Rota’s Umbral Calculus and Recursions
Heinrich Niederhausen Florida Atlantic University Boca Raton, FL 33431 November 15, 2001
Abstract Umbral Calculus can provide exact solutions to a wide range of linear recursions. We summarize the relevant theory and give a variety of examples from combinatorics in one, two and three variables. Keywords: Umbral Calculus, recursions
1Introduction
What kind of recursions can we hope to solve with Rota’s Umbral Calculus (UC)? We like to call them delta recursions; however this cannot be explained without some UC terminology. Therefore the answer is postponed until the end of this introduction. As a first approximation we may say that Rota’s Umbral Calculus is about an isomorphism between formal power series, linear functionals an a certain class of linear operators (shift invariant) on polynomials.
The isomorphisms Example j q c (q)= j 0 jq c (q)= b ≥ uj %.P -& aj uj %. j -& I j! = j ←− Pm(u)= j 0 j auj m (u) Eval j! = j! ←− B m (u)=m (u + ) | −→ ≥ Evaluation| −→ Shift operator ® P ® Common ground to the three concepts are special polynomial sequences, called Sheffer sequences. A poly- nomial sequence (pj (u)) is a sequence of polynomials pj (u) [u] such that deg pj = j, p =0. j N0 K 0 ∈ ∈ 6 It is convenient to define pj =0for negative j.Thecoefficient ring K isassumedtobeofcharacteristic 0. For this paper it suffices to choose K as R [φ], the ring of polynomials in some weight parameter φ.A formal power series d(q) of order 1, i.e. , d (0) = 0, d0 (0) =0, will be called a delta series. We substitute the derivative operator for q in a delta series, and obtain6 a shift-invariant linear operator N called a delta operator. The derivativeD itself is a delta operator, and like every delta operator N reduces degrees by one, and has its null-spaceD equal to the constant polynomials.D The solutions to the system of operator equations Npj (u)=pj 1(u) − are therefore only determined up to constants; every polynomial sequence (pj) solving that system is a N -Sheffer sequence. Initial values determine the constants, and if the initial values are pj (u)= 0j,this special Sheffer sequence is called the N - basic sequence. Umbral Calculus can be used as a tool for solving recursions, if the exact solutions to such recursions are Sheffer sequences. In that case, the recursion will lead to an operator equation involving the (unknown) delta operator N,andtheN-basic sequence is expanded with the help of the Transfer Formulas (1) or (3). It is rare that an interesting combinatorial problem can be solved by a basic sequence, but the following example is an exception.
Example 1 A staircase polygon (parallelogram polyominoe) can be viewed as a pair of lattice paths with steps and , starting and ending at a common point without intersections in between, thus forming a → ↑ staircase polygon. In order to count the number c2k of staircase polygons with circumference 2k,weactually
1 enumerate truncated (by a line with slope 1) staircase polygons which have a gap f,0 (from truncation) − and call their number bj (f) if the remaining staircase circumference is 2f +2j.Notethat2f is the smallest possible remaining circumference, thus b0 (f)=1. The picture below explains the terminology and shows what we mean by the “handles” of the truncated polygon.
Truncated staircase polygon (gap f =3,andj = f +10) . cut .. → . ···· .. upper handle · ···· → .······ .. · .····· .. ··· .···· .. ··· · ··lower handle ·····↑ If we move the truncation line one unit to the right, we lose the old handles, and the gap changes according to their direction, bj (f)= bj (f 1) + bj 2 (f +1) +2bj 1 (f) − − − both handles inwards both handles outwards handles parallel
(Conway, Delest, and Guttmann [2,| 1997]).{z } The initial| values{z b}j (0) = 0|for{zj,} 0 guarantee that there always is a gap. The above recursion can be interpreted as a system of difference equations, and from b0 (f)=1follows that the solution can be extended to a polynomial sequence (bj (u)).Define the operator B by linear extension of its action on the basis (bj), Bbj (u):=bj 1 (u) for all j.Therecursionbj (u)= − bj (u 1) + bj 2 (u +1)+2bj 1 (u) is equivalent to the operator identities − − − 1 1 2 1=B− + B B +2B or = B1B2 +2B ∇ 1 u 1+j where =1 B− is the backwards difference operator, a delta operator with basic sequence − . ∇ − j j 0 The Transfer Theorem 3 shows that B is a delta operator, and its basic sequence is our solution sequence≥ ¡¡ ¢¢ (bj (u)) because of the initial values bj (0) = 0j. By formulas (3) and (5)
j j 2g j j 1 2 f 1 f 1+u g 2 − u j + u bj (u)=u [B ] B B +2B − = u f j g j + u g f=1 µ ¶ g=0 µ − ¶ µ ¶ X ¡ u ¢ X 1 1 − b (u) qj = q + q j 2 4 Ã − r − ! X j j ([q ] c (q) stands for the coefficient of q in c (q)). The total number c2k of staircase polygons of circumference 2k equals the number of truncated staircase polygons of circumference 2k 2 and gap 1, − k 2 2g k+2 − g 2 − k 1 1 2k 2 c2k = bk 2 (1) = − = − − k 2 g k 1 g k k 1 g=1 X µ − − ¶ − µ ¶ µ − ¶ (Levine [5, 1959]), a Catalan number.
