https://doi.org/10.5831/HMJ.2017.39.2.217
GENERATING FUNCTIONS FOR LEGENDRE-BASED POLY-BERNOULLI NUMBERS AND POLYNOMIALS
N.U. Khan, T. Usman∗ and M. Aman
Abstract. In this paper, we introduce a generating function for a Legendre-based poly-Bernoulli polynomials and give some identities of these polynomials related to the Stirling numbers of the second kind. By making use of the generating function method and some functional equations mentioned in the paper, we conduct a further investigation in order to obtain some implicit summation formulae for the Legendre-based poly-Bernoulli numbers and polynomials.
1. Introduction
Generalized and multivariable forms of the special functions of mathe- matical physics have witnessed a significant evolution during the recent years. In particular, the special polynomials of more than one variable provided new means of analysis for the solution of large classes of par- tial differential equations often encountered in physical problems. Most of the special function of mathematical physics and their generalization have been suggested by physical problems.
To give an example, we recall that the 2-variable Hermite Kampe0 de Fe0riet polynomials Hn(x, y) [3] defined by the generating function
exp(xt + yt2) =
∞
X
n=0
Hn(x, y)tn
n! (1.1)
Received January 11, 2017. Accepted May 16, 2017.
2010 Mathematics Subject Classification. 05A10, 05A15, 33C45, 33C90.
Key words and phrases. Legendre polynomials, Hermite polynomials, poly- Bernoulli polynomials, Legendre poly-Bernoulli polynomials and summation formulae.
*Corresponding Author
are the solution of heat equation
∂
∂yHn(x, y) = ∂2
∂x2Hn(x, y), Hn(x, 0) = xn (1.2) The higher order Hermite polynomials, sometimes called the Kampe0 de Fe0riet polynomials of order m or the Gould-Hopper polynomials Hn(m)(x, y) defined by the generating function [[8], p.58 (6.3)]
exp(xt + ytm) =
∞
X
n=0
Hn(m)(x, y)tn
n! (1.3)
are the solution of the generalized heat equation [5]
∂
∂yf (x, y) = ∂m
∂xmf (x, y), f (x, 0) = xn (1.4) Also, we note that
Hn(2)(x, y) = Hn(x, y), (1.5) Hn(2x, −1) = Hn(x), (1.6) where Hn(x) are the classical Hermite polynomials [1].
Next, we recall that the 2-variable Legendre polynomials Sn(x, y) and Rn(x, y) are given by Dattoli et al. [6]
Sn(x, y) = n!
[n2]
X
k=0
xkyn−2k
[(n − 2k)!(k!)2] (1.7) and
Rn(x, y) = (n!)2
∞
X
k=0
(−1)n−kxn−kyk
[(n − 2k)!]2(k!)2 (1.8) respectively, and are related with the ordinary Legendre polynomials Pn(x) [21] as
Pn(x) = Sn
−1 − x2 4 , x
= Rn 1 − x 2 ,1 + x
2
. From equation (1.7) and (1.8), we have
Sn(x, 0) = n! x[n2]
[(n2)!]2, Sn(0, y) = yn, (1.9) Rn(x, 0) = (−x)n, Rn(0, y) = yn
((−n)!)2. (1.10)
The generating functions for two variable Legendre polynomials Sn(x, y) and Rn(x, y) are given by [6]
eytC0(−xt2) =
∞
X
n=0
Sn(x, y)tn
n! (1.11)
C0(xt)C0(−yt) =
∞
X
n=0
Rn(x, y) tn
(n!)2 (1.12)
where C0(x) is the 0-th order Tricomi function [21]
C0(x) =
∞
X
r=0
(−1)rxr
(r!)2 (1.13)
As it is well known, the Bernoulli polynomials are defined by their generating function
t
et− 1
ext =
∞
X
n=0
Bn(x)tn
n!, (see[1] − [22]) (1.14) When x = 0, Bn= Bn(0) are called the Bernoulli numbers.
