Twelfth degree spline with application to quadrature

Mohammed and Hamasalh SpringerPlus (2016)5:2096 DOI 10.1186/s40064-016-3711-2 Open Access RESEARCH Twelfth degree spline with application to quadrature P. O. Mohammed* and F. K. Hamasalh *Correspondence: pshtiwansangawi@gmail. com Department of Mathematics, College of Education, University of Sulaimani, Sulaimani, Kurdistan Region, Iraq Abstract In this paper existence and uniqueness of twelfth degree spline is proved with application to quadrature. This formula is in the class of splines of degree 12 and continuity order C 12 that matches the derivatives up to order 6 at the knots of a uniform partition. Some mistakes in the literature are pointed out and corrected. Numerical examples are given to illustrate the applicability and efficiency of the new method. Keywords: Interpolation, Spline approximation, Quadrature Mathematics Subject Classification: 65D05, 65D07, 41A15 Background In the last two decades, Clarleft et al. (1967) have constructed a direct cubic spline that fits the first derivatives at the knots together with the value of the function and its second derivative at the beginning of the interval. They used it for the solution quadrature formula. El Tarazi and Karaballi (1990) have constructed five types of even degree splines ( j = 2k, k = 1, 2, 3, 4, 5) that match the derivatives up to the order k at the knots of a uniform partition for each k = 1, 2, 3, 4, and 5. These splines are also applied to quadrature. Recently, Rathod et al. (2010) presented a formulation and study of an interpolatory cubic spline (named Subbotin cubic spline) to compute the integration over curved domains in the Cartesian two space and the integral approximations (quadrature). In this work, we construct a twelfth degree spline which interpolates the derivatives up to the order 6 of a given function at the knots and its value at the beginning of the interval. We obtain a direct simple formula for the proposed spline. Error bounds for the function is derived in the sense of the Hermite interpolation. Also, a mistakes in the literature was corrected. Finally, numerical examples and comparison with other available methods are presented to illustrate the usefullness of proposed method. Description of the spline (existence and uniqueness) We construct here a class of interpolating splines of degree 12. Error estimates for this spline is also represented. © The Author(s) 2016. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Mohammed and Hamasalh SpringerPlus (2016)5:2096 Page 2 of 12 Let 0 = x0 < x1 < · · · < xn−1 < xn = 1 be a uniform partition of [0, 1]. We denote by (6) Sn,12 the linear space of twelfth degree spline s(x) such that 1 s(x) ∈ C (6) [0, 1]; 2 s(x) is a polynomial of degree 12 in each subinterval [xi , xi+1 ]. Set the stepsize h = xi+1 − xi (i = 0(1)n). Note that if g is a real-valued function in [0, 1], then gi stands for g(xi ) (i = 0(1)n). Theorem 1 Let s(x) be the spline defined in section “Description of the spline (existence (k) (k) and uniqueness)”. Given the real numbers f0 and si = fi for i = 0(1)n, k = 1(1)6. (6) Then, there exist a unique s(x) ∈ Sn,12 such that  (k) (k) si = fi s 0 = f0 i = 0(1)n, k = 1(1)6; (1) The twelfth degree spline s(x) which satisfies (1) in [xi , xi+1 ] is