Sum of Interpolated Multiple Q-Zeta Values

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1 Introduction/Main theorem

l k

For any multi-index k =(k1,k2,...,kl) ∈ N , the quantities wt( )= k1 + k2 + ··· + kl,

k k dep(k) = l and i-ht( ) = ♯{j|kj ≥ i +1} are called weight, depth and i-height of , respectively.

For a formal parameter q and an index k =(k1,k2,...,kl) of positive integers with k1 ≥ 2 (which we call admissible), q-analogues of multiple zeta and zeta-star values (qMZVs and qMZSVs for short, respectively) are deﬁned by

q(k1−1)m1+(k2−1)m2+···+(kl−1)ml

ζq(k)= ζq(k1,k2,...,kl)= (∈ Q[[q]]), k1 k2 kl m >m >···>m ≥1 [m1] [m2] ··· [ml] 1 2X l (k1−1)m1+(k2−1)m2+···+(kl−1)ml ⋆ ⋆ q ζ (k)= ζ (k1,k2,...,kl)= (∈ Q[[q]]), q q k1 k2 kl m ≥m ≥···≥m ≥1 [m1] [m2] ··· [ml] 1 2X l 1 − qn where [n] denotes the q-integer [n]= . In the case of l = 1, qMZVs and qMZSVs 1 − q coincide and are reduced to q(k−1)m ζ (k)= . q [m]k m≥1 X The qMZ(S)Vs are investigated for example in Bradley[4], Okuda-Takeyama[10] and Zhao[15]. If q ∈ C, qMZVs and qMZSVs are absolutely convergent in |q| < 1. Taking the limit as q → 1, qMZ(S)Vs turn into ordinary multiple zeta(-star) values (MZ(S)Vs for short) given by 1

k R ζ( )= ζ(k1,k2,...,kl)= k k k (∈ ), m 1 m 2 ··· ml l m >m >···>m ≥1 1 2 1 2X l ⋆ ⋆ 1

k R ζ ( )= ζ (k1,k2,...,kl)= k k k (∈ ). m 1 m 2 ··· ml l m ≥m ≥···≥m ≥1 1 2 1 2X l The generating functions of sum of MZ(S)Vs are studied in [1, 2, 3, 6, 11]. Those for qMZ(S)Vs are studied in [4, 7, 10, 9, 12]. In this paper, we give a universal result including these as special cases. As is introduced in [14], with an additional parameter, t, we deﬁne the interpolation

of qMZVs (abbreviated as t-qMZVs) by Ô

t t k−wt(Ô) l−dep( ) Ô ζq(k)= ζq(k1,k2,...,kl)= (1 − q) ζq( )t (∈ Q[[q]][t]), XÔ

where k = k + k + ··· + kl and stands for the sum where Ô runs over all indices 1 2 Ô of the form Ô = (k1 ··· kl) in which each is ﬁlled by three candidates: “, ”, P 2 “+” or “ − 1 + ” (minus 1 plus). If q ∈ C, t-qMZVs are absolutely convergent in |q| < 1. Taking the limit as q → 1, t-qMZVs turn into t-MZVs which were introduced 0 1 ⋆ t in Yamamoto[13]. We notice that ζq = ζq, ζq = ζq and ζq(k) = ζq(k) for k ≥ 2 and arbitrary t. The q-shifted factorial (a; q)n is deﬁned by

1, n =0, (a; q) = n (1 − a)(1 − aq) ··· (1 − aqn−1), n =1, 2,....

For simplicity, we denote the product (a1; q)n(a2; q)n ··· (am; q)n by (a1, a2,...,am; q)n. a1,...,ar+1 For a positive integer r, the basic hypergeometric function r+1φr ; q, z b1,...,br is deﬁned by the series

∞ a1,...,ar+1 (a1,...,ar+1; q)n n r+1φr ; q, z = z , b1,...,br (q, b ,...,br; q)n n=0 1 X −m where bj =6 q for m =0, 1,... and j =1, 2,...,r. If0

t t ξ0(k,l,h1,...,hr)= ζq(k),

k∈I0(k,l,hX1,...,hr) where I0(k,l,h1,...,hr) denotes the subset of admissible multi-indices (to wit indices with additional requirement k1 ≥ 2) of weight k, depth l, 1-height h1,..., r-height hr. In the deﬁnition, the sum is treated as 0 whenever the index set is empty. We also deﬁne

t t Ψ0 =Ψ0(u1,u2,...,ur+2) k−l−Pr h t j=1 j l−h1 h1−h2 hr−1−hr hr = ξ0(k,l,h1,...,hr)u1 u2 u3 ··· ur+1 ur+2. k,l,h ,...,h 1X r≥0 In order to state our main theorem, we give some other notations below. For variables u1,u2,...,ur+2, we put

u1 x1 = (1) 1+(1 − q)u1 and

r 1 +1 k − 2 u x = ((q − 1)u )k−j(ur+2−ku − u )+ r+2 j r+2−j j − 2 1 1 k r+2 (1+(1 − q)u )j−1 u1 ( k j 1 ) X= (2) for j =2,...,r +2. Let α1,α2,...,αr+1 be parameters determined by

α1 + α2 + ··· + αr+1 = (1 − t)x2 − x1, α ··· α = (1 − t)(x − x x ), j =2,...,r +1,  i1 ij j+1 1 j 1≤i <···

Also let β1, β2,...,βr+1 be parameters determined by

β1 + β2 + ··· + βr+1 = −(x1 + tx2), β ··· β = −t(x − x x ), j =2,...,r +1,  i1 ij j+1 1 j 1≤i <···

Main theorem. Let r be a positive integer and a1, a2,...,ar+1, b1, b2,...,br+1 as above. Then we obtain (1+(1 − q)u )2 Ψt = 1 0 r+1 k−2 r 1 − qu1 − t(1 − qu1) k=2 q uk − tq ur+2 r−1 j+1 j j qP , a1q ,...,ar+1q b1 ··· br+1 × AjBj r+2φr+1 j j ; q, r , b1q ,...,br+1q q a ··· ar j=0 1 +1 X where f r−1 u u A = c S (m +1, j +1)+ 1 2 S (r, j + 1), j =0,...,r − 1, j m q (1+(1 − q)u )2 q m=j 1 X with r +1 k − 2 (1 − q)u k − 2 c = + 1 (−(1 − q))k−r+m−2 m r − m 1+(1 − q)u r − m − 1 k r m 1 =X− +1 u (1 − q)u × k − uk−r−2u + k+1 , 1+(1 − q)u 1 r+2 1+(1 − q)u 1 1 Sq(n, k) is the q-Stirling number of the second kind which is deﬁned recursively by k−1 q Sq(n − 1,k − 1) + [k]Sq(n − 1,k), 0

4 2 Special cases of main theorem

2.1 The case of r =1 Our main theorem mentioned in the previous section is reduced to the following corollary in the case of r = 1. Corollary 1. We obtain the generating function of t-qMZVs for 1-ht:

t qu3 q, a1, a2 q{1+(1 − q)(1 − t)u2} Ψ0 = 3φ2 ; q, , (5) (1 − qu )(1 − tu ) − tqu b1, b2 1 − (1 − q)tu 1 2 3 2 where q q2 ai = , bi = 1+(1 − q)αi 1+(1 − q)βi and α1,α2, β1, β2 are determined by

α + α = −u1+(1−t){u2+(1−q)(u1u2−u3)} , β + β = −u1−t{u2+(1−q)(u1u2−u3)} , 1 2 1+(1−q)u1 1 2 1+(1−q)u1 α α = (1−t)(u3−u1u2) , β β = t(u1u2−u3) . ( 1 2 1+(1−q)u1 ( 1 2 1+(1−q)u1

u1u2 proof. Set r = 1 in the main theorem. Since A = c + 2 Sq(1, 1), B = q, 0 0 (1+(1−q)u1) 0 u1u2 u3 c0 = − (1+(1−q)u )2 + (1+(1−q)u )2 , Sq(1, 1) = 1, we have 1 1 f u A 3 . 0 = 2 (1+(1 − q)u1) Hence we obtain

t qu3 q, a1, a2 b1b2 Ψ0 = 3φ2 ; q, , (6) (1 − qu )(1 − tu ) − tqu b1, b2 qa a 1 2 3 1 2 where q q2 ai = , bi = 1+(1 − q)αi 1+(1 − q)βi and α1,α2, β1, β2 are determined by

α1 + α2 = (1 − t)x2 − x1, β1 + β2 = −(x1 + tx2), α α = (1 − t)(x − x x ), β β = −t(x − x x ). 1 2 3 1 2 1 2 3 1 2 Because of the transformation (1) and (2), we conclude the corollary.

2.1.1 Reduction to Okuda and Takeyama’s result We ﬁnd that our main theorem includes Okuda-Takeyama[10, Theorem 3] by putting 2 t = 0 in Corollary 1. In fact, if t = 0 we have b1 = q (since we may assume β1 = 0 and β2 = −x1), and using Heine’s summation formula (see [5] for example)

b1 b1 a ; q a ; q a1, a2 b1 1 ∞ 2 ∞ 2φ1 ; q, = , (7) b1 a1a2 b1 (b1; q)∞ a a ; q 1 2 ∞ 5 ∞ n−1 where (a; q)∞ = (1 − aq ), and the formula n=1 Y ∞ qn 1 ∞ qn ∞ ∞ (q − 1)m log 1 − x = log{1+(1 − q)x} − ζ (n) xm+n, [n] q − 1 [n] q m + n n=1 n=1 n=2 m=0 Y X X X (8) we know that the equation (6) turns into a a 1 , 2 u q q qb Ψ0 = 3 φ ; q, 2 − 1 0 2 1  b2   u3 − u1u2 a1a2 q     u   ∞ ∞ (q − 1)m = 3 exp ζ (n) (um+n + um+n − zm+n − zm+n) − 1 , u − u u q m + n 1 2 1 2 3 1 2 ( n=2 m=0 ! ) X X where z1 + z2 = u1 + u2 + (1 − q)(u1u2 − u3), z z = u . 1 2 3 Moreover we ﬁnd that this is reduced to the result in Ohno-Zagier[11] as q → 1.

2.1.2 Reduction to Takeyama’s result Also our main theorem includes Takeyama[12, Theorem 1.1] by putting t = 1 in Corol- lary 1. In fact, if t = 1 we have a1 = q and a2 = q{1+(1 − q)u1} (since we may asumme α1 = 0 and α2 = −x1). Hence the equation (5) turns into q,q,q q u qu {1+(1 − ) 1} q Ψ1 = 3 φ q2 q2 ; q, , 0 (1 − qu )(1 − u ) − qu 3 2  , 1 − (1 − q)u  1 2 3 x y 2   where x =1+(1 − q)β1,y =1+(1 − q)β2. Therefore we have

2 2+(1 − q)(u1 − u2)+(1 − q) (u3 − u1u2) x + y =2+(1 − q)(β1 + β2)= , 1+(1 − q)u1  q u  2 1 − (1 − ) 2  xy =1+(1 − q)(β1 + β2)+(1 − q) β1β2 = . 1+(1 − q)u1  Moreover we ﬁnd that this is reduced to the result in Aoki-Kombu-Ohno[1] as q → 1.

2.1.3 Sum formula for t-qMZVs I As an application of Corollary 1, we establish the following sum formula for t-qMZVs. Theorem 2. For any positive integers k > n ≥ 1, we have

ζq(k) k wt(k)=k,dep( )=n

k:admissibleX 1 k − 1 j = (k − 1 − l)tj(1 − t)n−1−j(1 − q)lζ (k − l). k − 1 j l q l j n 0≤ ≤X≤ −1 6 Remark 1. Taking the limit as q → 1, this formula is reduced to the sum formula of t-MZVs

n−1 t k − 1 j n−1−j

ζ (k)= t (1 − t) ζ(k)

j k wt(k)=k,dep( )=n ( j=0 ) k:admissibleX X which is proved in Yamamoto[13]. If t =0, we have the sum formula for qMZVs

ζq(k)= ζq(k) k wt(k)=k,dep( )=n

k:admissibleX which is proved in Bradley[4]. And also, if t =1, we have the sum formula for qMZSV

n−1 ⋆ 1 k − 1 n − 1 l

ζ (k)= (k − 1 − l)(1 − q) ζ (k − l)

q k − 1 n − 1 l q k wt(k)=k,dep( )=n l=0 k:admissibleX X which is proved in Ohno-Okuda[9].

Proof of Theorem 2. Using the q-analogue of the Kummer-Thomae-Whipple formula (see [5] for example)

b1 b1b2 ; q ; q b2 b2 a , a , a b b a1 a2a3 a , , b 1 2 3 1 2 ∞ ∞ 1 a2 a3 1 3φ2 ; q, = 3φ2 b1b2 ; q, , b1, b2 a1a2a3 b1b2 b2, a a a1 (b1; q)∞ a a a ; q 2 3 1 2 3 ∞ we have

q, a1, a2 q{1+(1 − q)(1 − t)u2} 3φ2 ; q, b1, b2 1 − (1 − q)tu 2 q, a , a b b = φ 1 2 ; q, 1 2 3 2 b , b qa a 1 2 1 2 b1 b1b2 ; q ; q b2 b2 q a1a2 q, , b ∞ ∞ a1 a2 1 = 3φ2 b1b2 ; q, b1b2 b2, a a q (b1; q)∞ qa a ; q 1 2 1 2 ∞ b 1 b2 b2 1 − q q, , b1 = φ a1 a2 ; q, . (9) b1b2 3 2 b1b2 1 − b2, a a q qa1a2 1 2

Putting u3 = u1u2 in Corollary 1, we have

−u1 + (1 − t)u2 q{1+(1 − q)u1} α1 =0, α2 = , a1 = q, a2 = , 1+(1 − q)u1 1+(1 − q)(1 − t)u2

2 −u1 − tu2 2 q {1+(1 − q)u1} β1 =0, β2 = , b1 = q , b2 = , 1+(1 − q)u1 1 − (1 − q)tu2

(1 − qu1)(1 − tu2) − tqu3 =1 − qu1 − tu2.

