)2

= Eaθx

follows that the umbral shift is not a member of ΣD. We will write just x for θ inthis section. Note that the umbral shift is not translation invariant.

With the help of the umbral shift we define the Pincherle derivative T ′ ofany operator T ,

T ′ = Tx− xT.In which sense is T ′ a derivative? The product rule of differentiation holds, (ST )

′=

S′T +ST ′ (see Exercise 2.4.5), but there is more. If T ∈ ΣD, T = τ (D), say, thenddtτ (t) exists, and we can define the (ordinary) derivative of T as d

dDT = ddD τ (D),

again in ΣD. The two concepts agree if T ∈ ΣD: ddDT = Tx− xT . In other words,

the Pincherle derivative equals the ordinary derivative if T is translation invariant(see Exercise 2.4.4 for a proof). For example,

(En)′

= Enx− xEn = Enx− En (x− n) = nEn.

The way we defined the umbral shift and the Pincherle derivative above isfocused on the derivative operator D. We can switch to any other delta operatorA, say, with basic sequence (an), and define the umbral shift θA associated to Aas

θAan = (n+ 1) an+1 for n ≥ 0.

Note thatθAAan = nan for all n ≥ 1.

The corresponding Pincherle derivative of an operator T is defined as the commu-tator of θA and T ,

T ′A = TθA − θAT,


and we can easily prove that ddAT = T ′A 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 = θD′A.

Proof. We have

θAa (x, t) =d

dta (x, t) = M (x)M (α′) exα(t)

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

θAa (x, t) = M (x)α′ (A) exα(t) = M (x)d

dAα(α−1 (A)

)exα(t)

= M (x)d

dADexα(t).

Hence θA = xD′A.

Example 2.4.7. If A = ∇, then θ∇(n−1+xn

)= (n+ 1)

(n+xn+1

)= xE1

(n−1+xn

), thus

θ∇ = xE1. Note that E = D′∇. For T ′∇ we get (Tx− xT )E1 = T ′DE1 if T is

translation invariant. We can check this by calculating

dT

d∇ =dT

dDdDd∇ = − dT

dDd ln (1−∇)

d∇ =dT

dD1

1−∇ =dT

dDE1.

As a concrete example we find(Ek)′∇ = EkxE1 − xE1Ek =

(Ekx− Ek (x− k)

)E1 = kEk+1.

It follows that Ekθ∇ = (x+ k)Ek+1. In general,

EkθA − θAEk =(Ek)′A

=(Ek)′DdDdA

= kEkdDdA

hence

EkθA = θAEk + kEk

dDdA

= θAEk +

kEk

Ax− xA.

2.4.2 Proof of the Transfer Formula

Because B is of order 1, we can write B = DP−1, where P ∈ ΣD is invertible.Thus sn (x) = B′Pn+1xn/n! is a Sheffer polynomial for B, because

Bsn (x) = B′PnDxn

n!= B′Pn

xn−1

(n− 1)!= sn−1 (x) .


Here we only need commutativity in ΣD. What are the initial values of (sn (x))?We calculate B′Pn+1 in more details: For n ≥ 1 holds

B′Pn+1 =(DP−1

)′Pn+1 =

(P−1 +D

(P−1

)′)Pn+1 = Pn − P−2P ′Pn+1D

= Pn − 1

n(Pn)

′D.

All this can be done with the ordinary derivative, it does not need the Pincherlederivative. However, using this concept, we see that

B′Pn+1xn

n!= Pn

xn

n!− 1

n(Pnx− xPn)Dx

n

n!=x

nPnDx

n

n!

because 1nP

nxDxn/n!=Pnxn/n!. Note that this cancelling out effect only occurswhen B′Pn+1 is applied to xn/n! (the n in B′Pn+1 must agree with the n inxn/n!). Thus

sn (x) = B′Pn+1xn

n!= xPn

xn−1

n!= θPn

xn−1

n!

for all n ≥ 1. Now 〈Eval0 | θp〉 = 0 for any polynomial p (x) 6= 0, thus sn (0) = 0for all n ≥ 1. For n = 0 we get

s0 (x) = B′P 11 = P 01− P ′P−1D1 = 1.

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

bn (x) = B′Pn+1xn/n!. (2.34)

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

Theorem 2.4.8. If B = DP is a delta operator with basic sequence (bn), then

bn (x) = xPnxn−1

n!(2.35)

The first Transfer Theorem also shows how to transfer from the basic se-quence (an) of some delta operator A to the basic sequence (bn) for B.

Corollary 2.4.9. If A = V B, where V is invertible, then for all n ≥ 1 holds

bn (x) = xV nan (x)

x(2.36)

= θAVn an−1 (x)

n


Proof. Let B = DP−1 and A = DS−1, hence PS−1 = V . By the previous Theo-rem, an (x) = xSn x

n−1

n! . For n ≥ 1, the polynomials an (x) are 0 at 0, hence theyare divisible by x (we have seen that (n+ 1) an+1 (x) /x are Sheffer polynomials forA). Therefore an (x) /x = Snxn−1/n!, or equivalently S−n (an (x) /x) = xn−1/n!.Substituting this into (2.35) gives the Corollary.

We saw in Proposition 2.4.6 that θA = xD′A, hence the second expression forbn (x) follows,

θAVn an−1 (x)

n= xV nD′A

an−1 (x)

n= xV n

1

xθAan−1 (x)

n= xV n

an (x)

x.

Finally, the proof of the Transfer Theorem 2.4.2 follows by Lagrange Inversionof (2.36). We have to apply

(PS−1

)nto the Sheffer polynomial nan (x) /x for A,

thus we need an expansion of this operator in terms of powers of A. We only knowthat A = τ (B) in the Transfer Formula. We want to find(

PS−1)n

=∑i≥0

⟨Ai |

(PS−1

)n⟩Ai =

∑i≥0

⟨Ai | (A/B)

n⟩Ai.

In the last expression we think of B as a power series in A, B = A/φ (A), whereφ is of order 0. Hence(

PS−1)n

=∑i≥0

⟨Ai | φ (A)

n⟩Ai =

∑i≥0

[φn]iAi.

Now apply the inversion formula (1.11), n[γk]n

= k [φn]n−k, from right to left;we want to know [φn]i, where t/φ (t) is the compositional inverse of τ (t). Hence

[φn]i =n

n− i[τn−i

]n

and (PS−1

)n=∑i≥0

n

n− i[τn−i

]nAi and

bn (x) = x∑i≥0

n

n− i[τn−i

]nAian (x) /x = x

∑i≥0

[τn−i

]nan−i (x) /x.

2.4.3 Exercises

2.4.1. Use the Transfer Formula to show that the basic sequence for E−dA 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

∇ = B − E−2B2.


Use the Transfer Formula (Theorem 2.4.2) to show that

bn (x) =

n∑i=1

(i

n− i

)x

i(−1)

n−i(

3i− 2n− 1 + x

i− 1

),

and use formula (2.33) to find the generating function((1 +

√1− 4t2 + 4t3

)/ (2 (1− t))

)xfor (bn).

2.4.3. Show: θ∇ = xE−1

2.4.4. Show: If T =∑k≥0 τkDk then T ′ =

∑k≥0 τkkDk−1.

2.4.5. Prove the product rule of differentiation for the Pincherle derivative, (ST )′

=

S′T+ST ′. This implies (Pn)′

= P ′Pn−1+P(Pn−1

)′= P ′Pn−1+P (n− 1)P ′Pn−2.

2.4.6. Show: If A is a delta operator and T =∑k≥0 τkA

k ∈ ΣA then T ′A = ddAT .

2.4.7. Show:(Ek)′

∆= kEk−1.

2.4.8. Show directly, without the Pincherle derivative, that

bn (x) =

n∑k=0

[τk]nak (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 abovethe x-axis. Give the horizontal step the weight ω ∈ C. The number of pathsS (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 ↑,→, arecounted by dn (m), say, where dn+m (m− n) counts the corresponding Schröderpaths. We have dn (n) = S (n;ω). The path counts dn (m) can be extended to aSheffer sequence (dn). Use (2.33) to show that

∑n≥0

bn (x) tn =

(1 + ωt

1− t

)xis 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≥0

S (n;ω) tn =∑n≥0

bn (n+ 1) tn =2

1− ωt+√

1− 2t (ω + 2) + t2ω2.

Chapter 3

Applications

The binomial theorem (2.14) together with Abelization is at this time our only toolfor finding Sheffer sequences that satisfy certain initial conditions. With the helpof the Functional Expansion Theorem 3.1.4 we will enlarge the pool of problemsthat can be solved; the theorem contains the binomial theorem as a corollary. Todemonstrate 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 matrixS = (si,j)i,j≥0 has elements which are not only coeffi cients of formal power seriesgenerating Sheffer sequences, they are also values of Sheffer sequences. More pre-cisely, these are the matrices where s2,1 6= s1,0s1,1/s0,0. This is a consequence ofTheorem 2.2.11.

In the subsection on determinants of Hankel matrices, Finite Operator Cal-culus allows for a systematic description of a rather simple case. However, thegeneral case is diffi cult, and only some special cases have been considered.

Finally, we look at the relationship between Umbral Calculus and FiniteOperator Calculus. Both theories are similar; one can as well view Finite OperatorCalculus as an application of Umbral Calculus. Only an abbreviated introductioninto 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. Bylinear extension, functionals are defined on all of F [x] if we define them just on abasis of F [x]. For example, if we know 〈L | xn/n!〉 =: an for all n ≥ 0, then L isdefined as a functional on F [x],

〈L | p〉 =

deg p∑k=0

πk⟨L | xk/k!

⟩=

deg p∑k=0

πkak

80 Chapter 3. Applications

if p (x) =∑deg pk=0 πkx

k/k!. The sequence of coeffi cients (in F) a0, a1, . . . we will storeas λ(t) =

∑n≥0 ant

n = Lext, and we see immediately that F [x]∗ ' F [[t]] with

respect to addition. For convenience we will make this also into an isomorphismrespecting products by introducing the “right”product of functionals,

〈L ∗N | xn〉 :=

n∑k=0

(n

k

)⟨L | xk

⟩ ⟨N | xn−k

⟩,

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

L∗k for the k-th power under this product. The multiplicative unit in F [[t]] is1 = e0t, hence the evaluation at 0, Eval0, is the multiplicative unit in F [x]

∗. Wecall Eval0 also the identity functional. A linear functional L has a reciprocal (w.r.t.∗-multiplication), iff 〈L | 1〉 has a reciprocal in F.

Thus the functionals F [x]∗ and the translation invariant operators ΣD are

both isomorphic to F [[t]]. This means that everything we did with operators inthe preceding sections we could as well have achieved with functionals. Indeed,Roman’s Umbral Calculus [76] is based on functionals. However, we found theinterpretation of a recursion as an operator equation more natural, and chosethe operator approach. On the other hand, we like to view initial conditions onpolynomial 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 ΣD) out of it by consideringλ (D), the operator associated to L. Note that λ (t) ext = λ (D) ext, because tnext =Dnext. If L has a reciprocal, then λ (D) is invertible. For example, let L = Evalafor some a ∈ F. We first get λ (t) = Evala e

xt = eat, and then the associatedoperator λ (D) = eaD = Ea, the translation by a.

The notation λ (D) for the associated operator of L stresses the dependentson D more than necessary. It holds that λ (D) =

∑k≥0 〈L | bk〉Bk for any delta

operator B with basic sequence (bn) (Exercise 3.1.1). We will also use the notationop (L) for the operator associated to L. For example, if we want λ (D) = op (L) =∆, we must define

⟨L |(xk

)⟩= δ1,k for all k ≥ 0. Alternatively, from ∆ = ED − 1

follows 〈L | xn〉 = 1 for all n ≥ 1, and 〈L | 1〉 = 0.We now prove a more technical lemma, that sheds some light on the “pur-

pose”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〉

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 =∑n≥0

〈L ∗N | xn〉 tn/n! =∑n≥0

〈N | λ (D)xn〉 tn/n!

= Nλ (D) ext,

which follows from λ (t) ext = λ (D) ext.

Corollary 3.1.2. Let T ∈ ΣD, and J, L ∈ F [x]∗. If 〈J | p〉 = 〈L | Tp〉 for all

p ∈ F[x], then op (J) = op (L)T .

Proof. Let N be the functional such that T = op (N). The above Lemma showsthat 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 L, thetransform of L. Looking closer at

Lext =∑n≥0

⟨L | x

n

n!

⟩tn =

∑n≥0

xn

n!Ltn =

x0

0!λ (t)

shows that the t-operator L maps 1 to λ (t), and Ltn = 0 for all n > 0. However,we are using λ (t) as the multiplication operator M (λ). The transform of M (λ) isthe x-operator λ (D) = op (L). The transform of L is the functional L, of course.

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 initialconditions 〈L | sn〉 = yn, say, for some functional L ∈ F [x]

∗.Let us begin with the functional L and its isomorphic power series λ (t) =

Lext. Suppose λ (t) has a reciprocal 1/λ (t). Define the Sheffer sequence (ln (x))for the delta operator B = β−1 (D) by the generating function

∑n≥0 ln (x) tn =

exβ(t)/λ (β (t)). Thus∑n≥0 ln (0) tn = 1/λ (β (t)). Applying L to the generating

function gives

L1

λ (β (t))exβ(t) = C (β)

1

λ (t)Lext = C (β) 1 = 1

and therefore 〈L | ln〉 = δ0,n. The identity

ln (x) =∑k≥0

〈L | ln−k〉 lk (x) =∑k≥0

⟨L | Bkln

⟩lk (x)

is trivial, but it shows that every polynomials p ∈ F [x] can be written as p (x) =∑k≥0

⟨L | Bkp

⟩lk (x), because (ln) is a basis. Using the Sheffer operator S : bk 7→

lk for all k ≥ 0 we can write this as p (x) =∑k≥0

⟨L | Bkp

⟩Sbk (x). However,

λ (D)−1

= S, as can be seen from

λ (D)1

λ (β (t))exβ(t) = λ (D)C (β)

1

λ (t)ext = C (β)λ (D)

1

λ (D)ext = exβ(t),


hence λ (D) ln = bn. This means that

p (x) =∑k≥0

⟨L | Bkp

⟩λ (D)

−1bk (x) .

The special sequence (ln) is no longer needed! We apply this identity to the Sheffersequence (sn) for B,where the initial conditions are given in terms of L, and obtainthe functional expansion theorem, writing op (L) for λ (D):

Theorem 3.1.4. If L is a functional such that 〈L | 1〉 6= 0, and (sn) is a Sheffersequence and (bn) the basic sequence for the same delta operator, then

sn (x) =

n∑k=0

〈L | sn−k〉 op (L)−1bk (x) ,

where op (L) is the invertible operator

op (L) =∑n≥0

⟨L | x

n

n!

⟩Dn =

∑k≥0

〈L | ak〉Ak

for any delta operator A with basic sequence (an).

The first thing we want to do with this Theorem is showing that it impliesthe binomial theorem. Let 〈L | sn〉 = sn (y) for some y ∈ F. Thus L = Evaly, andwe have seen in section 3.1 that op (L) = Ey. Hence

sn (x) =∑k≥0

sn−k (y)E−ybk (x) =∑k≥0

sn−k (y) bk (x− y) .

Using the same notation as in the functional expansion Theorem, we find thegenerating function of (sn),

∑n≥0

sn (x) tn =exβ(t)

∑k≥0 〈L | sk〉 tk∑

n≥0 〈L | bn〉 tn(3.1)

Remark 3.1.5. In a typical application of the functional expansion theorem we havesome 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 insuch a recursion; for example, sn (b) = γn + a0sn (x0) + a1sn−1 (x1) becomes thefunctional

L = Evalb−a0 Evalx0 −a1 Evalx1 B

where 〈L | sn〉 = γn, but 〈L | 1〉 = 1−a0 has to be a unit so that L has a reciprocal.Hence the “defining recursion” for (sn) cannot be used if it is of the type wediscussed in the Transfer Theorem 2.4.2,

Asn =∑j≥1

TjBjsn


thusL = EvalbA− Evalb

∑j≥1

TjBj

for some b ∈ F, because

〈L | 1〉 = 〈Evalb | A1〉 − 〈Evalb | T1B1〉 = 0.

3.1.1 Some Applications of the Functional Expansion Theorem

We will look at some problems that require initial conditions different from theinitial value problems we have seen so far.

Example 3.1.6. Suppose we are asked to solve the system of differential equationssn (x) = s′n (x) under the condition

∫ 1

0sn (x) dx = 1 for all n ≥ 0. The integral

is a functional on F [x], we have∫ 1

0extdx = et−1

t =: λ (t), hence λ (D)−1

=

op (L)−1

= D/(eD − 1

)=∑n≥0

Bnn! D

n, where B0 = 1, B1 = −1/2, B2 = 1/6,B3 = 0, etc., are the Bernoulli numbers. Note that B2n+1 = 0 for all n ≥ 1

(because t (et − 1)−1

+ t/2 is an even function). The delta operator D has basicsequence (xn/n!), hence

sn (x) =

n∑k=0

λ (D)−1 x

k

k!=

n∑k=0

k∑j=0

Bjj!Dj x

k

k!=

n∑k=0

k∑j=0

Bjj!

xk−j

(k − j)!

=

n∑k=0

xk

k!

k∑j=0

(k

j

)Bjx

−j =

n∑j=0

Bjj!

n−j∑k=0

xk

k!.

For the first few n we get s0 = 1, s1 (x) = x + 1/2, s2 (x) =(x2 + x+ 7/6

)/2,

s3(x) = x3

6 + x2

2 +x+1− 12

(x2

2 + x+ 1)

+ 112 (x+ 1), etc. Using that

∑nk=0

(nk

)Bk =

(−1)nBn we find

sn (1) =

n∑k=0

(−1)k

k!Bk.

If we do the same for the forward difference operator, i.e., looking for asolution (rn (x)) such that ∆rn = rn−1, and

∫ 1

0rn (x) dx = 1 for all n ≥ 0, then

λ (·)−1= ln (1 + ∆) /∆ =

∑n≥0

(−1)n ·n/ (n+ 1)

and

rn (x) =

n∑k=0

λ (∆)−1

(x

k

)=

n∑k=0

k∑j=0

(−1)j

j + 1

(x

k − j

)=

n∑j=0

(x

n− j

) j∑k=0

(−1)k

k + 1.

Hence limn→∞ rn (0) = ln 2, and limn→∞ rn (1) = 2 ln 2.


Example 3.1.7. In chapter 2 we posed the problem of finding the Sheffer sequence(pn) that satisfies the initial condition pn (1− n) =

∑n−1i=0 pi (n− 2i) for all n ≥ 1,

p0 (−1) = 1, and follows Pascal’s recursion pn (x) = pn (x− 1) + pn−1 (x), whichimplies (see also Exercise 2.1.1) that we are looking for a Sheffer sequence for ∇with basic polynomials bn (x) =

(n−1+xn

).

m 1 4 12 35 107 3442 1 3 8 23 72 2371 1 2 5 15 49 1650 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

Pascal’s recursion with pn (1− n) =∑n−1i=0 pi (n− 2i)

We have for n ≥ 1 the information that 0 = pn (1− n)−∑n−1i=0 pi (n− 2i), which

implies that

0 = 〈Eval1 |pn (x− n)〉 −⟨

Eval0 |n∑i=1

E2i∇ipn (x− n)

⟩.

Hence we define a functional L such that

〈L | pn (x− n)〉 = 〈Eval1 | pn (x− n)〉 −⟨

Eval0 |E2∇

1− E2∇pn (x− n)

⟩= 0

for n ≥ 1, and 〈L | p0〉 = 1. The polynomials (pn (x− n))n≥0 are a Sheffer se-

quence for E1∇ = ∆, with op (L) = E1− E1∆1−E1∆ =

E1−(E2+E1)∆

1−E1∆ , and op (L)−1

=E−1(1−E1∆)1−(E1+1)∆ . Hence we derive from the Functional Expansion Theorem that

pn (x− n) = op (L)−1

(x

n

)= E−1

(1− E1∆

)∑j≥0

(E1 + 1

)j∆j

(x

n

)

=(1− E1∆

) n∑j=0

(E1 + 1

)j (x− 1

n− j

)

=

n∑j=0

j∑i=0

(j

i

)(i+ x− 1

n− j

)−

n∑j=1

j−1∑i=0

(j − 1

i

)(i+ x

n− j

)

=

(x− 1

n

)+

n−1∑j=0

j∑i=0

(j

i

)(i+ x− 1

n− 1− j

)


thus pn (x) =(n−1+xn

)+∑n−1j=0

∑ji=0

(ji

)(n−1+i+xn−1−j

).

Example 3.1.8. Also in chapter 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 exten-sion.

m 1 9 30 50 55 55 558 1 8 23 33 34 34 347 1 7 17 21 21 21 216 1 6 12 13 13 13 135 1 5 8 8 8 8 84 1 4 5 5 5 5 53 1 3 3 3 3 3 32 1 2 2 2 2 2 21 1 1 1 1 1 1 10 1 1 1 1 1 1 1

0 1 2 3 4 5 nFn (0) = 1

m 1 9 30 50 55 55 558 1 8 23 33 34 34 347 1 7 17 21 21 21 216 1 6 12 13 13 13 135 1 5 8 8 8 8 94 1 4 5 5 5 4 113 1 3 3 3 4 -2 262 1 2 2 1 6 -15 691 1 1 2 -2 13 -43 1670 1 0 3 -7 28 -98 364

0 1 2 3 4 5 nThe 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 Theorem2.1.1 shows that there is a sequence of polynomials such that fn (m) = Fn (m) forall m ≥ n, because b1 = −2 ≥ n−1−n−1 for all n ≥ 0 in that Theorem. It is easyto show that for the remaining 0 ≤ m < n also holds Fn(m) = Fn−1 (m). Thereforewe define the functional L on fn (x+ n), a Sheffer polynomial for E−1B, as

〈L | fn (x+ n)〉 = fn (n)− fn−1 (n)

= 〈Eval0 | fn (x+ n)〉 − 〈Eval0 | Bfn (x+ n)〉 ,

where ∇ = E−2B. The basic polynomials for E−1B = E1∇ = ∆ equal(xn

). We

find op (L) = 1−B, and from the Functional Expansion Theorem follows

fn (x+ n) =1

1− E1∆

(x

n

)=

n∑k=0

Ek∆k

(x

n

)=

n∑k=0

(x+ k

n− k

)The generating function is also easy to check: For x ≥ 0 holds∑

n≥0

Fn (n+ x) tn = (1 + t)x/(1− t− t2

).

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 patternrruu follow the recursion

d (n,m) = d (n,m− 1) + d (n− 1,m)− d (n− 2,m− 2)


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 handside belonging to the polynomial of degree n− 1.

m 1 8 30 77 163 306 519 794 7947 1 7 23 53 103 178 275 275 06 1 6 17 35 62 97 97 05 1 5 12 22 35 35 04 1 4 8 13 13 03 1 3 5 5 02 1 2 2 0 Avoiding the pattern1 1 1 0 rruu0 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)

= 〈Eval1 | dn (x+ n)〉 − 〈Eval0 + Eval1 | Bdn (x+ n)〉 ,

where Bdn (x) = dn−1(x) for all n ≥ 0. We know that 〈L | dn (x+ n)〉 = δ0,n, andop (L) = E1 −

(1 + E1

)B, thus

op (L)−1

=E−1

1− (1 + E1)E−1B=∑k≥0

k∑j=0

(k

j

)Ej−1E−kBk.

Apply the Functional Expansion Theorem 3.1.4 to (dn (n+ x))n≥0, a Sheffer se-

quence for E−1B, with (still unknown) basic polynomials b(1)n (x)

dn (n+ x) =∑k≥0

k∑j=0

(k

j

)Ej−1E−kBkb(1)

n (x) =

n∑k=0

k∑j=0

(k

j

)Ej−1b

(1)n−k (x) .

The basic polynomials for B we found in Exercise 2.4.2 ,

bn (x) = x

n∑i=1

(i

n− i

)(−1)

n−i(

3i− 2n− 1 + x

i− 1

)/i

hence dn (x) =∑nk=0

∑kj=0

(kj

)Ej−1b

(1)n−k (x)

=∑nk=0

∑kj=0

(kj

)Ej−1

∑n−ki=1

(i

n−k−i)x(−1)n−k−i

i

(3i−n+k−1+x

i−1

)


= 2n +∑n−1j=0 (x− n+ j − 1)

∑n−1k=j

(kj

)∑n−ki=1

(i

n−k−i) (−1)n−k−i

i

(3i−2n+k+j−2+x

i−1

)for all x ≥ n + 1. We find d (n) from dn−1 (n). The generating function ofdn (x+ n) can be calculated from (3.1),

∑n≥0

dn (x+ n) tn =exβ1(t)

∑k≥0 〈L | dk (x+ k)〉 tk∑n≥0

⟨L | b(1)

n

⟩tn

=exβ1(t)

Eval1 exβ1(t) − (Eval0 + Eval1)E1 (E−1B) exβ1(t)

=exβ1(t)

eβ1(t) − (Eval0 + Eval1) te(x+1)β1(t)=

e(x−1)β1(t)

1− t− teβ1(t).

Now we have to determine β1 (t) from e−tβ−1 (t) = β−11 (t). We have β (t) =

ln 1+√

1−4t2+4t3

2(1−t) (see Exercise 2.4.2) and find (by taking the root that gives us apower series of order 1)

β−1 (t) =1

2e2t − 1

2

√e4t + 4et − 4e2t.

Hence β−11 (t) = 1

2et − 1

2

√e2t + 4e−t − 4, and solving again a quadratic equation

we get

eβ1(t) =1

2t

(1 + t2 −

√(1 + t2)

2 − 4t

).

Finally,

∑n≥0

dn (x+ n) tn =

(12t

(1 + t2 −

√(1 + t2)

2 − 4t

))x−1

12 − t−

12 t

2 + 12

√(1 + t2)

2 − 4t

for all x ≥ 1. We find the generating function for dn (n) as∑n≥0

dn (n) tn = 1 + t∑n≥1

dn−1 (n) tn−1 = 1 +2t

1− 2t− t2 +

√(1 + t2)

2 − 4t

(see [84]).

3.1.2 Exercises

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

λ(D) =

∑k≥0 〈L | sk〉Bk∑

n≥0 〈Eval0 | sn〉Bn.


Especially for the basic sequence (bn) of B holds

λ(D) =∑k≥0

〈L | bk〉Bk.

This also shows that the Sheffer sequence (sn) is uniquely defined by L, if λ (D) isinvertible , because ∑

n≥0

sn (0) tn =

∑k≥0 〈L | sk〉 tk

λ (β (t)).

3.1.2. Prove the generating function (3.1).

3.1.3. Prove thatn∑k=0

〈L | bk〉 sn−k (x) =

n∑k=0

〈L | sk〉 bn−k (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(i) ≥ 0 for i = 1, . . . , k − 1, . . . , and

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

sn (x) =

n∑j=0

(−1)j

j + 1

n∑k=j

(x

k − j

)is the solution to the difference equation sn (x+ 1) − sn (x) = sn−1 (x) under thecondition

∫ 1

0sn (x) dx = 1 for all n ≥ 0.

3.1.6. Use the Functional Expansion Theorem to show that

sn (x) =x+ 1− nx+ 1

(n+ x

n

)is the solution to the difference equation sn (x) − sn (x− 1) = sn−1 (x) under thecondition

∑ni=0 sn (i) = 1 for all n ≥ 0. See also Example 2.3.6.

3.1.7. Use the Functional Expansion Theorem to show that

sn (x) =

(x

n

)−(x+ 1

n− 1

)is the solution to the difference equation sn (x+ 1) − sn (x) = sn−1 (x) under thecondition

∑ni=0 sn−i (n+ i) = 1 for all n ≥ 0.

3.2. Diagonals of Riordan Matrices as Values of Sheffer Sequences 89

3.2 Diagonals of Riordan Matrices as Values of ShefferSequences

The following Riordan matrix B has been constructed with a first column bn,0 =

δn,0, and the k + 1-th column according to bn,k+1 =∑n−1j=0 bj,kCn−j (Corollary

2.3.17), where Ck stands for the n-th Catalan number. We know (Theorem 2.3.13)that the rows as coeffi cients of polynomials, bn,k =

[xk/k!

]bn (x) = [tn]β (t)

k,where

∑n≥0 bn (x) tn = exβ(t). However, it is striking that the numbers on the

diagonal and subdiagonals seem to be values of polynomials too!

bn,k 0 1 2 3 4 5 6 7 k

0 11 0 12 0 1 13 0 2 2 14 0 5 5 3 15 0 14 14 9 4 16 0 42 42 28 14 5 17 0 132 132 90 48 20 6 1n 0 429 429 297 165 75 27 7 1bn,k+1 =

∑n−1j=0 bj,kCn−j by Corollary 2.3.17

Theorem 2.2.11 tells us that this must happen because b2,1 6= 0. In that case there

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 arevalues of the polynomials bn (x) with generating function

β (t) = lnβ (t)

t.

Example 3.2.1. The Riordan matrix B = (bn,k) above is generated by σ (t) = 1

and by bn,i+1 =∑n−1k=0 sk,iCn−k, hence β (t) =

∑n≥1 Cnt

n (Corollary 2.3.17), and

bn,k = [tn]β (t)k. Here Cn =

(2nn

)/ (n+ 1) is the n-th Catalan number. We saw in

Example 2.3.8 that β (t) =((

1−√

1− 4t)/2)k. The matrix shows that b2,1 = 1.

We find β (t) = ln((

1−√

1− 4t)/ (2t)

), and∑

n≥0

bn (x) tn/n! =((

1−√

1− 4t)/ (2t)

)x.

In Example 2.3.7 we saw that

bn (x) =

(2n+ x

n

)x+ 1

n+ x+ 1−n−1∑i=0

Cn−i−1

(2i+ x

i

)x+ 1

i+ x+ 1.


Therefore, bn,k = bn−k (k) =(2n− kn− k

)k + 1

n+ 1−n−k−1∑i=0

(2 (n− k − 1− i)n− k − 1− i

)1

n− k − i

(2i+ k

i

)k + 1

i+ k + 1.

In addition to the condition b2,1 6= 0, Theorem 2.2.11 requires that bn,n = 1

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 fromthe definition of a Riordan matrix B that bn,n = b0,0a

n0 for n > 0. If b0,0 = c 6= 0,

then divide every element of the matrix B by c. Now all main diagonal elements ofthe new matrix are an0 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

exβ(t) =1

c

(β (t)

t

)x(3.2)

(see Exercise 3.2.1). Next we assume that bn,n = an0 for all n, and that a0 6= 1.Depending on the matrix (see Example 3.2.4) we may choose one of the twomethods.

1. (Column standardization) Divide the elements of the k-th column by ak0 .Theresult is a new matrix with elements a−k0 bn,k, satisfying the condition thatthe main diagonal equals one everywhere. We find bn,k = ak0 bn−k (k), where

the basic sequence(bn (x)


exβ(t) =

(β (t)

a0t

)x(3.3)

(see Exercise 3.2.2).

2. (Row standardization) Divide the elements of the n-th row by an0 .The result isa new matrix with elements a−n0 bn,k, satisfying the condition that the maindiagonal is all ones. We find bn,k = an0 bn−k (x), where the basic sequence(bn (x)


exβ(t) =

(β (t/a0)

t

)x(3.4)

(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) ,


how does it help us finding bn,k? The relationship exβ(t) = (β (t) /t)x is not very

useful, as the example above shows, because expanding bn,k = [tn]β (t)k or bn (k) =

[tn] (β (t) /t)k is basically the same work. This will happen if we know c1, c2, . . .

in Corollary 2.3.17 such that bn,i+1 =∑n−1k=0 bk,icn−k. If the matrix is given by the

a-sequence, bn+1,i+1 =∑n−ik=0 bn,i+kak for all 0 ≤ i ≤ n, then we only get β−1 (t)

(Theorem 2.3.13), and β (t) must be obtained through an inversion process, whichmay not be possible or ugly. However, it can be easily checked that in this situation

∆ =

∞∑k=1

akEkBk (3.5)

(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

kind sn,k = [tn](σ (t)β (t)

k), but can be brought into a form where the diagonal

are values of a Sheffer sequence. Because

σ (t)β (t) = (σ0 + σ1t+ . . . )(β1t+ β2t

2 + . . .)

