Connection Problem for Sums of Finite Products of Chebyshev Polynomials of the Third and Fourth Kinds

S S symmetry

Article Connection Problem for Sums of Finite Products of Chebyshev Polynomials of the Third and Fourth Kinds

Dmitry Victorovich Dolgy 1, Dae San Kim 2, Taekyun Kim 3,4 and Jongkyum Kwon 5,∗ 1 Institute of National Sciences, Far Eastern Federal University, 690950 Vladivostok, Russia; [email protected] 2 Department of Mathematics, Sogang University, Seoul 04107, Korea; [email protected] 3 Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin 300160, China; [email protected] 4 Department of Mathematics, Kwangwoon University, Seoul 01897, Korea 5 Department of Mathematics Education and ERI, Gyeongsang National University, Jinju 52828, Gyeongsangnamdo, Korea * Correspondence: [email protected]; Tel.: +82-(0)10-8978-73576

Received: 8 October 2018; Accepted: 5 November 2018; Published: 9 November 2018 Abstract: This paper treats the connection problem of expressing sums of finite products of Chebyshev polynomials of the third and fourth kinds in terms of five classical orthogonal polynomials. In fact, by carrying out explicit computations each of them are expressed as linear combinations of Hermite, generalized Laguerre, Legendre, Gegenbauer, and Jacobi polynomials which involve some terminating hypergeometric functions 2F0, 2F1, and 3F2.

Keywords: sums of finite products of Chebyshev polynomials of the third and fourth kinds; Hermite; generalized Laguerre; Legendre; Gegenbauer; Jacobi

MSC: 11B83; 33C05; 33C20; 33C45

1. Introduction and Preliminaries

In this section, we will recall some basic facts about relevant orthogonal polynomials that will be needed throughout this paper. For this, we will ﬁrst ﬁx some notations. For any nonnegative integer n, the falling factorial polynomials (x)n and the rising factorial polynomials < x >n are respectively given by (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 evidently related by

n n (−1) (x)n =< −x >n, (−1) < x >n= (−x)n. (3)

2n−2s s 1 (2n − 2s)! 2 (−1) < >n = 2 , (n ≥ s ≥ 0). (4) (n − s)! 1 < 2 − n >s Z 1 Γ(x)Γ(y) B(x, y) = tx−1(1 − t)y−1dt = , (Re x, Re y > 0). (5) 0 Γ(x + y) √ 1 (2n)! π Γ(n + ) = , (n ≥ 0). (6) 2 22nn!

Symmetry 2018, 10, 617; doi:10.3390/sym10110617 www.mdpi.com/journal/symmetry Symmetry 2018, 10, 617 2 of 14

Γ(x + 1) Γ(x + n) = (x)n, =< x >n, (n ≥ 0), (7) Γ(x + 1 − n) Γ(x) where Γ(x) and B(x, y) are the gamma and beta functions respectively. The hypergeometric function is deﬁned by

pFq(a1, ··· , ap; b1, ··· , bq; x) ∞ n < a1 >n ··· < ap >n x (8) = . ∑ < > ··· < > n=0 b1 n bq n n!

We are now ready to state some basic facts about Chebyshev polynomials of the third kind Vn(x), those of the fourth kind Wn(x), Hermite polynomials Hn(x), generalized (extended) Laguerre α (λ) polynomials Ln(x), Legendre polynomials Pn(x), Gegenbauer polynomials Cn (x), and Jacobi (α,β) polynomials Pn . Chebyshev polynomials are diversely used in approximation theory and numerical analysis, Hermite polynomials appear as the eigenfunctions of the harmonic oscillator in quantum mechanics, Laguerre polynomials have important applications to the solution of Schrödinger’s equation for the hydrogen atom, Legendre polynomials can be used to write the Coulomb potential as a series, Gegenbauer polynomials play an important role in the constructive theory of spherical functions and Jacobi polynomials occur in the solution to the equations of motion of the symmetric top. All the necessary facts on those special polynomials can be found in [1–9]. For the full accounts of this fascinating area of orthogonal polynomials, the reader may refer to [10–13]. The above special polynomials are given in terms of generating functions by