7 Then by (9), we get

b2 b2 qu1u2 1 − q q, , Ψt = · φ q a2 ; q, q . 0 b2 3 2 qb2 1 − qu1 − tu2 1 − b2, a2 " a2 #

By the deﬁnition of 3φ2,

b b q(1 − q)u u ∞ 1 − 2 1 − 2 Ψt = 1 2 q · a2 qn 0 n−1 n b2 b2 1 − b2q 1 − q (1 − qu1 − tu2) 1 − n=0 a2 a2 X b 2 ∞ n q(1 − q)u1u2 1 − q q = . (10) (1 − qu1 −tu2) n−1 n b2 n=0 (1 − b2q ) 1 − q X a2 Since we have b (1 − q)(1 − qu − tu ) 1 − 2 = 1 2 q 1 − (1 − q)tu2 and b (1 − b qn−1) 1 − qn 2 2 a 2 {1 − qn+1 − (1 − q)(tu + qn+1u )}{1 − qn+1 − (1 − q)(tu + qn+1(1 − t)u )} 2 1 2 2 , = 2 {1 − (1 − q)tu2} we calculate the right-hand side of (10) as

∞ qn Ψt = u u {1 − (1 − q)tu } 0 1 2 2 {[n] − (tu + qnu )}{[n] − (t + qn(1 − t))u } n=1 2 1 2 X ∞ qn ∞ tu + qnu m t + qn(1 − t) l = u u {1 − (1 − q)tu } 2 1 ul 1 2 2 [n]2 [n] [n] 2 n=1 m,l X X=0 = u1u2{1 − (1 − q)tu2} ∞ qn i + j (tu )i(qnu )j l + m tl(qn(1 − t))m × 2 1 ul+m [n]2 i [n]i+j l [n]l+m 2 n=1 i,j≥0 l,m X X X≥0 = u1u2{1 − (1 − q)tu2} i + j l + m ∞ qn(j+m+1) × ti+l(1 − t)muj ui+l+m . (11) i l 1 2 [n]i+j+l+m+2 i,j,l,m n=1 X≥0 X Because of l l qnm = (1 − q)i[n]iqn(m+l−i) i i=0 X

8 for any l ≥ 0, the right-hand side of (11) turns into

i + j l + m u u {1 − (1 − q)tu } ti+l(1 − t)muj ui+l+m 1 2 2 i l 1 2 i,j,l,m X≥0 i l ∞ + i + l (1 − q)p[n]pqn(i+j+m+l+1−p) × p [n]i+j+l+m+2 n=1 p=0 X X i + j l + m = u u {1 − (1 − q)tu } ti+l(1 − t)muj ui+l+m 1 2 2 i l 1 2 i,j,l,m X≥0 i l + i + l × (1 − q)pζ (2 + i + j + l + m − p) p q p=0 X i + j l + m i + l = ti+l(1 − t)m(1 − q)pζ (2 + i + j + l + m − p) i l p q i,j,l,m≥0 0≤Xp≤i+l j+1 i+l+m+1 × u1 u2 i + j l + m i + l − ti+l+1(1 − t)m(1 − q)p+1ζ (2 + i + j + l + m − p) i l p q i,j,l,m≥0 0≤Xp≤i+l j+1 i+l+m+2 × u1 u2 .

k−n n Comparing the coeﬃcients of u1 u2 for k, n with k > n ≥ 1, we have

ζq(k) k wt(k)=k,dep( )=n

k:admissibleX i + k − n − 1 l + m i + l = ti+l(1 − t)m(1 − q)pζ (k − p) i l p q i+l+m=n−1 i,l,m≥0 0≤Xp≤i+l i + k − n − 1 l + m i + l − ti+l+1(1 − t)m(1 − q)pζ (k − p) i l p − 1 q i+l+m=n−2 i,l,m≥0 1≤pX≤i+l+1 j i + k − n − 1 j − i + m j = tj(1 − t)m(1 − q)pζ (k − p) i j − i p q j+m=n−1 i=0 j,m≥0 0X≤p≤j X j−1 i + k − n − 1 j − i + m − 1 j − 1 − tj(1 − t)m(1 − q)pζ (k − p). i j − i − 1 p − 1 q j+m=n−1 i=0 j≥1,m≥0 1X≤p≤j X

9 j i+k j−i+m k+m+j+1 Using i=0 i j−i = j ,

P t

ζq(k) k wt(k)=k,dep( )=n

k:admissibleX

k − n + m + j j k − n + m + j − 1 j − 1 =  −  j p j − 1 p − 1  j+m=n−1 j+m=n−1   j,m≥0 j≥1,m≥0  0X≤p≤j 1X≤p≤j  j m p  × t (1 − t) (1 − q) ζq(k − p)  k − 1 j k − 2 j − 1 = −  j p j − 1 p − 1   0≤j≤n−1 1≤j≤n−1   0X≤p≤j 1X≤p≤j  × tj(1 − t)n−1−j(1 − q)pζ (k − p)  q  k − 1 j p k − 1 j = −  j p k − 1 j p   0≤j≤n−1 1≤j≤n−1   0X≤p≤j 1X≤p≤j  × tj(1 − t)n−1−j(1 − q)pζ (k − p)  q  1 k − 1 j = (k − 1 − p) tj(1 − t)n−1−j(1 − q)pζ (k − p). k − 1 j p q 0≤p≤j≤n−1 X Hence we obtain the theorem.

2.1.4 Sum formula for t-qMZVs II As another application of Corollary 1, we establish the following sum formula (of full height) for t-qMZVs.

Theorem 3. We have

t k−2l l

ζ (k) u u  q  1 3

k,l k X≥0 ∈IX0(k,l,l)  ∞ ∞(q − 1)m = exp ζ (n) (wm+n + wm+n − zm+n − zm+n) − 1, q m + n 1 2 1 2 (n=2 m=0 ) X X where z1, z2,w1,w2 are determined by

z1 + z2 = u1 − (1 − t)(1 − q)u3, w1 + w2 = u1 + t(1 − q)u3, z z = (1 − t)u , w w = −tu . 1 2 3 1 2 3

t k

Remark 2. Theorem 3 implies that the sum of t-qMZVs ζq(k) with wt( )= k and k dep(k)=ht( )= l can be written as a polynomial in Riemann q-zeta values. If t = 1, this formula is reduced to a formula for qMZSVs with full height, which is proved in Takeyama[12]. Taking the limit as q → 1, we obtain Li-Qin[8, Corollary 3.6].

10 Proof of Theorem 3. Applying (9) to Corollary 1 and substituting u2 = 0, we have

b 1 − 1 b2 b2 t qu3 q q, , b1 Ψ = φ a1 a2 ; q, . 0 b1b2 3 2 2 1 − qu1 − tqu3 1 − b2, q q qa1a2 

By the deﬁnition of 3φ2,

b b b1 ∞ 2 2 n 1 − a ; q a ; q t qu3 q 1 n 2 n b1 Ψ0 = b b 2 1 − qu − tqu 1 2 (q ; q)n(b ; q)n q 1 3 1 − qa a n=0 2 1 2 X b b b b1 ∞ (1 − q) 1 − 2 2 ; q 2 ; q n−1 qu 1 − q qa1 qa2 b = 3 q n n 1 b1b2 1 − qu1 − tqu3 1 − b2 b2 b2 q qa1a2 n=1 1 − qa 1 − qa (q; q)n q ; q X 1 2 n b1 b2 b b qu 1 − q 1 − q (1 − q) q 2 , 2 b 3 qa1 qa2 1 = 2φ1 b ; q, − 1 . b b b b 2 1 − qu1 − tqu3 1 − 1 2 1 − 2 1 − 2 b1 ( " q q # ) qa1a2 qa1 qa2 Because of

2 4 2 a1a2 = q {1+(1 − q)u1}, b1b2 = q {1+(1 − q)u1} = q a1a2, 1 1 1 1+(1 − q)u − (1 − t)(1 − q)2u + = 1 3 , a1 a2 q 1+(1 − q)u1 and 1 1 1 2+(1 − q)u + t(1 − q)2u 1 3 , + = 2 b1 b2 q 1+(1 − q)u1 the quantity b1 b2 1 − q 1 − q (1 − q) b b b b 1 − 1 2 1 − 2 1 − 2 b qa1a2 qa1 qa2 1 is calculated as

1 1 1 1 1 1 1 1 − (b1 + b2)+ 2 b1b2 b b − q b + b + q2 q q = 1 2 1 2 b b b − 1 2 1 + 1 + b 1 + 1 − 1 1 + 1 1 q a1 a2 2 b1 b2 q a1 a2 2 (1−q) (1−qu 1−tqu3) 4 q (1+(1−q)u1) 1 − qu1 − tqu3 = 2 = . (1−q) u3 q2u 2 3 q (1+(1−q)u1) Therefore we have

b2 , b2 b t qa1 qa2 1 Ψ0 = 2φ1 b2 ; q, − 1. (12) " q q # Here using Heine’s summation formula (7) and qn 1 − qn{1+(1 − q)x} 1 − x = , [n] 1 − qn

11 the right-hand side of (12) is calculated as

∞ n−1 n−1 t (a1; q)∞ (a2; q)∞ (1 − a1q )(1 − a2q ) Ψ0 = − 1= − 1 b1 b2 b1 n−1 b2 n−1 q ; q q ; q n=1 1 − q q 1 − q q ∞ ∞ Y n n ∞ q q 1 − [n] z1 1 − [n] z2 = − 1, (13) qn qn n=1 1 − w 1 − w Y [n] 1 [n] 2 where αi βi zi = − , wi = − (i =1, 2) 1+(1 − q)αi 1+(1 − q)βi i.e. z1 + z2 = u1 − (1 − t)(1 − q)u3, w1 + w2 = u1 + t(1 − q)u3, z z = (1 − t)u , w w = −tu . 1 2 3 1 2 3 Using the formula (8), the right-hand side of (13) turns into

1 {1+(1 − q)z }{1+(1 − q)z } ∞ qn Ψt = exp log 1 2 0 q − 1 {1+(1 − q)w }{1+(1 − q)w } [n] ( 1 2 n=1 X ∞ ∞ (q − 1)m − ζ (n) (zm+n + zm+n − wm+n − wm+n − 1 q m + n 1 2 1 2 n=2 m=0 ) X X ∞ ∞ (q − 1)m = exp ζ (n) (wm+n + wm+n − zm+n − zm+n) − 1. q m + n 1 2 1 2 (n=2 m=0 ) X X Hence we obtain the theorem.

2.2 The case of t =0 When t = 0 our main theorem reduces to the result in Li[7] as follows. Note that, in the case of t = 0,

α1 + ··· αr+1 = x2 − x1, α ··· α = x − x x j =2,...,r +1,  i1 ij j+1 1 j 1≤i <···

β1 = ··· = βr =0, βr+1 = −x1. Theorem 4. (Li[7]) We have

∗ j ∗ j 2 r−1 a1q ,...,ar+1q ∗ 0 {1+(1 − q)u1} ∗ ∗ j+1 j+1 ∗ j b ∗ Ψ0 = Aj Bj r+1φr q ,...,q , b q ; q, ∗ ∗ − A0 , ur − u ur   a ··· a   +2 1 +1 j=0 1 r+1 X r−1     | {z }  12 where ∗ ai 1 ∗ br+1 ai = = , b = = q{1+(1 − q)u1}, q 1+(1 − q)αi q r−1 u u A∗ = c S (m, j)+ 1 2 S (r − 1, j), j m q {1+(1 − q)u }2 q m=j 1 X ∗ ∗ ∗ j (a ,...,a ; q)j b B∗ = 1 r+1 , j (q,...,q, b∗; q) (1 − q)a∗ ··· a∗ j 1 r+1 r−1 for 1 ≤ i ≤ r +1, 0 ≤ j ≤ r|−{z1 and} αi’s are as above. proof. Since β1 = ··· = βr =0, βr+1 = −x1, we see that

2 2 q 2 b1 = ··· = br = q , br+1 = = q {1+(1 − q)u1}. 1 − (1 − q)x1 Since j Sq(m +1, j +1) = q Sq(m, j) + [j + 1]Sq(m, j +1) (0 ≤ j ≤ m),

Aj’s in the main theorem are written as

j ∗ ∗ ∗ Aj = q Aj + [j + 1]Aj+1 (0 ≤ j ≤ r − 1, Ar = 0). We also ﬁnd q(1 − q)r+1(1 − b∗)a∗ ··· a∗ B 1 r+1 B∗ j = j+1 ∗ ∗ ∗ j+1 (1 − q )b (1 − a1) ··· (1 − ar+1) q(1 − q)r(1 − b∗) (1 − a∗qj) ··· (1 − a∗ qj) f 1 r+1 B∗. = j+1 ∗ ∗ j+1 r−1 ∗ j j (1 − q )(1 − a1) ··· (1 − ar+1) (1 − q ) (1 − b q ) Applying these to the main theorem, we have

