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Generalization of some hypergeometric functions
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ScienceDirect Indagationes Mathematicae 28 (2017) 711–720 www.elsevier.com/locate/indag
Generalization of some hypergeometric functions Mohamed Kerada a,∗ , Ali Boussayoud b , Abdelhamid Abderrezzak c a LMAM Laboratory and Department of Computer Science, University of MSB Jijel, BP 98 Ouled Aissa, Jijel 18000,
Algeria b LMAM Laboratory and Department of Mathematics, University of MSB Jijel, BP 98 Ouled Aissa, Jijel 18000, Algeria c University of Paris 7, LITP, Place Jussieu, Paris cedex 05, France
Received 14 November 2015; received in revised form 29 January 2017; accepted 24 March 2017 Communicated by F. Beukers
Abstract In this paper, we make use of the Lagrange interpolation to obtain some algebraic identities, involving one or two infinite sets of variables. c 2017 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved. ⃝
Keywords: Symmetric functions; Divided differences; Lagrange interpolation
1. Introduction One of the most important general expansion formulas for hypergeometric series is the Fields and Wimp expansion, described by [5]: ∞ (a R )n (α)n (β)n (−x)n a R , cT ; xw = r +t Fs+u r +2 Fs+1 b S, dU (b S )(γ + n)n n! n=0 n + α, n + β, n + a R × ;x 1 + 2n + γ , n + b S ∗ Corresponding author.
E-mail addresses: [email protected] (M. Kerada), [email protected] (A. Boussayoud), [email protected] (A. Abderrezzak). http://dx.doi.org/10.1016/j.indag.2017.03.002 c 2017 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved. 0019-3577/⃝
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M. Kerada et al. / Indagationes Mathematicae 28 (2017) 711–720
−n, n + γ , × t+2 Fu+2 α, β
cT ;w , dU
where the contract notations (a)n , a R , (a R )n , and n + a R represent a(a + 1) . . . (a + n − 1), a1 , . . . , ar , (a R )n , (a1 )n (a2 )n . . . (ar )n , and n + a1 , . . . , n + ar , respectively. Verma showed in [8] that this formula is a special case of expansion (6.1) and derived the q-analog (6.2). On the other hand, Al-Salam and Verma [3] showed that Euler’s transformation formula ∞
an bn x n =
n=0
∞ x k (k) f (x)∆k a0 , (−1)k k! k=0
(1.1)
where f (x) = b0 + b1 x + b2 x 2 + · · · , and ∆k a0 =
k k (−1) j ak− j , j j=0
has bibasic extension (5.2). For their part, Gessel and Stanton [6] obtained the following generating function ∞ (1 + x)a (1 + y)b a+ p b+k k p = x y . k p (1 − x y)a+b+1 k=0, p=0
(1.2)
In this paper, we make use of the divided differences [2] and the Lagrange interpolation to obtain algebraic identities, involving one or two infinite sets of variables (formulas (4.1)–(4.3)). By specializing variables sets, we recover formulas ((6.1) and (6.2)) provided by Verma [8], formula (5.2) given by Al-Salem and Verma [3], and generating function (7.1) of Gessel and Stanton [6]. We also give a new q-analog of results ((5.3), (5.4), (6.3), (6.4), (7.3), (7.4), (7.5)). For this purpose, the Lagrange interpolation is considered to describe the properties of a linear operator, sending function of one variable to a symmetric function [4]. It can be written as a summation on a set, or a product of divided differences. This later version will be used throughout this paper. 2. Multiple interpolations In this section, we consider the linear operator Λ(A) of the Lagrange interpolation defined by Λ(A)( f ) =
aϵ A
f (a) R(a, A \ a)
(2.1)
where R(a, B) = Rbϵ B (a − b) and R(a, ∅) = 1. Accordingly, this operator sends a polynomial of degree k to a symmetric polynomial in A of degree k − n, with car d(A) = n + 1. In particular, it annihilates polynomials of degree < n, and f (x) = x n on constant 1.
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Let Λ(E n+1 ) be the Lagrange operator of an alphabet E n+1 := {e1 , e2 , . . . , en+1 }. We have [1] Λ(E n+1 )( f ) =
n k=0
f (ek+1 ) . R(ek+1 , E n+1 \ ek+1 )
(2.2)
For our purpose, we make use of the Lagrange interpolation for two independent alphabets and functions of two variables, described by Λ(E m+1 )Λ(Bn+1 )( f ) =
m n k=0 p=0
f (ek+1 , b p+1 ) . R(ek+1 , E m+1 \ ek+1 )R(b p+1 , Bn+1 \ b p+1 )
(2.3)
3. Inversion identities Consider the alphabet E n+1 \ E s = {es+1 , . . . , en+1 }, then, Lagrange interpolation (2.2) takes the following form Λ(E n+1 \ E s )( f ) =
n p=s
f (e p+1 ) . R(e p+1 , E p \ E s )R(e p+1 , E n+1 \ E p+1 )
(3.1)
In the case n = s, we have Λ(E s+1 \ E s )( f ) = Λ(es+1 )( f ) = f . Take f (es+1 ) =
n (β +a e Π j=1 j j s+1 )
s+1 Π j=1 (β j +a j es+1 )
in (3.1). The simple fact that the Lagrange operator Λ(E n+1 )
decreases the polynomial by n implies the following identities n Π j=1 (β j + a j es+1 ) = δn,s , (βs+1 + as+1 es+1 )Λ(E n+1 \ E s ) s+1 Π j=1 (β j + a j es+1 ) n Π j=1 α j µ j n m Λ(E m+1 )Λ(Bn+1 )Π j=1 (β j + α j e1 )Π j=1 (γ j + µ j b1 ) = 0
(3.2) if m = n if m ̸= n,
(3.3)
where δn,s is the Kronecker’s delta symbol. By combining formulas (2.2) and (3.2), we deduce the identity δn,s = (βs+1 + as+1 es+1 ) ×
n p=s
n (β + a e Π j=1 j j p+1 ) 1 . s+1 R(e p+1 , E p \ E s )R(e p+1 , E n+1 \ E p+1 ) Π j=1 (β j + a j e p+1 )
Let us consider two infinite upper triangular matrices A and B such that 1 (βs+1 + as+1 es+1 ) if p > s s+1 Asp = R(e p+1 , E p \ E s ) Π j=1 (β j + a j e p+1 ) 0 if p < s, n Π j=1 (β j + a j e p+1 ) if p ≤ n B pn = R(e p+1 , E n+1 \ E p+1 ) 0 if p > n.