1 − t ∞ F(t, x) = = V (x)tn, (9) − + 2 ∑ n 1 2xt t n=0 1 + t ∞ G(t, x) = = W (x)tn, (10) − + 2 ∑ n 1 2xt t n=0 ∞ n 2xt−t2 t e = ∑ Hn(x) , (11) n=0 n! ∞ −α−1 xt α n (1 − t) exp(− 1−t ) = ∑ Ln(x)t , (α > −1), (12) n=0 ∞ 2 − 1 n (1 − 2xt + t ) 2 = ∑ Pn(x)t , (13) n=0 ∞ 1 ( ) 1 = C λ (x)tn ( > − 6= |t| < |x| ≤ ) 2 λ ∑ n , λ , λ 0, 1, 1 , (14) (1 − 2xt + t ) n=0 2 α+β ∞ 2 (α,β) = P (x)tn, (15) α β ∑ n R(1 − t + R) (1 + t + R) n=0 p (R = 1 − 2xt + t2, α, β > −1).

Explicit expressions for the above special polynomials are as in the following:

1 1−x Vn(x) = 2F1(−n, n + 1; 2 ; 2 ) n 2n − l = ∑ 2n−l(x − 1)n−l, (16) l=0 l Symmetry 2018, 10, 617 3 of 14

3 1−x Wn(x) = (2n + 1)2F1(−n, n + 1; 2 ; 2 ) n 2n−l 2n − l = (2n + 1) (x − 1)n−l, (17) ∑ − + l=0 2n 2l 1 l [ n ] 2 (−1)l H (x) = n! (2x)n−2l, (18) n ∑ ( − ) l=0 l! n 2l ! < α + 1 > Lα(x) = n F (−n, α + 1; x) n n! 1 1 n (−1)l(n+α) = ∑ n−l xl, (19) l=0 l! 1−x Pn(x) = 2F1(−n, n + 1; 1; 2 ) [ n ] 1 2 n2n − 2l = (− )l n−2l n ∑ 1 x , (20) 2 l=0 l n ( ) n + 2λ − 1 C λ (x) = F (−n, n + 2λ; λ + 1 ; 1−x ) n n 2 1 2 2 [ n ] 2 Γ(n − k + λ) = (−1)k (2x)n−2k, (21) ∑ ( ) ( − ) k=0 Γ λ k! n 2k ! (α,β) < α + 1 > P (x) = n F (−n, 1 + α + β + n; α + 1; 1−x ) n n! 2 1 2 n n + αn + β = ( x−1 )k( x+1 )n−k. (22) ∑ − 2 2 k=0 n k k

Next, we state Rodrigues-type formulas for Hermite and generalized Laguerre polynomials and Rodrigues’ formulas for Legendre, Gegenbauer and Jacobi polynomials.

n 2 d 2 H (x) = (−1)nex e−x , (23) n dxn 1 dn Lα(x) = x−αex (e−xxn+α), (24) n n! dxn 1 dn P (x) = (x2 − 1)n, (25) n 2nn! dxn n n 2 λ− 1 (λ) (−2) < λ >n d 2 n+λ− 1 ( − ) 2 ( ) = ( − ) 2 1 x Cn x n 1 x , (26) n! < n + 2λ >n dx n n (α,β) (−1) d (1 − x)α(1 + x)βP (x) = (1 − x)n+α(1 + x)n+β. (27) n 2nn! dxn The last thing we want to mention is the orthogonalities with respect to various weight functions enjoyed by Hermite, generalized Laguerre, Legendre, Gegenbauer and Jacobi polynomials.

Z ∞ √ −x2 n e Hn(x)Hm(x) dx = 2 n! πδn,m, (28) −∞ Z ∞ α −x α α 1 x e Ln(x)Lm(x) dx = Γ(α + n + 1)δn,m, (29) 0 n! Z 1 2 Pn(x)Pm(x) dx = δn,m, (30) −1 2n + 1 Z 1 1−2λ 2 λ− 1 (λ) (λ) π2 Γ(n + 2λ) (1 − x ) 2 Cn (x)Cm (x) dx = δn,m, (31) −1 n!(n + λ)Γ(λ)2 Symmetry 2018, 10, 617 4 of 14

Z 1 α β (α,β) (α,β) (1 − x) (1 + x) Pn (x)Pm (x) dx −1 2α+β+1Γ(n + α + 1)Γ(n + β + 1) = δn m. (32) (2n + α + β + 1)Γ(n + α + β + 1)Γ(n + 1) ,