From (1.14), we have
Bn(x) =
n
X
m=0
n m
Bn−m xm (1.15)
The classical polylogarithm function Lik(z) is Lik=
∞
X
m=1
zm
mk, |z| < 1 (see [9], [10]) (1.16) so for k ≤ 1,
Lik(z) = −ln(1 − z), Li0(z) = z
1 − z, Li−1(z) = z
(1 − z)2, · · · The poly-Bernoulli polynomials are given by
Lik(1 − e−t) et− 1 ext=
∞
X
n=0
Bn(k)(x)tn
n!, (see [9], [13] − [18]) (1.17) For k = 1, we have
Li1(1 − e−t)
et− 1 ext= t
et− 1ext=
∞
X
n=0
Bn(x)tn
n!, (1.18)
From (1.14) and (1.18), we obtain
Bn(1)(x) = Bn(x), (n ≥ 0)
Recently, Pathan et al. [19] introduced the generalized Hermite- Bernoulli polynomials of two variablesHBn(α)(x, y) is defined by
t
et− 1
α
ext+yt2 =
∞
X
n=0
HBn(α)(x, y)tn
n! (1.19)
which are essentially generalization of Bernoulli numbers, Bernoulli poly- nomials, Hermite polynomials and Hermite-Bernoulli polynomialsHBn(x, y) introduced by Dattoli et al. ([7], p.386 (1.6)) in the form
t et− 1
ext+yt2 =
∞
X
n=0
HBn(x, y)tn
n! (1.20)
The stirling number of the first kind is given by (x)n= x(x − 1) · · · (x − n + 1) =
n
X
l=0
S1(n, l)xl, (n ≥ 0) (1.21) and the stirling number of the second kind is defined by generating function to be
(et− 1)n= n!
∞
X
l=n
S2(l, n)tl
l! (1.22)
The special polynomials of more than one variable provide new means of analysis for the solutions of a wide class of partial differential equations often encountered in physical problems. It happens very often that the solution of a given problem in physics or applied mathematics requires the evaluation of infinite sum, involving special functions. Problem of this type arise, for example, in the computation of the higher-order mo- ments of a distribution or in evaluation of transition matrix elements in quantum mechanics. In [4], Dattoli showed that the summation formu- lae of special functions, often encountered in applications ranging from electromagnetic process to combinatorics, can be written in terms of Hermite polynomials of more than one variable.
In this paper, we first gives the definition of the Legendre-based poly- Bernoulli polynomialsSBn(k)(x, y, z) and we have given some formulae of those polynomials related to the Stirling numbers of second kind. Some implicit summation formulae are derived by using different analytical means and applying generating functions. These result extended some
known summations and identities of generalized Hermite-Bernoulli poly- nomials polynomials studied by Dattoli et al., Khan et al. and Pathan and Khan.
2. Generating functions for Legendre-based poly-Bernoulli numbers and polynomials
Now, we define the Legendre-based poly-Bernoulli polynomials as fol- lows:
Lik(1 − e−t)
et− 1 eyt+zt2C0(−xt2) =
∞
X
n=0
SB(k)n (x, y, z)tn
n!, (k ∈ z) (2.1) so that
SB(k)n (x, y, z) =
n
X
m=0 [n2]
X
r=0
Bn−m(k) Sm−2r(x, y)zrn!
(m − 2r)!r!(n − m)! (2.2) when x = y = z = 0, Bn(k)=LBn(k)(0, 0, 0) are called the poly-Bernoulli numbers. By (2.1), we easily get Bn(k)= 0. For k = 1, from (2.1), we have
Li1(1 − e−t)
et− 1 eyt+zt2C0(−xt2) =
∞
X
n=0
SBn(x, y, z)tn
n!, (2.3)
Thus by (2.1) and (2.3), we get
SBn(1)(x, y, z) =SBn(x, y, z), (n ≥ 0).
Theorem 2.1. For n ≥ 0, we have
SBn(2)(x, y, z) =
n
X
m=0
n m
Bm
m + 1 SBn−m(x, y, z) (2.4) Proof. Applying definition (2.1), we have
Lik(1 − e−t)
et− 1 eyt+zt2C0(−xt2) =
∞
X
n=0
SBn(k)(x, y, z)tn n!