(3.4)
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Then, relation (3.4) is simply the identity AB = I d∞ , where I d∞ is the infinite identity matrix. We have also n 1 1 δn, p = R(e , E \ E ) R(e , E n k n+1 p+1 k+1 \ E p+1 ) k= p ×
k (β + a e (βk+1 + ak+1 ek+1 )Π j=1 j j p+1 ) k+1 Π j=1 (β j + a j e p+1 )
.
(3.5)
Equality (3.5) is an explicit version of B A = I d∞ , which automatically follows from AB = I d∞ . 4. Principal formulas From formulas (3.4) and (3.5), we deduce the following theorem. Theorem 1. Let E = {e1, e2 , . . .} be a set of variables and {An }, {Bn }, {βn }, {an } four sequences. Then we have ∞
An Bn x n z n =
n=0
∞
p
Π j=1 (β j + a j e p+1 )x p
p=0
×
p s=0
× ∞
An Bn x n z n =
n=0
∞
1
n=0
R(e p+1 , E n+ p+1 \ E p+1 )
∞
xk
k=0
×
(βs+1 + as+1 es+1 ) 1 Bs z s s+1 R(e p+1 , E p \ E s ) Π j=1 (β j + a j e p+1 )
∞ n=0
n+ p
Π j= p+1 (β j + a j e p+1 )An+ p x n , (4.1)
k
(βk+1 + ak+1 ek+1 )
p=0
k (β + a e R(e p+1 , E k+1 \ E p+1 )Π j=1 j j p+1 )
Bpz p
1 1 An+k x n . k+1 R(en+k+1 , E n+k \ E k ) Π j=1 (β j + a j en+k+1 )
(4.2)
Denote by a the infinite row vector with components An x n and b the infinite row vector with components Bn z n . Then, identity (4.1) follows from the observation ∞
An Bn x n z n = b · at = bABat = (bA) · (Bat ),
n=0
and identity (4.2) from ∞
An Bn x n z n = b · at = bB Aat = (bB) · (Aat ).
n=0
On the other hand, formula (3.3) enables us to write this second theorem. Theorem 2. Let E = {e1, e2 , . . .}, B = {b1 , b2 , b3 , . . .} be two sets of variables, and {α} , {β} , {γ } , {µ} four sequences. Then we have ∞ n=0
n Π j=1 α j µ j x n yn
=
∞ k=0, p=0
p
k (γ + µ b Π j=1 (β j + α j ek+1 ) Π j=1 j j p+1 )
R(ek+1 , E k )
R(b p+1 , B p )
xk y p
M. Kerada et al. / Indagationes Mathematicae 28 (2017) 711–720
×
m+ p
∞
Π j= p+1 (β j + α j ek+1 )
n=0,m=0
R(ek+1 , E m+k+1 \ E k+1 )
×
715
n+k Π j=k+1 (γ j + µ j b p+1 )
R(b p+1 , Bn+ p+1 \ B p+1 )
x n ym .
(4.3)
5. Application of identity (4.1) (I) For E = {0, 1, 2, . . .}, β j = 1, and a j = −ab j−1 , identity (4.1) implies ∞
An Bn x n z n =
n=0
∞
p
Π j=1 (1 − ab j−1 p)x p
p=0
×
×
p s=0 ∞ n=0
(1 − abs s) 1 Bs z s s+1 R(e p+1 , E p \ E s ) Π j=1 (1 − ab j−1 p) 1
n+ p
R(e p+1 , E n+ p+1 \ E p+1 )
Π j= p+1 (1 − ab j−1 )An+ p x n .
Since we have p
Π j=1 (1 − ab j−1 p) = (abp, b) p−1 , s+1 Π j=1 (1 − ab j−1 p) = (abp, b)s , n+ p
n+1 Π j= p+1 (1 − ab j−1 ) = Π j=1 (1 − ab j+ p ) = (ab p p, b)n ,
R(e p+1 , E p \ E s ) = ( p − s)!, R(e p+1 , E n+ p+1 \ E p+1 ) = n!(−1)n , then we can derive the following relationship ∞
An Bn x n z n =
n=0
∞
(abp, b) p−1 x p
p=0
×
p s=0
∞ (ab p p, b)n
n!
n=0
1 − asbs Bs z s ( p − s)!(abp, b)s
An+ p (−x)n .
