Some remarks about q- Chebyshev polynomials and q- Catalan numbers and related results Johann Cigler Fakultät für Mathematik, Universität Wien Uni Wien Rossau, Oskar-Morgenstern-Platz 1, 1090 Wien
[email protected] http://homepage.univie.ac.at/johann.cigler/ Abstract The moments of the Lucas polynomials and of the Chebyshev polynomials of the first kind are (multiples of) central binomial coefficients and the moments of the Fibonacci polynomials and of the Chebyshev polynomials of the second kind are Catalan numbers. In this survey paper we present some generalizations of these results together with various q analogues. 0. Introduction The moments of the Fibonacci polynomials and of the Chebyshev polynomials of the second 1 2n kind are (multiples of) the Catalan numbers Cn and the moments of the Lucas n 1 n polynomials or equivalently of the Chebyshev polynomials of the first kind are (multiples of) 2n the central binomial coefficients Bn . We show first how these facts generalize to n various q analogues. The most natural q analogues of the monic Chebyshev polynomials (cf. [11] or [12]) are orthogonal polynomials and their moments are multiples of 2n 1 2n cn (q) b ( q ) and n n respectively. The same fact holds for a curious class [n 1] n (cf. [8]) of non-orthogonal q Fibonacci and q Lucas polynomials, whereas the moments Cn (q) of the Carlitz q Fibonacci polynomials - which are orthogonal - do not have explicit expressions. But the generating function Cq (u ) Cn (q)u n has a simple representation as n0
E ( qu ) for some q analogue E (u ) of the exponential series. Some of these E ( u ) results can be extended to q analogues of generalized Fibonacci polynomials f n( m ) ( x) Cq (u )
mn 1 and to (m 1)n 1 n q analogues of generalized Lucas polynomials whose moments are related to generalized mn central binomial coefficients . n
whose moments are multiples of generalized Catalan numbers Cn( m )
For the convenience of the reader I first recall some well-known background material about Fibonacci and Lucas polynomials and their generalizations f n( m ) ( x) and ln( m ) ( x ) and state some basic facts about q identities. 1
1. Background material 1.1. Let the special Fibonacci polynomials f n ( x ) be defined by n 2
nk k n2k f n ( x) (1) x . k 0 k
(1.1)
They satisfy the recurrence relation
f n ( x) xf n 1 ( x) f n 2 ( x)
(1.2)
with initial values f 0 ( x) 1 and f1 ( x) x and are orthogonal with respect to the linear functional f defined by f f n ( x ) [ n 0]. More precisely we have f f n ( x) f m ( x) [ n m].
(1.3)
The moments f x n can easily be deduced from the formula n 2
n n xn f n 2 k ( x). k 0 k k 1
(1.4)
f x 2 n Cn ,
(1.5)
This gives f x 2 n 1 0 and
2n 2n 1 2n where Cn is a Catalan number. n n 1 n 1 n The moment generating function is
x u 2n
n0
f
n
C (u )
1 1 4u 2u
(1.6)
and satisfies
C (u ) 1 uC (u ) 2 .
(1.7)
u 1 C 1. 1 u (1 u ) 2
(1.8)
Equivalently we have
x x2 4 x x2 4 If we set and , then n x n 1 n 2 and 2 2 n n 1 n2 x . Note that x and 1.
Since
n 1 n 1 satisfies recursion (1.2) and the initial values we get the Binet formulae f n ( x)
n 1 n 1 . 2
(1.9)
1.2. Let us also consider a variant ln ( x) of the Lucas polynomials defined by n 2
nk n (1) k x n 2 k ln ( x ) k 0 k nk l0 ( x) 1.
(1.10)
The polynomials ln ( x) satisfy the recurrence relation
ln ( x) xln 1 ( x) n 2ln 2 ( x)
(1.11)
with 0 2 and n 1 for n 0. We have ln ( x) f n ( x) f n 2 ( x) for n 2 and ln ( x) f n ( x) for n 0,1. This implies for n 0 the Binet formulae ln ( x ) n n .
(1.12)
The polynomials ln ( x) are orthogonal with respect to the linear functional l defined by l ln ( x ) [ n 0]. More precisely
l ln ( x )lm ( x ) 2[ n m]
(1.13)
for n 0 and l l0 ( x)2 1. From the representation n 2
n x n ln 2 k ( x ) k 0 k
(1.14)
we deduce that the moments l x n are l x 2 n 1 0 and
2n l x 2 n Bn , n
(1.15)
2n where Bn is a central binomial coefficient. n The moment generating function is
x u 2n
n0
l
n
B(u )
1 . 1 4u
(1.16)
Equivalently
un 1 Bn . 2 n 1 (1 u ) 1 u n0
3
(1.17)
1.3. Let us note some generalizations of the foregoing situation. Let m 1 and let n m
n (m 1)k k n mk f n( m ) ( x) . (1) x k k 0
(1.18)
f n( m ) ( x) xf n(m1) ( x) f n(mm) ( x)
(1.19)
These polynomials satisfy
for n 0 with initial values f 0( m ) ( x ) 1 and f (nm ) ( x) 0 for 0 n m. This implies f n( m ) ( x) x n for 0 n m.
We have n m
n n ( m) x n (m 1) f n mk ( x) k 0 k k 1
(1.20)
and therefore the linear functional f ( m ) defined by f ( m ) f n( m ) ( x) [ n 0] gives as moments the m Catalan numbers, which are also called Fuss-Catalan numbers, nm nm nm 1 (m) f ( m ) x nm (m 1) Cn . n n 1 (m 1)n 1 n
(1.21)
For some special cases cf. [14], OEIS A000108, A001764, A002293, A009294. The generating function satisfies m (u ) Cn( m )u n 1 u m (u ) m . n0
The most natural generalization of the Lucas polynomials are the polynomials ln( m ) ( x ) which satisfy ln( m ) ( x ) xln( m1) ( x ) n m ln( m m) ( x ) with initial values ln( m ) ( x) x n for 0 n m and
0 m and n 1 for n 0. This gives for n 0 n m
n (m 1)k n ln( m ) ( x) (1) k x n mk . k k 0 (n (m 1)k )
(1.22)
It is easily verified that ln( m ) ( x ) f n( m ) ( x) ( m 1) f n(mm) ( x)
for n 0.