Lik(1 − e−t)
et− 1 eyt+zt2C0(−xt2) =
1
et− 1
eyt+zt2C0(−xt2)
× Z t
0
1 ez− 1
Z t 0
1
ez− 1· · · 1 ez− 1
Z t 0
z
ez− 1dzdz · · · dz
| {z }
In particular k = 2, we have
SB(2)n (x, y, z) =
1
et− 1
eyt+zt2C0(−xt2) Z t
0
z ez− 1
SBn(2)(x, y, z) =
∞
X
m=0
Bmtm (m + 1)!
! t et− 1
eyt+zt2C0(−xt2)
SBn(2)(x, y, z) =
∞
X
m=0
Bmtm (m + 1)!
! ∞ X
n=0
SBn(x, y, z)tn n!
!
Replacing n by n − m in the above equation, we have
SBn(2)(x, y, z) =
∞
X
n=0 n
X
m=0
n m
Bm
m + 1 SBn−m(x, y, z)tn n!
On equating the coefficients of the like power of t in the above equation, we get the result (2.4).
Theorem 2.2 For n ≥ 1, we have
SBn(k)(x, y, z) =
n
X
m=0 [n2]
X
k=0
Bm(k)Sn−m−2k(x, y)zkn!
m!k!(n − m − 2k)! (2.5) Proof. From equation (2.1), we have
∞
X
n=0
SBn(k)(x, y, z)tn
n! = Lik(1 − e−t)
et− 1 eyt+zt2C0(−xt2)
∞
X
n=0
SBn(k)(x, y, z)tn n! =
∞
X
m=0
Bm(k)tm m!
!
∞
X
n=0 [n2]
X
k=0
Sn−2k(x, y)zk k!(n − 2k)! tn
Replacing n by n − m in the above equation and comparing the coefficients of tn, we get the result (2.5).
Theorem 2.3 For n ≥ 0, we have
SBn(k)(x, y, z) =
n
X
p=0 p+1
X
l=1
(−1)l+p+1l!S2(p + 1, l) lk(p + 1)
n p
SBn−p(x, y, z) (2.6)
Proof. From equation (2.1), we have
∞
X
n=0
SB(k)n (x, y, z)tn
n! = Lik(1 − e−t) t
t et− 1
eyt+zt2C0(−xt2)
(2.7) Now
1
tLik(1 − e−t) = 1 t
∞
X
l=1
(1 − e−t)l lk = 1
t
∞
X
l=1
(−1)l
lk (1 − e−t)l
= 1 t
∞
X
l=1
(−1)l lk l!
∞
X
p=l
(−1)pS2(p, l)tp p!
= 1 t
∞
X
p=1 p
X
l=1
(−1)l+p
lk l!S2(p, l)tp p!
=
∞
X
p=0 p+1
X
l=1
(−1)l+p+1
lk l!S2(p + 1, l) p + 1
!tp
p! (2.8)
From equation (2.7) and (2.8), we get
∞
X
n=0
SBn(k)(x, y, z)tn n! =
∞
X
p=0 p+1
X
l=1
(−1)l+p+1
lk l!S2(p + 1, l) p + 1
!tp p!
×
∞
X
n=0
SBn(x, y, z)tn n!
!
Replacing n by n − p in the r.h.s of above equation and comparing the coefficients of tn, we get the result (2.6).
Theorem 2.4 For n ≥ 1, we have
SBn(k)(x, y + 1, z) −SBn(k)(x, y, z) =
n
X
p=1 p
X
l=1 [n2]
X
k=0
(−1)l+p lk l!n!
×S2(p, l) Sn−p−2k(x, y)zk
p!k!(n − p − 2k)! (2.9)
Proof. Using the definition (2.1), we have
∞
X
n=0
SBn(k)(x, y + 1, z)tn n! −
∞
X
n=0
SBn(k)(x, y, z)tn n!
= Lik(1 − e−t)
et− 1 e(y+1)t+zt2C0(−xt2) −Lik(1 − e−t)
et− 1 eyt+zt2C0(−xt2)
∞
X
n=0
SB(k)n (x, y+1, z)tn n!−
∞
X
n=0
SBn(k)(x, y, z)tn
n! = Lik(1−e−t) eyt+zt2C0(−xt2)
=
∞
X
p=1 p
X
l=1
(−1)l+p
lk l!S2(p, l)
!tp
p! eyt+zt2C0(−xt2)
=
∞
X
p=1 p
X
l=1
(−1)l+p
lk l!S2(p, l)
! tp p!