= σ0β1t+ (σ1β1 + σ0β2) t2 + · · · = s1,1t+ s2,1t2 + . . .

we see that β2 6= 0 is equivalent to s2,1 6= σ1β1 = s1,0s1,1/s0,0 = s1,0a0 = l0a0.

Corollary 3.2.2. Let S = (sn,k) be a Riordan matrix such that


k)

for a delta series β and a power series σ that admits a reciprocal. There exists a

Sheffer sequence (sn (x)) such that sn,k =(s1,1s0,0

)nsn−k (k) with generating func-

tion∑n≥0 sn (x) tn = σ (t) exβ(t) iff s2,1 6= s1,0s1,1/s0,0. In this case,

σ (t) exβ(t) = σ

(s0,0

s1,1t

)(β (s0,0t/s1,1)

t

)x.

There also exists a Sheffer sequence (sn (x)) such that sn,k =(s1,1s0,0

)ksn−k (k)

with generating function∑n≥0 sn (x) tn = σ (t) exβ(t) iff s2,1 6= s1,0s1,1/s0,0. In

that case,

σ(t)exβ(t) = σ (t)

(s0,0β (t)

s1,1t

)x.

Remark 3.2.3. We obtain the same operator equation for (sn (x)) as for(bn (x)

),

∆ =

∞∑k=1

akak−10 EkBk. (3.6)


(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,or we can use the l-sequence to determine sn (0) =

∑n−1k=0 a

k0 lksn−1−k (k); in this

case we define the functional

〈L | sn〉 = 〈Eval0 | sn〉 −∞∑k=0

ak0 lk

⟨Evalk | Bk+1sn

⟩= 0

for all n > 0, and 〈L | s0〉 = s0,0 6= 0. The Functional Expansion Theorem tells usthat

sn (x) =s0,0

1−∑∞k=0 a

k0 lkE

kBk+1bn (x) . (3.7)

Example 3.2.4. Consider the following lattice path problem: Instead of taking stepsfrom →, ↑, 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).For example, D (0, 0) = 1, D (0,m) = 2m−1 for m ≥ 1, D (1, 1) = 3, D (2, 2) = 26,and D (n,m) = D (m,n). We have

D (n,m) =

n∑i=0

m−1∑j=0

D (i, j) +

n−1∑i=0

D (i,m)

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

D (n,m) = 2D (n,m− 1) +

n∑i=1

2iD (n− i,m− 1) (3.8)

for m ≥ n ≥ 0, m ≥ 2. We make a Riordan matrix S = (sn,k) out of D (n,m) bydefining sn,k = D (n− k, n) for all 0 ≤ k ≤ n, with the exception of s0,0 which wehave to define as 1/2 for making S Riordan (this fixes the problem that (3.8) doesnot hold for m = n = 1).

sn,k 0 1 2 3 4 5 k

0 1/21 3 12 26 8 23 252 76 20 44 2568 768 208 48 85 26 928 8016 2208 544 112 16n 287 648 85376 23776 6080 1376 256 32

sn,k = 2sn−1,k−1 +∑n−kj=1 2jsn−1,k−1+j

Obviously, the condition of Corollary 3.2.2 is satisfied, 8 = s2,1 6= s1,0s1,1/s0,0 = 6.Equation (3.8) implies that sn+1,i+1 = 2sn,i +

∑n−ik=1 2ksn,i+k for all 0 ≤ i ≤ n,


hence S has the a-sequence a0 = 2, ak = 2k for all k ≥ 1. This means that weknow β−1 (t), and it is easy to find β (t) from β−1 (t) (see Exercise 3.2.6), but wewant to follow the alternative route outlined in Remark 3.2.3. From (3.8) followsthat sn+1,0 = 6sn,0 +

∑nk=1 2k+2sn,k (see Exercise 3.2.5), and therefore S has the

l-sequence l0 = 6, lk = 2k+2 for k ≥ 1. We divide the columns by powers of 2,because we get all integers this way, except on the diagonal.

0 1 2 3 4 5 k

0 1/21 3 1/22 26 4 1/23 252 38 5 1/24 2568 384 52 6 1/25 26928 4008 552 68 7 1/2n 287648 42688 5944 760 86 8 1/2

sn,k/2k = sn−k (k)

Equation (3.6) shows that ∆ =∑∞k=1 aka

k−10 EkBk = 2EB/

(1− 4EB

), hence

b(−1)n (x) =

n∑i=1

[2iti

(1− 4t)i

]n

(x

i

)=

n∑i=0

(n− 1

n− i

)22n−i

(x

i

)

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 allm,n ≥ 0. This means for the Sheffer sequence (sn) that sn−k (k) = 2−2ksn (−k)for all k ≤ n. Hence 1

2 = s0 (n) = 2−2nsn (−n), and by the binomial theorem forSheffer sequences (2.14) follows

sn (x− n) =

n∑k=0

22k−1b(−1)n−k (x) =

n∑i=0

22n−i−1

(x

i

) n−i∑k=0

(n− k − 1

n− k − i

)

=

n∑i=0

22n−i−1

(x

i

)(n

i

).

Therefore,

sn,k = 2ksn−k (k) =

n−k∑i=0

22n−k−i−1

(n

i

)(n− ki

).

3.2.1 Exercises

3.2.1. Prove formula (3.2).





3.2.5. In Example 3.2.4 show that the a-sequence equals a0 = 2, ak = 2k for allk ≥ 1, and the l-sequence equals l0 = 6, lk = 2k+2 for k ≥ 1.

3.2.6. In Example 3.2.4 show that the a-sequence implies β−1 (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).

β−1 (t) =t∑

n≥0 antn

=t

2 +∑n≥1 2ntn

=1

2

t (1− 2t)

1− t .

By Theorem 2.3.13

σ(β−1 (t)

)=

s0,0

1− β−1 (t)∑n≥0 lnt

n=

s0,0

1− 6β−1 (t)− 8tβ−1 (t) / (1− 2t)

=1

2

1− t1− 4t+ 2t2

.

Now replace t by β (t).

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


k).

Eliminate the first m rows and columns of S and call the new matrix R, rn,k =sn+m,k+m. Show that R has the same a-sequence as S, and that

∑n≥0 rn,0t

n =

σ (t) (β (t) /t)m.

3.3 Determinants of Hankel Matrices

A Hankel matrix is constant along parallels to the second diagonal; more precisely(h′i,j)i,j=0,...,n−1

is a Hankel matrix iff h′i,j = h (i+ j), i.e., h′i,j depends only on

i+ j. If A is any square matrix, then AAT is a Hankel matrix.C. Radoux [71], [70], and Martin Aigner [1] considered matrices A = (am,k)m,k≥0such that am,m = 1, am,k = 0 for m > k, and

am+n,0 =∑k

am,kan,k (3.9)

for all m,n ≥ 0, i.e., the matrix is lower triangular, with ones on the diagonal, andfor all choices of m and n the inner product of the m-th and n-th row is the sameas long asm+n remains the same. We call the sequence an,0, n ≥ 0, an unweightedRadoux sequence. Not every sequence can be an unweighted Radoux sequence; we

3.3. Determinants of Hankel Matrices 95

have for even indices a2m,0 =∑mk=0 a

2m,k, which implies a dependence on previous

terms. For example, a1,0 can be freely selected, but then a2,0 = a21,0 + 1. Next,

a3,0 = a31,0 + a1,0 + a2,1 is arbitrary, because a2,1 can be chosen accordingly, but

a4,0 =(a2

1,0 + 1)2

+ a22,1 + 1 is determined.

Let An be the restriction of A to the (n+ 1)×(n+ 1) matrix (am,k)0≤m,k≤n,thus detAn = 1. Because of the inner product property (3.9), the Hankel matrixAnA

Tn = (ai+j,0)0≤i,j≤n has the determinant det (ai+j,0)0≤i,j≤n = 1. More inter-

esting 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 =

det

a10 1 · · · 0 01 a21 − a10 · · · 0 00 1 · · · 0 0· · · · · · · · · · · · . . .0 0 · · · 1 00 0 · · · an−1,n−2 − an−2,n−3 10 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 Aignerpoints out, the fact that det

(AnA

Tn

)= 1 together with the numbers dn ( n ≥ 1)

uniquely determine the matrix A. Therefore, we can ask for the matrix A, giventhe sequence of determinants d1, d2, . . .

Let cn (k) = an+k,k; in this notation ∇cn (k) = skcn−1 (k) + cn−2 (k + 1) forall k ≥ 1 (see (3.12) below), thus sk = ∇c1 (k) = ak+1,k − ak,k−1. For k = 0 wehave the initial values cn (0) = c1(0)cn−1(0)+ cn−2 (1). We assume that cn (k) = 0for all n < 0. The numbers c1 (k), k = 0, 1, . . . ,completely determine the matrixA.

A variation of the condition (3.9) appears in Zhang and Feng [103], introduc-ing a nonzero sequence fk such that f0 = 1. In an abuse of standard notation wewill define (only in this section!) fk! := Πk

i=0fi. Define the matrix A by am,k = 0for k > m, amm = 1, and

ai+j,0 =∑k≥0

ai,kaj,kfk! (3.11)

for all i, j ≥ 0 (a weighted inner product). If we define An := (ai,kfk!)0≤i,k≤n,then

det (ai+j,0)0≤i,j≤n = det(AnA

Tn

)= fn!!,


where fn!! = Πnk=0Πk

i=0fi = Πnk=1f

n+1−kk . 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 6= 0 for all k ≥ 0, such that an = an,0 =

∑k≥0 ai,kaj,kfk! for

all i+ j = n.

By defining the weights to be nonzero, we are still excluding some sequencesfrom being weighted Radoux sequences, because a2m −

∑m−1k=0 a2

m,kfk! = fm! 6= 0.

If we would allow fk = 0, we would reduce the nonzero columns of the matrix Ato the first k columns. For example, the Fibonacci sequence can be obtained thisway, 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 theweights are 1.

Lemma 3.3.1. A lower triangular matrix A = (ai,j)i,j≥0 with diagonal elementsequal 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 thei-th row of An, and ri the i-th row of An := (ai,kfk!)0≤i,k≤n. Then

rn rl =∑k≥0

an,kal,kfk! =∑k≥0

(fk+1an−1,k+1 + an−1,k−1 + skan−1,k) al,kfk!

=∑k≥0

(fk+1!an−1,k+1 + fk!an−1,k−1 + fk!skan−1,k) al,k

=∑k≥0

(al,k−1fk! + al,k+1fk+1! + fk!skal,k) an−1,k

=∑k≥0

(al,k−1 + al,k+1fk+1 + skal,k) an−1,kfk!

Hence the condition rn rl = rn−1 rl+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


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

fn−1!! det

s0 1 0 · · · 0f1 s1 1 0 · · · 00 f2 s2 1 · · · 0

· · ·fn−1 sn−1

= fn−1!!dn (3.14)

(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 ands0 = 1, s1 = −1, and sk = 0 for k ≥ 2, then

A3 =

1 0 01 1 0 02 0 1 02 2 + f2 0 1

Hence a4,0 = a2

2,0 + 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, andfk+1 equals a constant u 6= 0 for all k > 1. Hence c1(k) = sk + a, where we sets0 = a1,0 =: a. The weight f1 is another constant free to choose; for reasons thatwill appear later we set f1 = 2vu, for v 6= 0. We obtain the recursion

∇cn (k) = scn−1 (k) + ucn−2 (k + 1)

with initials conditions

cn (0) = acn−1(0) + 2uvcn−2 (1)


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

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) that1− E−1 = sB + uE1B2, thus

E =1

2uB2

(1− sB −

√(1− sB)

2 − 4uB2

)and therefore, by (2.33),

b (x, t) =∑n≥0

bn (x) tn =

1− st−√

(1− st)2 − 4ut2

2ut2

x

= 2x(

1− st+

√(1− st)2 − 4ut2

)−xwhere (bn) is the basic sequence for B. The initial conditions are described by thefunctional

L = Eval0−aEval0B − 2uvEval1B2

giving 〈L | bn〉 = δ0,n. It follows from the Functional Expansion Theorem 3.1.4that ∑

n≥0

cn (x) tn =b (x, t)

1− at− 2uvt2b (1, t).

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

∑n≥0

an,0tn =

1

1− at− 2uvt2b (1, t)=

1

1− v + t (vs− a) + v

√(1− st)2 − 4ut2

(3.16)

=1− v + t (vs− a)− v

√(1− st)2 − 4ut2

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

The generating function of the numbers dn can be calculated using Proposition1.1.4, ∑

n≥0

dn+1tn =

a− 2uvt

1− st+ ut2. (3.17)


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

dn+1 =2−n−1

√δ

(a

((s+√δ)n+1

−(s−√δ)n+1

)− 4uν

((s+√δ)n−(s−√δ)n))

=2−n−1

√δ

((a(s+√δ)− 4uν

)(s+√δ)n−(a(s−√δ)− 4uv

)(s−√δ)n)

if the discriminant δ := s2 − 4u 6= 0. If s2 = 4u then 1 − st + 4ut2 = (st− 2)2/4

anddn+1 = (as (n+ 1)− 4uνn) 2−nsn−1.

So we can identify the Radoux sequence (an,0) as values of a Sheffer sequence withgenerating function (3.16) if

det (ai+j,0)0≤i,j≤n = fn!! = 2nvnun(n+1)/2

and (dn) has a generating function of the above form.

Example 3.3.2 (Motzkin Numbers). We want to determine the Radoux sequence

(an,0) such that det (ai+j,0)0≤i,j≤n = (m+ 1)(n+12 ) 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 generatingfunction ∑

n≥0

dn+1tn =

1− (m+ 1) t

1 + (m+ 1) t (t− 1).

We find immediately from fn!! = (m+ 1)(n+1)n/2 that fk = m + 1 for all k ≥ 1.

Hence u = m+1 and 2v = 1. The generating function for dn is of the form a−2uvt1−st+ut2

if we choose a = 1, and s = m+ 1. The Radoux sequence (an,0) = (cn (0)) has thegenerating function

∑n≥0

an,0tn = 2

(1 + t (m− 1) +

√(1− (m+ 1) t)

2 − 4 (m+ 1) t2)−1

=1 + t (m− 1)−

√(1− (m+ 1) t)

2 − 4 (m+ 1) t2

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 se-quences. One such application, still closely related to Sheffer sequences, is givenin Exercises 3.3.6.


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

det (ai+j,0)0≤i,j≤n = (1 + µ)n

= 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

det (ai+j,0)0≤i,j≤n = 2nµn(n+1)/2

anddet (ai+j+1,0)0≤i,j≤n−1 = 2n−1µn(n−1)/2 (1 + µn) .

3.3.4. Determine the Radoux sequence (an,0) such that

det (ai+j,0)0≤i,j≤n = (−1)(n+12 ) (ω + 1)(

n+12 )

for some complex parameter ω 6= −1. Suppose

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

has the generating function∑n≥0

dn+1tn =

(ω + 1) (1 + t)

1− (ω + 2) t− (ω + 1) t2.

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 givensequence sn = an+1,n − an,n−1 and weights (fn), then an := anu

n is a Radouxsequence such that ai+j,0 =

∑k≥0 ai,kaj,k

(u2fk

)! and sn = an+1,n−an,n−1 = usn.

3.3.6. [71], [103] Show that the Bell numbers Bn =∑nk=0 S (n, k), the number of

partitions of an n-set , are a Radoux sequence. Let an,k = n!rn−k (k) /k!, where(rn) is the Sheffer sequence with generating function

∑n≥0

rn (x) tn = eet−1

(et − 1

t

)x

Show that n!rn (0) = Bn. Apply the operator x+ tDt to eet−1 ((et − 1) /t)

x to showthat

(n+ x) rn (x) = rn−2 (x+ 1) + xrn (x− 1) + (x+ 1) rn−1 (x) .


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

det (ai+j,0)0≤i,j≤n = n!!,

and the seconddet (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, givingthe Hankel determinants

det (ai+j,0)0≤i,j≤n = n!!µn(n+1)/2

anddet (ai+j+1,0)0≤i,j≤n−1 = (n− 1)!!µn(n+1)/2.

Show that an,0 =∑nk=0 S (n, k)µk for n ≥ 0 is the Radoux sequence in this case.

3.3.7. [21]Denote 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). Theweight of a path is the product of the weight of its steps (the diagonal steps haveweight 1). The weighted paths enumerated by D (n, 0;ω) are often called weightedgrand Motzkin paths. They have the generating function∑

n,m≥0

D (n, 0;ω) tn =1√

(1− ωt)2 − 4t2

If ω = 0, the path does not take horizontal steps; the counts are the central binomialcoeffi cients

(nn/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]∑k=0

(−1)k

(n− kk

)(xy)

k(x+ y)

n−2k=xn+1 − yn+1

x− y

we let ω = (x+ y) /√xy. Show that

|D (i+ j + 1, 0;ω)|0≤i,j≤n−1 = 2n−1 (xy)−n/2

(xn + yn) .

3.3.8. [21]The weighted paths in Exercise 3.3.7 become the Motzkin paths, if wemake the additional assumption that they stay weakly above the x-axis. Denotethe weighted Motzkin path to (n,m) by M (n,m;ω). TheMotzkin numbers have thegenerating function

∑n≥0

M (n, 0;ω) tn =

1− ωt−√

(1− ωt)2 − 4t2

2t2

.


Show that this generating function is of the type (3.16), and calculate the Hankeldeterminant |M (i+ j, 0;ω)|0≤i,j≤n−1. Show also that

|M (i+ j + 1, 0;ω)|0≤i,j≤n−1 = (xy)−n/2 x

n+1 − yn+1

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 UmbralCalculus, as Sylvester called it (see E.T. Bell’s “History of Blissard’s SymbolicCalculus”[11]). In Rota’s notation, an umbra α is a formal variable such that thefunctional E ∈ F [α]

∗ is defined by linear extension of

E [αn] = an

for some sequence of scalars a0 = 1, a1, a2, . . . in F. Here we follow Rota’s conven-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 differentumbra β evaluates differently, E [βn] = bn (we do not distinguish between dif-ferent evaluation functionals in our notation). We think of an umbral alphabetA = 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 form0 (c ∈ F). Two umbral polynomials are equivalent ('), if their evaluation (firstmoment) agrees.

The word “uncorrelated” is a statistical term, and there a more similaritiesbetween umbrae and random variables. Suppose X is a random variable such thatall moments of X exist, E [Xn] = xn, say, for all n ≥ 0. Then X “behaves” likethe umbra ξ with evaluation E [ξn] = xn. The converse is not true, of course; forexample, if x2 < x2

1, then X would have negative variance. However, we will keepthe language of random variables and their moments in this section, keeping inmind that random variables and umbrae just share a common intersection. Wetalk about the moment generating function mα (t) of an umbra α,

mα = mα (t) = E[eαt]

=∑n≥0

E [αn] tn/n! =∑n≥0

antn/n!,

even when α can not be interpreted as a random variable. We will reserve thesymbol E for the random variable case. A cursive E means evaluation in thissection.

3.4. Classical Umbral Calculus 103

We will follow the work of DiNardo and Senato [29] in this presentation. Alarge number of applications can be found in Gessel [37]. The connection betweenFinite 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 withthe same moments, there are sequences of (uncorrelated) umbrae α, α′, α′′, . . .all with the same moment generating function. Such umbrae are called similar ,α ≡ α′ in symbols,

α ≡ α′ ⇐⇒ αn ' (α′)n for all n ≥ 0.

All different and similar umbrae together make the (saturated) umbral alphabet,

which we call again A. For two similar umbrae α and α′ holds E[αi (α′)

j]

= aiaj ,

but E[αiαj

]= E

[αi+j

]= ai+j . Hence

E[(α+ α′)

n]= E

[n∑i=0

(n

i

)αi (α′)

n−i]

=

n∑i=0

(n

i

)E[αi]E[(α′)

n−i]

Both have the same moment sequence (an), hence

E[(α+ α′)

n]=

n∑i=0

(n

i

)aian−i.

The subscript n is the shadow (umbra) of the superscript n! Evaluating and distin-guishing between similar umbrae are new ideas Rota brought to Umbral Calculus.Blissard would have written (α+ α)

n when he meant∑ni=0

(ni

)aian−i. We write

k.α for the sum of k similar umbrae. This is different from kα, of course, becausemkα = E

[eαkt

], and

mk.α (t) = E[e(α+···+α′)t

]= E

[eαt]· · ·E

[eα′t]

= mα (t)k

= mkα

Note the analogy to the sum of k independent copies of a random variable! It isno surprise, that k.α has the same properties as an i.i.d. sequence X1, . . . , Xk; weonly mention j. (k.α) = (jk) .α. Of course, X1 · · ·Xk is also a random variable,with moments E [(X1 · · ·Xk)

n] = E [Xn

1 ] · · ·E [Xnk ], and we define the (auxiliary)

umbrae α.k to have moments

E[(α.k)n]

= akn

for all n ≥ 0.Umbrae are a strong symbolic language, but they are just in one-to-one cor-

respondence with formal power series. The strength is the result of the manydefinitions made for umbrae. We will give here only the most important ones. Forall n ≥ 0 we assume the existence of a unity umbra u,

E [un] = 1 hence mu = et,


and an augmentation ε,

E [εn] = δ0,n hence mε = 1.

We note that for any umbra γ holds

E [(ε+ γ)n] =

n∑i=0

(n

i

)E[εn−i

]E[γi]

= E [γn] .

Finally we need the singleton umbra χ,

E [χn] = 1 for n = 0, 1 and E [χn] = 0 else; hence ,mχ = 1 + t.

We saw that

mk.α = mα (t)k

=∑n≥0

1

n!

∑i1+···+ik=n

(n

i1, . . . , ik

)E[(tα1)

i1]· · ·E

[(tαk)

ik]

=∑n≥0

E [(k.α)n] tn/n!

(see also Remark 1.1.2). Hence for n ≥ 1

E [(k.α)n] =

∑i1+···+ik=n

(n

i1, . . . , ik

)E[αi11]· · ·E

[αikk]

=∑j≥1

(k

j

) ∑l1+···+lj=n; lν>0

(n

l1, . . . , lj

)al1 · · · alj (3.18)

Note that in the above sum for any j the largest possible lν such that l1+· · ·+lj = nis n− j + 1. Therefore,

E [(k.α)n] =

n∑j=1

(k − j + 1)j Bn,j (a1, . . . , an−j+1)

where

Bn,j (a1, . . . , an−j+1) =1

j!

∑l1+···+lj=n; lν>0

(n

l1, . . . , lj

)al1 · · · alj .

The functions Bn,j are called the partial Bell exponential polynomials in [73]. Wesee 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

mx.α (t) = mα (t)x.

The moments of x.α are in F [x]. If the first moment of α is different from 0 thenE [(x.α)

n] is a polynomial of degree n. More details are discussed in Exercises 3.4.3

- 3.4.5.


Example 3.4.1. (In this and the following examples we write terms from the FiniteOperator Calculus with a hat, like an (x), to distinguish them from umbral terms.)Let (an) be the basic sequence for the delta operator α−1 (D), with correspondingumbra α such that E [αn] = n!an (1), thus

mα (t) = eα(t).

It follows that E [(x.α)n/n!] = an(x) because

E[e(x.α)t

]= mα (t)

x= exa(t).

From xj = E[(x.u)

j]follows an (x.u) ' (x.α)

n/n!. The polynomial an (x.u) is a

simple example of an umbral polynomial. We will do more with this type of umbralpolynomials later.

Let γ be an umbra, and define the new factorial umbra γ(k) by its fallingfactorial moments

E[γ(k)

]= E [γ (γ − 1) · · · (γ − k + 1)] .

We defined

E [(x.α)n] =

∑j≥1

(x

j

) ∑i1+···+ij=n; il>0

(n

i1, . . . , ij

)E[αi1]· · ·E

[αij]

for all n > 0 as a consequence of substituting x for k in (3.18). We can take anumbra γ and substitute it for x, creating a new umbra γ.α with moment generatingfunction

mγ.α (t) =∑n≥0

E [(γ.α)n]tn

n!

= E

1 +∑n≥1

tn

n!

∑j≥1

(γ

j

) ∑i1+···+ik=n; il>0

(n

i1, . . . , ij

)αi1 · · ·αij

= 1 + E

∑k≥1

(γ

k

)∑m≥1

αmtm/m!

·k = E

∑k≥0

(γ

k

)(mα − 1)

k

(3.19)

= E[eγ log(1+mα(t)−1)

]= mγ (logmα (t)) . (3.20)

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(n)

]= 1(n) = δ1,n for n ≥ 1 we obtain

E[∑

n≥0 (u.α)n tn

n!

]= E [eαt], i.e., u acts a right and left identity in this product,

u.γ = γ = γ.u (3.21)


for all umbrae γ.If γ ≡ χ, the singleton umbra, then

E[χ(k)

]= (k − 1)! (−1)

k−1 , and (3.22)

E

∑k≥1

χ(k)tk/k!

= ln (1 + t)

for k ≥ 1, thus by (3.19)

E[(e(χ.α)t

)]=∑n≥0

E [(χ.α)n]tn

n!= 1 + E

∑k≥1

(−1)k−1

k

(eαt − 1

).k= 1 + logE

[1 + eαt − 1

]= 1 + logE

[eαt].

Of course we can get this much quicker from (3.20): mχ.α = mχ (logmα) = 1 +logmα.

Now let α ≡ χ in γ.α, then

E[(eχt).γ]

= E [(1 + t).γ

] = E [(1 + t)γ] or

mγ.χ = mγ (logmχ) = mγ (log (1 + t)) = E[eγ log(1+t)

]= E [(1 + t)

γ] ,

the γ-factorial umbra,(γ.χ)

n ' γ(n) (3.23)

for all n ≥ 0. Especially (χ.χ)n ' 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) 6≡ γ.α+ γ.η

This follows from (3.20), because for general γ

mγ.(α+η) (t) = mγ (logmα (t) + logmη (t)) 6= mγ (logmα (t))mγ (logmη (t))

= mγ.α+γ.η (t) .

For mγ = ect 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

mβ = E[eβt]

= E

∑n≥0

tn

n!

n∑k=0

S (n, k)β(k)

=∑n≥0

tn

n!

n∑k=0

S (n, k) = eet−1


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

mχ.β = 1+ logmβ = et

thusχ.β ≡ u ≡ β.χ (3.24)

We also have

mβ.α = mβ (logmα) = emα−1

mα.β = mα (logmβ) = mα

(et − 1

).

The Bell umbra can be used to define the composition umbra of α and γ asγ.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 inversecomposition umbra of α is denoted by α〈−1〉; it satisfies the relation α〈−1〉.β.α ≡ χ.

Lemma 3.4.2. If the inverse composition umbra of α is α〈−1〉, then

α〈−1〉.β.α ≡ χ ⇐⇒ α.β.α〈−1〉 ≡ χ

Proof. Let α〈−1〉.β.α. Then

α〈−1〉.β.α.β.α〈−1〉 ≡(α〈−1〉.β.α

).β.α〈−1〉 ≡ χ.β.α〈−1〉 ≡ (χ.β) .α〈−1〉

≡ u.α〈−1〉 ≡ α〈−1〉 ≡ α〈−1〉.u,

by (3.24) and (3.21), hence β.α.β.α〈−1〉 ≡ u (see Exercise 3.4.9) and

α.β.α〈−1〉 ≡ χ.β.α.β.α〈−1〉 ≡ χ.u ≡ χ.

So we have α.β.α〈−1〉 ≡ χ and α〈−1〉.β.α ≡ χ at the same time.We have

u〈−1〉 ≡ χ.χ (3.25)

for the inverse composition umbra of u, because

χ.χ =(u〈−1〉.β.u

).χ = u〈−1〉.β.χ = u〈−1〉.u = u〈−1〉

by (3.21) and (3.24).


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

E [(x.α)n] = n!an(x)

satisfies the generating function identity

mα (t)x

= exα(t).

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

an (x.u) ' (x.α)n/n!.

This is only possible if E [α] 6= 0, because a1 (1) 6= 0 for every basic sequence (an).Because of mα (t) = 1 +E [α] t+ . . . we know that logmα(t) exists, and there is aformal power series f (t) of order 0 such that 1 + logmα (t) = f (t). The form off (t) tells us that f (t) = mη (t) for some umbra η, and

E [η] = [t] logmα (t) = [t]mα (t) 6= 0.

Hence η undergoes the same restrictions as α. From

mα (t) = emη(t)−1

follows α ≡ β.η, the adjoint [30] of η〈−1〉, and therefore E [(x.β.η)n] = n!an(x).

We have 1+α (t) = mη (t). Note that η〈−1〉.α = χ. Let c 6= 0. If ρ is another umbra,then cρ + x.β.η has moment generating function cmρ (t) ex(mη(t)−1). This is thegenerating function of the Sheffer sequence (sn) for α−1 (D), with initial valuessn (0) = cE [ρn/n!]. The choice cρ ≡ ε brings us back to the umbral polynomialwith moments n!an(x). We write σx for the umbral polynomial cρ + x.α. Thebinomial theorem for Sheffer sequences becomes

σx+y ≡ sσx + y.α,

because

sn (x+ y) = E [σx+y/n!] =

n∑k=0

E

[(cσ + x.α)

k

k!

]E

[y.αn−k

(n− k)!

]

=

n∑k=0

sk(x)an−k (y) .

Example 3.4.4. Remember that the composition of two basic sequences gives an-other basic sequence. For example, we can find a basic sequence (gn) such thatan (g(x)) =

(xn

)(see section 2.3.2). This means that γ (α (t)) = log (1 + t), or

γ (t) = log(1 + α−1 (t)

)In umbral terms, mγ = eγ(t) and mα = eα(t), where

gn (x.u) ' (x.γ)n/n! and an (x.u) ' (x.α)

n/n!. We get

1 + t = mχ = mγ(logmα) = mγ.α


(see 3.20), hence χ ' γ.α. We solve this for γ by observing that

χ = α〈−1〉.β.α = γ.α

implies α〈−1〉.β = γ. Hence mγ = mα〈−1〉(logmβ) = mα〈−1〉 (et − 1), or

E

[γn

n!

]= E

∑k≥0

(α〈−1〉)kk!

[tn](et − 1

)k ,and therefore

γn '∑k≥0

(α〈−1〉

)kS (n, k) .

3.4.1 The Cumulant Umbra

Again we think of E [eαt] at the moment generating function of α. The momentgenerating function minus 1 of χ.α, mχ.α− 1, is the cumulant generating functionlogmα of α. Therefore, χ.α is called the α-cumulant umbra, also denoted as κα.Observe that

(χ.α)n '

∑λ`n

χ(|λ|)dλαλ

(see Exercise 3.4.5), where λ = (λ1, λ2 · · · ) ∼

1`1 , . . . , n`n, and

dλ =n!

(1!)`1 `1! (2!)

`2 `2! · · · (n!)`n `n!

.

Furthermore,αλ ≡

(α1

1

).`1 (α2

2

).`1 · · · (ann).`n

where E[αi] ≡ E [α], hence

E [αλ] = a`11 · · · a`nn .

The moments of the α-cumulant umbra κα are called the cumulants of α,

E [κnα] =∑λ`n

(|λ| − 1)! (−1)|λ|−1

dλ

n∏j=1

a`jj , (3.26)

written in terms of the moments of α. Hence

κnα '∑λ`n

(−1)(|λ|−1) dλαλ.