If necessary and possible, the initial conditions are formulated as a functional on K [u], and the Functional Expansion Theorem 5 expands the Sheffer polynomials pj in terms of the basic sequence. We call this the “bottom up” approach: Starting with simple “building blocks” (in combinatorics they usually are binomial coefficients) we construct the basic sequence, and from this more advanced material the exact solution is put together such that the initial conditions are satisfied. In the more commonly applied “top down” approach a generating function identity is solved, and the exact solution may be obtained by finding coefficients. We
2 will show that in many applications UC delivers the generating function too; the more interesting cases may be those where that is not the case (Theorem 3). While the theory behind the Umbral Calculus needed for this type of applications is so simple that we could as well omit it, the “bottom up” approach itself needs some examples to demonstrate its power. As always, the elegant examples are easy in most approaches, and the hard applications are too long to present. We hope that the rather long example in Section 4.1 gives a glimpse of the possible applications. The following examples are discussed in this paper. Two variables: Staircase polygons (Introduction and end of Section 2.1). Two variables: Tiling a 4 j + h rectangle with j squares of size 2 2 (Example 6 in Section 2.2). One variable: Enumeration× of generalized elevated Schröder paths by× area (Section 3.1). Two variables: Lattice paths with infinite step set and special access (Section 4.1). Three variables: A lattice walk with frequent stops (Section 5.2). All the above examples came from combinatorics and are discrete. However, recursions like a a dj (u)=dj 1 (u ")+φ dj 2 (u ) au − − au − − are also in the scope of this paper. Theorem 3 expands the basic solution (initial values dj (0) = ±0j)as
(j 1).2 h j 2h 1 − j h 1 uφ (u (j 2h) " h ) − − dj (u)= − − − − − : h (j 2h)! h=0 X µ ¶ − Returning to the question at the beginning of this introduction we can say that UC tools can be applied to delta recursions, i.e., recursions that are solved by polynomial sequences, and lead to operator identities where some (known) delta operator is expressed as a delta series in some other (unknown) delta operator. The coefficients of the delta series may be linear operators themselves, as long as they can be expanded as power series in ,andthefirst term in the delta series is invertible. D 2Theory
The Finite Operator Calculus (Rota, Kahaner, and Odlyzko [12, 1973]) contains almost all the Umbral Calculus we need for this paper, except for Theorem 5. We abbreviate this “classical” umbral calculus by UC. There are many generalizations (for example [3],[16], and[1]) and variants (e.g. Zeilberger [20, 2000]). The dynamic survey on umbral calculus, www.combinatorics.org/Surveys/ds3.pdf, by Di Bucchianico and Loeb, has more than 500 references. We cite from [12] that a N-Sheffer sequence (pj) has a generating function of the form
j u (q) pj (u) q =(q) b j 0 X≥ j where (q)= j 0 pj (0) q is a power series of order 0, ¯ (q) is a delta series, and the delta operator N ≥ 1 1 can be expanded in terms of the derivative as N = ¯− ( ),where¯− (q) is the compositional inverse of 1 P D D ¯, ¯− (¯ (q)) = q.IftheN - basic sequence (bj) is known, we can obtain ¯ (q) directly from
j ¯ (q)=ln bj (1) q : j 0 X≥ However in many applications none of these terms are explicitly given; only certain relationships among them are known.
2.1 The Basic Sequence
If we cannot guess the B-basic sequence (bj) of some delta operator B itmaybepossibletoconnect(bj) with a known N-basic sequence (nj), say. This is possible, because we need not expand delta operators in powers of ; a linear operator is a delta operator if it can be written as a delta series in terms of any other delta operator.D
3 Theorem 2 (Transfer Theorem) Let N be a delta operator with N-basic sequence (nj) .IfN =(B) j N0 for some delta series and linear operator B,thenB is also a delta operator, and the B∈ - basic sequence (bj) has for all j the expansion j N0 N0 ∈ ∈ j bj (u)= cjfnf (u) (1) f=0 X j f where cjf is the coefficient of B in (B) . For the generating function of the basic sequence (bj) holds
j f bj (u) q = nf (u)(q) : (2) j 0 f 0 X≥ X≥ The proof of this theorem is implicitly contained in [12]; because it is so short we reproduce it here. j u (q) 1 Proof. There exists a delta series ° (q) such that j 0 nj (u) q = b and N = °− ( ). Note that ≥ D B is a delta operator because B canbeexpandedasadeltaseriesinD,B = 1 (N)= 1 ° 1 ( ) . P − − − Hence D j ¡ ¢ j u ((q)) f j j j f bj (u) q = b = nf (u)(q) = q nj f (u) [q ](q) − : − j 0 f 0 j 0 f=0 X≥ X≥ X≥ X ³ ´ Denote the ring of shift invariant operators by Σ. Delta operators can be expanded as delta series Σ [[q]] if [q](q) is invertible; however this expansion is no longer unique and it often generates interesting identities.∈ With the help of Lagrange-Bürmann inversion the expansion (1) generalizes to the following theorem.