4
(1.23)
Here we have n m
n ) x n ln( m mk ( x) k 0 k
(1.24)
and thus for the linear functional l ( m ) defined by l ( m ) ln( m ) ( x ) [ n 0] nm l ( m ) x nm . n
(1.25)
mn The generating function m (u ) u n satisfies n 0 n m (u )
1 . 1 mu m (u ) m 1
It seems that these polynomials do not have interesting q analogues. Therefore we consider another generalization L(nm ) ( x ) of the Lucas polynomials which satisfy L(nm ) ( x) xL(nm1) ( x) n m L(nm)m ( x) with initial values L(nm ) ( x) x n for 0 n m and 0 2 and
n 1 for n 0. Here we get for n 0 n m
n (m 1)k n (m 2)k n mk L(nm ) ( x) (1) k x , k k 0 n (m 1)k
(1.26)
L(nm ) ( x ) f n( m ) ( x) f n(mm) ( x)
(1.27)
which implies that for n 0. For this is trivially true for 1 n m 1 since in this case L(nm ) ( x) x n x n 0. It is also true for n m, for in this case we have L(mm ) ( x ) xL(mm)1 ( x ) 2 L(0m ) ( x) x m 2 x m 1 1 f m( m ) ( x ) f 0( m ) ( x ).
For n m both sides satisfy the same recurrence relation. If we define the linear functional L( m ) by L( m ) L(nm ) ( x ) [ n 0] we get in this case n m
k 1 n n x n (m 2) L(nm)mk k 0 k j 0 j
5
(1.28)
and thus n 1 mn nm L( m ) x nm (m 2) . j 0 j n
(1.29)
1.4. Let us now state some well-known notations and results for q identities which will be needed later (cf. e.g. [5]).
We always assume that 0 q 1. Let x; q n 1 x 1 qx 1 q n 1 x and
x; q 1 q j x . j 0
q; q n n n 1 qn for 0 k n. Let [n] [n]q and 1 q k k q q; q k q; q n k The q binomial coefficients satisfy
n n 1 n n 1 n k n 1 k n 1 k q k k 1 and k k q k 1 . Let be the operator f ( x) f ( qx ). Let Dq be the q differentiation operator defined by Dq f ( x)
f ( x) f (qx) . x qx
Then 1 ( q 1) xDq since (1 (q 1) xDq ) x n x n q n 1 x n q n x n x n . A simple q analogue of the binomial theorem is the fact that the so called Rogers-Szegö n n polynomials rn ( x, s ) x k s n k can be represented as rn ( x, s ) ( x s ) n 1. This follows k 0 k by induction because n k n n n ( x s ) x k s n k x k 1s n k q k x k s n 1 k q k k k k k k k k 1
n k n 1 k k x s
n 1 k n 1 k . x s k k Note also that Dq rn ( x, s ) [ n]rn 1 ( x, s ) because
n n n 1 k 1 n k Dq rn ( x, s) Dq x k s n k [k ]x k 1s n k [n] x s [n]rn 1 ( x, s) k k k k k k 1 Therefore the Rogers-Szegö polynomials satisfy the recursion rn 1 ( x, s ) ( x s ( q 1) sxDq ) rn ( x, s ) ( x s ) rn ( x, s ) ( q n 1) xsrn 1 ( x, s ).
6
(1.30)
We shall also need the following q hypergeometric version of the binomial theorem (cf. e.g. [3]).
a; q k
q; q k 0
xk
k
ax; q x; q
(1.31)
for x 1. As special cases we will need the following q analogues of the exponential series xn 1 x; q n 0 q; q n
(1.32)
and
n0
q
n 2
( x) n x; q q; q n
(1.33)
and the simplest binomial theorems k
n (1) q 2 x k x; q n k 0 k
(1.34)
n k 1 k 1 x . k x; q n k 0
(1.35)
n
k
and
1.5. Notes
The polynomials f n ( x) are the special case f n ( x) Fibn 1 ( x, 1) of the bivariate Fibonacci n 1 2
n k 1 k n 1 2 k which satisfy s x k k 0 Fibn ( x, s) xFibn 1 ( x, s) sFibn 2 ( x, s) with initial values Fib0 ( x, s) 0 and Fib1 ( x, s) 1.
polynomials Fibn ( x, s )
For n 0 the polynomials ln ( x) are the special case ln ( x) Lucn ( x, 1) of the bivariate n 2
n n k k n2k for n 0 and s x k 0 n k k
Lucas polynomials which satisfy Lucn ( x, s )
Luc0 ( x, s ) 2. There is also a close connection with the Chebyshev polynomials in the usual form. The Chebyshev polynomials of the first kind Tn ( x) satisfy Tn ( x) 2 xTn 1 ( x) Tn 2 ( x) with 1 Lucn (2 x, 1) . initial values T0 ( x) 1 and T1 ( x) x and Tn ( x) 2n 1 Lucn x, 4 2
The monic Chebyshev polynomials of the first kind 7
n 2
n n k 1 n2k tn ( x) x k 0 n k k 4 k
for n 0 and t0 ( x) 1 satisfy tn ( x) xtn 1 ( x) n 2tn 2 ( x) with 0
(1.36) 1 1 and n . 2 4
The Chebyshev polynomials of the second kind U n ( x) satisfy U n ( x) 2 xU n 1 ( x) U n 2 ( x) 1 with initial values U 0 ( x) 1 and U1 ( x) 2 x and U n ( x) Fibn 1 (2 x, 1) 2n Fibn 1 x, . 4
The monic Chebyshev polynomials of the second kind n 2
n k 1 n2k un ( x ) x k 0 k 4 k
(1.37)
1 satisfy un ( x ) xun 1 ( x) un 2 ( x ). 4
Identities (1.4), (1.14), (1.20), (1.24) and (1.28) are special cases of the following situation: Let pn ( x) be polynomials satisfying
pn ( x) xpn 1 ( x) n pn m ( x)
(1.38)
for some integer m 1 and initial values pn ( x) x n for 0 n m. Then there are uniquely determined coefficients c ( n, k ) such that n
x n c(n, k ) pk ( x).