(5.1)
Similarly we proceed cases. in the remaining (II) For E = 1, q, q 2 , . . . , β j = 1, a j = −ab j−1 and by replacing An −→ (−1)n q
n(n−1) 2
An , Bn −→ (−1)n q − ∞
An Bn x n z n =
n=0
∞
n(n−1) 2
(abq p , b) p−1 x p
p=0
×
∞ n=0
with [n] =
1−q n 1−q .
Bn , we have p s=0
1 − aq s bs Bs z s (q, q) p−s (ab p q p , b)s
n(n−1) (ab p q p , b)n An+ p (−x)n q 2 , (q, q)n
(5.2)
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Remark 1. The case b = q is due to Jackson [7]. (III) For E = {[0], [1], [2], . . .}, β j = 1, a j = −ab j−1 and by replacing An −→ (−1)n q
n(n−1) 2
An , Bn −→ (−1)n q − ∞
An Bn x n z n =
∞
n(n−1) 2
Bn , we obtain
(ab[ p], b) p−1 x p
s=0
p=0
n=0
×
p
∞ n=0
1 − a[s]bs Bs z s [ p − s]!(ab p [ p], b)s
n(n−1) (ab p [ p], b)n An+ p (−x)n q 2 . [n]!
(5.3)
(IV) For E = {[0], [1], [2], . . .}, β j = γ + [ j − 1], a j = q j−1 and by replacing An −→ (−1)n q
n(n−1) 2
∞
An , Bn −→ (−1)n q −
An Bn x n z n =
n=0
∞
n(n−1) 2
Bn , we obtain
{γ + [ p + 1], b} p−1 x p
p=0
×
p {γ + [2s], b} p−1 s=0
×
[ p − s]!
{γ + [ p + 1], b}s Bs z s
∞ n(n−1) {γ + [2 p], b}n An+ p (−x)n q 2 , [n]! n=0
(5.4)
with {γ }k = γ (γ +[1])(γ +[2]) . . . (γ +[k −1]) and {a + [k]}n = (a +[k])(a +[k +1]) . . . (a + [k + n − 1]). 6. Application of identity (4.2) (I) For E = {0, 1, 2, . . .} , β j = γ + j − 1, a j = 1, and by replacing An −→ (α)n (β)n An and Bn −→ (α)n1(β)n Bn!n , identity (4.2) implies ∞
An Bn x n z n
n=0
=
∞
xk
k
k=0 p=0 ∞
×
n=0
(γ + k + j − 1) R(e p+1 , E k+1 \
k (γ E p+1 )Π j=1
+ p + j − 1)(α) p (β) p
1 (α)n+k (β)n+k An+k x n . k+1 R(en+k+1 , E n+k \ E k ) Π j=1 (γ + n + k + j)
Since we have k Π j=1 (γ + p + j − 1) = (γ + p)k , k+1 Π j=1 (γ + n + k + j) = (γ + n + k)k+1 ,
R(e p+1 , E k+1 \ E p+1 ) = (k − p)!, R(en+k+1 , E n+k \ E k ) = n!,
Bp
zp p!
M. Kerada et al. / Indagationes Mathematicae 28 (2017) 711–720
717
then we obtain ∞
An Bn x n
k=0
∞ k (−k) p (γ + k) p zn 1 (−x)k zp = Bp n! (γ + k)k k! p=0 (α) p (β) p p! k=0
×
∞ (α)n+k (β)n+k xn An+k . (γ + 2k + 1)n n! p=0
(6.1)
In the following cases, we proceed similarly. (II) For E = 1, q, q 2 , . . . , β j = 1, a j = −aq j−1 and by replacing An −→ (−1)n (α,q)n (β,q)n (a,q)n ∞
An q
n(n−1) 2
An Bn x n
n=0
(a,q)n , Bn −→ (−1)n (α) Bn q − n (β)n
n(n−1) 2
1 (q,q)n ,
we have
k ∞ −k , aq k , q) k(k−1) (q zn (−x)k p = q 2 B p (zq) p k (q, q)n (q, aq , q) (q, α, β, q) k p p=0 k=0
×
∞ (α, β; q)n+k An+k x n . 2k+1 ; q) (q, aq n n=0
(6.2)
(III) For E = {[0], [1], [2], . . .}, β j = γ + [ j − 1], a j = q j−1 and by replacing An −→ {α}n {β}n An q
n(n−1) 2
, Bn −→
1 {α}n {β}n
Bn q −
n(n−1) 2
1 [n]! ,
we obtain
k(k−1) ∞ zn (−x)k q 2 An Bn x = {γ + [k]}k [k]! [n]! n=0 k=0
∞
n
×
k [−k] {γ + [k]} (q −k , aq k , q) p p p
{α} p {β} p (q, α, β, q) p
p=0
×
∞ ({α}n+k {β}n+k , ; q)n+k n=0
{γ + [2k + 1]}n [n]!
B p (−zq) p
An+k x n .
