Expressing Sums of Finite Products of Chebyshev Polynomials ... - MDPI

mathematics Article

Expressing Sums of Finite Products of Chebyshev Polynomials of the Second Kind and of Fibonacci Polynomials by Several Orthogonal Polynomials Taekyun Kim 1,2 , Dae San Kim 3 , Jongkyum Kwon 4, * and Dmitry V. Dolgy 5 1 2 3 4 5

*

Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin 300160, China; [email protected] Department of Mathematics, Kwangwoon University, Seoul 139-701, Korea Department of Mathematics, Sogang University, Seoul 121-742, Korea; [email protected] Department of Mathematics Education and ERI, Gyeongsang National University, Jinju, Gyeongsangnamdo 52828, Korea Hanrimwon, Kwangwoon University, Seoul 139-701, Korea; [email protected] Correspondence: [email protected]

Received: 7 September 2018; Accepted: 16 October 2018; Published: 17 October 2018







Abstract: This paper is concerned with representing sums of the finite products of Chebyshev polynomials of the second kind and of Fibonacci polynomials in terms of several classical orthogonal polynomials. Indeed, by explicit computations, each of them is expressed as linear combinations of Hermite, generalized Laguerre, Legendre, Gegenbauer and Jacobi polynomials, which involve the hypergeometric functions 1 F1 and 2 F1 . Keywords: chebyshev polynomials of second kind; Fibonacci polynomials; sums of finite products; orthogonal polynomials

1. Introduction and Preliminaries In this section, we will fix some notations and recall some basic facts about relevant orthogonal polynomials that will be used throughout this paper. For any nonnegative integer n, the falling factorial polynomials ( x )n and the rising factorial polynomials < x >n are respectively defined by (see [1])

( x )n = x ( x − 1) · · · ( x − n + 1), (n ≥ 1), ( x )0 = 1,

(1)

< x >n = x ( x + 1) · · · ( x + n − 1), (n ≥ 1), < x >0 = 1.

(2)

The two factorial polynomials are related by:

(−1)n ( x )n =< − x >n , (−1)n < x >n = (− x )n . 22n−2s (−1)s < 12 >n (2n − 2s)! = , ( n ≥ s ≥ 0). (n − s)! < 12 − n >s √ 1 (2n)! π Γ(n + ) = , ( n ≥ 0). 2 22n n! Γ ( x + 1) Γ( x + n) = ( x )n , =< x >n , (n ≥ 0). Γ( x + 1 − n) Γ( x )

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(3) (4) (5) (6)

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B( x, y) =

Z 1 0

t x−1 (1 − t)y−1 dt =

Γ( x )Γ(y) , ( Re x, Re y > 0), Γ( x + y)

(7)

where Γ( x ) and B( x, y) are the gamma and beta functions, respectively. The hypergeometric function is defined by: p Fq ( a1 , · · · , a p ; b1 , · · · , bq ; x ) ∞ < a > ··· < a > n p n 1

=

∑

n =0

xn , ( p ≤ q + 1, | x | < 1). < b1 >n · · · < bq >n n!

(8)

We are now going to recall some basic facts about Chebyshev polynomials of the second kind Un ( x ), Fibonacci polynomials Fn ( x ), Hermite polynomials Hn ( x ), generalized (extended) (λ)

Laguerre polynomials Lαn ( x ), Legendre polynomials Pn ( x ), Gegenbauer polynomials Cn ( x ) and (α,β) Pn ( x ).

Jacobi polynomials All the necessary results on those special polynomials, except Fibonacci polynomials, can be found in [2–7]. Furthermore, the interested reader may refer to [8–11] for full accounts of the fascinating area of orthogonal polynomials. In terms of generating functions, the above special polynomials are given by: F (t, x ) =

∞ 1 = ∑ Un (x)tn , 1 − 2xt + t2 n =0

G (t, x ) =

∞ 1 = Fn+1 ( x )tn , ∑ 1 − xt − t2 n =0

2

e2xt−t =

∞

tn

∑ Hn (x) n! ,

(9) (10) (11)

n =0

(1 − t)−α−1 exp(− 1xt −t ) = 1

(1 − 2xt + t2 )− 2 =

∞

∑ Lαn (x)tn , (α > −1),

(12)

n =0

∞

∑ Pn (x)tn ,

(13)

n =0 ∞

1 1 (λ) = ∑ Cn ( x )tn , (λ > − , λ 6= 0, |t| < 1, | x | ≤ 1), 2 λ 2 (1 − 2xt + t ) n =0 ∞ 2α + β (α,β) = ∑ Pn (x)tn , R (1 − t + R ) α (1 + t + R ) β n =0 p 2 ( R = 1 − 2xt + t , α, β > −1).