2 0 {1+(1 − q)u1} Ψ0 = 1 − qu1 ∗ ∗ q, a1q,...,ar+1q ∗ ∗ 2 2 ∗ b × A0q r+2φr+1 q ,...,q , b q ; q, ∗ ∗   a1 ··· ar+1   r r −1  q(1 − q)r(1 − b∗) (1 −a∗qj) ··· (1 − a∗ qj) + qjA∗ | {z } 1 r+1 B∗ j (1 − qj+1)(1 − a∗) ··· (1 − a∗ ) (1 − qj+1)r−1(1 − b∗qj) j j=1 1 r+1 X j+1 ∗ j+1 ∗ j+1 q , a1q ,...,ar+1q ∗ j+2 j+2 ∗ j+1 b × r+2φr+1 q ,...,q , b q ; q, ∗ ∗  a1 ··· ar+1  r r−1 q(1 − q)r+1(1 − b∗)a∗ ··· a∗  + [j]A∗ | {z } 1 r+1 B∗ j (1 − qj)b∗(1 − a∗) ··· (1 − a∗ ) j j=1 1 r+1 X j ∗ j ∗ j q , a1q ,...,ar+1q ∗ j+1 j+1 ∗ j b ×r+2φr+1 q ,...,q , b q ; q, ∗ ∗ .  a1 ··· ar+1  r    | {z }  13 Let 1, 2, 3 be the ﬁrst, the second, and the third terms in the bracket of the right-hand side, respectively. Then we have P P P ∞ ∗ ∗ ∗ n ∗ (a1q,...,ar+1q; q)n b = A0q 1 (q2,...,q2, b∗q; q) a∗ ··· a∗ n=0 n 1 r+1 X X r (1 − q)r(1 − b∗) a∗ ··· a∗ ∞ (a∗,...,a∗ ; q) b∗ n A∗q | {z } 1 r+1 1 r+1 n = 0 ∗ ∗ ∗ ∗ ∗ ∗ (1 − a ) ··· (1 − a ) b (q,...,q, b ; q)n a ··· a 1 r+1 n=1 1 r+1 X r a∗,...,a∗ (1 − q)r(1 − b∗)a∗ ··· a∗ 1| {z }r+1 b∗ A∗q 1 r+1 φ q,...,q, b∗ q, , = 0 ∗ ∗ ∗ r+1 r ; ∗ ∗ − 1 b (1 − a1) ··· (1 − ar+1)   a1 ··· ar+1    r−1     | {z }  q(1 − q)r(1 − b∗)a∗ ··· a∗ = 1 r+1 2 ∗ ∗ ∗ b (1 − a1) ··· (1 − ar+1) X r −1 ∞ qj(qj+1; q) (a∗qj,...,a∗ qj; q) b∗ n × A∗B∗ n−1 1 r+1 n j j (q; q) (qj+1,...,qj+1, b∗qj; q) a∗ ··· a∗ j=1 n=1 n−1 n 1 r+1 X X r

∗ j ∗ j+1 ∗ j (using (1 − a q )(a q ; q)n =(a q ; q)n+1| ), and{z }

q(1 − q)r(1 − b∗)a∗ ··· a∗ = 1 r+1 3 ∗ ∗ ∗ b (1 − a1) ··· (1 − ar+1) X r−1 ∞ j ∗ j ∗ j ∗ n (q ; q)n(a q ,...,a q ; q)n b × A∗B∗ 1+ 1 r+1 . j j  j+1 j+1 ∗ j ∗ ∗   (q; q)n(q ,...,q , b q ; q)n a1 ··· ar+1  j=1  n=1  X X r   Since  | {z }  qj(qj+1; q) (qj; q) (qj+1; q) n−1 + n = n , (q; q)n−1 (q; q)n (q; q)n we have

+ 2 3 r ∗ ∗ ∗ Xq(1 −Xq) (1 − b )a1 ··· ar+1 = ∗ ∗ ∗ b (1 − a1) ··· (1 − ar+1)

r−1 ∞ ∗ j ∗ j ∗ n (a q ,...,a q ; q)n b × A∗B∗ 1+ 1 r+1 j j  j+1 j+1 ∗ j ∗ ∗   (q; q)n(q ,...,q , b q ; q)n a1 ··· ar+1  j=1  n=1  X X r−1 ∗ j ∗ j  r−1 a q ,...,a q  q(1 − q)r(1 − b∗)a∗ ··· a∗ | {z } 1 r+1  b∗ = 1 r+1 A∗B∗ φ qj+1,...,qj+1, b∗qj ; q, . b∗(1 − a∗) ··· (1 − a∗ ) j j r+1 r  a∗ ··· a∗  1 r+1 j=1 1 r+1 X r−1   | {z } 14 Therefore we obtain

2 r ∗ ∗ ∗ t {1+(1 − q)u1} q(1 − q) (1 − b )a1 ··· ar+1 Ψ0 = ∗ ∗ ∗ 1 − qu1 b (1 − a1) ··· (1 − ar+1) ∗ j ∗ j r−1 a q ,...,a q 1 r+1 b∗ × A∗B∗ φ qj+1,...,qj+1, b∗qj ; q, − A∗ .  j j r+1 r  a∗ ··· a∗  0 j=0 1 r+1 X r−1    We ﬁnd  | {z }  a∗ a∗ 1 1 r+1 , ∗ ··· ∗ = r+1 1 − a1 1 − ar+1 (1 − q) (xr+2 − x1xr+1) q(1 − q)r 1 − b∗ (1 − q)r+1 . ∗ = 1 − qu1 b 1+(1 − q)u1 Also by (1) and (2), we have

ur+2 − u1ur+1 xr+2 − x1xr+1 = . 1+(1 − q)u1 Hence we conclude the theorem.

3 t-qMPLs

l For any multi-index k =(k1,k2,...,kl) ∈ N and a parameter q, q-analogue of multiple polylogarithms (of non-star and star types, collectively abbreviated as qMPLs) are deﬁned by

zm1

Lik;q(z)= (∈ Q[[q, z]]), k1 k2 kl m >m >···>m ≥1 [m1] [m2] ··· [ml] 1 2X l zm1 Li⋆ z Q q, z .

k ( )= (∈ [[ ]]) ;q k1 k2 kl m ≥m ≥···≥m ≥1 [m1] [m2] ··· [ml] 1 2X l

These series converge if |z| < 1 for any k. We also deﬁne t-qMPLs by

′

t l−dep(Ô)

Li z LiÔ z t ∈ Q q, z t , k;q( )= ;q( ) ( [[ ]][ ]) XÔ

′

Ô where stands for the sum where Ô runs over all indices of the form =(k ··· kl) Ô 1 in which each is ﬁlled by two candidates: “, ” or “ + ”. P 3.1 Diﬀerence formula for t-qMPLs Lit (z) k;q

Denote by Dq the q-diﬀerence operator

f(z) − f(qz) (D f)(z)= . q (1 − q)z

15 Then by deﬁnition qMPLs satisfy 1 Li (z), k ≥ 2, z (k1−1,k2,...,kl);q 1  1 Li z Li (z), k =1,l ≥ 2, Dq (k1,k2,...,kl);q( )=  (k2,...,kl);q 1  1 − z  1 , k = l =1. 1 − z 1   Lemma 5. For positive inegers k1,k2,...,kl, t-qMPLs satisfy 1 Lit (z), k ≥ 2, z (k1−1,k2,...,kl);q 1 t  t 1 t DqLi (z)= + Li (z), k1 =1,l ≥ 2, (k1,k2,...,kl);q  (k2,...,kl);q  z 1 − z  1 , k = l =1. 1 − z 1  

proof. If l ≥ 2, we have the decomposition Ô

t l−dep(Ô) l−dep( ) Ô Li z LiÔ z t Li z t .

(k1,k2,...,kl);q( )= ;q( ) + ;q( )

Ô Ô=(k1,k2 ··· kl) =(k1+k2 ··· kl) =“X,”or“+” =“X,”or“+”

Hence if k1 ≥ 2, we have

Lit z Dq (k1,k2,...,kl);q( )

1 1 Ô

l−dep(Ô) l−dep( ) Ô = LiÔ (z)t + Li (z)t

z ;q z ;q

Ô Ô=(k1−1,k2 ··· kl) =(k1−1+k2 ··· kl) =“X,”or“+” =“X,”or“+” 1 = Lit (z), z (k1−1,k2,...,kl);q and otherwise

Lit z Dq (1,k2,...,kl);q( )

1 1 Ô

l−dep(Ô)−1 l−dep( ) Ô = LiÔ (z)t + Li (z)t

1 − z ;q z ;q

Ô Ô=(k2 ··· kl) =(k2 ··· kl) =“X,”or“+” =“X,”or“+” t 1 = + Lit (z). z 1 − z (k2,...,kl);q If l = 1, since Lit (z)= Li (z), we have (k1);q (k1);q 1 1 Li z Lit z , k , (k1−1);q( )= (k1−1);q( ) 1 ≥ 2 D Lit (z)= z z q (k1);q  1 , k =1.  1 − z 1 

16 t 3.2 Diﬀerence formula for sums of t-qMPLs Gj(z)

Let r be a positive integer. We deﬁne sets of indices Ij (−1 ≤ j ≤ r − 1),I for any integers k,l,h1, h2,...,hr ≥ 0 by

Ij(k,l,h1, h2,...,hr)

k k = {k =(k1,k2,...,kl) ∈ N |wt( )= k, i-ht( )= hi(i =1,...,r),k1 ≥ j +2} and

I(k,l,h1, h2,...,hr)= I−1(k,l,h1, h2,...,hr).

(I0 is nothing but that we deﬁned in §2.) We notice that:

(i) Ij(k,l,h1,...,hr) ⊆ Ij+1(k,l,h1,...,hr).

(ii) If I(k,l,h1, h2,...,hr) =6 φ, then

k ≥ l + hj,  j=1 X  l ≥ h1 ≥···≥ hr ≥ 0.

 (iii) If Ij(k,l,h1, h2,...,hr) =6 φ, then

k ≥ l + hj,  j=1 X  l ≥ h ≥···≥ hr ≥ 0,  1 hj+1 ≥ 1,  for j =0, 1,...,r − 1. 

For any integers k,l,h1,...,hr ≥ 0 and −1 ≤ j ≤ r − 1, we set

Gt z Gt k,l,h ,...,h z Lit z , j( )= j( 1 r; )= k;q( )

k∈Ij (k,l,hX1,...,hr) t t t G (z)= G (k,l,h1,...,hr; z)= G−1(k,l,h1,...,hr; z).

In the deﬁnition, the sum is regarded as 0 whenever the index set is empty except for Gt(0, 0,..., 0; z) := 1. t Applying the q-diﬀerence oparator Dq to the sum Gj(z), we obtain the following lemma.

r Lemma 6. (i) For k ≥ l + j=1 hj, l ≥ h1 ≥···≥ hr ≥ 1, we have

t P DqGr−1(k,l,h1,...,hr; z) 1 = Gt (k − 1,l,h ,...,h ; z) z r−1 1 r t t + Gr−2(k − 1,l,h1,...,hr−1, hr − 1; z) − Gr−1(k − 1,l,h1,...,hr−1, hr − 1; z) .

17 r (ii) For 0 ≤ j ≤ r − 2, k ≥ l + j=1 hj, l ≥ h1 ≥···≥ hr ≥ 0, hj+1 ≥ 1, we have

t P t Dq Gj(k,l,h1,...,hr; z) − Gj+1(k,l,h1,...,hr; z) 1 = Gt (k − 1,l,h ,...,h , h − 1, h ,...,h ; z) z j−1 1 j j+1 j+2 r t − Gj(k − 1,l,h1,...,hj, hj+1 − 1, hj+2,...,hr; z) .

r (iii) For k ≥ l + j=1 hj, l ≥ h1 ≥···≥ hr ≥ 0, l ≥ 2, we have

P t t Dq G (k,l,h1,...,hr; z) − G0(k,l,h1,...,hr; z) t 1 = + Gt(k − 1,l − 1, h ,...,h ; z). z 1 − z 1 r 

proof. (i) If k =(k1,...,kl) ∈ Ir−1(k,l,h1,...,hr), k1 ≥ r +1 ≥ 2. Then we have

t DqGr−1(k,l,h1,...,hr; z) 1 = Lit (z) z (k1−1,k2,...,kl);q k=(k1,...,kl) ∈Ir−1(k,l,hX1,...,hr) 1 1 = Lit (z)+ Lit (z)

z (k1−1,k2,...,kl);q z (k1−1,k2,...,kl);q k k=(k1,...,kl) =(k1,...,kl) ∈Ir−1(k,l,hX1,...,hr), ∈Ir−1(k,l,hX1,...,hr), k1≥r+2 k1=r+1 1 1

Gt k − ,l,h ,...,h z Lit z = ( 1 ; )+ k ( ) z r−1 1 r z ;q k=(k1,...,kl) ∈Ir−2(k−1,l,hX1,...,hr−1,hr−1), k1=r 1 = Gt (k − 1,l,h ,...,h ; z) z r−1 1 r 1 1

Lit z − Lit z k + k ( ) ( )

z ;q z ;q k k=(k1,...,kl) =(k1,...,kl) ∈Ir−2(k−1,l,hX1,...,hr−1,hr−1), ∈Ir−2(k−1,l,hX1,...,hr−1,hr−1), k1≥r k1≥r+1 1 = Gt (k − 1,l,h ,...,h ; z) z r−1 1 r 1 1 + Gt (k − 1,l,h ,...,h , h − 1; z) − Gt (k − 1,l,h ,...,h , h − 1; z). z r−2 1 r−1 r z r−1 1 r−1 r

(ii) For 0 ≤ j ≤ r − 2, if k =(k1,...,kl) ∈ Ij(k,l,h1,...,hr), k1 ≥ j +2 ≥ 2. Then

18 we have

t t Dq Gj(k,l,h1,...,hr; z) − Gj+1(k,l,h1,...,hr; z) Lit z = Dq k;q( )

k=(k1,...,kl) ∈Ij (k,l,hX1,...,hr), k1=j+2 1

Lit z = k ( ) z ;q k=(k1,...,kl) ∈Ij−1(k−1,l,h1,...,hXj ,hj+1−1,hj+2,...,hr), k1=j+1 1 = Gt (k − 1,l,h ,...,h , h − 1, h ,...,h ; z) z j−1 1 j j+1 j+2 r t − Gj(k − 1,l,h1,...,hj, hj+1 − 1, hj+2,...,hr; z) .