∞
X
n=0 [n2]
X
k=0
Sn−2k(x, y)zk k!(n − 2k)! tn
Replacing n by n − p in the above equation and comparing the co- efficients of tn, we get the result (2.9).
Theorem 2.5 For d ∈ N with d ≡ 1(mod2), we have
SBn(k)(x, y, z) =
n
X
p=0
n p
dn−p−1
p+1
X
l=0 d−1
X
a=0
(−1)l+p+1l!S2(p + 1, l) lk
×(−1)aSBn−p
x,a + y d , z
(2.10) Proof. From equation (2.1), we have
∞
X
n=0
SBn(k)(x, y, z)tn
n! = Lik(1 − e−t)
et− 1 eyt+zt2C0(−xt2)
∞
X
n=0
SB(k)n (x, y, z)tn
n! = Lik(1 − e−t) t
t
edt− 1
d−1
X
a=0
e(a+y)t+zt2C0(−xt2)
!
=
∞
X
p=0 p+1
X
l=1
(−1)l+p+1
lk l!S2(p + 1, l) p + 1
! tp p!
×
∞
X
m=0
dm−1
d−1
X
a=0
(−1)aSBn
x,a + y d , z tn
n!
!
Replacing n by n − p in the above equation and comparing the coefficients of tn, we get the result (2.10).
3. Implicit summation formulae involving Legendre-based poly-Bernoulli polynomials
This section of the paper is devoted to employing the definition of the Legendre-based poly-Bernoulli polynomials SBn(k)(x, y, z) to obtain fi- nite summations. For the derivation of implicit formulae involving the Legendre-based poly-Bernoulli polynomialsSBn(k)(x, y, z) the same con- sideration as developed for the ordinary Hermite and related polynomials in Khan Khan et al. [10]-[12] and Pathan [19]-[20] holds as well. First we prove the following results involving Legendre-based poly-Bernoulli polynomialsSBn(k)(x, y, z).
Theorem 3.1. For x, y, z ∈ R and n ≥ 0. The following implicit sum- mation formula for Legendre-based poly-Bernoulli polynomialsSBn(k)(x, y, z) holds true:
SBl+p(k)(x, v, z) =
l,p
X
m,n=0
l m
p n
(v − y)m+nSBl+p−m−n(k) (x, y, z) (3.1) Proof. We replace t by t + u and rewrite the generating function (2.1) as
Lik(1 − e−(t+u)) et+u− 1
!
ez(t+u)2C0(−x(t + u)2)
= e−y(t+u)
∞
X
l,p=0
SBl+p(k)(x, y, z) tl l!
up
p! (3.2)
Replacing y by v in the above equation and equating the resulting equa- tion to the above equation, we get
e(v−y)(t+u)
∞
X
l,p=0
SBl+p(k)(x, y, z)tl l!
up p! =
∞
X
l,p=0
SBl+p(k)(x, v, z)tl l!
up
p! (3.3) on expanding exponential function (3.3) gives
∞
X
N =0
[(v − y)(t + u)]N N !
∞
X
l,p=0
SB(k)l+p(x, y, z)tl l!
up p! =
∞
X
l,p=0
SB(k)l+p(x, v, z)tl l!
up p!
(3.4)
which on using formula ([22], p. 52(2))
∞
X
N =0
f (N )(x + y)N N ! =
∞
X
m,n=0
f (m + n)xn n!
ym
m! (3.5)
in the left hand side becomes
∞
X
m,n=0
(v − y)m+n tmun m!n!
∞
X
l,p=0
SBl+p(k)(x, y, z)tl l!
up p!
=
∞
X
l,p=0
SBl+p(k)(x, v, z)tl l!
up
p! (3.6)
Now replacing l by l − m, p by p − n and using the lemma ([22], p.
100(1)) in the left hand side of (3.6), we get
∞
X
l,p=0 l,p
X
m,n=0
(v − y)m+n
m!n! SBl+p−m−n(k) (x, y, z) tl (l − m)!
up (p − n)!
