International Journal of Scientific & Engineering Research, Volume 5, Issue 4, April-2014 1408

ISSN 2229-5518

Expansion of generalized polynomial set Qk {( x ), y} in terms of

Jocobi and Sister Celin’s Polynomials.

Rakesh Kumar Singh

M.Sc. Ph.D.

(J.P. University, Chapra)

Abstract :

Recently, we introduced “An unification of cirtain generalized

polynomials set

Qk {( x ), y}

with the help of generating relation which

contains the generalized Lauricella functions in the notation of Burchnall and Chaundy [2]. It is shown that this polynomial, set happen to be a generalization of as many as forty orthogonal and

non-orthogeenal polynomials. In this paper, we introduced the

polynomials set polynomials.

Introduction :

Qk {( x ), y} in terms Jocobi and Sister Celin’s

The generalized polynomial

of generating relation.

Qk {( x ), y} set is defined by means

 (ah )

− e e 

 ( Ar );(C p );(∝um ) et e e x e

( ) µ1 y t

× F µ x

 ( ) ( ) ( )

, µ2 x2 t

.......... µn xn t 

 bk

1 1 2 2 m n

  Bs Dq Brm 

Q K , µ ; µ1 µ2 µ3 ...............µn ;( Ar ); (C p ) (∝un );( ah ) {( x ) y,}t m m, ei e1 e2 e3 ...................en ( Bs ); ( Dq ); ( Bvn ); (bk ) n

......(1.1)

where µ, µ1, µ2 .................... µn are real and e. e1, e2, e3.......en

are non-negative integer.

The left hand side of (1.1) contains the product of generalized hypergeometric function and Lauricella function in the notation of

Burchnall and chaundy. The polynomial set contains a number of

parameters for simplicity. It is denoted by

order of the polynomial set.

Qk {( x ), y} where m is the

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ISSN 2229-5518

(1.2) Notations

1. (n) = 1, 2, 3......... n

2. (Ap) = A1, A2, ................ Ap

3. (Ap)m = (A1)m, (A2)m, (A3)m.................(Ap)m

4. (Ap)rn = (Ap)n, (A2)n, (A3)n.................(Ap)n

5. [(Ap), i] = A1, A2 ................... Ai–1 ............Ap

a

6. [(a, b)] =
 b + r + 1 
=  b   b + 1 
............ b + a −1 

∏        

r =1 

a k

 a k 

a k

 a k

where an empty product is treated as unity

Theorem (1) for e2 > 1,................. en > 1, we have

Qk {x n), y} = K

m

∑

i =0

(m + c)!(−1)i (2i + c + d + 1)

P( c ,d ) (1 + 2xe ) (n − i)!(c + d + i + 1) i

× F q + s + h +1;u1 ,u2 ,u3 ..........un

[(– m + i); e1e2 ..........en ]

 − +

P + r + k ;v1 ,v2.................................... vm

[( m c); e , e ........e ]
 1 2 n

[1 − (Bs ) − m); e1 , e2 −1,..........(en −1)], [1 − (c p ) − m); e1 , e2e3 ........en ]

[(1 − ( Ar ) − m) : e1 , e2 −1,.........(em −1)],[(1 − (Dq ) − m) : e1e2e3 ......en ]

[(−c − d − i − m −1) : e1 , e2 ..........en ] [(ah ) :1].
[(∝

1

)].[(∝

2

) :1]..........[(∝

n

) :1]
[( pk ) :1].[(αv
) :1] [(α

2

) :1]..........[(α

n

) :1]

e1 ( r + s + p −q +1)

− e2 ( r + s + p −q +1)+ r + s

− em ( r + s + p −q +1)+ r + s 

µ1 (−1) , µ2 ( 1) x2e2

un ( 1)x e

............ n n 

µ e µ e2

µ en 

where

[( A ) [(C )] (µ xe )m

K = r m p m

[(Bs )]m (Dq )m m!

Proof : After little simplification (1.1) gives

∞ ∞ ∞

[(a )] n (µ )n1 [( A )]

∑

m=0

S k {( x ), y}t m =

∑ ∑

m=0 n1 =0,n2 =0 ........ nn =0

h

[(bk )]

1 1

n1 ! y

r m+ n2 +.....nn

e n [( B )]

1 1 s m+n2 +....... nn

n1

[(C
)] [(∝
)] (µ xe )n (µ xe2 )n2 .......(µ

xen )nn t m+e1n1 +e2 n2 +....+en nn

× p n un en 2 2 m m

[(Dq )]m [(αv )] m!

n2 !

nn !

