SPLINE INTERPOLATION Spline Background • Problem: high degree interpolating polynomials often have extra oscillations. 1 Example: \Runge" function f(x) = 1+4x2 , x 2 [−1; 1]. 1/(1+4x2) and P (x) and P (x) 8 16 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 • Piecewise Polynomials provide alternative to high degree polynomials: approximation interval [a; b] is sub- divided into pieces [x1; x2]; [x2; x3];:::; [xn−1; xn], with a = x1 < x2 < ··· < xn = b, and a low degree polyno- mial is used to approximate f(x) on each subinterval. Example: piecewise linear approximation S(x) f(xj+1) − f(xj) S(x) = f(xj)+(x−xj) ; if x 2 [xj; xj+1] xj+1 − xj • Splines are piecewise polynomial approximations, con- nected at xj's with various continuity conditions. CUBIC SPLINE INTERPOLATION Cubic Interpolating Splines for a = x1 < ··· < xn = b with given data (x1; y1); (x2; y2);:::; (xn; yn). • Properties of Cubic Interpolating Spline S(x), a) S(x) is composed of cubic polynomial pieces Sj(x) S(x) = Sj(x) if x 2 [xj; xj+1]; j = 1; 2; : : : ; n − 1: b) S(xj) = yj, j = 1; : : : ; n: (interpolation) c) Sj−1(xj) = Sj(xj), j = 2; : : : ; n − 1 (S 2 C[a; b]). 0 0 1 d) Sj−1(xj) = Sj(xj), j = 2; : : : ; n − 1 (S 2 C [a; b]). 00 00 2 e) Sj−1(xj) = Sj (xj), j = 2; : : : ; n − 1 (S 2 C [a; b]). f) two end conditions: examples 00 00 i) S (x1) = S (xn) = 0 (natural or free spline); 0 0 0 0 ii) S (x1) = f (x1);S (xn) = f (xn) (complete or clamped spline); 000 000 iii) S1 = Sn−1 = 0 (parabolically terminated); 000 000 000 000 iv) S1 (x2) = S2 (x2);Sn−2(xn−1) = Sn−1(xn−1) (not-a-knot). Note: if 2 3 Sj(x) = aj +bj(x−xj)+cj(x−xj) +dj(x−xj) , condition a) provides 4(n − 1) free parameters; b)-f) give n + 3(n − 2) + 2 = 4(n − 1) constraints. 2 CUBIC SPLINE INTERPOLATION • Example: n = 3, natural, data (1,2), (2,3), (3,5). 3 1 3 S1(x) = 2 + 4(x − 1) + 4(x − 1) , 3 3 2 1 3 S2(x) = 3 + 2(x − 2) + 4(x − 2) − 4(x − 2) , 3 CUBIC SPLINE INTERPOLATION Cubic Interpolating Spline Construction • Spline Linear System: let hj = xj+1 − xj; start with 2 3 Sj(x) = aj + bj(x − xj) + cj(x − xj) + dj(x − xj) 2 3 = yj + bj(x − xj) + cj(x − xj) + dj(x − xj) : c) Sj+1(xj+1) = Sj(xj+1), implies 2 3 ∆yj 2 yj+1 = yj +bjhj +cjhj +djhj; = bj +cjhj +djhj; hj where ∆yj = yj+1 − yj. 00 Notice Sj (x) = 2cj + 6dj(x − xj), so 00 00 e) Sj+1(xj+1) = Sj (xj+1), implies 2cj+1 = 2cj + 6djhj; djhj = (cj+1 − cj)=3; 00 with extra unknown cn = Sn−1(xn)=2 added. Then ∆yj (cj+1 − cj)hj (cj+1 + 2cj)hj = bj +cjhj + = bj + ; hj 3 3 3∆yj+1 3∆yj − = 3∆bj+(cj+2+2cj+1)hj+1−(cj+1+2cj)hj: hj+1 hj 0 2 Also Sj(x) = bj + 2cj(x − xj) + 3dj(x − xj) , so 0 0 2 d) Sj+1(xj+1) = Sj(xj+1); bj+1 = bj + 2cjhj + 3djhj; ∆bj = 2cjhj + (cj+1 − cj)hj = (cj+1 + cj)hj 3∆yj+1 3∆yj − = cjhj + 2cj+1(hj + hj+1) + cj+2hj+1 hj+1 hj 4 CUBIC SPLINE INTERPOLATION 00 00 • Natural Splines: S (x1) = S (xn) = 0, so c1 = cn = 0 Linear system equations are a \tridiagonal" system c1 = 0 3∆y2 3∆y1 c1h1 + 2c2(h1+h2) + c3h2 = − h2 h1 3∆y3 3∆y2 c2h2 + 2c3(h2+h3) + c4h3 = − h3 h2 . 3∆yn−2 3∆y3−2 cn−3hn−3 + 2cn−2(hn−3+hn−2) + cn−1hn−2 = − hn−2 hn−3 3∆yn−1 3∆yn−2 cn−2hn−2 + 2cn−1(hn−2+hn−1) + cnhn−1 = − hn−1 hn−2 cn = 0: which can be solved uniquely for cj's with O(n) work; 2 dj = (cj+1 − cj)=(3hj); bj = ∆yj=hj − cjhj − djhj can be used to find remaining coefficients for Sj(x)'s. Note: if all hj = h, a simpler tridiagonal system. 