International Journal of Analysis and Applications Volume 19, Number 2 (2021), 180-192 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-19-2021-180 RADAU QUADRATURE FOR AN ALMOST QUASI-HERMITE-FEJÉR-TYPE INTERPOLATION IN RATIONAL SPACES SHRAWAN KUMAR1, NEHA MATHUR2, VISHNU NARAYAN MISHRA3,∗, PANKAJ MATHUR1,∗ 1Department of Mathematics and Astronomy, University of Lucknow, Lucknow 2Department of Mathematics, Career Convent Degree College, Lucknow 3Department of Mathematics, Indira Gandhi National Tribal University, Lalpur, Amarkantak, Anuppur, Madhya Pradesh 484 887, India ∗Corresponding authors: vishnunarayanmishra@gmail.com, pankaj mathur14@yahoo.co.in Abstract. In this paper, we have studied an almost quasi Hermite-Fejér-type interpolation in rational spaces. A Radau type quadrature formula has also been obtained for the same. 1. Introduction Hermite Fejér and Quasi-Hermite-Fejér-type interpolation processes has been a subject of interest for several mathematicians. In almost all the cases the interpolatory polynomials are considered on the nodes which are the zeros of certain classical orthogonal polynomials. The main idea of the present paper is to construct a rational interpolation process and its corresponding quadrature formula with prescribed nodes based on the Chebyshev Markov fractions. Received November 1st, 2019; accepted March 2nd, 2020; published February 1st, 2021. 2010 Mathematics Subject Classification. Primary 05C38, 15A15; Secondary 05A15, 15A18. Key words and phrases. Almost Quasi-Hermite-Fejér-type interpolation; Radau-type quadrature; rational space; prescribed poles; Chebyshev-Markov fractions. ©2021 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 180 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-19-2021-180 Int. J. Anal. Appl. 19 (2) (2021) 181 Chebyshev and Markov introduced rational cosine and sine fractions [3] which generalizes Chebyshev polynomials, possesses many similar properties ( [2, 10, 11]) and are called Chebychev–Markov rational frac- tions. Different aspects of the rational generalization of Chebyshev polynomials are discussed in many works ( [1, 12]). In 1962, Rusak [9] initiated the study of interpolation processes by means of rational functions on the interval [−1, 1]. The nodes were taken to be the zeros of Chebyshev–Markov rational fractions. In [6] rational interpolation functions of Hermite-Fejér-type were constructed [7]. Min [4] was the first to consider the rational quasi-Hermite-type interpolation. He constructed the interpolatory function and proved its uniform convergence for the continuous functions on the segment with the restriction that the poles of the approximating rational functions should not have limit points on the interval [−1, 1]. Recently, based on the