For convenience, let us put

n r − 1 + n − l γ (x) = V (x)V (x) ··· V (x), (n ≥ 0, r ≥ 1), n,r ∑ ∑ r − 1 i1 i2 ir+1 (33) l=0 i1+i2+···+ir+1=l

n r − 1 + n − l E (x) = (−1)n−l W (x)W (x) ··· W (x), (n ≥ 0, r ≥ 1). n,r ∑ ∑ r − 1 i1 i2 ir+1 (34) l=0 i1+i2+···+ir+1=l

We observe here that both γn,r(x) and En,r(x) have degree n. In this paper, we will consider the connection problem of expressing the sums of ﬁnite products α (λ) (α,β) in (33) and (34) as linear combinations of Hn(x), Ln(x), Pn(x), Cn (x), and Pn (x). These will be done by performing explicit computations based on Proposition 1. We observe here that the formulas in Proposition1 follow from their orthogonalities, Rodrigues’ and Rodrigues-type formulas and integration by parts. Our main results are the following Theorems1 and2.

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

n r − 1 + n − l V (x)V (x) ··· V (x) ∑ ∑ r − 1 i1 i2 ir+1 l=0 i1+i2+···+ir+1=l (2n + 2r)! n (−2)k = n+r 1 ∑ (n − k)! r!4 (n + r − 2 )n+r k=0 [ k ] 2 F (2j − k, 1 − n − r; −2n − 2r; 2) × 2 1 2 ( ) ∑ j Hn−k x (35) j=0 j!4 (k − 2j)! 1 n k (−2)n−l(2n + 2r − l)!(n + r − l)! = ∑ ∑ ( + − ) ( − ) r! k=0 l=0 l! 2n 2r 2l ! k l ! α × 2F0(l − k, n − k + α + 1; −; 1)Ln−k(x) (36) (−1)nn!(2n + 2r)! n (−1)k(2k + 1) = r 1 ∑ (n − k)!(n + k + 1)! r!4 (n + r − 2 )n+r k=0 1 × F (k − n, − n − r, −n − k − 1; −2n − 2r, −n; 1)P (x) (37) 3 2 2 k (−1)n(2n + 2r)!4λ−rΓ(λ)Γ(n + λ + 1 ) n (−1)k(k + λ) = √ 2 1 ∑ Γ(n + k + 2λ + 1)(n − k)! πr!(n + r − 2 )n+r k=0 1 1 ( ) × F (k − n, − n − r, −n − k − 2λ; −2n − 2r, −n − λ + ; 1)C λ (x) (38) 3 2 2 2 k (−1)n(2n + 2r)!Γ(n + α + 1) n (−1)k(2k + α + β + 1)Γ(k + α + β + 1) = r 1 ∑ (n − k)!Γ(α + k + 1)Γ(n + k + α + β + 2) r!4 (n + r − 2 )n+r k=0 1 (α,β) × F (k − n, − n − r, −n − k − α − β − 1; −2n − 2r, −n − α; 1)P (x). (39) 3 2 2 k Symmetry 2018, 10, 617 5 of 14

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

n r − 1 + n − l (−1)n−l W (x)W (x) ··· W (x) ∑ ∑ r − 1 i1 i2 ir+1 l=0 i1+i2+···+ir+1=l (2n + 1)(2n + 2r)! n (−2)k = 2n+2r+1 1 ∑ (n − k)! r!2 (n + r + 2 )n+r+1 k=0 [ k ] 2 F (2j − k, −n − r − 1 ; −2n − 2r; 2) × 2 1 2 ( ) ∑ j Hn−k x (40) j=0 j!4 (k − 2j)! (2n + 1) n k (−2)n−l(2n + 2r − l)!(n + r − l)! = ∑ ∑ ( + − + ) ( − ) r! k=0 l=0 l! 2n 2r 2l 1 ! k l ! α × 2F0(l − k, n − k + α + 1; −; 1)Ln−k(x) (41)