(6.3)
(IV) For E = {[0], [1], [2], . . .}, α j = 1, a j = −ab j−1 and by replacing An −→ (α, b)n (β, b)n An q ∞ n=0
n(n−1) 2
, Bn −→
An Bn x n
1 (α,b)n (β,b)n
Bn q −
n(n−1) 2
1 [n]! ,
we have
k(k−1) ∞ q 2 zn = (1 − abk [k])(−x)k [n]! [k]! k=0
×
×
k
[−k] p (a[ p], b)k B p (−zq) p (α, b) (β, b) (q, α, β, q) p p p p=0
∞ (α, b)n+k (β, b)n+k An+k x n . (a[n + k], b) [n]! k+1 n=0
(6.4)
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7. Application of identity (4.3) (I) For E = {0, 1, 2, . . .}, B = {0, 1, 2, . . .}, β j = b − j + 1, γ j = a − j + 1 and α j = µ j = 1, identity (4.3) implies ∞
n Π j=1 α j µ j x n yn
n=0 ∞
=
p
k (a + p − j + 1) Π j=1 (b + k − j + 1) Π j=1
R(ek+1 , E k )
k=0, p=0
R(b p+1 , B p )
xk y p
m+ p
∞
n+k Π j= p+1 (b + k − j + 1) Π j=k+1 (a + p − j + 1)
n=0,m=0
R(ek+1 , E m+k+1 \ E k+1 ) R(b p+1 , Bn+ p+1 \ B p+1 )
×
x n ym .
Since
b+k + k − j + 1) = (b + k) p = p! , p a+p k Π j=1 (a + p − j + 1) = (a + p)k = k! , k b+k− p m+ p Π j= p+1 (b + k − j + 1) = (b + k − p)m = m! , m a+ p−k n+k Π j=k+1 (a + p − j + 1) = (a + p − k)n = n! , n R(ek+1 , E k ) = k!, R(b p+1 , B p ) = p!,
p Π j=1 (b
R(ek+1 , E m+k+1 \ E k+1 ) = m!(−1)m , R(b p+1 , Bn+ p+1 \ B p+1 ) = n!(−1)n , then we obtain ∞
n x n yn Π j=1
n=0
∞ b+k p a+ p k = y x p k k=0, p=0 ∞ b+k− p a+ p−k (−x)n . (−y)m × n m n=0,m=0
∞ ω n 1 n ω n n On the other hand, we know that ∞ n=0 Π j=1 x y = 1−x y and that (1 + x) = n=0 n x . Therefore, ∞ 1 b+k p a+ p k = y x × (1 − x)a+ p−k (1 − y)b+k− p . 1 − xy p k k=0, p=0 Accordingly, ∞ 1 b+k a+ p 1−y k 1−x p = x y . p k 1−x 1−y (1 − x y) (1 − x)a (1 − y)b k=0, p=0
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Changing variables x and y as follows, x =
x 1+x
y 1+y ,
and y =
we get
k p ∞ (1 + x)a (1 + y)b b+k a+ p x y = . xy p k 1+y 1+x 1− k=0, p=0 (1+x)(1+y)
1+Y 1+X Taking x = X 1−X Y and y = Y 1−X Y , we finally derive the following formula
(1 + X )a (1 + Y )b (1 − X Y )a+b+1
∞ b+k a+ p k p X Y . = p k k=0, p=0
(7.1)
Similarly we proceed in the following cases. (II) For E = {0, 1, 2, . . .}, B = {0, 1, 2, . . .} and β j = a + j − 1, γ j = b + j − 1 and α j = µ j = 1, we obtain k p ∞ (a + k) p (b + p)k x y k! p! (1 + x)(1 + y) (1 + x)(1 + y) k=0, p=0 (1 + x)b (1 + y)a . (7.2) 1 − xy (III) For E = 1, q, q 2 , . . . , β j = 1 and α j = −aq j−1 , B = 1, q, q 2 , . . . , γ j = 1 and µ j = −bq j−1 , we get =
∞ (x, q)∞ (y, q)∞ q n(n−1) (bx)n (ay)n (bx, q)∞ (ay, q)∞ n=0 ∞
p( p−1)
k(k−1)
(−y) p q 2 (aq k , q) p (−x)k q 2 (bq p , q)k = . (q, q)k (ay, q)k (xq −k , q)k (q, q) p (bx, q) p (yq − p , q) p k=0, p=0
(7.3)
(IV) For E = {[0], [1], [2], . . .}, β j = 1 and α j = −at j−1 , B = {[0], [1], [2], . . .}, and γ j = 1, µ j = −bt j−1 , we have ∞
t n(n−1) (bx)n (ay)n =
n=0
∞
(a[k], t) p q − [k]! k=0, p=0
k(k−1) 2
(b[ p], t)k q − [ p]!
p( p−1) 2
xk y p
× V(a; t; p, k; −x)V(b; t; k, p; −y),
(7.4)
with V(a; t; p, k; −x) =
∞ (at p [k], t)n n=0
[n]!
(−x)n .
(V) For E = {[0], [1], [2], . . .}, β j = a + [ j − 1], α j = q j−1 , B = {[0], [1], [2], . . .}, γ j = b + [ j − 1], and µ j = q j−1 , we get ∞ n=0
q n(n−1) x n y n =
∞ k=0, p=0
q−
k(k−1) 2
{a + [k]} p [k]!
q−
p( p−1) 2
{b + [ p]}k k p x y [ p]!
× exp(a + [k + p]; −xq −k ) exp(b + [k + p]; −yq − p ),
(7.5)
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M. Kerada et al. / Indagationes Mathematicae 28 (2017) 711–720
with exp(a + [k]; x) =
∞ {a + [k]}n n=0
[n]!
xn.