ideas of [6] and using method that was different from that of [4], Rouba et. al. ( [5], [8]) revisited the rational interpolation functions of Hermite-Fejér-type. They also proved the uniform convergence of the interpolation process for the function f ∈ C[−1, 1] and obtained explicitly its corresponding Lobatto type quadrature formula. In this paper, we have considered an almost quasi-Hermite-Fejér-type interpolation process on the zeros of the rational Chebyshev-Markov sine fraction on the semi closed interval (−1, 1], that is, when the interpola- tory condition is prescribed only at one of the end points. A Radau type quadrature formula corresponding to the interpolation process has also been obtained. 2. Preliminaries Consider a set of points ak, k = 0, 1, · · · , 2n− 1 which are either real and ak ∈ (−1, 1) or be paired by complex conjugation. Also let Un(x) be the rational Chebyshev-Markov sine fraction, Un(x) = sin µ2n(x)√ 1 −x2 (2.1) where, µ2n(x) = 1 2 2n−1∑ k=0 arccos x + ak 1 + akx ,(2.2) µ′2n(x) = − λ2n(x)√ 1 −x2 ,(2.3) λ2n(x) = 1 2 2n−1∑ k=0 √ 1 −a2k 1 + akx , n ∈ N.(2.4) Also U′n(x) = −cos µ2n(x)λ2n(x) √ 1 −x2 + x sin µ2n(x) (1 −x2)3/2 (2.5) Int. J. Anal. Appl. 19 (2) (2021) 182 and U′n(xk) = − λ2n(xk) (1 −x2k) .(2.6) Let R2n−1(a0,a1, · · · ,a2n−1) be a rational space defined as R2n−1(a0,a1, · · · ,a2n−1) := { p2n−1(x)∏2n−1 k=0 (1 + akx) } (2.7) where p2n−1(x) is a polynomial of degree ≤ 2n−1 and {ak}2n−1k=0 are real and belong to [−1, 1] or are paired by complex conjugation. The rational fraction Un(x) can be expressed as Un(x) = Pn−1(x)√ Π2n−1k=0 (1 + akx) where Pn−1(x) is an algebraic polynomial of degree n−1 with real coefficient. The fraction Un(x) has n−1 zeros on the interval (−1, 1) given by, −1 < xn−1 < xn−2 < · · · < x2 < x1 < 1, µ2n(xk) = kπ, k = 1, 2, · · · ,n− 1, where µ2n(x) is given by (2.2). Also, the rational function λ2n(x), given by (2.4), can be expressed as λ2n(x) = q2n−1(x)∏2n−1 k=0 (1 + akx) where q2n−1(x) is a polynomial of degree atmost 2n− 1. It has no zeros on [−1, 1]. 3. Almost Quasi-Hermite-Fejér-type interpolation Let x0 = 1. Then for any function f ∈ C(−1, 1] the almost quasi type Hermite interpolation function Hn(x,f) satisfying the conditions Hn(xk,f) = f(xk), k = 0, 1, · · · ,n− 1,(3.1) H′n(xk,f) = yk, k = 1, 2, · · · ,n− 1,(3.2) is given by (3.3) Hn(x,f) = n−1∑ k=0 f(xk)Ak(x) + n−1∑ k=1 ykBk(x), where yk,k = 1, 2, · · · ,n−1 are arbitrarily given real numbers, {Ak(x)}n−1k=0 and {Bk(x)} n−1 k=1 are fundamental functions satisfying the conditions  Ak(xj) = δkj, j,k = 0, 1, · · · ,n− 1,A′k(xj) = 0, j = 1, 2, · · · ,n− 1,k = 0, 1, · · · ,n− 1(3.4) Int. J. Anal. Appl. 19 (2) (2021) 183 and   Bk(xj) = 0, j = 0, 1, · · · ,n− 1,k = 1, 2, · · · ,n− 1,B′k(xj) = δkj, j,k = 1, 2, · · · ,n− 1.