Bivariate generating functions for Rogers–Szegö polynomials

Applied Mathematics and Computation 217 (2010) 2209–2216

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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

Bivariate generating functions for Rogers–Szegö polynomials Jian Cao Department of Mathematics, Hangzhou Normal University, Hangzhou City, Zhejiang Province, 310036, PR China

a r t i c l e

i n f o

a b s t r a c t In the present paper, we utilize the general q-exponential operators to derive several Carlitz type bivariate generating functions for Rogers–Szegö polynomials. Moreover, we give an equivalent expansion formula of a certain bivariate generating functions for Rogers–Szegö polynomials and propose an open problem. Ó 2010 Elsevier Inc. All rights reserved.

Keywords: Rogers–Szegö polynomials Bivariate generating function Hahn polynomials General q-exponential operator Operator identity

1. Introduction The Rogers–Szegö polynomials [6,18]

hn ðxjqÞ ¼

n   X n k x k k¼0

and g n ðxjqÞ ¼

n   X n k¼0

k

qkðknÞ xk ¼ hn ðxjq1 Þ;

ð1:1Þ

are closely related to the continuous q-Hermite polynomials via [17]

Hn ðcos hjqÞ ¼ einh hn ðe2ih jqÞ:

ð1:2Þ

The bivariate Rogers–Szegö polynomials [10] are defined by

hn ðx; yjqÞ ¼

n   n   X X n n ðkþ1Þkn Pk ðx; yÞ and g n ðx; yjqÞ ¼ q 2 Pk ðy; xÞ; k k k¼0 k¼0

ð1:3Þ

where hn(x, yjq) = gn(x, yjq1) and the Eulerian family of polynomials Pn(b, a) is defined by [3, p. 347] n

Pn ðb; aÞ ¼ b ða=b; qÞn ¼ ðb  aÞðb  aqÞ    ðb  aqn1 Þ:

ð1:4Þ

The Hahn polynomials [2,7,19] are defined by

/ðaÞ n ðxjqÞ ¼

n   X n ða; qÞk xk k k¼0

and wðaÞ n ðxjqÞ ¼

n   X n kðknÞ k q x ðaq1k ; qÞk : k k¼0

ð1:5Þ

In fact, the bivariate Rogers–Szegö polynomials are equivalent to the Hahn polynomials via hn ðx; axjqÞ ¼ /ðaÞ n ðxjqÞ and g n ðx; axjqÞ ¼ wðaÞ n ðxjqÞ. The q-difference operator Dq and the q-shifted operator g are defined by

Dq ff ðaÞg ¼

f ðaÞ  f ðaqÞ a

and gff ðaÞg ¼ f ðaqÞ:

Chen and Liu [8,9] introduced h = g1Dq and the q-exponential operators E-mail addresses: [email protected], [email protected] 0096-3003/$ - see front matter Ó 2010 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2010.07.021

ð1:6Þ

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J. Cao / Applied Mathematics and Computation 217 (2010) 2209–2216

EðbhÞ ¼

n 1 X qð2Þ ðbhq Þn ; ðq; qÞn n¼0

TðbDq Þ ¼

1 X n¼0

1 ðbDq Þn ; ðq; qÞn

ð1:7Þ

then they obtained many useful operator identities, please refer to [8,9,16]. In [21,22], the authors deduced that: Proposition 1. For n 2 N, we have

 n    q ; q=ðasÞ Eðdha Þ an ðas; qÞ1 ¼ an ðas; ds; qÞ12 /1 ; q; ds ; 0    k n   n n X n a a d TðdDa Þ ¼ ; maxfjatj; jdtjg < 1; ðat; qÞk a ðat; qÞ1 ðat; dt; qÞ1 k¼0 k

ð1:8Þ ð1:9Þ

where for the definitions of Da and ha operators, refer to the next section. The authors [11,12] constructed the following more general q-exponential operators 1 X ða; qÞn n n b Dq ,Tða; bDq Þ; ðq; qÞn n¼0  X 1 a ða; qÞn ; q; bh ¼ ðbÞn hn ,Eða; bhÞ; 1 U0 ðq; qÞ  n n¼0