(1.39)
k 0
This implies
c(n, k ) p ( x) x c(n 1, k ) p ( x) c(n 1, k ) p k
k
k 1
k
k
( x) k m1 pk m 1 ( x)
k
pk ( x) c(n 1, k 1) k c(n 1, k m 1) k
Therefore we get
c(n, k ) c(n 1, k 1) k c(n 1, k m 1)
(1.40)
with initial values c (0, k ) [ k 0] and boundary values c ( n, 1) 0. If we apply the linear functional we get x n c ( n, 0).
8
(1.41)
Let us recall a well-known combinatorial interpretation of (1.40): The number c ( n, k ) is the weight of all lattice paths in 2 which initial point 0, 0 and endpoint n, k , where each step is either an up-step j , j 1, 1 or a down-step
j, m 1 j 1, . The weight of an up-step is 1 and the weight of a down-step with endpoint , k is k . The weight of a path is the product of the weights of its steps. The weight of a set of paths is the sum of their weights. The trivial path 0, 0 0, 0 has by definition weight 1. It is then clear that c mn k i, k 0 for 0 i m. Therefore (1.39) can also be written in the form n
x n c(n, n km) pn km ( x).
(1.42)
k 0
Let (u ) c(mn, 0)u n be the generating function of the moments. Each nontrivial path has n
a unique decomposition of the following form: For each i with 1 i m 1 there is an upstep from height i 1 to height i followed by a maximal path which ends at height i and never falls below height i, and finally a down-step which ends on height 0 followed by a non-negative path which ends on height 0. Therefore we get
(u ) 1 0u(u )1 (u ) 2 (u ) m1 (u )
(1.43)
where i (u ) denotes the generating function of the moments if in (1.38) n is replaced by
ni . For m 2 we get n 1
c 2n, 0 0 b(2k ,0)c 2n 2 2k , 0 ,
(1.44)
k 0
where b( n, m) is the corresponding weight when k is replaced by k 1. Let (u ) c(2n, 0)u n and (u ) b(2n, 0)u n . n
n
Then (1.44) is equivalent with (u ) 1 0u (u ) (u ).
For the polynomials f n ( x ) we have n 1 and therefore (1.45) reduces to (1.7).
9
(1.45)
2n k 2 n k Identity (1.4) is equivalent with c(2n k , k ) and c(2n k 1, k ) 0. n n 1 The corresponding matrix c(n, k ) is known as Catalan triangle (cf. [14], OEIS, A053121). For the polynomials ln ( x) we have 0 2 and n 1. Therefore (u ) C (u ) and (u ) 1 2u (u )C (u ), which gives (u )
1 1 . 1 2uC (u ) 1 4u
Formula (1.24) is easily proved by induction. It is trivially true for n m. For n m we have 1 m lm( m ) ( x ) x m m and (1.24) reduces to lm( m)mk ( x) lm( m ) ( x) ml0( m ) ( x) x m . k 0 k If it is true for n 1 m then n m
n
n
n
m n 1 m n 1 m n 1 n (m) (m) (m) (m) (m) l ( x ) l ( x ) l ( x ) n mk n mk n mk ln mk ( x) ln m mk ( x) k 0 k k 0 k k 0 k 1 k 0 k n 1 ( m ) n xln 1 mk ( x) x . k
From (1.23) we get (1.20) and from (1.27) we deduce (1.28).
2. The simplest q-analogues n m
n (m 1)k k n mk The simplest q analogues of f n( m ) ( x) are the polynomials (1) x k k 0 n m
f
(m) n
( x, q ) q
k m 2
k 0
n (m 1)k k n mk , (1) x k
(2.1)
which satisfy f n( m ) ( x, q ) xf n(m1) ( x, q ) q n m f n(mm) ( x, q )
(2.2)
with initial values f 0( m ) ( x, q ) 1 and f (nm ) ( x, q ) 0 for 0 n m. For m 2 these are orthogonal polynomials which are closely related to Carlitz’s q Fibonacci polynomials. What can be said about the moments of these polynomials? Denote the moments by Cn( m ) ( q ). The generating function Cq( m ) (u ) Cn( m ) (q )u n satisfies n0
C
(m) q
(u ) 1 uC
(m) q
(u )C
(m) q
10
( qu ) C
(m) q
(q
m 1
u ).
(2.3)
These q Catalan numbers have no simple closed formula, but their generating function can be represented (cf. [9]) as Cq( m ) (u )
E ( m ) (qu ) E ( m ) (u )
(2.4)
with n m 2
q un. q ; q n0 n
E ( m ) (u ) n m 2
(2.5)
n m 2
q q (1 q n )u n u u n 1 uE ( m ) (q mu ) implies (2.3). n 0 q; q n n 0 q; q n 1
For E ( m ) (u ) E ( m ) (qu )
For m 2 the generating function Cq (u ) Cn (q)u n satisfies n0
Cq (u ) 1 uCq (u )Cq (qu ).
(2.6)
Comparing coefficients this gives n 1
Cn (q ) q k Ck (q )Cn 1 k (q ) with C0 (q) 1. Further properties can be found in [13]. k 0
From (2.1) we see that 0 f ( m ) ,q f C
(m) n
( x, q ) q n
(m) nm
k m 2
k 0
nm (m 1)k (1) k Cn( m k) (q ). This can be used to compute k
( q ).
In [6] a shorter algorithm to compute these numbers has been given. We have
(1)
k
(2.7)
1 (m 1)(n k ) ( m ) Cn k (q) 0. k
(2.8)
q
k 0
k 2 2
n 1 k k Cn k ( q ) 0
n
and more generally n
(1) k 0
k
q
k m 2
Identity (2.7) follows immediately from orthogonality because n
(1) k 0
k
q
k 2 2
n 1 k n 1 (2) k Cn k (q) f ( m ) ,q x f n 1 ( x, q) 0.
