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Acta Mathematica Scientia 2014,34B(4):1157–1165 http://actams.wipm.ac.cn APPROXIMATION BY COMPLEX SZÁSZ-DURRMEYER OPERATORS IN COMPACT DISKS∗ Sorin G. GAL Department of Mathematics and Computer Science, University of Oradea Str. Universitatii No. 1, 410087 Oradea, Romania E-mail : galso@uoradea.ro Vijay GUPTA† Department of Mathematics, Netaji Subhas Institute of Technology Sector 3 Dwarka, New Delhi-110078, India E-mail : vijaygupta2001@hotmail.com Abstract In the present paper, we deal with the complex Szász-Durrmeyer operators and study Voronovskaja type results with quantitative estimates for these operators attached to analytic functions of exponential growth on compact disks. Also, the exact order of approximation is found. Key words complex Szász-Durrmeyer operators; Voronovksaja type result; exact order of approximation in compact disks; simultaneous approximation 2010 MR Subject Classification 1 41A25; 41A28; 30E10 Introduction Concerning the convergence of the Bernstein polynomials in the complex plane, Bernstein [10] proved that if f : G → C is analytic in the open set G ⊂ C, with D1 ⊂ G (with D1 = {z ∈ n P n k n−k C : |z| < 1}), then the complex Bernstein polynomials Bn (f )(z) = f (k/n), k z (1 − z) k=0 uniformly converges to f in D1 . Voronovskaja-type results with quantitative estimates for the complex Bernstein, complex q-Bernstein, complex Baskakov, complex Szász (more precisely Favard-Szász-Mirakjan), complex Bernstein-Kantorovich, complex Balázs-Szabados and complex Stancu-Kantorovich operators attached to analytic functions on compact disks and the exact order of simultaneous approximation by these complex operators were collected by the recent book Gal [1]. Recall that for f ∈ C[0, ∞), x ∈ [0, ∞) and n ∈ N, the classical Szász operator of real variable is defined as ∞ ν X Sn (f )(x) = sn,ν (x)f , n ν=0 ∗ Received March 25, 2013; revised December 9, 2013. author: Vijay GUPTA. † Corresponding 1158 ACTA MATHEMATICA SCIENTIA Vol.34 Ser.B ν where sn,ν (x) = e−nx (nx) ν! . Also, the Szász -Stancu operator of real variable x ∈ [0, ∞) is defined by ∞ X ν +α α,β Sn (f )(x) = sn,ν (x)f , n+β ν=0 where α, β are two given parameters satisfying the conditions 0 ≤ α ≤ β. For α = β = 0 we recapture the classical Szász operator. In Gal [2], [5], Gupta-Verma [9], the overconvergence properties of the complex variants for the above Szász type operators were studied. Also, in Gal [3], [4], Gal-Gupta [6], [7], Gal-Gupta-Mahmudov [8], the overconvergence properties of various complex variants for Durrmeyer-type operators were intensively studied. In the present paper, we deal with the Szász-Durrmeyer operators Z +∞ ∞ X SDn (f )(z) = n sn,ν (z) sn,ν (t)f (t)dt , 0 ν=0 introduced and studied in the case of real variable in Mazhar-Totik [11]. Throughout the article, we consider DR = {z ∈ C : |z| < R}. In the present paper, we study the rate of approximation of analytic functions and the Voronovskaja type result for the Szász-Durrmeyer operator SDn (f, z). Also, the exact order of approximation by this operator is obtained. 