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New trigonometric sums by sampling theorem

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New trigonometric sums by sampling theorem
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  J. Math. Anal. Appl. 339 (2008) 811–827www.elsevier.com/locate/jmaa New trigonometric sums by sampling theorem H.A. Hassan  Department of Mathematics, Faculty of Science, Cairo University, Giza, Egypt  Received 16 October 2006Available online 19 July 2007Submitted by D. Waterman Abstract We use a sampling theorem associated with second-order discrete eigenvalue problems to derive some trigonometric identitiesextending the results of Byrne and Smith [G.J. Byrne, S.J. Smith, Some integer-valued trigonometric sums, Proc. Edinburg Math.Soc. 40 (1997) 393–401]. We derive both integral and non-integral valued trigonometric sums. We give illustrative examplesinvolving representations of the trigonometric sums  nk = 0 cot 2 m (( 2 k  + 1 )π/ 2 ( 2 n + 1 ))  and  nk = 0 tan 2 m (kπ/( 2 n  +  1 ))  in anintegral-valued polynomial in  ( 2 n + 1 )  of degree 2 m ,  m =  1 , 2 ,... . © 2007 Elsevier Inc. All rights reserved. Keywords:  Difference operator; Finite sampling expansion; Lagrange’s interpolation expansion 1. Introduction The idea of using sampling theorems to obtain closed forms of infinite sums is old. One can find examples of usingsampling theorems to get an infinite sums in the work of Whittaker [14] and Krishnan [9]. One formula of the secondwork is the expansion ∞  n =−∞ sin [ a √  n 2 α 2 + λ 2 ]√  n 2 α 2 + λ 2 = παJ  0 (λa),  0 <α  2 π/a,  (1.1)where  J  0 (x)  is the Bessel function of order zero, see also [17, Chapter 7]. In the work of Zayed [15] many summationformulae which contain infinite series with trigonometric and special functions are given. Some of these formulaemay be found in the work of Gosper, Ismail and Zhang [8]. This work was extended in [16] for multivariate infiniteseries.Here we use a discrete form of samplingtheorem to derive finite sums of trigonometricfunctionsformulae. We startwith an interesting formula of M. Riesz. Let  S  n (θ)  be a trigonometric polynomial of degree  n . Then the followingidentity is satisfied, cf. [18, p. 10] S  ′ n (θ) = 14 n 2 n  k = 1 ( − 1 ) k − 1 S  n (θ   + θ  k ) sin 2 (θ  k / 2 ), θ  k  = ( 2 k  − 1 )π 2 n.  (1.2)  E-mail address:  hassanatef1@yahoo.com.0022-247X/$ – see front matter  © 2007 Elsevier Inc. All rights reserved.doi:10.1016/j.jmaa.2007.06.067  812  H.A. Hassan / J. Math. Anal. Appl. 339 (2008) 811–827  Putting  S  n (θ) =  sin θ  ,  S  n (θ) =  sin nθ   and letting  θ   =  0, we get [5] n  k = 1 ( − 1 ) k − 1 cot  ( 2 k  − 1 )π 4 n  = n, n  k = 1 cot 2  ( 2 k  − 1 )π 4 n  =  2 n 2 − n,  (1.3)respectively. A polynomial  p(x)  is said to be integral-valued if   p(x)  is an integer whenever  x  is an integer. In [5]Byrne and Smith introduced generalizations of the integral-valued identities (1.3) as follows. Theorem 1.1.  