11
Identity (2.8) is equivalent with f ( m ) ,q x n 1 f ((mm)1) n 1 ( x, q ) 0. Let r (n, k ) f ( m ) , q x k f mn( m)k ( x, q ) . By (2.2) r (n, k ) is a linear combination of r (n, k 1) and r (n 1, k 1). Therefore r (n, n 1) is a linear combination of r (n, n 2) and r (n 1, n 2) and therefore a linear combination of r (n, n 3), r (n 1, n 3) and r (n 2, n 3) and thus a
linear combination of r (n i, 0) for 0 i n 1. But r (n i, 0) f ( m ) ,q f m( m( n)i ) ( x, q ) 0. As q analogue of (1.8) we get (cf. [13])
q
n 2 2
n0
To prove this observe that by (1.35) n
q
k 2 2
k 0
xn x; q 1
Cn (q ) 1.
(2.9)
2 n 1
n k 1 k 1 x and that ( x; q) n 1 k 0 k
2n k k (2) k (1) Cn k (q) f ( 2 ) ,q f 2 n ( x, q) [n 0].
Therefore we get
q
j 2 2
j 0
xj x; q 1
C j (q) q
j 2 2
j 0
2 j 1
2 j k k C j (q) x j ( x) k k 0 q1 nk 2
2 2 j k 2 k 2 2 kn q Cn k ( q ) q x k q n j k n n
(1) x q k
n
n
n 2 n 2
q
k 2 2
k 0
2n k k k (1) Cn k (q ) 1.
In the same way we see that
q
n m 2
n 0
xn x; q 1
C mn 1
(m) n
(q) q
j m 2
j 0
mj k Cn( m ) (q) x j ( x)k k q1 k 0
nk 2
m mj k mjk ( m ) x q Cn k ( q ) q k q n j k n n
(1) x q k
n
n
n m n 2
q k 0
k m 2
2n k (m) k k (1) Cn k (q) 1.
and therefore
q n 0
n m 2
xn x; q 1
12
Cn( m ) (q) 1. mn 1
(2.10)
It seems that there are no q analogues of ln( m ) ( x ) with simple recurrence relations. But there is a rather curious class of polynomials which satisfies an operator recurrence relation. 3. Some curious q-analogues 3.1. Let us consider the polynomials (cf. [8]) n 2
Fn ( x, q) q
k 1 2
k 0
n k k n2k k (1) x .
(3.1)
They are not only q analogues of f n ( x) in the sense that lim Fn ( x, q ) f n ( x) but can be q 1
obtained from f n ( x) by first computing the operator f n ( x (1 q ) Dq ) and then applying it to the constant polynomial 1. Thus Fn ( x, q ) f n ( x (1 q ) Dq )1.
(3.2)
They satisfy the recurrence relation
Fn ( x, q) x (1 q) Dq Fn 1 ( x, q) Fn 2 ( x, q)
(3.3)
with initial values F0 ( x, q) 1 and F1 ( x, q) x. They also satisfy Fn ( x, q ) xFn 1 ( x, q ) q n 1 xFn 3 ( x, q ) q n 1 Fn 4 ( x, q ).
(3.4)
Let F ,q Fn ( x, q ) [n 0]. The polynomials Fn ( x, q) are not orthogonal. For example F ,q xF3 ( x, q ) ( q 1) q 3 0. Nevertheless there a very nice q analogue of (1.4): n 2
n n xn Fn 2 k ( x, q ). k 0 k k 1 n 2 k
(3.5)
This implies that the moments F ,q x 2 n are 2n 2n n F ,q x 2 n q cq (n), n n 1 where cq (n)
1 2n is a explicit q analogue of the Catalan numbers. [n 1] n
13
(3.6)
I do not know a simple q analogue of the generating function (1.7), but we have instead
cn (q)q
n 2
n 0
un q n u; q
1
(3.7)
2 n 1
which is a q analogue of (1.8). 3.2. Let now
ln ( x, q) Fn ( x, q) Fn 2 ( x, q)
(3.8)
for n 2 and l0 ( x, q) 1 and l1 ( x, q) x. Then n
ln ( x, q) (1) q k
k 2
k 0
[ n] n k n 2 k x [n k ] k
(3.9)
for n 0 and l0 ( x, q) 1. The polynomials ln ( x, q) satisfy
ln ( x, q) x (1 q) Dq ln 1 ( x, q) n 2ln 2 ( x, q)
(3.10)
with initial values l0 ( x, q) 1 and l1 ( x, q) x. Here 0 2 and n 1 for n 0. This can also be written as ln ( x, q ) ln ( x (1 q ) Dq )1.
(3.11)
The polynomials ln ( x, q) are not orthogonal. The identity n 2
n x n ln 2 k ( x, q) k 0 k
(3.12)
2n l ,q x 2 n , n
(3.13)
implies that
if we define the linear functional l ,q by l , q ln ( x, q ) n 0 .
2n I do not know a simple q analogue of (1.16) for bq (u ) u n . n0 n Instead of this n 1 2
2n q n0 n
is a q analogue of
un q n u; q
2n un 1 . 2 n 1 1 u n 0 n (1 u )
14
q 2 n 1
n0
n 1 2
un
(3.14)
3.3. Proofs and remarks
The polynomials Fn ( x, q) and ln ( x, q) have been systematically studied in [8]. To prove (3.1) it suffices to compare coefficients in (3.3). Since these polynomials are not orthogonal and thus do not satisfy a 3 term recurrence of the form (1.38) the above combinatorial interpretation fails. But formula (3.3) implies Binet-type formulae for these polynomials: Let A be the operator A x (1 q ) Dq . For each polynomial p ( x ) in x we define p ( x ) p A 1.
n 1 n 1 Thus Fn ( x, q ) f n ( x) and analogously
ln ( x, q) ln ( x) n n for n 0. This is an exact version of a symbolic method which I used in [10]. This implies n n k nk x n . k 0 k n
From (1.30) we get
n
k k 0
k
(3.15)
n 1 k n k n 2 k nk q n1 1 . k k k
n k x k
Since these are by induction polynomials in x we get again by induction n n k n k Ax n 1 q n 1 1 x n 2 x n . k 0 k To prove (3.12) observe that for odd n n n2 n2 2 n n n2k n nk k n2k nk k k ln 2 k ( x, q ) k k 0 k k 0 k 0 n n k nk x n . k 0 k If n 2m then m 1 2m 2 m 2 k 2 m 2m k 2 m k 2m 2m 2m2k 2m l ( x , q ) x . 2m2k m k k 0 k 0 k 0 k m
k
Since ln ( x, q ) Fn ( x, q ) Fn 2 ( x, q ) we also get (3.5). 15
Comparing coefficients we see that (3.7) is equivalent with q
n n 2
(1) c j
n j
j 0
(q)q
j 2
2n j j [n 0].