Theorem 3 (Generalized Transfer Formula) Let N be a delta operator with N - basic sequence (nj) . j N0 j f ∈ If ∂jf is the coefficient of B in (B) , where is the delta series N =(B) Σ [[B]],thenB is also a ∈ delta operator. The B - basic sequence (bj) has for positive j the expansion j N0 ∈ j 1 b (u)=u ∂ n (u) : (3) j jf u f f=1 X See [8] for the proof. The factor u in this formula will not cancel in general, because ∂jf is an operator. For example, we can write the forward difference operator ∆ : m (u) m (u +1) m (u) in terms of the backwards difference , 7→ − ∇ 1 1 1 ∆ = B 1 B− = B Σ [[ ]] : − ∇∈ ∇ u The ∆-basic sequence can be guessed as j¡ j 0.By3¢ ≥ j ¡¡ ¢¢ f 1 u 1 u + j j 1+u u [ j] B1 = u = − ∇ ∇ u f u + j j j f=1 µ ¶ µ ¶ µ ¶ X ³ ¡ ¢ ´ is the j-th basic polynomial which we used already in the staircase example in the Introduction. This j u°(q) example is a special case of a more general result. If N = B B and j 0 nj (u) q = b then for all j,0 ≥ j P j j f 1 j 1 u b (u)= u [B ](B B) − n (u)=uB n (u)= n (u + ®j) : (4) j u f u j u + ®j j f=1 X ³ ´ ( ) Remark 4 If N( ) := B− N then nj (u)=unj (u + j) . (u + j) is the N( ) - basic polynomial (see j 1 (4)). The Abel polynomials u (u + j) − .j! are the D( ) - basic polynomials. The process of moving from N to N( ) generates many identities; we call this process “Abelization”. Note that (pj (u + j))j 0 is a [ ] ≥ N( )-Sheffer sequence iff (pj) is a N -Sheffer sequence. We write pj (u) for pj ( j + u). In general, if ( ) nj (u)=unj (u + j) . (u + j) is the N( )-basic polynomial, then
( ) u j qj (u):=n (u j)= − nj (u) j − u
4 ( )[ ] is a Sheffer sequence for the delta operator B N( ) = N.Notethatpj ( j)=±0j.Insymbols, nj − is a N -Sheffer sequence. ³ ´
( ) [ ] u¯(q) The notation N( );nj and pj will be used later in the paper. If (nj) has the generating function b
( ) u¯ (q) we will denote the generating function of nj by b ( ) . j Unfortunately, Theorem 3 does not tell³ us´ the generating function j 0 bj (u) q as Theorem 2 does. However, in many applications the only operator occurring among the coeffi≥ cients of are shift operators, and it may be possible to express B1 as ° (B), say, using only scalar coeffiPcients. In that case =1 1.° (B), and ∇ − f j f 1+u 1 u bj (u) q = − 1 = ° (q) (5) f − ° (q) j 0 f 0 µ ¶µ ¶ X≥ X≥ 1 1 2 by Theorem 2. For example, in the introductory staircase example we found 1=B− + B B +2B, hence 1 1 B− = 1 2B + √1 4B . 2 − − ¡ ¢ 2.2 The Particular Functional
We are looking for the particular Sheffer sequence (pj) which satisfies some given “initial” conditions. We assume that these conditions can be expressed as values of a particular functional I,sothat I pj is known for all j. There may be many such functionals. With every choice of particular functionalh I| comesi a unique shift invariant (particular) operator
j ¹ := I bj B : (6) I h | i j 0 X≥ If I 1 =0then ¹ (I) is invertible. It is shown in [11] that the operator ¹ is invariant under the choice of h | i6 I the delta operator B and its basic sequence (bj). For example, if I =Eval it is convenient to choose the pair and (uj.j!) to show that D j j ¹ = = b D = B ; (7) Eval j! D j 0 X≥ h the shift operator by . It is also easy to show that a functional defined as I pj = Eval B pj for all h h | i | j 0 and given and h, has the associated operator ¹I = B B . ≥Now we are ready to state the Functional Expansion Theorem [9]. ®
Theorem 5 Suppose (pj) is a B -Sheffer sequence and I afunctionalsuchthat I 1 =0.The j N0 ∈ h | i6 polynomials pj (u) can be expanded in terms of the B - basic sequence (bj) as j N0 ∈ j 1 pj (u)= I ph ¹I− bj h (u) : h | i − h=0 X They have the generating function
h ∞ j h∞=0 I ph q ∞ j pj (u) q = h | i bj (u) q : ∞ I b qg j=0 g=0 g j=0 X P h | i X The Binomial Theorem for Sheffer SequencesP[12] j pj (u + )= ph ( ) bj h (u) (8) − h=0 X is a special case of the Functional Expansion Theorem if we choose I =Eval , j j 1 pj (u + )=B pj (u)=B Eval ph ¹Eval− bj h (u)=B ph ( ) B− bj h (u) : h | i − − h=0 h=0 X X
5 Example 6 (Tiling a rectangle with 2 2 squares) Denote by mj (h) the number of tilings of a 4 × × j + h rectangle with j squares of size 2 2.Obviouslym0 (h)=1for all h 0,andmj (0) = 1 for even h 0,and0 else. × ≥ ≥ Tiling a 4 7 rectangle with three 2 2 squares × × · · · · · · · · · · · · · · · · Heubach[4]foundtherecursion
h
mh (u +1)=mh (u)+3mh 1 (u)+mh 2 (u)+2 mh o (u) : − − − o=2 X If B is the delta operator mapping mj into mj 1, the above recursion is equivalent to the operator identity − 1+2B B3 B1 =1+3B + B2 +2 Bo = − : 1 B o 2 X≥ − By (5) the B-basic sequence (bj) has the generating function
3 u j 1+2q q bj (u) q = − : 1 q j 0 µ ¶ X≥ − We choose I =Eval0 as the particular functional and obtain from the Functional Expansion Theorem 5
h 3 u 3 u ∞ j h∞=0 I mh q 1+2q q 1 1+2q q mj (u) q = h | i − = − : ∞ I b qg 1 q 1 q2 1 q j=0 g=0 g X P h | i µ − ¶ − µ − ¶ Merlini, Sprugnoli, and Verri [6,P 2000] developed an algorithm transforming tiling problems like this into a regular grammar, and “automatically” produce the generating function from the tiling problem with the help of a computer algebra package.