(iii) If l ≥ 2, we have

t t Dq G (k,l,h1,...,hr; z) − G0(k,l,h1,...,hr; z) t 1

 Lit z Lit z k

= D k ( )= + ( )

q ;q z 1 − z ;q k k=(k1,...,kl) =(k2,...,kl) ∈I(k,l,hX1,...,hr), ∈I(k−1,lX−1,h1,...,hr), k1=1 t 1 = + Gt(k − 1,l − 1, h ,...,h ; z). z 1 − z 1 r 

t 3.3 Diﬀerence formula for generating functions Φj(z) t Let x1,...,xr+2 be variables. We deﬁne generating functions for Gj(z)(−1 ≤ j ≤ r−1) as follows:

t t t Φj = Φj(z) = Φj(x1,...,xr+2; z) r t k−l−Pi=1 hi n−h1 h1−h2 hr−1−hr hr = Gj(k,l,h1,...,hr; z)x1 x2 x3 ··· xr+1 xr+2, k,l,h1,...,hr≥0 t t X Φ = Φ−1.

t t Applying the q-diﬀerence oparator to the generating functions Φj, Φ , we obtain the following proposition.

Proposition 7. Let r be a positive integer. We have x x D Φt = 1 Φt + r+2 (Φt − Φt ), q r−1 z r−1 zx r−2 r−1 x r+1  D t − t j+3 t − t , j , ,...,r − ,  q(Φj Φj+1)= (Φj−1 Φj) =1 2 2  zxj+2  t t x3 t t  Dq(Φ0 − Φ1)= (Φ − Φ0 − 1),  zx2 x t 1 D (Φt − Φt )= 2 + + x (Φt − 1).  q 0 1 − z z 1 − z 2     19 t proof. By the deﬁnition of Φr−1, we have

k−l−Pr h t t j=1 j l−h1 h1−h2 hr−1−hr hr DqΦr−1 = DqGr−1(k,l,h1,...,hr; z)x1 x2 x3 ··· xr+1 xr+2. k≥l+Pr h j=1 j l≥h1≥···≥Xhr≥1

Applying the equation (i) of Lemma 6 to the right-hand side, we have

t DqΦr−1 r 1 k−l−P hj = Gt (k − 1,l,h ,...,h ; z)x j=1 xl−h1 xh1−h2 ··· xhr−1−hr xhr z r−1 1 r 1 2 3 r+1 r+2 k≥l+Pr h j=1 j l≥h1≥···≥Xhr≥1 1 + Gt (k − 1,l,h ,...,h , h − 1; z) z r−2 1 r−1 r k≥l+ r h Pj=1 j l≥h1≥···≥Xhr≥1 k−l−Pr h t j=1 j l−h1 h1−h2 hr−1−hr hr −Gr−1(k − 1,l,h1,...,hr−1, hr − 1; z) x1 x2 x3 ··· xr+1 xr+2

x 1 = 1 Φt +  −  z r−1 z  k−1≥l+Pr h −1 k−1≥l+Pr h −1   j=1 j j=1 j  l≥h1≥···≥Xhr−1≥hr−1≥0 l≥h1≥···≥Xhr−1=hr−1≥0 × Gt (k − 1,l,h ,...,h , h − 1; z)  r−2  1 r−1 r  k−l−Pr h t j=1 j l−h1 h1−h2 hr−1−hr hr −Gr−1(k − 1,l,h1,...,hr−1, hr − 1; z) x1 x2 x3 ··· xr+1 xr+2 x1 t xr+2 t t = Φr−1 + (Φr−2 − Φr−1). z zxr+1

Note that the second sum is 0 since no indices have the property hr−1 = hr − 1. t t We proceed to show the second equation. Since Φj = Φj+1 when hj+1 = hj+2, we see that

t t Dq Φj − Φj+1 t t = Dq Gj(k,l,h1,...,hr; z) − Gj+1(k,l,h1,...,hr; z) k≥l+Pr h i=1 i l≥h1≥···≥hj+1X>hj+2≥···≥hr≥0 (hj+1≥1) r k−l−Pi=1 hi l−h1 h1−h2 hr−1−hr hr × x1 x2 x3 ··· xr+1 xr+2.

20 By applying Lemma 6 (ii), the right-hand side turns into 1 Gt (k − 1,l,h ,...,h , h − 1, h ,...,h ; z) z j−1 1 j j+1 j+2 r k≥l+Pr h i=1 i l≥h1≥···≥hj+1X>hj+2≥···≥hr≥0 (hj+1≥1) t −Gj(k − 1,l,h1,...,hj, hj+1 − 1, hj+2,...,hr; z) r k−l−Pi=1 hi l−h1 h1−h2 hr−1−hr hr × x1 x2 x3 ··· xr+1 xr+2

1 xj+3 t = Gj−1(k,l,h1,...,hj, hj+1, hj+2,...,hr; z) z xj +2 k≥l+Pr h i=1 i l≥h1≥···≥hj >hjX+1≥hj+2≥···≥hr≥0 (hj ≥1) t −Gj(k,l,h1,...,hj, hj+1, hj+2,...,hr; z) r k−l−Pi=1 hi l−h1 h1−h2 hr−1−hr hr × x1 x2 x3 ··· xr+1 xr +2 xj+3 t t = (Φj−1 − Φj). zxj+2

t t For the third one, note that Φ0 = Φ1 when h1 = h2. Then, by using Lemma 6 (ii), we have

t t Dq(Φ0 − Φ1) 1 = Gt(k − 1,l,h − 1, h ,...,h ; z) z 1 2 r k≥l+ r h Pj=1 j l≥h1>hX2≥···≥hr≥0 (h1≥1) k−l−Pr h t j=1 j l−h1 h1−h2 hr−1−hr hr −G0(k − 1,l,h1 − 1, h2,...,hr; z) x1 x2 x3 ··· xr+1 xr+2

1 x3   = − Gt(k,l,h , h ,...,h ; z) z x   1 2 r 2  k≥l+Pr h k≥l+Pr h   j=1 j j=1 j  l≥h1≥···≥Xhr≥0 l=h1≥···≥Xhr≥0 l≥1 l≥1  k−l−Pr h t  j=1 j l−h1 h1−h2 hr−1−hr hr −G0(k,l,h1, h2,...,hr; z) x1  x2 x3 ··· xr+1 xr+2. The second sum is 0 because

I(k,l,h1,...,hr) − I0(k,l,h1,...,hr)= φ if l = h1 ≥ 1. (14)

Hence we have

t t Dq(Φ0 − Φ1) 1 x = 3 Gt(k,l,h , h ,...,h ; z) − Gt (k,l,h , h ,...,h ; z) z x 1 2 r 0 1 2 r 2 k≥l+Pr h j=1 j l≥h1≥···≥Xhr≥0 l≥1 k−l−Pr h j=1 j l−h1 h1−h2 hr−1−hr hr × x1 x2 x3 ··· xr+1 xr+2 x3 t t = (Φ − 1 − Φ0). zx2

21 t t We move on to show the last equation. First decompose Φ − Φ0 into three parts (for l =0, l = 1 and l ≥ 2):

t t t t Φ − Φ0 = G (k,l,h1,...,hr; z) − G0(k,l,h1,...,hr; z) k≥l+Pr h j=1 j l≥h1≥···≥Xhr≥0 k−l−Pr h j=1 j l−h1 h1−h2 hr−1−hr hr × x1 x2 x3 ··· xr+1 xr+2 t =1+ G (1, 1, 0,..., 0; z)x2 t t + G (k,l,h1,...,hr; z) − G0(k,l,h1,...,hr; z) k≥l+Pr h j=1 j l>h1≥···≥Xhr≥0 l≥2 k−l−Pr h j=1 j l−h1 h1−h2 hr−1−hr hr × x1 x2 x3 ··· xr+1 xr+2.

Note that we used (14) in the third term. Applying Lemma 6 (iii), we obtain

x t 1 D (Φt − Φt )= 2 + + x Gt(k,l,h ,...,h ; z) q 0 1 − z z 1 − z 2 1 r k≥l+Pr h j=1 j l≥h1≥···≥Xhr≥0 l≥1 k−l−Pr h j=1 j l−h1 h1−h2 hr−1−hr hr × x1 x2 x3 ··· xr+1 xr+2 x t 1 = 2 + + x Φt − 1 . 1 − z z 1 − z 2

Now we deﬁne another q-diﬀerence operator Θq :

f(z) − f(qz) (Θ f)(z) := z(D f)(z)= . q q 1 − q

Corollary 8. (i) We have

t t xr+2 t t ΘqΦr−1 = x1Φr−1 + (Φr−2 − Φr−1), xr+1  xj+3 t t Θ (Φt − Φt )= (Φ − Φ ), j =1, 2,...,r − 2,  q j j+1 x j−1 j  x j+2  t t 3 t t  Θq(Φ0 − Φ1)= (Φ − Φ0 − 1),  x2 t t z z + (1 − z)t t  Θq(Φ − Φ0)= x2 + x2(Φ − 1).  1 − z 1 − z   t t t t t (ii) Let y0 = Φr−1, yj = Φr−1−j − Φr−j (j =1, 2,...,r − 1), yr = Φ − Φ0. Then we have xr+2−j j x1xr+2−j j−1 yj = Θqy0 − Θq y0 + δj,r xr+2 xr+2 for j =1, 2,...,r, where δ stands for the Kronecker’s delta.

22 t (iii) The function y0 = Φr−1(z) satisﬁes the following q-diﬀerence equation r−1 r+1 r j Θq − (x1 + tx2)Θq − t (xr+2−j − x1xr+1−j)Θq ( j=0 X r−1 r+1 r j −z Θq + ((1 − t)x2 − x1)Θq + (1 − t) (xr+2−j − x1xr+1−j)Θq y0 j=0 !) X = xr+2z. proof. (i) is easily shown by using Θq = zDq. (ii) The relations in (i) can be written as x Θ y = x y + r+2 y , q 0 1 0 x 1 x r+1 x  Θ y = r+2−j y − r+2−j δ , j =1, 2,...,r − 1  q j x j+1 x r−1,j (15)  r+1−j r+1−j  z z + (1 − z)t Θ y = x + x (y + y + ··· + y − 1). q r 1 − z 2 1 − z 2 0 1 r   By the second relation of (15), we have xr+1−j yj+1 = Θqyj + δj,r−1, j =1, 2,...,r − 1. (16) xr+2−j Hence we have xr+2−j j−1 yj = Θq y1, j =2,...,r − 1. xr+1 xr+1 Since y = (Θqy − x y ), we have 1 xr+2 0 1 0

xr+2−j j j−1 yj = Θqy0 − x1Θq y0 xr+2 for j =2,...,r − 1. This is also valid for j = 1. By (16) we have

x2 yr = Θqyr−1 +1. x3 This completes the proof. (iii) By the diﬀerence equation (ii) for j =1, 2,...,r, we have r r xr+2−j j x1xr+2−j j−1 y0 + y1 + ··· + yr − 1= y0 + Θqy0 − Θq y0 xr xr j=1 +2 j=1 +2 X X r−1 xr+2−j − x1xr+1−j j x2 r = Θqy0 + Θqy0. (17) xr xr j=0 +2 +2 X x2 r x1x2 r−1 On the other hand, yr = Θ y − Θ y + 1 which is the case of j = r in (ii). xr+2 q 0 xr+2 q 0 Applying Θq to this equation and combining it with (17) and the third relation of (15), we have x2 r+1 x1x2 r Θq y0 − Θqy0 xr+2 xr+2 r−1 z z + (1 − z)t xr+2−j − x1xr+1−j j x2 r = x2 + x2 Θqy0 + Θqy0 . 1 − z 1 − z xr xr j=0 +2 +2 ! X

23 Hence we obtain

r+1 r (1 − z)Θq y0 − x1(1 − z)Θqy0 r−1 j r = xr+2z + {z + (1 − z)t} (xr+2−j − x1xr+1−j)Θqy0 + x2Θqy0 j=0 ! X and the conclusion.

t 3.4 Relation between Φj(z) and the basic hypergeometric functions t First, we prove the following relation between Φr−1(z) and the basic hypergeometric function r+2φr+1. Theorem 9. Let r be a positive integer. We have

t xr+2z Φr−1 = r−1 1 − (x1 + tx2) − t j=0(xr+2−j − x1xr+1−j) q, a ,...,a b ··· b φ 1P r+1 q, 1 r+1 z , × r+2 r+1 ; r+1 b1,...,br+1 q a ··· a 1 r+1 where ai, bi’s are given by (3) and (4). For the proof, we need the next lemmas.

−m Lemma 10. (i) Let r be a positive integer and bj =6 q for m = 0, 1,... and j =1, 2,...,r. For any non-negative integer n, a ,...,a Dn φ 1 r+1 ; q, az q r+1 r b ,...,b 1 r n n n (a1,...,ar+1; q)n a a1q ,...,ar+1q = n r+1φr n n ; q, az . (b ,...,b ; q) (1 − q) b1q ,...,brq 1 r n −m (ii) Let r be a positive integer and a1 ··· ar+1b1 ··· br =6 0. Let bj =6 q for m = a ,...,a 0, 1,... and j =1, 2,...,r. Then u(z)= φ 1 r+1 ; q, z satisﬁes r+1 r b ,...,b 1 r q − b q − b Θ Θ + 1 ··· Θ + r u(z) q q b (1 − q) q b (1 − q) 1 r a a ··· a 1 − a 1 − a = 1 2 r+1 qrz Θ + 1 ··· Θ + r+1 u(z), b b ··· b q a (1 − q) q a (1 − q) 1 2 r 1 r+1

a1,...,ar+1 b1···br and v(z)= r+1φr ; q, r z satisﬁes b ,...,b a1···ar+1q 1 r q − b q − b Θ Θ + 1 ··· Θ + r v(z) q q b (1 − q) q b (1 − q) 1 r 1 − a 1 − a = z Θ + 1 ··· Θ + r+1 v(z). q a (1 − q) q a (1 − q) 1 r+1 24 (iii) For a non-negative integer n,

n n m m Θq = Sq(n, m)z Dq . m=0 X proof. (i) and (iii) are shown by induction on n. (ii) is derived from a consequence of [5, Exercise 1.31].