(−1)nn!(2n + 1)(2n + 2r)! n (−1)k(2k + 1) = 2r+1 1 ∑ (n − k)!(n + k + 1)! r!2 (n + r + 2 )n+r+1 k=0 1 × F (k − n, −n − r − , −n − k − 1; −2n − 2r, −n; 1)P (x) (42) 3 2 2 k (−1)n(2n + 2r)!22λ−2r−1(2n + 1)Γ(λ)Γ(n + λ + 1 ) = √ 2 1 πr!(n + r + 2 )n+r+1 n (−1)k(k + λ) × ∑ ( + + + )( − ) k=0 Γ n k 2λ 1 n k ! 1 1 ( ) × F (k − n, −n − r − , −n − k − 2λ; −2n − 2r, −n − λ + ; 1)C λ (x) (43) 3 2 2 2 k (−1)n(2n + 2r)!(2n + 1)Γ(n + α + 1) = 2r+1 1 r!2 (n + r + 2 )n+r+1 n (−1)k(2k + α + β + 1)Γ(k + α + β + 1) × ∑ ( − ) ( + + ) ( + + + + ) k=0 n k !Γ α k 1 Γ n k α β 2 1 (α,β) × F (k − n, −n − r − , −n − k − α − β − 1; −2n − 2r, −n − α; 1)P (x). (44) 3 2 2 k

Before closing the section, we are going to mention some of previous results on the related connection problems. The papers [14–16] treat the connection problem of expressing sums of finite products of Bernoulli, Euler and Genocchi polynomials in terms of Bernoulli polynomials. In fact, they were carried out by deriving Fourier series expansions for the functions closely related to those sums of finite products. Moreover, the same was done for the sums of finite products of Chebyshev polynomials of the second kind and of Fibonacci polynomials in [17]. Along the same line as the present paper, sums of finite products of Chebyshev polynomials of the second kind and Fibonacci polynomials were expressed in [18] as linear combinations of the α (λ) (α,β) orthogonal polynomials Hn(x), Ln(x), Pn(x), Cn (x), and Pn (x). Also, the connection problem of expressing in terms of all kinds of Chebyshev polynomials were done for sums of finite products of Chebyshev polynomials of the second, third and fourth kinds and of Fibonacci, Legendre and Laguerre polynomials in [19–21]. Finally, we let the reader refer to [22,23] for some applications of Chebyshev polynomials.

2. Proof of Theorem 1 First, we will state Propositions1 and2 that will be needed in showing Theorems1 and2. Symmetry 2018, 10, 617 6 of 14

The results in (a), (b), (c), (d) and (e) in Proposition1 follow respectively from (3.7) of [ 4], (2.3) of [4] (see also (2.4) of [2]), (2.3) of [5], (2.3) of [3] and (2.7) of [7]. They can be derived from their orthogonalities in (28)–(32), Rodrigues-type and Rodrigues’ formulas in (23)–(27) and integration by parts.

Proposition 1. Let q(x) ∈ R[x] be a polynomial of degree n. Then we have the following.

n (a) q(x) = ∑ Ck,1 Hk(x), where k=0 k Z ∞ k (−1) d −x2 Ck,1 = √ q(x) e dx, 2kk! π −∞ dxk n α (b) q(x) = ∑ Ck,2Lk (x), where k=0 Z ∞ k 1 d −x k+α Ck,2 = q(x) (e x ) dx, Γ(α + k + 1) 0 dxk n (c) q(x) = ∑ Ck,3Pk(x), where k=0 Z 1 k 2k + 1 d 2 k Ck,3 = + q(x) (x − 1) dx, 2k 1k! −1 dxk n ( ) ( ) = (λ)( ) d q x ∑ Ck,4Ck x , where k=0 Z 1 k (k + λ)Γ(λ) d 2 k+λ− 1 Ck,4 = √ q(x) (1 − x ) 2 dx, k 1 − dxk (−2) πΓ(k + λ + 2 ) 1 n (α,β) (e) q(x) = ∑ Ck,5Pn (x), where k=0 (−1)k(2k + α + β + 1)Γ(k + α + β + 1) Ck,5 = 2α+β+k+1Γ(α + k + 1)Γ(β + k + 1) Z 1 dk × q(x) (1 − x)k+α(1 + x)k+βdx. −1 dxk

The following proposition will be used in showing Theorems1 and2. In fact, (a) is needed for (35) and (40), (b) for (39) and (44), and (b) or (c) for (37), (38), (42) and (43).