For   m =  1 , 2 ,...,  we have the following integral-valued polynomial in  n : n  k = 1 ( − 1 ) k − 1 cot 2 m − 1  ( 2 k  − 1 )π 4 n  = m  j  = 1 a m,j  n 2 j  − 1 ,  (1.4) where  a m, 1  = ( − 1 ) m − 1  , and the rest of the coefficients  a m,j   can be determined recursively as m  j  = 1 a m,j   =  1 , a m,j   = 12 2 (m − j) − 1 m − j   r = 1 ( − 1 ) r  2 m − 1 r  a m − r,j  , j <m.  (1.5) The leading coefficients of   ( 1 . 4 )  are given explicitly by a m,m  = 2 2 m − 2 ( 2 m − 2 ) ! E 2 m − 2 , a m,m − 1  = − ( 2 m − 1 ) 2 2 m − 4 3 ( 2 m − 4 ) ! E 2 m − 4 ,a m,m − 2  = ( 2 m − 1 )( 5 m − 6 ) 2 2 m − 6 45 ( 2 m − 6 ) ! E 2 m − 6 ,  (1.6) where  E 2 j   are the even-numbered Euler numbers defined by sec x  = ∞  j  = 0 E 2 j  x 2 j  ( 2 j) ! ,  | x | <π 2  .  Moreover  n  k = 1 cot 2 m  ( 2 k  − 1 )π 4 n  = ( − 1 ) m n + m  j  = 1 b m,j  n 2 j  ,  (1.7) where m  j  = 1 b m,j   =  1 + ( − 1 ) m − 1 , b m,j   = 12 2 (m − j) − 1 m − j   r = 1 ( − 1 ) r  2 mr  b m − r,j  , j <m.  (1.8) The leading coefficients of   ( 1 . 7 )  are given explicitly by b m,m  = 2 2 m − 1 ( 2 m − 1 ) ! E 2 m − 1 , b m,m − 1  = − m 2 2 m − 2 3 ( 2 m − 3 ) ! E 2 m − 3 , b m,m − 2  = m( 10 m − 7 ) 2 2 m − 5 45 ( 2 m − 5 ) ! E 2 m − 5 , (1.9) where  E 2 j  − 1  are the odd-numbered Euler numbers defined by tan x  = ∞  j  = 0 E 2 j  − 1 x 2 j  − 1 ( 2 j   − 1 ) ! ,  | x | <π 2  . The following lemma is useful in the sequel and also used in the proof of Theorem 1.1, see [5],   H.A. Hassan / J. Math. Anal. Appl. 339 (2008) 811–827   813 Lemma 1.2. (1)  Let   T  n (x) =  cos (n arccos x)  ,  − 1  x  1 ,  be the Chebyshev polynomial of degree  n . Then T  (r)n  ( 1 ) =  2 r − 1 (r  − 1 ) ! n  n + r  − 12 r  − 1  , r  =  1 , 2 ,...,n,  (1.10) T  (r)n  ( − 1 ) = ( − 1 ) n + r T  (r)n  ( 1 ), r  =  0 , 1 , 2 ,...,n,  (1.11) where  nk  =  n ! k ! (n − k) ! ,  cf. Rivlin  [11, p. 33] . (2)  The quantity nr  n + r  − 12 r  − 1  , r  n,  (1.12) is an even integral-valued polynomial in  n  of degree  2 r  , see  [10, pp. 129–130] . In [1,4,6,7] several sampling (interpolation) formulae were derived associated with difference equations. In [3],it is shown that generalized trigonometric sums may be derived using sampling theorems associated with differenceequations with separate-type conditions. In [2], the authors established sampling theorems associated with differenceoperators with mixed-type conditions. This connection gives new interpolation formulae, which are expected to leadto more general trigonometric identities. This is the aim of the present paper. In Section 2 we introduce briefly theresults of [2] and the associated generalized trigonometric sums. Section 3 is devoted to applications of the results of Section 2. In Section 4 we give genuine integral-valued trigonometric sums. 2. Generalized trigonometric identities Let Z N   := { 1 , 2 ,...,N  }  and  ℓ 2 ( Z N  )  denote the space of all finite sequences  x  = (x 1 ,...,x N  ) ,  x k  ∈ C , 1  k  N, with the inner product   x,y  =  N k = 1 x k y k ,  x,y  ∈ ℓ 2 ( Z N  ).  