But this is clear since n
q
n
(1) c j
n j
j 0
(q )q
j 2
j 1 n 2n j j 2 2n j 2 n 2 j ( 1) q x F ,q F2 n ( x, q ) [ n 0]. F ,q j j j 0
In the same way we prove identity (3.14): n 1 2
2n q n 0 n
un q n u; q
j 1 2
2 n 1
2 j q j 0 j
2 j k uj (1) k q jk u k k k 0
n k 1 ( n k ) k 2
2n 2k 2n k u (1) q n 0 k 0 nk k n
n
u q n
k
n 1 2
n 0
u q n
n 1 n 2
n0
(1)
k
k 0
q
k 1 2
2n k 2n 2k k nk
l ,q F2 n ( x, q ) n 2
Since Fn ( x, q) ln ( x, q) Fn 2 ( x, q) we get Fn ( x, q ) ln 2 k ( x, q ). This implies k 0
l ,q F2 n ( x, q ) 1.
3.4. The polynomials Fn ( x, q) can be generalized to n m
(m) n
F
( x, q) (1) q k
k 1 2
k 0
n (m 1)k n mk x k
which satisfy Fn( m ) ( x, q ) xFn(m1) ( x, q ) (1 q ) Dq Fn(mm) 1 ( x, q ) Fn(mm) ( x, q ).
Note that the operator (1 q ) Dq is applied to Fn(mm) 1 ( x, q ).
This follows since the coefficient of (1) q k
k 1 2
x n mk of the right-hand side is
16
(3.16)
n 1 (m 1)k k n (m 1)k n (m 1)k 1 1 q n mk 1 q k q k k 1 k 1 n 1 (m 1)k n 1 (m 1)k k k n ( m 1) k qk qk q 1 q k k k n (m 1)k k n (m 1)k n (m 1)k qk q 1 q k k k k For m 1 this reduces to n
F ( x, q) (1) q (1) n
k
k 1 2
k 0
n nk 2 n k x ( x q)( x q ) x q .
Let cn( m ) (q ) F ( m ) , q x mn .
(3.17)
Note that cn(2) ( q ) q n cn ( q ). (m) If we apply F ( m ) ,q to Fmn ( x, q ) we get a recurrence for cn( m ) ( q ).
n
(1)
k
q
k 0
k 1 2
mn (m 1)k ( m ) cn k (q) 0. k
(3.18)
The numbers cn( m ) (q ) satisfy
c n 0
(m) n
(q)q
n 1 2
xn q n x; q
1.
(3.19)
mn 1
For the left-hand side is
c j 0
(m) j
(q )q
x q n
n0
j 1 2
n k 1 k ( nk ) 2
n mj k k kj n k (m) x ( x ) q x ( 1) c ( q ) q nk k k 0 n0 k 0 j
n 1 n 2
(1)
k
q
k 1 2
k 0
n 1 2
mn (m 1)k ( m ) n c q x q ( ) nk k n0
F ( m ) , q Fmn( m ) ( x, q) 1.
In the same way the polynomials ln ( x, q) L(2) n ( x, q ) can be generalized to n m
L ( x, q ) (1) q (m) n
k
k 0
k 2
n (m 1)k [n (m 2)k ] n mk [n (m 1)k ] x k
17
mn (m 1)k k
which satisfy L(nm ) ( x, q ) xL(nm1) ( x, q ) (1 q ) Dq L(nm)m 1 ( x, q ) n m L(nm)m ( x, q )
(3.20)
with 0 2 and n 1. As above we have
L(nm ) ( x, q) Fn( m ) ( x, q) Fn(mm) ( x, q).
(3.21)
bn( m ) (q ) L( m ) ,q x mn .
(3.22)
Let
Then we get the recurrence n
(1)
k
q
k 2
k 0
mn (m 1)k [mn (m 2)k ] ( m ) [mn (m 1)k ] bn k (q) 0. k
(3.23)
Generalizing (3.14) we get
b n0
(m) n
(q)q
n 1 2
un q n u; q
u q n
mn 1
n 1 2
.
(3.24)
n0
n m
Since by (3.21) we have Fn( m ) ( x, q) L(nm)mk ( x, q) we have L( m ) ,q Fn( m ) ( x, q ) 1. k 0
Thus we get
b n 0
(m) n
u
(q)q
n 1 2
un q n u; q
n
n
n0
(1) b k
k 0
u q n
n 1 n 2
n0
(m) nk
mn 1
j 0
(q)q
j 1 2
mj k (1) k q jk u k uj k k 0
n k 1 ( n k ) k 2
mn (m 1)k (q) q k
(1) k 0
b
(m) j
k
q
k 1 2
n 1 2
mn (m 1)k ( m ) n b ( q ) u q n k k n 0
n 1
n j For m 1 we get L(1) n ( x, q ) ( x 1 q ) x q j 1
for n 1.
18
L( m ) ,q F
(m) nk
( x, q ) u q n
n 0
n 1 2
.
4. q-Chebyshev polynomials
Now we come to a class of orthogonal polynomials where almost all facts from the classical case have simple counterparts. 4.1. The polynomials n 2
n k k 2 x n2k k u n ( x, q ) q (1) q; q k q n1k ; q k k 0 k
(4.1)
will be called special q Chebyshev polynomials of the second kind. They satisfy the recurrence relation q n 1 un ( x, q ) xun 1 ( x, q ) u ( x, q ) 1 q n1 1 q n n2
(4.2)
with initial values u0 ( x, q) 1 and u1 ( x, q) x. The polynomials un ( x, q) are orthogonal with respect to the linear functional defined by u ,q (un ( x, q )) [ n 0].
(4.3)
More precisely we have
u ,q (un ( x, q)um ( x, q))
q
n 1 2
q; q n q 2 ; q n
[n m].