5 CUBIC SPLINE INTERPOLATION • Example: n = 3, natural, with data (1,2), (2,3), (3,5), so h1 = h2 = 1, ∆y1 = 1; ∆y2 = 2. Only one equation 2c2(1 + 1) = 3(2 − 1), so c2 = 3=4, d1 = 1=4; d2 = −1=4; b1 = 1 − 0 − 1=4 = 3=4; b2 = 2 − 3=4 + 1=4 = 3=2. 3 1 S (x) = 2 + (x − 1) + (x − 1)3; 1 4 4 3 3 1 S (x) = 3 + (x − 2) + (x − 2)2 − (x − 2)3: 2 2 4 4 1 • Example: \Runge" function f(x) = 1+4x2 , x 2 [−1; 1]. 1/(1+4x2), S(x) and P (x) 4 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 6 CUBIC SPLINE INTERPOLATION 0 0 00 0 • Clamped Splines: let S (x1) = y1;S (xn) = yn, 0 0 2 so y1 = b1, yn = bn−1 + 2cn−1hn−1 + 3dn−1hn−1 ∆yj (cj+1+2cj)hj Using = bj + , 3hjdj = (cj+1 − cj); hj 3 \1st" and \nth" equations become 3∆y1 0 2c1h1 + c2h1 = − 3y , and h1 1 0 3∆yn−1 cn−1hn−1 + 2cnhn−1 = 3y − . n hn−1 Linear system equations are a \tridiagonal" system 3∆y1 0 2c1h1 + c2h1 = − 3y1 h1 3∆y2 3∆y1 c1h1 + 2c2(h1+h2) + c3h2 = − h2 h1 . 3∆yn−1 3∆yn−2 cn−2hn−2 + 2cn−1(hn−2+hn−1) + cnhn−1 = − hn−1 hn−2 0 3∆yn−1 cn−1hn−1 + 2cnhn−1 = 3yn − : hn−1 which can be solved (uniquely) for cj's with O(n) work; 2 dj = (cj+1 − cj)=(3hj); bj = ∆yj=hj − cjhj − djhj can be used to find remaining coefficients for Sj(x)'s. 7 CUBIC SPLINE INTERPOLATION • Example: n = 3, clamped, with data (1,2), (2,3), (3,5), 0 0 and y1 = 1, y3 = 2. Three equations: 2c1 + c2 = 3(2 − 1) − 3 = 0, c1 + 4c2 + c3 = 3, c2 + 2c3 = 6 − 3(2) = 0, so c1 = c3 = −1=2; c2 = 1; d1 = 1=2; d2 = −1=2; b1 = 1; b2 = 2−1+1=2 = 3=2. 1 2 1 3 S1(x) = 2 + (x − 1) − 2(x − 1) + 2(x − 1) , 3 2 1 3 S2(x) = 3 + 2(x − 2) + (x − 2) − 2(x − 2) , 000 000 • Parabolically Terminated Splines: S1 = Sn−1 = 0, so d1 = dn−1 = 0, c1 = c2; cn−1 = cn. Linear system equations are a \tridiagonal" system c1 − c2 = 0 3∆y2 3∆y1 c1h1 + 2c2(h1+h2) + c3h2 = − h2 h1 . 3∆yn−1 3∆yn−2 cn−2hn−2 + 2cn−1(hn−2+hn−1) + cnhn−1 = − hn−1 hn−2 cn−1 − cn = 0: which can be solved (uniquely) for cj's with O(n) work; 2 dj = (cj+1 − cj)=(3hj); bj = ∆yj=hj − cjhj − djhj can be used to find remaining coefficients for Sj(x)'s. 8 CUBIC SPLINE INTERPOLATION • Not-a-Knot Splines: 000 000 000 000 S1 (x2) = S2 (x2);Sn−2(xn−1) = Sn−1(xn−1), so d1 = d2, dn−2 = dn−1, and S1 = S2;Sn−2 = Sn−1. Then (c2 − c1)=h1 = (c3 − c2)=h2, (cn−1 − cn−2)=hn−2 = (cn − cn−1)=hn−1, so \1st" and \nth" equations become c1h2 − c2(h1 + h2) + c3h1 = 0 cn−2hn−1 − cn−1(hn−2 + hn−1) + cnhn−2 = 0. Linear system equations are a \tridiagonal" system c1h2 − c2(h1 + h2) + c3h1 = 0 3∆y2 3∆y1 c1h1 + 2c2(h1+h2) + c3h2 = − h2 h1 . 3∆yn−1 3∆yn−2 cn−2hn−2 + 2cn−1(hn−2+hn−1) + cnhn−1 = − hn−1 hn−2 cn−2hn−1 − cn−1(hn−2 + hn−1) + cnhn−2 = 0: which can be solved (uniquely) for cj's with O(n) work; 2 dj = (cj+1 − cj)=(3hj); bj = ∆yj=hj − cjhj − djhj can be used to find remaining coefficients for Sj(x)'s. 9 CUBIC SPLINE INTERPOLATION Efficient Spline Evaluation • Setup: solve linear system for cj's in O(n) time • For each evaluation point x, find interval x 2 [xj; xj+1] in O(log(n)) time and evaluate Sj(x) in O(1) time. Alternate formula for Sj(x) (without bj's and dj's): cj 3 cj+1 3 Sj(x) = (xj+1 − x) + (x − xj) 3hj 3hj+1 yj cjhj yj+1 cj+1hj+1 + ( − )(xj+1 − x) + ( − )(x − xj): hj 3 hj+1 3 • Compare with polynomial interpolation, where setup time is O(n2) and evauation time is O(n). Error Theorem : 4 (4) if f 2 C [a; b], with maxx2[a;b] jf (x)j = M, and S(x) is the unique clamped spline for f(x) with nodes a = x1 < x2 ··· < xn = b, then 5M 4 jf(x) − S(x)j ≤ max hj: 384 1≤j<n 10.