Piecewise Polynomial Interpolation

Transcription

1 Piecewise Polynomial Interpolation 1

2 Piecewise linear interpolation Suppose we have data point (x k,y k ), k =0, 1,...N. A piecewise linear polynomial that interpolates these points is given by p(x) =p k (x), x 2 [x k,x k+1 ] where the polynomials p k (x) can be written as p k (x) =a 1 (x x k )+a 0, k =1, 2,...N 1 At this point, we could easily compute the coe cients a =[a 1,a 0 ], but let s go through a more general procedure that we can use for higher order piecewise polynomials. 2

3 Piecewise linear interpolation We can write this as a function of a local variable t which measures the distance along the interval. t = x x k x k+1 x k x = h k t + x k where h k = x k+1 x k. Then, for 0 apple t apple 1, we have p k (x) =ep k (t) =ea 1 t + ea 0 t =1 (x k+1,y k+1 ) t =0 Constraints to impose ep k (0) = y k (x k,y k ) 0 <t<1 ep k (1) = y k+1 3

4 Piecewise linear interpolation This leads to a linear system apple apple ea1 ea 0 = apple yk y k+1 which we write as Aea = p. To solve for the coe we use the matrix co-factor C to get A 1 A 1 = CT det(a) = 1 det(a) apple cients, We have det(a) = 1, and so ea = C T p 4

5 Piecewise linear interpolation To see what ep k (t) looks like, we let q(t) =[t, 1] and write ep k (t) = ea T q(t) = p T Cq(t) apple T = yk y k apple t 1 The advantage of this form is that we can clearly see how the polynomial ep k (t) depends on our data y k and y k+1 : ep k (t) =(1 t)y k + ty k+1 5

6 Piecewise linear interpolation To build a spline in Matlab, we need to have the coefficients a 0 and a 1 of our original polynomial. p k (x) =a 1 (x x k )+a 0 We can recover these coe cients by substituting t = (x x k )/h k into ep k (t) =ea 1 t + ea 0 to get p k (x) =ea 1 x h k xk + ea 0 = a 1 (x x k )+a 0 where a 1 = ea 1 /h k and ea 0 = a 0. 6

7 Piecewise linear interpolation We can also write a =[a 1,a 0 ]using a = p T apple 1 1 CS = y k y k " yk+1 y # k = h k y k apple 1/hk These are exactly the coefficients that Matlab needs to build a spline using mkpp. function pp_linear = build_linear_spline(xdata,ydata)! N = length(xdata)-1;! h = diff(xdata); %(x1-x0, x2-x1, x3-x2,...)! C = [1-1; -1 0];! for k = 1:N,! p = [ydata(k); ydata(k+1)];! S = diag([1./h(k) 1]);! coeffs(k,:) = -p'*c*s;! end! pp_linear = mkpp(xdata,coeffs);! end 7

8 Piecewise linear interpolation (x 4,y 4 ) (x 0,y 0 ) (x 2,y 2 ) (x 5,y 5 ) 8

9 Piecewise linear interpolation Piecewise linear interpolation has these properties Continuous at nodes (x k,y k ). Derivatives are not continuous (the function is not smooth!) Can we come design a method that matches derivatives at the nodes as well? 9

10 Hermite interpolation We can impose smoothness at the nodes to get a nicer looking curve. Suppose we match the function values and derivatives. p(x k+1 )=y k+1 p 0 (x k+1 )=d k+1 This approach assumes that we have derivatives dk at each of the nodes xk. p(x k )=y k Four conditions allow us to find the four coefficients of a cubic polynomial p 0 (x k )=d k p k (x) =a 3 (x x k ) 3 + a 2 (x x k ) 2 + a 1 (x x k )+a 0 10

11 Hermite interpolation As before, define t = x x k x k+1 x k,! x = h k t + x k where h k = x k+1 x k. Then, for 0 apple t apple 1, we have p k (x) ep k (t) =ea 3 t 3 + ea 2 t 2 + ea 1 t + ea 0 and our four conditions become ep k (0) = y k ep k (1) = y k+1 ep 0 k (0) = d kh k ep 0 k (1) = d k+1h k 11

12 Hermite interpolation This leads to the following linear system for the coe cients ea n : We write the above system as Aea = p. To invert A, weusethe co-factor matrix C to get A 1 = CT det(a) See Strang s Introduction to Linear Algebra for more information on the co-factor formula for the matrix inverse. 12

13 Hermite interpolation We can then write ep k (t) as ep k (t) =ea T q(t) = p T Cq(t) where q(t) =[t 3,t 2,t,1] T and we have used the fact that det(a) = 1. Computing C, we get t 3 ep k (t) = T y k y k+1 d k h k d k+1 h k t t which leads to a formula that clearly shows how the polynomial depends on our data y k, y k+1, d k and d k+1 : ep k (t) = 1 t 2 (3 2t) y k + t 2 (3 2t) y k+1 + t(t 1) 2 h k d k + t 2 (t 1) h k d k+1 13

14 Piecewise Cubic Hermite Interpolating Polynomial (PCHIP) ep k (t) = 1 t 2 (3 2t) y k + t 2 (3 2t) y k+1 + t(t 1) 2 h k d k + t 2 (t 1) h k d k+1 x k )/h k, we can recover our original poly- If we substitute t =(x nomial in x : p k (x) =a 3 (x x k ) 3 + a 2 (x x k ) 2 + a 1 (x x k )+a 0 where a n = ea n /h n k. We can write these coe a = p T CS cients as These are exactly the coefficients that Matlab needs to build a spline using mkpp. where S = diag(h 3 k,h 2 k,h 1 k, 1), p =[y k,y k+1,d k h k,d k+1 h k ], and C is the co-factor matrix of our original coe cient matrix A. 14

15 The derivatives? Where do we get the derivatives d k from? If we know f(x), we can set d k = f 0 (x k ). We can approximate derivatives using data (x k,y k ). d k = G ( k 1, k), k = y k+1 y k h k where, for example, we could take G(a, b) = a+b 2, G(a, b) = a, G(a, b) = b or some other reasonable function. Matlab s PCHIP chooses derivatives to avoid introducing new extrema in the spline. We can impose additional smoothness conditions and solve implicitly for the derivatives. 15

16 Derivatives? Given data points (x 0,y 0 ),...,(x N,y N ), impose continuity in the second derivative at the interior nodes, p 00 k(x k+1 )=p 00 k+1(x k+1 ), k =1, 2,...N 1. We can formulate this as a linear expression F(d) =Bd + c =0 where B 2R (N+1) (N+1), c 2R (N+1) 1 and F(d) =0expresses the equality of the second derivative at the internal nodes, plus two additional constraints. Solving for d, we have d = B 1 c, d 2R N+1 16

17 End conditions? 17

18 Cubic spline 18

19 Periodic spline 19