Unification of a class of trilateral generating functions

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Rev. Real Academia de Ciencias. Zaragoza. 59: 71–77, (2004).
Unification of a class of trilateral generating functions
Manik Chandra Mukherjee
Netaji Nagar Vidyamandir
Calcutta 700 092. India
Abstract
In this paper we have proved a general theorem in connection with the unification of a class of trilateral generating functions of certain special functions. Some
particular cases of interest are also pointed out.
Keywords: Unification, Generating functions, Group-theoretic method.
MSC: 33A65
1
Introduction
Theories in connection with the unification of generating functions are of greater im-
portance in the study of special functions. In this direction attempts have been made by
some researchers [1–8]. The aim at presenting this article is to state and prove a theorem
on the unification of a class of trilateral generating functions of certain special functions.
In fact we have proved the following theorem.
Theorem 1 For a set of functions {Sn (x)|n ≥ 0} generated by
∞
An (m)Sn+m (x)tn =
n=0
f (x, t)
Sm (h(x, t))
g(x, t)m
(1)
where m is a non-negative integer, An (m) are arbitrary constants and f, g, h are arbitrarily
chosen functions of x and t, let
F (x, y, t) =
∞
an (m)Sn+m (x)gn (t)tn ,
n=0
then the following trilateral generating relation holds:
∞
n=0
Sn+m (x)σn (y, z)tn =
f (x, t)
F (h(x, t), y, zt/g(x, t))
g(x, t)m
71
where
σn (y, z) =
n
ak An−k (m + k)gk (y)z k .
k=0
Proof.
∞
=
=
=
n=0
∞
Sn+m (x)σn (y, z)tn
Sn+m (x)
n=0
∞
∞ n
ak An−k (m + k)gk (y)z
k
tn
k=0
Sn+m+k (x)σk (x)An (m + k)gk (y)z k tn+k
n=0 k=0
∞
ak gk (y)(zt)k
k=0
f (x, t)
Sm+k (h(x, t))
g(x, t)m
∞
zt
f (x, t) a
S
(h(x,
t))g
(y)
=
k
m+k
k
g(x, t)m k=0
g(x, t)
f (x, t)
zt
=
F h(x, t), y,
.
m
g(x, t)
g(x, t)
k
The result given in [9] can be obtained directly from the theorem. Indeed, by putting
m = 0 in the above theorem, we get
F (x, y, t) =
∞
an Sn (x)gn (y)tn ,
n=0
then,
∞
zt
Sn (x)σn (y, z)t = f (x, t)F h(x, t), y,
,
g(x, t)
n=0
n
where
σn (y, z) =
n
ak An−k (k)gk (y)z k
k=0
which is found derived in [9].
2
Applications
We now apply our theorem in the derivation of generating functions of various special
functions
2.1
On Hermite polynomials
We first consider the following generating relation involving Hermite polynomials [9]
∞
n=0
Hn+m
tn
= exp(2xt − t2 )Hm (x − t).
n!
(2)
Now it is evident that the relation (2) being of the form (1), one can easily obtain the
following result with the help of theorem 1.
72
Proposition 1 If there exists a generating relation of the form
∞
F (x, y, t) =
an Hn+m (x)gn (y)
n=0
then
∞
Hn+m (x)σn (y, z)
n=0
tn
= exp(2xt − t2 )F ((x − t), y, zt),
n!
where
σn (y, z) =
n
k=0
2.2
tn
,
n!
n
ak gk (y)z k .
k
On Laguerre polynomials
Next we consider the following generating relation involving Laguerre polynomials
[10,11]
∞
n=0
m+n α
−xt
x
Ln+m (x)tn = (1 − t)−1−α−m exp
Lαn
.
n
1−t
1−t
(3)
The relation (3) being of the form (1), one can state the following result with the help of
theorem 1.
Proposition 2 If
∞
F (x, y, t) =
an Lαn+m (x)gn (y)tn ,
n=0
then
∞
Lαn+m (x)σn (y, z)tn
−1−α−m
= (1 − t)
n=0
where
σn (y, z) =
n
−xt
exp
1−t
F
zt
x
,
, y,
1−t
1−t
m+n
ak gk (y)z k .
m+k
k=0
2.3
On modified Laguerre polynomials
Next we consider the following generating relation involving modified Laguerre poly-
nomials [10]
exp(xt)(1 −
β
[x(1
t)−β−m fm
− t)] =
∞
(m + 1)n β
fm+n (x)tn .
n!
n=0
(4)
The relation (4) being of the form (1), one can obtain the following result with the help
of theorem 1.
Proposition 3 If there exist a generating relation of the form
F (x, y, t) =
∞
β
an fm+n
(x)gn (y)tn ,
n=0
73
then
∞
β
−β−m
fn+m (x)σn (y, z)t = exp(xt)(1 − t)
n
n=0
where
σn (y, z) =
2.4
zt
F x(1 − t), y,
,
1−t
n
(m + k + 1)n−k
ak gk (y)z k .