(14)

(15)

Explicit expressions of special polynomials can be given as in the following. Un ( x ) = (n + 1)2 F1 (−n, n + 2; 32 ; 1−2 x ) [n 2]

=

Fn+1 ( x ) =

∑ (−1)

l



 n−l (2x )n−2l , l

l =0   [n 2] n − l

∑

l =0

l

x n−2l ,

(16)

(17)

[n 2]

(−1)l (2x )n−2l , l!(n − 2l )! l =0

Hn ( x ) = n! ∑

(18)

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< α + 1 >n 1 F1 (− n, α + 1; x ) n! n (−1)l (n+α) n−l =∑ xl , l! l =0

Lαn ( x ) =

(19)

Pn ( x ) = 2 F1 (−n, n + 1; 1; 1−2 x )    [n] 2n − 2l n−2l 1 2 l n x , = n ∑ (−1) l n 2 l =0   n + 2λ − 1 (λ) 1 1− x Cn ( x ) = 2 F1 (− n, n + 2λ; λ + 2 ; 2 ) n [n 2]

=

Γ(n − k + λ)

∑ (−1)k Γ(λ)k!(n − 2k)! (2x)n−2k ,

(20)

(21)

k =0

(α,β)

Pn

< α + 1 >n 1− x 2 F1 (− n, 1 + α + β + n; α + 1; 2 ) n!   n  n+α n + β x −1 k x +1 n − k = ∑ ( 2 ) ( 2 ) . n−k k k =0

(x) =

(22)

Next, we recall Rodrigues-type formulas for Hermite and generalized Laguerre polynomials and Rodrigues’ formulas for Legendre, Gegenbauer and Jacobi polynomials. dn − x2 e , dx n 1 dn Lαn ( x ) = x −α e x n (e− x x n+α ), n! dx 1 dn 2 Pn ( x ) = n ( x − 1) n , 2 n! dx n 1 (λ) 1 (−2)n < λ >n dn (1 − x2 )λ− 2 Cn ( x ) = (1 − x 2 ) n + λ − 2 , n n! < n + 2λ >n dx (−1)n dn ( α,β ) (1 − x ) n + α (1 + x ) n + β . (1 − x )α (1 + x ) β Pn ( x ) = n 2 n! dx n Hn ( x ) = (−1)n e x

2

(23) (24) (25) (26) (27)

The following orthogonalities with respect to various weight functions are enjoyed by Hermite, generalized Laguerre, Legendre, Gegenbauer and Jacobi polynomials. Here, δn,m is Kronecker’s delta, so that δn,m = 1, for n = m, and δn,m = 0, for n 6= m. Z ∞ −∞

Z ∞ 0

Z 1 −1 Z 1 −1 Z 1 −1

=

√ 2 e− x Hn ( x ) Hm ( x ) dx = 2n n! πδn,m , x α e− x Lαn ( x ) Lαm ( x ) dx =

Pn ( x ) Pm ( x ) dx = 1

1 Γ(α + n + 1)δn,m , n!