3 Recursions in one variable
Suppose ¾0;¾1;:::;¾` 1 are given initial values for the recursion − j ¾j = ®g¾j g − g=1 X for j `,where®1;®2;::: is a sequence of given factors. We have to consider two case, ® =0and ® =0. ≥ 6 Suppose ®1 =0. To solve this type of problem with UC we define a polynomial sequence (pj (u)) such that 6 pj (u) pj (u 1) = ®gpj g (u) − − − g 1 X≥ and ¾j g 1 ®g¾j g for 0 j<` pj ( 1) = − − ≤ − ≥0 for j `: ½ P ≥ It is easy to check that ¾j = pj (0) for all j N0. If we denote the linear operator which maps pj (u) into g∈ pj 1 (u) by B,then = (B)= g∞=1 gB is a delta series in B (remember that ®1 =0), and (pj) is a − ∇ ∇ 6 B-Sheffer sequence. Let (bj) be the B-basic sequence. We apply the binomial formula (8) and find P j ` 1 j j j 1 − h ¾jq = pj (0) q = q ph ( 1) bj h (1) = ¾h ®g¾h g q (9) − − 1 (q) − − j 0 j 0 j 0 h=0 h=0 g 1 X≥ X≥ X≥ X −∇ X X≥ 6 (see (2) for bj h (1)). − If ®1 =0; we define (pj (u)) such that
j pj (u) pj (u 1) = pj 1 (u 1) + ®gpj g (u) : − − − − − g=2 X 1 g For the associated delta operator N holds = B− N + g 2 ®gN ,thus ∇ ≥ P 1 g 1 (N)=B− (N)= 1+ ®gN . (1 + N) : −∇ g 2 X≥ Let j ¿ j g=2 ®g¿ j g for 0 j<` − pj ( 1) = j 1 ` − ` 1 ≤ (10) − ( 1) − − ¿ ` 1 g−=2 ®g¿ ` 1 g for j `: ( − P− − − − ≥ j f ³ P ´ where ¿ j := f=0 ( 1) ¾j f. Note that ¿ j + ¿ j 1 = ¾j. It follows that pj (0) = ¾j for 0 j<` because of − − − ≤ P j j 1 − pj (0) = pj ( 1) + pj 1 ( 1) + ®gpj g (0) = ¿ j ®g¿ j g + ¿ j 1 ®g¿ j 1 g − − − − − − − − − − g 2 g=2 g=2 X≥ X X = ¾j ®g¾j g + ®gpj g (0) : − − − g 2 g 2 X≥ X≥
For j ` holds pj ( 1) = pj ( 1), and therefore by induction ≥ − − − j j pj (0) = ®gpj g (u)= ®g¾j g = ¾j: − − g=2 g=2 X X
As before, the numbers ¾j are expanded with the help of the binomial formula for Sheffer sequences,
j j j ∞ h ¾jq = ph ( 1) nj h (1) q = ph ( 1) q . (1 (q)) : − − − −∇ j 0 j 0 h=0 h=0 X≥ X≥ X X
Substitution for ph ( 1) shows that − ` 1 h h h−=0 ¾h g=2 ®g¾h g q j − − ¾jq = g : (11) ³1+ ®gq ´ j 0 P Pg 2 X≥ ≥ P 3.1 Example: Area under elevated Schröder paths The three different step vectors on a generalized Schröder path are the horizontal step (t; 0) of length t 2, and the diagonal steps (1; 1) and (1; 1).The(t; 0)-steps get the (multiplicative) weight φ. The paths start≥ at the origin, end on the u - axis, and− never go below the u - axis. Elevated Schröder paths stay strictly above the u - axis, with the exception of the first and last step. Sulanke [14],[15] found the following recursion for the total weighted area ¾j under all elevated Schröder paths from (0; 0) to (j +2; 0).
2j if j is even and 0 j t 1 0 if j is odd and 0 j j 1 t t 1 For 0 j t 1 h t t j h=0− (¾h 4¾h 2) q +(¾t 4¾t 2 2φ¾0) q 1+(φ (t +1) 2φ) q − − ¾jq = − −g − = 2 − 1+ ®gq (1 φqt) 4q2 j 0 P g 2 X≥ ≥ − − j.t P j 1 gt j gt g j g (t 1) + 1 2 − − (j +1) 1+( 1) − φ ¾j = − − − : g j g (t³ 1) + 1 ´ g=0 X µ ¶ − − For “ordinary” elevated Schröder paths (t =2) the sum simplifies, 1 2j+1 1 2j+1 ¾2j = 1+ (1 + φ) + 1 (1 + φ) : 2 2 − ³ p ´ ³ p ´ This (generating) function is familiar: The numbers ½j of non-selfintersecting (1; 0); ( 1; 0); (0; 1) -paths { − } j+1 1 starting at (0; 0) and having length j (Stanley [13, Example 4.1.2]) are enumerated by 2 1+ (1 + φ) + j+1 1 1 (1 + φ) , when the weight φ is given to the runs (maximal subwalks) of ³ - steps.p They´ solve 2 − → the³ recursion ½j´=2½j 1 + φ½j 2. p − − ½h and ¾h when φ =1 h =0123456 ½h = 1 3 7 17 41 99 239 ¾h =1070410239 4 Applying the Functional Expansion Theorem None of the above examples needed the power of the Functional Expansion Theorem 5, because the binomial formula suffices when explicit initial values are given. In [9] we investigated Sheffer sequences (pj) satis- fying initial conditions of a different form, starting with pj(u)=qj(u) for all j =0; 1;:::;` 1,where 1 − q0 (u) ;:::q` 1 (u) are given (initial) polynomials, continuing with conditions like pj(v + cj)av =0or − 0 j 1 f f+1 p (cj)= − ( 1) c p (cj) for all j `. Another common feature of the above examples is j f=0 j f 1 R that the existence of− polynomial− solutions− to the recursion≥ and initial values is rather obvious. The following example diffPersonbothcounts. 