(3.5) 4. Explicit Representation of the Fundamental Functions Lemma 4.1. The fundamental functions {Ak(x)}n−1k=0 satisfying the conditions (3.4) can be explicitly repre- sented as for k = 1, 2, · · · ,n− 1 Ak(x) = (1 + xk)(1 −x2k)(1 −x){1 − bk(x−xk)}U 2 n(x) λ2n(xk)(x−xk)2λ2n(x) ,(4.1) where bk = 2xk − 1 1 −x2k (4.2) and (4.3) A0(x) = U2n(x) λ2n(x)λ2n(1) . Proof. We will show that Ak(x), k = 0, 1, · · · ,n− 1 defined by (4.1) and (4.3) satisfy the conditions (3.4) Obviously for k = 1, 2, · · · ,n− 1, Ak(x0) = 0 and Ak(xj) = 0, j = 1, 2, · · · ,n− 1 when j 6= k. For j = k using the l’Hospital’s rule, we have lim x→xk Ak(x) = (1 −x2k) λ22n(xk) ( lim x→xk sin µ2n(x) (x−xk) )2 = (1 −x2k) λ22n(xk) ( lim x→xk −λ2n(x) cos µ2n(x)√ 1 −x2 )2 = 1. On differentiating (4.1) with respect x we get A′k(x) = (1 + xk)(1 −x2k) λ2n(xk) [ 2{1 − bk(x−xk)} (1 + x)λ2n(x) ( sin µ2n(x) x−xk )( sin µ2n x−xk )′ − bk(1 + x)λ2n(x) + {1 − bk(x−xk)}{(1 + x)λ′2n(x) + λ2n(x)} (1 + x)2λ22n(x) × ( sin µ2n(x) x−xk )2 ] then for j 6= k we have A′k(xj) = 0, j = 1, 2, · · · ,n− 1. For j = k, lim x→xk A′k(x) = (1 −x2k) λ22n(xk) [ 2 lim x→xk (( sin µ2n(x) x−xk )( sin µ2n(x) x−xk )′) − bk(1 + xk)λ2n(xk) + (1 + xk)λ ′ 2n(xk) + λ2n(xk) (1 + xk)λ2n(xk) × ( lim x→xk sin µ2n(x) x−xk )2 ] Int. J. Anal. Appl. 19 (2) (2021) 184 We know that lim x→xk sin µ2n(x) (x−xk) = µ′2n(xk) cos µ2n(xk) = − λ2n(xk)√ 1 −x2k (4.4) and lim x→xk ( sin µ2n(x) x−xk )′ = 1 2 cos µ2n(xk)µ ′′ 2n(xk)(4.5) where µ′′2n(x) = − (1 −x2)λ′2n(x) + xλ2n(x) (1 −x2)3/2 (4.6) then lim x→xk A′k(x) = [ 2xk − 1 (1 −x2k) − bk ] = 0 due to (4.2) which shows that Ak(x), k = 1, 2, · · · ,n− 1, given by (4.1), satisfy all the conditions given by (3.4). Similarly, for A0(x), given by (4.3), we have that A0(xj) = 0, j = 1, · · · ,n − 1. For j = 0 and using the fact that Un(1) = λ2n(1), we have A0(x0) = 1. Again by differentiating (4.3) with respect x, we get A′0(xj) = 0, j = 1, 2, · · · ,n − 1. Thus A0(x) given by (4.3) satisfy the conditions (3.4) for j = 0, which completes the proof of the lemma. � Lemma 4.2. The fundamental functions {Bk(x)}n−1k=1 satisfying the conditions (3.5) can be explicitly repre- sented as (4.7) Bk(x) = (1 −x)(1 + xk)(1 −x2k)U 2 n(x) λ2n(x)λ2n(xk)(x−xk) . Proof. Obviously, Bk(xj) = 0, k = 1, 2, · · · ,n − 1, j = 0, 1, · · · ,n − 1 and for j 6= k, B′k(xj) = 0, j,k = 1, 2, · · · ,n− 1. For j = k, lim x→xk B′k(x) = (1 −x2k) 2 λ22n(xk) lim x→xk ( U2n(x) (x−xk) ) = (1 −x2k) 2 λ22n(xk) lim x→xk ( 2Un(x)U ′ n(x) (x−xk) − ( Un(x) x−xk )2) = (1 −x2k) 2 λ22n(xk) (U′n(xk)) 2 = 1, due to (2.6), which proves that Bk(x), k = 1, 2, · · · ,n− 1 given by (4.7) satisfy all the conditions given by (3.5). � From Lemma 4.1 and Lemma 4.2 it follows that Hn(f,x) satisfying the conditions (3.1) is an almost quasi Hermite interpolation. Int. J. Anal. Appl. 19 (2) (2021) 185 Theorem 4.1. The function Hn(f,x) is a rational function of degree atmost 2n− 1 that is Hn(f,x) ∈R2n−1(a0,a1, · · · ,a2n−1).