A Generalization of Hermite Polynomials

International Journal of Contemporary Mathematical Sciences
Vol. 9, 2014, no. 13, 615 - 623
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ijcms.2014.4993
A Generalization of Hermite Polynomials
Ghulam Farid
Global Institute Lahore
New Garden Town, Lahore, Pakistan
G. M. Habibullah
Global Institute Lahore
New Garden Town, Lahore, Pakistan
Sundas Shahzeen
Global Institute Lahore
New Garden Town, Lahore, Pakistan
Copyright © 2014 Ghulam Farid, G. M. Habibullah and Sundas Shahzeen. This is an open
access article distributed under the Creative Commons Attribution License, which permits
unrestricted use, distribution, and reproduction in any medium, provided the original work is
properly cited.
Abstract


In this paper, we introduce a simple set H np ( x) , which is a generalized
form of Hermite polynomial H n ( x) . We establish series form, basic recurrence
relations, the pure recurrence relation and Rodrigues type formula for the
generalized polynomial.
Mathematics Subject Classification: 33C45, 11B37, 05A15
Keywords: Hermite polynomial, generalized Hermite polynomial, recurrence
relation, pure recurrence relation, Rodrigues formula
616
Ghulam Farid, G. M. Habibullah and Sundas Shahzeen
1 Introduction
Hermite polynomials are one of the most significant classical orthogonal
polynomials arisen from the expressions of the form

exp(2 xt  t 2 )   H n ( x)
n 0
tn
.
n!
(1.1)
Large dedicated literature is available to study the orthogonal polynomials. We
refer some of them beginning with [1], followed by [4], [7], [9] and [5]. We also
refer some recent work regarding properties, extensions, generalizations and
applications of Hermite and the related orthogonal polynomials; e.g. [2], [8], [6],
[10], [3] and [4].
Analogous to Lemma 11, pp 57 of [1], it is easy to see for each p  Z 


n 
  p 
 A(k , n)   A(k , n  pk ).
n 0 k 0
(1.2)
n 0 k 0
For each p  1 , if we define H ((pp)1) n  as
 p 1  p 
 
tn
exp     (t ) p k x k    H ((pp)1) n ( x) ,
n!
 k 0  k 
 n 0
(1.3)
then H np ( x) is a simple set.
One may observe that for p  2 , the relations (1.1) and (1.3) are identical.
That is, H n(2) ( x)  H n ( x) , which is the classical Hermite polynomial.
2 Main Results
We discus some results of the generalized Hermite polynomials generally
and for some particular values of the parameter p .
A generalization of Hermite polynomials
617
Theorem 2.1: The polynomial H n(3) ( x) can be written as
n
k
 2 k 3 j

2

3

3




n

4
k

6
j
 n  n 
x
   x
 4  6  

1
n
  2 
H n(3) ( x)  (  1)     
2
k!
 k 0 j 0 ( n  2k  3 j  1)
2


 3
 
 3
j!
j



.



(1.4)
Proof: For p  3 , using Lemma 11, pp 57 of [1] and (1.2) it follows from (1.3)
that

tn
(3)
H
(
x
)
 exp(3x 2t  3xt 2  t 3 )

2n
n!
n 0
n 
  2

n 0 k 0
(3x)k (3x 2 ) n 2 k n t 3
t e
k !(n  2k )!
 n 2  n 3  1 (3x) k (3x 2 ) n 3 j 2 k 
 tn.
  
n  0  k  0 j  0 j ! k !( n  3 j  2k )! 



Hence by comparison, we have
n  2 k 3 j
k

  3 2 
 3 
 n   n     x 
   x
  2   3   1  
  2 
(3)
H 2 n ( x)  n !   
(n  2k  3 j )!
k!
 k 0 j 0


n
2
We find the result (1.4) by replacing n by
 3
 
 3
j!
j



.



and observe that H n(3) ( x) is a
polynomial of degree precisely n in x .
Theorem 2.2: The Polynomial H n(4) ( x) can be written as
n  n  n
H
(4)
n

 6   9   12
n
( x)  (  1)    y,
3
k 0 j 0 i 0
n
 2 k 3 j  4 i
k
j
i
  4 2   4    4
 43
x n 6 k 9 j 12i )     x     x      
 