Tða; b; Dq Þ ¼

ð1:10Þ ð1:11Þ

and further generalized the corresponding results in [8,9]. For more information, please refer to [11,12]. Comparing to Proposition 1, we give the following more general operator identities. Theorem 2. For n 2 N and jbtj < 1, we have

 n    ðct; abt; qÞ1 cn q ; a; q=ðctÞ ; q; q : Eða; bhc Þ cn ðct; qÞ1 ¼ 3 /2 ðbt; qÞ1 0; q=ðbtÞ

ð1:12Þ

Theorem 3. For n 2 N and maxfjbtj; jctjg < 1, we have

 n   n q ; a; ct cn ðabt; qÞ1 cn bq ; ¼ ; q; Tða; bDc Þ 3 /1 ðct; qÞ1 ðbt; ct; qÞ1 c abt

ð1:13Þ

Remark 4. Replacing a by a/b, then letting b ? 0, formula (1.12) reduces to (1.8) and (1.13) becomes (1.9) by setting a = 0. When n = 0, formula (1.12) and (1.13) reduce to (2.8) and (2.9), respectively, see Lemma 15 below. The results of Carlitz type generating functions for Hahn polynomials are usually rewritten as the sum of two terms (see Propositions 19 and 20 below), see also [13], however, we derive another expression here by Theorems 2 and 3. In the last section, we point out that these two forms are equivalent. Proposition 5. For k 2 N and maxfjutj; jxtj; jtj; juxtjg < 1, we have

" # k   k ðy=x; ut; t; qÞj xj tn ðv t; yt; qÞ1 X v =u; tqj ; yqj =x hnþk ðx; yjqÞhn ðu; v jqÞ ¼ ; q; uxt : 3 /2 ðq; qÞn ðut; xt; t; qÞ1 j¼0 j ðv t; yt; qÞj v tqj ; ytqj n¼0

1 X

ð1:14Þ

Proposition 6. For k 2 N and maxfjv tqk j; jytqk jg < 1, we have 1 X n ð1Þn qð2Þ g nþk ðx; yjqÞg n ðu; v jqÞ n¼0

" # k ðqk ;x=y; q=t;q=ðutÞ; qÞj qu j u=v ;xqj =y; q1þj =t tn ðut; xt; t; qÞ1 X ;q;q : ¼ / ðq; qÞn ðv t; yt;qÞ1 j¼0 ðq; q=ðv tÞ;q=ðytÞ;qÞj v 3 2 q1þj =ðv tÞ;q1þj =ðytÞ ð1:15Þ

Remark 7. For k = 0, formula (1.14) and (1.15) reduce to q-Mehler formula of generating functions for bivariate Rogers– Szegö polynomials (or Hahn polynomials) [2, Eqs. (1.17) and (1.18)], respectively, see also [10, Eq. (2.1)] and [11, Eq. (5.8)]. There are many clever proofs of the generating functions for Rogers–Szegö polynomials, such as the moments method [1,15], the method of transformation theory [7,19] and the combinatorial method [14], which similar to the generating functions for Hahn polynomials, please refer to [2,19,20]. In the previous paper [5], we have shown how to use q-exponential operators to deduce generating functions for Rogers– Szegö polynomials in view of TðDx Þfxn g ¼ hn ðxjqÞ and Eðhx Þfxn g ¼ g n ðxjqÞ. For more information, please refer to [4,5,16].