Lemma 11. Suppose f(z) is holomorphic at z = 0. If (Θq − 1)(f(z)) = 0, we have f(z)= cz for certain constant c.

∞ n proof. By assumption we have f(qz)= qf(z). Substituting f(z)= n=0 anz into this equality, we have a0 = a2 = a3 = ··· = 0 while a1 is arbitrary. P Proof of Theorem 9. Put

r−1 r+1 r j L := Θq − (x1 + tx2)Θq − t (xr+2−j − x1xr+1−j)Θq j=0 X r−1 r+1 r j −z Θq + ((1 − t)x2 − x1)Θq + (1 − t) (xr+2−j − x1xr+1−j)Θq j=0 ! X By Corollary 8 (iii), we have

L(y0)= xr+2z, (18)

t where y0 = Φr−1. By the way, according to the deﬁnition of αi’s and βi’s (see §1), we have

L = (Θq + β1) ··· (Θq + βr+1) − z(Θq + α1) ··· (Θq + αr+1).

Put 1 1 L := Θ Θ + (1 + β ) ··· Θ + (1 + β ) q q q 1 q q r+1 1 1 1 e − zq Θ + Θ + (1 + α ) ··· Θ + (1 + α ) . q q q q 1 q q r+1 

Since Θq(zf(z)) = z(qΘq + 1)(f(z)), we ﬁnd

r+2 (Θq − 1)L(zf(z)) = zq L(f(z)).

By Lemma 10 (ii), we have e L(v(z))=0,

q, a1,...,ar+1 b1···br+1 where v(z)= r+2φr+1 e; q, r+1 z , and hence b ,...,b a1···ar+1q 1 r+1 

(Θq − 1)L(zv(z))=0.

25 In particular, by Lemma 11, we have L(zv(z)) = cz (19) for certain constant c. From (18) and (19), we have

L(cy0 − xr+2zv(z))=0.

Since both y0 and zv(z) have no constant terms, we have

cy0 = xr+2zv(z) due to [7, Lemma 3.4]. 2 n n Put y0 = Az + o(z ) and let us determine the constant A. Notice that Θqz = [n]z for any positive integer n. Then from the diﬀerence equation of Corollary 8 (iii), we get r−1 2 1 − (x1 + tx2) − t (xr+2−j − x1xr+1−j) Az + o(z )= xr+2z, ( j=0 ) X and hence

xr+2 xr+2 A = = r−1 . c 1 − (x1 + tx2) − t (xr+2−j − x1xr+1−j) j=0 This completes the proof. P

t Secondly, we prove the following relation for Φj(z). t Theorem 12. Put y0 = Φr−1. For −1 ≤ j ≤ r − 1, we have

r−1−j t 1 (j) i i Φj(z)= Ai z Dqy0 + δj,−1, xr +2 i=0 X where r−1−j (j) Ai = (xr+2−m − x1xr+1−m)Sq(m, i)+ x1xj+2Sq(r − 1 − j, i) m=i X for 0 ≤ i ≤ r − 1 − j.

t t t t t proof. If y0 = Φr−1, y1 = Φr−2 − Φr−1,..., yr−1−j = Φj − Φj+1, we have

r−1−j t Φj = y0 + ym. m=1 X Using Corollary 8 (ii), we have

r−1−j t xr+2−m m x1xr+2−m m−1 Φj = y0 + Θq y0 − Θq y0 + δj,−1 xr xr m=1 +2 +2 r−1−j X xr+2−m − x1xr+1−m m x1xj+2 r−1−j = Θq y0 + Θq y0 + δj,−1. xr xr m=0 +2 +2 X

26 By Lemma 10 (iii),

r−1−j m t xr+2−m − x1xr+1−m i i Φj = Sq(m, i)z Dqy0 xr m=0 +2 i=0 X r−1−j X x1xj+2 i i + Sq(r − 1 − j, i)z Dqy0 + δj,−1 xr +2 i=0 r−1−j r−1−jX xr+2−m − x1xr+1−m i i = Sq(m, i)z Dqy0 xr i=0 m=i +2 X Xr−1−j x1xj+2 i i + Sq(r − 1 − j, i)z Dqy0 + δj,−1 xr +2 i=0 r−1−jX 1 (j) i i = Ai z Dqy0 + δj,−1. xr +2 i=0 X

t Finally, we show the following relation between Φj and the basic hypergeometric function r+2φr+1. Theorem 13. For −1 ≤ j ≤ r − 1, we have

t 1 Φj(z)= r−1 1 − (x1 + tx2) − t i=0 (xr+2−i − x1xr+1−i) r−1−j qi+1, a qi,...,a qi b ··· b A(j)B zi+1 Pφ 1 r+1 q, 1 r+1 z δ , × i i r+2 r+1 i i ; r+1 + j,−1 b1q ,...,br+1q q a ··· ar i=0 1 +1 X g where

r−1−j (j) Ai = (xr+2−m − x1xr+1−m)Sq(m +1, i +1)+ x1xj+2Sq(r − j, i + 1) m=i g X for 0 ≤ i ≤ r − 1 − j, and

(q, a ,...,a ; q) b ··· b i B = 1 r+1 i 1 r+1 i (b ,...,b ; q) qr+1(1 − q)a ··· a 1 r+1 i 1 r+1 for 0 ≤ i ≤ r.

For the proof, we need the next lemma.

Lemma 14. For any non-negative integer i,

i i−1 i i Dq(zf(z)) = [i]Dq (f(z)) + q zDq(f(z)).

27 proof. We prove this lemma by induction on i. If i = 0, it holds because both sides become zf(z). If i = 1, by the deﬁnition of Dq, zf(z) − qzf(qz) D (zf(z)) = q (1 − q)z zf(z) − qzf(z)+ qzf(z) − qzf(qz) = (1 − q)z

= f(z)+ qzDq(f(z)). Therefore the identity holds. Assume that the identity holds for i (i ≥ 1). By the induction hypothesis, i+1 i i−1 i i Dq (zf(z)) = DqDq(zf(z)) = Dq [i]Dq (f(z)) + q zDq(f(z)) i i i = [i]Dq(f(z)) + q Dq(zDq(f(z))) i i i i+1 i+1 = [i]Dq(f(z)) + q Dq(f(z)) + q zDq (f(z)).

i 1−qi i 1−qi+1 By [i]+ q = 1−q + q = 1−q = [i + 1], we have i+1 i i+1 i+1 Dq (zf(z)) = [i + 1]Dq(f(z)) + q zDq (f(z)).

Proof of Theorem 13. Set 1 A = r−1 1 − (x1 + tx2) − t i=0 (xr+2−i − x1xr+1−i) and P q, a ,...,a b ··· b v z φ 1 r+1 q, 1 r+1 z . ( )= r+2 r+1 ; r+1 b1,...,br+1 q a ··· a 1 r+1 Applying Theorem 9 to Theorem 12, we have r−1−j t (j) i i Φj(z)= A Ai z Dq(zv(z)) + δj,−1. i=0 X By Lemma 14, r−1−j r−1−j r−1−j (j) i i (j) i i−1 (j) i+1 i i Ai z Dq(zv(z)) = Ai z [i]Dq (v(z)) + Ai z q Dq(v(z)) i=0 i=0 i=0 X rX−1−j X (j) (j) i i+1 i = Ai+1[i +1]+ Ai q z Dq(v(z)), i=0 X (j) (j) (j) where Ar−j = 0. According to the deﬁnition of Ai , Ai and the q-Stirling number Sq,

(j) (j) i g(j) Ai+1[i +1]+ Ai q = Ai . Also by Lemma 10 (i), we obtain g qi+1, a qi,...,a qi b ··· b i v z B φ 1 r+1 q, 1 r+1 z . Dq( ( )) = i r+2 r+1 i i ; r+1 b1q ,...,br+1q q a ··· a 1 r+1 This completes the proof.

28 For i =0, 1,...,r − 1, we set

i+1 i i q , a1q ,...,ar+1q b1 ··· br+1 vi = r+2φr+1 i i ; q, r b1q ,...,br+1q q a ··· a 1 r+1 and (q, a ,...,a ; q) b ··· b i B = B qi+1 = q 1 r+1 i 1 r+1 . i i (b ,...,b ; q) qr(1 − q)a ··· a 1 r+1 i 1 r+1 According tof Theorem 13, we obtain the following corollary. Corollary 15. The identity

r−1−j 1 Φt (q)= A(j)B v , j =0, 1,...,r − 1 j c i i i i=0 X gf holds, where

r−1

c =1 − (x1 + tx2) − t (xr+2−i − x1xr+1−i). i=0 X Or equivalently we have

t Φ0(q) B0v0 t Φ1(q) 1  B1v1   .  = A f. . c .     t q ]f  Φr−1( ) Br−1vr−1         with (0) (0) ](0) ](0) A0 A1 ··· Ar−2 Ar−1  (1) (1) ](1)  Ag0 Ag1 ··· Ar−2  . . ..  A =  . . .  .  g g  ^(r−2) ^(r−2)  A A   0 1  ^(r−1)  A   0    3.5 t-qMZVs as special values of t-qMPLs The t-qMPLs are related to t-qMZVs as follows.

Lemma 16. For positive integers k1,k2,...,kl with k1 ≥ 2, we have

k1 k2 kl l t k1 − 2 kj − 1 Pl (k −a ) t Li q ··· − q i=1 i i ζ a , a ,...,a . k;q( )= (1 ) q( 1 2 l) a − 2 aj − 1 a =2 a =1 a =1 1 j=2 X1 X2 Xl Y For its proof we here introduce the algebraic setup of t-qMZVs (see [14] for de- tails). Let ~ be a formal variable. Denote by H~,t = Q[~, t]hx, yi the non-commutative 1 0 polynomial algebra over Q[~, t] in two indeterminates x and y, and by H~,t and H~,t its

29 j−1 subalgebras Q[~, t]+H~,ty and Q[~, t]+xH~,ty, respectively. Put zj = x y (j ≥ 1). We define the weight and the depth of a word u = zk1 zk2 ··· zkl by wt(u)= k1 +k2 +···+kl t 0 and dep(u)= l, respectively. Define the Q[~, t]-linear map Zq : H~,t −→ Q[~, t][[q]] by t Zq(1) = 1 and t t b Zq(zk1 zk2 ··· zkl )= ζq(k1,k2,...,kl) (k1 ≥ 2). b We also define the substitution map f : Q[~, t][[q]] −→ Q[t][[q]] by f : ~ 7−→ 1 − q and b set t t Zq = f ◦ Zq. t t −1 t We let S = Ryγ Ry , where γ denotes the automorphism on H~,t characterized by t t b γ (x) = x, γ (y) = tx + y and the maps Ry and Lx are Q-linear maps, called right- multiplication of y and left–multiplication of x, defined respectively by Ry(w) = wy t t −1 t and Lx(w) = xw for any w ∈ H~,t. Also we put S~ = Ryγ~Ry , where γ~ denotes t t

the automorphism on H~,t characterized by γ~(x) = x, γ~(y) = tx + y + ~t, and k zk := zk1 zk2 ··· zkl for =(k1,...,kl). Proof of Lemma 16. This identity holds when t = 0 according to (7) in Okuda-Takeyama[10]. Let L~ be the Q[t]-linear map given by

k1 k2 kl l k1 − 2 kj − 1 l Pi=1(ki−ai) L~(zk1 zk2 ··· zkl ) := ··· ~ za1 za2 ··· zal . a − 2 aj − 1 a =2 a =1 a =1 1 j=2 X1 X2 Xl Y 

Let f~ denote the automorphism on H~,t characterized by f~(x) = x + ~, f~(y) = y. −1 t Then we have L~ = Lxf~Lx . By deﬁnition of S~, we ﬁnd that

t 0 t Zq = Zq S~,

0 t 0 where Zq (w) := Zq(w)|t=0 (w ∈ H~,t). Hence we have

0 t Zq S~L~(zk1 zk2 ··· zkl ) t = ZqL~(zk1 zk2 ··· zkl )

k1 k2 kl l k1 − 2 kj − 1 l Pi=1(ki−ai) t = ··· (1 − q) ζq(a1, a2,...,al). a − 2 aj − 1 a =2 a =1 a =1 1 j=2 X1 X2 Xl Y We also have

t k1−1 kl−1−1 kl−1 L~S (zk1 zk2 ··· zkl )= L~(x (tx + y) ··· x (tx + y)x y)

′ ′ Ô

l−dep(Ô) l−dep( ) Ô

= L~ t zÔ = t L~(z ) Ô Ô ! X X 0 Zq ′ l−dep(Ô)

7→ t LiÔ;q(q) (by using Lemma 16 when t = 0) XÔ Lit q . = k;q( )

30 Moreover we have

t −1 t −1 L~S = Lxf~Lx Ryγ Ry −1 t −1 −1 = Lxf~RyLx γ Ry (since [Lx , Ry]=0) t −1 −1 −1 t = LxRyf~γ Lx Ry (since [f~, Ry]=0, [Lx ,γ ]=0) t −1 −1 t t = RyLxγ~f~Ry Lx (since [Lx, Ry]=0, f~γ = γ~f~) t −1 −1 −1 t = Ryγ~LxRy f~Lx (since [f~, Ry ]=0, [Lx,γ~]=0) t −1 −1 −1 = Ryγ~Ry Lxf~Lx (since [Lx, Ry ]=0) t = S~L~, and hence the conclusion.