=SB(k)(l+p)(x, v, z)tl l!
up
p! (3.7)
Finally, on equating the coefficients of the like powers of t and u in the above equation, we get the required result.
Remark 1. By taking l = 0 in equation (3.1), we immediately deduce the following corollary.
Corollary 3.1. The following implicit summation formula for Legendre- based poly-Bernoulli polynomials SB(k)n (x, v, z) holds true:
SBp(k)(x, v, z) =
p
X
n=0
p n
(v − y)nSBp−n(k) (x, y, z) (3.8) Remark 2. On replacing v by v + y and setting x = z = 0 in The- orem (3.1), we get the following result involving Legendre-based poly- Bernoulli polynomial of one variable
SBl+p(k)(v + y) =
l,p
X
m,n=0
l m
p n
(v)m+nSBl+p−m−n(k) (y) (3.9)
whereas by setting v = 0 in Theorem (3.1), we get another result involv- ing Legendre-based poly-Bernoulli polynomial of one and two variable
SBl+p(k)(x, z) =
l,p
X
m,n=0
l m
p n
(−y)m+nSBl+p−m−n(k) (x, y, z) (3.10) Remark 3. Along with the above result we will exploit extended forms of Legendre-based poly-Bernoulli polynomialSBl+p(k)(x, v) by setting z = 0 in the Theorem (3.1) to get
SB(k)l+p(x, v) =
l,p
X
m,n=0
l m
p n
(v − y)nSBl+p−m−n(k) (x, y) (3.11) Theorem 3.2. For x, y, z ∈ R and n ≥ 0. The following implicit sum- mation formula for Legendre-based poly-Bernoulli polynomialsSBn(k)(x, y, z) holds true:
SBn(k)(x, y + u, z) =
n
X
j=0
n j
ujSB(k)n−j(x, y, z) (3.12) Proof. Since
∞
X
n=0
SBn(k)(x, y + u, z)tn
n! = Lik(1 − e−t)
et− 1 e(y+u)t+zt2C0(−xt2)
∞
X
n=0
SBn(k)(x, y + u, z)tn n! =
∞
X
n=0
SBn(k)(x, y, z)tn n!
!
∞
X
j=0
ujtj j!
Now, replacing n by n − j and comparing the coefficients of tn, we get the result (3.12).
Theorem 3.3. For x, y, z ∈ R and n ≥ 0. The following implicit sum- mation formula for Legendre-based poly-Bernoulli polynomialsSBn(k)(x, y, z) holds true:
SB(k)n (x, y + u, z + w) =
n
X
m=0
n m
SBn−m(k) (x, y, z)Hm(u, w) (3.13) Proof. By the definition of Legendre-based poly-Bernoulli polynomials and the definition (1.1), we have
Lik(1 − e−t)
et− 1 e(y+u)t+(z+w)t2C0(−xt2)
=
∞
X
n=0
SBn(k)(x, y, z)tn n!
! ∞ X
m=0
Hm(u, w)tm m!
!
Now, replacing n by n − m and comparing the coefficients of tn, we get the result (3.13).
Theorem 3.4. For x, y, z ∈ R and n ≥ 0. The following implicit sum- mation formula for Legendre-based poly-Bernoulli polynomialsSBn(k)(x, y, z) holds true:
SB(k)n (x, y, z) =
n−2j
X
m=0 [n2]
X
j=0
Bm(k)Sn−m−2j(x, y)zjn!
m!j!(n − m − 2j)! (3.14) Proof. Applying the definition (2.1) to the term Like(1−et−1−t) and ex- panding the exponential and tricomi function eyt+zt2C0(−xt2) at t = 0 yields
Lik(1 − e−t)
et− 1 eyt+zt2C0(−xt2) =
∞
X
m=0
Bm(k)tm m!
!
×
∞
X
n=0
Sn(x, y)tn n!
!
∞
X
j=0
zjt2j j!
∞
X
n=0
SBn(k)(x, y, z)tn n! =
∞
X
n=0 n
X
m=0
Bm(k)Sn−m(x, y) (n − m)!m!
! tn
∞
X
j=0
zjt2j j!
Now, replacing n by n − 2j and comparing the coefficients of tn, we get the result (3.14).