of the form: s(x) = 11    (j) (j) hj si Aj (t) + si+1 Aj (t) + h12 fi(12) A12 (t) (2) j=0 where   A0 (t) = (t − 1)6 462t 6 + 252t 5 + 126t 4 + 56t 3 + 21t 2 + 6t + 1 ,   A1 (t) = −t 7 462t 5 − 2520t 4 + 5544t 3 − 6160t 2 + 3465t − 792 ,   A2 (t) = t(t − 1)6 252t 5 + 126t 4 + 56t 3 + 21t 2 + 6t + 1 ,   A3 (t) = t 7 (t − 1) 210t 4 − 924t 3 + 1540t 2 − 1115t + 330 ,   1 A4 (t) = t 2 (t − 1)6 126t 4 + 56t 3 + 21t 2 + 6t + 1 , 2   1 A5 (t) = − t 7 (t − 1)2 84t 3 − 280t 2 + 315t − 120 , 2   1 3 A6 (t) = t (t − 1)6 56t 3 + 21t 2 + 6t + 1 , 6     1 1 4 t (t − 1)6 21t 2 + 6t + 1 , A7 (t) = t 7 (t − 1)3 28t 2 − 63t + 36 , A8 (t) = 6 24 1 5 −1 7 t (t − 1)4 (7t − 8), t (t − 1)6 (6t + 1), A10 (t) = A9 (t) = 24 120 1 7 1 6 A11 (t) = t (t − 1)5 , A12 (t) = t (t − 1)6 , 120 720 and x = xi + th, t ∈ [0, 1], with a similar expression for s(x) in [xi−1 , xi ]. The coefficient si in (2) are given by the recurrence formula: � �  � ′ � � � (3) 1 5 2 ′′ 1 3 (3) ′ ′′   si = si−1 + 2 h fi−1 + fi + 44 h fi−1 − fi + 66 h fi−1 + fi  � � � � � � (4) 1 1 1 4 (4) 5 f (5) + f (5) + 6 f (6) − f (6) h h h − f f + +  i−1 i i−1 i i−1 i 792 15840 665280   s0 = f 0 . (3) Mohammed and Hamasalh SpringerPlus (2016)5:2096 Page 3 of 12 Proof We can express any polynomial p(t) in [0, 1] of degree 12 in terms of its values and its derivatives upto order 5 at 0 and 1, and its sixth derivative at 0, p(t) = 5    (j) (j) p0 Aj (t) + p1 Aj+1 (t) + p0(6) A12 (t) j=0 and to determine the coefficients Aj , j = 0, 1, . . . , 12, we write the above equality for p(t) = 1, t, t 2 , . . . , t 12, we obtain the following system: A0 + A1 A1 + A2 + A3 + 2A5 A1 + 2A3 + 2A4 A1 + 3A3 + 6A5 + 6A6 + 6A7 A1 + 4A3 + 12A5 + 24A7 + 24A8 + 24A9 A1 + 5A3 + 20A5 + 60A7 + 120A9 + 120A10 + 120A11 A1 + 6A3 + 30A5 + 120A7 + 360A9 + 720A11 + 720A12 A1 + 7A3 + 42A5 + 210A7 + 840A9 + 2520A11 A1 + 8A3 + 56A5 + 336A7 + 1680A9 + 6720A11 A1 + 9A3 + 72A5 + 504A7 + 3024A9 + 15120A11 A1 + 10A3 + 90A5 + 720A7 + 5040A9 + 30240A11 A1 + 11A3 + 110A5 + 990A7 + 7920A9 + 55440A11 A1 + 12A3 + 132A5 + 1320A7 + 11880A9 + 95040A11 =1 =t = t2 = t3 = t4 = t5 = t6 = t7 = t8 = t9 = t 10 = t 11 = t 12 Solving this system, to obtain Aj , j = 0(1)12, above. Now for a fixed i ∈ {0, 1, . . . , n}, set x = xi + th, 0 < t < 1. In [xi , xi+1 ] the spline s(x) of degree 12 satisfying (1) is s(x) = 11    (j) (j) hj si Aj (t) + si+1 Aj (t) + h12 fi(12) A12 (t) j=0 We have a similar expression for s(x) in [xi−1 , xi ]. Since s(x) ∈ C (6) [0, 1], we have s(6) (xi− ) = s(6) (xi+ ), s (6) (6) − (xn+1 ) = fn+1 . for i = 0(1)n, This gives the above recurrence formula (3). Thus, the proof is completed.   Error bounds In this section, error estimates for the above interpolatory twelfth spline is considered. Note that � · � represents the L∞ norm. Theorem 2 (Birkhoff and Priver 1967; Clarleft et al. 1967; Varma and Howell 1983) Let g ∈ C 2m [0, h] be given. Let p2m−1 be the unique Hermite interpolation polynomial of degree 2m − 1 that matches g and its first m − 1 derivatives g (r) at 0 and h. Then   r m−r G  (r)  h [x(h − x)] , e (x) ≤ r!(2m − 2r)! r = 0(1)m; 0 ≤ x ≤ h, (4) Mohammed and Hamasalh SpringerPlus (2016)5:2096 Page 4 of 12 where      (r)   (r)  (r) e (x) = g (x) − p2m−1 (x) and     G = max g (2m) (x). 0≤x≤h (5) The bounds in (2) are best (...truncated)