J. Cao / Applied Mathematics and Computation 217 (2010) 2209–2216

2211

In this paper, we will use the general q-exponential operators to prove the generating functions for bivariate Rogers– Szegö polynomials, the reason are based on the following facts:

lim Tðy=x; xDc Þfcn g ¼ hn ðx; yjqÞ; c!1

lim Eðx=y; yhc Þfcn g ¼ g n ðx; yjqÞ: c!1

ð1:16Þ

Furthermore we deduce some Carlitz type generating functions as follows. Theorem 8. For maxfjxtj; jtj; jtzj; jxtzjg < 1, we have 1 X

hkþn ðx; yjqÞhn ðzjqÞ

n;k¼0

  k y=x; t; s=ðtzÞ ð1Þk qð2Þ sk tn ðyt; s; qÞ1 ; q; xtz : ¼ / 3 ðq; qÞk ðq; qÞn ðxt; t; tz; qÞ1 2 yt; s

ð1:17Þ

Theorem 9. For maxfjsj; jytjg < 1, we have 1 X

g kþn ðx; yjqÞg n ðzjqÞ

n;k¼0

  n tz=s; x=y; q=t ð1Þn qð2Þ t n sk ðxt; t; tz; qÞ1 ¼ ; q; q : 3 /2 ðq; qÞk ðq; qÞn ðs; yt; qÞ1 q=ðytÞ; q=s

ð1:18Þ

Theorem 10. For maxfjsj; jtj; jszj; jtxj; jtzj; jxtzjg < 1, we have 1 X

hn ðx; yjqÞhnþk ðzjqÞ

n;k¼0

  y=x; t; s tn sk ðyt; stz; qÞ1 ¼ ; q; xtz : 3 /2 ðq; qÞk ðq; qÞn ðs; t; sz; tx; tz; qÞ1 yt; stz

ð1:19Þ

Theorem 11. For maxfjytj; jstz=qjg < 1, we have 1 X

g n ðx; yjqÞg nþk ðzjqÞ

n;k¼0

  k n x=y; q=t; q=s ðtÞn ðsÞk qð2Þþð2Þ ðs; t; sz; tx; tz; qÞ1 ¼ ; q; q : 3 /2 ðq; qÞk ðq; qÞn ðyt; stz=q; qÞ1 q=ðytÞ; q2 =ðstzÞ

ð1:20Þ

Corollary 12. [11. Eq. (2.7)]. For maxfjxztj; jxtj; jtj; jztjg < 1, we have 1 X

hn ðx; yjqÞhn ðzjqÞ

n¼0

  y; xt tn ðyt; qÞ1 ; q; zt : ¼ / 2 ðq; qÞn ðxzt; xt; t; qÞ1 1 yt

ð1:21Þ

Corollary 13. For maxfjytj; jxztjg < 1, we have 1 X

g n ðx; yjqÞg n ðzjqÞ

n¼0

  n n 1=y; q=ðxtÞ ð1Þn qð2Þt ðxzt; xt; t; qÞ1 ¼ ; q; txz : 2 /1 ðq; qÞn ðyt; qÞ1 q=ðytÞ

ð1:22Þ

The structure of this paper is organized as follows. In Section 2, some notations and proofs of Theorems 2 and 3 are presented. In Section 3, proofs of Theorems 8–11 and Propositions are given. In Section 4, an additional proof of Proposition 5 is given and an open problem is proposed.

2. Notations and proofs of Theorems 2 and 3 In this paper, we follow the notations and terminology in [13] and suppose that 0 < q < 1. The q-series and its compact factorials are defined respectively by

ða; qÞ0 ¼ 1;

ða; qÞn ¼

n1 Y ð1  aqk Þ; k¼0

ða; qÞ1 ¼

1 Y ð1  aqk Þ

ð2:1Þ

k¼0

and (a1, a2, . . . , am; q)n = (a1;q)n(a2;q)n. . .(am;q)n, where m is a positive integer and n is a non-negative integer or 1. The operators Dq and h acting on the variable a will be denoted by Da and ha, respectively. LHS (or RHS) means the left (or right) hand side of certain equality, and N ¼ f0; 1; 2; . . .g. The basic hypergeometric series r/s [13, Eq. (1.2.22)] is given by

"

r /s

a1 ; . . . ; ar b1 ; . . . ; bs

#

; q; z ¼

1 X

sþ1r ða1 ; a2 ; . . . ; ar ; qÞn n z ð1Þn qnðn1Þ=2 ; ðq; b1 ; . . . ; bs ; qÞn n¼0