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we have

m

( xe )m = n!(m + c)!∑

(−1)i (2i + c + d + 1)

p( c.d ) (1 + 2 xe )

i =0 (n − i)!(c + d + i + 1)m+1

∞ ∞ ∞

( Ar )m+ m +.....n

(C p )

∑

m=0

Qk {( x ), y}t m =

∑ ∑

m=0 n1 =0,n2 =0 ...... nn =0

 2
[(Bs )]

n m

 Dq 
(∝
)
(∝
)
.... (∝
)
[(a )]

m+ n2 +.........nn  m

µ m (µ )n1 (µ xe2 )n2 .....(µ

xen )nn

× u1 n1

u2 n2

un nn

h n1

1 2 2 n n

α 
(α
)
...... α 
[(b )]

m ! ye1 n1 ....n ! n !

v1 n1

v2 n2

vn en

k n1 1 2 n

(−1)i (2i + c + d + 1)(n + 1)! p( c ,d ) (1 + 2 xe )t m+e1n1 +e2 n2 +.....en nn

× i
(m − i) (c + d + i + 1)m+1

∞ m m / e1

m−e1n1

e2

[ m−e1n1 −e2 n2 ..........−en−1nn−1 ]

= ∑ ∑ ∑ ∑
......

∑ ........2.2

m=0

i =0

n1 =0

n2 =0

nn =0

[ Ar )]m−e n −( e −1) ..........( e )

C p  m − e1n1 − e2 n2 ......em nn [(an )]

(αu ) (αu )

1 1 2 n2

n−1 nn  

 

n1 

1 n 

2 n

[(Bs )]n−e n −( e −1) n .........( e −1)

n

(Dq )

m−e n −e n ..........e n

(b )

(α

) (α )

1 1 2 2 n n

1 1 2 2

n n [

k ]n

1

v1 n1 

v2  n2

(∝
)
(µ )n1 (µ xe2 )n2 .........(µ

xcn )nn

(−1)i (2i + c + d + 1)µ n −e m −e m ........−e n

un nn

1 2 2 n n

1 1 2 2 n n

(αv )

n

m ! yc1 n1

n2 !

nn !

(m − i − e1n1 − e2 n2 ........... − en nn )!

(n + c − en − e n .......... − (e n )!P( c ,d ) (1 + 2 xe )t m
× 1 1 2 2 n n i

(c + d + i + 1)m+1 − e1n1 − e2 n2 ........ − en nn

........(2.3)

Equating the coefficient of tm from both sides we get if

e2 > 1 .... en > 1

Qm {( n), y} = K

m

∑

i =0

(m + c)!(−1)i (2i + c + d + 1) p( c.d ) (1 + 2xe ) (m + i)! (c + d + i + 1)m+1
∞ [1 − (Bs ) −

× ∑

m]e n + ( e −1) +.....( e −1)

2 n

n1 ,n2 ....nn =0 [1 − ( Ar ) − m]e n +( e −1) n +.....+( e −1)

[1 − (Dq ) − m]e n + e n +.........+ e [(ah )]n [(∝u )]n [(∝u )]n ....[(∝u )]n

× n

[1 − (c p ) − m]e n +..........e n +.......+ e n [(bk )]n [(αv )]n [(αv )]n ......[(αv )n

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(µ )n1 (µ xe2 )m2 .......(µ

×

em nn

n n e1m1 +e2 n2 +.....+en nn .

n ! ye1 n1 n2 !+...............+en nn [(b )] [(α

)] [(α
)] ......[(α )]

1 k n1

v1 n1

v2 n2

vn nn

(µ )n (µ , xe2 )n2 .........(µ

×

en nn

n n e1n1 +e2 n2 +.....en n

n ! ye1n1 n !

n !(−n − c)

+ .......e n

1 2 n e1n1 +e2 n2 n n

( c d i n

1) ( 1)e1 ( r + s + p + q +1) n1

− − − − −

×

e1n1 +e2 n2 +e3n3 +......en nn −

µ e1n1 + e

2

........ + en nn

(−1)e2 {( r + s + p + q +1)+ r + s}m2 .......(−1)en {( r + s + p + q +1)+ r + s}nn

×
1

where c is non-negativ integer

The single terminating factor summation in (2.4) runs up to ∞.