4 Proof of main theorem

Combining the following four propositions (Propositions 17, 21, 24, 27), we obtain our main theorem.

t t 4.1 Relation between Ψj and Φ0(q)

For any non-negative integers k,l,h1,...,hr (r ∈ N) and j =0, 1,...,r − 1, we set

t t ξj(k,l,h1,...,hr)= ζq(k),

k∈Ij (k,l,hX1,...,hr) t t Ψj =Ψj(u1,u2,...,ur+2) r t k−l−Pi=1 hi l−h1 h1−h2 hr−1−hr hr = ξj(k,l,h1,...,hr)u1 u2 u3 ··· ur+1 ur+2. k,l,h ,...,h 1X r≥0 We set

r r+1−i i − 1 i−p x1 xi+1 − xr+2 −p Xp = ((1 − q)x1) +(1 − (1 − q)x1) , p =1,...,r +1, p − 1 xr i=p +2 X and

r r+1−i i − 2 i−p x1 xi+1 − xr+2 1−p Yp = ((1 − q)x1) + (1 − (1 − q)x1) , p =2, . . . , r, p − 2 xr i=p +2 X where xi’s are given by (1), (2).

Proposition 17. We have

t t Φ0(q) − Φ1(q) t t t q Φ0(q) − Φ2(q) t Φ0( )   Ψ0 = + XM1 . , 1 − (1 − q)x1 .   Φt (q) − Φt (q)  0 r−1    31 where Y Y X = X − 2 ,...,X − r 2 1 − (1 − q)x r 1 − (1 − q)x 1 1 and

1 1 r−1 x1 x3 1 1 1 r−2 −1 1  x  x4 M x 1 T     1 = r+2 ......        1   1   −1 1   xr+1  x1            with the lower triangular matrix T =(tij)(r−1)×(r−1) deﬁned by

i−1 (q − 1)i−j, i ≥ j, t = j−1 ij 0, i < j. To prove this proposition, we show three lemmas. Set j r+1−i i − 2 i−p x1 xi+1 Zpj = ((1 − q)x1) , 2 ≤ p ≤ j ≤ r. p − 2 xr i=p +2 X Let W and D be r × r matrices deﬁned by

Y2 Y3 Y4 ··· Yr 1 Y − Z Y Y ··· Yr 1  2 22 3 4  Y2 − Z23 Y3 − Z33 Y4 ··· Yr 1 W =  ......   ......    Y − Z ,r Y − Z ,r Y − Z ,r ··· Yr 1  2 2 −1 3 3 −1 4 4 −1   Y − Z r Y − Z r Y − Z r ··· Yr − Zrr 1  2 2 3 3 4 4    and −1 −1 −1 D = diag(X2 ,X3 ,...,Xr , 1 − (1 − q)x1). Lemma 18. We have

t t t Ψ0(u1,u2,...,ur+2) − Ψ1(u1,u2,...,ur+2) Φ0(x1, x2,...,xr+2; q) t t t Ψ1(u1,u2,...,ur+2) − Ψ2(u1,u2,...,ur+2) Φ1(x1, x2,...,xr+2; q)     WD . , . = . .  t t   t  Ψr−2(u1,u2,...,ur+2) − Ψr−1(u1,u2,...,ur+2) Φr−1(x1, x2,...,xr+2; q)  t     Ψ (u ,u ,...,ur )     r−1 1 2 +2    where, for j =2,...,r +2,

u = x1 , 1 1−(1−q)x1 xr+2 uj = r+2−j Xj−1  x1  r  1 i−1 i−j+1 r+1−i xr+2  = r −j ((1 − q)x ) x x − x + j− .  x +2 j−2 1 1 i+1 r+2 (1−(1−q)x ) 1 1 (i=j−1 1 )  P  (20)  32 proof. Let j be an integer with 0 ≤ j ≤ r − 1. Step 1. By deﬁnitions and Lemma 16, we have

r t t k−l−Pi=1 hi l−h1 h1−h2 hr−1−hr hr Φj(q)= Gj(k,l,h1,...,hr; q)x1 x2 x3 ··· xr+1 xr+2 k,l,h ,...,h 1X r≥0 k k k l 1 2 l k − 2 k − 1 = ··· 1 i a − 2 ai − 1 k,l,h ,...,h a =2 a =1 a =1 1 (i=2 ) 1X r≥0 (k1,...,kl)∈XIj (k,l,h1,...,hr) X1 X2 Xl Y l P (ki−ai) r k−l−P hi hr− −hr i=1 t i=1 l−h1 h1−h2 1 hr × (1 − q) ζq(a1,...,al)x1 x2 x3 ··· xr+1 xr+2.

By changing the order of the sums, we get

′ l ′ t k1 − 2 ki − 1 Φj(q)= ′ a1 − 2 ai − 1 k,l,h1,...,hr≥0 (a1,...,al)∈I0(k,l,h1,...,hr) kp≥ap (i=2 ) X X p=1X,...,l Y k′ ≥j+2 1 l ′ r ′ ′ ′ ′ ′ ′ ′ ′ k −l−P h l−h h −h h −hr h Pi=1(ki−ai) i=1 i 1 1 2 r−1 r t × (1 − q) x1 x2 x3 ··· xr+1 xr+2ζq(a1,...,al),

′ ′ ′ ′ ′ ′ ′ ′ where k , h1,...,hr are non-negative integers such that (k1,...,kl) ∈ Ij(k ,l,h1,...,hr). Since

′ ′ ′ ′ ′ ′ ′ h1 + ··· + hr = rl − r(l − h1) − (r − 1)(h1 − h2) −···− (hr−1 − hr) and ′ ′ ′ ′ ′ ′ hr = l − (l − h1) − (h1 − h2) −···− (hr−1 − hr), we get

′ r ′ ′ ′ ′ h′ h′ ′ k −l−Pi=1 hi l−h1 h1−h2 r−1− r hr x1 x2 x3 ··· xr+1 xr+2 ′ ′ ′ ′ ′ r l−h r−1 h1−h2 h −hr ′ x x 1 x x x x r−1 = xk −(r+1)lxl 1 2 1 3 ··· 1 r+1 . 1 r+2 x x x r+2 r+2 r+2 ′ l ′ ′ l l h δ ′ h h δ ′ j ,...,r By using the facts that − 1 = i=1 ki,1, j−1 − j = i=1 ki,j, = 2 and k′ = k′ + ··· + k′, we have 1 l P P ′ r ′ ′ ′ ′ h′ h′ ′ k −l−Pi=1 hi l−h1 h1−h2 r−1− r hr x1 x2 x3 ··· xr+1 xr+2 δ l l r δk′ , r−1 k′ ,2 δk′ ,r ′ i 1 i i xr+2 kj x1x2 x1 x3 x1xr+1 = r+1 x1 ··· . x xr xr xr 1 i=1 +2 +2 +2 Y

33 Hence

′ t k1 − 2 Φj(q)= ′ a1 − 2 k,l,h1,...,hr≥0 (a1,...,al)∈I0(k,l,h1,...,hr) kp≥ap X X p=1X,...,l k′ ≥j+2 1 l ′ l k − 1 l ′ xr+2 i Pi=1(ki−ai) × (1 − q) r+1 ai − 1 x (i=2 ) 1 Y δ l r δk′ , r−1 k′ ,2 δk′ ,r ′ i 1 i i ki x1x2 x1 x3 x1xr+1 t × x1 ··· ζq(a1,...,al) xr xr xr i=1 +2 +2 +2 Y k′ − 2 = 1 ′ a1 − 2 k,l,h1,...,hr≥0 (a1,...,al)∈I0(k,l,h1,...,hr) kp≥ap X X p=1X,...,l k′ ≥j 1 +2

l l δ ′ ′ r δk′ ,1 r−1 k ,2 l ′ i i ki − 1 P (k −ai) x1x2 x1 x3 × ((1 − q)x1) i=1 i ai − 1 xr xr (i=2 ) i=1 +2 +2 Y Y δk′ ,r l x x i x ×···× 1 r+1 xk r+2 ζt(a ,...,a ). x 1 xr+1 q 1 l r+2 1

Step 2. For any ﬁxed (a1,...,al) ∈ I0(k,l,h1,...,hr), we set

′ l ′ k − 2 k − 1 l ′ (j) 1 i Pi (k −ai) S (a1,...,al)= ((1 − q)x1) =1 i ′ a1 − 2 ai − 1 kp≥ap (i=2 ) p=1X,...,l Y k′ ≥j 1 +2 δ l r δk′ , r−1 k′ ,2 δk′ ,r x x i 1 x x i x x i × 1 2 1 3 ··· 1 r+1 . xr xr xr i=1 +2 +2 +2 Y (j) (j) Then S (a1,...,al)= S1 S2 ··· Sl, where

i xrx δi,1 xr−1x δi,2 x x δi,r (j) − 2 i−a1 1 2 1 3 1 r+1 S1 = ((1 − q)x1) ··· a − 2 xr xr xr i≥a ,i≥j+2 1 +2 +2 +2 X1 and

r δi,1 r−1 δi,2 δi,r i − 1 i−am x1x2 x1 x3 x1xr+1 Sm = ((1 − q)x1) ··· am − 1 xr xr xr i≥a +2 +2 +2 Xm for m =2,...,l. (j) To compute S1 , we consider the following three cases. Case (11): a1 ≥ r + 1. In this case, we have

j i − 2 S( ) = ((1 − q)x )i−a1 = (1 − (1 − q)x )1−a1 . 1 a − 2 1 1 i≥a 1 X1 34 Case (12): a1 = p for some p with j +2 ≤ p ≤ r. In this case, we have

r δi,1 r−1 δi,2 δi,r (j) i − 2 i−p x1x2 x1 x3 x1xr+1 S1 = ((1 − q)x1) ··· p − 2 xr xr xr i≥p +2 +2 +2 X r r+1−i i − 2 i−p x1 xi+1 i − 2 i−p = ((1 − q)x1) + ((1 − q)x1) p − 2 xr p − 2 i=p +2 i≥r+1 X X r r+1−i i − 2 i−p x1 xi+1 − xr+2 1−p = ((1 − q)x1) + (1 − (1 − q)x1) p − 2 xr i=p +2 X = Yp.

Case (13): a1 = p for some p with 2 ≤ p ≤ j + 1. In this case,

r δi,1 r−1 δi,2 δi,r (j) i − 2 i−p x1x2 x1 x3 x1xr+1 S1 = ((1 − q)x1) ··· p − 2 xr xr xr i≥j+2 +2 +2 +2 X r r+1−i i − 2 i−p x1 xi+1 i − 2 i−p = ((1 − q)x1) + ((1 − q)x1) p − 2 xr p − 2 i=j+2 +2 i≥r+1 X X = Yp − Zp,j+1.

Now we compute Sm for 2 ≤ m ≤ l. We have two cases. Case (m1): am = p for some 1 ≤ p ≤ r. Then we have

r r+1−i i − 1 i−p x1 xi+1 i − 1 i−p Sm = ((1 − q)x1) + ((1 − q)x1) p − 1 xr p − 1 i=p +2 i≥r+1 X X r r+1−i i − 1 i−p x1 xi+1 − xr+2 −p = ((1 − q)x1) + (1 − (1 − q)x1) p − 1 xr i=p +2 X = Xp.

Case (m2): am ≥ r + 1. Then

i − 1 i−am −am Sm = ((1 − q)x1) = (1 − (1 − q)x1) . am − 1 i≥a Xm p (j) Setting Xp = Xp(1 − (1 − q)x1) for p =1,...,r, we obtain S (a1,...,al). There are three cases. Case (i): ae1 ≥ r + 1.

35 In this case,

(j) 1−a1 l−h1 h1−h2 hr−1−hr S (a1,...,al) =(1 − (1 − q)x1) X1 X2 ··· Xr l a (1 − (1 − q)x )− Pm=2 m × 1 −{(l−h )+2(h −h )+···+r(h −hr)} (1 − (1 − q)x1) 1 1 2 r−1

1 − (1 − q)x1 l−h1 = k {X1(1 − (1 − q)x1)} (1 − (1 − q)x1)

2 h1−h2 r hr−1−hr × X2(1 − (1 − q)x1) ···{Xr(1 − (1 − q)x1) } 1 − (1 − q)x 1 l−h1 h1−h2 hr−1−hr = k X1 X2 ··· Xr . (1 − (1 − q)x1) e e e Case (ii): a1 = p for some p with j +2 ≤ p ≤ r. Then

(j) S (a1,...,al)

l−h1 h1−h2 hp−2−hp−1 hp−1−hp−1 hp−hp+1 hr−1−hr = YpX1 X2 ··· Xp−1 Xp Xp+1 ··· Xr − Pl a (1 − (1 − q)x ) m=2 m × 1 −{(l−h )+2(h −h )+···+(p−1)(hp− −hp− )+p(hp− −hp−1)+(p+1)(hp−hp )+···+r(hr− −hr)} (1 − (1 − q)x1) 1 1 2 2 1 1 +1 1 −1 YpX p l−h1 h1−h2 hr−1−hr = k X1 X2 ··· Xr . (1 − (1 − q)x1) e e e Case (iii): a1 = p for some p with 2 ≤ p ≤ j + 1. Then

−1 (Yp − Zp,j+1)X (j) p l−h1 h1−h2 hr−1−hr S (a1,...,al)= k X1 X2 ··· Xr . (1 − (1 − q)x1) e e e Step 3. Notice that I0(k,l,h1,...,hr) equates to the disjoint union

r−2

Ir−1(k,l,h1,...,hr) ⊔ (Ii(k,l,h1,...,hr) \ Ii+1(k,l,h1,...,hr)) . i=0 G Therefore

r t Φj(q)=  +  a ,...,a ∈I k,l,h ,...,h k,l,h1,...,hr≥0 (a1,...,al)∈Ir−1(k,l,h1,...,hr) i=2 ( 1 l) i−2( 1 r)  X  X X Xa1=i  l  xr  × S(j)(a ,...,a )xk +2 ζt(a ,...,a ).  1 l 1 xr+1 q 1 l 1