ð2:2Þ

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such that the infinite series in (2.2) is convergent for either jqj < 1 and jzj < 1 when r 6 s or jqj < 1 and jzj < 1 when r = s + 1, provided that no zero appears in the denominator. Recall Heine’s q-Euler transformation [13, Eq. (III.3)]

 2 /1

a; b c

   c=a; c=b ðabz=c; qÞ1 abz ; q; z ¼ ; q; ; 2 /1 c ðz; qÞ1 c

maxfjzj; jabz=cjg < 1:

ð2:3Þ

The q-Chu–Vandermonde formula [13, Eqs. (II.6) and (II.7)] reads that

 2 /1

qn ; b c

 ðc=b; qÞn n ; q; q ¼ b ðc; qÞn

and

 2 /1

qn ; b c

; q;

 cqn ðc=b; qÞn : ¼ b ðc; qÞn

ð2:4Þ

We also need the three-term 3/2 transformation [13, Eq. (III.34)]

 3 /2

a; b; c d; e

; q;

     d=a; b; c e=b; e=c; de=ðabcÞ de ðe=b; e=c; qÞ1 ðd=a; b; c; de=ðbcÞ; qÞ1 ¼ ; q; q þ ; q; q ; 3 /2 3 /2 abc ðe; e=ðbcÞ; qÞ1 ðd; e; bc=e; de=ðabcÞ; qÞ1 d; bcq=e de=ðbcÞ; eq=ðbcÞ

ð2:5Þ

where jde/(abc)j < 1, and bc/e is not an integer power of q. In order to proof Theorems 2 and 3, the following lemmas are necessary. Lemma 14 (Leibnize formula). For n 2 N, we have n   X n k k h ff ðaÞghnk a fgðaq Þg; k a k¼0   n X n k k qkðknÞ D ff ðaÞgDnk Dna ff ðaÞgðaÞg ¼ a fgðaq Þg: k a k¼0

hna ff ðaÞgðaÞg ¼

ð2:6Þ ð2:7Þ

Lemma 15 [11, Eq. (2.3)] and [12, Eq. (10)]. We have

ðct; abt; qÞ1 Eða; bhc Þfðct; qÞ1 g ¼ ; jbtj < 1; ðbt; qÞ1   1 ðabt; qÞ1 ¼ ; maxfjbtj; jctjg < 1: Tða; bDc Þ ðct; qÞ1 ðbt; ct; qÞ1

ð2:8Þ ð2:9Þ

Proof of Theorem 2. In light of (2.6), LHS of (1.12) equals 1 k   X X  k j n kj  ða; qÞk hc fc ghc ðctqj ; qÞ1 ðbÞk ðq; qÞ j k j¼0 k¼0 1 k     X X  jþ1 k ða; qÞk ðq; qÞn cnj kj  q 2 nj ¼ ðbÞk hc ðctqj ; qÞ1 ðq; qÞ ðq; qÞ j k nj j¼0 k¼0 1     1 X X   jþ1 n ða; qÞ k q 2 nj cnj ðbÞk hkj ðctqj ; qÞ1 ¼ c ðq; qÞ j kj j¼0 k¼j n     X   jþ1 n q 2 nj cnj ða; qÞj ðbÞj Eðaqj ; bhc Þ ðctqj ; qÞ1 ; ¼ j j¼0

which is RHS of (1.12) by (2.8) and simplification. This completes the proof. Proof of Theorem 3. LHS of (1.12) is equal to

( ) 1 k   X k jðjkÞ ðq; qÞn nj kj ða; qÞk k X 1 q b c Dc ðq; qÞk ðq; qÞnj j ðctqj ; qÞ1 j¼0 k¼0 ( ) n   1 X n nj X qjðjkÞ 1 k kj ¼ ða; qÞk b Dc c ðq; qÞkj j ðctqj ; qÞ1 j¼0 k¼j ( )   n X n nj 1 j j ; c ða; qÞj b Tðaqj ; bq Dc Þ ¼ j ðctqj ; qÞ1 j¼0 which becomes RHS of (1.13) by (2.7) and (2.9). This achieves the proof.