(4.4)
The identity n n k k 1 un 2 k ( x, q ) x n n 2 2 k ;q k 0 q; q k q n 2
(4.5)
k
gives the moments
1 qn 1 2n n n2 u ,q x (1) q (1 q) 2 Cn (q) [n 1] n q; q n q 2 ; q n n 1 q2 2n
where Cn (q) is a q Catalan number in the sense of Andrews [2].
19
(4.6)
2
u u u As q analogue of C 1 C we get for the generating function 4 4 4 n C(u, q) Cn (q)u n0
C(u , q ) qC(qu , q ) qu 1 C(u , q )C(qu , q ). 1 q (1 q ) 2
u; q Let h(u ) : qu; q 2
(4.7)
u; q 2
2
. This is a q analogue of 1 u since h(u ) h( qu )
1 u.
q 2u; q 2
The formula C(u , q ) (1 q )
is a q analogue of
Cn
4 n0
n
un 2
1 h (u ) u
(4.8)
1 1 u . u
4.2. The special q Chebyshev polynomials of the first kind are the polynomials n 2
tn ( x, q ) ( 1) k q k k 0
2
[ n] n k 1 x n2k . nk [ n k ] k q; q k q ; q
(4.9)
k
The polynomials tn ( x, q) satisfy the recurrence relation
tn ( x, q ) xtn 1 ( x, q) n 2 (q)tn 2 ( x, q) with 0 (q )
(4.10)
q n 1 q and n (q) for n 0. 1 q 1 q n 1 q n1
It is easy to verify that t0 ( x, q) u0 ( x, q) 1, t1 ( x, q) u1 ( x, q) x and for n 2 q 2 n 1 t n ( x, q ) u n ( x , q ) u ( x, q). 1 q n1 1 q n n2
(4.11)
The polynomials tn ( x, q) are orthogonal with respect to the linear functional t , q defined by t , q tn ( x, q ) [ n 0].
(4.12)
More precisely we have for n 0 t ,q tn ( x, q )tm ( x, q )
q
n 1 2
q; q n1 q; q n
20
[n m].
(4.13)
The identity ([11], Theorem 4.3) n 2
n qk xn t n 2 k ( x, q ) n 2 k 1 ; q k 0 k q; q k q k
(4.14)
2n q n t ,q x 2 n . 2 n q; q n
(4.15)
q k k G (u, q) : 2 2 (1) k u k u; q 2 . k 0 q ; q
(4.16)
implies the moments
Let 2
k
By the q binomial theorem g (u )
2n G (qu, q ) 1 un. G (u, q ) n 0 n q; q 2
(4.17)
n
Note that g (u ) is a q analogue of g (u ) g (qu )
1 since G (u , q ) u; q 2 implies 1 u
1 . The generating function of the moments is 1 u
t ,q x 2n u n g (qu ) n0
G (q 2u , q ) . G (qu , q )
(4.18)
Notes
In [11] we introduced bivariate q Chebyshev polynomials Tn ( x, s, q) of the first kind
by Tn ( x, s, q) 1 q n 1 xTn 1 ( x, s, q) q n 1sTn 2 ( x, s, q) with initial values T0 ( x, s, q) 1 and T1 ( x, s, q) x and bivariate q Chebyshev polynomials U n ( x, s, q) of the second kind
by U n ( x, s, q) 1 q n xU n 1 ( x, s, q) q n 1sU n 2 ( x, s, q) with initial values U 0 ( x, s, q) 1 and U1 ( x, s, q) (1 q) x. We then have
tn ( x, q)
Tn ( x, 1, q) q; q n1
(4.19)
u n ( x, q )
U n ( x, 1, q) . q; q n
(4.20)
for n 0 and
21
Similar polynomials have also appeared in other publications, cf. [11] or [12] and the literature cited there. They are related to the Al-Salam and Ismail polynomials introduced in [1]. The recurrence relations can be easily verified by comparing coefficients. Proofs for (4.5) and (4.14) can be found in [11]. Formula (4.8) follows from the q binomial theorem (1.31) since
u; q h(u ) qu; q
2
1
q ; q q ; q 1
2
2
2
n 0
n
2
(qu ) 1 q n
n 1
n0
n
q ; q q ; q 1
2
2
2
n 1
u n 1
n 1
u 1 2n q u un 1 Cn (q)u n . 2 1 q n 0 [n 1] n q; q n q ; q 1 q n0 n
n
u 2 qC(u , q )C(qu , q ) u (4.8) implies 1 h(u ) 1 h(qu ) C(u, q) qC(qu, q) u 2 (1 q ) 1 q
and thus (4.7). The q binomial theorem gives
q u; q g ( qu ) qu; q 2
q; q q ; q 2
2
2
k
k 0
2
2
k
2k q k ( qu ) k uk . 2 k k 0 q; q k
1 1 uk Note that the binomial theorem gives 2 2 . G (u, q) u; q 2 k 0 q ; q
k
Thus by comparison of coefficients (4.17) is equivalent with the well-known formula 2
n 2n q . j 0 j n n
j2
(4.21)
5. A slight extension. 5.1. Consider the orthogonal polynomials (1.38) with k k ( z , q )
Calling them f n ( x, z , q) we get
22
q k 1 . (1 q k z )(1 q k 1 z )
n 2
2 n k ( 1) k f n ( x, z , q ) q k x n2k . nk k ; ; z q q z q k k 0 k
(5.1)
Note that f n ( x, q, q) un ( x, q). These polynomials are also related to the Al-Salam and Ismail polynomials (cf. [1,[11] or [12] and the literature cited there). By (1.45) we get for the generating functions f (u, z, q) c(2n, 0, z, q)u n and
f (u, z, q) b(2n, 0, z, q)u
n
n
n
f (u , z , q) 1
Since k 1 ( z , q )
qu f (u, z , q) f (u, z , q ). (1 z )(1 qz )
qk 2 qk (qz , q ) we have f (u , z , q ) f ( qu , qz , q ). (1 q k 1 z )(1 q k 2 z )
Therefore f (u , z , q ) satisfies f (u , z , q ) 1
qu f (u , z , q ) f (qu , qz , q ). (1 z )(1 qz )
For q 1 this gives f (u , z ,1) 1
(5.2)
qu f (u , z ,1) 2 and thus 2 (1 z )
u f (u, z ,1) C . 2 (1 z )
(5.3)
In the general case there are no simple formulae for c (2n, 0, z , q ), but there is a simple representation for their generating functions. Let G (u, z , q) q
n2 n
n0
un
z; q n q; q n
