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d2m THE DISCRETE ANALOGUE OF THE OPERATOR AND dx2m ITS PROPERTIES Kh.M.Shadimetov 0 Abstract 1 0 In this paper the discrete analogue Dm[β] of the differential operator d2m/dx2m is constructed 2 and its some new properties are proved. n a Key words and phrases: Discrete function, discrete analogue of the differential operator, Euler J polynomial. 4 ] 1 Main results. A N (m) First S.L.Sobolev [1] studied construction and investigated properties of the operator D [β], which . hH h is inverse of the convolution operator with function G(m)[β] = hnG (hHβ). The function D(m)[β] of t hH m hH a discrete variable, satisfying the equality m [ hnD(m)[β]∗G(m)[β] = δ[β] hH hH 1 v is called by the discrete analogue of the polyharmonic operator ∆m. S.L.Sobolev suggested an algo- 6 (m) 5 rithm for finding function DhH [β] and proved several properties of this function. In one dimensional 5 d2m 0 case, i.e. the discrete analogue of the operator was constructed by Z.Zh.Zhamalov [2, 3]. But dx2m . 1 there the form of this function was written with m+1 unknown coefficients. In works [4, 5] these 0 d2m 0 coefficients were found, hereunder the discrete analogue of the operator was constructed com- 1 dx2m pletely. : v In this paper we give the results of works [4-7], concerning to construction of the discrete analogue i X d2m D [β] of the operator , and discovery of its properties, which early were not known. r m dx2m a Following statements are valid. d2m Theorem 1. The discrete analogue of the differential operator have following form dx2m m−1 (1−λ )2m+1λ|β| k k for |β| ≥ 2, λ E (λ )  k 2m−1 k k=1 Dm[β] = (2mh2−m1)!  1X+mXk=−11 (1E−2mλ−k1)(2λmk+)1 for |β| = 1, (1) m−1 (1−λ )2m+1  −22m−1+ k for β = 0,  λkE2m−1(λk)  k=1  X   where E (λ) is the Euler polynomial of degree α, λ are the roots of the Euler polynomial E (λ), α k 2m−2 in module less than unity, i.e. |λ | < 1, h is the step of the lattice. k 1 Property 1. The discrete analogue D [β] of the differential operator of order 2m have represen- m tation m−1 |β|+m−2 (2m−1)! λ D [β] = ∆[m][β]∗ k , m h2m 2 E′ (λ ) k=1 2m−2 k X m 2m where ∆[m][β] = (−1)k+m δ[β−k] is symmetric difference of order 2m. 2 m+k k=−m (cid:18) (cid:19) X Property 2. The operator D [β] and monomials [β]k = (hβ)k are connected as m 0 for 0 ≤ k ≤ 2m−1, D [β][β]k = (2) m (2m)! for k = 2m, β (cid:26) X 0 for 2m+1 ≤ k ≤ 4m−1, D [β][β]k = h2m(4m)!B (3) m 2m  for k = 4m. Xβ  (2m)! Property 3. The operator D [β] and the function exp(2πihpβ) connected as m D [β]exp(2πihpβ) = m β X (−1)m22m(2m−1)!h−2msin2m(πhp) , m−2 (2m−2) (2m−2) 2 a cos2πhp(m−1−k)+a k m−1 k=0 P where k 2m a(2m−2) = (−1)j (k+1−j)2m−1 k j j=0 (cid:18) (cid:19) X are the coefficients of the Euler polynomial E (λ). 