References A. Abderrezzak, Quelques formules d. Inversion a` plusieurs variables, European J. Combin. 14 (1993) 507–512. A. Abderrezzak, G´en´eralisation de la transformation d’Euler d’une s´erie formelle, Adv. Math. 103 (1994) 180–195. W. Al-Salam, A. Verma, on quadratic transformations of basic series, SIAM J. Math. Anal. 15 (1984) 414–420. A. Boussayoud, A. Abderrezzak, M. Kerada, Some applications of symmetric functions, Integers 15 (2015) 1–7. J.L. Fields, J. Wimp, Exansions of hypergeometric functions in hypergeometric functions, Math. Comp. 15 (1961) 390–395. [6] I. Gessel, D. Stanton, Short proofs of Schaalsch¨utz’s and Dixon’s theorems, J. Combin. Theory Ser. A 38 (1985) 87–90. [7] F.H. Jackson, Transformations of q-series, Messenger Math. 39 (1910) 145–153. [8] A. Verma, Some transformations of series with arbitrary terms, Institulo Lombardo (Rend. Sci.) A 106 (1972) 342–353. [1] [2] [3] [4] [5]
ScienceDirect Indagationes Mathematicae 28 (2017) 711–720 www.elsevier.com/locate/indag
Generalization of some hypergeometric functions Mohamed Kerada a,∗ , Ali Boussayoud b , Abdelhamid Abderrezzak c a LMAM Laboratory and Department of Computer Science, University of MSB Jijel, BP 98 Ouled Aissa, Jijel 18000,
Algeria b LMAM Laboratory and Department of Mathematics, University of MSB Jijel, BP 98 Ouled Aissa, Jijel 18000, Algeria c University of Paris 7, LITP, Place Jussieu, Paris cedex 05, France
Received 14 November 2015; received in revised form 29 January 2017; accepted 24 March 2017 Communicated by F. Beukers
Abstract In this paper, we make use of the Lagrange interpolation to obtain some algebraic identities, involving one or two infinite sets of variables. c 2017 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved. ⃝
Keywords: Symmetric functions; Divided differences; Lagrange interpolation
1. Introduction One of the most important general expansion formulas for hypergeometric series is the Fields and Wimp expansion, described by [5]: ∞ (a R )n (α)n (β)n (−x)n a R , cT ; xw = r +t Fs+u r +2 Fs+1 b S, dU (b S )(γ + n)n n! n=0 n + α, n + β, n + a R × ;x 1 + 2n + γ , n + b S ∗ Corresponding author.
E-mail addresses: [email protected] (M. Kerada), [email protected] (A. Boussayoud), [email protected] (A. Abderrezzak). http://dx.doi.org/10.1016/j.indag.2017.03.002 c 2017 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved. 0019-3577/⃝
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−n, n + γ , × t+2 Fu+2 α, β
cT ;w , dU
where the contract notations (a)n , a R , (a R )n , and n + a R represent a(a + 1) . . . (a + n − 1), a1 , . . . , ar , (a R )n , (a1 )n (a2 )n . . . (ar )n , and n + a1 , . . . , n + ar , respectively. Verma showed in [8] that this formula is a special case of expansion (6.1) and derived the q-analog (6.2). On the other hand, Al-Salam and Verma [3] showed that Euler’s transformation formula ∞
an bn x n =
n=0
∞ x k (k) f (x)∆k a0 , (−1)k k! k=0
(1.1)
where f (x) = b0 + b1 x + b2 x 2 + · · · , and ∆k a0 =
k k (−1) j ak− j , j j=0
has bibasic extension (5.2). For their part, Gessel and Stanton [6] obtained the following generating function ∞ (1 + x)a (1 + y)b a+ p b+k k p = x y . k p (1 − x y)a+b+1 k=0, p=0
(1.2)
In this paper, we make use of the divided differences [2] and the Lagrange interpolation to obtain algebraic identities, involving one or two infinite sets of variables (formulas (4.1)–(4.3)). By specializing variables sets, we recover formulas ((6.1) and (6.2)) provided by Verma [8], formula (5.2) given by Al-Salem and Verma [3], and generating function (7.1) of Gessel and Stanton [6]. We also give a new q-analog of results ((5.3), (5.4), (6.3), (6.4), (7.3), (7.4), (7.5)). For this purpose, the Lagrange interpolation is considered to describe the properties of a linear operator, sending function of one variable to a symmetric function [4]. It can be written as a summation on a set, or a product of divided differences. This later version will be used throughout this paper. 2. Multiple interpolations In this section, we consider the linear operator Λ(A) of the Lagrange interpolation defined by Λ(A)( f ) =
aϵ A
f (a) R(a, A \ a)
(2.1)
where R(a, B) = Rbϵ B (a − b) and R(a, ∅) = 1. Accordingly, this operator sends a polynomial of degree k to a symmetric polynomial in A of degree k − n, with car d(A) = n + 1. In particular, it annihilates polynomials of degree < n, and f (x) = x n on constant 1.