Proposition 2. The following holds true. (a) For any nonnegative integer m,  Z ∞ 0, if m ≡ 1 (mod 2), m −x2  √ x e dx = m! π − , if m ≡ 0 (mod 2), ∞  ( m ) m 2 !2

(b) For any real numbers r, s > −1, we have

Z 1 Γ(r + 1)Γ(s + 1) (1 − x)r(1 + x)sdx = 2r+s+1 , −1 Γ(r + s + 2)

Z 1 Γ(r + s + 1)Γ(s + 1) (1 − x)r(1 − x2)sdx = 2r+2s+1 . −1 Γ(r + 2s + 2) Symmetry 2018, 10, 617 7 of 14

Proof. (a) This is an easy exercise. (c) This follows from (b) with r replaced by r + s. (b) This follows from the change of variable 1 + x = 2y and (5).

The following lemma can be obtained by differentiating (9), as was shown in [24].

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

n r − 1 + n − l 1 (r) Vi (x)Vi (x) ··· Vi (x) = V + (x), (45) ∑ ∑ r − 1 1 2 r+1 2rr! n r l=0 i1+i2+···+ir+1=l where the inner sum runs over all nonnegative integers i1, i2, ··· ir+1, with i1 + i2 + ··· + ir+1 = l.

From (16), we see that the rth derivative of Vn(x) is given by

n−r (r) 2n − l n−l n−l−r Vn (x) = ∑ 2 (n − l)r(x − 1) . (46) l=0 l

Especially, we have

n−k (r+k) 2n + 2r − l n+r−l n−k−l Vn+r (x) = ∑ 2 (n + r − l)r+k(x − 1) . (47) l=0 l

Now, we are ready to prove Theorem1. As (38) and (39) can be shown similarly to (43) and (44) in the next section, we will show only (35), (36) and (37). With γn,r(x) as in (33), we let

n γn,r(x) = ∑ Ck,1 Hk(x). (48) k=0

Then, from (a) of Proposition1,(45), (47), and integration by parts k times, we obtain

k Z ∞ k (−1) d −x2 Ck,1 = √ γn,r(x) e dx 2kk! π −∞ dxk k k (−1) Z ∞ d 2 = √ (r) ( ) −x + Vn+r x e dx 2k rk!r! π −∞ dxk 1 Z ∞ 2 = √ (r+k)( ) −x + Vn+r x e dx (49) 2k rk!r! π −∞ 1 n−k 2n + 2r − l = √ n+r−l(n + r − l) k+r ∑ 2 r+k 2 k!r! π l=0 l Z ∞ 2 × (x − 1)n−k−le−x dx. −∞ Symmetry 2018, 10, 617 8 of 14

Before proceeding further, by making use of (a) in Proposition2, we note that

Z ∞ 2 (x − 1)me−x dx −∞ m m Z ∞ 2 = ∑ (−1)m−s xse−x dx s −∞ s=0 √ m s! π = (−1)m−s (50) ∑ ( s ) s 0≤s≤m s 2 !2 s≡0 (mod 2) [ m ] √ 2 m (2j)! =(− )m (m ≥ ) 1 π ∑ 2j , 0 . j=0 2j j!2

From (48)–(50), and after simpliﬁcations, we have

[ k−l ] 1 n (−2)k k 2 (− 1 )l(2n + 2r − l)!(n + r − l)! γ (x) = 2 H (x) n,r ∑ ( − ) ∑ ∑ j n−k r! k=0 n k ! l=0 j=0 l!(2n + 2r − 2l)!(k − l − 2j)!j!4 [ k ] − 1 n (−2)k 2 1 k 2j (− 1 )l(2n + 2r − l)!(n + r − l)! = 2 H (x) ∑ ( − ) ∑ j ∑ ( + − ) ( − − ) n−k r! k=0 n k ! j=0 j!4 l=0 l! 2n 2r 2l ! k l 2j ! [ k ] (2n + 2r)! n (−2)k 2 1 = r n+r < 1 > ∑ (n − k)! ∑ j!4j(k − 2j)! !4 2 n+r k=0 j=0 (51) k−2j 2l < 2j − k > < 1 − n − r > × l 2 l H (x) ∑ < − − > n−k l=0 l! 2n 2r l (2n + 2r)! n (−2)k = n+r 1 ∑ (n − k)! r!4 (n + r − 2 )n+r k=0 [ k ] 2 F (2j − k, 1 − n − r; −2n − 2r; 2) × 2 1 2 ( ) ∑ j Hn−k x . j=0 j!4 (k − 2j)!