2 δn,m , 2n + 1

(λ)

(λ)

(α,β)

(α,β)

( x ) Pm

(29) (30)

(1 − x2 )λ− 2 Cn ( x )Cm ( x ) dx = (1 − x )α (1 + x ) β Pn

(28)

π21−2λ Γ(n + 2λ) δn,m , n!(n + λ)Γ(λ)2

(31)

( x ) dx

2α + β +1 Γ ( n + α + 1 ) Γ ( n + β + 1 ) δn,m . (2n + α + β + 1)Γ(n + α + β + 1)Γ(n + 1)

(32)

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For convenience, we put:

∑

γn,r ( x ) =

Ui1 ( x )Ui2 ( x ) · · · Uir+1 ( x ), (n, r ≥ 0),

i1 +i2 +···+ir+1 =n

∑

En,r ( x ) =

Fi1 +1 ( x ) Fi2 +1 ( x ) · · · Fir +1 ( x ), (n ≥ 0, r ≥ 1).

i1 +i2 +···+ir =n

(33)

(34)

We note here that both γn,r ( x ) and En,r ( x ) have degree n. The classical linearization problem in general consists of determining the coefficients cnm (k ) in the expansion of the product of two polynomials qn ( x ) and rm ( x ) in terms of an arbitrary polynomial sequence { pk ( x )}k≥0 : n+m

q n ( x )r m ( x ) =

∑

cnm (k) pk ( x ).

k =0

Here, we will study the sums of finite products of Chebyshev polynomials of the second kind in (33) and those of Fibonacci polynomials in (34). Then, we would like to express each of γn,r ( x ) (λ)

(α,β)

and En,r ( x ) as linear combinations of Hn ( x ), Lαn ( x ), Pn ( x ), Cn ( x ) and Pn ( x ). These will be done by performing explicit computations and exploiting the formulas in Proposition 1. They can be derived from their orthogonalities, Rodrigues’ and Rodrigues-like formulas and integration by parts. This may be viewed as a generalization of the above-mentioned linearization problem. Our main results are as follows: Theorem 1. Let n, r be integers with n ≥ 0, r ≥ 1. Then, we have the following.

∑

Ui1 ( x )Ui2 ( x ) · · · Uir+1 ( x )

i1 +i2 +···+ir+1 =n

(n + r )! = r! =

∑

j =0

1 F (− j; −n − r; −1) Hn−2j ( x ) j!(n − 2j)! 1 1

2n Γ ( α + n + 1 ) r! k [ n− 2 ]

×

∑

l =0

[ n2 ]

n

(−1)k Γ ( α + k + 1) k =0

1 (2n − 4j + 1) 2 F1 (− j; j − n − ; − n − r; 1) Pn−2j ( x ) 2 2 n− j j!

∑ (n − j + 1 )

j =0

(35)

∑

(− 41 )l (n + r − l )! Lα ( x ) l!(n − k − 2l )!(α + n)2l k

(n + r )! = r! =

[ n2 ]

(36)

(37)

Γ(λ)(n + r )! Γ(n + λ + 1)r! [ n2 ]

(n + λ − 2j)(n + λ) j (λ) 2 F1 (− j; j − n − λ; − n − r; 1)Cn−2j ( x ) j! j =0

×∑ =

(−2)n r! k [ n− 2 ]

× ×

∑

(38)

n

(−2)k Γ(k + α + β + 1) Γ(2k + α + β + 1) k =0

∑

(− 41 )l (n + r − l )! 2 F1 ( k + 2l − n, k + β + 1; 2k + α + β + 2; 2) l!(n − k − 2l )!

l =0 (α,β) Pk ( x ).

(39)

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Theorem 2. Let n, r be integers with n ≥ 0, r ≥ 1. Then, we have the following.

∑

Fi1 +1 ( x ) Fi2 +1 ( x ) · · · Fir +1 ( x )

i1 +i2 +···+ir =n

( n + r − 1) ! = n 2 (r − 1) ! =

×

∑

l =0

1

∑ (n − 2j)!j! 1 F1 (− j; 1 − n − r; 4) Hn−2j (x) n

(−1)k Γ ( α + k + 1) k =0

∑

( n + r − l − 1) ! Lα ( x ) l!(n − k − 2l )!(α + n)2l k

( n + r − 1) ! = (r − 1)!4n

(40)

j =0

Γ ( α + n + 1) (r − 1) ! k [ n− 2 ]

=

[ n2 ]

[ n2 ]

(2n − 4j + 1)4 j 1 2 F1 (− j; j − n − ; 1 − n − r; −4) Pn−2j ( x ) 2 2 n− j j!