h

h

J. Cao / Applied Mathematics and Computation 217 (2010) 2209–2216

2213

3. Proof of Propositions and Theorems 8–11 In this section, we will prove above results, the following lemmas are necessary. Lemma 16 [11, Eq. (2.8)]. We have

    a; cs; w=t ðcw; qÞ1 ðabs; cw; qÞ1 ¼ ; q; bt ; Tða; bDc Þ 3 /2 ðcs; ct; qÞ1 ðbs; cs; ct; qÞ1 abs; cw

ð3:1Þ

where maxfjcsj; jctj; jbsj; jbtjg < 1. For maxfjcwj; jbtjg < 1, we have [12, Eq. (6)]

Eða; bhc Þ

    s=w; a; q=ðctÞ ðcs; ct; qÞ1 ðabt; cs; ct; qÞ1 ¼ ; q; q : / 3 2 ðcw; qÞ1 ðcw; bt; qÞ1 q=ðbtÞ; q=ðcwÞ

ð3:2Þ

Lemma 17 [6, p. 366]. We have 1 X m;n¼0 1 X

hmþn ðxjqÞ

sm tn ðstx; qÞ1 ¼ ; ðq; qÞm ðq; qÞn ðs; sx; t; tx; qÞ1 m

n

g mþn ðxjqÞð1Þmþn qð 2 Þþð2Þ

m;n¼0

maxfjsj; jsxj; jtj; jtxjg < 1;

sm t n ðs; sx; t; tx; qÞ1 ¼ ; ðq; qÞm ðq; qÞn ðstx=q; qÞ1

ð3:3Þ

jstx=qj < 1:

ð3:4Þ

Lemma 18. For k 2 N, we have

" # qk ; y=x; t tn ðyt; qÞ1 n ; q; xq ; maxfjtj; jxtjg < 1; ¼ / 3 ðq; qÞn ðxt; t; qÞ1 1 yt n¼0 " # 1 X qk ; x=y; q=t tn ðxt; t; qÞ1 n ðn2Þ ; q; q ; jytj < 1: ð1Þ q g nþk ðx; yjqÞ ¼ 3 /2 ðq; qÞn ðyt; qÞ1 0; q=ðytÞ n¼0

1 X

hnþk ðx; yjqÞ

ð3:5Þ ð3:6Þ

Proof. By (1.16) and (1.12), LHS of (3.5) equals

" #   qk ; y=x; ct ck ðyt; qÞ1 ck xqn ; ; q; ¼ lim lim Tðy=x; xDc Þ / 3 1 c!1 c!1 ðxt; ct; qÞ1 ðct; qÞ1 c yt which is RHS of (3.5). Similarly, we obtain (3.6) by (1.13). The proof is complete. Now we are in a position to prove main results.

h

Proof of Proposition 5. According to (1.16), LHS of (1.14) is equivalent to

) ðctÞn c!1 ðq; qÞn n¼0 ( " k #) q ; y=x; ct ðyct; qÞ1 k ¼ lim Tðv =u; uDc Þ ; q; xq / 3 c!1 ðxct; ct; qÞ1 1 yct ( ) " # k n X ðyctqj ; qÞ1 ðy=x; qÞj xj lim Tðv =u; uDc Þ ¼ c!1 ðxct; ctqj ; qÞ1 k j¼0 " # " # k n X v =u; ctqj ; yqj =x ðv tqj ; yctqj ; qÞ1 / ; q; uxt ; ðy=x; qÞj xj lim ¼ 3 2 c!1 ðutqj ; xct; ctqj ; qÞ k v tqj ; yctqj 1 j¼0

lim Tðv =u; uDc Þ

( 1 X

hnþk ðx; yjqÞ

which equals RHS of (1.14). This completes the proof of Proposition 5.

h

ð3:7Þ

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J. Cao / Applied Mathematics and Computation 217 (2010) 2209–2216