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Let Λ(E n+1 ) be the Lagrange operator of an alphabet E n+1 := {e1 , e2 , . . . , en+1 }. We have [1] Λ(E n+1 )( f ) =
n k=0
f (ek+1 ) . R(ek+1 , E n+1 \ ek+1 )
(2.2)
For our purpose, we make use of the Lagrange interpolation for two independent alphabets and functions of two variables, described by Λ(E m+1 )Λ(Bn+1 )( f ) =
m n k=0 p=0
f (ek+1 , b p+1 ) . R(ek+1 , E m+1 \ ek+1 )R(b p+1 , Bn+1 \ b p+1 )
(2.3)
3. Inversion identities Consider the alphabet E n+1 \ E s = {es+1 , . . . , en+1 }, then, Lagrange interpolation (2.2) takes the following form Λ(E n+1 \ E s )( f ) =
n p=s
f (e p+1 ) . R(e p+1 , E p \ E s )R(e p+1 , E n+1 \ E p+1 )
(3.1)
In the case n = s, we have Λ(E s+1 \ E s )( f ) = Λ(es+1 )( f ) = f . Take f (es+1 ) =
n (β +a e Π j=1 j j s+1 )
s+1 Π j=1 (β j +a j es+1 )
in (3.1). The simple fact that the Lagrange operator Λ(E n+1 )
decreases the polynomial by n implies the following identities n Π j=1 (β j + a j es+1 ) = δn,s , (βs+1 + as+1 es+1 )Λ(E n+1 \ E s ) s+1 Π j=1 (β j + a j es+1 ) n Π j=1 α j µ j n m Λ(E m+1 )Λ(Bn+1 )Π j=1 (β j + α j e1 )Π j=1 (γ j + µ j b1 ) = 0
(3.2) if m = n if m ̸= n,
(3.3)
where δn,s is the Kronecker’s delta symbol. By combining formulas (2.2) and (3.2), we deduce the identity δn,s = (βs+1 + as+1 es+1 ) ×
n p=s
n (β + a e Π j=1 j j p+1 ) 1 . s+1 R(e p+1 , E p \ E s )R(e p+1 , E n+1 \ E p+1 ) Π j=1 (β j + a j e p+1 )
Let us consider two infinite upper triangular matrices A and B such that 1 (βs+1 + as+1 es+1 ) if p > s s+1 Asp = R(e p+1 , E p \ E s ) Π j=1 (β j + a j e p+1 ) 0 if p < s, n Π j=1 (β j + a j e p+1 ) if p ≤ n B pn = R(e p+1 , E n+1 \ E p+1 ) 0 if p > n.
(3.4)
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Then, relation (3.4) is simply the identity AB = I d∞ , where I d∞ is the infinite identity matrix. We have also n 1 1 δn, p = R(e , E \ E ) R(e , E n k n+1 p+1 k+1 \ E p+1 ) k= p ×
k (β + a e (βk+1 + ak+1 ek+1 )Π j=1 j j p+1 ) k+1 Π j=1 (β j + a j e p+1 )
.
(3.5)
Equality (3.5) is an explicit version of B A = I d∞ , which automatically follows from AB = I d∞ . 4. Principal formulas From formulas (3.4) and (3.5), we deduce the following theorem. Theorem 1. Let E = {e1, e2 , . . .} be a set of variables and {An }, {Bn }, {βn }, {an } four sequences. Then we have ∞
An Bn x n z n =
n=0
∞
p
Π j=1 (β j + a j e p+1 )x p
p=0
×
p s=0
× ∞
An Bn x n z n =
n=0
∞
1
n=0
R(e p+1 , E n+ p+1 \ E p+1 )
∞
xk
k=0
×
(βs+1 + as+1 es+1 ) 1 Bs z s s+1 R(e p+1 , E p \ E s ) Π j=1 (β j + a j e p+1 )
∞ n=0
n+ p
Π j= p+1 (β j + a j e p+1 )An+ p x n , (4.1)
k
(βk+1 + ak+1 ek+1 )
p=0
k (β + a e R(e p+1 , E k+1 \ E p+1 )Π j=1 j j p+1 )
Bpz p
1 1 An+k x n . k+1 R(en+k+1 , E n+k \ E k ) Π j=1 (β j + a j en+k+1 )
(4.2)
Denote by a the infinite row vector with components An x n and b the infinite row vector with components Bn z n . Then, identity (4.1) follows from the observation ∞
An Bn x n z n = b · at = bABat = (bA) · (Bat ),
n=0
and identity (4.2) from ∞
An Bn x n z n = b · at = bB Aat = (bB) · (Aat ).
n=0
On the other hand, formula (3.3) enables us to write this second theorem. Theorem 2. Let E = {e1, e2 , . . .}, B = {b1 , b2 , b3 , . . .} be two sets of variables, and {α} , {β} , {γ } , {µ} four sequences. Then we have ∞ n=0
n Π j=1 α j µ j x n yn
=
∞ k=0, p=0
p
k (γ + µ b Π j=1 (β j + α j ek+1 ) Π j=1 j j p+1 )
R(ek+1 , E k )
R(b p+1 , B p )
xk y p
M. Kerada et al. / Indagationes Mathematicae 28 (2017) 711–720
×
m+ p
∞
Π j= p+1 (β j + α j ek+1 )
n=0,m=0
R(ek+1 , E m+k+1 \ E k+1 )
×
715
n+k Π j=k+1 (γ j + µ j b p+1 )
R(b p+1 , Bn+ p+1 \ B p+1 )
x n ym .
(4.3)
5. Application of identity (4.1) (I) For E = {0, 1, 2, . . .}, β j = 1, and a j = −ab j−1 , identity (4.1) implies ∞
An Bn x n z n =
n=0
∞
p
Π j=1 (1 − ab j−1 p)x p
p=0
×
×
p s=0 ∞ n=0
(1 − abs s) 1 Bs z s s+1 R(e p+1 , E p \ E s ) Π j=1 (1 − ab j−1 p) 1
n+ p
R(e p+1 , E n+ p+1 \ E p+1 )
Π j= p+1 (1 − ab j−1 )An+ p x n .
