Definite Sums as Solutions of Linear Recurrences With Polynomial Coefficients
aa r X i v : . [ c s . S C ] A p r Definite Sums as Solutions of Linear RecurrencesWith Polynomial Coefficients
Marko PetkovˇsekFaculty of Mathematics and Physics, University of LjubljanaInstitute of Mathematics, Physics and Mechanics, [email protected]
Abstract
We present an algorithm which, given a linear recurrence operator L with polynomial coefficients, m ∈ N \ { } , a , a , . . . , a m ∈ N \ { } and b , b , . . . , b m ∈ K , returns a linear recurrence operator L ′ with rationalcoefficients such that for every sequence h , L ∞ X k =0 m Y i =1 a i n + b i k ! h k ! = 0if and only if L ′ h = 0. Keywords: definite hypergeometric sums; (formal) polynomial series;binomial-coefficient bases; solutions of linear recurrencesMSC (2010) 68W30; 33F10
Holonomic sequences are, by definition, given by a homogeneous linear recur-rence with polynomial coefficients (and suitable initial conditions). Often onewishes to find an explicit representation of a holonomic sequence, so algorithmshave been devised to find solutions of such recurrences within some class of ex-plicitly representable sequences (such as, e.g., polynomial [1], rational [2, 3], hy-pergeometric [7], d’Alembertian [4], or Liouvillian [6] sequences). These classesdo not exhaust all explicitly representable holonomic sequences. For example,every definite hypergeometric sum on which Zeilberger’s
Creative Telescoping algorithm [8, 9] succeeds is a holonomic sequence, but many such sequencesare not Liouvillian. Therefore it makes sense to consider what one might callthe inverse Zeilberger’s problem : given a homogeneous linear recurrence withpolynomial coefficients, find its solutions representable as definite sums from acertain class. 1ere we make a (modest) first step in this direction by providing an algo-rithm which, given a linear recurrence operator L with polynomial coefficientsand a fixed product of binomial coefficients of the form F ( n, k ) = m Y i =1 (cid:18) a i n + b i k (cid:19) (1)with a i ∈ N \ { } , b i ∈ K , returns a linear recurrence operator L ′ with rationalcoefficients such that for any sequence y of the form y n = P ∞ k =0 F ( n, k ) h k , wehave Ly = 0 if and only if L ′ h = 0. This enables us to find all such solutions y where h belongs to a class of holonomic sequences with a known algorithm forconverting from recursive to explicit representation (such as those mentioned inthe preceding paragraph). We regard P ∞ k =0 F ( n, k ) h k as a formal series, butnote that it is terminating if at least one of the b i is an integer.Throughout the paper, N = { , , , . . . } denotes the set of nonnegative in-tegers, K a field of characteristic zero, K N the set of all sequences with termsfrom K , K [ x ] the K -algebra of univariate polynomials over K , and L K [ x ] the K -algebra of linear operators L : K [ x ] → K [ x ]. Definition 1
Let m ∈ N \ { } and j ∈ { , , . . . , m − } . • A sequence c ∈ K N is called the j -th m -section of a ∈ K N if c k = a mk + j for all k ∈ N . We say that c is obtained from a by multisection , anddenote it by s mj a . • A sequence c ∈ K N is called the interlacing of a (0) , a (1) , . . . , a ( m − ∈ K N if c k = a ( k mod m ) k div m for all k ∈ N . (cid:3) The power series method is a time-honored approach to solving differential equa-tions by reducing them to recurrences satisfied by the coefficient sequences oftheir power series solutions. In [5] it was shown how, by generalizing the notionof formal power series to formal polynomial series , one can use this method tofind solutions of other linear operator equations such as q -difference equations,and recurrence equations themselves, which interests us here. In this section wesummarize some relevant definitions, examples and results from [5]. Definition 2