2 Auxiliary Results First, we need the following auxiliary results. S S Lemma 2.1 Suppose that f : [R, +∞) DR is continuous in [R, +∞) DR , analytic in DR and there exists B, C > 0 such that |f (x)| ≤ CeBx , for all x ∈ [R, +∞). Denoting ∞ ∞ P P f (z) = ck z k , z ∈ DR , we have SDn (f )(z) = ck SDn (ek )(z) for all z ∈ DR and n > B. k=0 Proof k=0 For any m ∈ N and 0 < r < R, let us define fm (z) = m X cj z j if |z| ≤ r and fm (x) = f (x) if x ∈ (r, +∞). j=0 Since |fm (z)| ≤ ∞ P |cj | · rj := Cr , for all |z| ≤ r and m ∈ N, f is continuous on [r, R], from the j=0 hypothesis on f it is clear that for any m ∈ N it follows |fm (x)| ≤ Cr,R eBx , for all x ∈ [0, +∞). This implies that for each fixed m, n ∈ N, n > B and z, Z ∞ ∞ j X (n|z|)j −nz −nt n j Bt |SDn (fm )(z)| ≤ Cr,R |e | n e · t e dt j! j! 0 j=0 = Cr,R |e−nz | ∞ X (n|z|)j j=0 j! · nj+1 < ∞, (n − B)j+1 since by the ratio criterium the last series is convergent. Therefore SDn (fm )(z) is well-defined. Denoting fm,k (z) = ck ek (z) if |z| ≤ r and fm,k (x) = f (x) m+1 m P fm,k is of exponential growth on [0, ∞) and that fm (z) = k=0 if x ∈ (r, ∞), it is clear that each fm,k (z). Since from the linearity No.4 S.G. Gal & V. Gupta: COMPLEX SZÁSZ-DURRMEYER OPERATORS of Sn we have SDn (fm )(z) = m X 1159 ck SDn (ek )(z), for all |z| ≤ r, k=0 it suffices to prove that lim SDn (fm )(z) = SDn (f )(z) for any fixed n ∈ N and |z| ≤ r. But m→∞ this is immediate from lim kfm − f kr = 0, from kfm − f kB[0,+∞) ≤ kfm − f kr and from the m→∞ inequality |SDn (fm )(z) − SDn (f )(z)| ≤ |e−nz | · en|z| · kfm − f kB[0,∞) ≤ Mr,n kfm − f kr , valid for all |z| ≤ r. Here k · kB[0,+∞) denotes the uniform norm on C[0, +∞)-the space of all complex-valued bounded functions on [0, +∞). Lemma 2.2 Denoting ek (z) = z k and Tn,k (z) = SDn (ek )(z), we have the recurrence formula n k+1 ′ Tn,k (z) = Tn,k+1 (z) − n + Tn,k (z). z z We have Proof Tn,k (z) = n =n ∞ X ν=0 ∞ X ν=0 sn,ν (z) Z sn,ν (z) +∞ e−nt 0 nν · ν! Z +∞ (nt)ν k t dt ν! e−nt tν+k dt 0 This immediately implies = ∞ 1 X (ν + k)! sn,ν (z) . nk ν=0 ν! ′ ∞ 1 X (ν + k)! −nz (nz)ν e · nk ν=0 ν! ν! ∞ (nz)ν (nz)ν−1 1 X (ν + k)! −ne−nz · + e−nz n · = k n ν=0 ν! ν! (ν − 1)! ′ Tn,k (z) = ∞ n 1 X (ν + k + 1)! −nz (nz)ν ν = −nTn,k (z) + · k+1 ·e · · z n ν! ν! ν + k+1 ν=0 = −nTn,k (z) + ∞ n n k + 1 X (ν + k + 1)! −nz (nz)ν 1 Tn,k+1 (z) − · k+1 ·e · · z z n ν! ν! ν +k+1 ν=0 n k+1 Tn,k+1 (z) − Tn,k (z), z z which proves the recurrence in statement. = −nTn,k (z) + Remark 2.3 Tn,k (z) is a polynomial of degree k. Indeed, by direct calculation we get Tn,0 (z) = SDn (e0 )(z) = 1 and since by Lemma 2.2 it follows z ′ nz + k + 1 (z) + Tn,k (z), Tn,k+1 (z) = Tn,k n n taking above step by step k = 1, 2, · · · , by mathematical induction we easily get that Tn,k (z) is a polynomial of degree k. 3 Main Results Our first main result is the following theorem for upper bound. 