Since we have p
Π j=1 (1 − ab j−1 p) = (abp, b) p−1 , s+1 Π j=1 (1 − ab j−1 p) = (abp, b)s , n+ p
n+1 Π j= p+1 (1 − ab j−1 ) = Π j=1 (1 − ab j+ p ) = (ab p p, b)n ,
R(e p+1 , E p \ E s ) = ( p − s)!, R(e p+1 , E n+ p+1 \ E p+1 ) = n!(−1)n , then we can derive the following relationship ∞
An Bn x n z n =
n=0
∞
(abp, b) p−1 x p
p=0
×
p s=0
∞ (ab p p, b)n
n!
n=0
1 − asbs Bs z s ( p − s)!(abp, b)s
An+ p (−x)n .
(5.1)
Similarly we proceed cases. in the remaining (II) For E = 1, q, q 2 , . . . , β j = 1, a j = −ab j−1 and by replacing An −→ (−1)n q
n(n−1) 2
An , Bn −→ (−1)n q − ∞
An Bn x n z n =
n=0
∞
n(n−1) 2
(abq p , b) p−1 x p
p=0
×
∞ n=0
with [n] =
1−q n 1−q .
Bn , we have p s=0
1 − aq s bs Bs z s (q, q) p−s (ab p q p , b)s
n(n−1) (ab p q p , b)n An+ p (−x)n q 2 , (q, q)n
(5.2)
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Remark 1. The case b = q is due to Jackson [7]. (III) For E = {[0], [1], [2], . . .}, β j = 1, a j = −ab j−1 and by replacing An −→ (−1)n q
n(n−1) 2
An , Bn −→ (−1)n q − ∞
An Bn x n z n =
∞
n(n−1) 2
Bn , we obtain
(ab[ p], b) p−1 x p
s=0
p=0
n=0
×
p
∞ n=0
1 − a[s]bs Bs z s [ p − s]!(ab p [ p], b)s
n(n−1) (ab p [ p], b)n An+ p (−x)n q 2 . [n]!
(5.3)
(IV) For E = {[0], [1], [2], . . .}, β j = γ + [ j − 1], a j = q j−1 and by replacing An −→ (−1)n q
n(n−1) 2
∞
An , Bn −→ (−1)n q −
An Bn x n z n =
n=0
∞
n(n−1) 2
Bn , we obtain
{γ + [ p + 1], b} p−1 x p
p=0
×
p {γ + [2s], b} p−1 s=0
×
[ p − s]!
{γ + [ p + 1], b}s Bs z s
∞ n(n−1) {γ + [2 p], b}n An+ p (−x)n q 2 , [n]! n=0
(5.4)
with {γ }k = γ (γ +[1])(γ +[2]) . . . (γ +[k −1]) and {a + [k]}n = (a +[k])(a +[k +1]) . . . (a + [k + n − 1]). 6. Application of identity (4.2) (I) For E = {0, 1, 2, . . .} , β j = γ + j − 1, a j = 1, and by replacing An −→ (α)n (β)n An and Bn −→ (α)n1(β)n Bn!n , identity (4.2) implies ∞
An Bn x n z n
n=0
=
∞
xk
k
k=0 p=0 ∞
×
n=0
(γ + k + j − 1) R(e p+1 , E k+1 \
k (γ E p+1 )Π j=1
+ p + j − 1)(α) p (β) p
1 (α)n+k (β)n+k An+k x n . k+1 R(en+k+1 , E n+k \ E k ) Π j=1 (γ + n + k + j)
Since we have k Π j=1 (γ + p + j − 1) = (γ + p)k , k+1 Π j=1 (γ + n + k + j) = (γ + n + k)k+1 ,
R(e p+1 , E k+1 \ E p+1 ) = (k − p)!, R(en+k+1 , E n+k \ E k ) = n!,
Bp
zp p!
M. Kerada et al. / Indagationes Mathematicae 28 (2017) 711–720
717
then we obtain ∞
An Bn x n
k=0
∞ k (−k) p (γ + k) p zn 1 (−x)k zp = Bp n! (γ + k)k k! p=0 (α) p (β) p p! k=0
×
∞ (α)n+k (β)n+k xn An+k . (γ + 2k + 1)n n! p=0
(6.1)
In the following cases, we proceed similarly. (II) For E = 1, q, q 2 , . . . , β j = 1, a j = −aq j−1 and by replacing An −→ (−1)n (α,q)n (β,q)n (a,q)n ∞
An q
n(n−1) 2
An Bn x n
n=0
(a,q)n , Bn −→ (−1)n (α) Bn q − n (β)n
n(n−1) 2
1 (q,q)n ,
we have
k ∞ −k , aq k , q) k(k−1) (q zn (−x)k p = q 2 B p (zq) p k (q, q)n (q, aq , q) (q, α, β, q) k p p=0 k=0
×
∞ (α, β; q)n+k An+k x n . 2k+1 ; q) (q, aq n n=0
(6.2)
(III) For E = {[0], [1], [2], . . .}, β j = γ + [ j − 1], a j = q j−1 and by replacing An −→ {α}n {β}n An q
n(n−1) 2
, Bn −→
1 {α}n {β}n
Bn q −
n(n−1) 2
1 [n]! ,
we obtain
k(k−1) ∞ zn (−x)k q 2 An Bn x = {γ + [k]}k [k]! [n]! n=0 k=0
∞
n
×
k [−k] {γ + [k]} (q −k , aq k , q) p p p
{α} p {β} p (q, α, β, q) p
p=0
×
∞ ({α}n+k {β}n+k , ; q)n+k n=0
{γ + [2k + 1]}n [n]!
B p (−zq) p
An+k x n .