2m−2 2 Lemmas. As known, Euler polynomials E (λ) have following form k λ λE (λ) = (1−λ)k+2Dk , (4) k (1−λ)2 where d d D = λ , Dk = λ Dk−1. dλ dλ (k) In [8] was shown, that all roots λ of the Euler polynomial E (λ) are real, negative and different: j k (k) (k) (k) λ < λ < ... < λ < 0. (5) 1 2 k 2 Furthermore, the roots, equal standing from the ends of the chain (5) mutually inverse: (k) (k) λ ·λ = 1. (6) j k+1−j k If we denote E (λ) = a(k)λs, then the coefficients a(k) of Euler polynomials, as this was shown by k s s s=0 Euler himself, are exprPessed by formula s k+2 a(k) = (−1)j (s+1−j)k+1. s j j=0 (cid:18) (cid:19) X From the definition E (λ) follow following statement. k Lemma 1. For polynomial E (λ) following recurrence relation is valid k E (λ) = (kλ+1)E (λ)+λ(1−λ)E′ (λ), (7) k k−1 k−1 where E (λ) = 1, k = 1,2,.... 0 Lemma 2. The polynomial E (λ) satisfies the identity k 1 E (λ) = λkE , (8) k k λ (cid:18) (cid:19) (k) (k) or otherwise a = a , s = 0,1,2,...,k. s k−s Proof of lemma 1. From (4) we can see, that λ E (λ) = λ−1(1−λ)k+1Dk−1 . (9) k−1 (1−λ)2 Differentiating by λ the polynomial E (λ), we get k−1 λ E (λ) E′ (λ) = −(1−λ)kλ−2(kλ+1)Dk−1 + k . k−1 (1−λ)2 λ(1−λ) Hence and from (9) we obtain, that λ (kλ+1)E (λ)+λ(1−λ)E′ (λ) = (kλ+1)λ−1(1−λ)k+1Dk−1 − k−1 k−1 (1−λ)2 λ −(1−λ)k+1λ−1(kλ+1)Dk−1 +E (λ) = E (λ). (1−λ)2 k k So, lemma 1 is proved. Proof of lemma 2. The lemma we will prove by induction method. When k = 1 from (4) we find E (λ) = λ+1. 1 (k−1) (k−1) We suppose, that when k ≥ 1 the equality a = a , n = 0,1,...,k−1 is fulfilled. We assume, n k−1−n (k−1) that a = 0 for n < 0 and n > k−1. n 3 From (7) we have a(k) = (s+1)a(k−1)+(k−s+1)a(k−1), s s s−1 then, using assumption of induction, we get a(k) = (k−s+1)a(k−1)+(s+1)a(k−1) = (k−s+1)a(k−1)+(s+1)a(k−1) = a(k), k−s k−s k−s−1 s−1 s s and lemma 2 is proved. 3 Proof of theorem 1 For this we will use function x2m−1signx G (x) = . m 2·(2m−1)! To this function we correspond following function of discrete argument: (hβ)2m−1sign(hβ) G [β] = . m 2·(2m−1)! Here we must find such function D [β], which satisfies the equality m hD [β]∗G [β] = δ[β]. (10) m m According to the theory of periodic generalized functions and Fourier transformation in them instead of function D [β] it is convenient to search harrow shaped function [1] m ↽⇁ Dm (x) = Dm[β]δ(x−hβ). β X The equality (10) in