A sequence of polynomials B = h P k ( x ) i ∞ k =0 from K [ x ] is a facto-rial basis of K [ x ] if for all k ∈ N : P1. deg P k = k , P2. P k | P k +1 . (cid:3) efinition 3 A factorial basis B of K [ x ] is compatible with an operator L ∈L K [ x ] if there are A, B ∈ N , and α k,i ∈ K for k ∈ N , − A ≤ i ≤ B , such that LP k = B X i = − A α k,i P k + i for all k ∈ N , (2)with P j = 0 when j <
0. To assert that (2) holds for specific
A, B ∈ N , we willsay that B is ( A, B ) -compatible with L . (cid:3) Proposition 1
A factorial basis B of K [ x ] is compatible with L ∈ L K [ x ] if andonly if there are A, B ∈ N such that C1. deg LP k ≤ k + B for all k ≥ , C2. P k − A | LP k for all k ≥ A .Proof: Necessity of these two conditions is obvious. For sufficiency, let LP k = deg LP k X j =0 λ j,k P j be the expansion of LP k w.r.t. B . By C1 , we can replace the upper summationbound by k + B . Rewriting the resulting equation as LP k − k + B X j = k − A λ j,k P j = k − A − X j =0 λ j,k P j , we see by C2 and P2 that P k − A divides the left-hand side, while the right-handside is of degree less than k − A = deg P k − A . Hence both sides vanish, and so LP k = k + B X j = k − A λ j,k P j = B X i = − A λ k + i,k P k + i = B X i = − A α k,i P k + i where α k,i := λ k + i,k . This proves compatibility of B with L . (cid:3) Example 1 [5, Ex. 1] Let P = (cid:28) x k (cid:29) ∞ k =0 be the power basis , and C = (cid:28)(cid:18) xk (cid:19)(cid:29) ∞ k =0 the binomial-coefficient basis of K [ x ], respectively. Clearly, both P and C are fac-torial bases. Further, let D , E , Q , X ∈ L K [ x ] be the differentiation , shift , q-shift ,3nd multiplication-by-the-independent-variable operators , respectively, acting on p ∈ K [ x ] by Dp ( x ) = p ′ ( x ) ,Ep ( x ) = p ( x + 1) ,Qp ( x ) = p ( qx ) ,Xp ( x ) = xp ( x )where q ∈ K ∗ is not a root of unity. Then: • P is (1,0)-compatible with D (take α k, − = k , α k, = 0), • C is (1,0)-compatible with E (take α k, − = α k, = 1), • P is (0,0)-compatible with Q (take α k, = q k ), • every factorial basis is (0,1)-compatible with X . (cid:3) Let B = h P k ( x ) i ∞ k =0 be a factorial basis, and let ℓ k : K [ x ] → K for k ∈ N be linear functionals such that ℓ k ( P m ) = δ k,m for all k, m ∈ N (i.e., ℓ k ( p ) is thecoefficient of P k in the expansion of p ∈ K [ x ] w.r.t. B ). Property P2 impliesthat ℓ k ( P j P m ) = 0 when k < max { j, m } . Therefore K [ x ] naturally embeds intothe algebra K [[ B ]] of formal polynomial series of the form y ( x ) = ∞ X k =0 c k P k ( x ) ( c k ∈ K ) , (3)with multiplication defined by ∞ X k =0 c k P k ( x ) ! ∞ X k =0 d k P k ( x ) ! = ∞ X k =0 e k P k ( x ) ,e k = X max { i,j }≤ k ≤ i + j c i d j ℓ k ( P i P j ) . Let B be compatible with L ∈ L K [ x ] . Using (2), extend L to K [[ B ]] by defining L ∞ X k =0 c k P k ( x ) := ∞ X k =0 c k LP k ( x ) = ∞ X k =0 c k B X i = − A α k,i P k + i ( x )= B X i = − A ∞ X k =0 α k,i c k P k + i ( x ) = B X i = − A ∞ X k = i α k − i,i c k − i P k ( x )= ∞ X k = − A min( k,B ) X i = − A α k − i,i c k − i P k ( x ) = ∞ X k =0 B X i = − A α k − i,i c k − i P k ( x )= ∞ X k =0 A X i = − B α k + i, − i c k + i ! P k ( x ) (4)with A, B, α k,i as in (2), using P k ( x ) = 0 for k < c k − i = 0 for i > k inthe next to last equality. By this definition we have:4 roposition 2 A formal polynomial series y ( x ) = P ∞ k =0 c k P k ( x ) ∈ K [[ B ]] sat-isfies Ly = 0 if and only if its coefficient sequence h c k i k ∈ Z satisfies the recurrence A X i = − B α k + i, − i c k + i = 0 (for all k ≥
0) (5) where c k = 0 when k < . Hence L naturally induces a recurrence operator R B L := A X i = − B α n + i, − i E in (6)where E n is the shift operator w.r.t. n ( E in c n = c n + i for all i ∈ Z ). Example 2
Using (6) we read off from Example 1 that R P D = ( n + 1) E n , R C E = E n + 1 , R P Q = q n , while x x n = x n +1 and x (cid:0) xn (cid:1) = ( n + 1) (cid:0) xn +1 (cid:1) + n (cid:0) xn (cid:1) imply by (2) and (6) that R P X = E − n , R C X = n ( E − n + 1) . (cid:3) Notation: • For any factorial basis B of K [ x ], denote by σ B the map K [[ B ]] → K Z assigning to y ( x ) = P ∞ k =0 c k P k ( x ) ∈ K [[ B ]] its coefficient sequence c = h c k i k ∈ N extended to h c k i k ∈ Z by taking c k = 0 whenever k <