(6.3)
(IV) For E = {[0], [1], [2], . . .}, α j = 1, a j = −ab j−1 and by replacing An −→ (α, b)n (β, b)n An q ∞ n=0
n(n−1) 2
, Bn −→
An Bn x n
1 (α,b)n (β,b)n
Bn q −
n(n−1) 2
1 [n]! ,
we have
k(k−1) ∞ q 2 zn = (1 − abk [k])(−x)k [n]! [k]! k=0
×
×
k
[−k] p (a[ p], b)k B p (−zq) p (α, b) (β, b) (q, α, β, q) p p p p=0
∞ (α, b)n+k (β, b)n+k An+k x n . (a[n + k], b) [n]! k+1 n=0
(6.4)
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7. Application of identity (4.3) (I) For E = {0, 1, 2, . . .}, B = {0, 1, 2, . . .}, β j = b − j + 1, γ j = a − j + 1 and α j = µ j = 1, identity (4.3) implies ∞
n Π j=1 α j µ j x n yn
n=0 ∞
=
p
k (a + p − j + 1) Π j=1 (b + k − j + 1) Π j=1
R(ek+1 , E k )
k=0, p=0
R(b p+1 , B p )
xk y p
m+ p
∞
n+k Π j= p+1 (b + k − j + 1) Π j=k+1 (a + p − j + 1)
n=0,m=0
R(ek+1 , E m+k+1 \ E k+1 ) R(b p+1 , Bn+ p+1 \ B p+1 )
×
x n ym .
Since
b+k + k − j + 1) = (b + k) p = p! , p a+p k Π j=1 (a + p − j + 1) = (a + p)k = k! , k b+k− p m+ p Π j= p+1 (b + k − j + 1) = (b + k − p)m = m! , m a+ p−k n+k Π j=k+1 (a + p − j + 1) = (a + p − k)n = n! , n R(ek+1 , E k ) = k!, R(b p+1 , B p ) = p!,
p Π j=1 (b
R(ek+1 , E m+k+1 \ E k+1 ) = m!(−1)m , R(b p+1 , Bn+ p+1 \ B p+1 ) = n!(−1)n , then we obtain ∞
n x n yn Π j=1
n=0
∞ b+k p a+ p k = y x p k k=0, p=0 ∞ b+k− p a+ p−k (−x)n . (−y)m × n m n=0,m=0
∞ ω n 1 n ω n n On the other hand, we know that ∞ n=0 Π j=1 x y = 1−x y and that (1 + x) = n=0 n x . Therefore, ∞ 1 b+k p a+ p k = y x × (1 − x)a+ p−k (1 − y)b+k− p . 1 − xy p k k=0, p=0 Accordingly, ∞ 1 b+k a+ p 1−y k 1−x p = x y . p k 1−x 1−y (1 − x y) (1 − x)a (1 − y)b k=0, p=0
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Changing variables x and y as follows, x =
x 1+x
y 1+y ,
and y =
we get
k p ∞ (1 + x)a (1 + y)b b+k a+ p x y = . xy p k 1+y 1+x 1− k=0, p=0 (1+x)(1+y)
1+Y 1+X Taking x = X 1−X Y and y = Y 1−X Y , we finally derive the following formula
(1 + X )a (1 + Y )b (1 − X Y )a+b+1
∞ b+k a+ p k p X Y . = p k k=0, p=0
(7.1)
Similarly we proceed in the following cases. (II) For E = {0, 1, 2, . . .}, B = {0, 1, 2, . . .} and β j = a + j − 1, γ j = b + j − 1 and α j = µ j = 1, we obtain k p ∞ (a + k) p (b + p)k x y k! p! (1 + x)(1 + y) (1 + x)(1 + y) k=0, p=0 (1 + x)b (1 + y)a . (7.2) 1 − xy (III) For E = 1, q, q 2 , . . . , β j = 1 and α j = −aq j−1 , B = 1, q, q 2 , . . . , γ j = 1 and µ j = −bq j−1 , we get =
∞ (x, q)∞ (y, q)∞ q n(n−1) (bx)n (ay)n (bx, q)∞ (ay, q)∞ n=0 ∞
p( p−1)
k(k−1)
(−y) p q 2 (aq k , q) p (−x)k q 2 (bq p , q)k = . (q, q)k (ay, q)k (xq −k , q)k (q, q) p (bx, q) p (yq − p , q) p k=0, p=0
(7.3)
(IV) For E = {[0], [1], [2], . . .}, β j = 1 and α j = −at j−1 , B = {[0], [1], [2], . . .}, and γ j = 1, µ j = −bt j−1 , we have ∞
t n(n−1) (bx)n (ay)n =
n=0
∞
(a[k], t) p q − [k]! k=0, p=0
k(k−1) 2
(b[ p], t)k q − [ p]!
p( p−1) 2
xk y p
× V(a; t; p, k; −x)V(b; t; k, p; −y),
(7.4)
with V(a; t; p, k; −x) =
∞ (at p [k], t)n n=0
[n]!
(−x)n .
(V) For E = {[0], [1], [2], . . .}, β j = a + [ j − 1], α j = q j−1 , B = {[0], [1], [2], . . .}, γ j = b + [ j − 1], and µ j = q j−1 , we get ∞ n=0
q n(n−1) x n y n =
∞ k=0, p=0
q−
k(k−1) 2
{a + [k]} p [k]!
q−
p( p−1) 2
{b + [ p]}k k p x y [ p]!
× exp(a + [k + p]; −xq −k ) exp(b + [k + p]; −yq − p ),
(7.5)
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with exp(a + [k]; x) =
∞ {a + [k]}n n=0
[n]!
xn.
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