8 4.1 Example: Privileged Access Suppose a lattice walker can choose from the infinite number of steps in the step set S := (1; 1) ; (1; 1) f { − } ∪ (f; 0) f N1 . The horizontal steps (f; 0) are weighted with . Suchweightsareusefulifwewantto study{ the| ∈ corresponding} random walk with geometric probabilities for the horizontal steps. We also require that the -stepkeepsthewalkinthefirst quadrant; it cannot be chosen when the path has reached the u - axis.& However, after leaving the origin the walker has special access to the boundary v = 1 by selecting − the fitting access step vector. In this example we let (0; k) k N1 be the set of special access steps, andwegivethemtheweight®. The walker can vertically{ “jump− | down”∈ } to the boundary from any position. It is implied that the path must leave the boundary in the next step. Sample path to (13; 1) with access steps (0; k) k,0 k { − | } 2 ◦↓ ◦& . 1 % . % ◦ 99K ◦ ◦ ◦ . 0 % . 99K % ◦→ ◦↓ ◦ ◦ ◦ 1 % % − ¥ ◦ ¤¤¤¤◦ ¤¤¤ ¤ ¤ ¤ ¤ 0 1 23456789101112 j → The (weighted) numbers > (j; k) of such paths from the origin to (j; k) have no polynomial extension such that j (k)=> (j; k) on the support of > for any polynomial sequence ( j (u)). k > (j; k) 3 1 2 13 1 12 52 + ® +2+® 0 122 + ® +1+® 4 3 +3® 2 + 3+3® + ®2 +2® + ®2 1 ® (1 + ) ® 2 2 +( +1)(2+®) ® 4 3 +(5+3®) 2 + 6+4® + ®2 +3+3® + ®2 − ¡ ¢ 01 2 3 j ¡ ¢ ¡ ¡ ¢ → ¢ 1 1 However the transformation J := 10− maps the given problem into an equivalent walk where such an extension exists. The steps , , (f; 0) are mapped into (2; 1) ; (0; 1) and (f; f) ;f N1, the latter taken with weight f. The transformed& special% ¡ access¢ steps (0; k) are the horizontal steps (k;∈ 0), k,0,withweight®. Let A (j; k) be the weighted number of transformed− paths to (j; k), A (j; k)=> (k; k j). We denote − the polynomial extension of A (j; k) by aj (k) aj (k)=A (j; k) for all k j 1: ≥ − The polynomial extension aj (k) (special access values in boxes) k 13 52 + ® +2+® 4 3 +3® 2 + 3+3® + ®2 +2® + ®2 2 3 2 2 2 2 12 2+ ® +1+® 0 +2® +¡ 0+2® + ¢® +2® + ® 1 1 0 2 + ® +0+® φ3+αφ2+φ ¡1 + α + α2 ¢+2α + α2 − − 0 10 φ2+αφ 1 + α 0φ3+0φ2+φ 0 + 0α + α2 +2α + α2 − − ¡ ¢ 2 . 1 1 φ φ +αφ 2 + α . ¡ ¢ − − − − 01 2 3 j (the bold face values are extensions outside the support of >) → Thepolynomialsaresolutionstothesystemofdifference equations g aj (u) aj (u 1) = aj 2 (u 1) + aj g (u g) − − − − − − g 1 X≥ 9 and must satisfy the initial conditions a0 (u)=1, a1 (0) = 0, and the transformed access condition aj (j 1) = ®aj f (j 1) (12) − − − f 1 X≥ for all j,1.DenotebyB the linear operator that maps aj (u) into aj 1 (u). The recursion for (aj) shows that − 1 1 2 g g g 1 2 B− B = B− B + B− B = B− B + (13) ∇ 1 B 1 B g 1 − X≥ − The basic sequence (bj (u)) can be expanded according to the transform formula (3) j 1 f j 1 2 B− B 1 f + u 1 bj (u)= u [B ] B− B + − 1+B 1 B u f f=1 µ − ¶ µ ¶ Xj f f j 2g 1 j 2g u u + f j + g 1 = − − − − − g j f g f f 1 f=1 g=0 X X µ ¶µ − − ¶ µ − ¶ for j,1. The operator identity (13) “simplifies”tothefinite recursion 1 1 1 2 1 = B− B 2 B− + B− B 1 B− B : ∇ − − After solving for B1 we obtain from (5) ¡ ¢ ¡ ¢ u j 1 2 1 2 2 2 bj (u) q = q +2 q +1 + (q +2 q +1) q (1 + q ) 2 4 − j 0 à r ! ≥ X ¡ ¢ u 1 2 q (1 2 q)2 4q2 ( q 1)2 b(1) (u) qj = − − − − − : j q 2q2 (1 q) j 0 X≥ − [1] 1 It is convenient to switch to the polynomials ak (u)=ak (k + u) with delta operator B(1) = B− B and basic (1) [1] polynomials bj = ubj (u + j) . (u + j) (see Remark 4). In terms of aj the privilege access condition [1] [1] (12) becomes aj ( 1) = f 1 ®aj f (f 1).Wedefine the particular functional³ ´ I by linear extension of − ≥ − − P [1] [1] f [1] I aj = Eval 1 aj ® Evalf 1 B(1)aj (14) | − | − − | f 1 D E D E X≥ D E and obtain the particular operator 1 1 f 1 f 1 ®B B(1) ¹ = B− ®B − B = B− 1 : I − (1) − 1 B1B f 1 µ (1) ¶ X≥ − [1] [1] [1] [1] Note that I a =1, I a = a ( 1) ® = ®,and I aj =0for all j,1.BytheFunctional | 0 | 1 1 − − − | ExpansionD TheoremE 5, D E D E [1] 1 (1) (1) aj (u)=aj (u j)=¹ (I)− bj (u j) ®bj 1 (u j) − − − − − ®B1B ³ ´ (1) b(1) u j ®b(1) u j = 1+ 1 j ( +1) j 1 ( +1) 1 (1 + ®) B B(1) − − − − µ − ¶ ³ ´ bj (u +1) ®bj 1 (u) =(u j +1) − − u +1 − (1 + ®) u j µ ¶ i 1 u + i j +1 ® (u + i j) +® (1 + ®) − − bj i (u +1) − bj i (u) u +1 − − (1 + ®) u − i=1 X µ ¶ 10 5 Bivariate Umbral Calculus For the remaining part of the paper we will focus on the bivariate case; generalizations of the following theorems to higher dimensions are straight forward. More on multivariate umbral calculus can be found in [10, 1979] and [17], [18]. A bivariate Sheffer sequence (mjk (u; v)) has a generating function of the jk N0 form ∈ j k u¯ (pq)+v¯ (pq) mjk (u; v) p q = ½ (p; q) b 1 2 jk 0 X≥ where ½ (p; q) K [[p; q]] has order 0,i.e.,½ (0; 0) =0,and¯1 (p; q) ;¯2 (p; q) is a pair of delta series in K [[p; q]], ∈ 6 which means that ¯1 (p; q) .p and ¯2 (p; q) .q are both power series of order 0 in K [[p; q]].If½ (p; q)=1the resulting Sheffer sequence is a bivariate basic sequence. For every pair of delta series ¯1 (p; q) ;¯2 (p; q) there exists a compositional inverse °1 (p; q) ;°2 (p; q), also a delta pair, such that ¯1 (°1 (p; q) ;°2 (p; q)) = p and ¯2 (°1 (p; q) ;°2 (p; q)) = q: Let 1 := /./u and 2 := /./v. Using the above notation it can be shown that D D °1 ( 1; 2) mjk (u; v)=mj 1k (u; v) and °2 ( 1; 2) mjk (u; v)=mjk 1 (u; v) : D D − D D − We call ° ( 1; 2) ;° ( 1; 2) a pair of delta operators,and(mjk)the associated Sheffer sequence. Of 1 D D 2 D D course, all operators in K [[ 1; 2]] commute. Again we will state two transfer theorems. D D Theorem 7 (Bivariate Transfer Theorem I) Let N1;N2 be a delta operator with N1;N2 - basic sequence (njk (u; v)) .IfN1 = (B1;B2) and N2 = (B1;B2) for some pair ; of delta series and linear jk N0 1 2 1 2 ∈ operator B1;B2,thenB1;B2 is also a pair of delta operators, and the B1;B2-basic sequence (bjk) jk N0 ∈ has for all j; k N0 the expansion ∈ j k bjk (u; v)= cjkfg nfg (u; v) (15) f=0 g=0 X X j k f g where cjkfg is the coefficient of B1 B2 in 1 (B1;B2) 2 (B1;B2) . For the generating function of the basic sequence (bjk) holds j k f g bjk (u; v) p q = nfg (u; v) 1 (p; q) 2 (p; q) : (16) jk 0 fg 0 X≥ X≥ The proof is analogous to the proof of Theorem 2. The Jacobian G (½1;½2) of a pair ½1 (p; q) ;½2 (p; q) of power series is defined as /M1./p /M2./p G (½1;½2)(p; q):= : /M1./q /M2./q ¯ ¯ ¯ ¯ ¯ ¯ Theorem 8 (Bivariate Transfer Theorem II) Let¯ N1;N2 be a delta¯ operator with N1;N2 -basicse- quence (njk (u; v)) .IfB1 = ½ (N1;N2) N1 and B2 = ½ (N1;N2) N2 for some pair ½ ;½ of power series of jk N0 1 2 1 2 ∈ order 0,thenB1;B2 is also a pair of delta operators, and the B1;B2-basic sequence (bjk) has for all j N0 ∈ j; k N0 the expansion ∈ j 1 k 1 bjk (u; v)=½1 (N1;N2)− − ½2 (N1;N2)− − G (N1½1;N2½2)(N1;N2) njk (u; v) (17) 1 1 Proof. Suppose Nf = ¾− ( 1; 2) and Bf = ¯− ( 1; 2) for some pairs of delta series ¾1;¾2 and f D D f D D ¯1;¯2. For notational simplicity we define 1 φf (p; q):=¾f− (¯1 (p; q) ;¯2 (p; q)) : 11 Hence j k u¯ (pq)+v¯ (pq) j k bjk (u; v) p q = b 1 2 = njk (u; v) φ1 (p; q) φ2 (p; q) jk 0 jk 0 X≥ X≥ f g f g j k = p q p q φ1 (p; q) φ2 (p; q) njk (u; v) f 0g 0 0 j f0 k g ≥X≥ ≤ ≤X≤ ≤ £ ¤ f g f g f j g k j k = p q p q φ1 (p; q) − φ2 (p; q) − N1 N2 nfg (u; v) : f 0g 0 j 0k 0 ≥X≥ ≥X≥ £ ¤ By Lagrange - Bürmann inversion f 1 g 1 1 − − 1 − − f g f j g k j k φ1− (p; q) φ2− (p; q) 1 1 p q φ (p; q) − φ (p; q) − =[p q ] G φ ;φ (p; q) : 1 2 p q 1− 2− µ ¶ µ ¶ Noting£ that¤ ¡ ¢ 