In the last expression, replace the −1 by any umbra δ,∑λ`n

δ(|λ|−1)dλαλ.


Lemma 3.4.5. For all umbrae α, δ and all positive integers n holds∑λ`n

δ(|λ|−1)dλαλ ' α (α+ δ.α)n−1

.

Proof. We have

α (α+ δ.α)n−1 '

n∑i=1

(n− 1

i− 1

)αi (δ.α)

n−i

'n∑i=1

(n− 1

i− 1

)αi

∑λ`n−i

δ(|λ|)dλαλ.

To show why this last equivalence equals the equivalence in the Lemma, we haveto write the partitions λ ` n − i in the multiset form

1`1 , 2`2 , . . .

. Hence

α (α+ δ.α)n−1 '

n∑i=1

∑1`1+2`2+···=n−i

(n− 1

i− 1

)αiδ(`1+`2+... ) (n− i)!

(α1

1

).`1 · · · (αii).ì+1 · · ·(1!)

`1 `1! (2!)`2 `2! · · ·

'n∑i=1

∑1`1+...i(ì+1)+···=n

i (ì + 1)

n

δ(`1+`2+... )n!(α1

1

).`1 · · · (αii).ì+1 · · ·(1!)

`1 `1! · · · (i!)ì+1(ì + 1)! · · ·

'∑

1¯1+2¯

2+···=n and ¯ik>0

δ(−1+¯1+¯

2+... )n!(α1

1

).¯1 · · · (αii).¯i · · ·(1!)

¯1 ¯

1! · · · (i!)¯i ¯i! · · ·

∑k

ik ¯ik

n.

From∑k ik

¯ik =

∑i i

¯i = n follows the Lemma.

As a first application of the above Lemma we see that

κnα ' α (α− 1.α)n−1

. (3.27)

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 eval-uating

αn ' (β.κa)n '

∑λ`n

β(|λ|)dλ (κα)λ '∑λ`n

dλ (κα)λ .

This is the inverse relation to (3.26). Applying the same Lemma as before,

αn ' κα (κa + β.κα)n−1

. (3.28)

Remark 3.4.6. (a) The pair (3.27) and (3.28) of relationships between cumulantsand moments can be extended to boolean and to free cumulants ([28]).(b) We did not dwell on the connections between Umbral Calculus and probabil-ity theory, except for motivating some of the definitions. Most of the material insection 2.2.3 also has a natural formulation in umbral terms.


3.4.2 Exercises

3.4.1. Verify that E [(x.u)n] = xn and E [(x.χ)

n/n!] =

(xn

).

3.4.2. Substitute x for k in (3.18) and show that E[e(x.α)t

]= mα (t)

x.

3.4.3. The partial Bell polynomials are defined as

Bn,j (α1, α2, . . . , αn−j+1) =1

j!

∑i1+···+ij=n; il>0

(n

i1, . . . , ij

)ai1ai2 · · · aij

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

frequency of every integer occurring in λ, then∑nj=1 j`j = n and

∑nj=1 `j = |λ|.

We also write λ as a multiset, λ =

1`1 , 2`2 , . . .. Show that

Bn,j (α1, α2, . . . , αn−j+1) =∑1`1+2`2+···+n`n=n`1+`2+···+`n=j

n!

(1!)`1 `1! (2!)

`2 `2! · · · (n!)`n `n!

a`11 a`22 · · · a`nn

=∑

λ`n, |λ|=j

dλaλ

where aλ = a`11 a`12 · · · a`nn , and dλ = n!

(1!)`1`1!(2!)`2`2!···(n!)`n`n!.

3.4.4. Let α be an umbra with moments E [αn] = an for all n ≥ 1. Show that

E [(k.α)n/n!] =

∑λ`n

(k)|λ|

`1! · · · `n!

n∏i=1

(aii!

)`i3.4.5. Let α be an umbra with moments E [αn] = an for all n ≥ 1. The (auxiliary)umbra x.α has moments

E [(x.α)n] =

∑λ`n

(x)(|λ|) dλaλ

(Exercise 3.4.3).Show that E [(x.α)n] is obtained by substituting x for k in E [(k.α)

n].

Verify that the degree of E [(x.α)n] is n iff a1 6= 0. Show that for the unity umbra

u holds E [(x.u)n] = xn.

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 ∈ FIn F [x, y] holds

3. (x+ y) .α ≡ x.α+ y.α (analogue of the binomial theorem!)


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 momentgenerating function mξ (t), i.e., E [ξn] = E [Xn] or all n ≥ 0. Furthermore, let Nbe a random variable on the natural numbers, with moment generating functionmν (t), where ν is an umbra such that E [νn] = E [Nn] for all n. Show that therandom sum X1 + · · ·+XN has moment generating function ν.ξ.

3.4.11. Prove Theorem 2.2.11 with Umbral Calculus. Let n!an (1) = E [αn], themoments of an umbra α. We assume that E [α] = 1. The basic polynomials an (x)have polynomial coeffi cients, an,i = an−i(i) for some basis (an). Let n!an (1) =E [αn]. Show that nαn−1 ' (χ.α)

n for all n ≥ 1. The umbra with n-th momentequal to nαn−1 is also written as αD. Hence αD ≡ χ.α.

Chapter 4

Finite Operator Calculus inSeveral Variables

We will present the higher dimensional Finite Operator Calculus in two variablesonly, with a few exceptions; generalizations to a larger number of variables requirea more streamlined notation, but no additional insight. An exception to this ruleis the bivariate transfer formula (Lemma 4.2.12). More than two variables wouldgive 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 thediscussion in this section to the minimum needed for solving recursions. In thischapter 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 fewhave 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 expressionof the form

p (u, v) = a0,0 + a1,0u+ a0,1v + · · ·+ ai,juivj .

We define for all k, l ≥ 0 the (linear) coeffi cient functional[ukvl

]on F [u, v] as[

ukvl]p = ak,l. We will also use the shorter notation [p]k,l, especially when we do

not really care about the variables u and v. Hence, any matrix with finitely manynonzero 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 thescalars a, b ∈ F are substituted for the formal variables u and v. A polynomial p hasdegree m in u if

[umvl

]p (u, v) 6= 0 for some l ≥ 0, and

[ukvl

]p (u, v) = 0 for all

k > m and all l. The degree in v is defined correspondingly. Finally, a polynomialp (u, v) has degree deg p = (m,n) if it has degree m in u and degree n in v. Forinstance, u5v3 + u2v7 has degree 5 in u, degree 7 in v, hence deg

(u5v3 + u2v7

)=

(5, 7). We call (bm,n)m,n∈N0 a polynomial sequence, iff deg bm,n = (m,n). Notethat the degree of m in u may change if we substitute a specific value (scalar) forv. For instance, the polynomial sequence

bm,n (u, v) =

m∑j=0

(n+m− j − 1

j

)(n+m− 2j

m− j

)um−jvn−j/ (n+m− 2j)! (4.1)

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 abasis of F [u, v], if bm,n (u, v) contains the term umvn (Exercise 4.1.2). The sequencein (4.1) is a basis.

The partial derivative operators Duand Dv on F [u, v] are defined as ∂∂up (u, v)

and ∂∂vp (u, v), respectively. Hence we obtain the ring of (linear) finite operators

ΣDu,Dv on F [u, v] by letting

G ∈ ΣDu,Dv iff G = γ (Du,Dv) for some γ ∈ F [[s, t]] .

The isomorphism ΣDu,Dv ' F [s, t] we get from

GS ' γσ if G = γ (Du,Dv) and S = σ (Du,Dv) .

The operators in ΣDu,Dv commute. We call an operator G partial , if G has apower series expansion only in Du, or Dv. The (partial) translation operators Eauand Ebv are defined as e

aDu and ebDv , respectively. A linear operator T on F [u, v]is translation invariant iff TEauE

bv = EauE

b2T 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 ∈ΣDu,Dv . In that case,

T =∑i,j≥0

⟨Eval0,0 | T

uivj

i!j!

⟩DiuDjv

The proof is left as Exercise 4.1.3.Similar to multi-series, we have multi-operators (T1, T2); a delta pair (B1, B2) ∈

Σ2Du,Dv is multi-operator such that B1 = γ1 (Du,Dv), and B2 = γ2 (Du,Dv), where

(γ1, γ2) is a delta multi-series (see section 1.3). Note that we denote pairs of oper-ators by (T1, T2) or T1, T2, but write Fu and Fv for the partial operators, actingon u and on v only. We will make that distinction throughout this chapter, exceptunder 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 Dρ for thepartial derivatives (ρ = 1, 2, 3).

By Lemma 4.1.1 delta pairs consist of translation invariant operators. Thefollowing Lemma is useful and easy to prove.

Lemma 4.1.2. If (β1, β2) is a delta multi-series and m,n ≥ 0 then [βm1 βn2 ]k,l 6= 0

implies k ≥ m and l ≥ n. Furthermore, [βm1 βn2 ]m,n =

([β1]1,0

)m ([β2]0,1

)n6= 0.

Finally, we define a (bivariate) Sheffer sequence (sm,n) as a basis, i.e., deg sm,n =(m,n) and [sm,n]m,n 6= 0, solving the system

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 orn < 0). The Sheffer sequence (bn,m) with initial values bn,m (0, 0) = δm,0δn,0 isthe basic sequence for (B1, B2). The standard example is of course the basic se-quence (umvn/ (m!n!)) for (Du,Dv). Note that the coeffi cient of umvn in a Shefferpolynomial sm,n (u, v) is never 0. This excludes lots of ‘easy’cases like um + vn

from the Sheffer family! The quest for solutions to initial value problems needs theobservation that in the intersection of the kernel of B1 and the kernel of B2 onlythe 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(Du,Dv) with initial values sm,n (m+ n,m− n) = 1/ (m+ n)!.

In general, the operators B1 and B2 in a delta pair (B1, B2) act both on bothvariables u and v; they are usually not the partial operators acting only on onevariable, as it is the case for Du and Dv in the (special) delta pair (Du,Dv). Inaddition, we write ∆u = eDu − I = Eu − I, and ∇u = I − e−Du = I −E−1

u (samefor v) for the partial difference operators. Furthermore, taking any two univariatedelta operators A and B on F [x], we can make them into the delta pair (A,B), andA will be the partial operator with respect to u, B for v. The basic sequence factorsin this case, (am (u) bn (v)). Of course, this is the case we are not really interestedin, but it is often the starting point for getting to more useful constructions.


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

Theorem 4.1.3. The basic sequence (bm,n) has generating function∑m,n≥0 bm,n (u, v) smtn = euβ1(s,t)+vβ2(s,t) iff β−1

1 (Du,Dv) : bm,n 7→ bm−1,n andβ−1

2 (Du,Dv) : bm,n 7→ bm,n−1 for all m,n ≥ 0.

In Exercise 4.1.6 we verify that the basic sequence as defined in the Theoremabove really is a basis of the vector space of polynomials in two variables. Abivariate Sheffer sequence (sm,n) has generating function∑

m,n≥0

sm,n (u, v) smtn = σ (s, t) euβ1(s,t)+vβ2(s,t)

for some unit σ (s, t) ∈ F [[s, t]] (note that F [[s, t]] is not a field; if σ (0, 0) 6= 0 thenσ is a unit in F [[s, t]].). The generating function shows that

sm,n (u+ u′, v + v′) =

m∑i=0

n∑j=0

si,j (u, v) bm−i,n−j (u′, v′) , (4.2)

the multivariate version of the binomial theorem.

Example 4.1.4. To show the general structure of this example, we need at leastthree variables u, v, and w, say. The reader should have no problem generalizingthe above setting to the trivariate case. Suppose that for ρ = 1, 2, 3 we have aunivariate basic sequence (aρ,n)n≥0 with generating function

∑n∈N0 aρ,n (x) tn =

γρ (t)x where γρ (t) is a univariate power series such that γρ (0) = 1, i.e., γρ (t) =

eαρ(t) 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)) .

The polynomial sequence (bl,m,n) has the generating function ωu1ωv2ω

w3 , because∑

l,m,n≥0 bl,m,n (u, v, w) rlsmtn =∑l≥0

∑m≥0

∑n≥0

a1,l (u) a2,m (v + l) a3,n (w + l +m) rlsmtn

= ωw3∑l≥0

∑m≥0

a1,l (u) a2,m (v + l) (rω3)l(sω3)

m

= ωw3∑l≥0

a1,l (u) (rω3)lγ2 (sω3)

v+l= ωw3 ω

v2

∑l≥0

a1,l (u) (rω3ω2)l

= ωw3 ωv2ω

u1 = euβ1(r,s,t)+vβ2(r,s,t)+wβ3(r,s,t)

4.1. Polynomials and Operators in Several Variables 117

where βρ (r, s, t) = lnωρ (r, s, t). We verify (without deriving it!) thatβ−1

1 (r, s, t) = α−11 (r) e−s−t, β−1

2 (r, s, t) = α−12 (s) e−t, and β−1

3 (r, s, t) = α−13 (t).

We get

β3

(β−1

1 , β−12 , β−1

3

)= lnω3

(β−1

1 , β−12 , β−1

3

)= α3

(α−1

3 (t))

= t

β2

(β−1

1 , β−12 , β−1

3

)= α2

(α−1

2 (s)γ3

(α−1

3 (t))

et

)= α2

(α−1

2 (s))

= s

β1

(β−1

1 , β−12 , β−1

3

)= α1

(α−1

1 (r)γ2

(α−1

2 (s) e−tγ3

(α−1

3 (t)))

esγ3

(α−1

3 (t))

et

)

= α1

(α−1

1 (r)γ2

(α−1

2 (s))

es

)= α3

(α−1

1 (r))

= r.

We also check that β−1ρ (D1,D2,D3), ρ = 1, 2, 3, really is the corresponding delta

triple; β−1ρ (D1,D2,D3) bl,m,n (u, v, w)

= α−1ρ (Dρ) e−Dρ+1−···−D3bl,m,n (u, v, w)

= α−1ρ (Dρ)E−1

ρ+1 · · ·E−13 bl,m,n (u, v, w)

hence we get for ρ = 1

α−11 (Du)E−1

2 E−13 bl,m,n (u, v, w)

= 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 gen-

eral way. See Exercise 4.1.1 for an application to

bl,m,n (u, v, w) =

(u− 1 + l

l

)(v − 1 + l +m

m

)(w − 1 + l +m+ n

n

). (4.3)

4.1.1 Exercises

4.1.1. Apply Example 4.1.4 to find the polynomials (bl,m,n (u, v, w)) generated byaρ,n (x) =

(x+n−1n

)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]m,n 6= 0 (m,n ≥ 0) for any Sheffer sequence (sm,n).

4.1.5. Show that the basic sequence (bm,n (u, v)) can be reconstructed from the twounivariate sequences (bm,n (u, 0))m,n≥0 and (bm,n (0, v))m,n≥0. Is (bm,n (u, 0))m≥0for fixed n a univariate basic sequence?


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

m∑l=0

n∑k=0

ul

l!

vk

k!

m∑i=l

n−k∑j=0

[βl1]i,j

[βk2]m−i,n−j

and that this coeffi cient has degree (m,n). Hence the sequence (bm,n) in Theorem4.1.3 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 anotherbasic sequence, when a relation (the transfer) between the corresponding deltapairs of operators is known. This is not so straightforward in the multivariatesetting 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 thecoeffi cients of (τ1, τ2). Now all depends on the coeffi cients: If they are in the samering F as the coeffi cients of the polynomials, the result follows from Lagrange-Goodinversion, and is stated in Lemma 4.2.1. In this case we write (τ1, τ2) ∈ F [[s, t]]

2.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 thefirst subsection. An elementary example is A1 = Ev∆u, A2 = ∆v and τ1 (s, t) = s,τ2 (s, t) = (t− tα) / (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. Inthis case we write (τ1, τ2) ∈ ΣDu,Dv [[s, t]]

2. For example, τ1 (s, t) = Evs andτ2 (s, t) = EuEvt − t2 (Example 4.2.14). We consider such transfer functions inthe last subsection. In between, there is the operator based transfer formula, whenAρ = VρBρ, where Vρ is invertible in ΣDu,Dv . This can be seen as the “extremecase”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 havecoeffi cients from the infinite field F. We can write Bρ = τ−1

ρ (A1, A2), hence Bρis also a delta operator. From Bρ = β−1

ρ (Du,Dv) and Aρ = α−1ρ (Du,Dv), say,

follows τ−1ρ

(α−1

1 , α−12

)= β−1

ρ .Therefore∑

m,n≥0

bm,n (u, v) smtn = euβ1(s,t)+vβ2(s,t) = euα1(τ1,τ2)+vα2(τ1,τ2)

=∑k,l≥0

ak,l (u, v) τ1 (s, t)kτ2 (s, t)

l.

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 asa delta multi-series (τ1, τ2) ∈ F [[s, t]]

2 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 hasthe generating function∑

m,n≥0

bm,n (u, v) smtn = euα1(τ1,τ2)+vα2(τ1,τ2) (4.4)


and the expansion

bm,n (u, v) =

m∑k=0

n∑l=0

[τk1 τ

l2

]m,n

ak,l (u, v) .

Note that the above Transfer Formula only holds if the multi-series τ1, τ2connecting (A1, A2) and (B1, B2) has coeffi cients free of operators; we must have(τ1, τ2) ∈ F [[s, t]]

2. Similar to Corollary 2.4.4 in the univariate case, we havespecial cases where the generating function (4.4) simplifies. For example, if EuEdv =Edv + τ1 (B1, B2) for some d ∈ F, and Ev = 1 + τ2 (B1, B2), then∑

m,n≥0

bm,n (u, v) smtn =

(1 +

τ1 (s, t)

(1 + τ2 (s, t))d

)u(1 + τ2 (s, t))

v (4.5)

(Exercise 4.2.2). Such cases can arise when α1 and α2 are of a very simpleform.

Example 4.2.2. Suppose

(Ev∆u,∆v) =

(B1,

B2 −Ba21−B2

),

hence τ1 (s, t) = s and τ2 (s, t) = (t− tα) / (1− t) in Lemma 4.2.1. The deltaoperator (E2∆u,∆v) has the basic sequence

((um

)(v−mn

))m,n≥0

; this can either bechecked directly or has to wait until Example 4.2.10. We obtain from Lemma 4.2.1

bm,n (u, v) =

m∑k=0

n∑l=0

[τk1 τ

l2

]m,n

(u

k

)(v − kl

)

=

(u

m

) n∑l=0

(v −ml

)[tn]

(t− ta1− t

)l.

The coeffi cient of tn in ((t− tα) / (1− t))l has been written as(l

n−l)α−1

by Euler[32], (

l

n− l

)α−1

=

b(n−l)/(a−1)c∑j=0

(−1)j(l

j

)(n− (a− 1) j − 1

l − 1

)for l ≥ 1. In this notation,

bm,n (u, v) =

(u

m

) n∑l=0

(v −ml

)(l

n− l

)α−1

We can also arrive at the generating function of this basic sequence via formula(4.5), letting d = 1:∑

m,n≥0

bm,n (u, v) smtn =

(1 + s

1− t1− tα

)u(1− tα1− t

)v.


4.2.2 Operator based transfer

Suppose T = φ (Du,Dv) is an operator in ΣDu,Dv . We define the partial derivativesof T as ∂

∂DuT = φs (Du,Dv) and ∂∂Dv T = φt (Du,Dv), where φs = ∂

∂sφ (s, t) andφt = ∂

∂tφ (s, t), the partial derivatives defined in section 1.3. Similar, the Jacobiandeterminant of T1 = φ1 (Du,Dv) and T2 = φ2 (Du,Dv) is defined as

|J (φ1, φ2)| =∣∣∣∣ ∂(φ1, φ2)

∂ (Du,Dv)

∣∣∣∣ =

∣∣∣∣ ∂∂Duφ1 (Du,Dv) ∂

∂Duφ2 (Du,Dv)∂∂Dv φ1 (Du, t) ∂

∂Dv φ2 (Du,Dv)

∣∣∣∣=

(∂

∂Duφ1

)(∂

∂Dvφ2

)−(

∂

∂Dvφ2

)(∂

∂Dvφ1

).

If (T1, T2) is in ΣDu,Dv , then its derivative (Jacobian determinant) will be inΣDu,Dv . Note that the s- and t-derivative are now denoted by ∂

∂s and∂∂t, re-

spectively. This slightly cumbersome notation has the advantage that it is clearwhat happens under the above isomorphism: If Tρ = τρ (Du,Dv) then ∂

∂Dv T1 isisomorphic to ∂

∂tτ1 (s, t), for example.The choice of coeffi cient ring is very important: For a simple case consider∣∣∣∂(∇u,∇v)

∂(Du,Dv)

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

|J (∇u,∇v)| =∂(1− e−Du

)∂Du

∂(1− e−Dv

)∂Dv

−∂(1− e−Dv

)∂Du

∂(1− e−Du

)∂Dv

= e−Due−Dv − 0 = E−1u E−1

v .

If however ∇u = ι1 (Du,Dv) = ∇uD0u, and ∇v = ι2 (Du,Dv) = ∇vD0

v, henceτρ ∈ ΣDu,Dv [[s, t]], then |J (∇u,∇v)| = 0. Note that in the first version (τ1, τ2) isa delta pair. In the second case, (ι1, ι2) is of order 0.

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

β1 (s, t) = a−1s1− a−1bt

1− a−2b2stand β2 (s, t) = a−1t

1− a−1bs

1− a−2b2st.

We saw that b does not have to be invertible; if we take ΣDu,Dv as the coeffi cientring, we can choose a = I and b = Du, obtaining the inverses

β1 (s, t) = s1−Dut1−D2

ust∈ ΣDu,Dv [[s, t]] and

β2 (s, t) = t1−Dus1−D2

ust∈ ΣDu,Dv [[s, t]] .


Hence the delta pair γ1 (Du,Dv) = Du/ (1−DuDv) ∈ ΣDu,Dv [[s, t]] and γ2 (Du,Dv) =Dv/

(1−D2

u

)∈ ΣDu,Dv [[s, t]] has the inverses

β1 (Du,Dv) =Du −D2

uDv1−D3

uDvand β2 (Du,Dv) =

Dv −D2uDv

1−D3uDv

.

As written above, we do not distinguish between Du as “formal variable” and Duas coeffi cient. This will be a mistake when we try to verify the inverses, because notevery occurrence Du in γ1 (Du,Dv) can be replaced by β1.Let us write γ1 (Du,Dv) =

s/ (1−Dut)||(s,t)=(Du,Dv) and γ2 (Du,Dv) = t/ (1−Dus)||(s,t)=(Du,Dv). Thus wemust calculate γ1 (β1, β2) = β1/ (1−Duβ2) = Du and γ2 (β1, β2) = β2/ (1−Duβ1) =Dv.

The following Lemma shows the easiest case of an operator equation Aρ =τρ (B1, B2) where τρ ∈ ΣD1,D2 [[s, t]] is a delta multiseries with operator coeffi -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

bm,n (u, v) = V m+11 V n+1

2

∣∣∣∣∂ (B1, B2)

∂ (A1, A2)

∣∣∣∣ am,n (u, v) . (4.6)

Proof. Because (A1, A2) and (B1, B2) are both delta pairs, there must exist a deltapair of multiseries (τ1, τ2) ∈ F [[s, t]]

2 such that Aρ = τρ (B1, B2). Therefore

∑m,n≥0

bm,n (u, v) smtn =

m∑k=0

n∑l=0

[τm−k1 τn−l2

]m,n

Ak1Al2am,n (u, v) (4.7)

as in Lemma 4.2.1. FromBρ = τ−1ρ (A1, A2) = V −1

ρ Aρ we see that Vρ = υρ (Au, Av)

if we define τ−11 (s, t) = s/υ1 (s, t) and τ−1

2 (s, t) = t/υ2 (s, t). By Lagrange-Goodinversion [

τm−k1 τn−l2

]m,n

=

[υm+1

1 υn+12

∣∣∣∣∣∂(τ−11 , τ−1

2

)∂ (s, t)

∣∣∣∣∣]k,l

.

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

υ1 (A1, A2)m+1

υ2 (A1, A2)n+1

∣∣∣∣∣∂(τ−11 , τ−1

2

)∂ (A1, A2)

∣∣∣∣∣ = V m+11 V n+1

2

∣∣∣∣∂ (B1, B2)

∂ (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.


4.2.3 The multivariate Pincherle derivative

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

θA1: am,n (u, v) 7→ (m+ 1) am+1,n (u, v)

θA2: am,n (u, v) 7→ (n+ 1) am,n+1 (u, v) .

Note that θA1A2 = A2θA1 , and

θA1Aj1am,n (u, v) = (m− j + 1)Aj−1

1 am,n

for j > 0. The umbral shifts are not translation invariant! From

θA1

∑m,n≥0

am,n (u, v) smtn =∂

∂s

∑m,n≥0

am,n (u, v) smtn =∂

∂seuα1(s,t)+vα2(s,t)

(which means that θA1= ∂

∂s ) follows θA1

∑m,n≥0 am,n (u, v) smtn =(

u∂

∂sα1(s, t) + v

∂

∂sα2 (s, t)

) ∑m,n≥0

am,n (u, v) smtn

=

(u∂

∂A1α1(A1, A2) + v

∂

∂A1α2 (A1, A2)

) ∑m,n≥0

am,n (u, v) smtn

=

(u∂Du∂A1

+ v∂Dv∂A1

) ∑m,n≥0

am,n (u, v) smtn.

Hence

θA1= u

∂Du∂A1

+ v∂Dv∂A1

and (4.8)

θA2 = u∂Du∂A2

+ v∂Dv∂A2

.

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

∂∂t = ∂

∂t∂∂s

thatθA1

θA2= θA2

θA1. (4.9)

This result follows also from the observation that θA1θA2

am,n (u, v) = θA2θA1

am,n (u, v)for all natural numbers m and n. Note that when A1 is a function of Du only, thenDu is a function of A1 only, but Dv can be a power series in A1 and A2, thus

θA2= v

∂Dv∂A2

(4.10)

in this special case. We demonstrate this behavior in the following example.


Example 4.2.5. Let A1 = ∇u, and A2 = E−1u ∇v. For finding the umbral shifts with

respect to A1and A2 we need to express Du and Dv in terms of A1 and A2. Wehave A1 = I − e−Du, hence Du = − ln (I −A1). From A2 = (I −A1)

(I − e−Dv

)follows Dv = − ln (I −A2/ (I −A1)). Hence

θA1= u

∂Du∂A1

+ v∂Dv∂A1

= uI

1−A1+ v

A2/ (I −A1)2

I −A2/ (I −A1)= uEu + vEuEv∇v

θA2= −u∂ (I −A1)

∂A2− v ∂ ln (I −A2/ (I −A1))

∂A2= vEuEv

We check the commutation rule (4.9): θA1θA2

= ((u− v)Eu + vEuEv) vEuEv = v (u− v)E2uEv + vEuEvvEuEv

= vEuEv (u+ v)Eu − vEuEvvEuEv = vEuEv ((u+ v)Eu − vEuEv)= θA2θA1 .

The basic sequence for this delta pair A1, A2 is am,n (u, v) =(m+n−1+u

m

)(n−1+vn

),

and we check that θA1am,n (u, v)

= (uEu + vEvEu∇v)(m+ n− 1 + u

m

)(n− 1 + v

n

)=

(m+ n+ u

m

)(u

(n− 1 + v

n

)+ v

(n− 1 + v

n− 1

))=

(m+ n+ u

m

)(u+ n)

(n− 1 + v

n− 1

)= (m+ 1)

(m+ n+ u

m+ 1

)(n− 1 + v

n

)= (m+ 1) am+1,n (u, v)

and θA2am,n (u, v)

= vEvEu

(m+ n− 1 + u

m

)(n− 1 + v

n

)= v

(m+ n+ u

m

)(n+ v

n

)= (n+ 1) am,n+1 (u, v) .

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

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

T ′A1= TθA1 − θA1T and

T ′A2= TθA2 − θA2T.

Of course we want to know when T ′Aρ = ∂T∂Aρ

. This is answered in the followingLemma.


Lemma 4.2.7. Let T be a translation invariant operator on F [u, v], and A1, A2 adelta pair. Then

T ′Aρ =∂T

∂Aρ.

and T ′Aρ is translation invariant.

Proof.

T ′Aρ = T

(u∂Du∂Aρ

+ v∂Dv∂Aρ

)−(u∂Du∂Aρ

+ v∂Dv∂Aρ

)T

= (Tu− uT )∂Du∂Aρ

+ (Tv − vT )∂Dv∂Aρ

= T ′θu∂Du∂Aρ

+ T ′θv∂Dv∂Aρ

=∂T

∂Du∂Du∂Aρ

+∂T

∂Dv∂Dv∂Aρ

=∂T

∂Aρ

where θu and θv are the univariate umbral shifts defined in section 2.4.1,

θuumvn

m!n!= u

umvn

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

T ′Bρ = T ′A1

∂A1

∂Bρ+ T ′A2

∂A2

∂Bρ.

Special results for the bivariate case

We can rephrase the Jacobian determinant of translation invariant operators interms of umbral shifts. In∣∣∣∣ ∂ (R,S)

∂ (A1, A2)

∣∣∣∣ = (RθA1− θA1

R) (SθA2− θA2

S)− (RθA2− θA2

R) (SθA1− θA1

S)

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

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)

∣∣∣ =

RθA1SθA2

− θA1RSθA2

+ θA1RθA2

S −RθA2SθA1

+ θA2RSθA1

− θA2RθA1

S.

This can be further simplified with the help of the Pincherle derivatives.


Proof.∣∣∣ ∂(R,S)∂(A1,A2)

∣∣∣ =

(RθA1− θA1

R) (SθA2− θA2

S)− (RθA2− θA2

R) (SθA1− θA1

S)

= RθA1SθA2

− θA1RSθA2

+ θA1RθA2

S −RθA2SθA1

+ θA2RSθA1

− θA2RθA1

S

because θA1θA2

= θA2θA1

.If we exchange RθA1

− θA1R with SθA2

− θA2S, and RθA2

− θA2R with

SθA1− θA1

S in the above proof, we get∣∣∣ ∂(R,S)∂(A1,A2)

∣∣∣ =

SθA2RθA1−θA2SRθA1 +θA2SθA1R−SθA1RθA2 +θA1SRθA2−θA1SθA2R (4.11)

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 thenbivariate transfer formula holds,

bm,n (u, v) =1

nθA2

V m1 V n2 am,n−1 +1

mn(θA1

V m1 θA2− θA2

V m1 θA1)V n2 am−1,n−1.

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

V m+11 V n+1

2

∣∣∣∣∂ (B1, B2)

∂ (A1, A2)

∣∣∣∣= V m1 V n2 −

1

nV m1 A2

∂V n2∂A2

− 1

mV n2 A1

∂V m1∂A1

+1

mnA1A2

∣∣∣∣∂ (V m1 , V n2 )

∂ (A1, A2)

∣∣∣∣ .In this proof we will write R for V m1 and S for V n2 , slightly shortening the longexpressions. Replacing the partial derivatives by the Pincherle derivatives we get

V m+11 V n+1

2

∣∣∣∣∂ (B1, B2)

∂ (A1, A2)

∣∣∣∣ am,n=

(RS − 1

mS (RθA1

− θA1R)A1 −

1

nR (SθA2

− θA2S)A2

)am,n

+1

mn

∣∣∣∣ ∂ (R,S)

∂ (A1, A2)

∣∣∣∣A1A2am,n

=1

mSθA1

Ram−1,n −RSam,n +1

nRθA2

Sam,n−1

+1

mn

∣∣∣∣ ∂ (R,S)

∂ (A1, A2)

∣∣∣∣ am−1,n−1.


