SPLINE INTERPOLATION

Transcription

1 Spline Background SPLINE INTERPOLATION Problem: high degree interpolating polynomials often have extra oscillations. Example: Runge function f(x = 1 1+4x 2, x [ 1, 1]. 1 1/(1+4x 2 and P 8 (x and P 16 (x Piecewise Polynomials provide alternative to high degree polynomials: approximation interval [a, b] is subdivided into pieces [x 1, x 2 ], [x 2, x 3 ],..., [x n 1, x n ] with a = x 1 < x 2 < < x n = b, and a low degree polynomial is used to approximate f(x on each subinterval.

2 Piecewise Linear Approximation S(x: S(x = f(x j + (x x j f(x j+1 f(x j x j+1 x j, if x [x j, x j+1 ] 2 Splines are piecewise polynomial approximations, connected at x j s with various continuity conditions.

3 Cubic Interpolating Splines for a = x 1 < < x n = b with given data (x 1, y 1, (x 2, y 2,..., (x n, y n. Properties of Cubic Interpolating Spline S(x, a S(x is composed of cubic polynomial pieces S j (x S(x = S j (x if x [x j, x j+1 ], j = 1, 2,..., n 1; b S(x j = y j, j = 1,..., n. (interpolation; c S j 1 (x j = S j (x j, j = 2,..., n 1 (S C[a, b]; d S j 1 (x j = S j (x j, j = 2,..., n 1 (S C 1 [a, b]; e S j 1 (x j = S j (x j, j = 2,..., n 1 (S C 2 [a, b]; f two end conditions: some possible choices i S (x 1 = S (x n = 0 ( natural or free spline; or ii S (x 1 = f (x 1, S (x n = f (x n (complete or clamped spline; or iii S 1 (x 2 = S 2 (x 2, S n 2(x n 1 = S n 1(x n 1 (not-a-knot; or iv S 1 = S n 1 = 0 (parabolically terminated. Notes: i if S j (x = a j + b j (x x j + c j (x x j 2 + d j (x x j 3, condition a provides 4(n 1 free parameters; b-f give n + 3(n = 4(n 1 constraints; ii spline word from thin metal strips that were used for drafting. 3

4 4 Example: n = 3, natural, data (1,2, (2,3, (3,5. S 1 (x = (x (x 13, S 2 (x = (x (x (x 23.

5 Cubic Interpolating Spline Construction Spline Linear System: let h j = x j+1 x j, and start with S j (x = a j +b j (x x j +c j (x x j 2 +d j (x x j 3 = y j +b j (x x j +c j (x x j 2 +d j (x x j 3. c S j+1 (x j+1 = S j (x j+1, implies (with y j = y j+1 y j y j+1 = y j + b j h j + c j h 2 j + d j h 3 y j j; = b j + c j h j + d j h 2 h j. j Notice S j (x = 2c j + 6d j (x x j, so e S j+1 (x j+1 = S j (x j+1, implies 2c j+1 = 2c j + 6d j h j ; d j h j = c j /3, with extra unknown c n = S n 1(x n /2 added. Then y j = b j + c j h j + c jh j = b j + (c j+1 + 2c j h j ; h j 3 3 = 3 y j+1 3 y j = 3 b j + (c j+2 + 2c j+1 h j+1 (c j+1 + 2c j h j. h j+1 h j Also S j (x = b j + 2c j (x x j + 3d j (x x j 2, so d S j+1 (x j+1 = S j (x j+1, implies b j+1 = b j + 2c j h j + 3d j h 2 j ; rewritten b j = 2c j h j + (c j+1 c j h j = (c j+1 + c j h j 5 = 3 y j+1 h j+1 3 y j h j = c j h j +2c j+1 (h j +h j+1 +c j+2 h j+1, for j = 1, 2,..., n 2.