1 1 φ1− (p; q)=p½1 (p; q) and φ2− (p; q)=q½2 (p; q) finishes the proof. Example 9 Suppose we have the pair of operator identities 1 1 = B1− B1 ∇ 1 1 ∆2 = B1− B2 + φB1B1− B2 1 1 We apply the above theorem with B1 = ∆1 and B2 = ∆2. B1− + φB1B1− = ∆2 (1 + ∆1) . (1 + φ∆1). Hence ½ (p; q)=1, ½ (p; q)=(1+p) . (1 + φp),and 1 2 ¡ ¢ j 1 k 1 u v b (u; v)=½ (∆ ; ∆ )− − ½ (∆ ; ∆ )− − G (∆ ½ ; ∆ ½ )(∆ ; ∆ ) jk 1 1 2 2 1 2 1 1 2 2 1 2 j k µ ¶µ ¶ k 1 /½2 1+∆1 − − 1 ∆2 u v = /∆1 1+∆1 1+φ∆1 0 j k ¯ 1+∆1 ¯ µ ¶ ¯ ¯ µ ¶µ ¶ ¯ ¯ k k ¯u v ∞¯ k f u k v = B− (1 + φ∆1) ¯ = ¯ φ − : 1 j k f j f k f=0 µ ¶µ ¶ X µ ¶ µ − ¶µ ¶ For comparison we apply the first Transfer Theorem to the same problem, with 1 (B1;B2)=B1 and B2 (1 + φB1) B2 (1 + φB1) B2 (1 + φB1) 2 (B1;B2)= 1 = = : B1 1+∆1 1+B1 Hence the B - basic sequence has the generating function u v q (1 + φp) k 1+φp v b (u; v) pjqk = pj =(1+p)u 1+q : jk j k 1+p 1+p jk 0 jk 0 µ ¶µ ¶ µ ¶ µ ¶ X≥ X≥ If we apply the bivariate transfer formula 16 we get the expression j k B (1 + φB ) g u v b (u; v)= [BjBk] Bf 2 1 jk 1 2 1 1+B f g f=0 g=0 à 1 ! X X µ ¶ µ ¶µ ¶ j k h v j f k (φ 1) B1 u = B1 − − k h 1+B1 f µ ¶ f=0 à h=0 µ ¶µ ¶ ! µ ¶ X £ ¤ X v ∞ k h u h = (φ 1) − : k h − j h h=0 µ ¶ X µ ¶ µ − ¶ The two solutions we found for bjk (u; v) reflect the well-known hypergeometric function identity u j; k u k j; k 2C1 − − ;1 φ = − 2C1 − − ; φ : j u − j 1 k j u µ ¶ µ − ¶ µ ¶ µ − − − ¶ 12 The Functional Expansion Theorem 5 generalizes as expected, with ¹I defined as j k ¹ := I bjk B B (18) I h | i 1 2 jk 0 X≥ for any B1;B2 - basic sequence (bjk). Theorem 10 Suppose (pjk) is a B1B2 -Sheffer sequence and I a linear functional on [[p; q]] such jk N0 K ∈ that I 1 =0. The polynomials pjk (u; v) can be expanded in terms of the B1B2 - basic sequence (bjk) as h | i6 j k 1 pjk (u)= I pgh ¹I− bj gk h (u; v) : h | i − − g=0 h=0 X X They have the generating function g h j k gh 0 I pgh p q j k ≥ h | i pjk (u; v) p q = g h bjk (u; v) p q : I bgh p q jk 0 Pgh 0 jk 0 X≥ ≥ h | i X≥ P 5.1 Bi-indexed Umbral Calculus Three variable recursions like mjk (»)=mjk (» 1) + mj 1k (»)+mjk 1 (») are not part of the above bivariate theory. Very basic combinatorial objects,− like multinomial− coeffi−cients, follow recursions of this type. An obvious approach to a polynomial theory for this type of recursions equates the variable u and v in a bivariate polynomial sequence to give (mjk (»)) . Abusing the notation we we write mjk (»)= jk N0 ∈ mjk (»; »), and we introduce the diagonalization operator Ξ : K [u; v] K [»] → Ξm (u; v)=m (») : u¯ (pq)+v¯ (pq) Suppose (pjk) is a Sheffer sequence with generating function ½ (p; q) b 1 2 .ByK [[p; q]] - linear extension we define j k j k »(¯ (pq)+¯ (pq)) Ξ pjk (u; v) p q := Ξpjk (u; v) p q = ½ (p; q) b 1 2 : jk 0 jk 0 X≥ X≥ The following lemma is easy to verify. Lemma 11 For all h N0 holds ∈ h h Ξ ( 1 + 2) = Ξ D D D A bi-indexed Sheffer sequence (pjk (»)) is a double sequence of univariate polynomials such that deg pjk (»)=j + k for all j; k N0, p00 (») =0,and ∈ 6 ½ (p; q) b»¯(pq) where ½ (p; q) is of order 0,and¯ (p; q)=¯1 (p; q)+¯2 (p; q) for some pair of bivariate delta series. Obviously this pair is not unique. If (pjk (u; v)) is a bivariate N1;N2 - Sheffer sequence then (Ξpjk (u; v)) is a bi-indexed Sheffer sequence, and Ξ (N1 + N2) pjk (u; v)=Ξ (pj 1k (u; v)+pjk 1 (u; v)) : − − We say that (pjk (»)) is a N -Sheffer sequence iff Npjk (»)=pj 1k (»)+pjk 1 (»),andcallN adelta − − operator associated to (pjk (»)). Trivial examples of bi-indexed Sheffer polynomials are products of two univariate Sheffer polynomials. The following example shows another important class. 