0. We willomit the subscript B if it is clear from the context. • For any factorial basis B of K [ x ], denote by L B the set of all operators L ∈ L K [ x ] such that B is compatible with L . • Denote by E the K -algebra of recurrence operators of the form L ′ = P Ri = − S a i ( n ) E in with R, S ∈ N and a i : Z → K for − S ≤ i ≤ R , act-ing on the K -algebra of all two-way infinite sequences K Z . Proposition 3 [5, Prop. 1]
For any L ∈ L B and y ∈ K [[ B ]] , we have σ B ( Ly ) = ( R B L ) σ B y. heorem 1 [5, Prop. 2 & Thm. 1] L B is a K -algebra, and the transformation R B : L B → E , defined in (6) , is an isomorphism of K -algebras. By Theorem 1 and Examples 1 and 2, every linear recurrence operator L ∈ K [ x ] h E i is compatible with the basis C = (cid:10)(cid:0) xn (cid:1)(cid:11) ∞ n =0 , and to compute theassociated operator L ′ = R C L ∈ E , it suffices to apply the substitution E E n + 1 ,x n ( E − n + 1)to all terms of L . Every h ∈ ker L ′ then gives rise to a solution y n = P ∞ k =0 (cid:0) nk (cid:1) h k of Ly = 0. Example 3
In this example we list some operators L ∈ K [ n ] h E i , their associ-ated operators L ′ = R C L ∈ E , and some of the elements of their kernels.1. L = E − c where c ∈ K ∗ : Here L ′ = E n − ( c − y n = ∞ X k =0 (cid:18) nk (cid:19) ( c − k = c n is indeed a solution of Ly = 0.2. L = E − E + 1: Here L ′ = E n , and by (5), any h ∈ ker L ′ satisfies h n +2 = 0 for all n ≥
0, or equivalently, h n = 0 for all n ≥
2. Hence y n = ∞ X k =0 (cid:18) nk (cid:19) h k = h + h n is indeed a solution of Ly = 0.3. L = E − E −
1: Here L ′ = E n + E n −
1, and any h ∈ ker L ′ is of the form h n = ( − n ( C F n + C F n +1 ) where C , C ∈ K and F = h , , , , . . . i isthe sequence of Fibonacci numbers. Hence every y ∈ ker L is of the form y n = ∞ X k =0 (cid:18) nk (cid:19) ( − k ( C F k + C F k +1 ) . In particular, by setting y = F and y = F , we discover the identity F n = n X k =0 (cid:18) nk (cid:19) ( − k +1 F k . L = E − ( n +1): Here L ′ = E n − n − nE − n , and the general solution of L ′ h = h n +1 − nh n − nh n − = 0 is of the form h n = n !( C + C P nk =1 ( − k /k !)where C , C ∈ K . The equation at n = 0 implies h = 0, forcing C = C =: C . Hence every y ∈ ker L is of the form y n = C ∞ X k =0 (cid:18) nk (cid:19) k ! k X j =1 ( − j j ! . In particular, by setting y = 0! = 1, we discover the identity n ! = n X k =0 (cid:18) nk (cid:19) k ! k X j =1 ( − j j ! . or equivalently, n X k =0 k ! n − k X j =1 ( − j j ! = 1 . L = E − ( n + 6 n + 10) E + ( n + 2)(2 n + 5) E − ( n + 1)( n + 2): Unlike inthe preceding four cases, the equation Ly = 0 has no nonzero Liouvilliansolutions. Here L ′ = E n − ( n + 6 n + 7) E n − (2 n + 8 n + 7) E n − ( n + 1) ,and equation L ′ h = 0 has a hypergeometric solution h n = n ! . So y n = ∞ X k =0 (cid:18) nk (cid:19) k ! = n X k =0 (cid:18) nk (cid:19) k ! is a non-Liouvillian, definite-sum solution of equation Ly = 0. (cid:3) To be able to use formal polynomial series to find other definite-sum solutionsof linear recurrence equations, not just those of the form P ∞ k =0 (cid:0) nk (cid:1) h k , we needto construct more factorial bases, compatible with the shift operator E . Definition 4
For a ∈ N \ { } , b ∈ K , and for all k ∈ N , let P ( a,b ) k ( x ) := (cid:0) ax + bk (cid:1) .We denote the polynomial basis D P ( a,b ) k ( x ) E ∞ k =0 by C a,b , and call it a generalizedbinomial-coefficient basis of K [ x ]. (cid:3) Proposition 4
Any generalized binomial-coefficient basis C a,b is a factorial ba-sis of K [ x ] , which is ( a, -compatible with the shift operator E .Proof: Clearly deg x P ( a,b ) k ( x ) = k and P ( a,b ) k +1 ( x ) = ax + b − kk + 1 P ( a,b ) k ( x ) for all k ∈ N , P ( a,b ) k ( x ) (cid:12)(cid:12) P ( a,b ) k +1 ( x ), and C a,b is factorial. By Chu-Vandermonde’s identity, EP ( a,b ) k ( x ) = P ( a,b ) k ( x + 1) = (cid:18) ax + a + bk (cid:19) = a X i =0 (cid:18) ai (cid:19)(cid:18) ax + bk − i (cid:19) = X i = − a (cid:18) a − i (cid:19)(cid:18) ax + bk + i (cid:19) = X i = − a (cid:18) a − i (cid:19) P ( a,b ) k + i ( x ) , so C a,b is ( a, E ( α k,i = (cid:0) a − i (cid:1) for i = − a, − a + 1 , . . . , (cid:3) Definition 5