Now apply Lemma 4.2.8 to expand |∂ (R,S) /∂ (A1, A2)|. Thus bm,n (u, v) =

1

mSθA1

Ram−1,n (u, v) +1

nRθA2

Sam,n−1 (u, v)−RSam,n (u, v)

+1

mn(−θA1

RSθA2−RθA2

SθA1+RθA1

SθA2) am−1,n−1 (u, v)

+1

mn(θA2

RSθA1+ θA1

RθA2S − θA2

RθA1S) am−1,n−1 (u, v)

=1

m(SθA1

− θA1S)Ram−1,n +

1

mRθA1

Sam−1,n −RSam,n

+1

nθA2

RSam,n−1 +1

mn(θA1

RθA2− θA2

RθA1)Sam−1,n−1.

The first line of the last two-line equation equals 0, because of translation invari-ance of the Pincherle derivatives.

Note that formula (4.6) contains 2 terms (from the 2× 2 determinant), eachcontaining 2 partial derivatives. Replacing them by Pincherle derivatives gives us 8terms. Theorem 4.2.9 reduces the 8 terms to 3. In the case of three dimensions, the3× 3 determinant will give us 6 terms, and the Pincherle derivatives will expandthem by a factor of 23, giving 48 terms. They can be reduced to 13 terms.

Example 4.2.10. Let B1 = EcvA1, and B2 = EfuA2, where the delta pair (A1, A2)has the basic polynomials am,n (u, v). We find V1 = E−cv and V2 = E−fu . We arelooking for the basic sequence for (B1, B2). If we apply Theorem 4.2.9 we get

bm,n (u, v) =1

nθA2

E−cmv E−fnu am,n−1 +1

mn

(θA1

E−cmv θA2− θA2

E−cmv θA1

)E−fnu am−1,n−1

=1

nθA2

E−cmv E−fnu am,n−1 −cu

n

∣∣∣∣∂ (Du,Dv)∂ (A1, A2)

∣∣∣∣E−cmv E−fnu am−1,n−1

If A2 has an expansion in Dv only, then ∂Dv∂A1

= 0, and θA1 = u∂Du∂A1(see (4.10)),

hence

bm,n (u, v) =1

n

(θA2− cmu∂Dv

∂A2

1

u− fn

)E−cmv E−fnu am,n−1

If, in addition, A1 is power series in Du only, meaning that A1 and A2 are partialoperators, then am,n (u, v) = pm (u) qn (v), where A1pm = pm−1 and A2qn = qn−1.In this case,

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

u− fnqn (v − cm)

v − cm .

For example, if A1 = Du and A2 = Dv then

bm,n (u, v) = (uv − vfn− ucm)(u− fn)

m−1(v −m)

n−1

m!n!


If A1 = ∆u and A2 = ∆v, then

bm,n (u, v) =vu− fnv − cmu

(u− fn) (v − cm)

(u− fnm

)(v − dmn

).

The sequence sm,n := Efnu Ecmv bm,n is a Sheffer sequence for (A1, A2), because

A1Efnu Ecmv bm,n = Efnu Ec(m−1)

v EcvA1bm,n = Efnu Ec(m−1)v B1bm,n = sm−1,n,

and the same holds for A2. If we choose Aρ = ∆ρ then

sm,n (u, v) =uv − fndm

uv

(u

m

)(v

n

)with generating function∑

m,n≥0

sm,n (u, v) smtn = (1 + s)u

(1 + t)v − fdst (1 + s)

u−1(1 + t)

v−1

= (1 + s)u−1

(1 + t)v−1

(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 termsof (am,n), but the operators Aρ and Bρ were connected by an invertible operatorVρ, not by a power series τρ. The following Corollary holds for any number ofvariables. 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 forthe 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) ∈ΣDu,Dv [[s, t]]

2 is a delta multi-series with translation invariant operator coeffi -cients. If Aρ = Bρυρ (A1, A2), then

υ1 (A1, A2)mυ2 (A1, A2)

n=∑i≥0

∑j≥0

[τm−1−i1 τn−1−j

2

∣∣∣∣∂ (τ1, τ2)

∂ (s, t)

∣∣∣∣]m−1,n−1

Ai1Aj2

=∑i≥0

∑j≥0

[εi+1−m

1 εj+1−n2

∣∣∣∣∂ (τ1, τ2)

∂ (s, t)

∣∣∣∣]i,j

Ai1Aj2

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


υm1 υn2 in terms of the coeffi cient of the inverse (τ1, τ2) by applying the Lagrange-

Good inversion formula (1.13),

[υm1 υn2 ]i,j =

[εi+1−m

1 εj+1−n2

∣∣∣∣ ∂τ1∂s

∂τ2∂s

∂τ1∂t

∂τ2∂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 withconnecting bivariate series (τ1, τ2) ∈ ΣDu,Dv [[t]] that may contain translation in-variant operators among its coeffi cients.

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

(τ1, τ2) ∈ ΣDu,Dv [[s, t]]2

is a bivariate delta series with translation invariant coeffi cients, then bm,n (u, v)

=1

nθA2υ

m1 υ

n2 am,n−1 +

1

mn(θA1υ

m1 θA2 − θA2υ

m1 θA1) υ

n2 am−1,n−1.

where υm1 υn2 is the power series in A1 and A2 given in Corollary 4.2.11, Aρ =

Bρυρ (A1, A2), and

υm1 υn2 (A1, A2) =

∑i,j≥0

[τm−1−i1 τn−1−j

2

∣∣∣∣∂ (τ1, τ2)

∂ (s, t)

∣∣∣∣]m−1,n−1

Ai1Aj2

=∑i,j≥0

[εi+1−m

1 εj+1−n2

∣∣∣∣∂ (τ1, τ2)

∂ (s, t)

∣∣∣∣]i,j

Ai1Aj2

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

mθA1

υm1 υn2 an,m−1 +

1

mn(θA2

υn2 θA1− θA1

υn2 θA2) υm1 am−1,n−1

(Exercise 4.2.4). The Lemma simplifies, of course, if A1 = B1, say. In this caseυ1 = 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) ∈ΣDu,Dv [[s, t]]

2 is a delta multi-series with translation invariant coeffi cients, then

bm,n (u, v) =1

nθA2

υn2 (A1, A2) am,n−1 (u, v)


where

υn2 (A1, A2) =∑i,j≥0

[sm−1−iτn−1−j

2

∂τ2∂t

]m−1,n−1

Ai1Aj2

=∑i,j≥0

[ε1+j−n

2

∂τ2∂t

]i,j

Ai1Aj2

(ε2 (s, t) = t/τ2 (s, t)).

Proof.θA1

υm1 θA2− θA2

υm1 θA1= θA1

θA2− θA2

θA1= 0

(see (4.9)).Of course there must be a connection between Lemma 4.2.12 and the transfer

formula

bm,n (u, v) =

m∑k=0

n∑l=0

[τk1 τ

l2

]m,n

ak,l (u, v)

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.

Example 4.2.14. Let A1 = E−1v ∇u, and A2 = ∇v, hence

am,n (u, v) =

(m− 1 + u

m

)(n+m− 1 + v

n

).

If A1 = B1 and A2 = (EuEv −B2)B2, then by Corollary 4.2.13

bm,n (u, v) =1

nθA2υ

n2 am,n−1.

We calculated the Pincherle derivative θA2in Example 4.2.5,

θA2= uEuEv − uEv + vEv.

Note that τ2 is a power series in t only; therefore we get in Corollary 4.2.13[τn−1−j2

∂τ2∂t

]n−1

=[(EuEv − t)n−j−1

(EuEv − 2t)]j

=n

n− j

(n− jj

)En−2ju En−2j

v (−1)j

for 0 ≤ j < n. Hence

υn2 (A2) =

∞∑j=0

n

n− j

(n− jj

)(−1)

j(EuEv)

n−2jAj2

= (EuEv)n

2−n(

1 +

√1− 4E−2

u E−2v A2

)n.


From(n−jj

)nn−j =

(n−j−1j−1

)nj follows [An2 ] υn2 = (−1)

nE−nu E−nv if n > 0. Thus

bm,n (u, v) =

1

n(uEuEv + (v − u)EvEv) υ

n2 am,n−1

= (uEu − u+ v)

n−1∑j=0

(−1)j

n− j

(n− jj

)(m+ n− 1− 2j + u

m

)(2n+m− 1− 3j + v

n− 1− j

)

=

n−1∑j=0

(−1)j

n− j

(n− jj

)(m+ n− 1− 2j + u

m

)

×(v +

um

n− 2j + u

)(2n+m− 1− 3j + v

n− 1− j

)This formula for bm,n also holds if n = 0; of course, b0,0 (u, v) = 1. A gneratingfunction for (bm,n) can be obtained as follows. The equations E−1

v ∇u = B1, and∇v = (EuEv −B2)B2 can be solved as Eu = 1/ (1− EvB1) and

Ev = 1 + τ2 (B1, B2) =1 +B2

2 +B1 +

√2 (B2 − 1)

2 − 2B1 − 1 + (B1 −B22)

2

2 (B2 +B1 +B1B22)

hence ∑m,n≥0

bm,n (u, v) smtn = (1− s (1 + τ2 (s, t)))u

(1 + τ2 (s, t))v

4.2.5 Exercises

4.2.1. [39] Suppose (bm,n) is the basic sequence for the delta pair(β−1

1 (Du,Dv) , β−12 (Du,Dv)

). Show that for any multi-series η ∈ F [[s, t]] holds

[η (β1 (s, t) , β2 (s, t))]m,n = 〈Eval0,0 | η (Du,Dv) bm,n〉 .

4.2.2. Show that the generating function (4.5) holds, if (τ1, τ2) ∈ F [[s, t]]2 and

A1 = Edv∆ = τ1 (B1, B2), A2 = ∆ = τ2 (B1, B2).

4.2.3. Show that the operator in Lemma ?? can be written as

V m+11 V n+1

2

∣∣∣∣∂ (B1, B2)

∂ (A1, A2)

∣∣∣∣ = V m1 V n2 −1

nV m1 A2

∂V n2∂A2

− 1

mV n2 A1

∂V m1∂A1

+1

mnA1A2

∣∣∣∣∂ (V m1 , V n2 )

∂ (A1, A2)

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

bm,n (u, v) =1

mθA1

V m1 V n2 an,m−1 +1

mn(θA2

V n2 θA1− θA1

V n2 θA2)V m1 am−1,n−1.


4.2.5. Suppose τρ and υρ are given by Aρ = τρ (B1, B2) = Bρυρ (A1, A2). Show

that[τ1 (s, t)

kτ2 (s, t)

l]m,n

= [υ1 (s, t)mυ2 (s, t)

n− tnυ1 (s, t)

m ∂υn2∂t− s

mυ2 (s, t)

n ∂υm1

∂s+

1

mnst

∣∣∣∣∂ (υm1 , υn2 )

∂ (s, t)

∣∣∣∣]m−k,n−l.4.2.6. Suppose υρ and τρ are defined as in Exercise 4.2.5, but now assume thatboth have coeffi cients in F only (no operators). Show that in this case1

nθA2

υm1 υn2 am,n−1 +

1

mn(θA1

υm1 θA2− θA2

υm1 θA1) υn2 am−1,n−1

= υm1 υn2 am,n −

1

nA2υ

m1

(∂

∂A2υn2

)am,n −

1

mA1υ

n2

(∂

∂A1υm1

)am,n +

A1A2

mn

∣∣∣∣∂ (υm1 , υn2 )

∂ (A1, A2)

∣∣∣∣ am,nTogether with Exercise 4.2.5 this implies the equivalence of Lemma 4.2.1 andLemma 4.2.12 in the bivariate case when υρ and τρ have no operators as coef-ficients.

4.2.7. Show that ∂En

∂∆ = nEn−1 and ∂En

∂∇ = nEn+1.

4.2.8. Find the basic sequence for the delta pair(EcuE

dv∆u, E

fuE

gv∆v

)4.2.9. Suppose we have the operator equations ∆u = E−1

v B1 (1 + αB2) and ∆v =B2. Use Lemma 4.2.1 to show that the basic sequence for (B1, B2) has the gener-ating function ∑

m,n≥0

bm,n (u, v) smtn =

(1 +

s (1 + αt)

1 + t

)u(1 + t)

v.

Derive from the generating function or otherwise that

bm,n (u, v) =

(u

m

) n∑k=0

(m

k

)(v −mn− k

)αk =

(u

m

) n∑k=0

(m

k

)(α− 1)

k

(v − kn− k

).

In terms of hypergeometric functions this means(x−mn

)2F1

[−n,−m; z

x−m− n+ 1

]=

(x

n

)2F1

[−n,−m; 1− z

−x

]for all positive integers n and m, where

2F1

[a, b; z

c

]=∑n≥0

(a)n (b)bn! (c)n

zn

and (a)n = a (a+ 1) · · · (a+ n− 1). This is part of the general identity

2F1

[a, b; z

c

]=

Γ (c) Γ (c− a− b)Γ (c− a) Γ (c− b) 2F1

[a, b; 1− z

a+ b+ 1− c

]+ (1− z)c−a−b Γ (c) Γ (a+ b− c)

Γ (a) Γ (b)2F1

[c− a, c− b; z1 + c− a− b

]


for all complex numbers a, b, c and z, which converges when |arg (1− z)| < π. Thereader interested in hypergeometric functions may consult the “Special Functions”by Andrews, Askey, and Roy [3]. A classic reference to this topic is the small bookby Bailey [8, 1935].

4.2.10. The bivariate basic sequence (am,n) with generating function eus1−t+v t

1−s

can be expanded as

am,n (u, v) =

n∑i=0

m∑j=0

ui

i!

vj

j!

(n+ j − i− 1

n− i

)(m+ i− j − 1

m− j

)

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).


4.3 The Multivariate Functional Expansion Theorem

The Functional Expansion Theorem 3.1.4 is easily adopted to the multivariatecase. A functional on F [u, v] is a special x-operator mapping F [u, v] to F. Afunctional L is invertible if 〈L | 1〉 6= 0. The bivariate power series λ (s, t) =Leus+vt becomes a translation invariant operator λ (Du,Dv) (in ΣDu,Dv ) by defin-ing λ (Du,Dv) eus+vt = λ (s, t) eus+vt. We call λ (Du,Dv), or op (L), the operatorassociated to L.

Theorem 4.3.1 (Bivariate Functional Expansion Theorem). Suppose L is an in-vertible linear functional on F[x, y], and (bm,n) is the basic sequence for the deltapair (B1, B2) =(β−1

1 (Du,Dv) , β−12 (Du,Dv)

). Any polynomial in F [u, v] can be expanded in the

formp(u, v) =

∑k,l≥0

⟨L | Bk1Bl2p

⟩op(L)−1bk,l(u, v).

If (sm,n) is a Sheffer sequence for (B1, B2), then

sm,n(u, v) =

m∑k=0

n∑l=0

〈L | sk,l〉 op(L)−1bm−k,n−l(u, v)

∑m,n≥0

sm,n (u, v) smtn =

∑k,l≥0

〈L | sk,l〉 sktl euβ1(s,t)+vβ2(s,t)

Leuβ1(s,t)+vβ2(s,t)and

m∑k=0

n∑l=0

〈L | bk,l〉 sm−k,n−l (u, v) =

m∑k=0

n∑l=0

〈L | sk,l〉 bm−k,n−l (u, v) .

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 the point (n,m) in k (discretetime) steps, under the following restriction: To reach any point (i, j) on the paththe random walker needs l > ai + bj steps, where a and b are given nonnegativeintegers. Because we require that the number of steps is large at each point, we aretalking about a “slow”walker, respecting a speed limit (see Exercise 5.1.8). Supposethe slow walker gets a ticket whenever her speed exceeds the limit ai+bj+1. Denoteby D (n,m; k; l) the number of such walks from (0, 0) to (n,m) in k steps and ltickets. Let us take the generating function of those tickets,

D (n,m; k) :=

∞∑l=0

D (n,m; k; l) tl.

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 conditionsare

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) 6= (0, 0).

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...

...... τ4−τ3

3... (τ−1)3

...... τ3−τ2 τ3

2 τ2−3τ+3 (τ−1)2... τ(τ−1) 3τ3−5τ2 τ2

... τ2+τ+1

...

1 τ−2 2(τ−1)(τ−2) τ−1 2τ2−4τ τ 2τ2−2τ−2 τ+1 2τ2−2 τ+2 2τ2+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→ nk = 0 k = 1 k = 2 k = 3 k = 4

At each k, the values D (n,m; k) can be extended to polynomials of degreem+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

bm,n (u, v) =

(u

m

)(v

n

).

This basic sequence does not contain the variable τ at all. The initial values willbring τ to the solution, such that D (n,m; k) is of degree n+m. In view of Theorem4.3.1 we let rm,n (u, v) = sm,n (u+ an+ bm, v + an+ bm), a Sheffer sequence for(E−bu E−bv ∆u, E

−au E−av ∆v

). This delta pair has the basic polynomials

b(b,a)m,n (u, v) =

uv + anu+ bmv

(u+ an+ bm) (v + an+ bm)

(u+ an+ bm

m

)(v + an+ bm

n

).

(see Exercise 4.2.8). In terms of (rm,n) we can write the (recursive) initial valuesas (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)

= (τ Eval0,0− (τ − 1) Eval1,1) rm,n = δm,0δn,0.


The operator associated to L is

µL = τ − (τ − 1)E1uE

1v = E1

uE1v

(τE−1

u E−1v − τ + 1

)with inverse

µ−1L =

E−1u E−1

v

1− τ(1− E−1

u E−1v

) .The bivariate Functional Expansion Theorem 4.3.1 implies

rm,n (u, v) = µ−1L b(b,a)

m,n (u, v) = E−1u E−1

v

∑l≥0

τ l(1− E−1

u E−1v

)lb(b,a)m,n (u, v) .

We find

D (n,m; ξ) = δsm,n (u, v) = δE−an−bmu E−an−bmv rm,n (u, v)

= δ (EuEv)−an−bm−1

∑l≥0

τ ll∑i=0

(l

i

)(−1)

iE−iu E−iv b(b,a)

m,n (u, v)

=∑l≥0

τ ll∑i=0

(l

i

)(−1)

i ξ − i− 1− an− bmξ − i− 1

(ξ − i− 1

m

)(ξ − i− 1

n

),

hence

D (n,m; k; l) =

l∑i=0

(l

i

)(−1)

i k − i− 1− an− bmk − i− 1

(k − i− 1

m

)(k − i− 1

n

)

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) =

= E−1v

∑l≥0

τ lE−l−1u (∆u +∇v)l

uv − bmu− anvuv

(u

m

)(v

n

).

Expand this expression to show that D (n,m; k; l) =

l∑i=0

(l

i

)(−1)

i k − i− 1− an− bmk − i− 1

(k − i− 1

m

)(k − i− 1

n

)4.3.3. Find the solution to the system of differential equations

∂

∂usm,n (u, v) = sm−1,n (u, v) and

∂

∂vsm,n (u, v) = sm,n−1 (u, v)

such that sm,n (m+ n,m− n) = 1/ (m+ n)!

Chapter 5

Special Constructions in SeveralVariables

Three “degenerated” types of multivariate polynomial sequences will be consid-ered in this chapter. First we study sequences (bn1,...,nr (ξ)) which can be thoughtas “diagonalizations”of multivariate Sheffer sequences (bn1,...,nr (x1, . . . , xr)), set-ting 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 themultinomial coeffi cient

(ξ+n1+···+nrn1,...,nr

). A different kind of multi-indexed polynomi-

als is considered in section 5.2, where we set all but one of the variables equal to0.

The third type reduces to one the number of indices but keeps the variables,bn (x1, . . . , xr) =

∑i1+···+ir=n bi1,...,ir (x1, . . . , xr). The sequence (bn (x1, . . . , xr))

is called a Steffensen sequence [83].

5.1 Multi-indexed Sheffer Sequences

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

σ (s, t) eξ(β1(s,t)+β2(s,t)) =∑m,n≥0

sm,n (ξ, ξ) smtn.

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 order0, then the coeffi cient of ξm+n in [smtn] eξ(β1(s,t)+β2(s,t)) equals

m+n∑i=0

(m+ n

i

)[sm−iti−m

]φ1 (s, t)

iφ2 (s, t)

m+n−i

=

(m+ n

m

)[s0t0

]φ1 (s, t)

mφ2 (s, t)

n=

(m+ n

m

)φ1 (0, 0)

mφ2 (0, 0)

n

which is always different from 0. Hence sm,n (ξ, ξ) has degreem+n in ξ. We call theoperator 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 diagonalizationoperator we write δ : sn1,...,nr (x1, . . . , xr) → sn1,...,nr (ξ); there should be noconfusion with the Kronecker delta δi,j . The polynomials sn1,...,nr (ξ) are calleda multi-indexed Sheffer sequence. From the generating function it is clear thatmulti-indexed Sheffer sequences and their basic sequence (i.e., σ = 1) satisfy thebinomial theorem.

For any delta pair (β1, β2) holds that β (s, t) := β1 (s, t)+β2 (s, t) is a bivariatepower series such that β (0, 0) = 0, and [β]1,0 6= 0, [β]0,1 6= 0. Vice versa, any βwith these properties can be decomposed as a sum of two delta series β1 andβ2, but β1 and β2 are not uniquely defined! This ambiguity should be seen as astrength, not a weakness of the theory!

Example 5.1.1. In Example 4.1.4 we found the trivariate basic sequence

bl,m,n (u, v, w) =

(u− 1 + l

l

)(v − 1 + l +m

m

)(w − 1 + l +m+ n

n

).

Diagonalization gives

bl,m,n (ξ) =

(ξ − 1 + l +m+ n

l,m, n

).

Note that

∇ξbl,m,n (ξ) :=

(ξ − 1 + l +m+ n

l,m, n

)−(ξ − 2 + l +m+ n

l,m, n

)= bl−1,m,n (ξ) + bl,m−1,n (ξ) + bl,m,n−1 (ξ) ,

thus∇ξδ = δ (B1 +B2 +B3) = δ

(E−1

2 E−13 ∇1 + E−1

3 ∇2 +∇3

).

The polynomials bl,m,n (ξ) have the generating function

∑l,m,n≥0

(ξ − 1 + l +m+ n

l,m, n

)rlsmtn = (1− r − s− t)−ξ .

5.1. Multi-indexed Sheffer Sequences 139

The same polynomials could be obtained by diagonalizing the trivariate basic se-quence (

u− 1 + l +m+ n

l

)(v − 1 +m

m

)(w − 1 +m+ n

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 adelta triple of basic operators. In general, we call the (univariate) delta operatorBξ on F [ξ] that satisfies Bξsm,n (ξ) = δ (B1 +B2) sm,n (u, v) the delta operatorof the multi-indexed Sheffer sequence (sm,n (ξ)), if degBξsm,n = m+ n− 1. Thislast 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 op-erator (but still is of binomial type), even if we know a multivariate basic se-quence (bm,n (u, v)) such that bm,n (ξ) = δbm,n (u, v). In this case, all we cansay is that δ (B1 +B2) bm,n (u, v) = bm−1,n (ξ) + bm,n−1 (ξ). For example, let(B1, B2) = (Du,−Dv), thus bm,n (u, v) = um

m!(−v)n

n! . Then

δ (B1 +B2) bm,n (u, v) = δ

(um−1

(m− 1)!

(−v)n

n!+um

m!

(−v)n−1

(n− 1)!

)

=ξm+n−1 (−1)

n−1

m!n!(n−m) .

The mapping bm,n (ξ) 7→ bm−1,n (ξ) + bm,n−1 (ξ) has bn,n (ξ) in its kernel, for alln > 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]; weneed 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≥0

bm,n (ξ) smtn = eξβ(s,t),

where β (0, 0) = 0, and [β]1,0 6= 0, [β]0,1 6= 0. There exists a delta operator Bξ onF [ξ] such that

Bξbm,n (ξ) = bm−1,n (ξ) + bn,m−1 (ξ)

for all m,n ≥ 0 iff there is a univariate delta series β, such that Bξ = β−1 (D)

and β (s+ t) = β (s, t).


Proof. We want

(s+ t) eξβ(s,t) = β−1 (D) eξβ(s,t) = C(β)β−1 (D) eξw = C(β)β−1 (w) eξw

= β−1 (β (s, t)) eξβ(s,t)

hence (s+ t) = β−1 (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 =β−1ρ (Du,Dv), and β1 (s, t) + β2 (s, t) = β (s, t).

In Example 5.1.2 we have β (s, t) = s− t, and we would need a delta series βsuch that β (s− t) = s + t, which does not exist. Note that β (s+ t) = β (s, t) iffs + t = β−1 (β (s, t)). In Example 5.1.1 we found β (r, s, t) = − ln (1− r − s− t),and β−1 (D) = ∇ξ = 1− e−D, hence β−1 (− ln (1− r − s− t)) = r + s+ t.

5.1.2 Translation invariance of diagonalization, and some examples.

The diagonalization δ is translation invariant in the following sense.

Lemma 5.1.4. Let Eaξ be the translation by a ∈ F on F [ξ]. The diagonalization δis translation invariant,

Eaξ δ = δEauEav .

Proof. Let p (u, v) ∈ F [u, v]. Then

δEauEavp (u, v) = δp (u+ a, v + a) = p (ξ + a) = Eaξ p (ξ) = Eaξ δp (u, v) .

Of course, δ is not invertible, hence there can be infinitely many bivariateSheffer sequences and delta pairs giving the same multi-indexed Sheffer sequenceand delta operator.

Example 5.1.5. Let α1 (s, t) = s+ st and α2 (s, t) = t− st, thus eus(1+t)+vt(1−s) =

∑k≥0

(us+ vt+ st (u− v))k

k!=

∞∑n,m=0

smtnn+m∑k=0

uk−nvk−m (u− v)n−k+m

(k − n)! (k −m)! (n+m− k)!.

Hence

am,n (u, v) =

m+n∑k=0

(uk−nvk−m (u− v)

m+n−k)/ ((k − n)! (k −m)! (n+m− k)!) .

If we proceed to the multi-indexed polynomials am,n (ξ) = δam,n (u, v), we are leftwith am,n (ξ) = ξm+n/ (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 = Dρand get bm,n (ξ) = δ

(um

m!vn

n!

)= am,n (ξ). The delta operator Bξ certainly exists in


this case, because β (s, t) = s+ t, thus Bξ = Dξ,and therefore Dξδ = δ (Du +Dv).In other words,

Dξξm+n/ (m!n!) = ξm+n−1/ ((m− 1)!n!) + ξm+n−1/ (m! (n− 1)!) .

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

α−11 (s, t) =

1

2

(s+ t+ 1−

√(s+ t+ 1)

2 − 4s

)α−1

2 (s, t) =1

2

(s+ t− 1 +

√(1 + s+ t)

2 − 4s

)hence α−1

1 (s, t) + α−12 (s, t) = s + t, and therefore A1 + A2 = B1 + B2. This will

happen only if Dξ = β−1 (Dξ), because the condition β−1 (α1 + α2) = s+ t is alsoequivalent to s + t = β

(α−1

1 + α−12

). In general, only δ (A1 +A2) = δ (B1 +B2)

will hold.

More applied is the following example, counting paths with k occurrences ofa certain pattern.

Example 5.1.6. Let D (n,m; k) enumerate the number of →, ↑-paths that stayabove the diagonal, reach the point (n,m), and contain the pattern u2r2 exactly ktimes. 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 1767 1 7 21 40 63 906 1 6 15 24 33 405 1 5 10 13 15 134 1 4 6 6 5 03 1 3 3 2 02 1 2 1 01 1 1 00 1 0

n→ 0 1 2 3 4 5

7 48 150 2856 35 92 1625 24 51 724 15 24 233 8 8 02 3 01 00

2 3 4 5

15 84 22210 45 936 20 273 6 01 00

4 5 6

k = 0 k = 1 k = 2D (n,m; k) for k = 0, 1, 2


Let tk,n (m) = D (n+ 2k,m+ 2k + n; k), hence tk,n(−1) = δn+k,0 anddeg 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,

I = E−1u E−1

v + EuEvB2 −B22 +B1

hence

I − E−1v + E−1

v

(1− E−1

u

)= EuEvB2 −B2

2 +B1, or

∇v + E−1v ∇u = EuEvB2 −B2

2 +B1.

We let ∇v = EuEvB2 − B22 , 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=0

(n− 2k − j

j

)(m− n+ 1) (−1)

j

m− k − 2j + 1

(m− k − 2j + 1

k

)×

×(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 6= 0, [β]0,1 6=0. The bivariate power series ∂

∂sβ (s, t) is of order zero 0, and so is ∂∂tβ (s, t).

Let (sm,n (ξ)) be the multi-indexed Sheffer sequence with generating function(∂∂sβ (s, t)

)eξβ(s,t), hence∑m,n≥0

ξsm,n (ξ) smtn = ξ

(∂

∂sβ (s, t)

)eξβ(s,t)

=∂

∂seξβ(s,t) =

∂

∂s

∑m,n≥0

bm,n (ξ) smtn

=∑m,n≥0

(m+ 1) bm+1,n (ξ) smtn


and therefore

sm,n (ξ) = (m+ 1)bm+1,n (ξ)

ξ.

In the same way we find tm,n (ξ) = (n+ 1) bn+1 (ξ) /ξ if we define the Sheffersequence (tm,n (u, v)) with generating function

(∂∂tβ (s, t)

)eξβ(s,t). We combine

both results in the following theorem.

Theorem 5.1.7. Let (bm,n (ξ)) be a multi-indexed basic sequence with generatingfunction eξβ(s,t), and let c, d ∈ F such that c [β]1,0 +d [β]0,1 6= 1. The multi-indexedSheffer sequence (rm,n (ξ)) with generating function∑

m,n≥0

rm,n (ξ) smtn =

(1− c

(∂

∂sβ (s, t)

)− d

(∂

∂tβ (s, t)

))e(ξ−η)β(s,t)

has roots in ξ = η + cm+ dn (η ∈ F),

rm,n (ξ) =ξ − η − cm− dn

ξ − η bm,n (ξ − η) .

Proof. By superposition, any linear combination of multi-indexed Sheffer sequencesof total degreem+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

rm,n (ξ) = bn,m (ξ − η)− (cm+ dn)bm,n (ξ − η)

ξ − η .

The polynomials b(c,d)m,n (ξ) are basic for E−c−dBξ because b

(c,d)m,n (0) = δ0,m+n and

E−c−dBξrm,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))

= δ(E−cu E−dv B1 + E−cu E−dv B2

)rm,n (u+ cm+ dm, v + cm+ dn) ,

thus one way of arriving at E−c−dBξ is by diagonalizing(E−cu E−dv B1, E

−cu E−dv 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(E−cu E−cv B1, E

−du E−dv B2

),

thus (sm,n (ξ + cm+ dn)) is a multi-indexed Sheffer for

δ(E−cu E−cv B1 + E−du E−dv B2

).


Choosing for (sm,n (ξ)) the special Sheffer sequence (rm,n (ξ)) from Theorem 5.1.7,with η = 0, we obtain

rm,n (ξ + cm+ dn) =ξ

ξ + cm+ dnbm,n (ξ + cm+ dn) ,

the basic sequence for δ(E−cu E−cv B1 + E−du E−dv 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 pat-tern 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 sub-tract are those where the pattern occurs at the end, and is preceded by an ↑-step. Ifk > 0 then we may have subtracted the k-th occurrence of the pattern, so we haveto add it back in. However, the pattern rα may already occur once at the end of thepath 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 comefromD (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 15729 1 9 44 154 423 9518 1 8 35 110 270 5367 1 7 27 75 161 2736 1 6 20 48 87 1185 1 5 14 28 40 364 1 4 9 14 13 03 1 3 5 5 02 1 2 2 01 1 1 0 k = 00 1 0

n→ 0 1 2 3 4 5

7 60 2756 45 1805 32 1084 21 563 12 212 5 01 00

k = 1

4 5 6

15 90 2259 36 04 00

8 9 10k = 0 k = 1 k = 2D (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) .


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

Ev (Eu − I) + Ev − I = Ev∆u + ∆v =B2 −Bα21−B2

+B1.