13 1 Example 12 Let ° be a univariate delta series, and let (nj) be the ordinary univariate N = ° ( ) - j N0 − j+k ∈ D basic sequence. Define mjk (»):= nj+k (») for all j; k N0.Frommkj (0) = ±j0±k0 and j ∈ Nmjk (»)=mj 1k (»)+mjk 1 (») follows that (mjk (»)) is a bi-indexed M -basicsequencewithM = N. − − ¡ ¢ What are M1;M2 such that M Ξ = Ξ (M1 + M2) ? How does the associated basic sequence (mjk (u; v)) look like?The generating function h h m (») pjqk = n (») pjqh j = b»°(p+q): jk h j − jk N0 h N0 j=0 µ ¶ X∈ X∈ X motivates us to define ¯ (p; q)=° (p + q) ° (q) and ¯ (p; q)=° (q) 1 − 2 because ¯1 (p; q) ;¯2 (p; q) is a delta pair, and ¯1 (p; q)+¯2 (p; q)=° (q).NotethatM Ξ = Ξ (M1 + M2) if 1 1 1 M1 = °− ( 1 + 2) °− ( 2) and M2 = °− ( 2) D D − D D 1 1 We can check this by applying Lemma 11: Ξ (M1 + M2)=Ξ°− ( 1 + 2)=°− ( ).TheM1;M2 -basic D D D polynomials diagonalize to mjk (»)=Ξmjk (u; v),and j k u°(p+q)+(v u)°(q) mjk (u; v) p q = b − (19) jk 0 X≥ k j + k f mjk (u; v)= − mj+k f (u) mf (v u) : j − f=0 − X µ ¶ Properties of the bivariate behind the bi-indexed Sheffer sequences can be used to solve certain recursions. For example, if (njk (»)) is a N - basic sequence, then ckj (»):=(j +1)nj+1k (») .» and dkj (»):= (k +1)njk+1 (») .» are N - Sheffer polynomials, and so is the shifted linear combination njk (» ´) pjk (»):=njk (» ´) cj 1k (» ´) bdjk 1 (» ´)=(» ´ j bk) − (20) − − − − − − − − − − » ´ − for some given constants ; b and ´.ThisN - Sheffer sequence satisfies the initial condition pjk (´ + j + bk)= ±0j±0k.If(ojk (»)) is any N - Sheffer sequence, and NΞ = Ξ (N1 + N2),then(ojk (» + j + bk)) is a b b N( b) - Sheffer sequence, where N( b)Ξ = Ξ B1− B2− N1 + B1− B2− N2 . The bi-indexed polynomial se- njk(»+ j+bk) quence » »+ j+bk is the N( b) ¡- basic sequence, the Abelization¢ of (njk (»)).Moreabout jk N0 multi-indexed³ UC can be´ found∈ in [7, 1980] and Watanabe [19, 1986]. 5.2 Example: A Slow Walk Suppose the lattice walker can choose at every tick of the clock one of the steps ; ; ; © ,where©=(0; 0) isthe“pausestep”.Giveweightφ to the - step vector. Let A (j; k; h) be{ the→ number↑ % } of weighted paths % from (0; 0) to (j; k) in h steps (seconds) under the restriction A (f; g; i)=0if (time) i 6 f + bg where and b are positive integers, and (f; g) =(0; 0). In other words, the lattice point (f; g) can not be reached in ashorttimei; the walk is slow. 6 A (k; j; h) for =2and b =1; φ =1 4 6 4 6 4 6 8 4 6 8 04 06 08 010 · · · · · · · · · · 3 5 3 5 3 5 7 03 05 07 13 05 07 09 · · · · · · · 2 4 2 4 02 04 06 12 04 06 32 04 06 08 · · · · 1 3 01 03 11 03 05 21 03 05 31 43 05 07 · · 1↑% 2 10 02 10 02 04 10 12 04 10 22 04 06 −→h =0· h =1 h =2 h =3 h =4 The subscripts show the values of f + bg. The levels h = f + bg are in bold. 14 The solution to the difference equation A (k; j; h) A (k; j; h 1) = A (k 1;j; h 1) + A (k; j 1; h 1) + φA (k 1;j 1; h 1) − − − − − − − − − can be extended to a polynomial sequence (ajk (»)) because A (0; 0; h)=1for all h 0.Ofcourse, A (j; k; h)=0if any of the three parameters is negative. In order to calculate the nonzero≥ values of A (j; k; h) the only additional initial condition is A (j; k; j + bk)=±j0±k0. The recursion and operator equation for the polynomial extension are akj (» +1) = akj (»)+ak 1j (»)+akj 1 (»)+φak 1j 1 (») − − − − ∆Ξ = Ξ (B1 + B2 + φB1B2) where Bakj (»)=akj (» 1) + ak 1j (» 1) and BΞ = Ξ (B1 + B2). In addition to a00 (»)=1the − − − solution must have the initial values akj ( k + bj)=±(kj)(00). It follows from Lemma 11 that ∆Ξ = 1 1 1 1 Ξ B B 1 , hence it is sufficient to determine B such that B1 + B2 + φB1B2 = B B 1,or 1 2 − 1 2 − ¡ ¢ 1 1 1 B− (B1 + B2 + φB1B2)=B B− = 1 + ∆2: 1 2 − 1 ∇ 1 1 1 We decide to split this sum of operators into 1 = B1− B1 and ∆2 = B1− B2 + φB1B1− B2,becausewe ∇ v j k u k f found already the B1;B2 - basic sequence (bjk (u; v)) in Example 9, bjk (u; v)= k f=0 f j− f φ . The numbers R (j; k; h) of unrestricted paths to (j; k) in h steps have initial values R (j; k;0)=0 except− ¡ ¢ P ¡ ¢¡ ¢ for R (0; 0; 0) = 1. 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