(5.4)
n
1 u which is a q analogue of the exponential series . n0 n ! 1 z Then
(u , z , q )
G (qu , qz , q ) G (u , z , q )
satisfies 23
(5.5)
(u , z , q ) 1
u (u , z , q ) (qu , qz , q ). (1 z )(1 qz )
(5.6)
Therefore we get G ( q 2u , qz , q ) f (u , z , q ) (qu , z , q ) . G ( qu , z , q )
(5.7)
This follows from G (qu , qz , q ) G (u , z , q ) q n
2
n
n
q
n2 n
u n q n (1 z ) (1 q n z
n
z; q n1 q; q n
q nu n un qz; q n q; q n z; q n q; q n
u q 2 nu n n2 n q q 2 z ; q q; q (1 z ) 1 qz n n n
u G q 2u , q 2 z , q (1 z ) 1 qz
which implies G q 2u , q 2 z , q G qu , qz , q u (u , z , q ) 1 1 (1 z )(1 qz ) G u , z , q G u, z, q
G q 2u , q 2 z , q G qu , qz , q u u (u , z , q ) (qu , qz , q ). (1 z )(1 qz ) G qu , qz , q G u , z , q (1 z )(1 qz )
It is clear that (u, z, q) (uz , z, q) which satisfies
(u , z , q ) 1
uz (u , z , q ) (u , qz , q ) (1 z )(1 qz )
(5.8)
is equivalent with
(u , z , q )
G ( quz , qz , q ) . G (uz , z , q )
(5.9)
5.2. If we choose
0 ( z, q)
q 1 z
(5.10)
q n 1 n ( z, q) 1 q n1 z 1 q n z we get the polynomials
n k n k 1 n k 1 z q k 2 k 1 n2k k k ln ( x, z, q ) (1) q x . z; q k q n1k z; q k 0 n 2
k
24
(5.11)
Note that tn ( x, q) ln ( x, q, q). For the generating function we get qu l (u , z , q ) l (u , z , q ) 1 z
(5.12)
q 2u l (u , z , q ) l (qu , qz , q ). (1 z )(1 qz )
(5.13)
l (u , z , q ) 1
and l (u , z , q ) 1
This implies
l (u, z , q) f (qu, z, q , q) 2
G q 3u, qz , q G q 2u , z , q
.
(5.14)
Since G qu, z , q G u , z , q q
k 2 k
k
u k q k 1
z ; q k q; q k
u ( q 2u ) k u k 2 k q G q 2u, qz , q 1 z k qz; q k q; q k 1 z
we get for (u, z , q)
G q 2u , z , q G qu, z , q
G q 2u , z , q
3 3 2 qu G q u , qz , q qu G q u , qz , q G q u , z , q (u , z, q ) 1 1 G qu , z , q 1 z G qu , z , q 1 z G q 2u , z , q G qu , z , q
1
qu l (u, z , q) (u , z , q ). 1 z
(5.12) implies l (u, z , q) (u, z, q). Thus l (u, z , q)
G q 2u , z , q G qu, z , q
For the special case tn ( x, q) ln ( x, q, q) we get l (u, q, q) same as (4.18).
25
(5.15)
. G q 2 u , q, q G qu, q, q
which is the
5.3. Consider the series F (u , z , q ) n 0
un
(5.16)
z ; q n q; q n
and let
(u , z , q )
F (u , qz , q ) . F (u , z , q )
(5.17)
Then
(u, z, q ) 1
uz (u, z, q ) (u, qz , q). (1 z )(1 qz )
(5.18)
This follows from
(u, z , q) 1
F (u, qz , q ) F (u, z , q ) 1 1 u n 1 z 1 q n z F (u, z , q) F (u , z , q ) n0 ( z; q ) n 1 (q; q) n
F u, q 2 z, q zu zu (u, z, q) (u, qz, q ). (1 z )(1 qz ) F u, z , q (1 z )(1 qz )
Comparing (5.18) and (5.8) we that F (u , qz , q ) G ( quz , qz , q ) . F (u , z , q ) G (uz , z , q )
(5.19)
In fact we have more precisely
F (u, qz , q )G (uz , z , q) F (u, z , q)G (quz , qz , q ) n
q z ; q n 2
n
z; q n qz; q n q; q n
un.
(5.20)
Comparing coefficients this is equivalent with n n k 2 k z; q n qz; q n n k 2 k k z; q n qz; q n qn z2 ; q . q z q z k n z; q nk qz; q k k 0 k qz; q nk z; q k k 0 n
This follows e.g. from the q Zeilberger algorithm. We use the Mathematica implementation of PeterPaule and Axel Riese [15]: qZeilq ^ k ^ 2 z ^ k qPochhammerz, q, n k qBinomialn, k, q qPochhammerz, q, n qPochhammerq z, q, n qPochhammerq z, q, k, k, 0, n, n, 1
SUMn
1 q22 n z2 1 q12 n z2 SUM1 n 1 q1n z2
and
26
qZeilq ^ k ^ 2 k z ^ k qPochhammerq z, q, n k qBinomialn, k, q qPochhammerz, q, n qPochhammerq z, q, n qPochhammer z, q, k, k, 0, n, n, 1
SUMn
1 q22 n z2 1 q12 n z2 SUM1 n 1 q1n z2
By (5.19) we can express the generating function f (u , z , q ) also using F (u, z, q).
qu F , qz, q G (q 2u, qz, q) z f (u, z, q) . G (qu, z, q ) qu F , z, q z
(5.21)
For the special case z q we get another representation of the generating function of the Andrews q Catalan numbers: F u, q 2 , q
G ( q 2u , q 2 , q ) . C(u , q ) f u , q, q F (u , q, q ) G ( qu , q, q )
(5.22)
Since by (1.32) and (1.33) F (u, q q)G (u, q, q) 1 this can be written as un q n n ( u ) n q nu n C(u, q) 2 n q ; q q; q n n q 2 ; q 2 n q 2 ; q 2 2
n
n
q n n (u ) n n q 2 ; q q; q n 2
n
n
This is equivalent with the following two (different) expressions for Cn (q). 1 q q2 ; q2
n
(1) n
(1 q)q n q2 ; q2
nk
k 0
n 1 k 1 q k 1 Cn (q ), q2
n 1 (1) k q k Cn (q). k 1 k 0 k q2 1 q n
n
q
nk 2 2
(5.23)
2
Remark In their paper [4] M. J. Cantero and A. Iserles prove that the rational functions an ( z , q ) defined by a0 ( z , q ) 1 and n
an j ( z , q)
q; q z; q j 0
j
j
qn
q; q n z ; q n
(5.24)
for n 0 satisfy lim an ( z , q ) (1) n Cn 1 q 1
This result also follows from our considerations.