∑ (n − j + 1 )

j =0

(41)

(42)

Γ(λ)(n + r − 1)! − 1)!Γ(n + λ + 1)

2n ( r

[ n2 ]

(n + λ) j (n + λ − 2j) (λ) 2 F1 (− j; j − n − λ; 1 − n − r; −4)Cn−2j ( x ) j! j =0

×∑ =

(43)

(−1)n n Γ(k + α + β + 1)(−2)k (r − 1)! k∑ Γ(2k + α + β + 1) =0 k [ n− 2 ]

× ×

∑

(n + r − 1 − l )! 2 F ( k + 2l − n, k + β + 1; 2k + α + β + 2; 2) l!(n − k − 2l )! 1

l =0 (α,β) Pk ( x ).

(44)

The sums of finite products of Bernoulli, Euler and Genocchi polynomials have been expressed as linear combinations of Bernoulli polynomials in [12–14]. These were done by deriving Fourier series expansions for the functions closely related to those sums of finite products. Further, the same were done for the sums of finite products γn,r ( x ) and En,r ( x ) in (33) and (34) in [15]. Along the same line as the present paper, sums of finite products of Chebyshev polynomials of the second, third and fourth kinds and of Fibonacci, Legendre and Laguerre polynomials were expressed in terms of all kinds of Chebyshev polynomials in [16–18]. Finally, we let the reader refer to [19,20] for some applications of Chebyshev polynomials and to [21–25] for some similar iteration methods. 2. Proof of Theorem 1 Here, we are going to prove Theorem 1. First, we will state two results that will be needed in showing Theorems 1 and 2. The results (a), (b), (c), (d) and (e) in Proposition 1 follow respectively from (3.7) of [3], (2.3) of [7] (see also (2.4) of [6]), (2.3) of [4], (2.3) of [2] and (2.7) of [5]. They can be derived from their orthogonalities in (26)–(30), Rodrigues-like and Rodrigues’ formulas in (21)–(25) and integration by parts. Proposition 1. Let q( x ) ∈ R[ x ] be a polynomial of degree n. Then, we have the following.

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n

( a) q( x ) =

∑ Ck,1 Hk (x),

where

k =0

(−1)k √ 2k k! π

Ck,1 =

Z ∞ −∞

q( x )

dk − x2 e dx, dx k

n

(b) q( x ) =

∑ Ck,2 Lαk (x),

where

k =0

1 Γ ( α + k + 1)

Ck,2 =

Z ∞ 0

q( x )

dk − x k+α (e x )dx, dx k

n

(c) q( x ) =

∑ Ck,3 Pk (x),

where

k =0

2k + 1 2k+1 k!

Ck,3 =

n

(d) q( x ) =

Z 1 −1

q( x )

∑ Ck,4 Ck

(λ)

dk 2 ( x − 1)k dx, dx k

( x ), where

k =0

(k + λ)Γ(λ) √ (−2)k πΓ(k + λ + 21 )

Ck,4 =

n

(e) q( x ) =

∑ Ck,5 Pk

(α,β)

Z 1 −1

q( x )

1 dk (1 − x2 )k+λ− 2 dx, k dx

( x ), where

k =0

Ck,5 =

×

(−1)k (2k + α + β + 1)Γ(k + α + β + 1) 2α + β + k +1 Γ ( α + k + 1 ) Γ ( β + k + 1 )

Z 1 −1

q( x )

dk (1 − x )k+α (1 + x )k+ β dx. dx k

Proposition 2. Let m, k be nonnegative integers. Then, we have the following.

( a)

Z ∞

=



(b)

2

x m e− x dx  0, if m ≡ 1 (mod 2),

−∞

√ m! π m m, ( 2 )!2

if m ≡ 0 (mod 2),

Z 1

x m (1 − x2 )k dx  0,

−1

=

=2

k!