Hence we can assume that we have the relation (Ev∆u,∆v) =(B1,

B2−Bα21−B2

)be-

tween delta pairs. The delta operator (Ev∆u,∆v) has the basic sequence((uk

)(v−kn

))k,n≥0

(Example 4.2.10). We found the basic sequence (bk,n) for (B1B2)

in Example 4.2.2,

bk,n (u, v) =

(u

k

) n∑l=0

(v − kl

)(l

n− l

)α−1

.

For sk,n (ξ) we need the Abelization of bk,n (ξ) where sk,n(αk + n − 1) = δk+n,0,hence

sk,n (ξ) =ξ + 1− n− αk

ξ + 1bk,n (ξ + 1)

by Theorem 5.1.7. Finally, D (n,m; k)

= sk,n−αk (m) =m+ 1− nm+ 1

bk,n−αk (m+ 1)

=m+ 1− nm+ 1

(m+ 1

k

) n−αk∑l=0

(m+ 1− k

l

)(l

n− αk − l

)α−1

=m+ 1− nm+ 1

(m+ 1

k

) n−αk∑l=0

(m+ 1− k

l

)×

×b(n−αk−l)/(α−1)c∑

j=0

(−1)j(l

j

)(n− αk − (α− 1) j − 1

l − 1

)In Example 4.2.10 we found the basic sequence (bm,n) for B1 = EcuE

dvA1,

B2 = EfuEgvA2, but only when the basic polynomials for (A1, A2) factor, am,n (u, v) =

aI1 (u) aII2 (v). Hence sm,n (u, v) = bm,n (u+ cm+ fn, v + dm+ gn) is the Shefferpolynomial 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 beapplied after diagonalization.


Example 5.1.9. We consider lattice paths in N30 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-mial coeffi cient

(ξ+m+nm,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 atany time candidate C to get at least as many votes as candidate A. For the numberof paths dm,n (ξ) this means that dm,n (m− 1) = 0 for all m > 0. In addition, wewant candidate C to have at least as many votes as candidate B has more votesthan A; for example, if B gets n = 3 votes and A gets m = 1, than C must get atleast n−m = 2 votes. Thus a coalition of A and C can balance or defeat B at anytime. 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 Example2.3.6?)

m ξ = 1 ξ = 2 ξ = 34 0 0 0 0 0 0 03 0 0 0 0 0 0 5 30 100 240 450 660 6602 0 0 0 0 0 2 8 18 28 28 0 5 25 70 140 210 210 01 1 2 2 0 2 6 10 10 0 3 12 27 42 42 00 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) = 0and dm,n (n−m− 1) = 0 (m+ n > 0)

Remember that

∇ξ = δ(E−1v ∇u +∇v

)= δ

(E−1u E−1

v ∆u + E−1v ∆v

),

where(E−1v ∇u,∇v

)has basic polynomials bm,n (u, v) =

(u+m−1m

)(v+m+n−1

n

), which

satisfy the condition bm,n (u, v −m) = δm,0. We now need the basic polynomialsrm,n (u, v) for

(E−1u E−1

v ∇u, E−1v ∇v

), which equal

rm,n (u, v) =u

u+ 2m

v +m

v +m+ 2n

(u+ 2m

m

)(v +m+ 2n

n

)by Example 4.2.10. They allow us to find the Sheffer polynomials

dm,n (u, v) = E−m+1u E−n+1

v rm,n (u, v)

for(E−1v ∇u,∇v

), with roots in u = m− 1 and v = n− 1−m. Hence

dm,n (ξ) = δE−m+1u E−n+1

v rm,n (u, v)

=(ξ −m+ 1) (ξ +m− n+ 1)

(ξ +m+ 1) (ξ +m+ n+ 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.


5.1.4 Exercises

5.1.1. We begin with a univariate basic sequence (an (ξ))n∈N0 for A with generating

function∑n≥0 an (ξ) tn = eξα(t), and define pl,m,n (ξ) := (l+m+n)!

l!m!n! al+m+n (ξ) forall n ∈ Nr0. Check that the generating function of (pl,m,n (ξ)) equals eξα(t1+t2+t3),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 trivariatebasic sequence (bl,m,n (x1, x2, x3)) such that pl,m,n (ξ) = δbl.m,n (x1, x2, x3). Let

βρ (t1, t2, t3) = α(∑3

σ=ρ tσ

)− α

(∑3σ=ρ+1 tσ

). Show that

1. (β1, β2, β3) is a delta multiseries,

2. the basic sequence (bl,m,n (x1, x2, x3)) for(β−1

1 , β−12 , β−1

3

)has the generating

functionδ∑

l,m,n≥0

bl,m,n (x1, x2, x3) tl1tm2 t

n3 = eξα(t1+t2+t3),

3. α−1 (t1 + t2 + t3) = β−11 (t1, t2, t3) + β−1

2 (t1, t2, t3) + β−13 (t1, t2, t3),

4. β−1ρ (t) = α−1

(∑3σ=ρ tσ

)− α−1

(∑3σ=ρ+1 tσ

),

5. the delta operator Bξ for (bl,m,n (ξ)) satisfies the condition

Bξδ = α−1 (D) δ

and therefore Bξ = Aξ.

As an example, show that((l+m+n+ξ−1

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 polynomialbn1,n2,n3 (x1, x2, x3) =

∏3ρ=1 anρ (xρ + n1 + · · ·+ nρ−1) where

(anρ (xρ)

)nρ∈N0

is

the same univariate basic sequence for A for all ρ, hence∑nρ≥0 anρ (xρ) t

nρρ =

γ (tρ)xρ and A = (ln γ)

−1(D).

1. Let pn1,n2,n3 (ξ) := δbn1,n2,n3 (x1, x2, x3) for all nρ ≥ 0. Find the gener-ating function

∑n1,n2,n3≥0 pn1,n2,n3 (ξ) tn11 tn22 tn33 . Show that the projections

pn1,0,0 (ξ) , p0,n2,0 (ξ) and p0,0,n3 (ξ) are all equal. What does that imply forthe delta operator for (pn1,n2,n3 (ξ)), if such an operator exists?

2. Consider the generating function∑n1,n2,n3≥0 pn1,n2,n3 (ξ) tn11 tn22 tn33 in the

special case A = D, and show that there is no delta operator for (pn1,n2,n3 (ξ))in this case.


3. Let A = ∇. Check the generating function of (pn1,nl,n3 (ξ)) to show that thedelta operator Pξ for (pn1,n2,n3 (ξ)) is equal to ∇ξ. In this case, pn1,n2,n3 (ξ)

is the “right” factorization to give(ξ−1+n1+n2+n3

n1,n2,n3

)5.1.4. Show that Dnξ δ = δ (Du + · · ·+Dr)n for all n ≥ 0.

5.1.5. Show that the components A1 = ∇1E−12 E−1

3 , A2 = ∇2E−13 , and A3 = ∇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 β (s+ t) = β (s, t) is equivalentto s+ t = β

(β−1

1 + β−12

), 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 therea combinatorial interpretation for this relationship to the Catalan numbers?

5.1.8. Suppose we join the “pause step” 〈0, 0〉 to the steps →, ↑, and . Considera 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. We wantto find the probability Pr (n,m; k) that this random walk starts at (0, 0) and reachesthe point (n,m) in k (discrete time) steps, under the following restriction: To reachany point (i, j) on the path the random walker needs l > ai+ bj steps, where a andb are given nonnegative integers. Because we require that the number of steps islarge at each point, we are talking about a “slow”walker, respecting a speed limitPr (i, j; ai+ bj) = 0 when (i, j) 6= (0, 0).

Pr (n,m; k) when a = 2 and b = 1m 03 0 p00p

301

2 0 p00p201 3p2

00p201 0

1 0 p00p01 2p200p01 0 3p3

00p01 p200 (3p10p01 + p00p11)

0 1 p00 p200 p3

00 p200p10 p

400 2p3

00p10

n→ 0 0 0 0 1 0 1k = 0 k = 1 k = 2 k = 3 k = 4

Let sm,n (k) := pm+n−k00 p−n10 p

−m01 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) =p−k00 Pr (0, 0; k) = 1 for k ∈ N0, and sm,n (k) = 0 if k = an + bm when (m,n) ∈N2

0\ (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-indexedpolynomial 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) ,


Show that

sm,n (ξ) =ξ − an− bm

ξ

(ξ

m

) m∑j=0

(m

j

)(ξ −mn− j

)αj

=

(ξ

m

) m∑j=0

(m

j

)(ξ −mn− j

)αj − (an+ bm)

m

(ξ − 1

m

) m∑j=0

(m

j

)(ξ −mn− j

)αj

by Theorem 5.1.7. In terms of probabilities,

Pr (n,m; k) = pk−m−n00 pn10pm01sm,n (k)

=k − an− bm

k

(k

m

) m∑j=0

(m

j

)(k −mn− j

)pj+k−m−n00 pj11p

n−j10 pm−j01

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 + nsteps. In addition, assume that a = b = 0. Then

Pr (n,m; k) =

(k

m+ n− k, k −m, k − n

)pm+n−k

11 pk−m10 pk−n01 .


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 obtaina bi-indexed sequence of polynomials sm,n (u) := sm,n (u, v), say. The binomialtheorem (4.2) tells us that

sm,n (x+ y) =

m∑i=0

n∑j=0

si,j (x) bm−i,n−j (y, 0)

where (bm,n (u, v)) is the basic sequence corresponding to (sm,n (u, v)). The gen-erating function of (sm,n (u)) is of the form∑

m,n≥0

sm,n (u) smtn = σ (s, t) euβ1(s,t).

If we write bm,n (u) for bm,n (u, 0), then∑m,n≥0 bm,n (u) smtn = euβ1(s,t), where∑

m,n≥0

bm,n (u, v) smtn = euβ1(s,t)+vβ2(s,t)

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 β2 (s, t) = t. The selected deltaoperator pair

(β−1

1 (Du,Dv),Dv)has the basic sequence

(bm,n (u, v)

), where

bm,n (u, v) =

n∑j=0

bm,j (u)vn−j

(n− j)! = Bmvn

n!and

bm,n (u) = bm,n (u, 0) =

[Bm

vn

n!

]v=0

We see that bm,n (u, v) = B(m) vn

n! where B(m) is the operator B(m) =

∑∞j=0 bm,j (u)Djv :

F [v]→ F [u, v] for all m ≥ 0, translation invariant in v. Therefore,

B1B(m) = B(m−1).

Note that B1 = β−11 (Du,Dv), where β−1

1 (β1 (s, t) , t) = s, which shows that B1

depends on our choice of B2 = Dv (but β1 (s, t) is independent of that choice).

Example 5.2.1. Let

bm,n (u, v) =

m∑j=0

(n+m− 2j

m− j

)(n+m− j − 1

j

)um−jvn−j

(n+m− 2j)!

=

min(m,n)∑j=0

(n+m− j − 1

j

)um−jvn−j

(m− j)! (n− j)! ,

5.2. Polynomials with all but one variable equal to 0 151

the polynomial sequence in the introduction to this chapter. We obtain the gener-ating function∑m,n≥0

bm,n (u, v) smtn =∑j≥0

sjtt∑m,n≥j

(n+m+ j − 1

j

)(us)

m(tv)

n

m!n!

=∑m,n≥0

(us)m

(tv)n

m!n!(1− st)−m−n = e

us+vt1−st = euβ1(s,t)+vβ2(s,t),

and we see that bm,n (u) =(m−1n

)um−n/ (m− n)! if m ≥ n, and 0 otherwise. We

find

bm,n (u, v) =

n∑j=0

bm,j (u)vn−j

(n− j)! =

m∑j=0

(m− 1

j

)um−j

(m− j)!Djv

vn

n!

=∑j≥0

(m− 1

j

)DjuDjv

um

m!

vn

n!= (1 +DuDv)m−1 u

m

m!

vn

n!= B(m) v

n

n!,

thus B(m) = (1 +DuDv)m−1um/m! for m ≥ 1. If m = 0 then b0,n (u, v) = vn/n!,

and therefore B(0) = I. From β1 (s, t) = s/ (1− st) and β2 (s, t) = t follows B1 =β−1

1 (Du,Dv), where β−11 (s, t) = s/ (1 + st), hence

B1 =Du

1 +DuDv.

We check our calculations by applying B1 to B(m) for m ≥ 1,

B1B(m) =

Du1 +DuDv

(1 +DuDv)m−1 um

m!= (1 +DuDv)m−2 um−1

(m− 1)!= B(m−1).

Suppose we are given B1 as the solution of the operator equations A1 =τ1 (B1,Dv), A2 = Dv, and we would like to expand bm,n (u) in terms of the basicsequence (am,n) for (A1,Dv),

am,n (u, v) =

n∑l=0

am,l (u, 0)vn−l

(n− l)! = A(m) vn

n!

From Corollary 4.2.13 we know that

bm,n (u, v) =1

mθA1

υm1 (A1,Dv) am−1,n (u, v)

=1

mθA1

m−1∑i=0

n∑j=0

[ε1+i−m

1

∂τ1∂s

]i,j

am−1−i,n−j (u, v)

where ε1 (s, t) = s/τ1 (s, t)).


Example 5.2.2. Suppose we want to know the number of ballot paths (taking stepsr =→ and u =↑, and staying weakly above the diagonal y = x) containing thepattern 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) =

D(n−1,m; k)+D(n,m−1; k)−k∑i=0

∑l≥0

(−1)i−lD(n−2− l,m−1− l; k−i)(l + 1

i

).

m 1 10 45 128 2739 1 9 36 91 1748 1 8 28 62 1057 1 7 21 40 596 1 6 15 24 305 1 5 10 13 134 1 4 6 6 43 1 3 3 2 02 1 2 1 01 1 1 0 k = 00 1 0

n→ 0 1 2 3 4

0 9 72 2730 8 86 1860 7 42 1200 6 30 720 5 20 390 4 12 180 3 6 60 2 2 00 1 00 0 k = 10

1 2 3 4

0 8 840 7 630 6 450 5 300 4 180 3 90 2 30 1 00 00

2 3 4k = 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 extendedto a polynomial of degree n for all k ≥ 0, and bk,n (0) = δkδn. The recursion forbk,n (v) equals

bk,n (v) = bk,n−1 (v + 1)+bk,n (v − 1)−k∑i=0

∑l≥0

(−1)i−lbk−i,n+i−2−l (v + 1)

(l + 1

i

),

and for bk,n (u, v) =∑ki=0 bi,n (v) uk−i

(k−i)! it remains the same, bk,n (u, v) =

bk,n−1 (u, v + 1) + bk,n (u, v − 1)−k∑i=0

∑l≥0

(l + 1

i

)(−1)i−lbk−i,n+i−2−l (u, v + 1) .

In terms of operators,

∇v = B2Ev −EvB2

(B2 −Du

)1 +B2 −Du

, hence

τ2 (s, t) = Evt

(1− t− s

1 + t− s

)= Evt

1

1 + t− s .

5.3. Cross-Sequences and Steffensen Sequences 153

We find for n > 0

bk,n (u, v) =1

nθ∇vυ

n2 (Du,∇v) ak,n−1 (u, v)

=1

nθ∇v

k∑i=0

n−1∑j=0

[ε1+j−n

2

∂τ2∂t

]i,j

ak−i,n−1−j (u, v)

=1

nθ∇v

k∑i=0

n−1∑j=0

(−1)j n

i+ n

(i+ n

i+ j

)(i+ j

j

)En−jv ak−i,n−1−j (u, v)

for n > 0, where θ∇2= (uEu − u+ v)Ev and ak,n (u, v) = uk

k!

(n−1+vn

). Hence

bk,n (v) = v

n−1∑j=0

(−1)j

i+ n

(k + n

k + j

)(k + j

j

)(2 (n− j) + v − 1

n− 1− j

),

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≥0

pm,n (u, v) smtn = ρ (s, t) euβ(s,t)+vα(s,t).

The polynomials

s[λ]m (x) :=

m∑n=0

pm−n,n (x, λ)

m ≥ 0, are a Steffensen sequence [83, Section 8] (note that in [83, Section 8] aSteffensen sequence may be of degree less than m in λ). Our Steffensen sequencehas the generating function

∞∑m=0

s[λ]m (x) rm :=

∞∑m=0

m∑n=0

pm−n,n (x, λ) rm−nrn = ρ (r, r) exβ(r,r)+λα(r,r)

Of course, ρ (r, r) is a power series of order 0 in r, and β (r, r) and α (r, r) are both

univariate delta series in r. Hence(s

[λ]m (x)

)m≥0

is for

• fixed λ a Sheffer sequence for β−1 (Dx) in the variable x, and for

• fixed x a Sheffer sequence for α−1 (Dλ) in the variable λ.


If ρ (s, t) = 1, the resulting Steffensen sequences(c[λ]m (x)

)is called a cross-

sequence.If (bm (x)) is the basic sequence for β−1 (Dx), and (am (λ)) the basic sequence

for α−1 (Dλ), then

s[λ]m (x+ y) =

m∑k=0

s[λ]k (x) bm−k (y) =

∑k≥0

bk (y)(α−1 (Dλ)

)ks[λ]m (x)

= eyβ(α−1(Dλ))s[λ]m (x)

s[λ+µ]m (x) =

m∑k=0

s[λ]k (x) am−k (µ) =

∑k≥0

ak (µ)(β−1 (Dx)

)ks[λ]m (x)

= eµα(β−1(Dx))s[λ]m (x)

s[λ+µ]m (x+ y) =

m∑k=0

s[λ]k (x) c

[µ]m−k (y) .

Thus the operator eµα(β−1(Dx)) on F [x] acts as the translation by µ in the para-

meter λ, and the operator eyβ(α−1(Dλ)) on F [λ] acts as the translation by y in theparameter x. Seen as operators on F [x, λ] we get

eµα(β−1(Dx)) = eµDλ = Eµλ

and

eyβ(α−1(Dλ)) = eyDx = Eyx ,

or β−1 (Dx) = α−1 (Dλ). Hence

s[λ]m (x) = Tλs[0]

m (x) = P xs[λ]m (0)

where T := eα(β−1(Dx)), and P := eβ(α−1(Dλ)). Because T (and P ) are invertible,the operators Tβ−1 (Dx) and Pα−1 (Dλ) are delta operators (on F [x] and F [y],respectively). According to (2.36) the delta operator Tβ−1 (Dx) has for m > 0the basic polynomials xTmbm (x) /x. Similar, Pα−1 (Dλ) has basic polynomialsλPm 1

λam (λ).

Suppose(p

[λ]m (x)

)m≥0

is the Steffensen sequence with generating function

β′ (r) exβ(r)+λα(r), thus p[0]m (x) = (m+ 1) bm+1 (x) /x for m > 0, and p[0]

0 (x) ≡β′ (0). For this Abelization type we find (Exercise 5.3.1) x+y

m+1p[m+1]m (x+ y) =

y

m+ 1p[m+1]m (y) +

x

m+ 1p[m+1]m (x) +

m∑k=1

x

kp

[k]k−1 (x)

y

m+ 1− kp[m+1−k]m−k (y) (5.2)

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

sequence(t[λ]m (x)

)m≥0

with generating function α′ (r) exβ(r)+λα(r), thus t[λ]m (0) =

(m+ 1) am+1 (λ) /λ for m > 0, and t[λ]0 (0) ≡ 1. Now λ+µ

m+1 t[λ+µ]m (m+ 1) =

λ

m+ 1t[λ]m (m+ 1)+

µ

m+ 1t[µ]m (m+ 1)+

m∑k=1

λ

kt[λ]k−1 (k)

µ

m+ 1− k t[µ]m−k (m+ 1− k)

for all m ≥ 0 .

Example 5.3.1. Suppose the basic polynomials lm,n (x, y) have the generating func-tion ∑

m,n≥0

lm,n (u, v) smtn = eus/(s−1) (1− t)−v ,

thus

lm,n (u, v) =

(n+ v − 1

n

) m∑i=0

(−u)i

i!

(m− 1

m− i

).

The cross-sequence with generating function exr/(r−1) (1− r)−α−1 is the sequenceof Laguerre polynomials

m∑n=0

lm−n,n (x, α+ 1) =

m∑i=0

(−x)i

i!

(m+ α

m− i

)= L(α)

m (x) . (5.3)

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

L(α)m (x− y) =

m∑i=0

yi

i!

m−i∑k=0

L(α)m−i−k (x)

(k + i− 1

k

)5.3.3. Show that for the Laguerre polynomials holds

L(α+β)m (x+ y) =

m∑k=0

L(α)k (x)L

(β−1)m−k (y)

5.3.4. Show

L(α+1)m (x) =

m∑k=0

L(α)k (x) and

L(α+1)m (x)− L(α+1)

m−1 (x) = L[α]m (x)


5.3.5.

yL(m+2)m (y) + xL(m+2)

m (x)− (x+ y)L(m+2)m (x+ y)

= xy

m∑k=1

m+ 1

k (m+ 1− k)L

(k+1)k−1 (x)L

(m+2−k)m−k (y)

5.3.6. The Hermite polynomials of variance σ2 are defined as

H[1/σ2]n (x) = e−D

2/(2σ2)xn/n!

for positive σ. The special case H [1]n is called “Hermite polynomial” later (see

(6.36)). Show that(H

[λ]n

)is not a cross sequence as we defined it, but has similar

properties. For example,

H [λ+µ]m (x) =

m/2∑k=0

(−1)k

2−kµkH[λ]m−2k (x)

and

H [λ+µ]m (x+ y) =

m∑k=0

H[λ]k (x)H

[µ]m−k (y)

Chapter 6

A General Finite OperatorCalculus

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 section2.2.2. Other approaches are closely related [97], or more general [9]; however, webelieve that this setting is exactly at the right level for our purpose, providing abetter understanding of the Finite Operator Calculus. In the following we give uptranslation invariance, but retain the commutativity of the operators under study.Of course, no translation invariance means no binomial theorem! From now on wewill 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(in x) as coeffi cients. The polynomial pn (x) = [tn] 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]],

f (x, t) =∑n≥0

pn (x) tn =∑n≥0

tnn∑k=0

pn,kxk =

∑k≥0

xk∑n≥0

pn,ktn, (6.1)

but only finitely many of the coeffi cients pn,k =[xk]pn (x) are different from

zero for every n. For example, the series (1− x)−1

(1− t)−1 ∈ k [[t]] [[x]] is notin k[x][[t]]. More can be said if we consider the special case when deg pn (x) = nfor all n ≥ 0. The following two closely related results of Freeman [35] give someinsight into the structure of this kind of series.

158 Chapter 6. A General Finite Operator Calculus

Lemma 6.1.1. Suppose s(x, t) =∑n≥0 sn (x) tn where (sn (x)) is a basis of k[x]

(i.e., deg sn = n and s0 6= 0). If for some sequence (pn) of polynomials and (βn)of power series holds

s(x, t) =∑n≥0

pn(x)βn (t) (6.2)

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 matricesof (sn (x)), (pn (x)), and (βn (t)), respectively, thus sn,k =

[xk]sn (x), etc. The first

statement says that if in the matrix product (sn,k) = (pn,k) (βn,k)T the matrix

(sn,k) and one of the matrices on the right hand side are lower triangular andinvertible, then the remaining one is of the same type.

Example 6.1.2. The standard basis r0(x) := 1, rn(x) := x (1− xn) /(1 − x) forn > 0, generates the series 1 + xt

(1−t)(1−xt) in k[x][[t]]. From

1 +xt

(1− t) (1− xt) = 1 +∑n≥1

xntn

1− t

follows that the basis (xn) corresponds to the pseudobasis β0 (t) = 1, βn (t) =tn/ (1− t) for n ≥ 1, in k [[t]], where deg βn = n. In matrix notation,

10 10 1 10 1 1 1...............

=

10 10 0 10 0 0 1...............

1 0 0 0 0

1 1 1 11 1 1

1 1...

T

Lemma 6.1.3. Suppose (pn) is a given basis of k[x]. If for some sequence (sn) ofpolynomials and (βn) of power series holds∑

n≥0

sn(x)tn =∑n≥0

pn(x)βn(t) (6.3)

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) =

(pn,k) (βn,k)T , where (pn,k) is lower triangular and invertible. The matrix (sn,k) is

of the same kind iff (βn,k)T 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 andmultiplicative structure of formal power series. Therefore, we restrict ourselves tothe case βn (t) = β (t)

n, where β (t) is of order 1. Other choices are possible, andindeed 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 6= 0. If the matrix(pn,k) is again lower triangular and invertible, then (sn,k) = (pn,k) (βn,k)

T , wheredeg (sn) = bn/wc in Lemma 6.1.3. For example, when w = 2, β (t) = t2/ (1− t),and pn (x) = xn (compare to example 6.3.3), then sn (x) = x

∑bm/2c−1k=0

(m−k−2

k

)xk.

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

X∑n≥0

pn (x) tn =∑n≥0

(Xpn (x)) tn

for all∑pn (x) tn ∈ k[x][[t]]. The k[[t]]-linear operators on k[x][[t]] are called x-

operators; their ring is denoted by Lx. For an x-operator X we only have todescribe its action on some basis of k [x]. The linear operators on k [x] can beidentified with Lx, if we remember the k [[t]]-linearity.

Example 6.1.5. (1) D, (2) M (x) (multiplication by x); (3) η := M (x)D : xn 7→nxn (note that η + 1 is invertible); (4) χ : p (x) 7→ (p (x)− p (0)) /x (divisionoperator), with χc = 0 for all c ∈ k. We have χ = (η + 1)

−1D, but χ is also theleft inverse of M (x), χM (x) = I.

Let s (x, t) =∑n≥0 sn (x, t) tn 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 anydegree). Hence

Xs (x, t) =∑n≥0

qn (x) tn =∑n≥0

tn∑k≥0

qn,ksk (x) =∑k≥0

sk (x)∑n≥0

tnqn,k

where qn,k = [sk (x)] qn (x) is different from 0 for only finitely many k for very n.

Therefore, the sequence(∑

n≥0 qn,ktn)n≥0

is a sequence of elements from k[x][[t]]

that is exactly of the type occurring in the presentation of any f(x, t) ∈ k[x][[t]]as an element of k[[t]][x] (see (6.1)). We call the mapping

Xs

∑k≥0

sk (x, t) tk =∑k≥0

sk (x)∑n≥0

tnqn,k

Xstk =

∑n≥0

tnqn,k

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 Lt =Xs : X ∈ Lt

is the set of all

t-operators on k[x][[t]]. It is easy to see that Lt does not depend on s (x, t).Let f (x, t) =

∑n≥0 pn (x) tn be an arbitrary element from k[x][[t]]. Define


the x-operator X : xn 7→ pn (x). Then

f (x, t) =∑n≥0

pn (x) tn = X (1− xt)−1= X1/(1−xt) (1− xt)−1

=∑k≥0

xkX1/(1−xt)tk

=∑k≥0

xk∑n≥0

pn,ktn

hence the (1− xt)−1-transform ofX : xn 7→ pn (x) is X1/(1−xt) : tk 7→∑n≥0 pn,kt

n.

Vice versa, we can take any T ∈ Lt and define the s-transform of T , Ts, say,as the x-operator

Tss (x, t) = Ts (x, t) .

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

n≥0

qn (x) tn = Ts(x, t).

Hence Tssn (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. Forexample, if Ttk =

∑n≥0 pn,kt

n is of order 0 for all k ≥ 0, then for any basis (sn)we will get

T∑k≥0

sk (x) tk =∑k≥0

sk (x)∑n≥0

pn,ktn =

∑n≥0

tn∑k≥0

pn,ksk (x)

where the coeffi cient of t0 equals∑k≥0 p0,ksk (x), which is an infinite sum of basis

polynomials, and therefore not in k[x]. Operators T where ordTtn ≥ n − k for afixed integer k, and for all n ≥ 0, are in Lt.

Example 6.1.6. The t-operator Dt, and multiplication by t, are examples for oper-ators in Lt. We denote differentiation with respect to the power series variable tby Dt, while we write D for the (more commonly used) derivative with respect tothe variable of the polynomials.

We now give two important classes of t-operators where the image of tn is aformal power series of order larger than or equal to n. Let σ, β ∈ k [[t]] such thatord (β) > 0. The k[x]-linear operators composition C(β) and multiplication M(σ)are defined on k[x][[t]] such that

M(σ)C(β)s(x, t) = M(σ)s(x, β(t)) =∑n≥0

sn(x)σ(t)β(t)n. (6.4)

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 isclear from the context, and whenever possible, we write just T and X, withoutsubscripts. Transforming is an anti-isomorphism,

(T1T2)`

= T2T1 and X1X2 = X2X1. (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.

1. The t-operators Dt andM (t) have the ext-transforms Dt = M (x),M (t)`

=

D, (M (t)Dt)`

= M (x)D = η.

2. If ηtis the t-analog to η, i.e., ηt : tn 7→ ntn, then ηt = η is the ext-transform.

3. The division operator χ has the ext-transform

χ =(

(η + 1)−1D

)∧= D

((η + 1)

−1)∧

= M (t) (ηt + 1)−1.

4. If the basic sequence (bn) is the r-image of β (t)n (the basic sequence with

respect to R), thenC(β)`rn (x) = bn (x) (6.8)

is an example of an umbral operator.

5. Examples for (1− xt)−1-transforms are Dt = ηM (x), M (t)`

= χ, and η =M (t)Dt.


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

The standard basis (xn/n!), the exponential series ext, and the derivative operatorD are all interrelated through formal power series, and we are now looking forsimilar triples (rn(x)), r(x, t) :=

∑n≥0 rn(x)tn, and R : rn(x) 7→ rn−1(x). We

will call any standard basis (rn(x)) a reference sequence, i.e., deg rn(x) = n, andrn(0) = δ0,n. The operator R ∈ ⊗ will be called the reference operator, r(x, t) thereference series, and all three 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 R defines the sequence (rn) uniquely, because rn (0) = δ0,n.For the reference operator R holds that R is the r-transform ofM(t), R = M(t)`.Every linear operator that commutes with R can be written as a power series inR (see Exercise 6.2.4 for details).Example 6.2.1. 1. rn(x) = xn/n! with reference series ext gives the Finite Op-

erator Calculus we discussed in the previous chapters. We call this referenceframe exponential.

2. The sequence(xn/ (n!)

2)is standard; if we take it as our reference sequence,

we will get the reference operator DM (x)D,

DM (x)D xn

n!n!= D xn

n! (n− 1)!=

xn−1

(n− 1)! (n− 1)!.

Note that DM (x)D = D +M (x)D2, and

(DM (x)D)k

=

k∑i=0

(k

i

)2

(k − i)!M(xi)Dk+i. (6.9)

(the coeffi cients(ki

)2(k − i)! occur in the Laquerre polynomials, m!L

(0)m (x),

Example 5.3.1). See Exercises 6.2.2 and 6.2.3 for details. Identy (6.9) showsthe “normal ordering”of the k-th power of the operator. The coeffi cients arecalled “Generalized Stirling Numbers” in [12]. For the reference series weobtain

∞∑n=0

xntn

n!n!=

1

π

∫ π

0

e2√xt cosu du = I0

(2√xt),

the (modified, real valued) Bessel function of the first kind. Hence tI0(2√xt)

=1πDM (x)D

∫ π0e2√xt cosu du. If we change the reference operator to

(m+ 1)D +M (x)D2,

6.2. Reference Frames, Sheffer Sequences, and Delta Operators 163

where m ∈ N0 is fixed, then the reference series equals Im(2√xt)/ (xt)

m/2

(Exercise 6.2.5). More on operators of the form∑n≥0

(∑k≥0 cn,kDk

)M (xn)

can be found in [25].

3. The operator DM (x)D−D3 is in Ω, and therefore a reference operator. TheHermite polynomials Hn (x) with generating function ext−t

2/2 are a basis thatfollows the recurrence

(DM (x)D −D3

)Hn (x) /n! = Hn−1 (x) / (n− 1)!, but

(Hn (x) /n!) is not the reference sequence, because Hn (0) 6= δ0,n (see (6.36)).