6 Natural Splines: S (x 1 = S (x n = 0, so c 1 = c n = 0 Linear system equations are a tridiagonal system (0h 1 + 2c 2 (h 1 +h 2 + c 3 h 2 = 3 y 2 h 2 3 y 1 h 1 c 2 h 2 + 2c 3 (h 2 +h 3 + c 4 h 3 = 3 y 3 h 3 3 y 2 h 2.. c n 3 h n 3 + 2c n 2 (h n 3 +h n 2 + c n 1 h n 2 = 3 y n 2 h n 2 3 y 3 2 h n 3 c n 2 h n 2 + 2c n 1 (h n 2 +h n 1 + (0h n 1 = 3 y n 1 3 y n 2. h n 1 h n 2 which can be solved uniquely for c j s with O(n work. Equations d j = c j /(3h j, and b j = y j /h j c j h j d j h 2 j can be used to find remaining coefficients for S j (x s. If all h j = h, a simpler tridiagonal system has c j + 4c j+1 + c j+2 = 3 2 y j /h 2. Example: n = 3, natural, with data (1,2, (2,3, (3,5, so h 1 = h 2 = 1, y 1 = 1, y 2 = 2, 2 y 1 = 1. Only one equation 4c 2 = 3(1, so c 2 = 3/4, d 1 = 1/4, d 2 = 1/4; b 1 = 1 0 1/4 = 3/4, b 2 = 2 3/4 + 1/4 = 3/2. 6 S 1 (x = (x (x 13, S 2 (x = (x (x (x 23.

7 7 CUBIC SPLINE INTERPOLATION Alternate form without b s and d s: S j (x = c j (x j+1 x 3 + c j+1 (x x j 3 + ( y j c jh j 3h j 3h j+1 h j 3 (xj+1 x+ ( y j+1 c j+1h j+1 (x xj. h j+1 3 For the natural spline example: S 1 (x = (x 1 3 /4 + 2(2 x + 11(x 1/4; S 2 (x = (3 x 3 /4 + 11(3 x/4 + 5(x 2. Another Example: Runge function f(x = 1, x [ 1, 1], 1+4x 2 with n = 5, x i s = 1,.5, 0,.5, /(1+4x 2, S(x and P 4 (x

8 Clamped Splines: given S (x 1 = y 1, S (x n = y n, so y 1 = b 1, y n = b n 1 + 2c n 1 h n 1 + 3d n 1 h 2 n 1 Using y j h j = b j + (c j+1+2c j h j 3, 3h j d j = (c j+1 c j, 1 st and n th equations are: 3y 1, and c n 1 h n 1 + 2c n h n 1 = 3y n 3 y n 1 h n 1. 2c 1 h 1 + c 2 h 1 = 3 y 1 h 1 Linear system equations are a tridiagonal system 8 2c 1 h 1 + c 2 h 1 = 3 y 1 h 1 3y 1 c 1 h 1 + 2c 2 (h 1 +h 2 + c 3 h 2 = 3 y 2 h 2 3 y 1 h 1.. c n 2 h n 2 + 2c n 1 (h n 2 +h n 1 + c n h n 1 = 3 y n 1 h n 1 3 y n 2 h n 2 c n 1 h n 1 + 2c n h n 1 = 3y n 3 y n 1 h n 1. Example: n = 3, clamped, with data (1,2, (2,3, (3,5, and y 1 = 1, y 3 = 2. Three equations: 2c 1 + c 2 = 3(2 1 3, c 1 + 4c 2 + c 3 = 3, c 2 + 2c 3 = 6 3(2, with solution c 1 = c 3 = 1/2, c 2 = 1; S 1 (x = (2 x 3 /6 + (x 1 3 /3 + 13(2 x/6 + 8(x 1/3, S 2 (x = (3 x 3 /3 (x 2 3 /6 + 8(3 x/3 + 31(x 2/6.