36 (j) Using the results of S (a1,...,an) obtained in Step 2, we have

1 − (1 − q)x Φt (q)= 1 Xl−h1 Xh1−h2 ··· Xhr−1−hr j (1 − (1 − q)x )k 1 2 r k,l,h ,...,h 1 1X r≥0 (a1,...,al)∈IrX−1(k,l,h1,...,hr) l j+1 e e e k xr+2 t × x1 r+1 ζq(a1,...,al)+ x1 i=2 k,l,h1,...,hr≥0 (a1,...,al)∈Ii−2(k,l,h1,...,hr) X X Xa1=i −1 l (Yi − Zi,j+1)X xr+2 × i Xl−h1 Xh1−h2 ··· Xhr−1−hr xk ζt(a ,...,a ) (1 − (1 − q)x )k 1 2 r 1 xr+1 q 1 l 1 1 r −1 e e e YiXi l−h1 h1−h2 + k X1 X2 (1 − (1 − q)x1) i=j+2 k,l,h1,...,hr≥0 (a1,...,al)∈Ii−2(k,l,h1,...,hr) X X Xa =i 1 e e x l hr−1−hr k r+2 t ×···× X x ζ (a ,...,al) r 1 xr+1 q 1 1 k l e t x1 xr+2 =(1 − (1 − q)x ) ξ (k,l,h ,...,hr) 1 r−1 1 1 − (1 − q)x xr+1 k,l,h ,...,h 1 1 1X r≥0 j +1 Y Z l−h1 h1−h2 hr−1−hr i − i,j+1 t × X1 X2 ··· Xr + ξi−2(k,l,h1,...,hr) Xi i=2 k,l,h1,...,hr≥0 X X e e e x k x l t 1 r+2 l−h1 h1−h2 hr−1−hr −ξ (k,l,h ,...,hr) X X ··· X i−1 1 1 − (1 − q)x xr+1 1 2 r 1 1 r Yi t t e e e + ξi−2(k,l,h1,...,hr) − ξi−1(k,l,h1,...,hr) Xi i=j+2 k,l,h1,...,hr≥0 X X k l x xr × 1 +2 Xl−h1 Xh1−h2 ··· Xhr−1−hr . 1 − (1 − q)x xr+1 1 2 r 1 1 Step 4. Since e e e l =(l − h1)+(h1 − h2)+ ··· +(hr−1 − hr)+ hr and l + h1 + h2 + ···+ hr =(l − h1)+2(h1 − h2)+3(h2 − h3)+ ···+ r(hr−1 − hr)+(r +1)hr, and (20), we have

k l x1 xr+2 Xl−h1 Xh1−h2 ··· Xhr−1−hr 1 − (1 − q)x xr+1 1 2 r 1 1 l−h h −h h −hr 1 2 1 2 r r−1 r+1 hr k l l h u xr X e u xer X e u xr Xr u xr =u − −Pi=1 i 1 +2 1 1 +2 2 ··· 1 +2 1 +2 1 xr+1 xr+1 xr+1 xr+1 1 ! 1 ! 1 ! 1 l e e e k−l−Pi=1 hi l−h1 h1−h2 hr−1−hr hr =u1 u2 u3 ··· ur+1 ur+2.

37 Thus

j+1 t t Yi − Zi,j+1 t Φj(q) =(1 − (1 − q)x1)Ψr−1(u1,...,ur+2)+ Ψi−2(u1,...,ur+2) Xi i=2 r X t Yi t t −Ψi−1(u1,...,ur+2) + Ψi−2(u1,...,ur+2) − Ψi−1(u1,...,ur+2) . Xi i=j+2 X Hence we complete the proof of Lemma 18.

Remark 3. We ﬁnd that (20) is equivalent to (1) and (2) as shown in the proof of Lemma 20 below.

Lemma 19. Let T =(tij) be the n × n lower triangular matrix deﬁned by

i−1 xi−j, i ≥ j, t = j−1 ij 0, i < j. ′ then the invertible matrix of T is the lower triangular matrix (tij) deﬁned by

i−1 (−x)i−j, i ≥ j, t′ = j−1 ij 0, i < j. ′ proof. This is easily shown by checking that T (tij) is the identity matrix. Lemma 20. Let M be a r × r matrix given by

Y2 Y3 Yr−1 Yr 1 X − X − ··· Xr − Xr − 2 1−(1−q)x1 3 1−(1−q)x1 −1 1−(1−q)x1 1−(1−q)x1 1−(1−q)x1 Y2 Y3 Yr−1 Yr 1 − X3 − ··· Xr−1 − Xr −  1−(1−q)x1 1−(1−q)x1 1−(1−q)x1 1−(1−q)x1 1−(1−q)x1  Y2 Y3 Yr−1 Yr 1 − − ··· Xr−1 − Xr −  1−(1−q)x1 1−(1−q)x1 1−(1−q)x1 1−(1−q)x1 1−(1−q)x1  M =  . .  .  ······ .. .. ······    Y2 Y3 Yr−1 Yr 1  − − ··· − Xr −   1−(1−q)x1 1−(1−q)x1 1−(1−q)x1 1−(1−q)x1 1−(1−q)x1   Y2 Y3 Yr−1 Yr 1   − − ··· − −   1−(1−q)x1 1−(1−q)x1 1−(1−q)x1 1−(1−q)x1 1−(1−q)x1    Let M1 be the (r − 1) × (r − 1) matrix deﬁned in Proposition 17. Then we have

t t Φ0(x1,...,xr+2; q) Ψ0(u1,...,ur+2) t t t Φ0(x1,...,xr+2; q) − Φ1(x1,...,xr+2; q) Ψ (u ,...,ur ) 1 1 +2 0 M1  t t   .  = M Φ0(x1,...,xr+2; q) − Φ2(x1,...,xr+2; q) . . 1 0 . .  .   t   .  Ψr−1(u1,...,ur+2)  t t    Φ (x ,...,xr ; q) − Φ (x ,...,xr ; q)    0 1 +2 r−1 1 +2    proof. The lemma follows from Lemma 18. Step 1. We ﬁrst remark that from (20), we can represent x1,...,xr+2 by u1,...,ur+2. In fact, by (20), it is trivial that u u x 1 , x r+2 . 1 = r+2 = r+1 1+(1 − q)u1 (1+(1 − q)u1)

38 We rewrite the formulas in (20) in matrix form

xr+2 1 x2 − r r u2 x1 x1(1−(1−q)x1) xr+2 1 x r− u3 3 − xr−1 x 1(1−(1−q)x )2   T  1  x  1 1  , . = 1 . + r+2 . . . .    .    u  xr+2   1   r+1 xr+1 − r    x1   x1(1−(1−q)x1)        (1)     where T1 =(tij ) is a r × r upper triangular matrix with j−1 (1 − q)j−i, i ≤ j, t(1) = i−1 ij 0, i > j. Hence u1 ur+2 q u u2 − r 1+(1− ) 1 x2 u1 2 ur+2 u1 x3 u3 − r−1 −1  u  ur+2  1+(1−q)u1    = T 1 + . . 1 . r+1 . . . u1  .     .     ur+2  r xr+1 ur −  u1   +1 u1       1+(1−q)u1        −1   We see that T1 is obtained from Lemma 19. A direct computation gives the repre- sentations of x2,...,xr+1 by u1,u2,...,ur+2 as we state at (1) and (2). Step 2. If we use the same notations as in Lemma 18, then t t Ψ0(u1,...,ur+2) 1 1 ··· 1 Φ0(x1,...,xr+2; q) t t Ψ (u1,...,ur+2) 1 ··· 1 Φ (x1,...,xr+2; q) 1 −1 −1 1  .  =  . . D W  .  ......  t     t  Ψ (u ,...,ur )  1 Φ (x ,...,xr ; q)  r−1 1 +2     r−1 1 +2  Now      1 1 1 −1 1    .  0 1 1 −1 W .. = , . .    T2 0 . ..   1      1 −1 −Y −Y ··· −Yr 1    2 3  where     Z22 Z23 Z33 Z Z Z  T2 = 24 34 44  . .   . ..    Z r Z r Z r ··· Zrr  2 3 4  is a (r − 1) × (r − 1) matrix. Hence  1 1 1 1 −1   −1   −1 . 0 T2 W = .. 1 −1 .   1 0 . .   1  . ..      −Y −Y ··· −Yr 1 1 −1  2 3        39 On the other hand, direct computations give that 1 1 1 ··· 1 1 1 ··· 1   −1 . M =  .  D .. . ..      1   1     −Y −Y ··· −Yr 1    2 3  −1   Thus we only need to show that T2 = M1. Step 3. By deﬁnition, 1 j − 1 Z − Z = (1 − q)j+1−pxr+1−px p,j+1 pj x p − 2 1 j+2 r+2 for 2 ≤ p ≤ j ≤ r − 1 and 1 r+1−p Zpp = x1 xp+1 xr+2 for 2 ≤ p ≤ r. Hence we get

1 r−1 x3 x1 −1 1 r−2   1 x4 x1 −1 1 T   T   , 2 = .. 3 ..  . .  xr+2 . .  .. ..         xr   x   −1 1  +1  1        (3)  where T3 =(tij ) is the lower triangular matrix deﬁned by

i−1 i−j (3) j−1 (1 − q) , i ≥ j, tij = ( 0, i < j. −1 From Lemma 19, T3 = T (which is given in Proposition 17), thus we obtain this lemma. Proof of Proposition 17. This is nothing but the ﬁrst component of the identity in Lemma 20.

t 4.2 Relation between Ψ0 and r+2φr+1

(i) We recall that Aj is given in Theorem 13 and c, Bj, vj are in Corollary 15. Under these notations, we have the following proposition. g Proposition 21. We have f r 1 1 −1 Ψt = A(0)B v 0 1 − (1 − q)x c j j j 1 j=0 X r−1 r g f xr+2 1 Yi + Xi − ci ,j Bjvj c xr+1−i 1 − (1 − q)x −1 +1 j=0 ( i=2 1 1 ) X X r 1 −1 f = A B v c(1 − (1 − q)x ) j j j 1 j=0 X f 40 where r (0) ci−1,j+1 Aj = A + xr {Xi (1 − (1 − q)x ) − Yi} j +2 1 xr+1−i i=2 1 g X with 0 ≤ j ≤ r − 1 and ci,j’s are given in Lemma 23 below.

Before we prove this proposition, we prepare two lemmas.

Lemma 22. We have

t t Φ0(q) − Φ1(q) t t Φ0(q) − Φ2(q) XM1  .  .   Φt (q) − Φt (q)  0 r−1   −(r−1)  −1 x1 x3 −(r−2) −1 x x x4 r+2  1    = X . T . c .. ..      x−1  x−1   1   r+1     1 −1 B0v0 1 −1  B1v1  ×  . .  A f. , .. .. .    f  1 −1 ]    Br−1vr−1         where X, T and M1 are same as those in Proposition 17 and A is same as that in Corollary 15. proof. Since

t t t 1 Φ0(q) − Φ1(q) 1 −1 Φ0(q) t t t −1 1 Φ0(q) − Φ2(q) 1 −1 Φ1(q)  . .   .  =  . .   .  ......          −1 1 Φt (q) − Φt (q)  1 −1 Φt (q)    0 r−1     r−1          and Corollary 15, we have the lemma.

Lemma 23. We have the following identities.

(i) ^(j) Ar−1−j = xj+3Sq(r − j, r − j) for j =0, 1,...,r − 1.

(ii) (j) ^(j+1) Ai − Ai = xj+3 {Sq(r − j, i + 1) − x1Sq(r − j − 1, i + 1)} for j =0,g1,...,r − 2 and i =0, 1,...,r − 2 − j.

41 (iii) −1 x3 1 −1 x−1 1 −1  4    . . . A =(aij)(r−1)×r ......      x−1   1 −1  r+1   with     S (r +1 − i, j) − x S (r − i, j), i + j ≤ r +1, a = q 1 q ij 0, otherwise, where A is given in Corollary 15.

(iv) −1 x3 1 −1 −1 x4 1 −1 T  .   . .  A =(cij)(r−1)×r ......      x−1   1 −1  r+1   with     min{i,r+1−j} i − 1 c = (q − 1)i−m a . ij m − 1 mj m=1 X (j) proof. (i) Considering the case of i = r − 1 − j in the deﬁnition Ai (in Theorem 13), we have g ^(j) Ar−1−j =(xj+3 − x1xj+2)Sq(r − j, r − j)+ x1xj+2Sq(r − j, r − j)

= xj+3Sq(r − j, r − j).

(ii) By deﬁnition we get

(j) ^(j+1) Ai − Ai r−1−j g = (xr+2−m − x1xr+1−m)Sq(m +1, i +1)+ x1xj+2Sq(r − j, i + 1) m=i Xr−2−j

− (xr+2−m − x1xr+1−m)Sq(m +1, i + 1) − x1xj+3Sq(r − 1 − j, i + 1) m=i X =(xj+3 − x1xj+2)Sq(r − j, i +1)+ x1xj+2Sq(r − j, i + 1)

− x1xj+3Sq(r − 1 − j, i + 1)

= xj+3 (Sq(r − j, i + 1) − x1Sq(r − 1 − j, i + 1)) .