(4.8) Proof. Since Un ∈Pn−1(a0,a1, · · · ,a2n−1), we can express it as Un(x) := Sn−1(x) (S∗n(x)) 1/2 where S∗n(x) := (x − a0)(x − a1) · · ·(x − a2n−1), Sn−1(x) := cn−1(x − x1)(x − x2) · · ·(x − xn−1) and cn−1 depends on n and {ak}2n−1k=0 . So, we have `k(x) = ( S∗n(xk) S∗n(x) )1/2 qk(x), k = 1, 2, · · · ,n− 1,(4.9) where qk(x) := Sn−1(x) S′n−1(xk)(x−xk) , k = 1, 2, · · · ,n− 1.(4.10) Thus, `k(x) ∈Pn−1(a0,a1, · · · ,a2n−1), thus by (3.3), (4.1) and (4.7) we easily find that Hn(f,x) = t2n−1(x) q2n−1(x) (4.11) where t2n−1(x) is a polynomial of degree ≤ 2n− 1, which proves the lemma. � Let yk = 0, k = 1, 2, · · · ,n− 1 then (3.3) reduces to Hn(f,x) = n−1∑ k=1 f(xk)Ak(x) + f(1)A0(x)(4.12) which is an almost quasi Hermite Fejér interpolation function for f ∈ C[−1, 1]. 5. Radau-type quadrature formula For a given function f defined on [−1, 1], we define the function (5.1) Gn(x,f) = n−1∑ k=0 f(xk)hk(x) where, hk(x) = 1 −x 1 −xk [ 1 − ( U′′n (xk) U ′ n(xk) − 1 (1 −xk) ) (x−xk) ] `2k(x), k = 1, 2, · · · ,n− 1 and h0(x) = U2n(x) U2n(1) . Int. J. Anal. Appl. 19 (2) (2021) 186 We have that Gn(f,x) ∈ R2n−1(a0,a1, · · · ,a2n−1). Also the rational function Gn(f,x) is an almost quasi Hermite Fejér interpolation function. Let (5.2) Ak = ∫ 1 −1 1 √ 1 −x2 hk(x)dx, k = 1, 2, · · · ,n− 1 and (5.3) A0 = ∫ 1 −1 1 √ 1 −x2 U2n(x) U2n(1) dx then the Radau-type quadrature formula is given by (5.4) ∫ 1 −1 f(x) √ 1 −x2 dx = A0f(1) + n−1∑ k=1 Akf(xk). With respect to this quadrature formula, we have the following Theorem 5.1. The quadrature formula (5.4) can be expressed as (5.5) ∫ 1 −1 f(x) √ 1 −x2 dx = 2π λ2n(1) f(1) + n−1∑ k=1 π λ2n(xk) f(xk). Lemma 5.1. For k = 1, 2, · · · ,n− 1, (5.6) ∫ 1 −1 (1 −x)(x−xk)`2k(x)dx = 0. Proof. We have that for k = 1, 2, · · · ,n− 1, `2k(x) = U2n(x) (U′n) 2(xk)(x−xk)2 = (1 −x2k) 2 sin2 µ2n(x) λ22n(xk)(1 −x2)(x−xk)2 .(5.7) Also, (5.8) Un(1) = lim x→1 sin µ2n(x)√ 1 −x2 = λ2n(1) and (5.9) Un(−1) = (−1)n+1λ2n(−1). By these equalities, the left hand side of (5.6) can be represented as (5.10) Ik = ∫ 1 −1 sin2 µ2n(x) (1 + x) √ 1 −x2(x−xk) dx Consider the transformation (5.11) x = 1 −y2 1 + y2 which gives (5.12) dx = − 4y (1 + y2)2 dy, Int. J. Anal. Appl. 19 (2) (2021) 187 (5.13) x−xk = − 2(y2 −y2k) (1 + y2)(1 + y2k) , (5.14) 1 + x = 2 1 + y2 , (5.15) √ 1 −x2 = 2y 1 + y2 . We know that, (5.16) sin µ2n ( 1 −y2 1 + y2 ) = sin φ2n(y) where sin φ2n(y) is a Bernstein sine fraction (5.17) sin φ2n(y) = 1 2i ( χn(y) −χ−1n (y) ) where χn(y) = ∏2n−1 j=0 y−zj y−z̄j and zk are the roots of the equations y 2 + (1 + ak)(1 −ak)−1 = 0, Izk > 0, k = 0, 1, · · · , 2n− 1. Taking into account