Let m ∈ N \ { } , and for i = 1 , , . . . , m , let B i = h P ( i ) k ( x ) i ∞ k =0 be a basis of K [ x ]. For all k ∈ N and j ∈ { , , . . . , m − } , let P ( π ) mk + j ( x ) := j Y i =1 P ( i ) k +1 ( x ) · m Y i = j +1 P ( i ) k ( x ) . Then the sequence Q mi =1 B i := h P ( π ) n ( x ) i ∞ n =0 is the product of B , B , . . . , B m . (cid:3) Theorem 2
Let B , B , . . . , B m be factorial bases of K [ x ] , and L ∈ L K [ x ] .1. Q mi =1 B i is a factorial basis of K [ x ] .2. Let L be an endomorphism of K [ x ] , and let each B i be ( A i , B i ) -compatiblewith L . Write A = max ≤ i ≤ m A i and B = min ≤ i ≤ m B i . Then Q mi =1 B i is ( mA, B ) -compatible with L .Proof:
1. Clearly deg P ( π ) mk + j = j ( k + 1) + ( m − j ) k = mk + j .If n = mk + j with 0 ≤ j ≤ m −
2, then n + 1 = mk + ( j + 1) and P ( π ) n +1 P ( π ) n = Q j +1 i =1 P ( i ) k +1 Q ji =1 P ( i ) k +1 · Q mi = j +2 P ( i ) k Q mi = j +1 P ( i ) k = P ( j +1) k +1 P ( j +1) k ∈ K [ x ]as B j +1 is factorial. If n = mk + ( m − n + 1 = m ( k + 1) + 0 and P ( π ) n +1 P ( π ) n = 1 Q m − i =1 P ( i ) k +1 · Q mi =1 P ( i ) k +1 P ( m ) k = P ( m ) k +1 P ( m ) k ∈ K [ x ]because B m is factorial. Hence Q mi =1 B i is factorial as well.2. Let p ∈ K [ x ] be arbitrary. For i = 1 , , . . . , m , let p = P deg pk =0 c ( i ) k P ( i ) k bethe expansion of p w.r.t. B i . Then Lp = P deg pk =0 c ( i ) k LP ( i ) k , and by condition C1 of Proposition 1,deg Lp ≤ max ≤ k ≤ deg p deg LP ( i ) k ≤ max ≤ k ≤ deg p ( k + B i ) = deg p + B i . i , we have deg Lp ≤ deg p + B for all p ∈ K [ x ]. Inparticular, deg LP ( π ) k ≤ k + B , so Q mi =1 B i satisfies C1 .Condition C2 of Proposition 1 and our definition of A imply that P ( i ) k +1 − A (cid:12)(cid:12) LP ( i ) k +1 and P ( i ) k − A (cid:12)(cid:12) LP ( i ) k for all k ≥ A and i ∈ { , . . . , m } , so P ( π ) m ( k − A )+ j = j Y i =1 P ( i ) k +1 − A · m Y i = j +1 P ( i ) k − A (cid:12)(cid:12)(cid:12)(cid:12) j Y i =1 LP ( i ) k +1 · m Y i = j +1 LP ( i ) k , or equivalently, since L is an endomorphism of the ring K [ x ], P ( π ) m ( k − A )+ j (cid:12)(cid:12)(cid:12)(cid:12) L j Y i =1 P ( i ) k +1 · m Y i = j +1 P ( i ) k = LP ( π ) mk + j . For n = mk + j ≥ mA , this turns into P ( π ) n − mA (cid:12)(cid:12) LP ( π ) n , so Q mi =1 B i satisfies C2 as well. By Proposition 1, this proves the claim. (cid:3) Definition 6
Let m ∈ N \ { } , and let a = ( a , a , . . . , a m ), b = ( b , b , . . . , b m )where a i ∈ N \ { } , b i ∈ K for i = 1 , , . . . , m . We denote the product ofgeneralized binomial-coefficient bases Q mi =1 C a i ,b i by C a , b , and call it a productbinomial-coefficient basis of K [ x ]. (cid:3) Corollary 1
Any product binomial-coefficient basis C a , b is a factorial basis of K [ x ] which is ( mA, E , where A = max ≤ i ≤ m a i . Proof:
Use Theorem 2 and Proposition 4. (cid:3)
By Corollary 1, we now have at our disposal a rich family of factorial bases C a , b , compatible with any operator L ∈ K [ x ] h E i , which, given L and F ( n, k ) ofthe form (1), can be used to find solutions of the form y n = P ∞ k =0 F ( n, k ) h k of Ly = 0. To this end, we need to compute expansions of EP n ( x ) and XP n ( x ) inthe basis C a , b = h P n ( x ) i ∞ n =0 . Example 4