42 (iii) By using (i) and (ii), we have

(0) (1) (0) (1) ](0) ](1) ](0) 1 −1 A0 − A0 A1 − A1 ··· Ar−2 − Ar−2 Ar−1 1 −1  (1) (2) (1) (2) ](1)  A0 − A0 A1 − A1 ··· Ar−2  . .  A = g g g g . .  . . ..  . .  . . .     g g g g   1 −1 ^(r−2) ^(r−1) ^(r−2)    A − A A     0 0 1   x3a11 x3a12 ··· x3a1,r−1 x3a1r  x4a21 x4a22 ··· x4a2,r−1 =  . . .  . . . ..   xr ar , xr ar ,   +1 −1 1 +1 −1 2    (iv) By using (iii), we have

r min{i,r+1−j} −1 i − 1 c = t a = t a = (q − 1)i−ma . ij im mj im mj m − 1 mj m=1 1≤m≤r−1 m=1 i≥m X m+Xj≤r+1 X

Proof of Proposition 21. Using Lemma 22 and Lemma 23 (iv), we have

t t −(r−1) Φ0(q) − Φ1(q) x1 B0v0 t t −(r−2) Φ0(q) − Φ2(q) x x B1v1   r+2  1    XM1 . = X . (cij) f. . . c .. .      f  Φt (q) − Φt (q)  x−1 ]   0 r−1   1  Br−1vr−1       The right-hand side is equal to  

r−1 r xr+2 1 Yi Xi − ci ,j Bjvj. c xr+1−i 1 − (1 − q)x −1 +1 j=0 ( i=2 1 1 ) X X Applying this and Corollary 15 to Proposition 17, we conclude Propositf ion 21.

4.3 Expression of Aj in terms of q-Stirling numbers

Put q1 =1 − q. We have the following expression of Aj in Proposition 21. Proposition 24. For 0 ≤ j ≤ r − 1, we have r−1 u u A = c S (m +1, j +1)+ 1 2 S (r, j + 1), j m q (1 + q u )2 q m=j 1 1 X where r +1 k − 2 q u k − 2 c = + 1 1 (−q )k−r+m−2 m r − m 1+ q u r − m − 1 1 k r m 1 1 =X− +1 u q u × k − uk−r−2u + 1 k+1 . 1+ q u 1 r+2 1+ q u 1 1 1 1 43 For the proof, we need two lemmas.

Lemma 25. The identity

r (−q x )r+1−p Y = (−q x )k−pX + 1 1 (21) p 1 1 k (1 − q x )r k p 1 1 X= holds for p =2,...,r, where Xj’s and Y j’s are same as those in Proposition 17. proof. For any 2 ≤ p ≤ r,

r r+1−i i − 2 i − 2 i−p x1 xi+1 − xr+2 −p Xp = + (q1x1) + (1 − q1x1) p − 2 p − 1 xr i=p +2 X r r+1−i 1−p −p i − 2 i−p−1 x1 xi+1 − xr+2 =Yp − (1 − q1x1) + (1 − q1x1) + q1x1 (q1x1) p − 1 xr i=p+1 +2 1−p −p X −p =Yp − (1 − q1x1) + (1 − q1x1) + q1x1Yp+1 − (1 − δp,r)q1x1(1 − q1x1) −p =Yp + q1x1Yp+1 + δp,rq1x1(1 − q1x1) , where Yr+1 := 0. Rewriting these formulas in matrix forms, we get

1 q x X 1 1 Y 0 2 1 q x 2 X 1 1 Y . 3  . .  3 .  .  = .. ..  .  +   . .   . 0    1 q1x1    q1x1  Xr   Yr  r     1     (1−q1x1)            Hence −1 1 q1x1 Y2 X2 1 q1x1 . Y3   .   .. ..  .  . . = . . .   Xr−1    1 q1x1  q1x1  Yr   Xr − r     1   (1−q1x1)        It is known that if we set 

−1 1 q1x1 1 q1x1  . .  (dij)= .. .. ,    1 q x   1 1  1      then (−q x )j−i, i ≤ j, d = 1 1 ij 0, i > j. Thus we obtain the identity.

Lemma 26. We have the following identities.

44 (i)

r−1 r+1 (0) k − 2 q1u1 k − 2 Aj = + r − m 1+ q1u1 r − m − 1 m=j (k=r−m+1 g X X k−r+m−2 k−r−2 r−k+2 × (−q1) u1 u1 uk − ur+2 Sq(m +1, j + 1) ) r u +1 (−q )ru u + 1 (−q )k−2u + 1 1 r+2 S (r, j + 1) 1+ q u 1 k (1 + q u )2 q 1 1 k 1 1 ! X=2 (0) for 0 ≤ j ≤ r − 1, where Aj ’s are same as those in Theorem 13. (ii) g r+1−p p−1 r+1 u1 (1 + q1u1) k−p Xp(1 − q1x1) − Yp = (−q1) (u1uk − uk+1) ur +2 k p X= +1 for 2 ≤ p ≤ r, where Xp’s and Yp’s are same as those in Proposition 17.

(iii)

r cp−1,j+1 xr+2 {Xp(1 − q1x1) − Yp} r+1−p p=2 x1 X r r −1 +1 k − 2 q u k − 2 = + 1 1 r − m 1+ q u r − m − 1 m=j (k r m 1 1 X =X− +1 (−q )k−r+m−1 × 1 (u u − u ) S (m +1, j + 1) 1+ q u 1 k k+1 q 1 1 u r+1 + 1 (−q )k−1(u u − u )S (r, j + 1) (1 + q u )2 1 1 k k+1 q 1 1 k X=2 for 0 ≤ j ≤ r − 1, where ci,j’s are same as those in Lemma 23.

(0) proof. (i) We compute Aj , which by deﬁnition is

r−1 g (0) Aj = (xr+2−m − x1xr+1−m)Sq(m +1, j +1)+ x1x2Sq(r, j + 1). m=j g X By (1) and (2), we have

r+1 u1 1 k−2 r+2−k ur+2 x1x2 = r (−q1u1) (u1 uk − ur+2)+ 1+ q1u1 u 1+ q1u1 1 (k=2 ) r X u +1 (−q )ru u = 1 (−q )k−2u + 1 1 r+2 , 1+ q u 1 k (1 + q u )2 1 1 k 1 1 X=2

45 and

xr+2−m − x1xr+1−m r 1 +1 k − 2 u = (−q u )k−r+m−2(ur−k+2u − u )+ r+2 um r − m 1 1 1 k r+2 (1 + q u )r−m+1 1 (k r m 1 1 ) =X− +2 r u 1 +1 k − 2 u − 1 (−q u )k−r+m−1(ur−k+2u − u )+ r+2 1+ q u um+1 r − m − 1 1 1 1 k r+2 (1 + q u )r−m 1 1 1 (k r m 1 1 ) =X− +1 r +1 k − 2 q u k − 2 = + 1 1 (−q )k−r+m−2uk−r−2(ur−k+2u − u ). r − m 1+ q u r − m − 1 1 1 1 k r+2 k r m 1 1 =X− +1 Hence we obtain the result. (ii) By (20), (1) and (2),

r+1−j r+1−p p x1 u1 up+1(1 + q1u1) Xp = up+1 = . xr+2 ur+2 Using (1) and (21), we have

r k−p r+1−k k r+1−p −q1u1 u1 uk+1(1 + q1u1) −q1u1 r Yp = + (1 + q1u1) 1+ q u ur 1+ q u k p 1 1 +2 1 1 X= r+1−p p−1 r u1 (1 + q1u1) k−p r+1−p = (−q1) uk+1(1 + q1u1)+(−q1) ur+2 . ur +2 (k p ) X= Hence we have

Xp Xp(1 − q1x1) − Yp = − Yp 1+ q1u1 ur+1−pu (1 + q u )p−1 = 1 p+1 1 1 ur+2 r+1−p p−1 r u1 (1 + q1u1) k−p r+1−p − (−q1) uk+1(1 + q1u1)+(−q1) ur+2 . ur +2 (k p ) X= Thus

ur+2 (Xp(1 − q1x1) − Yp) r+1−p p−1 u1 (1 + q1u1) r k−p r+1−p =up+1 − (−q1) uk+1(1 + q1u1) − (−q1) ur+2 k=p r X r k−p k+1−p r+1−p = − (−q1) uk+1 + (−q1) u1uk+1 − (−q1) ur+2 k p k p X= +1 X= r+1 k−p = (−q1) (u1uk − uk+1). k p X= +1

46 (iii) By (ii), (1) and (2), we have

r cp−1,j+1 xr+2 (Xp(1 − q1x1) − Yp) r+1−p p=2 x1 X r r u ur+1−p(1 + q u )p−1 +1 1+ q u r+1−p r+2 1 1 1 q k−p u u u c 1 1 = r+1 (− 1) ( 1 k − k+1) p−1,j+1 (1 + q u ) ur u 1 1 p=2 +2 k p 1 X X= +1 r r 1 +1 = (−q )k−p(u u − u )c . 1+ q u 1 1 k k+1 p−1,j+1 1 1 p=2 k p X X= +1

By deﬁnition of ci,j in Lemma 23:

min(p−1,r−j) p − 2 c = (−q )p−1−ma , p−1,j+1 m − 1 1 m,j+1 m=1 X we have

r cp−1,j+1 xr+2 (Xp(1 − q1x1) − Yp) r+1−p p=2 x1 X r r min(p−1,r−j) 1 +1 p − 2 = (−q )k(u u − u ) (−q )−m−1a . 1+ q u 1 1 k k+1 m − 1 1 m,j+1 1 1 p=2 k p m=1 X X= +1 X Changing the orders of the sums, we have

r cp−1,j+1 xr+2 (Xp(1 − q1x1) − Yp) r+1−p p=2 x1 X r−j r r 1 +1 p − 2 = (−q )k−m−1(u u − u ) a 1+ q u 1 1 k k+1 m − 1 m,j+1 1 1 m=1 p=m+1 k p X X X= +1 r−j r k 1 +1 −1 p − 2 = (−q )k−m−1(u u − u )a 1+ q u m − 1 1 1 k k+1 m,j+1 1 1 m=1 k m (p=m+1 ) X =X+2 X r−j r 1 +1 k − 2 = (−q )k−m−1(u u − u )a . 1+ q u m 1 1 k k+1 m,j+1 1 1 m=1 k m X =X+2

Using (1) and the deﬁnition of ai,j in Lemma 23:

am,j+1 = Sq(r +1 − m, j + 1) − x1Sq(r − m, j + 1),

47 the right-hand side is calculated as

r−j r 1 +1 k − 2 (−q )k−m−1(u u − u )S (r +1 − m, j + 1) 1+ q u m 1 1 k k+1 q 1 1 m=1 k m X =X+2 r−j r u +1 k − 2 − 1 (−q )k−m−1(u u − u )S (r − m, j + 1) (1 + q u )2 m 1 1 k k+1 q 1 1 m=1 k m X =X+2 r r 1 −1 +1 k − 2 = (−q )k−r+m−1(u u − u )S (m +1, j + 1) 1+ q u r − m 1 1 k k+1 q 1 1 m=j k r m X =X− +1 r r q u −2 +1 k − 2 + 1 1 (−q )k−r+m−1(u u − u )S (m +1, j + 1) (1 + q u )2 r − m − 1 1 1 k k+1 q 1 1 m=j k r m X =X− +1 r r −1 +1 k − 2 q u k − 2 (−q )k−r+m−1 = + 1 1 1 (u u − u ) r − m 1+ q u r − m − 1 1+ q u 1 k k+1 m=j (k r m 1 1 1 1 ) X =X− +1 q u r+1 × S (m +1, j + 1) − 1 1 (−q )k−2(u u − u )S (r, j + 1). q (1 + q u )2 1 1 k k+1 q 1 1 k X=2

Proof of Proposition 24. By Lemma 26, Aj in Proposition 21 is expressed as required.

4.4 The constant c

Finally, we write the constant c in Proposition 21 in terms of uj’s. As before, put q1 =1 − q. Proposition 27. We have

r u 1 − qu +1 qr c =1 − 1 − t 1 qk−2u − t u . 1+ q u 1+ q u k 1+ q u r+2 1 1 1 1 k 1 1 X=2 proof. Recall that

r−1

c =1 − (x1 + tx2) − t (xr+2−i − x1xr+1−i). i=0 X r u1 r+1 k−2 (−q1) ur+2 By (1) and (2), we have x = , x = (−q ) uk + , and 1 1+q1u1 2 k=2 1 1+q1u1 P xr+2−i − xixr+1−i r +1 k − 2 q u k − 2 = + 1 1 (−q )k−r+i−2(u − uk−r−2u ). r − i 1+ q u r − i − 1 1 k 1 r+2 k r i 1 1 =X− +1

48 We have

r−1

(xr+2−i − xixr+1−i) i=0 Xr r +1 −1 k − 2 q u k − 2 = + 1 1 (−q )k−r+i−2(u − uk−r−2u ) r − i 1+ q u r − i − 1 1 k 1 r+2 k i r k 1 1 X=2 =X+1− r+1 k−1 k−2 k − 2 k−2−i q1u1 k − 2 k−i−3 k−r−2 = (−q1) + (−q1) (uk − u1 ur+2) i 1+ q1u1 i k=2 ( i=1 i=0 ) Xr X X +1 1 − qu = 1 qk−2 − (−q )k−2 (u − uk−r−2u ). 1+ q u 1 k 1 r+2 k 1 1 X=2 Thus

r−1

− x2 − (xr+2−i − xixr+1−i) i=0 X r r r (−q )ru 1 − qu +1 1 − qu +1 +1 = − 1 r+2 − 1 qk−2u + 1 qk−2uk−2 − (−q )k−2uk−2 u−ru 1+ q u 1+ q u k 1+ q u 1 1 1 1 r+2 1 1 1 1 k 1 1 k k ! X=2 X=2 X=2 r r+1 r r (−q1) ur+2 1 − qu1 k−2 −q +(−q1) = − − q uk + ur+2 1+ q1u1 1+ q1u1 1+ q1u1 k=2 r X 1 − qu +1 qr = − 1 qk−2u − u . 1+ q u k 1+ q u r+2 1 1 k 1 1 X=2 This completes the proof.

Acknowledgments Thanks are due to Tatsushi Tanaka for his helpful comments and advice. The ﬁrst author is supported by the National Natural Science Foundation of China No. 11471245. The second author is supported by the Grant-in-Aid for Young Scientists (B) No. 15K17523, Japan Society for the Promotion of Science.

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