∑

s

s =0

(c)

if m ≡ 1 (mod 2),

m 2k +2  2 m k!m!(k+ 2 +1)! , ( 2 )!(2k+m+2)! m   m s 2k +1

Z 1

if m ≡ 0 (mod 2),

2 (−1)m−s 1

(k + s)! , (2k + s + 1)!

x m (1 − x2 )k+λ− 2 dx −1  0, if m ≡ 1 (mod 2), = Γ(k+λ+ 1 )Γ( m + 1 ) 2 2 2  , if m ≡ 0 (mod 2), Γ ( k + λ + m +1) 2

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Z 1

x m (1 − x )k+α (1 + x )k+ β dx m   m = 22k+α+ β+1 ∑ (−1)m−s 2s s s =0

(d)

−1

×

Γ ( k + α + 1) Γ ( k + β + s + 1) . Γ(2k + α + β + s + 2)

Proof. (a) This is trivial. (b) The first equality follows from (c) with λ = Z 1

(c)

1 2

and the second from (d) with α = β = 0.

1

−1

x m (1 − x2 )k+λ− 2 dx

=(1 + (−1)m )

Z 1 0

1

x m (1 − x2 )k+λ− 2 dx

1 1 m +1 1 = (1 + (−1)m ) (1 − y)k+λ+ 2 −1 y 2 −1 dy 2 0 1 1 m+1 m = (1 + (−1) ) B(k + λ + , ). 2 2 2

Z

The result now follows from (7).

(d)

Z 1 −1

x m (1 − x )k+α (1 + x )k+ β dx

=22k+α+ β+1

Z 1

(2y − 1)m (1 − y)k+α yk+ β dy   m 2k +α+ β+1 =2 ∑ s 2s (−1)m−s s =0 0 m

Z 1

(1 − y)k+α+1−1 yk+ β+s+1−1 dy m   m s =22k+α+ β+1 ∑ 2 (−1)m−s B(k + α + 1, k + β + s + 1). s s =0 ×

0

The result again follows from (7). Even though the following lemma was shown in [26], we will show it for the sake of completeness. Lemma 1. Let n, r be nonnegative integers. Then, we have the following identity.

∑

Ui1 ( x )Ui2 ( x ) · · · Uir+1 ( x ) =

i1 +i2 +···+ir+1 =n

1 2r r!

(r )

Un+r ( x ),

(45)

where the sum runs over all nonnegative integers i1 , i2 , · · · ir+1 , with i1 + i2 + · · · + ir+1 = n. ∂ r Proof. Noting that the degree of Un ( x ) has degree n and taking the partial derivative ( ∂x ) on both sides of (9), we have:

 −(r+1)  ∞  r +1 r!(2t)r 1 − 2xt + t2 = r!(2t)r ∑ Ui ( x ) i =0

∞

=

∑ Un

n =r

from which our result follows.

(r )

( x )tn ,

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It is immediate to see from (16) that the r-th derivative of Un ( x ) is equal to: (r ) Un ( x )

r [ n− 2 ]

∑ (−1)

=

l =0

l



 n−l (n − 2l )r 2n−2l x n−2l −r . l

(46)

Thus, in particular, we have: (r + k ) Un+r ( x )

k [ n− 2 ]

∑ (−1)

=

l



l =0

 n+r−l (n + r − 2l )r+k 2n+r−2l x n−k−2l . l

(47)

Here, we will show only (35), (37) and (38) in Theorem 1, leaving the proofs for (36) and (39) as an exercise, as they can be proved analogously to those for (41) and (44) in the next section. With γn,r ( x ) as in (33), we let: n

∑ Ck,1 Hk (x).

γn,r ( x ) =

(48)

k =0

Then, from (a) of Proposition 1, (45), (47) and integration by parts k times, we have: ∞ (−1)k dk − x2 √ γ ( x ) e dx, n,r dx k 2k k! π −∞ Z ∞ 2 (−1)k dk (r ) √ = k +r Un+r ( x ) k e− x dx, dx 2 k!r! π −∞ Z ∞ 2 1 (r + k ) √ U ( x )e− x dx, = k +r 2 k!r! π −∞ n+r k   [ n− 2 ] 2n − k 1 l n+r−l √ ∑ (− ) = (n + r − 2l )r+k 4 l k!r! π l =0

Z

Ck,1 =

×

Z ∞

−∞

(49)

2

x n−k−2l e− x dx.