This shows (35) of Theorem1. Next, we let

n α γn,r(x) = ∑ Ck,2Lk (x). (52) k=0

Then, from (b) of Proposition1,(45), (47) and integration by parts k times, we get Symmetry 2018, 10, 617 9 of 14

1 Z ∞ dk = (r) ( ) ( −x k+α) Ck,2 r Vn+r x e x dx 2 r!Γ(α + k + 1) 0 dxk (−1)k Z ∞ = (r+k)( ) −x k+α r Vn+r x e x dx 2 r!Γ(α + k + 1) 0 (−1)k n−k 2n + 2r − l = 2n+r−l(n + r − l) r ( + + ) ∑ r+k 2 r!Γ α k 1 l=0 l Z ∞ × (x − 1)n−k−le−xxk+αdx 0 (−1)k n−k 2n + 2r − l = 2n+r−l(n + r − l) r ( + + ) ∑ r+k 2 r!Γ α k 1 l=0 l n−k−l n − k − l × ∑ (−1)n−k−l−sΓ(s + k + α + 1) s=0 s (53) (−1)k n−k 2n + 2r − l = n+r−l( + − ) r ∑ 2 n r l r+k 2 r! l=0 l n−k−l n − k − l n−k−l−s × ∑ (−1) < k + α + 1 >s s=0 s 1 n−k (2n + 2r − l)!(−2)n−l(n + r − l)! = ∑ ( + − ) ( − − ) r! l=0 l! 2n 2r 2l ! n k l ! n−k−l 1 × ∑ < k + l − n >s< k + α + 1 >s s=0 s! 1 n−k (2n + 2r − l)!(−2)n−l(n + r − l)! = ∑ ( + − ) ( − − ) r! l=0 l! 2n 2r 2l ! n k l !

× 2F0(k + l − n, k + α + 1; −; 1).

Combining (52)–(53), we ﬁnally have

1 n k (2n + 2r − l)!(−2)n−l(n + r − l)! γ (x) = n,r ∑ ∑ ( + − ) ( − ) r! k=0 l=0 l! 2n 2r 2l ! k l ! (54) α × 2F0(l − k, n − k + α + 1; −; 1)Ln−k(x).

This completes the proof for (36) of Theorem1. Finally, let us put

n γn,r(x) = ∑ Ck,3Pk(x). (55) k=0

Then, from (c) of Proposition1,(45), (47) and integration by parts k times, we have

k Z 1 2k + 1(−1) (r+k) 2 k Ck,3 = + + Vn+r (x)(x − 1) dx 2k r 1k!r! −1 (2k + 1)(−1)k n−k 2n + 2r − l = n+r−l(n + r − l) (56) k+r+1 ∑ 2 r+k 2 k!r! l=0 l Z 1 × (x − 1)n−k−l(x2 − 1)kdx. −1 Symmetry 2018, 10, 617 10 of 14

By making use of (c) in Proposition2 and after simpliﬁcations, from (56) we obtain

(−1)n+k(2k + 1) n−k Ck,3 = ∑ r! l=0 (−1)l4n−l(2n + 2r − l)!(n + r − l)!(n − l)! × l!(2n + 2r − 2l)!(n − k − l)!(n + k − l + 1)! (−1)n(2n + 2r)!n! (−1)k(2k + 1) = r!4r(n + r − 1 ) (n − k)!(n + k + 1)! 2 n+r (57) n−k < k − n > < 1 − n − r > < −n − k − 1 > × l 2 l l ∑ < − − > < − > l=0 l! 2n 2r l n l (−1)n(2n + 2r)!n! (−1)k(2k + 1) = r 1 (n − k)!(n + k + 1)! r!4 (n + r − 2 )n+r 1 × F (k − n, − n − r, −n − k − 1; −2n − 2r, −n; 1). 3 2 2 From (55) and (57), we get

(−1)n(2n + 2r)!n! n (−1)k(2k + 1) γn,r(x) = r r(n + r − 1 ) ∑ (n − k)!(n + k + 1)! !4 2 n+r k=0 (58) 1 × F (k − n, − n − r, −n − k − 1; −2n − 2r, −n; 1)P (x). 3 2 2 k This proves (37) of Theorem1.