Hence, the theorem (2.1)

particular cases (2.1) (i) Hermite Polynomials

(−n + c)

e1n1 +e2 n2 +........+en nn

make all

It we set r = 0 = s = p = q = µ1 = v1 = h = k; e1 = 2 = µ

x1 = x, x2 = 1 = e = y;

µ1 = −4 ,

µ1 = −4 , we have

H n ( x) =

m

∑

i =0

2m (−1)i (m + c)! (2i + c + d + 1) (m − i)!(c + d + i + 1)m+1

p( c ,d ) (1 + 2xe )

∆ (2; − m + i), ∆ (2; − c − d − i − m −1);

F ∆ (2; − m + c)


−1
 

(ii) Legnedre Polynomials

on making the substitution

r = 0 = s = p = q = h = µ1 = v1 ;

k = 1 = e = µ = µ1 = y = b1, e1 = 2 and

x

x2 − 1

for x we set

Pm ( x) =

m

∑

i =0

2m (−1)i (m + c)!(2i + c + d + 1)P( c ,d ) (1 + 2xe ) (m − i)! (c + d + i + 1)m+1

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∆ (2; −m + i), ∆(2; − c − d − i − m −1);
× F ∆ (2; −m + c),1;

−1
 

Similarly, specializing the parameters in (2.1) all the polynomials defined by authours [8] can also be decluced in term of Jocobi also.

Theorem (2) for e2 > 1 ...................en > 1, we obtain

1

k n! 

( f ' )

   
2  h m ∞

(−1) j ( 2 j +1)

 

Q {( xn ), y} =

m

( ' )

∑ ( −

)!(

c ( xe )

+ + 1)! j

m

 d g m

m=0

m j n j

[(−m + j); e1 , e2 .........en ][−m − j −1)e1 , e2 ......en ]

× q + s + h +1;u1 ,u2 ........um   

Fp + r + k ; v , v .............v

[(−m), e , e .........e ]
 1 − m  ; e , e .......e
 1 2

n  2

 1 2 n 
   
[(1 − (Bs ) − m); e1 e2 −1...........(en −1)], (1 − (Dq ) − m)e1 , e2 ......en 

[(1 − ( Ar ) − m]; e1 , e2 −1.........(en −1), (1 − (C p ) − m); e1e2 .........en 

(1 − (d ' ) − m); e , e ......e

 [(a );1][∝

,1] (∝

);1) ....... (∝

);1

2 1 2

n h u u2 un

(1 − ( f ' ) − m); e , e .......e
 [(b ) :1] (α
);1 (α
) :1 ..... (α
) :1

h 1 2

n k v1 v2 vn

µ1 (−1)

e1 ( r + s + p + q + f '+ h '+1)

µ e1 ye1

µ xe2 (−1)

. 2 2

e2 ( r + s + p + q + f '+ h '+1) + r + s

µ e2

........

µ xen (−1)en ( r + s + p + q + f '+ h '+1) + r + s

.................3.1

Proof

(n!)2  =1  ( f ' )

µ en

  

h m m j

( xe )n =  2 m

∑ (−1) (2 j + 1)

e ( xe )

(d g )m
j =0 (m − j)!(n + j + 1)!
putting the value of ( xe )m in equation (2.2) we get

∞ ∞ ∞

Qk {( x ), y}t m

m [( Ar )]m+ n +.......n [(C p )]m

∑ m n

= ∑ ∑ ∑

m 0 m 0 n

0,n

0 ........ n

0 j 0 [(

s )]

[( q )]m

= = = = = = B D

1 2 n

m+ n +.........n

2 n

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(∝u1

×
)n (∝u2
)n
....... (∝un
)n
[(ah )]

1

µ m (µ )n1 (µ xe2 )n2 ......( µ

en nn

m n

(α
) (α )
....... (α )
[(b )]

n ! ye1n1

n2 !

mn !

 1 n 

2 n

 n e

n1 1

v v v k

(−1) j (2 j + 1)  =1 

n!( f ' )

t m+ e1n1 + e2 n2 +........+ en nn

 2  

h m

×  m
(n − j)! (n + j + 1)!
(d ' )

g m

m

∞ m e1

m−e1n1

e2

m − en − e n ......e n

= ∑ ∑ ∑ ∑
..................[

1 1 2 2

en

n −1

n −1 ]

m=0

j =0

n1 =0

n2 =0

∑

nn =0

[( Ar )]
− (e2 −1)n ........(en −1)n
(C p )

× m−e1n1 2

n m−e1n1 −e2 n2 ............en nn

[(Bs )]

........(en −1)n [(Dq)]m−e n −e n ........e n

m−e1n1 −( e2 −1)n2

n 1 1 2 2 n n

[(ah )]n [(∝u )]n [(∝u )]n ........[(∝u )]n
(µ )n1 (µ xe2 )n2 ......(µ

xen )nn

× 1 1 1 2 2

n n × 1 2 2 n n

e1n1

[(bk )]n [(αv )]n [(αv )]n .........[(αv )]n

n1 ! y n2 !

nn !

1 1 1 2 2

(−1) j (2 j + 1)  =1 

×  2 m−e n −e n .........e n

n n

µ n −e1n1 −e2 n2 ........en nn

(m− j −e n −e n

− .......... − en

nn )!