Proof of Proposition 6. LHS of (1.15) equals

( ) 1 X n ðctÞn ð1Þn qð2Þ g nþk ðx; yjqÞ c!1 ðq; qÞn n¼0 ( " #) k q ; x=y; q=ðctÞ ðct; xct; qÞ1 ; q; q / ¼ lim Eðu=v ; v hc Þ 3 2 c!1 ðyct; qÞ1 0; q=ðyctÞ ( ) j k k X ðq ; x=y; qÞj ðqyÞ ðctqj ; xct; qÞ1 lim Eðu=v ; v hc Þ ¼ c!1 ðq; qÞj ðyctqj ; qÞ1 j¼0 ( " #) j k X ðqk ; x=y; qÞj ðqyÞ u=v ; xqj =y; q1þj =ðctÞ ðctqj ; xct; utqj ; qÞ1 lim / ; q; q ; ¼ 3 2 c!1 ðq; qÞj ðyctqj ; v tqj ; qÞ1 q1þj =ðv tÞ; q1þj =ðyctÞ j¼0

lim Eðu=v ; v hc Þ

which is equivalent to RHS of (1.15). This achieves the proof of Proposition 6.

h

Proof of Theorem 8. LHS of (1.17) is equal to

( lim Tðy=x; xDc Þ c!1

1 X

k

c

nþk

n;k¼0

ð1Þk qð2Þ sk t n hn ðzjqÞ ðq; qÞk ðq; qÞn

) ¼ lim Tðy=x; xDc Þ c!1



 ðcs; qÞ1 ; ðctz; ct; qÞ1

which becomes RHS of (1.17) by letting (a, b, w, s, t) = (y/x, x, s, t, tz) in (3.1). The proof is complete.

h

Proof of Theorem 9. LHS of (1.18) is equivalent to

( lim Eðx=y; yhc Þ c!1

1 X

n

cnþk g n ðzjqÞ

n;k¼0

ð1Þn qð2Þ sk tn ðq; qÞk ðq; qÞn

) ¼ lim Eðx=y; yhc Þ c!1

  ðctz; ct; qÞ1 ; ðcs; qÞ1

which reduces to RHS of (1.18) by setting (a, b, w, s, t) = (x/y, y, s, tz, t) in (3.2). The proof is ended.

h

Proof of Theorem 10. LHS of (1.19) is equal to

(

1 X

sk t n lim Tðy=x; xDc Þ c hnþk ðzjqÞ c!1 ðq; qÞk ðq; qÞn n;k¼0 n

)

  ðcstz; qÞ1 ¼ lim Tðy=x; xDc Þ c!1 ðctz; ct; sz; s; qÞ1   1 ðcstz; qÞ1 ; lim Tðy=x; xDc Þ ¼ ðs; sz; qÞ1 c!1 ðctz; ct; qÞ1

which becomes RHS of (1.19) by letting (a, b, w, s, t) = (y/x, x, stz, t, tz) in (3.1). We complete the proof.

h

Proof of Theorem 11. LHS of (1.20) equals

(

1 X

ðsÞk ðtÞn c!1 ðq; qÞk ðq; qÞn n;k¼0   ðctz; ct; qÞ1 ; ¼ ðs; sz; qÞ1 lim Eðx=y; yhc Þ c!1 ðcstz=q; qÞ1

lim Eðx=y; yhc Þ

k

n

cn g nþk ðzjqÞqð2Þþð2Þ

)

  ðctz; ct; sz; s; qÞ1 ¼ lim Eðx=y; yhc Þ c!1 ðcstz=q; qÞ1

which is RHS of (1.20) by setting (a, b, w, s, t) = (x/y, y, stz/q, tz, t) in (3.2). We achieve the proof.

h

Proof of Corollaries 12 and 13. For s = 0, Theorems 8 and 10 reduce to 1 X

hn ðx; yjqÞhn ðzjqÞ

n¼0

  y=x; t tn ðyt; qÞ1 ; q; xtz ; ¼ / 2 ðq; qÞn ðxt; t; tz; qÞ1 1 yt

ð3:8Þ

and Theorems 9 and 11 become 1 X n¼0

g n ðx; yjqÞg n ðzjqÞ

  n x=y; q=t ðtÞn qð2Þ ðxt; t; tz; qÞ1 ¼ ; q; tz : 2 /1 ðq; qÞn ðyt; qÞ1 q=ðytÞ

By using (2.3), formula (3.8) and (3.9) become (1.21) and (1.22), respectively. Therefore we conclude the proof.