1160 ACTA MATHEMATICA SCIENTIA Vol.34 Ser.B Theorem 3.1 Let 1 < R < +∞ and suppose that f : [R, +∞) ∪ DR → C is continuous ∞ P in (R, +∞) ∪ DR , analytic in DR i.e., f (z) = ck z k , for all z ∈ DR , and suppose that there k=0 k 1 A exist M > 0 and A ∈ ( R , 1), with the property that |ck | ≤ M (2k)! , for all k = 0, 1, · · · (which A|z| Bx implies |f (z)| ≤ M e for all z ∈ DR ) and |f (x)| ≤ Ce for all x ∈ [R, +∞). 1 . Then for all |z| ≤ r and n ∈ N, with n > B, we have (i) Let 1 ≤ r < A |SDn (f )(z) − f (z)| ≤ where Cr,A = M 2r ∞ P Cr,A , n (rA)k < ∞; k=1 (ii) If 1 ≤ r < r1 < 1 A are arbitrary fixed, then for all |z| ≤ r and n, p ∈ N with n > B, |SDn(p) (f )(z) − f (p) (z)| ≤ p!r1 Cr1 ,A , n(r1 − r)p+1 where Cr1 ,A is given as at the above point (i). Proof (i) By using the recurrence relation of Remark 2.3, we have nz + k + 1 z ′ Tn,k (z) + Tn,k (z) n n for all z ∈ C, k ∈ {0, 1, 2, · · · }, n ∈ N . From this we immediately get the recurrence formula Tn,k+1 (z) = Tn,k (z) − z k = nz + k 2k − 1 k−1 z [Tn,k−1 (z) − z k−1 ]′ + [Tn,k−1 (z) − z k−1 ] + z n n n for all z ∈ C, k, n ∈ N . Now for 1 ≤ r < R, if we denote the norm-k.kr in C(Dr ), where Dr = {z ∈ C : |z| ≤ r}, then by a linear transformation, the Bernstein’s inequality in the closed unit disk becomes |Pk′ (z)| ≤ kr kPk kr , for all |z| ≤ r, where Pk (z) is a polynomial of degree ≤ k. Thus from the above recurrence relation, we get r k − 1 nr + k 2k − 1 k−1 .kTn,k−1 − ek−1 kr + kTn,k−1 − ek−1 kr + r , n r n n which implies the relation 2k − 1 k−1 2k − 1 kTn,k − ek kr ≤ r + .kTn,k−1 − ek−1 kr + r . n n kTn,k − ek kr ≤ In what follows we prove the result by mathematical induction with respect to k (with n ≥ 1 supposed to be fixed arbitrarily), that this recurrence implies (2k)! k−1 r for all k ≥ 1, n ≥ 1. 2n Indeed for k = 1 and n ∈ N, the left hand side is 1/n and the right hand side is 1/n. Suppose that it is true for k, the above recurrence relation implies that 2k + 1 (2k)! k−1 2k + 1 k kTn,k+1 − ek+1 kr ≤ r + . r + r . n 2n n kTn,k − ek kr ≤ It remains to prove that 2k + 1 (2k)! k−1 2k + 1 k (2(k + 1) k r+ . r + r ≤ r , n 2n n 2n or after simplifications, equivalently to 2k + 1 r+ .(2k)! + 2r(2k + 1) ≤ (2(k + 1))!r. n No.4 1161 S.G. Gal & V. Gupta: COMPLEX SZÁSZ-DURRMEYER OPERATORS It is easy to see by mathematical induction that this last inequality holds true for all k ≥ 1 and n ∈ N. From the hypothesis on f (i.e. |f (x)| ≤ max{M, C}emax{A,B}x , for all x ∈ R+ ), it follows that SDn (f )(z) is analytic in DR . Thus, by Lemma 2.1 we can write SDn (f )(z) = ∞ X ck SDn (ek )(z) = k=0 ∞ X ck Tn,k (z), for all z ∈ DR , n > B, k=0 which from the hypothesis on ck immediately implies for all |z| ≤ r |SDn (f )(z) − f (z)| ≤ ∞ X |ck |.