From (49) and invoking (a) of Proposition 2, we get:

Ck,1

k   [ n− 2 ] 2n − k 1 n+r−l √ ∑ (− )l = (n + r − 2l )r+k 4 l k!r! π l =0  0, if k 6≡ n (mod 2), √ × (n−k−2l )! π  n−k−2l n−k , if k ≡ n (mod 2).

(

2

2

(50)

−l )!

Now, from (48) and (50), and after some simplification, we obtain: 1 γn,r ( x ) = r! 1 = r!

=

∑

0≤ k ≤ n k ≡n (mod 2)

[ n2 ]

1 k!

k [ n− 2 ]

∑

l =0

(−1)l (n + r − l )! Hk ( x ) k l!( n− 2 − l )! j

1 (−1)l (n + r − l )! ∑ (n − 2j)! Hn−2j (x) ∑ l!( j − l )! j =0 l =0

(n + r )! r!

(n + r )! = r!

[ n2 ]

∑

j =0

[ n2 ]

∑

j =0

j

(−1)l

1 < − j >l H (x) ∑ j!(n − 2j)! n−2j l! < − n − r >l l =0 1 F (− j, −n − r; −1) Hn−2j ( x ). j!(n − 2j)! 1 1

(51)

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This shows (35) of Theorem 1. Next, we let: n

γn,r ( x ) =

∑ Ck,3 Pk (x).

(52)

k =0

Then, from (c) of Proposition 1, (45), (47) and integration by parts k times, we get: 1 dk 2k + 1 (r ) Un+r ( x ) k ( x2 − 1)k dx k + r + 1 2 k!r! −1 dx Z 1 k (−1) (2k + 1) (r + k ) = Un+r ( x )( x2 − 1)k dx 2k+r+1 k!r! −1   [ n−k ] (2k + 1)2n−k−1 2 1 l n+r−l = (n + r − 2l )r+k ∑ (− 4 ) k!r! l l =0

Z

Ck,3 =

×

Z 1 −1

(53)

x n−k−2l (1 − x2 )k dx.

From (52) and making use of the first equality of (b) in Proposition 2, we have:

Ck,3

  [ n−k ] 1 l n+r−l (2k + 1)2n−k−1 2 ) (n + r − 2l )r+k (− = ∑ 4 k!r! l l =0  0, if k 6≡ n (mod 2), × 22k+2 k!(n−k−2l )!( n+2 k −l +1)!  , if k ≡ n (mod 2). n−k (

2

(54)

−l )!(n+k−2l +2)!

From (52), (54), and using (4), we finally obtain: 22n+1 γn,r ( x ) = r!

[ n2 ]

2n − 4j + 1 Pn−2j ( x ) 22j j =0

∑

j

(− 14 )l (n + r − l )!(n − j + 1 − l )! l!( j − l )!(2n − 2j + 2 − 2l )! l =0

×∑

(n + r )! = 2r!

[ n2 ]

∑

(2n − 4j + 1) Pn−2j ( x )

j =0

(n − j + 21 )n− j+1 j!

(55)

j

< − j >l < j − n − 21 >l < −n − r >l l! l =0

×∑

(n + r )! = r!

[ n2 ]

(2n − 4j + 1)2 F1 (− j, j − n − 12 ; −n − r; 1) Pn−2j ( x ). (n − j + 12 )n− j j! j =0

∑

This shows (37) of Theorem 1. Remark 1. In the step of (54), if we use the second equality of (b) in Proposition 2 instead of the first, we would have the expression: γn,r ( x ) =

(−2)n r! [ n−2 k ]

×

∑

j =0

n

(−2)k k! (2k)! k =0

∑

(− 14 )l (n + r − l )! F (2l + k − n, k + 1; 2k + 2; 2) Pk ( x ). l!(n − k − 2l )! 2 1

(56)

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(0,0)

We note here that (56) is (39), with α = β = 0. This is what we expect, as Pn ( x ) = Pn Finally, we let: n

∑ Ck,4 Ck

γn,r ( x ) =

(λ)

( x ).

( x ).

(57)

k =0

Then, from (d) of Proposition 1, (45), (47) and integration by parts k times, we obtain: Ck,4 =

(k + λ)Γ(λ) × √ k + r 2 πΓ(k + λ + 21 )r!