ð3:9Þ h

J. Cao / Applied Mathematics and Computation 217 (2010) 2209–2216

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4. Additional proof and an open problem For the importance of generating functions, we give the equivalent proof of Proposition 5 and propose an open problem in this section. Verma and Jain defined the more general polynomial [20, Eq. (1.10)]

Xn ðx; a; bjqÞ ¼

n   X n r¼0

r

ða; qÞr ðb; qÞnr xr ;

ð4:1Þ

and studied its generating functions systematically, in which including the following Carlitz type generating functions. For more information, please refer to [20]. Proposition 19 [20, Eq. (2.8)]. For s 2 N, we have 1 X n¼0

zn Xnþs ðx; a; bjqÞXnþm ðy; c; djqÞ ðq; qÞn ðab; qÞnþs

" # m m a; b=x; yz; zqr ðb; ax; d; cyz; dzq ; qÞ1 X ðqm ; c; z; qÞr yq r q1þs ¼ ; q; 4 /3 m m x ðx; ab; z; yz; dq ; qÞ1 r¼0 ðq; q1m =d; cyz; qÞr d q=x; cyzqr ; dzq m m xs ða; b=x; d; cxyz; dxzq ; qÞ1 X ðqm ; c; xz; qÞr yq r m ð1=x; ab; xz; xyz; dq ; qÞ1 r¼0 ðq; q1s =d; cxyz; qÞr d " # ax; b; xyz; xzqr q1þs  4 /3 ; q; : m x xq; cxyzqr ; dxzq

þ

ð4:2Þ

Proposition 20 [20, Eq. (4.5)]. We have n 1 X ð1Þn ðatÞnþs qð2Þ ðaÞ wnþs ðxjqÞwnðbÞ ðyjqÞ ðq; qÞ n n¼0

" # 1=a; 1=ðaxÞ; byt t s ðat; axt; q=ðaxtÞ; q=ðaxÞ; qÞ1 1þs ¼ 2 ; q; q 3/ ða xt; q=ða2 xtÞ; q=t; byt; ;qÞ1 2 t; yt " # q=ðatÞ; q=ðaxtÞ; byq qs ðat; axt; 1=a; 1=ðaxÞ; yq; qÞ1 1þs : þ ; q; q 3 /2 ðt=q; a2 xt; q=ða2 xtÞ; byq; qÞ1 q2 =t; yq

ð4:3Þ

In fact, the Hahn polynomial is a special case of the Verma–Jain polynomial via /nðaÞ ðxjqÞ ¼ Xn ðx; a; 0jqÞ, so if letting (a, b, c, d, y, z, m) = (y/x, 0, v/u, 0, u, t, 0) in (4.2), we have another form of Proposition 5 as follows. Proposition 21. For k 2 N, we have 1 X

hnþk ðx; yjqÞhn ðu; v jqÞ

n¼0

    y=x; t; ut y; xt; uxt tn ðy; v t; qÞ1 xk ðy=x; v xt; qÞ1 ; q; q1þk þ ; q; q1þk : ¼ 3 /2 3 /2 ðq; qÞn ðx; t; ut; qÞ1 ð1=x; uxt; xt; qÞ1 q=x; v t xq; v xt

ð4:4Þ

For the main purpose of this section, we will deduce (4.4) from RHS of (1.14). Proof of (4.4) Setting (a, b, c, d, e) = (v/u, tqj, yqj/x, vtqj, ytqj) in (2.5), RHS of (1.14) equals