|Tn,k (z) − ek (z)| ≤ k=1 ∞ X k=1 ∞ M X Cr,A = (rA)k = , 2nr n M Ak (2k)! k−1 r (2k)! 2n k=1 where Cr,A = M 2r ∞ P (rA)k < ∞ for all 1 ≤ r < k=1 1 A, taking into account that the series ∞ P uk is k=1 uniformly convergent in any compact disk included in the open unit disk. (ii) Denoting by γ the circle of radius r1 > r and center 0, since for any |z| ≤ r and v ∈ γ, we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N with n > B, we have Z 2πr1 p! SDn (f )(v) − f (v) Cr1 ,A p! (p) (p) |SDn (f )(z) − f (z)| = dv ≤ p+1 2π (v − z) n 2π (r1 − r)p+1 γ Cr ,A p!r1 = 1 , n (r1 − r)p+1 which proves (ii) and the theorem. The following Voronovskaja type result holds. Theorem 3.2 Let 2 < R < +∞ and suppose that f : [R, +∞) ∪ DR → C is continuous ∞ P ck z k , for all z ∈ DR , and that there exist in (R, +∞) ∪ DR , analytic in DR i.e., f (z) = k=0 k 1 A M > 0 and A ∈ ( R , 1), with the property that |ck | ≤ M (2k)! , for all k = 0, 1, · · · (which implies A|z| Bx |f (z)| ≤ M e for all z ∈ DR ) and |f (x)| ≤ Ce , for all x ∈ [R, +∞). n o Ar If 1 ≤ r < r + 1 < A1 then for all |z| ≤ r and n ∈ N with n > max 1−Ar , B , we have SDn (f )(z) − f (z) − 1 f ′ (z) − z f ′′ (z) ≤ Cr,A,M (f ) , n n n2 where Cr,A,M (f ) = 2M r(1−Ar) + 4M (r+1)2 ∞ P (k − 1)[A(r + 1)]k < ∞. k=2 We denote ek (z) = z k , k = 0, 1, 2, · · · and Tn,k (z) = SDn (ek , z). By Lemma 2.1, ∞ P we can write SDn (f )(z) = ck Tn,k (z) for all n > B. Also Proof k=0 ′′ ′ zf (z) + f (z) z = n n Thus ∞ X k=2 ck k(k − 1)z k−2 + ∞ ∞ k=1 k=1 1X 1X ck kz k−1 = ck [k(k − 1) + k]z k−1 . n n ∞ X k 2 z k−1 SDn (f )(z) − f (z) − 1 f ′ (z) − z f ′′ (z) ≤ |c | T (z) − e (z) − k n,k k n n n k=1 1162 ACTA MATHEMATICA SCIENTIA Vol.34 Ser.B for all z ∈ DR , n ∈ N, n > B. By Remark 2.3, for all n ∈ N, z ∈ C and k = 0, 1, 2, · · · , we have Tn,k+1 (z) = z ′ nz + k + 1 Tn,k (z) + Tn,k (z). n n If we denote k 2 z k−1 , n then it is obvious that Ek,n (z) is a polynomial of degree less than or equal to k and by simple computation and the use of above recurrence relation, we are led to Ek,n (z) = Tn,k (z) − ek (z) − z ′ nz + k 2z k−2 (k − 1)3 Ek−1,n (z) + Ek−1,n (z) + n n n2 for all k ≥ 1, n ∈ N and |z| ≥ r. Using the estimate in the proof of Theorem 3.1, we have Ek,n (z) = |Tn,k (z) − ek (z)| ≤ ((2k)!)rk−1 2n for all k, n ∈ N, |z| ≤ r, with 1 ≤ r. For all k, n ∈ N, k ≥ 1 and |z| ≤ r, it follows r ′ k 2|z|k−2 (k − 1)3 |Ek,n (z)| ≤ |Ek−1,n (z)| + r + |Ek−1,n (z)| + . n n n2 ′ Now we shall find the estimation of |Ek−1,n (z)| for k ≥ 1. Taking into account the fact that Ek−1,n (z) is a polynomial of degree ≤ k − 1, we have (k − 1)2 ek−2 k−1 k−1 ′ |Ek−1,n (z)| ≤ kEk−1,n kr ≤ kTn,k−1 (z) − ek−1 (z)kr + r r n r k − 1 (2(k − 1))!rk−2 rk−2 (k − 1)2 2(k − 1)(2(k − 1))!rk−3 ≤ + ≤ . r 2n n n Thus and r ′ 2(k − 1)(2(k − 1))!rk−2 |Ek−1,n (z)| ≤ n n2 2(k − 1)(2(k − 1))!rk−2 k 2rk−2 (k − 1)3 |Ek,n (z)| ≤ + r + |E (z)| + , k−1,n n2 n n2 for all |z| ≤ r, k ≥ 1, n ∈ N, which implies 4rk−2 (2k)! k + r+ |Ek−1,n (z)| for all |z| ≤ r. |Ek,n (z)| ≤ n2 n For 2 ≤ k ≤ n and |z| ≤ r, taking into account that r + k/n ≤ r + 1, we get 4rk−2 (2k)! + (r + 1)|Ek−1,n (z)|. n2 But E0,n (z) = E1,n = 0, for any z ∈ C and therefore by writing last inequality for 2 ≤ k ≤ n, we easily obtain step by step the following |Ek,n (z)| ≤ |Ek,n (z)| ≤ k 4(r + 1)k−2 X 4(r + 1)k (k − 1)(2k)! (2j)! ≤ . n2 (r + 1)2 n2 j=2 No.4 S.G. Gal & V. Gupta: COMPLEX SZÁSZ-DURRMEYER OPERATORS 1163 It follows that SDn (f )(z) − f (z) − 1 f ′ (z) − z f ′′ (z) n n n ∞ X X ≤ |ck | · |Ek,n (z)| + |ck | · |Ek,n (z)| k=2 ≤ k=n+1 4M (r + 1)2 n2 But ∞ X |ck | · |Ek,n | ≤ k=n+1 ≤ ∞ X k=n+1 ∞ X ∞ X (k − 1)(A(r + 1))k + k=2 ∞ X |ck | · |Ek,n |. k=n+1 |ck | · [|Tn,k (z) − ek (z)| + k 2 |z|k−1 /n] |ck | · ∞ ∞ X k 2 rk−1 (2k)!rk−1 2M X + |ck | · ≤ (Ar)k n n rn k=n+1 k=n+1 k=n+1 2M (Ar)n+1 2M 1 = · ≤ · 2 r(1 − Ar) n r(1 − Ar) n o n Ar . for all |z| ≤ r and n > max B, 1−Ar Concluding, we obtain SDn (f )(z) − f (z) − 1 f ′ (z) − z f ′′ (z) n n ≤ ∞ X 4M 2M 1 (k − 1)(A(r + 1))k + · 2, 2 2 (r + 1) n r(1 − Ar) n k=2 where for (r + 1)A < 1 we obviously have the series convergent. The following exact order of approximation can be obtained. Theorem 3.3 (i) In the hypothesis of Theorem 3.2, if f is not a polynomial of degree ≤ 0 then for all 1 ≤ r < r + 1 < R we have 1 Ar kSDn (f ) − f kr ∼ , for all n > max ,B , n 1 − Ar where the constants in the equivalence depend only on f and r. (ii) In the hypothesis of Theorem 3.2, if r < r1 < r1 + 1 < 1/A and if f is not a polynomial of degree ≤ p, (p ≥ 1) then 1 Ar (p) (p) kSDn (f ) − f kr ∼ , for all n > max ,B , n 1 − Ar where the constants in the equivalence depend only on f , r, r1 and p. o n Ar , B , we can write Proof (i) For all |z| ≤ r and n ∈ N with n > max 1−Ar 1 ′ 1 f ′ (z) + zf ′′ (z) SDn (f )(z) − f (z) = f (z) + zf ′′ (z) + · n2 SDn (f )(z) − f (z) − n n n Applying the inequality kF + Gk ≥ | kF k − kGk | ≥ kF k − kGk, we obtain 1 1 2 f ′ + e1 f ′′ ′ ′′ . kSDn (f ) − f kr ≥ kf + e1 f kr − · n SDn (f ) − f − n n n r 1164 ACTA MATHEMATICA SCIENTIA Vol.34 Ser.B Since f is not a polynomial of degree ≤ 0 in DR , we get kf ′ + e1 f ′′ kr > 0. Indeed, supposing the contrary, it follows that f ′ (z) + zf ′′ (z) = 0, for all |z| ≤ r. The last equality is equivalent to [zf ′ (z)]′ = 0. Therefore zf ′ (z) = C, with C is a constant, that is f ′ (z) = Cz , for all |z| ≤ r with z 6= 0. But since f is analytic in Dr and r ≥ 1, we necessarily have C = 0 (contrariwise, we would get that f ′ (z) is not differentiable at z = 0, which is impossible because f ′ too is analytic in Dr , with r ≥ 1), which implies f ′ (z) = 0 and f (z) = c for all z ∈ Dr , a contradiction with the hypothesis. Now by Theorem 3.2, we