2n − k ( k + λ ) Γ ( λ ) = √ πΓ(k + λ + 12 )r!

×

Z 1 −1

x

n−k −2l

k [ n− 2 ]

∑

l =0

Z 1

1

(r + k )

−1

Un+r ( x )(1 − x2 )k+λ− 2 dx

  1 l n+r−l (n + r − 2l )r+k (− ) l 4

2 k +λ− 21

(1 − x )

(58)

dx.

From (58), and exploiting (c) in Proposition 2 and (5), we have:   [ n−k ] 1 l n+r−l 2n − k ( k + λ ) Γ ( λ ) 2 ) (− (n + r − 2l )r+k Ck,4 = √ ∑ 4 l πΓ(k + λ + 12 )r! l =0  0, if k 6≡ n (mod 2), √ × Γ(k +λ+ 12 )(n−k −2l )! π  n+k , if k ≡ n (mod 2). n−k −2l n−k Γ(

2

+ λ − l +1)2

(

2

(59)

−l )!

Making use of (6), and from (57) and (59), we finally derive: Γ(λ) γn,r ( x ) = r!

∑

∑

l =0 0≤ k ≤ n k ≡n (mod 2)

Γ(λ)(n + r )! = r!

×

k [ n− 2 ]

(−1)l (k + λ)(n + r − l )! (λ) Ck ( x ) n+k n−k l!Γ( 2 + λ − l + 1)( 2 − l )! k [ n− 2 ]

∑

∑

0≤ k ≤ n l =0 k ≡n (mod 2)

(k + λ) k k ( n− ) !Γ ( n+ 2 2 + λ + 1)

k n+k (−1)l ( n− 2 )l ( 2 + λ )l (λ) Ck ( x ) l!(n + r )l

Γ(λ)(n + r )! = r!

×

j

∑∑

j =0 l =0

(n − 2j + λ) j!Γ(n − j + λ + 1)

(−1)l ( j)

l ( n + λ − j )l (λ) Cn−2j ( x ) l!(n + r )l

Γ(λ)(n + r )! = r!

×

[ n2 ]

[ n2 ]

j

∑∑

j =0 l =0

(n − 2j + λ) j!Γ(n − j + λ + 1)

< − j >l < j − n − λ >l (λ) Cn−2j ( x ) l! < −n − r >l

Γ(λ)(n + r )! = Γ(n + λ + 1)r!

[ n2 ]

(n − 2j + λ)(n + λ) j j! j =0

∑

(λ)

× 2 F1 (− j, j − n − λ; −n − r; 1)Cn−2j ( x ) This completes the proof for (38) in Theorem 1.

(60)

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3. Proof of Theorem 2 Here, we will show only (41) and (44) in Theorem 2, leaving the proofs for (40), (42) and (43) as an exercise, as they can be shown similarly to those for (35), (37) and (38). The following lemma is stated in Equation (9) of [27] and can be derived by differentiating (10). Lemma 2. Let n, r be integers with n ≥ 0, r ≥ 1. Then, we have the following identity.

∑

Fi1 +1 ( x ) Fi2 +1 ( x ) · · · Fir +1 ( x ) =

i1 +i2 +···+ir =n

1 (r −1) F ( x ), (r − 1) ! n +r

(61)

where the sum runs over all nonnegative integers i1 , i2 , · · · , ir , with i1 + i2 + · · · + ir = n. From (17), it is easy to show that the r-th derivative of Fn+1 ( x ) is given by: r  [ n− 2 ]

(r ) Fn+1 ( x )

=

∑

l =0

 n−l (n − 2l )r x n−r−2l . l

(62)

Thus, especially, we have: (r + k −1)

Fn+r

k  [ n− 2 ]

(x) =

∑

l =0

 n+r−l−1 (n + r − 2l − 1)r+k−1 x n−k−2l . l

(63)

With En,r ( x ) as in (34), we let: n

γn,r ( x ) =

∑ Ck,2 Lk

(α)

( x ).

