Piecewise Polynomial Interpolation, cont d

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1 Jim Lambers MAT 460/560 Fall Semester Lecture 2 Notes Tese notes correspond to Section 4 in te text Piecewise Polynomial Interpolation, cont d Constructing Cubic Splines, cont d Having determined our constraints tat must be satisfied by s(x), we can set up a system of linear equations Ax = b based on tese constraints, and ten solve tis system to determine te coefficients a i, b i, c i, d i for i = 0,,, n In te case of free boundary conditions, A is an (n + ) (n + ) matrix is defined by ( 0 + ) A = 0 2( + 2 ) 2 0 n 2 2( n 2 + n ) n and te (n + )-vectors x and b are c 0 c x =, b = c n 0 (a 2 a ) 0 (a a 0 ) n (a n a n ) n 2 (a n a n 2 ) 0 were c n = s (x n ) In te case of clamped boundary conditions, we ave ( 0 + ) A = 0 2( + 2 ) 2 0 n 2 2( n 2 + n ) n 0 0 n 2 n,

2 and c 0 c x = c n, b = 0 (a a 0 ) z 0 (a 2 a ) 0 (a a 0 ) n (a n a n ) z n n 2 (a n a n 2 ) n (a n a n ) Once te coefficients c 0, c,, c n ave been determined, te remaining coefficients can be computed as follows: Te coefficients a 0, a,, a n ave already been defined by te relations a i = y i, for i = 0,,, n 2 Te coefficients b 0, b,, b n are given by b i = i (a i+ a i ) i (2c i + c i+ ), i = 0,,, n, Te coefficients d 0, d,, d n can be obtained using te constraints d i i + c i = c i+, i = 0,,, n Example We will construct a cubic spline interpolant for te following data on te interval [0, 2] j x j y j 0 0 / / Te spline, s(x), will consist of four pieces {s j (x)} j=0, eac of wic is a cubic polynomial of te form s j (x) = a j + b j (x x j ) + c j (x x j ) 2 + d j (x x j ), j = 0,, 2, We will impose free, or natural, boundary conditions on tis spline, so it will satisfy te conditions s (0) = s (2) = 0, in addition to te essential conditions imposed on a spline: it must fit te given data and ave continuous first and second derivatives on te interval [0, 2] Tese conditions lead to te following system of equations tat must be solved for te coefficients c 0, c, c 2, c, and c 4, were c j = s (x j )/2 for j = 0,,, 4 We define = (2 0)/4 = /2 to be 2

3 te spacing between te interpolation points c 0 = 0 (c 0 + 4c + c 2 ) = y 2 2y + y 0 (c + 4c 2 + c ) = y 2y 2 + y (c 2 + 4c + c 4 ) = y 4 2y + y 2 c 4 = 0 Substituting = /2 and te values of y j, and also taking into account te boundary conditions, 6 (4c + c 2 ) = 2 6 (c + 4c 2 + c ) = 40 6 (c 2 + 4c ) = 48 Tis system as te solutions c = 56/7, c 2 = 720/7, c = 684/7 Using te relation a j = y j, for j = 0,, 2,, and te formula Finally, using te formula b j = a j+ a j (2c j + c j+ ), j = 0,, 2,, b 0 = 84/7, b = 74/7, b 2 = 4, b = 46/7 d j = c j+ c j, j = 0,, 2,, d 0 = 44/7, d = 824/7, d 2 = 96/7, d = 456/7 We conclude tat te spline s(x) tat fits te given data, as two continuous derivatives on [0, 2], and satisfies natural boundary conditions is 44 7 x 84 7 x2 + if x [0, 05] 824 s(x) = 7 (x /2) (x /2) (x /2) 4 if x [05, ] 96 7 (x ) (x )2 4(x ) + 5 if x [, 5] (x /2) (x /2) (x /2) 6 if x [5, 2] Te grap of te spline is sown in Figure

4 Figure : Cubic spline tat passing troug te points (0, ), (/2, 4), (, 5), (2, 6), and (, 7) 4

5 Well-Posedness and Accuracy For bot boundary conditions, te system Ax = b as a unique solution, wic leads to te following results Teorem Let x 0, x,, x n be n+ distinct points in te interval [a, b], were a = x 0 < x < < x n = b, and let f(x) be a function defined on [a, b] Ten f as a unique cubic spline interpolant s(x) tat is defined on te nodes x 0, x,, x n tat satisfies te natural boundary conditions s (a) = s (b) = 0 Teorem Let x 0, x,, x n be n + distinct points in te interval [a, b], were a = x 0 < x < < x n = b, and let f(x) be a function defined on [a, b] tat is differentiable at a and b Ten f as a unique cubic spline interpolant s(x) tat is defined on te nodes x 0, x,, x n tat satisfies te clamped boundary conditions s (a) = f (a) and s (b) = f (b) Te following result provides insigt into te accuracy wit wic a cubic spline interpolant s(x) approximates a function f(x) Teorem Let f be four times continuously differentiable on [a, b], and assume tat f (4) (x) M on [a, b] for some constant M Let s(x) be te unique clamped cubic spline interpolant of f(x) on te nodes x 0, x,, x n, were a = x 0 < x < < x n < b Ten for x [a, b], were i = x i+ x i B-splines f(x) s(x) 5M max 84 0 i n 4 i, An alternative metod of computing splines to fit given data involves constructing a basis for te vector space of splines defined on te interval [a, b], and ten solving a system of linear equations for te coefficients of te desired spline in tis basis Te basis functions are known as B-splines, were te letter B is due to te fact tat tese splines form a basis, and te fact tat tey tend to ave bell-saped graps One advantage of using B-splines is tat te system of linear equations tat must be solved for te coefficients of a spline in te basis is banded, and terefore can be solved very efficiently Furtermore, because eac B-spline as compact support, it follows tat a cange in te data value y i only causes te coefficients of a few B-splines to be canged, wereas in cubic spline interpolation, suc a cange forces all of te coefficients of eac polynomial s i (x) to be recomputed 5