(n − k)!
k=0
On ultra spherical polynomials
Next we consider the following generating relation involving ultra spherical polynomi-
als [11]
∞
m+n λ
x−t
Pn+m (x)tn = ρ−m−2λ Pmλ
,
ρ
n
n=0
(5)
where ρ = (1 − 2xt + t2 )1/2 .
The relation (5) being of the form (1), one can state the following result with the help
of theorem 1.
Proposition 4 If
F (x, y, t) =
∞
λ
an Pn+m
(x)gn (y)tn ,
n=0
then
∞
λ
Pn+m
(x)σn (y, z)tn
=ρ
−2λ−m
n=0
where
σn (y, z) =
n
k=0
2.5
F
x−t
zt
,
, y,
ρ
ρ
m+n
ak gk (y) z k .
m+k
On modified Jacobi polynomials
Next we consider the following generating function involving modified Jacobi polyno-
mials [12].
∞
n=0
m + n (α−m−n,β−m−n)
Pn+m
(x)tn
n
α−m = 1 + 12 (x + 1)
α−n
1 + 12 (x − 1)
(6)
Pm(α−m,β−n) x + 12 (x2 − 1)t .
The relation (6) being of the form (1), one can easily state the following result with
the help of theorem 1.
Proposition 5 If
F (x, y, t) =
∞
(α−m−n,β−m−n)
an Pn+m
n=0
74
gn (y) tn ,
then
∞
(α−m−n,β−m−n)
Pn+m
(x) σn (y, z) tn
n=0
α−m = 1 + 12 (x + 1)
α−n
1 + 12 (x − 1)


×F 1 + 12 (x2 − 1)t, y, where
zt
1 + 12 (x + 1)t
n
σn (y, z) =
 ,
1 + 12 (x − 1)t
m+n
ak gk (y) z k .
m+k
k=0
If instead of (6) we consider the following generating realtion [13]
∞
(r + 1)n (k−r−n,β)
(x) tn
Pn+r
n!
n=0
−1−β−k
= (1 + t)k−r 1 + 12 (1 − x)t
Pr(k−r,β)
(7)
x − 12 (1 − x)t
,
1 + 12 (1 − x)t
then we get the following result
Proposition 6 If
∞
F (x, y, t) =
(k−r−n,β)
an Pn+r
gn (y) tn ,
n=0
then
∞
(k−r−n,β)
Pn+r
(x) σn (y, z) tn
n=0
k−r
= (1 + t)
1+
1
(1
2
− x)t
where
σn (y, z) =
2.6
−1−β−k
×F
x − 12 (1 − x)t
zt
, y,
,
1
1+t
1 + 2 (1 − x)t
n
(r + k + 1)n−k
ak gk (y) z k .
(n − k)!
k=0
On Bessel functions
Next we consider the following generating relation involving Bessel functions [14].
∞
tn
2t
Jn+m (x) = 1 −
n!
x
n=0
−m/2
√
Jm ( x2 − 2xt).
(8)
The relation (8) being of the form (1), one can easily state the following result with
the help of theorem 1.
Proposition 7 If
F (x, y, t) =
∞
an Jn+m (x) gn (y)
n=0
75
tn
,
n!
then
∞
tn
2t
Jn+m (x) σn (y, z) = 1 −
n!
x
n=0
where
σn (y, z) =
F
n
k=0
2.7
−m/2
√
xtz
x2 − 2xt, y, √ 2
,
x − 2xt
n
ak gk (y) z k .
k
On Bessel polynomials
Next we consider the following generating relation involving Bessel polynomials [10].
√
∞
1 − 1 − 2xt
x
tn
−(m+1)/2
Yn+m (x) = (1 − 2xt)
exp
Ym √
.
(9)
n!
x
1 − 2xt
n=0
The relation (9) being of the form (1), one can easily state the following result with
the help of theorem 1.
Proposition 8 If
∞
tn
an Yn+m (x) gn (y) ,
F (x, y, t) =
n!
n=0
then
∞
Jn+m (x) σn (y, z)
n=0
tn
n!
−(m+1)/2
= (1 − 2xt)
exp
1−
√
1 − 2xt
F
x
where
σn (y, z) =
n
k=0
2.8
x
zt
√
, y, √
1 − 2xt
1 − 2xt
,
n
ak gk (y) z k .
k
On modified generalised Bessel polynomiasl
Next we consider the following generating relation involving modified generalised
Bessel polynomiasl [15,16].
∞
β n α−m−n
x
.
Yn+m (x)tn = (1 − xt)1−α exp(βt)Ymα−m
1 − xt
n=0 n!
(10)
The relation (10) being of the form (1), one can easily state the following result with
the help of theorem 1.
Proposition 9 If
∞
F (x, y, t) =
α−m−n
an Yn+m
(x) gn (y) tn ,
n=0
then
∞
α−m−n
Yn+m
(x) σn (y, z) tn
= (1 − xt)
1−α
exp(βt)F
n=0
where
σn (y, z) =
n
β n−k
ak gk (y) z k .
(n
−
k)!
k=0
76
x
, y, zt ,
1 − xt
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