For the simplest nontrivial example, take F ( n, k ) = (cid:0) nk (cid:1) . Thepolynomial basis to be used here is C (1 , , (0 , = h P n ( x ) i ∞ n =0 where for all k ∈ N , P k ( x ) = (cid:18) xk (cid:19) , P k +1 ( x ) = (cid:18) xk + 1 (cid:19)(cid:18) xk (cid:19) . According to Corollary 1, C (1 , , (0 , is a factorial basis of K [ x ] with m = 2 and A = max { , } = 1, so it is (2 , E . In particular, this meansthat P k ( x + 1) can be expressed as a linear combination of P k ( x ), P k − ( x )9nd P k − ( x ), and P k +1 ( x + 1) as a linear combination of P k +1 ( x ), P k ( x ) and P k − ( x ), with coefficients depending on k . In the case of P k ( x + 1) this is justan application of Pascal’s rule: P k ( x + 1) = (cid:18) x + 1 k (cid:19) = (cid:20)(cid:18) xk (cid:19) + (cid:18) xk − (cid:19)(cid:21) = (cid:18) xk (cid:19) + 2 (cid:18) xk (cid:19)(cid:18) xk − (cid:19) + (cid:18) xk − (cid:19) = P k ( x ) + 2 P k − ( x ) + P k − ( x ) . (7)In the case of P k +1 ( x + 1), we can use the method of undetermined coefficients.Dividing both sides of P k +1 ( x + 1) = u ( k ) P k +1 ( x ) + v ( k ) P k ( x ) + w ( k ) P k − ( x ) , or (cid:18) x + 1 k + 1 (cid:19)(cid:18) x + 1 k (cid:19) = u ( k ) (cid:18) xk + 1 (cid:19)(cid:18) xk (cid:19) + v ( k ) (cid:18) xk (cid:19) + w ( k ) (cid:18) xk (cid:19)(cid:18) xk − (cid:19) where u ( k ) , v ( k ) , w ( k ) are undetermined functions of k , by (cid:0) xk (cid:1)(cid:0) xk − (cid:1) , yields( x + 1) k ( k + 1) = u ( k ) ( x − k + 1)( x − k ) k ( k + 1) + v ( k ) x − k + 1 k + w ( k ) , (8)which is an equality of two quadratic polynomials from K ( k )[ x ]. Plugging inthe values x = − , k, k −
1, we obtain a triangular system of linear equations u ( k ) − v ( k ) + w ( k ) = 0 k v ( k ) + w ( k ) = k +1 k w ( k ) = kk +1 whose solution is u ( k ) = 1 (as expected), v ( k ) = k +1 k +1 , w ( k ) = kk +1 , and so P k +1 ( x + 1) = P k +1 ( x ) + 2 k + 1 k + 1 P k ( x ) + kk + 1 P k − ( x ) . (9)For the expansion of XP n ( x ), recall that every factorial basis is (0 , X . Indeed, as x (cid:0) xk (cid:1) = k (cid:0) xk (cid:1) + ( k + 1) (cid:0) xk +1 (cid:1) , we have xP k ( x ) = (cid:18) xk (cid:19) (cid:20) x (cid:18) xk (cid:19)(cid:21) = ( k + 1) P k +1 ( x ) + kP k ( x ) ,xP k +1 ( x ) = (cid:18) xk + 1 (cid:19) (cid:20) x (cid:18) xk (cid:19)(cid:21) = ( k + 1) P k +2 ( x ) + kP k +1 ( x ) . (cid:3)
10n the general case F ( n, k ) = Q mi =1 (cid:0) a i n + b i k (cid:1) , we use the basis C a , b = h P ( π ) n ( x ) i ∞ n =0 which, by Corollary 1, is ( mA, E where A =max ≤ i ≤ m a i . In order to compute α k,j,i ∈ K ( k ) such that P ( π ) mk + j ( x + 1) = mA X i =0 α k,j,i P ( π ) mk + j − i ( x )for all k ∈ N and j ∈ { , , . . . , m − } , we divide both sides of this equation by P ( π ) mk + j − mA ( x ) which turns it into an equality of two polynomials of degree mA from K ( k )[ x ]. From this equality a system of mA + 1 linear algebraic equationsfor the mA +1 undetermined coefficients α k,j,i , i = 0 , , . . . , mA , can be obtainedby equating the coefficients of like powers of x on both sides, or (as in Example4) by substituting mA + 1 distinct values from K ( k ) for x in this equality. Notethat for each j ∈ { , , . . . , m − } , this system is uniquely solvable since C a , b is abasis of K [ x ], that the α k,j,i will be rational functions of k , and that, as the shiftoperator preserves leading coefficients and degrees of polynomials, α k,j, = 1. Example 5
Here we give some additional examples of expansions of a shiftedbasis element in the basis C a , b , computed by the procedure just described.(a) C a , b = C (2 , , (0 , = h P n ( x ) i ∞ n =0 where for all k ∈ N , P k ( x ) = (cid:18) xk (cid:19)(cid:18) xk (cid:19) , P k +1 ( x ) = (cid:18) xk + 1 (cid:19)(cid:18) xk (cid:19) Here m = 2, A = 3, mA = 6 and P k ( x + 1) = P k ( x ) + 6 P k − ( x )+ 3(7 k − k P k − ( x ) + 131 k − k P k − ( x )+ 211 k − k + 12036( k − k P k − ( x ) + 2(2 k − k − P k − ( x ) ,P k +1 ( x + 1) = P k +1 ( x ) + 2(2 k + 1) k + 1 P k ( x ) + 17 k + 72( k + 1) P k − ( x )+ 131 k − k − k ( k + 1) P k − ( x ) + 2 (cid:0) k − k − (cid:1) k ( k + 1) P k − ( x )+ 4( k − k − k − k + 1) P k − ( x ) − k (2 k − k − k + 1) P k − ( x ) . (b) C a , b = C (2 , , ( − , = h P n ( x ) i ∞ n =0 where for all k ∈ N , P k ( x ) = (cid:18) x − k (cid:19)(cid:18) x + 4 k (cid:19) , P k +1 ( x ) = (cid:18) x − k + 1 (cid:19)(cid:18) x + 4 k (cid:19) m = 2, A = 3, mA = 6 and P k ( x + 1) = P k ( x ) + 6 P k − ( x )+ 21 k + 132 k P k − ( x ) + 131 k − k P k − ( x )+ 211 k + 330 k + 79136( k − k P k − ( x ) + 2( k − k − k − k P k − ( x ) ,P k +1 ( x + 1) = P k +1 ( x ) + 2(2 k + 1) k + 1 P k ( x ) + 17 k − k + 1) P k − ( x )+ 131 k − k + 21418 k ( k + 1) P k − ( x ) + 4 (cid:0) k − k + 104 (cid:1) k ( k + 1) P k − ( x )+ 4( k − k − k − k − k ( k + 1) P k − ( x ) − k − k + 11)(2 k − k − k ( k + 1) P k − ( x ) . (c) C a , b = C (4 , , (0 , = h P n ( x ) i ∞ n =0 where for all k ∈ N , P k ( x ) = (cid:18) xk (cid:19) , P k +1 ( x ) = (cid:18) xk + 1 (cid:19)(cid:18) xk (cid:19) Here m = 2, A = 4, mA = 8 and P k ( x + 1) = P k ( x ) + 8 P k − ( x ) + 4(7 k − k P k − ( x )+ 28(2 k − k P k − ( x ) + 2 (cid:0) k − k + 22 (cid:1) ( k − k P k − ( x )+ 8 (cid:0) k − k + 5 (cid:1) ( k − k P k − ( x ) + 4 (cid:0) k − k + 32 k − (cid:1) ( k − k − k P k − ( x )+ 4(2 k − (cid:0) k − k + 1 (cid:1) ( k − k − k P k − ( x ) + P k − ( x ) ,P k +1 ( x + 1) = P k +1 ( x ) + 4(2 k + 1) k + 1 P k ( x ) + 4(7 k + 3) k + 1 P k − ( x )+ 8 (cid:0) k − (cid:1) k ( k + 1) P k − ( x ) + 2 (cid:0) k − k − (cid:1) k ( k + 1) P k − ( x )+ 4(2 k − (cid:0) k − k − (cid:1) ( k − k ( k + 1) P k − ( x ) + 4 (cid:0) k − k + 4 k + 1 (cid:1) ( k − k ( k + 1) P k − ( x )+ 8( k − k + 1 P k − ( x ) + k − k + 1 P k − ( x ) . (cid:3) As every factorial basis, C a , b = h P ( π ) n ( x ) i ∞ n =0 is (0 , X :12 roposition 5 For k ∈ N and j ∈ { , , . . . , m − } , let P ( π ) mk + j ( x ) := j Y i =1 (cid:18) a i x + b i k + 1 (cid:19) · m Y i = j +1 (cid:18) a i x + b i k (cid:19) . (10) Then xP ( π ) mk + j ( x ) = k + 1 a j +1 P ( π ) mk + j +1 ( x ) + k − b j +1 a j +1 P ( π ) mk + j ( x ) . Proof: k + 1 a j +1 P ( π ) mk + j +1 ( x ) + k − b j +1 a j +1 P ( π ) mk + j ( x ) = P ( π ) mk + j ( x ) · f ( x )where f ( x ) = k + 1 a j +1 · P ( π ) mk + j +1 ( x ) P ( π ) mk + j ( x ) + k − b j +1 a j +1 = k + 1 a j +1 · (cid:0) a j +1 x + b j +1 k +1 (cid:1)(cid:0) a j +1 x + b j +1 k (cid:1) + k − b j +1 a j +1 = k + 1 a j +1 · a j +1 x + b j +1 − kk + 1 + k − b j +1 a j +1 = x. (cid:3) Now we can use Proposition 2 and the associated operator R B L where B = C a , b to find h such that y n = P ∞ k =0 F ( n, k ) h k , with F as in (1), satisfies Ly = 0.Notice however that for m >
1, the coefficients α k,i expressing the actions of E resp. X on B are not rational functions of k anymore, but conditional expressionsevaluating to m generally distinct rational functions, depending on the residueclass of k mod m (cf. Example 5 and Proposition 5). So the coefficients of R B L ,obtained by composing the operators R B E and R B X repeatedly, will containquite complicated conditional expressions. In addition, ord R B L may exceedord L by a factor of mA which can be exponential in input size. Finally, onlythose h ∈ ker R B L that satisfy h k = 0 whenever k m ) (i.e., those thatare interlacings of an arbitrary sequence with m − L that have the desired form.To overcome these inconveniences, for a fixed m > R B L directly but represent it by a matrix [ R B L ] = [ L r,j ] m − r,j =0 ofoperators where L r,j ∈ E expresses the contribution of the j -th m -section s mj σ B y of the coefficient sequence of y to the r -th m -section s mr σ B ( Ly ) of the coefficientsequence of Ly . 13 roposition 6 Let L ∈ L K [ x ] , B = h P n ( x ) i ∞ n =0 ( a factorial basis of K [ x ]) , m ∈ N \ { } , and A, B ∈ N be such that for all k ∈ N and j ∈ { , , . . . , m − } , LP mk + j ( x ) = B X i = − A α k,j,i P mk + j + i ( x ) . (11) Furthermore, for all r, j ∈ { , , . . . , m − } define L r,j := X − A ≤ i ≤ Bi + j ≡ r (mod m ) α k + r − i − jm ,j,i E r − i − jm k ∈ E (12)( to keep notation simple, we do not make the dependence of L r,j on m explicit ) .Then for every y ∈ K [[ B ]] and r ∈ { , , . . . , m − } , s mr σ B ( Ly ) = m − X j =0 L r,j s mj σ B y. (13) Proof:
Write y ( x ) = P ∞ n =0 c n P n ( x ) as the sum of its m -sections y ( x ) = m − X j =0 ∞ X k =0 c mk + j P mk + j ( x ) = m − X j =0 ∞ X k =0 ( s mj c ) k P mk + j ( x ) . (14)Then Ly ( x ) = m − X j =0 ∞ X k =0 ( s mj c ) k LP mk + j ( x ) = m − X j =0 ∞ X k =0 ( s mj c ) k B X i = − A α k,j,i P mk + j + i ( x )(15)= m − X j =0 m − X r =0 X − A ≤ i ≤ Bi + j ≡ r (mod m ) ∞ X k =0 α k,j,i (cid:0) s mj c (cid:1) k P mk + i + j ( x ) (16)= m − X r,j =0 X − A ≤ i ≤ Bi + j ≡ r (mod m ) ∞ X k = i + j − rm α k + r − i − jm ,j,i (cid:0) s mj c (cid:1) k + r − i − jm P mk + r ( x ) (17)= m − X r,j =0 X − A ≤ i ≤ Bi + j ≡ r (mod m ) ∞ X k =0 α k + r − i − jm ,j,i (cid:16) E r − i − jm k s mj c (cid:17) k P mk + r ( x ) (18)= m − X r =0 ∞ X k =0 (cid:18) m − X j =0 L r,j s mj c (cid:19) k P mk + r ( x ) (19)where in (15) we used (11), in (16) we reordered summation on i with respect tothe residue class of i + j mod m , (17) was obtained by replacing k with k − i + j − rm ,(18) by noting that k < ⇒ mk + r < ⇒ P mk + r = 0 ,k < i + j − rm = ⇒ (cid:0) s mj c (cid:1) k + r − i − jm = 0 , Ly ( x ) and the series in (19) can berestated as (13). (cid:3) Corollary 2
Under the assumptions of Proposition 6, Ly = 0 ⇐⇒ ∀ r ∈ { , , . . . , m − } : m − X j =0 L r,j s mj σ B y = 0 . Note that for m = 1, Corollary 2 and Proposition 6 turn into Proposition 2and Proposition 3, respectively (with α k,i = α k, ,i and R B L = L , ). Proposition 7
Let B = C a , b . Then for all r, j ∈ { , , . . . , m − } , X r,j = [ r = j ] k − b j +1 a j +1 + [ r = 0 ∧ j = m − ka j +1 E − k + [ r = j + 1] k + 1 a j +1 where [ ϕ ] = (cid:26) , if ϕ is true , , otherwise is Iverson bracket.Proof: From Proposition 5 we read off that in this case α k,j, = k − b j +1 a j +1 , (20) α k,j, = k + 1 a j +1 . (21)From (12) with A = 0, B = 1 it follows that X r,j = [ j ≡ r (mod m )] α k + r − jm ,j, E r − jm k + [ j ≡ r − m )] α k + r − j − m ,j, E r − j − m k . (22)Combining (20) – (22) with 0 ≤ r, j ≤ m − (cid:3) Notation: [ R B L ] := [ L r,j ] m − r,j =0 ∈ M m ( E ) where L r,j is as given in (12). Proposition 8
Let L (1) , L (2) ∈ L B . Then h R B (cid:16) L (1) L (2) (cid:17)i = h R B L (1) ih R B L (2) i . Proof:
Write L = L (1) L (2) . By (13), s mt σ ( Ly ) = m − X j =0 L t,j s mj σy = m − X j =0 [ R B L ] t,j s mj σy.
15n the other hand, by (13) applied to L (1) and L (2) , s mt σ ( Ly ) = s mt σ ( L (1) L (2) y ) = m − X r =0 L (1) t,r s mr σ ( L (2) y )= m − X r =0 L (1) t,r m − X j =0 L (2) r,j s mj σy = m − X j =0 m − X r =0 L (1) t,r L (2) r,j ! s mj σy = m − X j =0 (cid:16)h R B L (1) ih R B L (2) i(cid:17) t,j s mj σy. Hence m − X j =0 [ R B L ] t,j s mj σy = m − X j =0 (cid:16)h R B L (1) ih R B L (2) i(cid:17) t,j s mj σy for any y ∈ K [[ B ]], which implies the claim. (cid:3) It follows that to compute [ R B L ] for an arbitrary operator L ∈ K [ x ] h E i , itsuffices to apply the substitution E [ R B E ] ,x [ R B X ] , I m (23)where I m is the m × m identity matrix, to all terms of L . Since we are interestedin finding y ∈ ker L of the form y ( x ) = ∞ X k =0 h k m Y i =1 (cid:18) a i x + b i k (cid:19) = ∞ X k =0 h k P mk ( x ) , we have s m σy = h and s mj σy = 0 for all j = 0. For such y , Corollary 2 implies Ly = 0 ⇐⇒ ∀ r ∈ { , , . . . , m − } : L r, h = 0 ⇐⇒ gcrd( L , , L , , . . . , L m − , ) h = 0 . (24)So any nonzero element of the first column of [ R B L ] = [ L r,j ] m − r,j =0 may serve asa nontrivial annihilator L ′ of h , and taking their greatest common right divisormight yield L ′ of lower order. The fact that we only need the first column[ R B L ] e (1) of [ R B L ] (where e (1) = (1 , , . . . , T is the first standard basis vectorof length m ) can also be used to advantage in its computation.In summary, we have the following algorithm: Algorithm
DefiniteSumSolutions