the class of harrow shaped functions goes to equation ↽⇁ ↽⇁ h Dm (x)∗ Gm (x) = δ(x), (11) where ↽⇁ Gm (x) = Gm[β]δ(x−hβ). β X It is known [1], that the class of harrow shaped functions and the class of functions of discrete variables are isomorphic. So instead of function of discrete argument D [β] it is sufficiently to investigate m ↽⇁ the function Dm (x), defining from equation (11). Later on we need following well known formulas of Fourier transformation: F[f(p)] = f(x)exp(2πipx)dx, Z F−1[f(p)] = f(x)exp(−2πipx)dx, Z 4 F[f(x)∗ϕ(x)] = F[f(x)]·F[ϕ(x)], F[δ(x)] = 1. Applying to both parts of (11) Fourier transformation, we get ↽⇁ ↽⇁ F[Dm (x)]·F[h Gm (x)] = 1. (12) ↽⇁ Fourier transform of h Gm (p) is well known periodic function, given in R with period h−1 ↽⇁ (−1)m 1 F[h Gm (x)] = (2π)2m |p−h−1β|2m, p 6= h−1β. (13) β X This formula is obtained from the equalities (−1)m 1 F[G (p)] = ( [1, p. 729]) m (2π)2m|p|2m and ↽⇁ Gm (x) = Gm(x) δ(x−hβ). β X Hence, taking into account (12), we get −1 ↽⇁ (−1)m 1 F[Dm (p)] = (2π)2m |p−h−1β|2m . (14) β X   Themainpropertiesofthisfunctioninmultidimensionalcase,appearinginconstructionofdiscrete analogue of the polyharmonic operator, were investigated in [1]. We give some of them, which we will use later on. ↽⇁ 1. Zeros of the function F[Dm (p)] are the points p = h−1β. ↽⇁ 2. The function F[Dm (p)] is periodic with period h−1, real and analytic for all real p. ↽⇁ The function F[Dm (p)] can be represented in the form of Fourier series ↽⇁ F[Dm (p)] = Dˆm[β]exp(2πihβp), (15) β X where h−1 ↽⇁ Dˆm[β] = F[Dm (p)]exp(−2πihβp)dp. (16) Z 0 Applying inverse Fourier transformation to the equality (15), we get harrow shaped function ↽⇁ Dm (x) = Dˆm[β]δ(x−hβ). (17) β X 5 Thus, Dˆ [β] is searching function D [β] of discrete argument or discrete analogue of the operator m m d2m . For finding the function Dˆ [β] calculation of the integral (16) inadvisable. We will find it by dx2m m following way. By virtue of known formula 1 π2 = (p−β)2 sin2πp β X and from the formula (13) we get ↽⇁ −1 1 −h2 F[h G1 (p)] = (2π)2 (p−h−1β)2 = 4sin2πph. β X Hence by differentiating we have d ↽⇁ 2 1 dpF[h G1 (p)] = (2π)2 (p−h−1β)3. β X Thus continuing further, we obtain d2m−2 ↽⇁ (2m−1)! 1 dp2m−2F[h G1 (p)] = − (2π)2m (p−h−1β)2m = β X ↽⇁ = (−1)m−1(2m−1)!