1
  2  3     4 .
y 
n
k!
j!
i!
(  2k  3 j  4i  1)
3
(1.5)
618
Ghulam Farid, G. M. Habibullah and Sundas Shahzeen
Proof: For p  4 , it follows from (1.3) that

 H3(4)n ( x)
n 0
tn
 exp(4 x3t ) exp(6 x 2t 2 ) exp(4 xt 3 ) exp(t 4 ).
n!
Now, using Lemma 11, pp 57 of [1] and (1.2), it follows from (1.3) that
n2k
k
 4 3 
  4 2 
x 
n  
   x 
  2  1 
  

  2  tn,
3
2 2
exp(4 x t ) exp(6 x t )   
k !(n  2k )!
n 0 k 0
e4 x t e 6 x t e 4 xt
3
2 2
3
j

 4 
n  
  x
 2
   3 
  
j!
k 0 n 0 j 0



j
n2k
k
 4 3 
  4 2  
  x 
   x  
1




  2  tn 

k !(n  2k )!



n 3 j  2 k
k
 4   4 3 
  4 2 
n 
n     x     x 
   x 
 2   3 
 3    1  

  2  tn,
 
j!
k !(n  3 j  2k )!
k 0 n 0 j 0
and thus
e
4 x3t 6 x 2t 2
e
e
4 xt 3  t 4
e
n  n 
 2  3 
   s,
k 0 j 0
i
n  4 i 3 j  2 k
j
k
  4 4   4   4 3 
  4 2 
n      t     x     x 
   x 
  4
 4     3    1  

  2   t n  4i .
s 
i!
j!
k !(n  4i  3 j  2k )!
n 0 i 0
So, we get

H
n 0
(4)
3n
( x)

tn
 u tn,
n ! n 0
i
j
n  4 i 3 j  2 k
k
  4  4   4 3 
  4 2 
n  n  n          x     x 
   x 
 2  3   4
4
3
1






 
 

  2  .
u
i!
j!
k !(n  4i  3 j  2k )!
k 0 j 0 i 0
A generalization of Hermite polynomials
619
Hence, by comparison, we obtain
n  2 k 3 j  4 i
k
j
i
 4 3 
  4 2   4    4
n  n  n     x 
   x    x   
 2  3   4
1  

  2  3     4 .
(4)
H 3n ( x)  n !   
k!
j!
i!
k  0 j  0 i  0 ( n  2k  3 j  4i )!
We obtain the result (1.5) by replacing n by
n
and observe that H n(4) ( x) is a
3
polynomial of degree precisely n in x .
Theorem 2.3: It follows by induction that
n  n 
 2  3 
n 
 p 
i1  0 i2  0
i p1  0
H ((pp)1) n ( x)  n !   ...  y,
n  2 i1 3i2 ... pi p1
i1
  p  p 1 
  p  p  2    p  p 3  2 
p 1  p  
x
 

   x    x 
 (1)   
1 
 p

 2 
 3 
 ... 
y 
(n  2i1  3i2  ...  pi p 1 )!
i1 !
i2 !
i p 1 !
i
i p1
and hence
H n( p ) ( x)  n !
 n  n 



 2( p 1)   3( p 1) 
 
i1  0
n
i2  0
 n 


 p( p 1) 
...

i p1  0
z1 z2 ,
 2( p 1) i1 3( p 1) i2 ... p ( p 1) i p1
  p  p 1  p 1
  x 
1 

z1  
,
n
(
 2i1  3i2  ...  pi p 1  1)!
p 1
i1
i2
  p  p  2    p  p 3 

p 1  p  
  x    x 
 (1)   
2 
 p
 3 
 ... 
z2  
i1 !
i2 !
i p 1 !
i p1
.
Theorem 2.4: The polynomial H 2(3)n ( x) can be written as
H 2(3)n ( x)  (1)n exp( x3 )
dn
 exp( x3 ) .
dx n
(1.6)
,
620
Ghulam Farid, G. M. Habibullah and Sundas Shahzeen

Proof: Let f ( x, t )   H 2(3)n ( x)
n 0
tn
,
n!
then it follows by Maclaurin’s theorem that
n
H 2(3)n ( x)  n  exp(3x 2t  3xt 2  t 3 )  .
t
t 0
Since


exp(3x 2t  3xt 2  t 3 )   exp( x3 )  exp(( x  t )3 )  ,

t
t
using w  x  t , we have

d
exp(3x 2t  3xt 2  t 3 )    exp( x3 )
exp(w3 )  .