Theorem 3.5. For x, y, z ∈ R and n ≥ 0. The following implicit sum- mation formula for Legendre-based poly-Bernoulli polynomialsSBn(k)(x, y, z) holds true:
SB(k)n (x, y + 1, z) =
[n2]
X
j=0 n−2j
X
m=0
n!(−1)j(−x)jHB(k)n−m−2j(y, z)
(n − m − 2j)!m!(j!)2 (3.15) Proof. By the definition of Legendre-based poly-Bernoulli polynomials, we have
∞
X
n=0
SBn(k)(x, y + 1, z)tn
n! = Lik(1 − e−t)
et− 1 e(y+1)t+zt2C0(−xt2)
∞
X
n=0
SBn(k)(x, y + 1, z)tn n! =
∞
X
n=0 n
X
m=0
HBn−m(k) (y, z) (n − m)!n!
! tn
!
×
∞
X
j=0
(−1)j(−xt2)j (j!)2
∞
X
n=0
SB(k)n (x, y+1, z)tn n! =
∞
X
n=0
∞
X
j=0 n
X
m=0
(−1)j(−x)jHBn−m(k) (y, z) (n − m)!n!(j!)2
tn+2j
Replacing n by n − 2j, we have
∞
X
n=0
SB(k)n (x, y+1, z)tn n! =
∞
X
n=0
[n2]
X
j=0 n−2j
X
m=0
(−1)j(−x)jHBn−m−2j(k) (y, z) (n − m − 2j)!n!(j!)2
tn
on comparing the coefficients of tn, we get the result (3.15).
whereHBn(k)(y, z) is the generalized Hermite poly-Bernoulli polynomials defined by means of the generating function as follows:
Lik(1 − e−t) et− 1
eyt+zt2 =
∞
X
n=0
HBn(k)(y, z)tn
n!, (k ∈ z) (3.16) Theorem 3.6. Let x, y, z ∈ R and n ≥ 0. The following implicit sum- mation formula for Legendre-based poly-Bernoulli polynomialsSBn(k)(x, y, z) holds true:
SBn(k)(x, y + 1, z) =
n
X
m=0
n m
SB(k)n−m(x, y, z) (3.17) Proof. By the definition of Legendre-based poly-Bernoulli polynomials, we have
∞
X
n=0
SBn(k)(x, y + 1, z)tn n! +
∞
X
n=0
SBn(k)(x, y, z)tn n!
= Lik(1 − e−t) et− 1
(et+ 1)eyt+zt2C0(−xt2)
=
∞
X
n=0
SBn(k)(x, y, z)tn n!
∞
X
m=0
tm m!+ 1
!
=
∞
X
n=0
SBn(k)(x, y, z)tn n!
∞
X
m=0
tm m!+
∞
X
n=0
SBn(k)(x, y, z)tn n!
=
∞
X
n=0 n
X
m=0
SBn−m(k) (x, y, z) tn
m!(n − m)! +
∞
X
n=0
SBn(k)(x, y, z)tn n!
Finally equating the coefficients of the like powers of tn, we get the result (3.17).
4. Concluding remarks
The paper aims at presenting the study of multivariables Legendre-based poly-Bernoulli polynomials which plays an important role in several di- verse field of physics, applied mathematics and engineering. Some ex- pressions, representation and summations of these polynomials are given in terms of well-known multidimensional classical special functions. The formulas we have given in this paper confirm the usefulness of the se- ries rearrangement technique used to deal with the theory of special functions. We have derived several implicit summation formulas for Legendre-based poly-Bernoulli polynomials by using different analyti- cal means on their respective generating functions. This process can be extended to derive new relations for conventional and generalized polynomials.
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Nabiullah Khan
Department of Applied Mathematics,
Faculty of Engineering and Technology, Aligarh Muslim University, Aligarh-202002, India.
E-mail: nukhanmath@gmail.com Talha Usman
Department of Applied Mathematics,
Faculty of Engineering and Technology, Aligarh Muslim University, Aligarh-202002, India.
E-mail: talhausman.maths@gmail.com Mohd Aman
Department of Applied Mathematics,
Faculty of Engineering and Technology, Aligarh Muslim University, Aligarh-202002, India.
E-mail: mohdaman.maths@gmail.com