the assumptions on the parameters ak, k = 0, 1, · · · , 2n− 1, we have the following: 1) z0 = i, 2) if ak and al are paired by complex conjugation, then the corresponding numbers zk and zl are symmetric with respect to the imaginary axis. Besides, the function sin φ2n(y) has zeros at ±yk, yk = √ (1 −xk)/(1 + xk), k = 1, 2, · · · ,n− 1. Thus, Ik = − 1 + y2k 4 ∫ ∞ −∞ (1 + y2) sin2 φ2n(y) y2 −y2k dy = − 1 + y2k 4 lim z→yk,Izk>0 Jk(z)(5.18) where (5.19) Jk(z) = ∫ ∞ −∞ (1 + y2) sin2 φ2n(y) y2 −z2 dy. From (5.17) we get (5.20) sin2 φ2n(y) = − 1 4 ( χ2n(y) − 2 + χ −2 n (y) ) due to which, we have (5.21) Jk(z) = − 1 4 (Jk1(z) − 2Jk2(z) + Jk3(z)) where Jk1(z) = ∫ ∞ −∞ (1 + y2)χ2n(y) y2 −z2 dy, Jk2(z) = ∫ ∞ −∞ (1 + y2) y2 −z2 dy Int. J. Anal. Appl. 19 (2) (2021) 188 and Jk3(z) = ∫ ∞ −∞ (1 + y2)χ−2n (y) y2 −z2 dy. Since Jk1(z) has only singular point y = z in the upper half plane. Thus by the residue theorem we have Jk1(z) = 2πi lim y→z (1 + y2)χ2n(y) (y + z) = 1 + z2 z χ2n(z)πi.(5.22) Similarly, (5.23) Jk3(z) = 1 + z2 z χ−2n (z)πi. Also, Jk2(z) has only singular point y = z in the upper half plane, therefore by the residue theorem, we have Jk2(z) = 2πi lim y→z (1 + y2) (y + z) = 1 + z2 z πi.(5.24) Putting the value of Jk1(z),Jk3(z) and Jk2(z) from (5.22), (5.23) and (5.24) respectively in (5.21) we get (5.25) Jk(z) = − 1 4 ( 1 + z2 z χ2n(z)πi− 2 1 + z2 z πi + 1 + z2 z χ−2n (z)πi ) which by (5.18) gives (5.26) Ik = 1 + y2k 16 lim z→yk,Izk>0 ( 1 + z2 z χ2n(z)πi + 1 + z2 z χ−2n (z)πi− 2 1 + z2 z πi ) . Since χn(yk) = 1, thus it follows that Ik = 0, which completes the proof of the lemma. � Lemma 5.2. For k = 1, 2, · · · ,n− 1, (5.27) Ak = π λ2n(xk) . Proof. Due to Lemma 5.1 and by putting the value of `2k(x),k = 1, 2, · · · ,n−1, from (5.7) in (5.2), we have (5.28) Ak = (1 + xk)(1 −x2k) λ22n(xk) ∫ 1 −1 sin2 µ2n(x) (x−xk)2(1 + x) √ 1 −x2 dx. We find the integrals (5.29) I∗k = ∫ 1 −1 sin2 µ2n(x) (x−xk)2(1 + x) √ 1 −x2 dx, k = 1, 2, · · · ,n− 1, by using the transformation (5.11), (5.12), (5.13), (5.14), (5.15) and (5.16), we have I∗k = (1 + y2k) 2 8 ∫ ∞ −∞ (1 + y2)2 sin2 φ2n(y) (y2 −y2k)2 dy. Consider the auxiliary integral J∗k (z) = ∫ ∞ −∞ (1 + y2)2 sin2 φ2n(y) (y2 −z2)2 dy Int. J. Anal. Appl. 19 (2) (2021) 189 then I∗k can be written as I∗k = (1 + y2k) 2 8 lim z→yk,=zk>0 J∗k (z).(5.30) which due to (5.20) can be expressed as I∗k = (1 + y2k) 2 32 lim z→yk,=zk>0 (I∗1k(z) − 2I ∗ 2k(z) + I ∗ 3k(z))(5.31) where, I∗1k(z) = ∫ ∞ −∞ (1 + y2)2 (y2 −z2)2 χ2n(y)dy, I∗2k(z) = ∫ ∞ −∞ (1 + y2)2 (y2 −z2)2 dy and I∗3k(z) = ∫ ∞ −∞ (1 + y2)2 (y2 −z2)2 χ−2n (y)dy. Since z0 = i the integrand of I ∗ 1k(z) has only singular point y = z in the upper half plane. Thus by the residue theorem, we have I∗1k(z) = 2πi lim y→z d dy ( (1 + y2)2 (y + z)2 χ2n(y) ) = 2πi lim y→z [ χ2n(y) d dy (y2 + 1)2 (y + z)2 + (y2 + 1)2 (y + z)2 d dy χ2n(y) ] .