3. Proof of Theorem 2 Here we will show only (43) and (44) in Theorem 2, as (40)–(42) can be shown analogously to the proofs for (35)–(37), respectively. The following can be derived by differentiating the Equation (10) and is stated in [24].

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

n n−l r − 1 + n − l 1 (r) (−1) Wi (x)Wi (x) ··· Wi (x) = W + (x), (59) ∑ ∑ r − 1 1 2 r+1 2rr! n r l=0 i1+i2+···+ir+1=l where the inner sum runs over all nonnegative integers i1, i2, ··· , ir+1, with i1 + i2 + ··· + ir+1 = l.

From (17), the rth derivative of Wn(x) is given by

n−r n−l (r) 2 2n − l W (x) = (2n + 1) (n − l) (x − 1)n−l−r. (60) n ∑ + − r l=0 2n 1 2l l

In particular, we have

n−k n+r−l (r+k) 2 2n + 2r − l W (x) = (2n + 1) (n + r − l) (x − 1)n−k−l. (61) n+r ∑ + + − r+k l=0 2n 2r 1 2l l

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

n E ( ) = (α)( ) n,r x ∑ Ck,4Ck x . (62) k=0 Symmetry 2018, 10, 617 11 of 14

Then, from (d) of Proposition1,(59), (61) and integration by parts k times, we get

(k + λ)Γ(λ) C = √ k,4 k+r 1 2 r! πΓ(k + λ + 2 ) Z 1 (r+k) 2 k+λ− 1 × Wn+r (x)(1 − x ) 2 dx −1 (k + λ)Γ(λ)(2n + 1) = √ k+r 1 2 r! πΓ(k + λ + 2 ) n−k 2n+r−l 2n + 2r − l × ∑ (n + r − l)r+k = 2n + 2r + 1 − 2l l l 0 (63) Z 1 n−k−l 2 k+λ− 1 × (x − 1) (1 − x ) 2 dx −1 (k + λ)Γ(λ)(2n + 1)(−2)n−k = √ 1 r! πΓ(k + λ + 2 ) n−k (− 1 )l(2n + 2r − l)!(n + r − l)! × 2 ∑ ( + − + ) ( + − ) ( − − ) l=0 2n 2r 2l 1 l! 2n 2r 2l ! n k l ! Z 1 n−k−l 2 k+λ− 1 × (1 − x) (1 − x ) 2 dx. −1

Invoking (c) of Proposition2 and after simpliﬁcations, from (63) we obtain

(−1)n−k(k + λ)Γ(λ)(2n + 1)22n+2λ+1Γ(n + λ + 1 )(2n + 2r)! C = √ 2 k,4 Γ(n + k + 2λ + 1)(n − k)!r! π n−k (− 1 )l(2n + 2r − l)!(n + r − l + 1)!(n − k)!(n + k + 2λ) × 4 l ∑ 1 l=0 l!(2n + 2r)!(2n + 2r − 2l + 2)!(n − k − l)!(n + λ − 2 )l (−1)k(k + λ)Γ(λ)(2n + 1)22λ−2r−1(−1)nΓ(n + λ + 1 )(2n + 2r)! = √ 2 Γ(n + k + 2λ + 1)(n − k)!r! π(n + r + 1 ) 2 n+r+1 (64) n−k < k − n > < −n − r − 1 > < −n − k − 2λ > × l 2 l l ∑ 1 l=0 l! < −2n − 2r >l< −n − λ + 2 >l (−1)k(k + λ)Γ(λ)(2n + 1)22λ−2r−1(−1)nΓ(n + λ + 1 )(2n + 2r)! = √ 2 1 Γ(n + k + 2λ + 1)(n − k)!r! π(n + r + 2 )n+r+1 1 1 × F (k − n, −n − r − , −n − k − 2λ; −2n − 2r, −n − λ + ; 1). 3 2 2 2 From (62) and (64), we have

Γ(λ)(2n + 1)22λ−2r−1(−1)nΓ(n + λ + 1 )(2n + 2r)! E ( ) = √ 2 n,r x 1 r! π(n + r + 2 )n+r+1 n (−1)k(k + λ) × (65) ∑ ( + + + )( − ) k=0 Γ n k 2λ 1 n k ! 1 1 ( ) × F (k − n, −n − r − , −n − k − 2λ; −2n − 2r, −n − λ + ; 1)C α (x). 3 2 2 2 k This shows (43) of Theorem2. Next, we let

n (α,β) En,r(x) = ∑ Ck,5Pn (x). (66) k=0 Symmetry 2018, 10, 617 12 of 14

Then, from (e) of Proposition1, and (59), (61), and integrating by parts k times, we obtain