have f ′ + e1 f ′′ Ar 2 n SDn (f ) − f − ≤ Cr,A,M (f ), for all n > max B, 1 − Ar . n r n o Ar Thus, there exists n0 > max B, 1−Ar (depending on f and r only) such that for all n ≥ n0 , we have f ′ + e1 f ′′ 1 2 ′ ′′ ≥ 1 kf ′ + e1 f ′′ k , kf + e1 f kr − · n SDn (f ) − f − r n n 2 r which implies that kSDn (f ) − f kr ≥ 1 kf ′ + e1 f ′′ kr 2n for all n ≥ n0n . o M (f ) Ar For max B, 1−Ar < n ≤ n0 − 1, we get kSDn (f ) − f kr ≥ r,n with Mr,n (f ) = n n · kSDn (f ) − f kr > 0 (since kSDn (f ) − f kr = 0 for a certain n is valid only for f a constant function, contradicting the hypothesis on f ). Therefore, finally we have Cr (f ) kSDn (f ) − f kr ≥ n n o Ar for all n > max B, 1−Ar , where 1 Cr (f ) = min Mr,n (f ), · · · , Mr,n0 −1 (f ), kf ′ + e1 f ′′ kr , n0 −1≥n>Ar/(1−Ar) 2 which combined with Theorem 3.1, (i), proves the desired conclusion. (ii) The upper estimate is exactly Theorem 3.1, (ii), therefore it remains to prove the lower estimate. Denote by Γ the circle of radius r1 and center 0. By the Cauchy’s formulas for all |z| ≤ r and n ∈ N we get Z p! SDn (f )(v) − f (v) SDn(p) (f )(z) − f (p) (z) = dv, 2πi Γ (v − z)p+1 where |v − z| ≥ r1 − r for all |z| ≤ r and v n ∈ Γ. o Ar For all v ∈ Γ and n ∈ N with n > max B, 1−Ar , we get 1 1 2 f ′ (v) + vf ′′ (v) SDn (f )(v) − f (v) = f ′ (v) + vf ′′ (v) + n Sn (f )(v) − f (v) − , n n n which replaced in the Cauchy’s formula implies SDn(p) (f )(z) − f (p) (z) No.4 1165 S.G. Gal & V. Gupta: COMPLEX SZÁSZ-DURRMEYER OPERATORS Z n2 SDn (f )(v) − f (v) − f ′ (v)+vf ′′ (v) n 1 p! f (v) + vf (v) 1 p! = dv + · dv n 2πi Γ (v − z)p+1 n 2πi Γ (v − z)p+1 Z n2 SDn (f )(v) − f (v) − f ′ (v)+vf ′′ (v) n 1 ′ 1 p! (p+1) [zf (z)] + · dv . = n n 2πi Γ (v − z)p+1 Z ′ ′′ Passing to the norm k · kr , for all n ∈ N we obtain kSDn(p) (f ) − f (p) kr Z n2 SDn (f )(v) − f (v) − 1 1 p! (p+1) ≥ [e1 f ′ ] − n n (v − z)p+1 r 2π Γ f ′ (v)+vf ′′ (v) n dv , r where by Theorem 3.2, it follows Z n2 SDn (f )(v) − f (v) − f ′ (v)+vf ′′ (v) p! n dv 2π p+1 (v − z) Γ r 2 ′ ′′ 2πr1 n f + e f p! 1 SDn (f ) − f − ≤ · 2π (r1 − r)p+1 n r1 ! ∞ 2M 4M X p!r1 k ≤ + (k − 1)[A(r + 1)] · . 2 r(1 − Ar) (r + 1) (r1 − r)p+1 k=2 (p+1) > 0. Indeed, supposing the contrary it follows Now, by hypothesis on f we have [e1 f ′ ] r ′ that zf (z) is a polynomial of degree ≤ p, which by the analyticity of f obviously implies that f is a polynomial of degree ≤ p, a contradiction with the hypothesis. For the rest of the proof, reasoning exactly as in the proof of the above point (i), we immediately get the required conclusion. References [1] Gal S G. Approximation by Complex Bernstein and Convolution Type Operators. Singapore, Hong Kong, London, New Jersey: World Scientific Publ Co, 2009 [2] Gal S G. Approximation and geometric properties of complex Favard-Szász-Mirakjan operators in compact disks. 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