(5.32) Since, χn(y) = 2n−1∏ j=0 y −zj y − z̄j which by logarithmic differentiation gives d dy χn(y) = χn(y) 2n−1∑ j=0 zj − z̄j (y −zj)(y − z̄j) . Also, d dy ( (y2 + 1)2 (y + z)2 ) = 2y4 + 4y3z + 4yz − 2 (y + z)3 . Therefore, I∗1k(z) = 2πiχ 2 n(z) [ 3z4 + 2z2 − 1 4z3 (5.33) + (z2 + 1)2 2z2 2n−1∑ j=0 zj − z̄j (z −zj)(z − z̄j) ] . Similarly, I∗3k(z) = 2πiχ −2 n (z) [ 3z4 + 2z2 − 1 4z3 (5.34) + (z2 + 1)2 2z2 2n−1∑ j=0 zj − z̄j (z −zj)(z − z̄j) ] . Int. J. Anal. Appl. 19 (2) (2021) 190 Again by residue theorem we have I∗2k(z) = 2πi lim y→z 2y4 + 4y3z + 4yz − 2 (y + z)3 = 2πi ( 3z4 + 2z2 − 1 4z3 ) .(5.35) By (5.33), (5.34), (5.35), (5.31) and (5.30) and taking into account that χ2n(yk) = 1 it follows that I∗k = − πi(1 + y2k) 4 32y2k 2n−1∑ j=0 zj − z̄j (yk −zj)(yk − z̄j) .. Since, yk = √ (1 −xk)/(1 + xk) and zk = i √ (1 + ak)/(1 −ak), thus by simple calculation, we have, 2n−1∑ j=0 zj − z̄j (yk −zj)(yk − z̄j) = 2n−1∑ j=0 ( 1 yk −zj − 1 yk − z̄j ) = 2n−1∑ j=0 i √ (1 + aj) √ (1 −aj) 1 + ajxk ( 2 1 + y2k ) = 4iλ2n(xk) (1 + y2k) (5.36) thus I∗k = πλ2n(xk)(1 + y 2 k) 3 8y2k . Therefore by (5.28), the lemma follows. � Lemma 5.3. For A0, defined by (5.3), we have A0 = ( π λ2n(1) ) .(5.37) Proof. By using the transformation (5.11), (5.12), (5.15) and (5.16), we have A0 = 1 4U2n(1) ∫ ∞ −∞ 1 + y2 y2 sin2 φ2n(y)dy which, due to (5.20), can expressed as (5.38) A0 = − 1 16U2n(1) (I1 − 2I2 + I3) where I1 = ∫ ∞ −∞ 1 + y2 y2 χ2n(y)dy, I2 = ∫ ∞ −∞ 1 + y2 y2 dy and I3 = ∫ ∞ −∞ 1 + y2 y2 χ−2n (y)dy. Int. J. Anal. Appl. 19 (2) (2021) 191 Since for I1, y = 0 is the only singular point in the upper half plane. Thus by the residue theorem, we have I1 = 2πi lim y→0 d dy { (1 + y2)χ2n(y) } (5.39) = 4πi 2n−1∑ j=0 ( 1 z̄j − 1 zj ) . Similarly, (5.40) I3 = 4πi 2n−1∑ j=0 ( 1 z̄j − 1 zj ) . The integrand of I2(z) has only singular point y = 0 in the upper half plane. Thus by the residue theorem, we have (5.41) I2 = 2πi lim y→0 d dy (1 + y2) = 0. Hence using (5.40), (5.40) and (5.41) in (5.38) we get A0 = − πi 2λ22n(1) 2n−1∑ j=0 ( 1 z̄j − 1 zj ) and since 2n−1∑ j=0 ( 1 z̄j − 1 zj ) = − 4 i λ2n(1) hence (5.42) A0 = 2π λ2n(1) which in turn proves the lemma. � By Lemma 5.2, Lemma 5.3 and (5.4), Theorem 5.1 follows. Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication of this paper. References [1] P. Borwein and T. Erdélyi, Polynomials and Polynomial Inequalities, Graduate Texts in Mathematics 161, Springer-Verlag, New York (1995). [2] A. L. Lukashov, Inequalities for the derivatives of rational functions on several intervals, Izv. Math. 68(3) (2004), 543–565. [3] A. A. Markov(1951), Izbrannye trudy, Teoriya cisel. Teoriya veroyatnostei, Izdat. Akad. 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