Consider the difference operator Ay  := ∇   p(n)y(n)  +  2 + q(n)  y(n) = ty(n), n =  1 ,...,N, y  ∈ ℓ 2 ( Z N  ),  (2.1) U  1 (y) := α 11 y( 0 ) + α 12 y( 1 ) + β 11 y(N) + β 12 y(N   + 1 ) =  0 ,U  2 (y) := α 21 y( 0 ) + α 22 y( 1 ) + β 21 y(N) + β 22 y(N   + 1 ) =  0 ,  (2.2)where    is the forward difference operator,  y(n) := y(n + 1 ) − y(n),  and  ∇   is the backward one,  ∇  y(n) := y(n) − y(n − 1 ) . The coefficients of the difference expression are assumed to be finite real-valued functions and  p(n)> 0 for n  0. Moreover  α ij  ,β ij  , 1  i,j    2 are real numbers and  α 11  β 12 α 21  β 22  =  0 , α 11 α 22  − α 12 α 21  = β 11 β 22  − β 12 β 21 .  (2.3)Problem (2.1)–(2.3) is self-adjoint, see [2], and the eigenfunctions generate orthogonal basis in  ℓ 2 ( Z N  ) . Let  φ(n,t) and  ψ(n,t)  be two linearly independent solutions of (2.1) satisfying the initial conditions φ( 0 ,t) =  0 , φ( 1 ,t) =  1 , ψ( 0 ,t) =  1 , ψ( 1 ,t) =  0 .  (2.4)The eigenvalues of (2.1)–(2.2) are the zeros of  D(t) :=  U  1 (φ) U  1 (ψ)U  2 (φ) U  2 (ψ)  =  0 .  (2.5)Let  { t  k } sk = 0  be the set of all different eigenvalues. Assume that all eigenfunctions are orthonormal. Let the eigenfunc-tions corresponding to the eigenvalue  t  k  be  { φ l (n,t  k ) } µ k l = 1 ,  where  µ k  is the multiplicity of   t  k  which does not exceedtwo. Then Green’s function of  (A − t)y  = f(n), U  1 (y) = U  2 (y) =  0  814  H.A. Hassan / J. Math. Anal. Appl. 339 (2008) 811–827  is [2] G(n,j,t) = s  k = 0 µ k  l = 1 φ l (j,t  k )φ l (n,t  k )t  k  − t , t   = t  k .  (2.6)The last equation shows that  G(n,j,t)  has simple poles exactly at the eigenvalues. Let  n 0  be a fixed point in  Z N  .Define the function  G(j,t)  to be  G(j,t) := Π(t)G(n 0 ,j,t),  where Π(t) = s  k = 0  1 − t t  k  .  (2.7)Then we have the following sampling result, see [2]. Theorem 2.1.  Let   F(t)  be the discrete transform F(t) = F(t) = N   n = 1 f(n)G(n,t), f(n) ∈ ℓ 2 ( Z N  ), t   ∈ C .  (2.8) Then  F(t)  has the sampling expansion F(t) = s  k = 0 F(t  k )Π(t)(t   − t  k )Π  ′ (t  k ).  (2.9)We note that (2.9) is a Lagrange interpolation formula which is valid for any polynomial of degree less than  s  + 1,see e.g. [13, p. 39]. Away from the eigenvalues, (2.9) becomes F(t)Π(t) = s  k = 0 F(t  k )Π  ′ (t  k ) 1 (t   − t  k ), t   = t  k , k  =  0 , 1 ,...,s.  (2.10)We assume without loss of generality that all eigenvalues of (2.1)–(2.2) lie in  [− 2 , 2 ] , and we put t  k  =  2cos θ  k .  (2.11)This assumption makes no restriction, since if the eigenvalues lie in  [ a,b ]  we can then take  ( 2 t  k  − b  − a)/(b  − a) = cos θ  k . We will use formula (2.10) to derive trigonometric sums by suitable choices of the polynomial  F(t) . In ourinvestigation we use only two choices of   F(t)  as different choice of the boundary-value problem (2.1)–(2.3). Puttingthat  t   = ± 2 if they are not eigenvalues and letting  F(t) = Π  ′ (t) , we obtain s  k = 0 csc 2  θ  k 2  =  4 Π  ′ ( 2 )Π( 2 ), s  k = 0 sec 2  θ  k 2  = − 4 Π  ′ ( − 2 )Π( − 2 ),  (2.12)respectively, or s  k = 0 cot 2  θ  k 2  =  4 Π  ′ ( 2 )Π( 2 ) − (s  + 1 ), s  k = 0 tan 2  θ  k 2  = − 4 Π  ′ ( − 2 )Π( − 2 ) − (s  + 1 ).  (2.13)More generally we have the following. Theorem 2.2.  Suppose that   t   = ± 2  are not eigenvalues of problem  ( 2 . 1 )  –  ( 2 . 3 ).  Then s  k = 0 csc 2 r  θ  k 2  = ( − 1 ) r − 1 4 r Ω r − 1 ( 2 ), s  k = 0 sec 2 r  θ  k 2  = − 4 r Ω r − 1 ( − 2 ),  (2.14) and    H.A. Hassan / J. Math. Anal. Appl. 339 (2008) 811–827   815 s  k = 0 cot 2 m  θ  k 2  = ( − 1 ) m  s  + 1 − m  r = 1  mr  4 r Ω r − 1 ( 2 )  ,  (2.15) s  k = 0 tan 2 m  θ  k 2  = ( − 1 ) m  s  + 1 − m  r = 1 ( − 1 ) r  mr  4 r Ω r − 1 ( − 2 )  ,  (2.16) where Ω r − 1 (t) := 1 (r  − 1 ) ! d  r − 1 dt  r − 1  Π  ′ (t)Π(t)  .  (2.17) Proof.  Differentiating (2.10)  r  − 1 times, putting  F(t) = Π  ′ (t)  and  t   =  2,  t   = − 2, one gets (2.14). Now s  k = 0 cot 2 m  θ  k 2  = ( − 1 ) ms  k = 0  1 − csc 2  θ  k 2  m = ( − 1 ) ms  k = 0 m  r = 0  mr  ( − 1 ) r csc 2 r  θ  k 2 = ( − 1 ) m  s  + 1 − m  r = 1  mr  4 r Ω r − 1 ( 2 )  . Similarly, s  k = 0 tan 2 m  θ  k 2  = ( − 1 ) ms  k = 0  1 − sec 2  θ  k 2  m = ( − 1 ) ms  k = 0 m  r = 0  mr  ( − 1 ) r sec 2 r  θ  k 2 = ( − 1 ) m  s  + 1 − m  r = 1 ( − 1 ) r  mr  4 r Ω r − 1 ( − 2 )  .  ✷ Remark 2.3.  If   t  ∗  =  2, or  − 2 are eigenvalues, we put  F(t) = Π  ′ (t) , then write (2.10) as s  k = 0 t  k = t  ∗ 1 (t   − t  k ) = Π  ′ (t)Π(t) − 1 t   − t  ∗  = (Π(t)/(t   − t  ∗ )) ′ Π(t)/(t   − t  ∗ ) = Π  ′ t  ∗ (t)Π  t  ∗ (t),  (2.18)where  Π  t  ∗ (t) := Π(t)/(t   − t  ∗ ) . Note that  Π  t  ∗ (t)  is well defined (as a limit) at  t  ∗  with non-zero value. Thus in eithercase (2.12)–(2.17) hold if the following satisfied.(1) Putting  Π  t  ∗ (t)  instead of   Π(t) .(2) Neglecting the term which contains  t  ∗  in the sum (in the left side).(3) Putting “ s ” instead of “ s  + 1” in the right sides whenever it exists.In this case we will write Ω r − 1 t  ∗  (t) = 1 (r  − 1 ) ! d  r − 1 dt  r − 1  Π  ′ t  ∗ (t)Π  t  ∗ (t)  .  (2.19)In the second choice we put  F(t)  =  1 and  t   = ± 2, if they not eigenvalues. We obtain similar results to (2.12)–(2.17); namely s  k = 0 1 Π  ′ (t  k ) csc 2  θ  k 2  = 4 Π( 2 ), s  k = 0 1 Π  ′ (t  k ) sec 2  θ  k 2  =− 4 Π( − 2 ),  (2.20) s  k = 0 1 Π  ′ (t  k ) csc 2 r  θ  k 2  = ( − 1 ) r − 1 4 r ω r − 1 ( 2 ), s  k = 0 1 Π  ′ (t  k ) sec 2 r  θ  k 2  = − 4 r ω r − 1 ( − 2 ),  (2.21)
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