Input: L ∈ K [ x ] h E i , m ∈ N \ { } , a , a , . . . , a m ∈ N \ { } , b , b , . . . , b m ∈ Z utput: L ′ ∈ K ( k ) h E k i such that L ∞ X k =0 m Y i =1 (cid:18) a i x + b i k (cid:19) h k ! = 0if and only if L ′ h = 01. A := max ≤ i ≤ m a i .2. For j = 0 , , . . . , m − P mk + j ( x ) := j Y i =1 (cid:18) a i x + b i k + 1 (cid:19) · m Y i = j +1 (cid:18) a i x + b i k (cid:19) . For j = 0 , , . . . , m − α k,j,i ∈ K ( k ) such that P mk + j ( x + 1) = X i = − mA α k,j,i P mk + j + i ( x )as explained below Example 4 on p. 10.3. For r, j = 0 , , . . . , m − E r,j := X − mA ≤ i ≤ i + j ≡ r (mod m ) α k + r − i − jm ,j,i E r − i − jm k ,X r,j := [ r = j ] k − b j +1 a j +1 + [ r = 0 ∧ j = m − ka j +1 E − k + [ r = j + 1] k + 1 a j +1 . Let [ R B E ] = [ E r,j ] m − r,j =0 , [ R B X ] = [ X r,j ] m − r,j =0 .4. Let [ R B L ] = [ L r,j ] m − r,j =0 be the matrix of operators obtained by applyingsubstitution (23) to L .5. Return L ′ := gcrd( L , , L , , . . . , L m − , ). Remark 1
In step 4 it suffices to compute [ R B L ] e (1) , the first column of [ R B L ] .We do so by proceeding from right to left through the expression obtained from L by substitution (23), multiplying a matrix with a vector at each point. Example 6
Consider again the basis B = C (1 , , (0 , from Example 4. Here m = 2, a = a = 1, b = b = 0. To compute [ R B E ], comparing (11) with (7)and (9) yields α k, , = 1 α k, , = 1 α k, , − = 2 α k, , − = k +1 k +1 α k, , − = 1 α k, , − = kk +1 , R B E ] = (cid:20) E , E , E , E , (cid:21) = (cid:20) E k + 1 k +1 k +1 E k k +1 k +2 E k + 1 (cid:21) . (25)For [ R B X ] it follows from Proposition 7 that[ R B X ] = (cid:20) X , X , X , X , (cid:21) = (cid:20) k kE − k k + 1 k (cid:21) . (26)By way of example we now look at two operators L ∈ K [ x ] h E i and computetheir associated operators [ R B L ] (only the first column) and L ′ . The 2 × I .1. L = ( n + 1) E − n + 1): Using (25) and (26), we obtain[ R B L ] e (1) = (([ R B X ] + I )[ R B E ] − R B X ] + I )) e (1) = (cid:20) ( k + 1)( E k − k + 1)( E k − (cid:21) , so we can take L ′ = E k −
1. This comes as no surprise, since y n = P ∞ k =0 (cid:0) nk (cid:1) = (cid:0) nn (cid:1) satisfies Ly = 0.2. L = 4(2 n +3) (4 n +3) E − n +5) (cid:0) n + 50 n + 27 (cid:1) E +9(4 n +7)( n +1) :Applying substitution (23) to L , we obtain[ R B L ] e (1) = (cid:20) L , L , (cid:21) where L , = 4(2 k + 3) (4 k + 3) E k + 2 (cid:0) k + 1388 k + 1254 k + 519 k + 81 (cid:1) k + 1 E k + 676 k − k − k − − (244 k + 41) k E − k ,L , = 8(2 k + 3) (cid:0) k + 108 k + 132 k + 51 (cid:1) k + 2 E k + 4 (cid:0) k + 720 k + 451 k + 82 (cid:1) E k − k + 1) (cid:0) k + 377 k + 133 (cid:1) − k + 1) k E − k . Taking L ′ = gcrd ( E k L , , E k L , ) = E k − k + 12(2 k + 1) ,
18e see that h k = ( kk ) satisfies L ′ h = 0, so y n = ∞ X k =0 (cid:0) nk (cid:1) (cid:0) kk (cid:1) is a definite-sum solution of equation Ly = 0. (cid:3)
1. The algorithm
DefiniteSumSolutions can be used recursively to pro-duce solutions of Ly = 0 in the form of nested definite sums.2. We only used the first column of [ R B L ], but this matrix provides in-formation on all definite-sum solutions of Ly = 0 having the form y ( x ) = P ∞ n =0 c n P n ( x ) with P n ( x ) = P ( π ) mk + j ( x ) as given in (10).3. It seems likely that our approach can be generalized to summands of theform F ( n, k ) h k where F ( n, k ) = m Y i =1 (cid:18) a i n + b i c i k + d i (cid:19) (27)with a i , c i ∈ N \ { } , b i ∈ Z , d i ∈ N .4. Could Galois theory of difference equations help to determine, given L ,for which values of m, a i , b i , c i , d i in (1) or (27) may the associated equa-tion L ′ h = 0 possess nonzero explicitly representable (e.g., Liouvillian)solutions? Acknowledgements
The author acknowledges financial support from the Slovenian Research Agency(research core funding No. P1-0294). The paper was started while he was attend-ing the thematic programme “Algorithmic and Enumerative Combinatorics” atthe Erwin Schr¨odinger International Institute for Mathematics and Physics inVienna, Austria. He thanks the Institute for its support and hospitality.
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