" # (  ) k   y;xt;uxt k ðy=x;ut;t;qÞj xj ðv t;yt;qÞ1 X ðy;xt;qÞ1 ðutqj ;tqj ;yqj =x;xv t;qÞ1 utqj ;tqj ;yqj =x ;q;q þ ;q;q : 3 /2 3 /2 ðut;xt;t;qÞ1 j¼0 j ðv t;yt;qÞj xv t;xq1j ðytqj ;xqj ;qÞ1 ðv tqj ;ytqj ;qj =x;uxt;qÞ1 v tqj ;qjþ1 =x ð4:5Þ

The proof is divided into two parts, the first term of (4.5) is equal to k X 1 ðy=x; ut; t; qÞlþj ðkþ1Þjþl ðqk ; qÞj ðv t; yt; qÞ1 X q ðut; xt; t; qÞ1 j¼0 l¼0 ðv t; q=x; qÞlþj ðq; qÞj ðq; qÞl 1 m ðqk ; qÞj ðq; qÞm mþkj ðv t; yt; qÞ1 X ðy=x; ut; t; qÞm X q ðut; xt; t; qÞ1 m¼0 ðq; v t; q=x; qÞm j¼0 ðq; qÞj ðq; qÞmj " # 1 qm ; qk ðv t; yt; qÞ1 X ðy=x; ut; t; qÞm m ; q; qmþk : ¼ q 2 /0 ðut; xt; t; qÞ1 m¼0 ðq; v t; q=x; qÞm 

¼

ð4:6Þ

2216

J. Cao / Applied Mathematics and Computation 217 (2010) 2209–2216

Letting c ? 1 in the second formula of (2.4), we have

 2 /0

qn ; b 

; q;

 qn n ¼b : b

So (4.6) is equivalent to

  y=x; t; ut ðy; v t; qÞ1 ; q; q1þk : 3 /2 ðx; t; ut; qÞ1 q=x; v t On the other hand, the second term of (4.5) equals

  k   y; xt; uxt k j ðy=x; xv t; qÞ1 X x ð1=x; qÞj 3 /2 ; q; q ðxt; 1=x; uxt; qÞ1 j¼0 j xv t; xq1j 1 k   k ð1=x; qÞj xj ðy=x; xv t; qÞ1 X ðy; xt; uxt; qÞl l X q ¼ ðxt; 1=x; uxt; qÞ1 l¼0 ðq; xv t; qÞl j ðxq1j ; qÞl j¼0 " # 1 ðy=x; xv t; qÞ1 X ðy; xt; uxt; qÞl l 1 qk ; ðxql Þ1 kþl ; q; xq ¼ q 2/ ðxq; qÞl 0  ðxt; 1=x; uxt; qÞ1 l¼0 ðq; xv t; qÞl   y; xt; uxt ðy=x; xv t; qÞ1 xk ¼ ; q; q1þk : 3 /2 ðxt; 1=x; uxt; qÞ1 xv t; xq Hence, we obtain RHS of (4.4). The proof is complete.

h

Open Problem 22. Can we prove the equivalence of Propositions 6 and 20?. In other words, how to prove the following result

  k X ðqk ; x=y; q=t; q=ðutÞ; qÞj qu j u=v ; xqj =y; q1þj =t ; q; q 3 /2 q1þj =ðv tÞ; q1þj =ðytÞ ðq; q=ðv tÞ; q=ðytÞ; qÞj v j¼0    k ðv t; q=ðxtÞ; q=t; qÞ1 xt x=y; 1=y; xt v =y 1þk / ; q; q ¼ 3 2 xt=y; xtu=y ðut; q=ðytÞ; yq=ðxtÞ; xt v =y; qÞ1 y   qk ðx=y; 1=y; v t; uq; qÞ1 q=t; q=ðxtÞ; v q þ ; q; q1þk : 3 /2 2 yq =ðxtÞ; uq ðut; v q; xt=ðyqÞ; q=ðytÞ; qÞ1

ð4:7Þ

Acknowledgements The author would like to thank the referee for valuable comments and corrections. The work was supported by PCSIRT and Innovation Program of Shanghai Municipal Education Commission. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22]

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