27
z n 1 . (1 z ) 2 n 1
(5.25)
For the equations (5.24) are equivalent with 1
1
a ( z, q)u ( z; q) (q; q) u ( z; q) (q; q) k
k 0
k
0
n0
n
(qu ) n
(5.26)
n
and thus with
a ( z, q)u k 0
k
k
F (qu , z , q ) . F (u , z , q )
From q nu n un un F (qu, z , q) F (u, z , q) n 0 z; q n q; q n n 0 z; q n q; q n n 0 z ; q n q; q n 1
u un u F (u, qz , q ) 1 z n0 qz; q n q; q n 1 z
follows F (qu , z , q ) u F (u , qz , q ) u 1 1 (u, z, q ). F (u , z , q ) 1 z F (u , z , q ) 1 z
By (5.18) (u , z ,1) 1
(5.27)
uz uz which implies (5.25). (u , z ,1) 2 C 2 2 (1 z ) (1 z )
Final remarks Some results hold also in a slightly more general version by introducing a third parameter s. The polynomials f n( m ) ( x, q, s ) which satisfy f n( m ) ( x, q, s ) xf n(m1) ( x, q, s ) q n m sf n(mm) ( x, q, s ) are n m
k m 2
n (m 1)k ( s ) k x n mk . Both the recurrences k k 0 and the coefficients are given by closed formulae. For m 1 we also have a nice product representation f n(1) ( x, q ) ( x s )( x qs ) x q n 1s .
given by the formulae f
(m) n
( x, q , s ) q
As already mentioned there is no interesting q analogue of ln( m ) ( x ) with simple recurrences whose coefficients are given by a closed formula. But if we define ln( m ) ( x, q, s ) f n( m ) ( x, q, s ) f n(mm) ( x, q, q m 1s ) then we get at least a nice formula for the polynomials n m
l
(m) n
( x, q , s ) q k 0
k m 2
[n (m 2)k ] n (m 1)k k n mk . ( s ) x k [n (m 1)k ]
For m 2 these polynomials reduce to the Carlitz q Lucas polynomials n 2
q k 0
k 2 2
[ n] n k ( s )k x n 2 k . [n k ] k
28
n m
(m) n
The polynomials F n m
L ( x, q) (1) q (m) n
k
k 0
( x, q) (1) q
k 2
k
k 0
k 1 2
n (m 1)k n mk and x k
n (m 1)k [n (m 2)k ] n mk [n (m 1)k ] x k
have closed formulae for the coefficients and curious q analogues of the recurrence relations. For m 2 they have moreover simple closed formulae for the moments. The most important special cases are the monic q Chebyshev polynomials which are orthogonal, have simple recurrence relations and closed formulae for both their coefficients and for the moments. It would be interesting if there exist m extensions of these formulae. Finally we have considered the polynomials f n ( x, z , q) and ln ( x, z, q) which also have closed formulae for their coefficients and which for z q reduce to special Chebyshev polynomials and for z 0 to the Carlitz q Fibonacci polynomials f n(2) ( x, q, q ). Their generating functions can be expressed as f (u , z , q ) l (u, z , q)
G q 2u , z , q G qu, z , q
G ( q 2u , qz , q ) and G ( qu , z , q )
respectively if G (u , z , q ) q n n0
2
n
un
z ; q n q; q n
.
References [1] W.A. Al-Salam and M.E.H. Ismail, Orthogonal polynomials associated with the RogersRamanujan continued fraction, Pacific J. Math. 104 (1983), 269 – 283 [2] G. E. Andrews, Catalan numbers, q-Catalan numbers and hypergeometric series, J. Comb. Th. A 44 (1987), 267-273 [3] G. E. Andrews, R. Askey and R. Roy, Special functions, Encyclopedia of Math. and Appl. 71, 1999 [4] M. J. Cantero and A. Iserles, On a curious q-hypergeometric identity, Nonlinear Analysis, Springer Optimization and Its Applications 68 (2012) 121-126. [5] J. Cigler, Elementare q-Identitäten, Sém. Loth. Comb. B05a (1981) [6] J. Cigler, Operatormethoden für q-Identitäten V: q-Catalanbäume, Sitzungsber. ÖAW 205 (1996), 175-182 [7] J. Cigler, q-Fibonacci polynomials, Fib. Quart. 41 (2003), 31-40 [8] J. Cigler, A new class of q-Fibonacci polynomials, Electronic J. Comb. 10 (2003), R 19 [9] J. Cigler, q-Catalan numbers and q-Narayana polynomials, arXiv:math/0507225
29
[10] J. Cigler, q-Lucas polynomials and associated Rogers-Ramanujan type identities, arXiv:0907.0165 [11] J. Cigler, A simple approach to q-Chebyshev polynomials, arXiv:1201.4703 [12] J. Cigler, q-Chebyshev polynomials. arXiv:1205.5383 [13] J. Fürlinger and J. Hofbauer, q-Catalan numbers, J. Comb. Th. A 40 (1985), 248-264 [14] OEIS, The Online-Encyclopedia of Integer Sequences, http://oeis.org/ [15] P. Paule and A. Riese, A Mathematica q-analogue of Zeilberger’s algorithm based on an algebraically motivated approach to q-hypergeometric telescoping, Fields Inst. Commun. 14 (1997), 179-210
30