(2k + α + β + 1)Γ(k + α + β + 1) Ck,5 = 2α+β+k+r+1r!Γ(α + k + 1)Γ(β + k + 1) Z 1 (r+k) k+α k+β × Wn+r (x)(1 − x) (1 + x) dx −1 (2k + α + β + 1)Γ(k + α + β + 1)(2n + 1) = 2α+β+k+r+1r!Γ(α + k + 1)Γ(β + k + 1) (67) n−k 2n+r−l 2n + 2r − l × (n + r − l) (−1)n−k−l ∑ + − + r+k l=0 2n 2r 2l 1 l Z 1 × (1 − x)n+α−l(1 + x)k+βdx. −1

By exploiting (b) in Proposition2 and after simpliﬁcations, from (67) we get

(−1)n−k(2k + α + β + 1)Γ(k + α + β + 1)22n+1(2n + 1)Γ(n + α + 1) C = k,5 Γ(α + k + 1)Γ(n + k + α + β + 2)r! n−k (− 1 )l(2n + 2r − l)!(n + r − l + 1)!(n + k + α + β + 1) × 4 l ∑ ( + − + ) ( − − ) ( + ) l=0 l! 2n 2r 2l 2 ! n k l ! n α l (−1)n−k(2k + α + β + 1)Γ(k + α + β + 1)(2n + 1)Γ(n + α + 1)(2n + 2r)! = Γ(α + k + 1)Γ(n + k + α + β + 2)(n − k)!r!22r+1(n + r + 1 ) 2 n+r+1 (68) n−k < k − n > < −n − r − 1 > < −n − k − α − β − 1 > × l 2 l l ∑ < − − > < − − > l=0 l! 2n 2r l n α l (−1)n−k(2k + α + β + 1)Γ(k + α + β + 1)(2n + 1)Γ(n + α + 1)(2n + 2r)! = 2r+1 1 Γ(α + k + 1)Γ(n + k + α + β + 2)(n − k)!r!2 (n + r + 2 )n+r+1 1 × F (k − n, −n − r − , −n − k − α − β − 1; −2n − 2r, −n − α; 1). 3 2 2 Thus, from (66) and (68), we have

(−1)n(2n + 1)Γ(n + α + 1)(2n + 2r)! En r(x) = , 2r+1 1 r!2 (n + r + 2 )n+r+1 n (−1)k(2k + α + β + 1)Γ(k + α + β + 1) × ∑ ( + + ) ( + + + + )( − ) k=0 Γ α k 1 Γ n k α β 2 n k ! 1 (α,β) × F (k − n, −n − r − , −n − k − α − β − 1; −2n − 2r, −n − α; 1)P (x). 3 2 2 n

4. Conclusions In this paper, we considered sums of finite products of Chebyshev polynomials of the third and fourth kinds and expressed each of them in terms of five orthogonal polynomials. Indeed, by explicit computations we expressed each of them as linear combinations of Hermite, generalized Laguerre, Legendre, Gegenbauer and Jacobi polynomials which involve some terminating hypergeometric functions. This can be viewed as a generalization of the classical linearization problem. In general, the linearization problem deals with determining the coefficients in the expansion of the products of two polynomials am(x) and bn(x) in terms of an arbitrary polynomial sequence {pk(x)}k≥0:

m+n am(x)bn(x) = ∑ cmn(k)pk(x). k=0 Symmetry 2018, 10, 617 13 of 14

Those sums of finite products were also represented by all kinds of Chebyshev polynomials in [20]. In addition, the same had been done for sums of finite products of Chebyshev polynomials of the first and second kinds, Fibonacci polynomials, Legendre polynomials, Laguerre polynomials and Lucas polynomials.

Author Contributions: T.K. and D.S.K. conceived the framework and structured the whole paper; T.K. wrote the paper; J.K. typed; D.V.D. checked for typos; D.S.K.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). Acknowledgments: We would like to thank the referees for their valuable comments and suggestions. Conﬂicts of Interest: The authors declare no conﬂict of interest.

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