4. In Example 6.1.2 we saw the reference sequence r0(x) := 1,

rn(x) := x (1− xn) /(1− x)

for n > 0, with reference series

r(x, t) = 1 +x

1− x

(t

1− t −xt

1− xt

)= 1 +

xt

(1− t) (1− xt) . (6.10)

The reference operator R is

Rxn = R (rn(x)− rn−1(x)) = rn−1(x)− rn−2(x) = xn−1

for all n ≥ 3, Rx2 = x− 1, and Rx = Rr1(x) = r0(x) = 1 = x0. Hence

Rxn =

xn−1 if n ≥ 1, n 6= 2x− 1 if n = 2.

We denoted the operator M (x)D by η, thus ηxn = nxn. If d (n) is anyk-valued function on N0 (see [65], [35]), then we define

d(η) (xn) := d(n)xn. (6.11)

If d (n) 6= 0 for all n ≥ 0, and d (0) = 1, the polynomial sequence (d (n)xn)n≥0

is an example of a reference sequence. Such sequences and their reference framesare called diagonal. If (rn) is diagonal we write r (xt) =

∑n≥0 d (n)xntn instead

of r (x, t). With the help of the division operator χ,

χp (x) =p (x)− p (0)

xfor all p (x) ∈ k [x] ,

we can explicitly write down the reference operator R,

Rd(η)xn = d (η)xn−1 =d (η)

d (η + 1)χd(η)xn

hence R = d(η)d(η+1)χ = d (η)χ/d (η). (In Kwasniewski’s notation [54], R = ∂ψ,

when d (n) = ψn.)


Example 6.2.2. 1. If d(n) = n+ 1, we obtain d (n)xn = (η + 1)xn. Hence∑n≥0

d (n)xntn = Dx (x/ (1− xt)) = (1− xt)−2

R (η + 1)xn = ηxn−1 =

(η

η + 1χ

)(η + 1)xn.

This is an example from the family of 2-binomial reference frames. More onbinomial reference frames in section 7.1.

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

R =η!−1

(η + 1)!−1χ = (η + 1)χ = χη = D

3. Let (z; q)n :=∏n−1k=0

(1− zqk

), and rn (x) = xn/ (q; q)n. The reference oper-

ator equals R = Dq := χ (1− qη), because

Dqrn (x) = Dqxn

(q; q)n= χ (1− qn)

xn

(q; q)n=

χxn

(q; q)n−1

=xn−1

(q; q)n−1

.

This reference frame is called q-differential by Andrews [5]. The referenceseries ∑

n≥0

xn

(q; q)ntn =

1

(xt; q)∞

is derived in Exercise 6.2.1. The q-differential reference frame follows fromthe exponential reference frame by substituting 1− qη for η.

Diagonal reference sequences have been studied, among others, by Roman[78] and Kwasniewski [54].

Three diagonal reference series and their reference framesBinomial (−α /∈ N0) Exponential q-differential

d(η) =(η+α−1

η

)1/η! 1/ (q; q)η

r(xt) = (1− xt)−α ext 1/ (xt; q)∞

rn(x) =(n+α−1

n

)xn xn/n! xn/ (q; q)n

R = (η + α)−1D = η+1

η+αχ D = χη D(q) := χ (1− qη)

(6.12)

(continued in Example 6.3.2).


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≥0

sn(x)tn =∑n≥0

rn(x)βn(t).

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, andγn (t) satisfies some multiplication rule, like γi (t) γj (t) = qf(i+j)γi+j (t), where fis some function on N0, and q is some suitable constant or formal variable. For thefollowing, we choose f identical 0.

Definition 6.2.3. A (generalized) Sheffer sequence (sn) is the r - image of (σ(t)β(t)n)where β is a delta series and σ is invertible,∑

n≥0

sn(x)tn = σ(t)∑n≥0

rn(x)β(t)n.

The pair (σ, β) is in the umbral group; only the reference sequence has changed. Ifσ(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 deltaseries β (t) and invertible series σ (t) the series M(σ)C(β)r(x, t) is the generatingfunction of a Sheffer sequence for the corresponding reference operator R. Weusually omit the qualifier “generalized”. A Sheffer sequence with respect to adiagonal reference series is a Boas-Buck sequence [14]; its generating function istherefore of the form σ (t)

∑n≥0 d (n)xnβ (t)

n.

Example 6.2.4. Let rn(x) = (n+ 1)xn, ρ (t) = (1− t)−2, and β (t) = t/ (1− t).The reference frame is therefore binomial with α = 2 (2-binomial). The series∑

n≥0

(n+ 1)xn

(1− t)2

(t

1− t

)n=

1

(1− t2)

(1− x t

1− t

)−2

= (1− t (1 + x))−2

=

∞∑n=0

(n+ 1) tn (1 + x)n

is the generating function of the 2-binomial Sheffer polynomial (n+ 1) (1 + x)n

=E1rn(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

M (2t− 1)C

(t

1− t

)r(x, t) = C

(t

1− t

)M

(t− 1

t+ 1

)r(x, t)

= 2t− 1 +xt (1− t)

(1− t− xt) ∈ k[x][[t]].


hence s0 (x) = −1, s1 (x) = 2 + x, and sn (x) = x2 (1 + x)n−2 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 se-quence (bn) holds

sn (x) =

n∑i=0

si (0) bn−i (x)

in any reference frame for all n ≥ 0.

Proof. By Definition 6.2.3,∑n≥0 sn (0) tn = σ (t) and∑

n≥0

bn (x) tn =∑n≥0

rn(x)β(t)n.

As in the exponential reference frame, a delta operator is isomorphic to adelta series, but with respect to a general reference operator R.Definition 6.2.7. Suppose B is an x-operator. We call B the R-delta operatorassociated to the delta series β provided that B is the C (β) r-transform of M (t),

BC (β) r(x, t) = M(t)C (β) r(x, t). (6.13)

Hence R is also a delta operator, when we choose β (t) = t. By Exercise 6.2.8,

B = β−1(M(t)`) = β−1(R) (6.14)

where β−1 (t) is the compositional inverse of β, i.e., β−1(β(t)) = t. This shows thatB ∈ ΣR, the set of operators that have a power series expansion in R. Any R-delta operator depends on the delta series β−1 (t) and the reference operator R. Ifthe frame of reference is understood from the context, we call B a delta operator.Note that every delta operator is also a reference operator, but for the framer(x, β (t)). In other words, if B ∈ ΣR and B is a delta operator, then R ∈ ΣB .This means that the set Ω of operators reducing the degree by one, is partitionedinto equivalence classes Ω ∩ ΣR, where R and B are equivalent iff there exists adelta series β such that β (B) = R.By Exercise 6.2.4, dividing Ω into equivalenceclasses of commuting operators (centralizers) gives the same partition.

Example 6.2.8. The Catalan operator C : xn 7→∑nk=1 Ck−1x

n−k, is a χ-deltaoperator associated to β (t) = t− t2, because

CC(t− t2

) 1

1− xt =∑n≥1

(t− t2

)n n∑k=1

Ck−1xn−k =

1

2

1−√

1− 4 (t− t2)

1− x (t− t2)

= t(1− x

(t− t2

))−1= M (t)C

(t− t2

) 1

1− xt .


Hence β−1 (t) =(1−√

1− 4t)/2, and

C =

∞∑k=1

Ck−1χk = β−1(χ) =

1

2− 1

2

√1− 4χ.

We saw in Definition 6.2.3 that for a Sheffer sequence (sn) holds

s (x, t) =∑n≥0

sn (x) tn = M (σ)C (β) r (x, t) ,

where β is a delta series, and σ has a reciprocal in k [[t]]. Note that

s (0, t) =∑

sn (0) tn = σ(t). (6.15)

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 oneargument for every n must be identical, i.e., if there is a sequence x0, x1, . . . suchthat 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.,

b (x, t) =∑n≥0

bn (x) tn = C (β) r (x, t)

for some delta series β and the reference series r. Again, B = β−1(R) is the associ-ated R-delta operator. Of course, we can make B into the new reference operatorwith reference series b (x, t) and reference sequence bn (x), because B = M (t)w.r.t. C (β) r (x, t). This shows that bn (0) = δ0,n. However, not every referenceframe can be obtained by substitution of a delta series. For example, the two ref-erence series ext and 1/ (1− xt) are related by tn ↔ tn/n!, an operation that isnot a substitution of a delta series. Every reference frame defines a “universe”ofother reference frames, but there are infinitely many “parallel universes”!

We call two reference frames r (x, t) and r′ (x, t) equivalent , if there exists deltaseries β (t) such that r (x, β (t)) = r′ (x, t). If both frames are diagonal, r (x, t) =∑n≥0 d (n)xntn and r′ (x, t) = r′ (x, t), then we also say that d (η) and d′ (η) are

equivalent. We saw above that 1/η! and 1 are not equivalent. The operators d (η)and d′ (η) are equivalent iff d (η) = aηd′ (η) for some a ∈ k, because∑

n≥0

d (n)xnβ (t)n

=∑n≥0

d′ (n)xntn

iff d (n) /d′ (n) = tn/β (t)n.


Sheffer operator Let B = β−1(R) be a delta operator, and (sn) a Sheffer se-quence for B with generating function s (x, t) = σ(t)r (x, β (t)). The Sheffer opera-tor of (sn) is the invertible operator S := σ(β−1(R)) = σ(B). The Sheffer operatoris 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, andtherefore

s (x, t) = M (σ)C (β) r (x, t) = C (β)M(σ(β−1

))r (x, t) = C (β)Sr (x, t)

= SC (β) r (x, t)

hencesn(x) = Sbn (x) . (6.16)

Sheffer operators commute with all other operators in ΣR. Every invertible oper-ator τ (R) in ΣR is a Sheffer operator,

τ (R) = τ(β(β−1 (R)

))(6.17)

thus τ (R) is the Sheffer operator for the Sheffer sequences with generating functionτ(β (t))r (x, β (t)).

Superposition of Sheffer sequences If (sn) and (tn) are Sheffer sequences for thesame R-delta operator B, then sn (x) + tn−k (x) is a Sheffer polynomial for B, aslong 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-zeroconstant polynomial. It can be shown by straightforward verification that (tn)defined by

tn(x) =

n∑j=0

tj(νj)sj,n−j(x), (6.18)

is a Sheffer sequence for B, if (sj,n)n∈N0 denotes for each j ∈ N0 the generalizedSheffer 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 canexpand tn (x) provided the Sheffer sequences (sj,n)n∈N0 can be found.

The umbral group Suppose B = β−1 (R) and A = α−1 (R) are both R-deltaoperators, with basic sequences (bn) and (an), respectively, where

an (x) =

n∑i=0

an,iri (x) .


The operator B (A) = α−1(β−1 (R)

)is also an R-delta operator. The basic se-

quence of this operator is called the umbral composition of (bn) with (an), withassociated basic polynomials an (b (x)) :=

∑ni=0 an,ibi (x), because

C (α)C (β) r = C (α)C (β)`r = C (β)

`C (α) r = C (β)

`∑n≥0

an (x) tn (6.19)

=∑n≥0

n∑i=0

an,iC (β)`ri (x) tn =

∑n≥0

n∑i=0

an,ibi (x) tn

(see (6.8)). The sequence (an (b (x)))n≥0 is basic for B (A). The operator C (β)`

:an (x) 7→ an (b (x)) is called an umbral operator. In exponential finite operator cal-culus we denoted umbral operators by Uβ−1(D). Umbral operators do not commutewith R except when β (t) = t (Exercise 2.3.13).

Umbral 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) forall n, we say that (bn) is inverse to (an) (see section 2.3.2). The details of theumbral group are exactly the same as in the exponential reference frame (section2.3.2). 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 elementhas the representation

tn (s (x)) =

n∑k=0

rk (x)

n∑i=k

tn,isi,n.

Hence (σ, β) is inverse to (τ, α) if tn (s (x)) = rn (x), i.e.,

β−1 (t) = α (t) and 1/σ (α (t)) = τ (t) . (6.20)

6.2.3 Exercises

6.2.1. Let 0 < q < 1. Show that limn→∞1∏n

k=0(1−xtqk)converges for small xt. Show

that ∑n≥0

tnxn

(q; q)n=

1

(xt; q)∞

(Euler).

6.2.2. Show identity (6.9) by applying (DM (x)D)k to the reference sequence.

6.2.3. For any 0 6= a ∈ F, the operator Ua := eaDM(x)D can be seen as an analogto eaD, the translation by a. Show that Ua : xn/n! 7→ anL

(0)n (−x/a) (see 5.3). As

an x-operator, Uaext = 11−ate

xt/(1−at). More on operators of the form Ua can befound in [31].


6.2.4. Suppose T is a linear operator that commutes with R. Show:

T =∑k≥0

〈Eval0 | T rk〉Rk

The case R = D is equivalent to Lemma 2.2.2.6.2.5. Show that for any given integer m ∈ N0 the reference sequence for theoperator (m+ 1)D +M (x)D2 is the Bessel function Im

(2√xt)/ (xt)

m/2

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 exponentialreference 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 thatB = β−1(R).

6.2.9. Let (sn) be a polynomial sequence with generating function

s (x, t) =∑n≥0

sn (x) tn.

Show: (sn) is a Sheffer sequence with generating function s (x, t) = M (σ)C (β) r (x, t)iff Rs (x, t) = M (β) s (x, t). Note that the latter condition is equivalent to Bsn (x) =sn−1(x), i.e.,

β−1 (R) s (x, t) = M (t) s (x, t) .

6.2.10. Let B = β−1(R) be a delta operator, and (sn) a Sheffer sequence for B withSheffer operator S. Show: If pn := Rsn+1 for all n ∈ N0, then (pn) is a Sheffersequence for B. Let φ (t) := β (t) /t. 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) areSheffer 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 R. SupposeB = β−1 (R) is a delta operator, and (bn(x)) the basic sequence for B. Show : Ifa is a scalar different from 0, then (bn(ax)) is the basic sequence for the deltaoperator β−1 (R/a).

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 operatorτ (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 aformal power series τ ∈ k [[t]], we have in analogy to section 2.4

B = β−1 (R) = τ−1 (A) = τ−1(α−1 (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

bn (x) =

n∑i=0

[τ i]nai (x) (6.22)

As before, we are mainly interested in the case where τ has coeffi cients in ΣR. Inthis case, the transfer formulas are expressed in terms of the Pincherle derivative,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 R is the linear operator denotedby θR such that

θRrn(x) = (n+ 1)rn+1(x) for n ≥ 0.

The degree reducing operator θ−R is the left inverse of θR,

θ−Rrn(x) =1

nrn−1(x) =

1

nRrn(x) for n ≥ 1, and

θ−R1 = 0.

Note thatRiθ−Rrn(x) = Ri+1rn(x)/n for all n ≥ 1.

In transform notation,

θR = Dt and θ−R =

(D−1t

)^where D−1

t : tn 7→ tn+1/ (n+ 1). We obtain

θnR1 = n!rn(x) for all n ≥ 0.

Example 6.3.2. Let R be a diagonal reference operator, R = d (η)χ/d (η), andrn (x) = d (η)xn (see (6.11)). We calculate the diagonal umbral shift as

θRd (η)xn = ηd (η)xn+1 = ηd (η)M(x)xn = ηM(x)d (η + 1)xn


hence

θR = ηM(x)d (η + 1)

d (η)

We calculate the umbral shifts for the three basic examples (6.12) of diagonalreference frames.

Three important reference frames and their umbral shiftsBinomial Exponential q-differential

d (η) =(η+α−1

η

)1/η! 1/ (q; q)η

r(xt) = (1− xt)−α ext 1/ (xt; q)∞

rn(x) =(n+α−1

n

)xn xn/n! xn/ (q; q)n

R = Ba = (η + α)−1D = η+1

η+αχ D = χη D(q) = χ (1− qη)

θR = M(x) (η + α) M(x) η1−qηM(x)

θ−R = 1η+αχ χ χ 1−qη

η

In all three example we expressed θR and θ−R in terms of operators applicable toany polynomial p, not just the reference sequence.

Example 6.3.3. Define the reference sequence (rn) by r0(x) := 1,

rn(x) := x (1− xn) /(1− x)

for n > 0. We saw in Example 6.2.1 that the reference operator R equals

Rxn =

xn−1 if n ≥ 1, n 6= 2x− 1 if n = 2.

We calculate the umbral shift as

θRxn = θR (rn(x)− rn−1(x)) = (n+ 1) rn+1(x)− nrn(x)

= rn(x) + (n+ 1)xn+1 = rn+1(x) + nxn+1

for n ≥ 2, and θRx = θRr1(x) = 2r2(x), θR1 = x. Hence

θRxn =

rn+1(x) + nxn+1 if n ≥ 0, n 6= 1

2x+ 2x2 if n = 1.

The inverse shift can be represented as

θ−Rxn = θ−R (rn(x)− rn−1(x)) =

1

nrn−1(x)− 1

n− 1rn−2(x)

=(n− 1) rn−1(x)− nrn−2(x)

n(n− 1)=nxn−1 − rn−1(x)

n(n− 1)

for n ≥ 3, θ−Rx2 = 1

2x− 1, and θ−Rx = 1.


If T is an operator then the Pincherle derivative T ′R of T is defined as

T ′R = TθR − θRT. (6.23)

If k is a ring of scalars, then T ∈ ΣR implies T ′R = ddRT (Exercise 6.3.2); if

k contains operators they must commute with θR for this result to hold true.Different umbral shifts produce different Pincherle derivatives. Note that we markPincherle derivatives by a prime, and write d

dRT if T has a power series seriesexpansion in R.

6.3.2 Equivalent Transfer Formulas

Let B be a delta operator. As a power series in R, the operator B is of order1, hence the linear operator P−1 := B/R is invertible. In terms of P , PB = 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 allpositive integers n holds

bn(x) =

(d

dRB)Pn+1rn(x) (6.24)

bn(x) = Pnrn(x)− 1

n

(d

dRPn

)rn−1(x) (6.25)

bn(x) = θRPnθ−Rrn(x) =

1

nθRP

nrn−1(x) (6.26)

(see Exercise 6.3.6). Combining the first and the last of the three formulas provesthe Rodrigues’type formula

bn(x) =1

nθR

(d

dRB)−1

bn−1(x) =1

nθR

dRdB

bn−1(x)

=1

nθR

d

dBβ (B) =

1

n!(θRβ

′ (B))n

1.

The transfer formulas can also transfer from any basic sequence (an) to (bn).

Lemma 6.3.4. Let A,B ∈ ΣR be delta operators with basic sequences (an) and (bn),respectively. Define the invertible operator T ∈ ΣR by TB = A. For all positiveintegers n holds

bn(x) =

(d

dRB)(

d

dRA)−1

Tn+1an(x) (6.27)

= θRTnθ−Ran(x).

Proof. In the same way as we defined P through PB = R we define S throughSA = R. We saw above that an(x) =

(ddRA

)Sn+1rn(x), and therefore rn(x) =


(ddRA

)−1S−n−1an(x). Note that T = PS−1. Substituting for rn into (6.24) gives

the first result. The third formula above shows that θRS−nθ−Ran(x) = rn(x) for

all n ≥ 1. Hence

bn(x) = θRPnθ−Rrn(x) = θRP

nS−nθ−Ran(x)

shows the second equation.Remember that

[ti]φn is long-hand for [φn]i, the coeffi cient of t

i in the powerseries φ (t)

n. Combining transfer with Lagrange-Bürmann inversion gives the fol-lowing useful expansion in situations where T cannot be explicitly calculated.

Corollary 6.3.5. If the delta operator A ∈ ΣR can be written as A = τ(B) =∑i≥1 TiB

i ∈ ΣR [[B]] for some linear operator B, such that Ti ∈ ΣR for all i ≥ 1,and T1 is invertible, then B is also a delta operator in ΣR, and the basic sequence(bn) of B can be expressed in terms of the basic sequence (an) of A as

bn (x) = θR

n−1∑i=0

n

n− i[τn−i

]nAiθ−Ran (x) .

for all n > 0.

Proof. We saw in (6.27) that bn = θR(PS−1

)nθ−Ran and we obtain(

PS−1)n

=∑i≥0

n

n− i[τn−i

]nAi

exactly in the same way as in section 2.4.2.When A and B are delta operators in ΣR, then they are also delta operators

in ΣA. Hence we can think of (an) as the reference sequence, with θ−Aan(x) =an−1 (x) /n for positive n, and expand bn as

bn (x) = θA

n−1∑i=0

1

n− i[τn−i

]nan−1−i (x) (6.28)

= θR

n−1∑i=0

1

n− i[τn−i

]n

(d

dRA)−1

an−1−i (x)

(see Exercise 6.3.1). If τ ∈ k [[t]], we arrive at the simple expansion (6.22).If A is a polynomial in B, A = τ (B) =

∑kj=1 TjB

j , then the coeffi cient ofBn in τn−i (B) equals

∑l1+···+lk=n−i

l1+2l2+···+klk=n

(n− i

l1, . . . , lk

) k∏j=1

Tljj .

This proves the following Corollary.


Corollary 6.3.6. If the delta operator A ∈ ΣR can be written as τ(B) =∑kj=1 TjB

j

such that Tj ∈ ΣR, T1invertible, then B is also a delta operator, and the basicsequence (bn) of B can be expressed in terms of the basic sequence (an) of A asfollows

bn = θR

n−1∑i=0

n

n− i∑

l1+···+lk=n−il1+2l2+···+klk=n

(n− i

l1, . . . , lk

) k∏j=1

Tljj

Aiθ−Ran

= θR∑

l1+2l2+···+klk=n

n

l1 + · · ·+ lk

(l1 + · · ·+ lkl1, . . . , lk

) k∏j=1

Tljj

An−(l1+···+lk)θ−Ran.

6.3.3 Exercises

6.3.1. Suppose we make a (minor) change to the reference frame by substituting adelta series β (t) for t, i.e., our new reference series is b (x, t) := r(x, β (t)), andour new reference operator is the R-delta operator B = β−1 (R). Find the umbralshift θB associated to B, and show that θ

−B =

(β−1

)′(R) θ−R.

6.3.2. Show: If T =∑k≥0 τkRk ∈ ΣR then

T ′R =d

dR∑k≥0

τkRk =∑k≥0

τkkRk−1.

This will not hold in general if T has operators from ΣR as coeffi cients.

6.3.3. Show that ddDE

1 is different from dd∆E

1.

6.3.4. Show the product rule of differentiation for the general Pincherle derivative,(ST )

′R = S′RT + ST ′R, for S, T ∈ ΣR. This implies (Tn)

′R = nTn−1T ′R.

6.3.5. In Example 6.3.3 express the umbral shifts θR and θ−R in terms of χ, η, andM (x).

6.3.6. Show that the equalities (6.24) - (6.26) hold by verifying that the right handsides are the same, and generate a Sheffer sequence for B, which has the correctinitial values.


6.4 Functionals

The transform of a functional is defined with respect to the reference series r(x, t),

Lrr(x, t) = Lr(x, t) =∑n≥0

〈L | rn〉 tn.

For any t-operator T holds T r(x, t) =∑n≥0 rn (x)Ttn, hence Lrtn = δ0,n, and Lr

maps t0 = 1 onto the power series λ (t) :=∑n≥0 〈L | rn〉 tn. More important for

us will be the application of λ (t) as multiplication operator M (λ (t)). We definethe product of two functionals L and N ,

〈L ∗r N | rn〉 :=

n∑k=0

〈L | rk〉〈N | rn−k〉 ,

i.e.,

(L ∗r N) r(x, t) = (Lr(x, t)) (Nr(x, t)) =(Lr1)Nr1 = λ (t) ν (t) . (6.29)

The evaluation at 0, Eval0, is the multiplicative unit. A linear functional L isinvertible (w.r.t. ∗r-multiplication), iff 〈L | 1〉 is a unit in k, the same as in section3.1. The associated operator to the functional L ∈ K [x]

∗ is the transform withrespect to r (x, t) of the multiplication operator M(λ), i.e.

λ(R)r (x, t) = M(λ)r (x, t)

for all L ∈ K[x]∗. We also write op (L) for λ(R). We have the isomorphismsK [x]

∗←→K [[t]]←→ ΣR. Note that λ(R) commutes with delta operators, becausethey are all in ΣR. In Exercise 6.4.7 we will prove that

op (L) =∑n≥0

〈L | rn〉Rn =

∑k≥0 〈L | sk〉Bk∑

n≥0 〈Eval0 | sn〉Bn(6.30)

for any Sheffer sequence (sn) with delta operator B in the reference frame r.

Example 6.4.1. Let c ∈ k [x]. We get

op (Evalc) = M

∑n≥0

rn (c) tn

`

=∑n≥0

rn (c)M (tn)`

=∑n≥0

rn (c)Rn = r (c,R) .

We can check this result applying (6.30),

op (Evalc) =

∑k≥0 sk (c)Bk∑n≥0 sn(0)Bn

= r (c, β (B)) = r (c,R) .

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) and J = L ∗r N above, we obtain

op (J) = op (L)T

if 〈J | p〉 = 〈L | Tp〉 for all p ∈ k[x].Now we have 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(sn) is a Sheffer sequence and (bn) the basic sequence for the same r-delta operator,then

sn (x) =

n∑k=0

〈L | sn−k〉 op (L)−1bk (x) (6.31)

and ∑n≥0

sn (x) tn =

∑k≥0 〈L | sk〉 tk∑n≥0 〈L | bn〉 tn

r (x, β (t))

Proof. The proof is very much the same as the proof of Theorem 3.1.4, and weonly give a sketch. First, for every Sheffer sequence (ln) for the r-delta operator Bwith basic sequence (bn) holds that

p (x) =∑k≥0

⟨L | Bkp

⟩lk (x)

for all p (x) ∈ k [x]. With the help of the Sheffer operator S for (ln) this can bewritten as

p (x) =∑k≥0

⟨L | Bkp

⟩Sbk (x) .

In Exercise 6.4.1 it is shown that S = λ (R)−1.

Example 6.4.3. [65] Suppose L is an invertible linear functional, and ` a positiveinteger. 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, . . .

According to (6.31) we have

sn(x) =

n∑k=0

〈L | sn−k〉 op (L)−1bk (x) =

`−1∑i=0

〈L | pi〉 op(L)−1bn−i(x).


We can also apply (6.31) to pn (x),

pn (x) =

n∑k=0

〈L | pn−k〉 op (L)−1bk (x) ,

hence

sn(x) = pn(x)−n∑i=`

〈L | pi(x)〉 op(L)−1bn−i.

For the generating function we get∑n≥0

sn (x) tn =

∑`−1k=0 〈L | pk〉 tk∑k≥0 〈L | bk〉 tk

∑j≥0

bj (x) tj .

Suppose we take the functional⟨∫10 | pn

⟩=∫ 1

0pn(x)dx, obtaining

sn(x) = pn(x)−n∑i=`

(∫ 1

0

pi(x)dx

)bn−i(x)∫ 1

0r(x,R)dx

.

We discussed this functional in Example 3.1.6 for the exponential reference frame.If r (x, t) = (1− xt)−α, the binomial framework, we must distinguish two cases. Ifα = 1 then R = χ : f(x) 7→ (f(x)− f(0)) /x, and

1

∫0r (x, t) =

∫ 1

0

(1− xt)−1dx = −1

tln (1− t) .

Hence

op

(1

∫0

)−1

xn =−χ

ln (1− χ)xn = −

∑j≥0

bjxn−j (6.32)

where b0, b1, . . . are the Bernoulli numbers of the second kind [47, (9) § 97] (Ex-ercise 2.3.3).

If α 6= 1, 0,−1,−2, . . . then

1

∫0r (x, t) =

∫ 1

0

(1− xt)−α dx =1− (1− t)1−α

t(1− α).

For example, if α = 1/2, then ∫10 r (x, t) = 2(1−√

1− t)/t, which equals the Cata-

lan generating function c (t) in (1.2) at t/4. Hence(∫10 r (x, t)

)−1= 1 − t

4c (t/4).In this case

op

(1

∫0

)−1

= 1−∞∑k=0

4−k−1Ck

(η + 1

η + 1/2χ

)k+1

=1

2+

1

22F1

[− 12 , η + 1;χ

η + 12

].


6.4.1 Augmentation

The functional I∗ is the identity functional with respect to the reference seriesr (x, t) iff 〈I∗ | rn〉 = δ0,n, i.e., I∗ is the identity in the ∗r -multiplication. Becausewe require that rn (0) = δ0,n we get I∗ = Eval0, independent of the referenceseries. Another name for the identity functional is augmentation. The isomorphicpower series is the multiplicative identity in the ring of formal power series, andthe 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〉 6= 0, then

LM (σ)C (β) r (x, t) = 1

iffL = I∗σ

(β−1 (R)

)−1.

Proof. LM (σ)C (β) r (x, t) =

I∗M (σ)C (β)σ(β−1 (R)

)−1r (x, t)

= I∗M (σ)C (β)M(σ(β−1 (t)

)−1)r (x, t) = I∗C (β) r (x, t) = 1

For any delta operator B = β−1 (R) we can define the functional IB :=I∗β−1 (R)

〈IB | rn〉 = 〈I∗ | Brn〉 = 〈I∗ | rn−1〉 = δ1,n

such that B = op (IB) (Exercise 6.4.8). Writing I∗mB for IB ∗r · · · ∗r IB (m factors)we get

op (Evala ∗I∗mB ) = r (a,R)Bm.

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 wasfirst seen by Kholodov [49]. We follow Freeman’s approach because of its simplicity.We will show two applications, the orthogonal polynomials that are exponentialSheffer sequences at the end of this section, and those that are binomial Sheffersequences later in section 7.1.1.

Let (µn)n≥0 be a sequence of (complex) numbers and Λ the functional definedby 〈Λ | xn〉 = µn for all n ≥ 0. The sequence is called the (formal) momentsequence, 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 6= 0that are nonnegative for all real x. For all positive definite functionals Λ exists abounded nondecreasing function ψΛ such that 〈Λ | xn〉 =

∫∞−∞ xndψΛ (x) (Stieltjes

integral).


Definition 6.4.5. A polynomial sequence (pn) is orthogonal if a functional Λ onC [x] and a sequence λ (n) 6= 0, n ≥ 0, exist such that

〈Λ | pnpm〉 = λ(n)δn,m (6.33)

for all non-negative integers n,m. In that case (pn) is called the orthogonal poly-nomial system (OPS) corresponding to Λ. We will assume that 〈Λ | p0〉 = 1.

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

q (x) =∑j≥0

⟨Λ | pjq

λ(j)

⟩pj (x)

for all polynomials q. This property explains the importance of finding an OPSand its corresponding functional. The condition (6.33) implies

⟨Λ | xkpn

⟩= 0 for

all 0 ≤ k < n, and 〈Λ | xnpn〉 6= 0.Let p (x, t) =

∑n≥0 pn (x) tn, where (pn) is an OPS. We have

Λxkp (x, t) =

k∑n=0

⟨Λ | xkpn

⟩tn =: fk (t) ,

where (fk)k≥0 is a sequence of polynomials in C [t] such that deg fk = k, andf0 ≡ 1 = λ (0). Taking transforms with respect to p (x, t) gives

fk (t) = ΛM(xk)p (x, t) = M (xk)Λp (x, t) = M (x)

k1.

If we write xp for M (x) we have the simple condition xkp1 = fk (t) for all k ≥ 1,which says that xp (a linear operator on C [[t]]) increases the degree by one whenrestricted to C [t], deg xpg (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

deg xptn = n+ 1 (6.34)

for all n ≥ 0.

Lemma 6.4.6. If p (x, t) is the generating function for an OPS, then

xptn = αn−1t

n−1 + βntn + γn+1t

n+1

for some sequences (αn)n≥0, (βn)n≥0, (γn)n≥1, γn 6= 0 for all n ≥ 1.