(2π)2m−2F[h Gm (p)]. So, ↽⇁ (−1)mh2 d2m−2 1 F[h Gm (p)] = 22mπ2m−2(2m−1)!dp2m−2 sin2πhp . (cid:18) (cid:19) Consider, the expression d2m−2 1 . dp2m−2 sin2πhp (cid:18) (cid:19) Using exp(πihp)−exp(−πihp) sinπhp = , 2i we have d2m−2 −4 d2m−2 exp(2πihp) = −4 . dp2m−2 (exp(πihp)−exp(−πihp))2 dp2m−2 (exp(2πihp)−1)2 (cid:18) (cid:19) (cid:18) (cid:19) We will do change of variables λ = exp(2πihp), then in view of that d dλ d d d = and = 2πihλ , dp dpdλ dp dλ we get d2m−2 = (2πih)2m−2D2m−2, dp2m−2 6 where d d D = λ , D2m−2 = λ D2m−3. dλ dλ Thus, ↽⇁ h2m λ F[h Gm (p)] = (2m−1)!D2m−2(1−λ)2. Hence in virtue of (4) we have ↽⇁ h2m λE2m−2(λ) F[h Gm (p)] = (2m−1)! (1−λ)2m . (18) From (18), according to (12), we obtain ↽⇁ (2m−1)! (1−λ)2m F[Dm (p)] = h2m λE (λ). (19) 2m−2 Now in order to obtain Fourier-series expansion, we will do following. We divide the polynomial (1−λ)2m to the polynomial λE (λ): 2m−2 (1−λ)2m P (λ) (2m−2) 2m−2 = λ−2m−a + , (20) 2m−2 2m−3 λE (λ) λ a(2m−2)λs 2m−2 s s=0 P where P (λ) is a polynomial of degree 2m−2. It is not difficult to see, that the rational fraction 2m−2 P (λ) 2m−2 is proper fraction, i.e. degree of the polynomial P (λ) is less than degree of the 2m−2 λE (λ) 2m−2 polynomial λE (λ). Since the roots of the polynomial E (λ) are real and different, then the 2m−2 2m−2 P (λ) 2m−2 rational fraction is expanded to the sum of elementary fractions. Searching expansion has λE (λ) 2m−2 following form m−1 m−1 P (λ) A A A 2m−2 0 1,k 2,k = + + , (21) λE (λ) λ λ−λ λ−λ 2m−2 1,k 2,k k=1 k=1 X X where A , A , A are unknown coefficients, λ are the roots of the polynomial E (λ), in 0 1,k 2,k 1,k 2m−2 modulus less than unity, and λ are the roots of the polynomial E (λ), in modulus greater than 2,k 2m−2 unity. By (21) the equality (20) takes the form (1−λ)2m A (2m−2) 0 = λ−2m−a + + 2m−2 2m−3 λ λ a(2m−2)λs s s=0 P m−1 m−1 A A 1,k 2,k + + . (22) λ−λ λ−λ 1,k 2,k k=1 k=1 X X Reducing to the common denominator and omitting it, we get (1−λ)2m = λ2E (λ)−λ(2m+a(2m−2))E (λ)+ 2m−2 2m−3 2m−2 7 m−1 m−1 A λE (λ) A λE (λ) 1,k 2m−2 2,k 2m−2 +A E (λ)+ + . (23) 0 2m−2 λ−λ λ−λ 1,k 2,k k=1 k=1 X X Assuming in the equality (23) consequently λ = 0, λ = λ and λ = λ , we find 1,k 2,k 1 = E (0)A ; (1−λ )2m = λ E′ (λ )A ; 2m−2 0 1,k 1,k 2m−2 1,k 1,k (1−λ )2m = λ E′ (λ )A . 2,k 2,k 2m−2 2,k 2,k Hence (1−λ )2m (1−λ )2m 1,k 2,k A = 1; A = ; A = . 0 1,k λ E′ (λ ) 2,k λ E′ (λ ) 1,k 2m−2 1,k 2,k 2m−2 2,k Using (6), we have A = (1− λ21,k)2m = (λ2,k −1)2m . 1,k λ−1E′ ( 1 ) λ2m−1E′ ( 1 ) 2,k 2m−2 λ2,k 2,k 2m−2 λ2,k In virtue of (7) we obtain 1 1 1 1 1− E′ ( ) = E ( ), λ λ 2m−2 λ 2m−1 λ 2,k (cid:18) 2,k(cid:19) 2,k 2,k λ (1−λ )E′ (λ ) = E (λ ), 2,k 2,k 2m−2 2,k 2m−1 2,k hence E ( 1 )λ2 E′ ( 1 ) = 2m−1 λ2,k 2,k, 2m−2 λ λ −1 2,k 2,k E (λ ) E′ (λ ) = 2m−1 2,k . 