9 Parabolically Terminated: S 1 = S n 1 = 0; d 1 = d n 1 = 0, c 1 = c 2, c n 1 = c n. Linear system equations are a tridiagonal system c 1 c 2 = 0 c 1 h 1 + 2c 2 (h 1 +h 2 + c 3 h 2 = 3 y 2 3 y 1 h 2 h 1.. c n 2 h n 2 + 2c n 1 (h n 2 +h n 1 + c n h n 1 = 3 y n 1 3 y n 2 h n 1 h n 2 c n 1 c n = 0. 9 Example: n = 3, parabolically terminated, with data (1,2, (2,3, (3,5. One equation: 2c 2 = 3(2 1, with solution c 2 = 3/2 = c 1 = c 3 ; S 1 (x = (2 x 3 /2 + (x 1 3 /2 + 3(2 x/2 + 5(x 1/2, from S 2 (x = (3 x 3 /2 + (x 2 3 /2 + 5(3 x/2 + 9(x 2/2, S j (x = c j (x j+1 x 3 + c j+1 (x x j 3 + ( y j c jh j 3h j 3h j+1 h j 3 (xj+1 x+ ( y j+1 c j+1h j+1 (x xj. h j+1 3

10 Not-a-Knot Splines: S 1 (x 2 = S 2 (x 2, S n 2(x n 1 = S n 1(x n 1, so d 1 = d 2, d n 2 = d n 1, and S 1 = S 2, S n 2 = S n 1. Then (c 2 c 1 /h 1 = (c 3 c 2 /h 2, (c n 1 c n 2 /h n 2 = (c n c n 1 /h n 1, so 1 st and n th equations become c 1 h 2 c 2 (h 1 + h 2 + c 3 h 1 = 0, and c n 2 h n 1 c n 1 (h n 2 + h n 1 + c n h n 2 = 0. Linear system equations are a tridiagonal system c 1 h 2 c 2 (h 1 + h 2 + c 3 h 1 = 0 c 1 h 1 + 2c 2 (h 1 +h 2 + c 3 h 2 = 3 y 2 3 y 1 h 2 h 1.. c n 2 h n 2 + 2c n 1 (h n 2 +h n 1 + c n h n 1 = 3 y n 1 3 y n 2 h n 1 h n 2 c n 2 h n 1 c n 1 (h n 2 + h n 1 + c n h n 2 = 0. Example: n = 3, not-a-knot, with data (1,2, (2,3, (3,5. Three equations: c 1 2c 2 + c 3 = 0, c 1 + 4c 2 + c 3 = 3, c 1 2c 2 + c 3 = 0? but c 1 = c 2, so also = c 3 and therefore c 1 = c 2 = c 3 = 1/2; S 1 (x = (2 x 3 /6 + (x 1 3 /6 + 11(2 x/6 + 17(x 1/6, or S 2 (x = (3 x 3 /6 + (x 2 3 /6 + 17(3 x/6 + 29(x 2/6. Also S 1 (x = 2 + (x 1/2 + (x 1 2 /2 or = 3 + 3(x 2/2 + (x 2 2 /2 = S 2 (x. 10

11 11 Efficient Spline Evaluation Setup: solve linear system for c j s in O(n time For each evaluation point x, find interval x [x j, x j+1 ] in O(log(n time and evaluate S j (x in O(1 time. Alternate formula for S j (x (without computing b j s and d j s: S j (x = c j (x j+1 x 3 + c j+1 (x x j 3 + ( y j c jh j (xj+1 x+ ( y j+1 c j+1h j+1 (x xj. 3h j 3h j+1 h j 3 h j+1 3 Compare with poly. interpolation, with O(n 2 setup and O(n evaluation time. Error Theorem : if f C 4 [a, b], with max x [a,b] f (4 (x = M, and S(x is the unique clamped spline for f(x with a = x 1 < x 2 < x n = b, then f(x S(x 5M 384 max 1 j<n h4 j. Smoothness Theorem : if f C 2 [a, b], and S(x is the unique clamped spline for f(x with a = x 1 < x 2 < x n = b, and U C 2 [a, b], with U S, satisfying the same interpolation and deriative endpoint conditions, then b a (U (x 2 > b a (S (x 2.