(64)

k =0

Then, from (b) of Proposition 1, (61), (63), (6) and integration by parts k times, we have: Ck,2 =

1 Γ(α + k + 1)(r − 1)!

(−1)k = Γ(α + k + 1)(r − 1)! (−1)k = Γ(α + k + 1)(r − 1)! ×

Z ∞ 0

=

e

− x n+α−2l

x

Γ ( α + n + 1) (r − 1) ! k [ n− 2 ]

×

∑

l =0

Z ∞ (r −1) 0

Fn+r ( x )

Z ∞ (r + k −1) 0

Fn+r

k  [ n− 2 ]

∑

l =0

dk − x k+α (e x )dx dx k

( x )e− x x k+α dx

 n+r−l−1 (n + r − 2l − 1)r+k−1 l

(65)

dx

n

(−1)k Γ ( α + k + 1) k =0

∑

( n + r − l − 1) ! (α) L ( x ). l!(n − k − 2l )!(α + n)2l k

Next, we let: n

γn,r ( x ) =

∑ Ck,5 Pn

k =0

(α,β)

( x ).

(66)

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Then, from (e) of Proposition 1, and (61), (63) and integration by parts k times, we obtain: Ck,5 =

(2k + α + β + 1)Γ(k + α + β + 1) 2k+α+ β+1 Γ(k + α + 1)Γ(k + β + 1)(r − 1)! ×

=

Z 1 −1

(r + k −1)

Fn+r

( x )(1 − x )k+α (1 + x )k+ β dx

(2k + α + β + 1)Γ(k + α + β + 1) 2k+α+ β+1 Γ(k + α + 1)Γ(k + β + 1)(r − 1)! k   [ n− 2 ] n+r−l−1 × ∑ (n + r − 2l − 1)r+k−1 l l =0 ×

Z 1 −1

(67)

x n−k−2l (1 − x )k+α (1 + x )k+ β dx.

Now, from (67), and using (d) in Proposition 2 and (6), we have: Ck,5 =

(−1)n (2k + α + β + 1)Γ(k + α + β + 1)(−2)k (r − 1)!Γ(k + β + 1) k [ n− 2 ]

×

∑

l =0

( n + r − l − 1) ! l!

n−k −2l

×

∑

s =0

=

(−1)n Γ(k + α + β + 1)(−2)k (r − 1)!Γ(2k + α + β + 1) k [ n− 2 ]

×

∑

l =0

(n + r − 1 − l )! l!(n − k − 2l )!

n−k −2l

× =

(−2)s Γ(k + β + s + 1) s!(n − k − 2l − s)!Γ(2k + α + β + s + 2)

∑

< 2l + k − n >s < k + β + 1 >s 2s < 2k + α + β + 2 >s s!

s =0 (−1)n Γ(k + α +

(68)

β + 1)(−2)k

(r − 1)!Γ(2k + α + β + 1) k [ n− 2 ]

×

∑

l =0

(n + r − 1 − l )! l!(n − k − 2l )!

× 2 F1 (2l + k − n, k + β + 1; 2k + α + β + 2; 2). As we desired, we finally obtain: γn,r ( x ) = k [ n− 2 ]

×

∑

l =0

(−1)n n Γ(k + α + β + 1)(−2)k (r − 1)! k∑ Γ(2k + α + β + 1) =0 (n + r − 1 − l )! (α,β) ( x ). 2 F ( k + 2l − n, k + β + 1; 2k + α + β + 2; 2) Pn l!(n − k − 2l )! 1

Author Contributions: T.K. and D.S.K. conceived the framework and structured the whole paper; T.K. wrote the paper; J.K. and D.V.D. checked the results of the paper; D.S.K., J.K. and D.V.D. completed the revision of the article. Funding: This work was supported by the National Research Foundation of Korea (NRF)grant funded by the Korea government (MEST) (No. 2017R1E1A1A03070882).

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Acknowledgments: The first author has been appointed a chair professor at Tianjin Polytechnic University by Tianjin City in China from August 2015 to August 2019. The authors would like to express their sincere gratitude to the referees for their valuable comments which have significantly improved the presentation of this paper. Conflicts of Interest: The authors declare no conflict of interest.

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