2m−2 2,k λ (1−λ ) 2,k 2,k From here application of the lemma 2 gives −A (1−λ )2m+1 2,k 1,k A = , A = . (24) 1,k λ2 1,k E (λ ) 2,k 2m−1 1,k Since |λ | < 1 and |λ | > 1, then 1,k 2,k m−1 m−1 A A 1,k 2,k and λ−λ λ−λ 1,k 2,k k=1 k=1 X X can be represented as Laurent series on the circle |λ2| = 1: m−1 m−1 m−1 ∞ β A 1 A 1 λ 1,k 1,k 1,k = = A , (25) k=1 λ−λ1,k λ k=1 1− λ1λ,k λ k=1 1,kβ=0(cid:18) λ (cid:19) X X X X m−1 m−1 m−1 ∞ β A A A λ 2,k 2,k 2,k = − = − . (26) Xk=1 λ−λ2,k Xk=1 λ2,k(1− λ2λ,k) Xk=1 λ2,k βX=0(cid:18)λ2,k(cid:19) 8 Putting (25), (26) to (22) and taking into account λ = exp(2πihp) from (19), (20), we obtain ↽⇁ (2m−1)! (2m−2) F[Dm (p)] = h2m exp(2πihp)−2m−a2m−3 + " m−1 ∞ β +exp(−2πihp)+ A λ exp(−2πihp(β +1))− 1,k 1,k k=1 β=0 X X ∞ β 1 −A λ−1 exp(2πihpβ) . 2,k 2,k λ β=0(cid:18) 2,k(cid:19) !# X ↽⇁ Thus, searching Fourier series for F[Dm (p)] have following form ↽⇁ F[Dm (p)] = Dm[β]exp(2πihβp), β X where m−1 −β−1 A λ for β ≤ −2, 1,k 1,k  k=1  P m−1  1+ A1,k for β = −1,   k=1  (2m−1)!  P m−1 Dm[β] = h2m  −22m−1− k=1 A2,kλ−2,k1 for β = 0,  m−1 P 1− A λ−2 for β = 1, 2,k 2,k   k=1   −m−P1A2,kλ−2,kβ−1 for β ≥ 2.  k=1   P With the help (24) the function D [β]we rewrite in the form m  m−1 (1−λ )2m+1λ|β| 1,k 1,k for |β| ≥ 2,  λkE2m−1(λ1,k) Dm[β] = (2mh2−m1)!  1Xk=+1mXk=−11 (1E−2mλ−11,k(λ)21m,k+)1 for |β| = 1, m−1 (1−λ )2m+1  −22m−1+ 1,k for β = 0.  λ1,kE2m−1(λ1,k)  Xk=1    We note, that  D [β] = D [−β]. m m Theorem 1 is proved completely. 9 4 Proofs of properties. Proof of property 1. Following is takes placed ↽⇁[1] F[∆ (p)] = −4sin2πph. 2 Indeed, ↽⇁[1] ↽⇁ ↽⇁ ↽⇁ ∆ (x) = δ (x+1)−2 δ (x)+2 δ (x−1) = δ(x+h)−2δ(x)+δ(x−h). 2 By definition of Fourier transformation we have ↽⇁[1] ↽⇁[1] F[∆ (p)] = exp(2πipx) ∆ (x)dx = −4sin2πph. 2 2 Z Hence consequently we obtain m times ↽⇁[m] ↽⇁[1] ↽⇁[1] ↽⇁[1] F[∆ (p)] = F[∆ (p)∗ ∆ (p)∗...∗ ∆ (p)] = (−4)msin2mπph. (27) 2 2 2 2 z }| { Immediately we have (1−λ)2 m = sin2mπhp. (28) −4λ (cid:20) (cid:21) By virtue of (27) and (28) the formula (19) takes form ↽⇁ (2m−1)! λm−1 ↽⇁[m] F[Dm (p)] = h2m E (λ)F[∆2 (p)]. (29) 2m−2 Now expanding rational fraction λm−1 to the sum of elementary fractions, we have E2m−2(λ) λm−1 m−1 B B 1,k 2,k = + , (30) E (λ) λ−λ λ−λ 2m−2 k=1 (cid:20) 1,k 2,k(cid:21) X where λm−1 λm−1 1,k 2,k B = , B = . 1,k E′ (λ ) 2,k E′ (λ ) 2m−2 1,k 2m−2 2,k Since |λ | < 1 and |λ | > 1, then expanding B1,k and B2,k to the Laurent series on the circle 1,k 2,k λ−λ1,k λ−λ2,k |λ| = 1, we find ∞ β B 1 B B λ 1,k 1,k 1,k 1,k = · = , (31) λ−λ1,k λ 1− λ1λ,k λ β=0(cid:18) λ (cid:19) X ∞ β B B B λ 2,k 2,k 2,k = − = − . (32) λ−λ2,k λ2,k(1− λ2λ,k) λ2,k βX=0(cid:18)λ2,k(cid:19) 10