(n − e1n1 − e2 n2 .........en nn )!

(n + j + 1 − e1n1 − e2 n2 ........en nn )!

( f ' )

× h m−e1n1 −e2 n2 −.......−en nn c j ( xe )t m

(d ' )

g m−e1n1 −e2 n2 −........−en nn

......(3.2)

Equating the coefficient of tm from both side we get

k  =1  ( f ' ) m!
  

h m ∞ j e

Qk {( x ), y} =  2 m ∑ (−1) (2 j + 1)c j ( x )
m n (d ' )
(n − j)!(n + j + 1)!
 g m

j =0

∞ [1 − (Bs ) −

× ∑

m]e n + ( e −1) +.........+ ( e −1) n

2

n1 ,n2 ,n3 .....nn =0 [1 − ( Ar ) − m]e n t + (e2 − 1)n + ......... + (en −1 )nn

[1 − (D ) − m] [1 − (d ' ) − m]

× q e1n1 + p2 n2 +.........+en nn .

g m+e1n1 +e2 n2 +....en nn

[1 − (C
) − m] .[1 − ( f ' ) − m]

p e1n1 + p2 n2 +.....+en nn

h m+e1n1 +e2 n2 +.....+en nn

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[(−m + j)e n +e n +.........+e n (−m − j −1)e n +e n +.........+e n .

× 1 1 2 2 n n 1 1 2 2 n n

 1 − m 
(−m)

 2 

e1n1 +e2 n2 +.........+en nn .

 e n +e n +.........+e n

[(a )] [(∝
)] [(∝
)] .......[(∝

)] (µ )n1 (−1)e1 ( r + s + p + q + f '+ h '+1) n1

× h m1

u1 n1

u2 n2

un nn 1

[(bk )]n [(αv
)] [(α )]

1 2 2

+ .......[(α )]

n n

n ! ye1n1

(µ xe2 )n2 (−1)e2 ( r + s + p + q + f '+ h '+1) n2

×

n2 !

......

( µ xen )nn (−1)en ( r + s + p + q + f '+ h '+1) nn

nn !

where j is non negative integer .......(3.3)

The single terminating factor (–m + j) e1 n1 + e2 n2 +.....+en nn

make all summation in (3.3) runs upto ∞

particular cases of (3.1)

1. Sylvester Polynomials

On setting r = 0 s = p = q = µ1 = v1 = k

h1 = 1 = y = e = e1 = µ1, a1 = x, we optain

 =1 

 

m

[( f ' )] (−1) j (2 j + 1)c ( xe )

'

φ ( x )

j =0

 2 n

h m j

[(d g )]m (m − j)!(m + j + 1)!
−m + j − m − j −1, 1 − (d ' ) − m, x;
× F  
−m, 1 − m,1 − ( f ' ) − m ) 1 

 2 h 

2. Bedient Polynomials :

On taking s = 0 = un = vn ; r = 1 = p = q = e = µn , µn = 4

en = 2 ; µ = 2, D1 = ∝ + α1 A1 = α1 ; c1 = α , we get

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 =1 

m  

[( f ' )] (−1) j (2 j + 1)(α ) (β ) 2m c ( xe )
Gm (α , β , x) =

∑

j =0

 2 n

[d ' )] [(m − j)!(m + j + 1)!(α + β )
∆(2; −m + j), ∆(2; −m − j −1), D(2;1 − (d ' ) − m), ∆(2;1 − α + −β − m); 
× F  2 1
∆(2 − m), ∆(2;1 / 2 − m), ∆(2;1 − ( f ' ) − m), ∆(2;1+ ∝ −m), ∆(2;1 − β − m); 
 h 

Similaly, specializing the parameters of (3.1) the polynomials of Sah, Khanna, Krawtchonk gonld, Humbrt, Lomenel pradhan, etc can also be deduced in terms of sister celin’s polynomials.

References :

1. Askey. R.A. (1975); Theory and application of special function, Academic Press.

2. Burchnall, J.L. and Chaundy. T.W. (1941); Expansion of Appell’s double hyper geomaric function (ii) 1941 Quart. J math, oxford, sec 12 p. 112-128.

3. Rail Ville, E.D. (1960) special functions, Macmillan, Co. New

Yoak.

4. Shah, Manilal; Expansion formule for generalised hypergeometric polynomial in series of Jocobi Polynomials.

5. Rainville E.P. Special functics the in Macmillan Co., New York

1967.

6. Srivastava, B.M. and F.Singh; On some new generating relations, Ranchi, Univ. Math J.Vol. 5, 1974

7. Srivastava H.M. and Manocha, HL (1984); A tneatise on generating functions, Halsted press, John willy and Sons, New Yoark.

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