Proof. From deg xptn = n + 1 follows that xptn =

∑n+1i=0 an,it

i for some scalarsan,i. Hence

xpp (x, t) =∑n≥0

pn (x) xptn =

∑n≥0

pn (x)

n+1∑i=0

an,iti =

∞∑i=0

ti∞∑

n=i−1

an,ipn (x) .

Of course, the last inner sum has to terminate, and because

xpp (x, t) = xp (x, t) =∑i≥0

xpi (x) ti,

it has to terminate already at i+ 1.The Lemma shows that

xpn (x) = αnpn+1 (x) + βnpn (x) + γnpn−1 (x) , (6.35)

orpn+1 (x) = α−1

n (x− βn) pn (x)− α−1n γnpn−1 (x) .

The latter is called the three term recurrence. Substituting for pn (x) in 〈Λ | pnpm〉 =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).The first and the third of the three relations give αnλ (n+ 1) = γn+1λ(n). Fromdeg (xpn (x)− αnpn+1 (x)) = n we find αn = kn/kn+1, where kn = [xn] pn (x).

Remark 6.4.7. (a) Let Evalt=0 be the t-operator on C [[x]] [t] that evaluates at 0.From fn(t) = 〈Λ | xnp (x, t)〉 follows

fn (0) = ΛM (xn) Evalt=0 p (x, t) = 〈Λ | xn〉 = µn,

the n-th moment of Λ. It can be shown [20] that in an OPS the determinant ofthe Hankel matrix Mn := (µi+j)i,j=0,...,n is different from 0 for all n ≥ 0. Moreprecisely,

〈Λ | xnpn (x)〉 =kn |Mn||Mn−1|

where kn = [xn] pn (x) as before.

(b) It is usually assumed that λ (n) is real and positive for all n. In this caseαnγn+1 > 0.

The Meixner classification In 1934, Meixner classified all orthogonal polynomialsystems that are also Sheffer sequences [59]. Freeman showed how to do this easilywith the transform approach [33]. The following is based on his ideas. We willconsider another application, to binomial reference frames, in section 7.1.1,


Suppose (pn) is a Sheffer sequence in the exponential reference frame, hencethe generating function is

p (x, t) =∑n≥0

pn (x) tn = σ (t) exβ(t) = eα(t)+xβ(t)

where we assume that σ (0) = 1, hence α (t) = lnσ (t). We will also assume thatβ′ (0) = 1. From

Dteα(t)+xβ(t) = M (α′ (t) + xβ′ (t)) eα(t)+xβ(t)

follows

M (x) p (x, t) = (M (1/β′ (t))Dt −M (α′ (t) /β′ (t))) p (x, t) ,

hence xp = M (1/β′ (t))Dt −M (α′ (t) /β′ (t)). From

1 = deg (xp1) = deg (−M (α′ (t) /β′ (t)))

we get that α′ (t) /β′ (t) must be linear,

α′ (t) /β′ (t) = ct+ d,

and c 6= 0. From

2 = deg (xpt) = deg

(1

β′ (t)− (ct+ d) t

)it follows that (pn) is a OPS, deg xpt

n = n+ 1, iff 1/β′ (t) is at most a quadraticpolynomial,

1/β′ (t) = (1− at) (1− bt) .

In this case

xptn = (1− at) (1− bt)ntn−1 − (ct+ d) tn

= (nab− c) tn+1 − (n (a+ b) + d) tn + ntn−1

must be of degree n+1 for all n ≥ 1, thus c/ (ab) is not a positive integer if ab 6= 0.From

α (t) + xβ (t) =

∫ t

0

(xβ′ (u) + α′ (u)) du =

∫ t

0

x+ α′ (u) /β′ (u)

β′ (u)−1 du

follows that for an OPS we can write∑n≥0

pn (x) tn = p (x, t) = exp

(∫ t

0

x+ cu+ d

(1− au) (1− bu)du

),


where abn 6= c for all n ≥ 1. The term d in the above generating function onlyshifts x, and can be set to 0. If we consider real polynomials only, we must choosea 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

xtn = (1− at) (1− bt)ntn−1 − ctn+1 = (abn− c) tn+1 − (a+ b)ntn + ntn−1.

Note that we are interested in any OPS up to linear transformations in xand scaling in t.It can be shown that the functional is positive definite if ab ≥ 0and c < 0. The following types occur:

1. a = b = 0, and c = −1 :

exp

(∫ t

0

(x− u) du

)= ext−

12 t

2

.

This is the generating function of the Hermite polynomialsHn (x). Theyare orthogonal with respect to the normal probability measure

µn = 〈Λ | xn〉 =1√2π

∫ ∞−∞

xne−x2/2dx,

(see Example 2.2.9), because

1√2π

∫ ∞−∞

e−x2/2ext−

12 t

2

exs−12 s

2

dx =1√2πest∫ ∞−∞

e(−(s+t)2+2x(s+t)−x2)/2dx

= est.

The three term recursion becomes

(n+ 1)Hn+1 (x) = xHn (x)−Hn−1 (x)

= xDHn+1(x)−D2Hn+1 (x) .

Therefore,

(n+ 1)Hn (x) = DxDHn+1(x)−D3Hn+1 (x) . (6.36)

Hence the polynomials Hn (x) = Hn (x) /n! follow the recursion(DxD −D3

)Hn =

Hn−1 for all n ≥ 0, with initial values H2n (0) = H2n (0) / (2n)! = (−1)n

2−n/ (n! (2n)!),and H2n+1 (0) = 0. The more common variation for the Hermite polynomials[3] is obtained from choosing c = −2 and replacing x by 2x,

exp

(2

∫ t

0

(x− u) du

)= e2xt−t2 .


2. a = b 6= 0, and c = −1 :

exp

(∫ t

0

x− u(1− au)

2 du

)= exp

((ax− 1) t

a (1− ta)

)(1− ta)

−1/α2.

Transform (ax− 1) /a2 to x, and rescale ta as t to get (1− t)−1/α2ext/(t−1).

Writing a+ 1 for 1/α2 gives the generating function of the Laguerre polyno-mials (Example 5.3.1).

3. a 6= 0, b = 0, and c < 0 :

exp

(∫ t

0

x+ cu

1− audu)

= e−ct/a (1− at)−(c+ax)/a2.

Transform − (c+ ax) /a2 to x, rescale t as −t/a2and let c = −∣∣a3∣∣ to get the

generating function of the Poisson-Charlier polynomials e−t(a+ta

)x(see also

Exercise 2.3.14).

4. ab > 0, a 6= b, a and b real, c = −1, gives Meixner type I, and a and bconjugate complex gives Meixner type II.

6.4.3 Exercises

6.4.1. Let (ln (x)) be the Sheffer sequence for the delta operator B = β−1 (D) withgenerating function

∑n≥0 ln (x) tn = r (x, β (t)) /λ (β (t)), where Lr (x, t) = λ (t)

for some invertible functional L. Show that λ (R)−1 is the Sheffer operator of (ln).

6.4.2. Let L ∈ K∗ [x]. For every delta series β(t), and every series φ (t) of order0 holds that LC (β)

`φ (R) is also a functional, and therefore

κ (t) := LC (β)`φ (R) r (x, t)

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]∗ → ΣR, where op (L) =λ (R), is a ring isomorphism .

6.4.5. Show that for all p ∈ K [x] and L,N ∈ K[x]∗,

〈L ∗r N | p〉 = 〈N | λ (R) p〉

(see [35, Proposition 6.1] and [65, Lemma 8]).


6.4.6. Let (bn) be the basic sequence for the delta operator B = β−1 (R). Suppose,(sn) is the Sheffer sequence with generating function M(σ(t))C (β) r(x, t). Definethe invertible linear functional L by λ(t) = σ(β−1(t)), and show that λ(R) is theSheffer operator for (sn), sn(x) = λ(R)bn(x). Let L∗−1 be the reciprocal of L inthe ∗-multiplication of functionals, and show that

⟨L∗−1 | sn

⟩= δ0,n.

6.4.7. Let B be any delta operator and (sn) any B - Sheffer sequence. Show that

λ (R) =

∑k≥0 〈L | sk〉Bk∑

n≥0 〈Eval0 | sn〉Bn.

Especially for the B - basic sequence (bn) holds

λ (R) =∑k≥0

〈L | bk〉Bk

6.4.8. Suppose (bn)is the basic sequence belonging to the delta operator B = (R).Show that for the functional IB = I∗β

−1 (R) holds

〈IB | bn〉 = δn,1.

6.4.9. Suppose (bn) is the basic sequence for the r-delta operator B. Show that forall m ∈ N0

op (Evala ∗I∗mB ) = r (a,R)Bm.

Chapter 7

Applications of the GeneralTheory

We pursue only two applications of the general theory. In the first case we selectthe “binomial reference”, i.e., we look at delta operators obtained from delta seriesβ (t) substituted into (1− xt)−α. We show the Freeman classification of orthog-onal polynomials for this case, and we give some other examples, like Dicksonpolynomials.

The second application goes back to George Andrews in 1971 [5]. It appliesto all diagonal reference frames, but it were the Eulerian differential operatorsAndrews had in mind. The delta operators in diagonal reference frames are ingeneral not translation invariant. A new invariant is introduced, referring to scal-ing. The binomial theorem is replaced by a multiplicative version (7.22), holdingfor Eulerian sequences.

We only mention here another view of q-polynomials and operators, intro-duced by the Askey-Wilson divided difference operator Dq,x [7, 1985]. This op-erator is a q-analog of the differentiation operator D. The operator Dq,x maps(1− q)n Tn (x) /

(qn(n−1)/4 (q; q)n

)into (1− q)n−1

Un−1 (x) /(q(n−1)(n−2)/4 (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 Dq,xcommutes with this translation. He then defines q-Delta operators and Sheffersequences in the Dq,x reference frame.

188 Chapter 7. Applications of the General Theory

7.1 The Binomial Reference Frame

The elements of a binomial reference frame are the reference sequence((n+α−1

n

)xn),

the reference series

(1− xt)−α =∑n≥0

(n+ α− 1

n

)xntn = χα−1Dα−1M

(xα−1

)(1− xt)−1

(−α /∈ N0), and the reference operator

Bα =η + 1

η + αχ = (η + α)

−1D

Bα(n+ α− 1

n

)xn =

(n+ α− 2

n− 1

)xn−1

for all n ≥ 1. To emphasize α, we also say α-binomial reference frame. Mostnotable is the fact that a binomial reference series also becomes an exponentialreference series if we view x as fixed, x ∈ C, say, and α as the polynomial variable.This explains immediately the “binomial theorem”

sn(x; ν + µ) =

n∑k=0

sk(x; ν)bn−k(x;µ). (7.1)

that holds for every Sheffer sequence (sn (x; ν)) and basic sequence (bn (x;µ)) suchthat

∑n≥0 bn (x;µ) tn = (1− xβ (t))

µ and∑n≥0 sn (x; v) tn = σ (t) (1− xβ (t))

ν

(delta series β, and invertible σ). Of course, this is not the binomial theorem wewant. If we construct (tn (x)) according to convolution identity

tn(x) =

n∑k=0

tk(y)bn−k(x− y;α)

we obtain a binomial Sheffer sequence for the delta operator β−1 (Bα/ (1 + yBα))(Exercise 7.1.1). At least, we can expand a binomial Sheffer sequence for β−1 (Bα)when we know the initial values at 0,

tn(x;α) =

n∑k=0

tk(0;α)bn−k(x;α). (7.2)

We can still find the Sheffer sequence with initial values sn (c) = δ0,n in thegeometric case, i.e., the binomial case with α = 1 (see Exercise 7.1.4). Note thatbn (x− c;α) is not a binomial Sheffer polynomial in general, because we do nothave translation invariance. An operator “similar” to the translation operatorEc =

∑k≥0 c

kDk/k! is the invertible

Hα,c :=∑k≥0

ckBkα =1

1− cBα∈ ΣBα ,

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 formp (x, t) = (φ (t)− xφ (t)β (t))

−λ for −λ /∈ N0. We assume in this subsection thatφ (0) = 1, β (0) = 0, and β′ (0) = 1. Let xp be the p (x, t)-transform of M (x). Forxp 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 thatβ′ (t) = 1, and define η (t) = φ (t)− 1. Then xpη (t)

n(φ (t)β (t))

k

=

(2n+ k

2n+ k + λ− 1+

2n+ k + 2λ

2n+ k + λ+ 1η (t)

)η (t)

n(φ (t)β (t))

k−1 (7.3)

for all n, k ≥ 0, where p (x, t) = (φ (t)− xφ (t)β (t))−λ and −λ /∈ N0.

Proof. Expanding p (x, t) shows that

p (x, t) =∑k≥0

(k + λ− 1

k

)xk∑n≥0

(−1)n

(n+ k + λ− 1

n

)ηn (φβ)

k.

Substituting the right-hand side of (7.3) for ηn (φβ)k above gives

∑k≥0

(k + λ− 1

k

)xk∑n≥0

(−1)n

(n+ k + λ− 1

n

)ηn (φβ)

k−1 (2n+ k)

(2n+ k + λ− 1)

−∑k≥0

(k + λ− 1

k

)xk∑n≥1

(−1)n

(n+ k + λ− 2

n− 1

)ηn (φβ)

k−1 (2n+ k + 2λ− 2)

(2n+ k + λ− 1)

which simplifies to xp (x, t).If p (x, t) is the generating function of an OPS, then xptk = αk−1t

k−1+βktk+

γk+1tk+1 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 thatφ (t)β (t) = t is the only possibility! We begin with an obvious simplification:Lemma 7.1.1 directly gives us xpt0,

xp1 =2λ

λ+ 1

η (t)

φ (t)β (t)= b+ ct = β0 + γ1t (7.4)

hence η (t) = λ+12λ (β0 + γ1t)φ (t)β (t). Note that γ1 6= 0. If n = 0 in Lemma 7.1.1,

then

xp (φβ)k

=

(k

k + λ− 1+

k + 2λ

k + λ+ 1

λ+ 1

2λ(β0 + γ1t)φ (t)β (t)

)(φ (t)β (t))

k−1

=k

k + λ− 1(φ (t)β (t))

k−1+

k + 2λ

k + λ+ 1

λ+ 1

2λ(β0 + γ1t) (φ (t)β (t))

k


Suppose, (φ (t)β (t))k

= tk +∑m>k Ck,mt

m, hence xp (φβ)k

=

αk−1tk−1 + (βk + Ck,k+1αk) tk + (γk+1 + Ck,k+1βk+1 + Ck,k+2αk+2) tk+1

+∑

m≥k+2

(Ck,m+1αm + Ck,mβm + Ck,m−1γm) tm

Comparing coeffi cients in both expansions shows that αk, βk, and γk are dependenton 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

αk =k + 1

k + λ(7.5)

βk =(k + 2λ) (λ+ 1)

2λ (k + λ+ 1)β0 −

k (k + 2λ− 1)

(k + λ− 1) (k + λ)C1,2

and γk+1 =

(k + 2λ) (λ+ 1) γ1

2λ (k + λ+ 1)+

k (λ+ 1) (λ− 1)

2λ (k + λ+ 1) (k + λ+ 2)C1,2 −

k (k + 4λ− 1)

(k + λ− 1) (k + λ+ 2)C1,3

+

(k (k − 1) (k − 2)

2 (k − 1 + λ)− k2 (k + 1)

k + λ+k (k + 1) (k + 2)

k + 1 + λ− (k − 1) k (k + 3)

2 (k + 2 + λ)

)C2

1,2

Remember we want to show that∑m>k Ck,mt

m = 0. Substituting for αk, βk,

and γk and comparing terms for m = k+2 in the two representations for xp (φβ)k

gives us infinitely many equations in the unknowns C1,2, C1,3, and C1,4. Thissystem only has the solutions C1,2 = C1,3 = C1,4 = 0, which in turn impliesCk,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>k Ck,mt

m = 0, and therefor φ (t)β (t) = t. More details can be found inExercise 7.1.6.

Gegenbauer polynomials We have shown that φ (t)β (t) = t, thus φ (t) − 1 =η (t) = λ+1

2λ t (b+ ct), where we write b for β0 and c for γ1. From (7.5) we obtain

αk = k+1k+λ , βk = b (λ+1)(k+2λ)

2λ(k+λ+1) , and γk = c (λ+1)(k+2λ−1)2λ(k+λ) . We find

β (t) =t

1 + λ+12λ t (b+ ct)

and by rescaling t and replacing x by 2x we get the generating function of theGegenbauer polynomials P (−λ)

n (x),∑n≥0

P (−λ)n (x)tn =

(1 + t2 − 2xt

)−λ. (7.6)


The three term recursion for the Gegenbauer OPS equals

2x (n+ λ)P (−λ)n (x) = (n+ 2λ− 1)P

(−λ)n−1 + (n+ 1)P

(−λ)n+1 (x) (7.7)

From the generating function it is easy to expand the Gegenbauer polynomials as

P (−λ)n (x) =

n/2∑k=0

(n− k + λ− 1

n− k

)(n− kn− 2k

)(−1)

k(2x)

n−2k.

A different recursion can be derived by taking the x-derivative of(1 + t2 − 2xt

)−λ(writing Pn (x) for P (−λ)

n (x)),

P ′n (x) + P ′n−2 (x)− 2xP ′n−1 (x) = 2λPn−1 (x) . (7.8)

Both recursions need the initial values

P(−λ)2n (0) =

(−λn

)and P (−λ)

2n+1 = 0, (7.9)

following from (7.6). Note that P (−1)n = Un (x) is called the Chebychev polynomial

of the second kind, and P (−1/2)n (x) is the Legendre polynomial.

Remark 7.1.2. We have found just one important example for a binomial OPS, byletting 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 simplecondition

αk−1tk−1 + βkt

k + γk+1tk+1 =

∑j≥k

cj (1 + η (t)) (φ (t)β (t))j−1

=tk

β (t).

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

p (x, t) =1− bt+ (1− c) t2

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,

Un (x) =

n/2∑k=0

(−1)k

(n− kk

)(2x)

n−2k,


because∑n≥0

tnn/2∑k=0

(−1)k

(n− kk

)(2x)

n−2k=

1

1− 2xt

∞∑k=0

(−t2

1− 2xt

)k=(1− 2xt+ t2

)−1.

The delta operator β−1 (χ) =(

1−√

1− χ2)/χ maps Un to Un−1. The 3-term

recursion (??) specializes to

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

Un (cos θ) =

n/2∑k=0

(−1)k

(n− kk

)(2 cos θ)

n−2k=

sin ((n+ 1) θ)

sin θ.

for all n ≥ 0. In Exercise 7.1.8 we show that

Un (x) = 2nn∏k=1

(x− cos

kπ

n+ 1

)

=

∣∣∣∣∣∣∣∣∣∣∣

2x 1 0 . . . 01 2x 1 . . . 00 1 2x . . . 0...

...... . . .

...0 0 0 1 2x

∣∣∣∣∣∣∣∣∣∣∣n×n

. (7.11)

The Chebychev polynomials of the first kind are defined as Tn (x) = Un (x)−xUn−1 (x). Hence that

1 + 2∑n≥1

Tn (x) tn = 2(1− 2xt+ t2

)−1 − 1− 2xt(1− 2xt+ t2

)−1

=1− t21 + t2

(1− 2xt

1 + t2

)−1

and

Tn (cos θ) =sin ((n+ 1) θ)− cos θ sinnθ

sin θ= cosnθ.

Now consider the general case, p (x, t) = 1−bt+(1−c)t21+t2−2xt (D. Stanton [91, 1983]).

In this casepn (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, becauseof initial conditions.


Remark 7.1.3. Anshelevich [6] studies the orthogonal Sheffer sequences for the caseλ = 1, but in several non-commuting variables. The recursion relation is describedin 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 geometricseries 1/ (1− xt) with reference operator χ and reference sequence (xn). For giveny ∈ C, say, the operator H1,y = (1− yχ)

−1 is a Sheffer operator for χ (see (6.17)),hence sn (x) := H1,yx

n is a 1-binomial Sheffer polynomial for χ and has the initialvalues generated by

∑n≥0 sn (y) tn = (1− yt)−2, thus

sn (y) = (n+ 1) yn. (7.12)

In detail,

sn (x) = H1,yxn =

n∑k=0

xn−kyk =xn+1 − yn+1

x− y .

By induction we can show that

sn (x) =

n∑k=0

(−1)k

(n− kk

)xkyk (x+ y)

n−2k.

Hence the identity

xn+1 − yn+1

x− y =

n∑k=0

(−1)k

(n− kk

)xkyk (x+ y)

n−2k (7.13)

follows (identity 1.60 in Gould’s list [38]). This is a special case of the identity

xn+1 − yn+1

x− y =

bn/(m+1)c∑k=0

(−1)k

(n−mk

k

)( xm

)kymk

( xm

+ y)n−(m+1)k

+

n∑k=dn/me

(−1)k

(n−mk

k

) n−k∑j=0

( xm

)j+kyn−k−j

(n− (m+ 1) k

j

)


which holds for allm ∈ N1 . We show this identity by expanding(xm + y

)n−(m+1)k=

bn/(m+1)c∑k=0

(−1)k

(n−mk

k

) n−k∑j=0

( xm

)k+j

yn−k−j(n− (m+ 1) k

j

)

+

n∑k=dn/me

(−1)k

(n−mk

k

) n−k∑j=0

( xm

)k+j

yn−k−j(n− (m+ 1) k

j

)

=

n∑k=0

(−1)k

(n−mk

k

) n−k∑j=0

( xm

)k+j

yn−k−j(n− (m+ 1) k

j

)

=

n∑l=0

xlyn−l (−m)−l

l∑k=0

(n−mk

k

)(l − n− 1 +mk

l − k

).

By identity (2.16),

(−m)l

=

l∑k=0

(c− kmk

)(l − 1− c+ km

n− k

)for all c.

If α is any positive integer, then

(1− xt)−α = Dα−1M(xα−1

)(1− xt)−1

.

Therefore,Hα,y (1− xt)−α = Hα,yDα−1M

(xα−1

)(1− xt)−1

,

and from

Hα,yDα−1 =1

1− y (η + α)−1D

Dα−1 = Dα−1 1

1− y (η + 1)−1D

(7.14)

follows

Hα,y

(n+ α− 1

n

)xn = Dα−1H1,yx

n+α−1 = Dα−1xn+α − yn+α

x− y .

The relationship Hα,yDα−1 = Dα−1H1,y in (7.14) implies

Hkα,yDα−1 = Dα−1Hk

1,y

for all natural numbers k. Note that the Sheffer polynomial Hα,y

(n+α−1

n

)xn taken

at x = y gives n!yn(n+αn

), a generalization of (7.12).


The evaluation functional Let c ∈ k and α be a positive integer. The functional〈Evalc | xn〉 = cn has the associated operator

op (Evalc) = (1− cBα)−α

= Hαα,c

in the α-binomial reference frame, according to Example 6.4.1. Hence

op (Evalc)−1

= (1− cBα)α

=

α∑i=0

(α

i

)(−1)

iciBiα.

Suppose we want to find the Sheffer sequence (sn (x))for Ba satisfying sn (c) =yn ∈ k, where y0 is a unit in k. The general functional expansion theorem 6.4.2tells us that

sn (x) =

n∑k=0

〈L | sn−k〉 op (L)−1

(k + α− 1

k

)xk

=

n∑k=0

yn−k

k∑i=0

(α

i

)(−1)

ici(k − i+ α− 1

k − i

)xk−i

=

n∑i=0

(α

i

)(−1)

ici

n∑k=i

yn−k

(k − i+ α− 1

k − i

)xk−i

=

n∑i=0

(α

i

)(−1)

icin−i∑k=0

yn−i−k

(k + α− 1

k

)xk

In terms of generating function,

∑n≥0

sn (x) tn =

∑k≥0 ykt

k

(1− ct)α (1− xt)α

If c = 0, then

sn (x) =

n∑k=0

yn−k

(k + α− 1

k

)xk

in agreement with Lemma 6.2.6. If α = 1 and yn = yn, then

sn (x) = yn + (x− c) yn − xny − x = yn + (x− c)H1,yx

n−1.

7.1.2 Generalized Catalan Operators

Let Cα be the generalized Catalan operator

Cαxn :=

n∑k=1

(n

k

)Ck−1x

n−k/

(n− 1 + α

k

)


(−α /∈ N0), where Ck is the k-th Catalan number. From

Cαxn =

n∑k=1

Ck−1

(1

η + αD)k

xn (7.15)

follows that Cα = 12 −

12

√1− 4Bα (see Example 6.2.8), a delta operator for the α-

binomial reference operator Bα = (η + α)−1D. Therefore we find Cα = β−1 (Bα),

where β−1 (t) = 12 −

12

√1− 4t, and by inversion, β (t) = t− t2. The basic polyno-

mials (cn) for Cα therefore have generating function∑n≥0

cn (x;α) tn =(1− x

(t− t2

))−α.

We expand cn (x;α) as

cn (x;α) =

n∑k=0

(k + α− 1

k

)(k

n− k

)(−1)

n−kxk. (7.16)

From Cα = β−1 (Bα) follows Bα = β (Cα) = Cα − C2α. Any Sheffer sequence (sn)

for Cα must satisfy this recursion,

(η + α)−1Dsn (x) = sn−1 (x)− sn−2 (x) .

In terms of the coeffi cient sn,i =[xi]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

cn (x; 1) =

n∑k=0

(k

n− k

)(−1)

n−kxk,

an interesting sequence indeed. On one side we observe that the identity (7.13)implies

(−z/4)ncn (−4/z; 1) =

an+1 − bn+1

2n (a− b)where a = 1 +

√z + 1 and b = 1 −

√z + 1. For z = 4 we get (−1)

ncn (−1; 1) =∑n

k=0

(n−kk

)= Fn, the n-th Fibonacci number. On the other hand, let

Un (x) := (2x)−n

cn

((2x)

2; 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

second kind (Exercise 7.1.13), hence cn(

(2x)2

; 1)

= (4x)n∏n

k=1

(x− cos kπ

n+1

).


7.1.3 Dickson Polynomials

Let a be a complex number. The Dickson polynomials (of the first kind) are theunique polynomial sequence defined by

Dn (y + a/y; a) = yn + (a/y)n (7.18)

for all n ≥ 0.In Exercise 7.1.16 we show that (Dn (x; a)) is a basis; it holds

xn =(nn/2

)an/2 +

(n−1)/2∑k=0

(nk

)akDn−2k (x; a) ,

where(nn/2

)= 0 for odd n. Obviously D0 (x) ≡ 2,D1 (x) = x,andD2 (x) = x2−2a.

We get the generating function in terms of y as∑n≥0

Dn (y + a/y; a) tn =1

1− yt +1

1− at/y

In terms of x = y + a/y (it does not matter which solution we choose!) we get∑n≥0

Dn (x; a) tn =1

1− yt +1

1− (x− y) t=

2− xt1− xt+ t2a

This generating function becomes a Sheffer sequence in the B1 = χ reference frame,if we change D0 from 2 to 1,

∑n≥0

Dn (x; a) tn − 1 =1− t2a1 + t2a

(1− xt

1 + t2a

)−1

.

=1− t21 + t2

(1− 2xt

1 + t2

)−1

Note that

∑n≥0

Dn (x; a) (−a)−n/2

tn − 1 =1 + t2

1− t2

(1− xt/

√−a

1 + t2

)−1

showing that we can restrict the discussion to the Sheffer polynomials

dn (x) := Dn (x;−1)− δ0,n

becauseDn (x; a) = (−a)

n/2Dn

(x/√−a;−1

).


The Dickson polynomials with a = −1 are closely related to the Pell polynomialsPn (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∑n≥0

Tn (x) tn − 1 =1− t21 + t2

(1− 2xt

1 + t2

)−1

hence 2Tn (x) = Dn (2x; 1) .The definition (7.18)

Dn (y − 1/y;−1) = yn + (−1)n/yn

shows that for x = y − 1/y holds

Dn (x;−1) =

(x+√x2 + 4

)n+(x−√x2 + 4

)n2n

.

What properties does the basic sequence (bn) for the delta operator β−1 (χ)have, where β−1 (t) is the compositional inverse of β (t) = t/

(1− t2

), thus β−1 (t) =(√

1 + 4t2 − 1)/ (2t)? In Exercise 7.1.19 it is shown that

bn (x) = xbn−1 (x) + bn−2 (x) (7.19)

for all n > 2, with initial values b0 (x) ≡ 1, b1 (x) = x, and b2 (x) = x2. Thisrecursion also holds for dn (x) and Dn (x;−1), but with different initial values. Weget an explicit expression for bn (x) directly from the generating function,

bn (x) =

n/2∑j=0

(n− j − 1

j

)xn−2j .

From

dn (0) = [tn]1 + t2

1− t2 = 2

for even n ≥ 0, and 0 else, we get

Dn (x;−1) =

n/2∑j=0

(n− jj

)n

n− j xn−2j

andbn (x) =

x

nDDn (x;−1) =

η

n(Dn (x;−1)−D (0;−1))


for n ≥ 1. The Sheffer operator for (dn) equals

1 + β−1 (χ)2

1− β−1 (χ)2 = 1 + 2β−1 (χ)

β−1 (χ)

1− β−1 (χ)2 = 1 + 2β−1 (χ)β

(β−1 (χ)

)= 1 + 2β−1 (χ)χ,

and thereforeDn (x;−1) = bn (x) + 2bn−1 (x) /x

for all n ≥ 2.Applying (7.19) we can write

Dn (x;−1) = 2bn+1 (x)

x− bn (x) (7.20)

for all n ≥ 1.We can also apply the factoring method (section 1.1.1) to the recursion (7.19),

and obtain

bn (x) = x

(x+√x2 + 4

)n − (x−√x2 + 4)n

2n√x2 + 4

for n ≥ 1 (Exercise 7.1.20). Combining this formula with (7.20) shows again that

Dn (x;−1) =

(x+√x2 + 4

)n+(x−√x2 + 4

)n2n

for all n ≥ 0. If k is a positive integer, then

bnk (x) = x

(x+√x2 + 4

)nk − (x−√x2 + 4)nk

2n√x2 + 4

= bn (x) pn,k

(x+√

4 + x2

2,x−√

4 + x2

2

), (7.21)

where

pn,k (u, v) =unk − vnkun − vn =

k−1∑i=0

univn(k−1−i)

is a symmetric bivariate polynomial of degree n (k − 1) in u and also in v. If we lety =

(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 ofSymmetric Functions the symmetric function pn,k (y, y) can be expressed as apolynomial 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), wherepn,k (y, y) must be of degree n (k − 1). For k = 2 we obtain the doubling formula

b2n (x) = 2−n((x+

√4 + x2

)n+(x−

√4 + x2

)n)bn (x) = Dn (x;−1) bn (x) .


n bn (x) Dn (x;−1) n bn (x)

0 1 2

1 x x 7 x(x6 + 5x4 + 6x2 + 1

)2 x2 x2 + 2 8 x2

(x2 + 2

) (x4 + 4x2 + 2

)3 x

(x2 + 1

)x(x2 + 3

)9 x

(x2 + 1

) (x6 + 6x4 + 9x2 + 1

)4 x2

(x2 + 2

)x4 + 4x2 + 2 10 x2

(x4 + 3x2 + 1

) (x4 + 5x2 + 5

)5 x

(x4 + 3x2 + 1

)x(x4 + 5x2 + 5

)11 x

(x10 + 9x8 + 28x6 + 35x4 + 15x2 + 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 seriesr(xt) = (1− xt)−α.

1. Find the reference polynomials rn(x) and the reference operator Bα.

2. Suppose (bn(x;α))n≥0 is the basic sequence for the delta operator

B = β−1 (Bα) where α ∈ R, but −α /∈ N0, hence b (x, t) = (1− xβ (t))−α.

Show (7.1) directly, without using the exponential reference frame.