t
dw
Eventually, we obtain (1.6).
Similarly, the polynomial H 3(4)
n ( x) can be written as
dn
H ( x)  (1) exp( x ) n  exp( x 4 )  .
dx
(4)
3n
n
4
(1.7)
And by induction or otherwise following the same procedure of that of the above
theorem we have
Theorem 2.5:
H ((pp)1) n ( x)  (1)n exp( x p )
dn
 exp( x p ) .
dx n
 p 1  p 

exp     (t ) p k x k   f ( x, t ),
 k 0  k 

then it follows by Maclaurin’s theorem that


tn
n
tn
( p)
H
(
x
)

f
(
x
,
t
)
,




( p 1) n
n ! n0 t n
n!
n 0
t 0
Proof: Let
implying that
H ((pp)1) n ( x) 
 p 1  p 
n
p k k 
exp
    (t ) x  .
n
t
 k 0  k 
 t 0
(1.8)
A generalization of Hermite polynomials
621
Since

 p 1  p 

p k k 
p 
p
 exp     (t ) x    exp( x )  exp(( x  t )  ,
k
t 

t
 k 0  

using w  x  t , we have

 p 1  p 

d
pk k 
p
 exp     (t ) x     exp( x )
 exp(w p )  .
k
t 
dw
 k 0  

Eventually, we obtain (1.8).
Theorem 2.6: The generalized polynomial H 2(3)n ( x) satisfies the recurrence
relations
(i) H (3) ( x)   3x 2 H (3) ( x)  H (3) ( x),

(ii)  H
2n
(3)
2n

( x)   6nx H
2( n 1)
2n
(3)
2( n 1)
( x)  3n(n  1) H 2((3)n2) ( x),
(iii) H 2((3)n1) ( x)  3x 2 H 2(3)n ( x)  6nx H 2((3)n1) ( x)  3n(n 1) H 2((3)n2) ( x).
(1.9)
(1.10)
(1.11)
Proof:

H 2(3)n ( x)
.
n!
n 0
Then differentiating partially with respect to x and t , we have
n
(3)*
Let F ( x, t )  exp(3x 2t  3xt 2  t 3 )   H 2(3)*
n ( x )t , H 2 n ( x) 

 n
Fx ( x, t )  (6 xt  3t 2 ) exp(3x 2t  3xt 2  t 3 )    H 2(3)*
n ( x)  t ,
(1.12)
n 0

n 1
Ft ( x, t )  (3x 2  6 xt  3t 2 ) exp(3x 2t  3xt 2  t 3 )   n H 2(3)*
.
n ( x) t
n 0
Since Fx ( x, t )  Ft ( x, t )  3x 2 F ( x, t ), we have

 H
n 0
(3)*
2n


n 0
n 0
n 1
n
( x)  t n   n H 2(3)*
 3x 2  H 2(3)*
n ( x)t
n ( x)t ,
implying that
H
(3)*
2n
2
(3)*
( x)   (n  1) H 2((3)*
n 1) ( x)  3 x H 2 n ( x),
and ultimately we obtain (1.9).
(1.13)
622
Ghulam Farid, G. M. Habibullah and Sundas Shahzeen
From relation (1.12) we have

(3)

H 2(3)n ( x) n 1
H 2(3)n ( x) n  2   H 2 n ( x)  n
6x 
t  3
t 
t .
n!
n!
n!
n 0
n 0
n 0
As H 0(3) ( x)   H 0(3) ( x)    H 2(3) ( x)   0, we obtain


6x 
H 2((3)n 1) ( x)

t  3
n
H 2((3)n 2) ( x)
(n  1)!
n  2 ( n  2)!
and eventually we get (1.10).
n2

t 
n
n2
H
(3)
2n
( x) 
n!
tn,
Relation (1.9) and (1.10) yield (1.11), which is pure recurrence relation satisfied
by the polynomial H 2(3)n ( x) .
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E. D. Rainville, Special Functions, The Macmillan Company, New York,
1960.
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G. Andrews, R. Askey and R. Roy, Special Functions, Cambridge
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[4]
G. Farid and G. M. Habibullah, Extensions of Legendre polynomials, int.
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Received: September 21, 2014; Published: October 31, 2014

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