3. Let y ∈ R and

sn(x) :=

n∑k=0

sk(y;α)bn−k(x− y;α).

Show that (sn) is a Sheffer sequence for the α-binomial delta operatorβ−1 (Bα/ (1 + yBα)).

7.1.2. Expand Ec in terms of η and Bα = η+1η+αχ, the α-binomial reference operator,

as Ec =∑n≥0 c

n(η+α+n−1

n

)Bnα, and verify that Ec

(k+α−1

k

)xk =

(k+α−1

k

)(x+ c)

k

for all c ≥ 0. Of course, EcBα 6= BαEc, because Ec /∈ ΣBα .

7.1.3. Let

tn (x;α+ 1) :=n+ 1

xbn+1(x;α),

where∑n≥0 bn (x) tn = (1− xβ (t))

−α. Show that (tn) is a Sheffer sequence forthe α+ 1-binomial delta operator β−1 (Bα+1).


7.1.4. Let α = 1 and gn (x) = bn+1 (x; 1) /x, n = 0, 1, . . . , where∑n≥0 bn (x) tn =

(1− xβ (t))−1. Show that (gn) is a Sheffer sequence for B = β−1 (B1) with gener-

ating function ∑n≥0

gn(x)tn =β (t) /t

1− xβ (t).

Therefore, sn (x) =(cx − 1

)bn (x; 1) is a Sheffer sequence for B with initial values

sn (c) = δ0,n.

7.1.5. Show that

1

1− y (η + α)−1D

Dα−1 = Dα−1 1

1− y (η + 1)−1D

(Equation (7.14)).

7.1.6. Show that in the orthogonal binomial case the three equations (7.5) hold forαk, βk, and γk. Show that φ (t)β (t) = t.

7.1.7. Show that the delta operator G : P(−λ)n → P

(−λ)n−1 solves the operator equation

Bλ =2G

(1 +G2)

where Bλ = (η + λ)−1D. Find the basic sequence (gn) for G. Using the initial

values (7.9), we can expand

P(−λ)2n (x) =

n/2∑k=0

(−λk

)g2n−k (x)

with the help of (7.2).

7.1.8. Show that the determinant in (7.11) follows the Chebychev recursion (7.10).

7.1.9. Show that sin((n+1)θ)sin θ follows the Chebychev recursion (7.10).

7.1.10. Show that (Un) is a Sheffer sequence for(

1−√

1− χ2)/χ.

7.1.11. Find the basic sequence (bn (x;α)) (formula (7.16)) for the general Catalandelta operator with the help of Corollary 6.3.6.

7.1.12. Let C1 =(C2

1 + 1)B + B2. Find the basic sequence (bn) of B in the geo-

metric framework (α = 1).

7.1.13. Show that for the Chebychev polynomials Un (x) holds

Un (x) =

(x+√x2 − 1

)n+1 −(x−√x2 − 1

)n+1

2√x2 − 1


7.1.14. Show that the polynomials sn (x) =∑nk=0 sn,kx

k follow the recursionBαsn = sn−1− sn−2 if the coeffi cients satisfy sn+1,i+1 = i+α

i+1 (sn,i − sn−1,i) for all0 ≤ i ≤ n. Verify that the basic polynomials cn (x;α)in (7.16) have this property.

7.1.15. Suppose we have the following recursion for sn (x) =∑ni=0 sn,ix

i

n∑i=1

sn,i

i∑k=1

Ck−1xi−k =

n−1∑i=1

sn−1,i

(xi +

i∑k=1

Ck−1xi−k

)+

n−2∑i=0

sn−2,ixi

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 andhas the initial values bn (0) = δn,0, hence the recursion gives

bn−2 (x) + bn−1 (x) =

n∑k=1

Ck−1

(n∑i=k

(bn,i − bn−1,i)xi−k

)= C1 (bn (x)− bn−1 (x))

if C = C1 is the Catalan operator in (7.15). Show that(1− xt

(1 + t)(2t+ t2 − 1

)(1− t)2

)−1

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 polynomialin y + a/y of degree n, and that

xn =(nn/2

)an/2 +

(n−1)/2∑k=0

(nk

)akDn−2k (x; a) .

where(nn/2

)= 0 for odd n.

7.1.17. Show that 2nDn (cos θ; 1/4) = 2 cosnθ. This implies

cosn θ =(nn/2

)2−n + 21−n

(n−1)/2∑k=0

(nk

)cos (n− 2k) θ

7.1.18. For the Sheffer polynomials Dn (x; a)− δ0,n find the inverse in the umbralgroup (6.20). Compare to Exercise 7.1.16.

7.1.19. Show that the recursion (7.19) leads to the generating function(1− xt/

(1− t2

))−1for bn (x).


7.1.20. For given x we can solve the recursion bn (x) = xbn−1 (x) + bn−2 (x) forn > 2 with the factoring method (Exercise 1.1.4), using the initial values σ0 =b1 (x) = x and σ1 = b2 (x) = x2. Show that bn (x) =

x

(x+√x2 + 4

)n − (x−√x2 + 4)n

2n√x2 + 4

= 2−n+1

(n−1)/2∑k=0

(n

2k+1

)xn−2k

(x2 + 4

)k.

7.1.21. Prove the existence of a unique polynomial sequence (En (x; a)) such that

En (y + a/y; a) =(yn+1 − (a/y)

n+1)/ (y − a/y) .

Show that

En (x; a) = Dn (x; a) + aEn−2 (x; a)

= (nmod 2) an/2 +

(n−1)/2∑k=0

akDn−2k (x; a) .


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 Andrewspointed 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 orderaccording to our theme. We give only a glimpse into the vast subject of EulerianDifferential Operators, showing how it ties in with our presentation of diagonaloperators. We omit completely the combinatorial interpretation, the number oftransformations on finite vector spaces.

Suppose we have a diagonal reference frame. In Exercise 6.2.6 we men-tioned that a Sheffer sequence (sn) for β−1 (R) has initial values sn (1) = δ0,niff∑n≥0 sn (x) tn = r (xβ (t)) /r (β (t)). We will see that such Sheffer sequences for

the case β (t) = at (0 6= a ∈ k) play an important role in the theory of Euleriandifferential operators. The fact that we concentrate on sn (1) instead of sn (0) isusually expressed by changing the variable to qx for q 6= 1, so that sn (1) = sn

(q0).

We will follow Andrews in writing sn (X), keeping in mind that X = qx. Insteadof the translation operator Ey we will introduce the scaling operator mappingXn to (XY )

n, and in view of Y n = eyn we denote this operator by qyη. Henceqyηp (X) = p (XY ) for any polynomial p ∈ k [X]. Writing qx for X shows whyqyη : qxn 7→ q(x+y)n is the analogue to Ey. Note that Andrews writes η instead ofFreeman’s qη. If for an operator T on k [X] holds

Tqyη = 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 somefunction f : N0 → k, i.e.,

TXn = f (η)Xn = f (n)Xn

for all n ≥ 0.

Proof. Let pn (X) = TXn, where pn is of any degree. Then

TqyηXn = T (Xqy)n

= qynpn (X) and

qaηTXn = qyηpn (X) = pn (qyX)

for all y ∈ k and n ≥ 0. If we think of Y = qy as the variable and X as a parameter,then Y npn (X) = pn (Y X) implies that

TXn = pn (X) = f (n)Xn

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 shiftinvariant operators. Because we reserved the word shift for a change of degree,and used scaling instead, we deviate from Andrews’terminology in this point. Forbrevity, we also omit the word Eulerian in this connection.

Definition 7.2.3. The operator U on k [X] is an Eulerian differential operator iffand

Uqyη = qy(η+1)U

for all y ∈ k, and UXn 6≡ 0 for all n > 0.

The operator χ : p (X) 7→ (p (X)− p (0)) /X is the standard example of anEulerian differential operator

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

pn (XY ) =

n∑k=0

pk (X)Y kpn−k (Y ) (7.22)

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 7→ pn−1.

Lemma 7.2.5. If (pn) is an Eulerian sequence, and P : pn 7→ pn−1, then P is anEulerian differential operator.

Proof.

Pqyηpn (X) = P

n∑k=0

pn−k (X)Y n−kpk (Y ) = Y

n−1∑k=0

pn−1−k (X)Y n−1−kpk (Y )

= Y pn−1 (XY ) = qy(η+1)Ppn (X) .

Because pn is of degree n, we have that PXn 6≡ 0 for all n > 0.The Eulerian differential operators and 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) 6= 0 for alln > 0, then χd (η) is an Eulerian differential operator. Vice versa, if U is anEulerian differential operator, thenM (X)U is scaling invariant,M (X)U = d (η),satisfying d (n) 6= 0 for all n > 1, and d (0) = 0.

Proof. That χ is an Eulerian differential operator shows the first direction of theLemma. Next suppose U is an Eulerian differential operator. We show for n > 0that M (X)UXn is translation invariant:

M (X)Uqyη = M (X) qy(η+1)U = qyηM (X)U.


Therefore, all Eulerian differential operators are of the form χd (η), whered is a nonzero function on N1, and d (0) can be chosen as 0. Remember that wedefined a diagonal reference operator R as R = d′ (η)χ/d′ (η), where d′ is anynonzero function on N0 (section 6.2.1). Letting d′ (n) = 1/

∏nk=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. We will call the functiond also diagonal, i.e., a diagonal function is 0 at 0, and d (n) 6= 0 for all n > 0. Wesummarize the above in the following Corollary.

Corollary 7.2.7. The polynomial sequences (pn) is Eulerian iff (pn) is the Sheffersequences with pn (1) = δ0,n in some diagonal reference frame.

The diagonal reference operators are called Eulerian differential operatorsby Andrews, and the corresponding Sheffer sequences are called Eulerian basicpolynomials. Finding Eulerian basic polynomials is easy in principle, because weknow there generating function,

∑n≥0 pn (X) tn = r (Xt) /r (t), where

r (Xt) =∑n≥0

Xntn/

n∏k=1

d (k) ,

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((X − 1)

n/n!)n≥0 must be Eulerian. We check this by defining pn (X) = (X − 1)

n/n!

and calculating

pn (XY ) =(XY − 1)

n

n!=

n∑k=0

(XY − Y )n−k

(n− k)!

(Y − 1)k

k!=

n∑k=0

pn−k (X)Y n−kpk (Y ) .

(b) The q-differentiation operator

Dq = χ (1− qη)

was defined in section 6.2.1 . Its Eulerian basic sequence has the generating func-tion (t; q)∞ / (Xt; q)∞ (Exercise 6.2.1). The identity∑

n≥0

(X − 1)(X − q) · · · (X − qn−1)

(q; q)ntn =

(t; q)∞(Xt; q)∞

is due to Heine, and shown by many authors [3, Theorem 10.2.1]. We can writethe Eulerian basic sequence for Dq as pn (X) = Xn

(X−1; q

)n/ (q; q)n. Phrasing

equation (7.22) for these polynomials gives

(XY ; q)n(q; q)n

=

n∑k=0

(Y ; q)k(q; q)k

(X; q)n−k(q; q)n−k

Y n−k, (7.23)

the q-binomial theorem (Andrews [4, (3.3.10)]).


If d is diagonal, where do the values d (n) occur in the Eulerian basic poly-nomials pn (X) for χd (η)? We show in Exercise 7.2.2 that the leading coeffi cientof pn (x) equals d′ (n) = 1/

∏nk=1 d (k), where d′ (η)χ/d′ (η) = χd (η). Hence

∑n≥0

pn (X) tn =r (Xt)

r (t)=

∑n≥0X

ntn [Xn] pn (X)∑n≥0 t

n [Xn] pn (X). (7.24)

We will now derive a second generating function for (pn), involving the derivativeof the polynomials.

Theorem 7.2.9. If (pn) is an Eulerian basic sequence, then∑n≥0

pn (X) tn = e∑n≥1 p

′n(1)tn(Xn−1)/n

(Andrews [5, Theorem 8]).

Proof. We have∑n≥0 pn (X) tn = r (Xt) /r (t), thus

∑n≥1

p′n (Y ) (Xt)n

= Xtr′ (Y Xt)

r (Xt).

Letting Y = 1 we get

Xtr′ (Xt) = r (Xt)∑n≥1

p′n (1) (Xt)n.

Seeing this as a differential equation in y = r with independent variable Xt weobtain the solution

y = C (t) e∑n≥1 p

′n(1)(Xt)n/n = r (Xt)

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

basic sequence for D. From p′n (X) = (X − 1)n−1

/ (n− 1)! we see that p′n (1) =δ1,n, and therefore we confirm that∑

n≥0

(X − 1)n/n! = e(X−1)t

by Theorem 7.2.9. We get the same generating function, eXt/et, from (7.24).

(b) The q-differentiation operator Dq = χ (1− qη) has Eulerian basic sequencepn (X) = Xn

(X−1; q

)n/ (q; q)n. We saw that in Example 7.2.8(b) that∑

n≥0

pn (X) tn =(t; q)∞

(Xt; q)∞.


We give a proof for this result by applying Theorem 7.2.9. First we have to find

p′n (X) = D(X − 1) (X − q) · · ·

(X − qn−1

)(q; q)n

=

n−1∑k=0

(X − 1) (X − q) · · ·(X − qn−1

)(X − qk) (q; q)n

hence p′n (1) = (q; q)n−1 / (q; q)n = 1/ (1− qn) for all n ≥ 1. Thus

exp∑n≥1

p′n (1) tn/n = exp∑n≥1

tn

n (1− qn)= exp

∞∑k=0

∞∑n=1

tn

nqnk

= exp

−∞∑k=0

ln(1− tqk

)= 1/

∞∏k=0

(1− tqk

)= 1/ (t; q)∞ .

Andrews [5] gives more results in the flavor of part (b) of the above. Theorem7.2.9 shows how an infinite sum becomes an infinite product,∑

n≥0

pn (X) tn =∏n≥1

ep′n(1)tn(Xn−1)/n.

We will finish this short introduction with a result (Corollary 7.2.12) thatshows 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))kXn = d (η − k + 1) · · · d (η)Xn

if n ≥ 0. If 0 ≤ n < k then M(Xk)

(χd (η))kXn = 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 =∑k≥0

〈Eval1 | Tpk〉 d (η − k + 1) · · · d (η)

Proof. Let x be any element of k [X], and write X for qx. An Eulerian basicsequence satisfies condition (7.22), hence

pn(XX

)=

n∑k=0

pk(X)Xkpn−k (X)

Keeping X fix, this is a polynomial in X, and we obtain

Tpn(XX

)=

n∑k=0

(Tpk

(X))Xkpn−k (X)


for any operator T . Only if T is scaling invariant, the left hand side equalsqxηTpn

(X). Letting X = 1 we have

Tpn (X) =

n∑k=0

〈Eval1 | Tpk〉Xkpn−k (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 seriesin the Eulerian operator χd (η) by

χc (η) = χ∑k≥0

〈Eval1 | c (η) pk〉M(Xk)

(χd (η))k

= χ∑k≥0

(c (η) pk) (1) d (η − k + 1) · · · d (η) ,

where (pn) is the Eulerian basic sequence for χd (η).

7.2.1 Exercises

7.2.1. Show that pn (X) = (X − 1)Xn−1 for n > 0, p0 (X) = 1, satisfies thecondition (7.22) of an Eulerian sequence. In which reference frame is (pn) a Sheffersequence?

7.2.2. Let χd (η) be a diagonal operator. Show that [Xn] pn (X) = 1/∏nk=1 d (k), if

(pn) is the Eulerian basic sequence for χd (η). Find an expression for[Xn−1

]pn (X).

7.2.3. Let c (η) = 1− qη and d (η) = η. Show that

c (η) =∑k≥0

(c (η)πk) (1) d (η − k + 1) · · · d (η) ,

where πk (X) is the Eulerian basic sequence for D. Show also the other direction,

η =∑k≥0

(ηpk) (1) c (η − k + 1) · · · c (η)

where pk (X) is the Eulerian basic sequence for Dq.

Bibliography

[1] Aigner, M. (1999). Catalan-like numbers and determinants, J. Combin. The-ory A 87, 33 —51.

[2] Al-Salam, W.A. (1967). q-Appell polynomials, Annali di Matematica, 77,31-45.

[3] Andrews, G.E., Askey, R., Roy, R. (1999) Special Functions. CambridgeUniversity Press, NY

[4] Andrews, G.E. (1976) The Theory of Partitions. Addison-Wesley, Reading,MA.

[5] Andrews, G.E. (1971) On the foundations of combinatorial theory: V.Eulerian differential operators, Stud. Appl. Math. 50, 345-375.

[6] Anshelevich, M. (2008) Orthogonal polynomials with a resolvent-type ge-neating function, Transactions Amer. Math. Soc. 360, 4125—4143.

[7] Askey, R., and Wilson, J. (1985) Some basic hypergeometric polynomials thatgeneralize Jacobi polynomials. Mem. Amer. Math. Soc. 319.

[8] Bailey, W. N. (1935) Generalized Hypergeometric Series. Cambridge Univer-sity Press (reprinted by Hafner Publishing Company, New York 1972).

[9] Barnabei, M., Brini, A., and Nicoletti, G. (1982). Recursive Matrices andUmbral Calculus, J. Algebra 75, 546-573.

[10] Bergeron, F., Labelle, G., and Leroux, P. (1998). Combinatorial Species andTree-like Structures. Cambridge University Press, Cambridge, UK.

[11] Bell, E.T. (1938) The history of Blissard’s symbolic calculus, with a sketchof the inventor’s life. Amer. Math. Monthly 45, 414—421.

[12] Blasiak, P., Penson, K.A., and Solomon, A.I. (2003) The bosom normalordering problem and generalized Bell numbers. Ann. Combinatorics 7, 127—139.

[13] Blissard, J. (1861) Theory of generic equations. Quart. J. Pure Appl. Math.,4, 279-305.

258 BIBLIOGRAPHY

[14] Boas, R.P., and Buck, R.C., Polynomials Expansions of Analytic Functions.Academic Press, New York, 1964.

[15] Bousquet-Mélou, M. and Schaeffer, G., Counting paths on the slit plane, inMathematics and Computer Science: Algorithms, trees, combinatorics andprobabilities, Trends in Mathematics, Birkäuser, 2000.

[16] Bousquet-Mélou, M. and Petkovšek, M (2000) Linear recurrences with con-stant coeffi cients: the multivariate case. Discrete Math. 225 , 51—75.

[17] Bousquet-Mélou, M. (2002). Counting walks in the quarter plane, Trends inMathematics, 49-67, Birkhäuser, Basel.

[18] Brenke, W.C. . On generating functions of polynomial systems, Amer. Math.Monthly 52 (1945) 297-301.

[19] Cartier, P. (2000). Mathemagics (a tribute to L. Euler and R. Feynmann).Séminaire Lotharingien de Combinatoire, 44, Article B44d

[20] Chihara, T. S. An Introduction to Orthogonal Polynomials. Gordon andBreach, New York, 1978.

[21] Cigler, J. and Krattenthaler, C. (2010). Some determinants of path gener-ating functions. To appear

[22] Di Bucchianico,A. and Soto y Koelemeijer, G. (2000) Solving linear recur-rences using functionals. Algebraic Combinatorics and Computer Science.A Tribute to Gian-Carlo Rota (Eds: Crapo, H. and Senato, D.), 461-472,Springer, Milano, Italy.

[23] Di Bucchianico,A. and Loeb, D.E. (1998) Umbral Calculus and natural ex-ponential families. Mathematical Essays in Honor of Gian-Carlo Rota (Eds:B. Sagan and R.P. Stanley), 195-211, Birkhäuser, Boston.

[24] Di Bucchianico,A. (1997) Probabilistic and Analytic Aspects of the UmbralCalculus. CWI Tract, Amsterdam, ISBN 90 6196 471 7.

[25] Di Bucchianico, A. and Loeb, D.E. (1996) Operator expansions in the deriv-ative and multiplication by x. Integral Transforms and Special Functions, 4,49—68.

[26] Di Bucchianico, A. and Loeb, D.E. (1995, last updated 2000) A selected sur-vey of umbral calculus, Dynamical Survey 3, Electronic J. of Combinatorics

[27] Di Bucchianico, A. and Loeb, D. (1994) A simpler characterization of Shefferpolynomials. Stud. Appl. Math. 92, 1—15.

[28] Di Nardo, E., Petrullo, P., and Senato, D. (2009) Cumulants and convolu-tions via Abel polynomials. Submitted.

[29] Di Nardo, E., and Senato, D. (2001) Umbral nature of the Poisson variables,in Algebraic Combinatorics and Computer Science, pp. 245-266, Springer,Milan.

BIBLIOGRAPHY 259

[30] DiNardo, E., Senato, D., and Niederhausen, H. (2008) The classical UmbralCalculus: Sheffer Sequences. Submitted.

[31] Duchamp, G.H.E., Poinsot, L., Solomon, A.I., Penson, K.A., Blasiak, P., andHorzela, A. (2009) Ladder operators and endomorphisms in combinatorialphysics. Submitted.

[32] Euler, L. (1801) De evolutione potestatis polynomialis cuiuscunque(1 + x+ x2 + x3 + x4 + etc.

)n. Nova Acta Academiae Scientarum Imperi-

alis Petropolitinae 12, 47—57

[33] Freeman, J.M. (1998) A strategy for determining polynomial orthogonality.Mathematical essays in honor of Gian-Carlo Rota (Cambridge, MA, 1996),239—244, Progr. Math., 161, Birkhäuser Boston, Boston, MA.

[34] Freeman,J.M. (1987) Orthogonality via transforms. Studies in Appl. Math.77, 119-127.

[35] Freeman, J.M. (1985) Transforms of operators on K[x][[t]]. Congr. Numer-antium 48, 115-132.

[36] Garsia, A. M. and Joni, S. A. (1980) Composition sequences. Comm. Algebra8, 1195—1266.

[37] Gessel, I. M. (2003) Applications of the classical umbral calculus. Dedicatedto the memory of Gian-Carlo Rota. Algebra Universalis 49, 397—434.

[38] Gould, H.W. (1972). Combinatorial Identities. Morgantown, Va.

[39] Gzyl, H. (1988) Umbral Calculus via Integral Transforms. J. Math. Anal.Appl. 129, 315-325.

[40] He, T.-X., Hsu, L. C., Shiue, P. J.-S. (2007) The Sheffer group and theRiordan group. Disc Appl. Math 155, 1895—1909.

[41] Henrici, P. (1988), Applied and Computational Complex Analysis, Vol. 1.Wiley, New York.

[42] Hickerson, D. (1999) Counting horizontally convex poly-ominoes. J. Integer Sequences 2 (1999) Article 99.1.8,http://www.research.att.com/~njas/sequences/JIS/HICK2/chcp.html.

[43] Hofbauer, J. (1979) A short proof of the Lagrange-Good formula. Disc. Math.25, 135-139.

[44] Ismail, M.E.H. (2001) An operator calculus for the Askey-Wilson operator,Annals of Comb. 5, 347 —362.

[45] Ismail, M.E.H. (1978) Polynomials of binomial type and approximation the-ory, J. of Approximation Th. 23, 177-186.

[46] Ismail, M.E.H. and May, C.P. (1978) On a family of approximation opera-tors. J. Math. Anal. Appl. 63, 446-462.

http://www.research.att.com/~njas/sequences/JIS/HICK2/chcp.html

260 BIBLIOGRAPHY

[47] Jordan, C. (1939). Calculus of Finite Differences. Chelsea Publ. Co., NewYork, 3rd edition 1979.

[48] Joyal, A, (1981). Une théorie combinatoire des séries formelles. Adv. in Math.42, 1-82.

[49] Kholodov, A. N. (1990) The Umbral Calculus and orthogonal polynomials.Acta Applicandae Math. 19, 1-54.

[50] Krattenthaler, C. (2001) Permutations with restricted patterns and Dyckpaths, Adv. Appl. Math, 27, 510-530

[51] Krattenthaler, C., Guttmann, A.J. and Viennot, X. G. (2000). Vicious walk-ers, friendly walkers and Young tableaux II: with a wall, J. Phys. A: Math.Gen. 33, 8835-8866.

[52] Krattenthaler, C. (1997) The enumeration of lattice paths with respect totheir number of turns, Advances in Combinatorial Methods and Applica-tions to Probability and Statistics (Ed.: N. Balakrishnan), 19-58, Boston,Birkhäuser.

[53] Kreweras, G. (1965). Sur une classe de problèmes liés au treillis des partitionsd’entiers, Cahiers du B.U.R.O. 6, 5-105.

[54] Kwasniewski, A. K. and Borak, E. (2005) Extended finite operator calculus- an example of algebraization of analysis, Central European J. of Math. 2,767-792.

[55] Lukacs, E. (1960) Characteristic Functions. Second edition 1970, Griffi n,London.

[56] Markowsky, G. (1978). Differential operators and the Theory of BinomialEnumeration. J. Math. Anal. Appl., 63,145-155.

[57] Matthews, L. (2001). Dissertation: Combinatorial Interpretations of Han-kel Matrices and Further Combinatorial Uses of Riordan Group Methods,Howard University.

[58] Merlini, D.,Rogers, D. G., Sprugnoli, R., and Verri, M.C. (1997). On somealternative characterizations of Riordan arrays. Canadian J. Math. 49, 301-320.

[59] Meixner, J. (1934). Orthogonale Polynomsysteme mit einer besonderenGestalt der erzeugenden Funktion. J. London Math. Soc. 9, 6-13.

[60] Mohanty, S.G. (1979). Lattice Path Counting and Applications. AcademicPress, New York.

[61] Morris, R.A., editor (1982). Umbral calculus and Hopf algebras, volume 6 ofContemporary Mathematics. Amer. Math. Soc., 1982.

BIBLIOGRAPHY 261

[62] Mullin, R., Rota, G. C. (1970). On the Foundations of Combinatorial TheoryIII: Theory of Binomial Enumeration, in B. Harris (ed.). Graph Theory andIts Applications, Academic Press, New York, pp. 167-213.

[63] Narayana, T.V. (1955). Sur les treillis formés par les partitions d’un entier.Comptes Rendus Acad. Sci. Paris. Ser. I 240, 1188-89.

[64] Niederhausen, H. (2002). Planar walks with recursive initial conditions, J.of Statistical Planning and Inference, 101, 229-253

[65] Niederhausen, H. (2001). Generalized Sheffer Sequences Satisfying PiecewiseFunctional Conditions, Computers & Math. with Appl. 41, 1155-1171.

[66] Niederhausen, H. (1999). Recursive initial value problems for Sheffer se-quences, Discrete Math. 204, 319-327.

[67] Niederhausen, H. (1992). Fast Lagrange inversion, with an application tofactorial numbers, Discr. Math. 104, 99-110.

[68] Niederhausen, H. (1980). Linear recurrences under side conditions. EuropeanJ. Combin. 1, 353-368.

[69] Nkwanta, A., and Shapiro, L. W. (2005). Pell walks and Riordan matrices,Fibonacci Quarterly 43, 170—180.

[70] Radoux, C. (2000) Addition formulas for polynomials built on classical com-binatorial sequences, J. Comput. Appl. Math., 115, 471—477.

[71] Radoux, C. (1979) Calcul effectif de certains determinants de Hankel, Bull.Soc. Math. Belg. Ser. B, 31, 49—55.

[72] Riordan, J. (1968). Conbinatorial Identities, Wiley, New York.

[73] Riordan, J. (1958). An Introduction to Combinatorial Analysis, Wiley, NewYork.

[74] Pincherle, S. and Amaldi, U.: Le operazioni distributive e le loro applicazioniall‘analisi, N. Zanichelli, Bologna, 1901.

[75] Rogers, D.G. (1978). Pascal triangles, Catalan numbers and renewal arrays,Discrete Math. 22, 301-310.

[76] Roman, S. (1984). The Umbral Calculus. New York: Academic Press, 1984

[77] Roman, S., De Land, P., Shiflett, R., and Shultz, H. (1983). The umbralcalculus and the solution to certain recurrence relations, J. Combin. Inform.System Sci. 8 , 235—240.

[78] Roman, S. (1982). The theory of the umbral calculus. I, J. Math. Anal. Appl.87, 58-115.

[79] Roman, S.M. (1980). The algebra of formal series II, J.Math. Anal. Appl.74, 120-143.

262 BIBLIOGRAPHY

[80] Roman, S.M. (1979). The algebra of formal series III: Several variables, J.Approximation Theory 26, 340-381.

[81] Roman, S.M., and Rota,G.-C. (1978). The Umbral Calculus, Adv. in Math.27, 95 - 188.

[82] Rota, G.-C., Shen, J., and Taylor, B. (1997). All polynomials of binomialtype are represented by Abel polynomials. Ann. Scuola Norm. Sup. Pisa Cl.Sci. (4) 25, 731 - 738

[83] Rota, G.-C., Kahaner, D., and Odlyzko, A. (1973). On the Foundations ofCombinatorial Theory VIII: Finite operator calculus, J. Math. Anal. Appl.42, 684-760.

[84] Sapounakis, A., Tasoulas, I., and Tsikouras, P. (2007) Counting strings inDyck paths. Discrete Math. 307, 2909 - 2924.

[85] Sharma, A., and Chak, A. (1954). The basic analog of a class of polynomials,Revista di Matematica della Universita di Parma, 5, 325-337.

[86] Sheffer, I.M. (1945) Note on Appell polynomials, Bull. Amer. Math. Soc. 51,739-744.

[87] Schröder,E. (1870) Vier Kombinatorische Probleme, Z. Math. Phys. 15, 361-376.

[88] Stanley, R.P. (2008) Catalan addendum (version of 29 February 2008; 69pages). http://www-math.mit.edu/~rstan/ec/catadd.pdf

[89] Stanley, R.P. (1986) Enumerative Combinatorics, Vol. I, Wadsworth &Brooks/Cole, Monterey, California.

[90] Stanley, R.P. (1999) Enumerative Combinatorics, Vol. II, Cambridge Uni-versity Press, Cambridge, UK.

[91] Stanton, D. (1983) Generalized n-gons and Chebychev polynomials, J.Comb. Th. A 34, 15—27.

[92] Sulanke, R. A. (2001). Bijective recurrences for Motzkin paths, Adv. in Appl.Math. 27, 627—640.

[93] Sulanke, R.A. (2000) Counting lattice paths by Narayana polynomials, Elec-tronic J. of Combinatorics, Vol. 7(1), R40.

[94] Sulanke, R.A. (2000) Moments of generalized Motzkin paths. J. Integer Se-quences, 3, Article 00.1.1.

[95] Sulanke, R.A. (1998). Bijective recurrences concerning Schröder paths. Elec-tronic J. Combinatorics 5, # R47.

[96] Taylor, B. (1998). Difference equations via the classical umbral calculus.Mathematical essays in honor of Gian-Carlo Rota (Cambridge, MA, 1996),Progr. Math., 161, Birkhäuser Boston, MA, 397 - 411.

BIBLIOGRAPHY 263

[97] Verde—Star, L. (1985). Dual operators and Lagrange inversion in severablevariables, Adv. Math. 58, 89-108.

[98] Wald, A. and Wolfowitz, J. (1939) Confidence limits for continuous distrib-ution functions. Ann. Math. Statist. 10, 105-118

[99] Waring, E. (1779). Problems concerning interpolation, Philos. Trans. RoyalSoc. London, 69, 59-67.

[100] Watanabe, T. (1984). On dual relations for addition formulas for additivegroups, I., Nagoya Math. J. 94, 171-191.

[101] Watanabe, T. (1985). On dual relations for addition formulas for additivegroups, II., Nagoya Math. J. 97, 95-135.

[102] Wilf, H. S. (1990). generatingfunctionology, Academic Press. 3rdedition 2006by A. K. Peters, Ltd.

[103] Zhang, Z. Z., and Feng, H. (2006). Two kinds of numbers and their applica-tions, Acta Math. Sinica, 22, 999 —1006.

finite operator calculus with applications to linear recursions

Documents