Lecture 10 Polynomial interpolation

Transcription

1 Lecture 10 Polynomial interpolation Weinan E 1,2 and Tiejun Li 2 1 Department of Mathematics, Princeton University, weinan@princeton.edu 2 School of Mathematical Sciences, Peking University, tieli@pku.edu.cn No.1 Science Building, 1575

2 Outline Examples Polynomial interpolation Piecewise polynomial interpolation

3 Basic motivations Plotting a smooth curve through discrete data points Suppose we have a sequence of data points Coordinates x 1 x 2 x n Function y 1 y 2 y n Try to plot a smooth curve (a continuous differentiable function) connecting these discrete points.

4 Basic motivations Representing a complicate function by a simple one Suppose we have a complicate function y = f(x), we want to compute function values, derivatives, integrations,... very quickly and easily. One strategy 1. Compute some discrete points from the complicate form; 2. Interpolate the discrete points by a polynomial function or piecewise polynomial function; 3. Compute the function values, derivatives or integrations via the simple form.

5 Polynomial interpolation Polynomial interpolation is one the most fundamental problems in numerical methods.

6 Outline Examples Polynomial interpolation Piecewise polynomial interpolation

7 Method of undetermined coefficients Suppose we have n + 1 discrete points (x 0, y 0), (x 1, y 1),..., (x n, y n) We need a polynomial of degree n to do interpolation (n + 1 equations and n + 1 undetermined coefficients a 0, a 1,..., a n p n(x) = a nx n + a n 1x n a 0 Equations p n(x 0) = y 0 p n(x 1) = y 1 p n(x n) = y n

8 Method of undetermined coefficients The coefficient matrix V n = x n 0 x n 1 0 x 0 x n 1 x n 1 1 x 1 x n n x n 1 n x n is a Vandermonde determinant, nonsingular if x i x j (i j). Though this method can give the interpolation polynomial theoretically, the condition number of the Vandermonde matrix is very bad! For example, if x 0 = 0, x 1 = 1 n, x2 = 2 n,, xn = 1 then V n 1 n n!

9 Lagrange interpolating polynomial Consider the interpolation problem for 2 points (linear interpolation), one type is the point-slope form Another type is as p(x) = y1 y0 y0x1 y1x0 x + x 1 x 0 x 1 x 0 p(x) = y 0l 0(x) + y 1l 1(x) where satisfies l 0(x) = x x1 x 0 x 1, l 1(x) = x x0 x 1 x 0 l 0(x 0) = 1, l 0(x 1) = 0; l 1(x 0) = 0, l 1(x 1) = 1 l 0(x), l 1(x) are called basis functions. They are another base for space spanned by functions 1, x.

10 Lagrange interpolating polynomial Define the basis function l i(x) = then we have (x x0)(x x1) (x xi 1)(x xi+1) (x xn) (x i x 0)(x i x 1) (x i x i 1)(x i x i+1) (x i x n) l i(x j) = δ ij = { 1 i = j 0 i j The functions l i(x) (i = 0, 1,..., n) form a new basis in P n instead of 1, x, x 2,..., x n.

11 Lagrange interpolating polynomial General form of the Lagrange polynomial interpolation L n(x) = y 0l 0(x) + y 1l 1(x) + + y nl n(x) then L n(x) satisfies the interpolation condition. The shortcoming of Lagrange interpolation polynomial: If we add a new interpolation point into the sequence, all the basis functions will be useless!

12 Newton interpolation Define the 0-th order divided difference f[x i] = f(x i) Define the 1-th order divided difference f[x i, x j] = f[xi] f[xj] x i x j Define the k-th order divided difference by k 1-th order divided difference recursively f[x i0, x i1,..., x ik ] = f[xi 0, x i1,..., x ik 1 ] f[x i1, x i2,..., x ik ] x i0 x ik

13 Newton interpolation Recursively we have the following divided difference table Coordinates 0-th order 1-th order 2-th order x 0 f[x 0] x 1 f[x 1] f[x 0, x 1] x 2 f[x 2] f[x 1, x 2] f[x 0, x 1, x 2] x 3 f[x 3] f[x 2, x 3] f[x 1, x 2, x 3]....

14 Newton interpolation Divided difference table: an example Discrete data points x f(x) Divided difference table i x i f[x i ] f[x i 1, x i ] f[x i 2, x i 1, x i ] f[x 0, x 1, x 2, x 3 ]

15 Newton interpolation The properties of divided difference 1. f[x 0, x 1,..., x k ] is the linear combination of f(x 0), f(x 1),..., f(x n). 2. The value of f[x 0, x 1,..., x k ] does NOT depend on the order the coordinates x 0, x 1,..., x k. 3. If f[x, x 0,..., x k ] is a polynomial of degree m, then f[x, x 0,..., x k, x k+1 ] is of degree m If f(x) is a polynomial of degree n, then f[x, x 0,..., x n] = 0

16 Newton interpolation From the definition of divided difference, we have for any function f(x) f(x) = f[x 0] + f[x 0, x 1](x x 0) + f[x 0, x 1, x 2](x x 0)(x x 1) + + f[x 0, x 1,..., x n](x x 0)(x x 1) (x x n 1) +f[x, x 0, x 1,..., x n](x x 0)(x x 1) (x x n) Take f(x) as the Lagrange interpolation polynomial L n(x), because L n[x, x 0, x 1,..., x n] = 0 we have L n(x) = f[x 0] + f[x 0, x 1](x x 0) + f[x 0, x 1, x 2](x x 0)(x x 1)+ + + f[x 0, x 1,..., x n](x x 0)(x x 1) (x x n 1) This formula is called Newton interpolation formula.

17 Hermite interpolation Hermite interpolation is the interpolation specified derivatives. Formulation: find a polynomial p(x) such that p(x 0) = f(x 0), p (x 0) = f (x 0), p(x 1) = f(x 1), p (x 1) = f (x 1) Sketch of Hermite interpolation Hermite interpolation

18 Hermite interpolation We need a cubic polynomial to fit the four degrees of freedom, one choice is p(x) = a + b(x x 0) + c(x x 0) 2 + d(x x 0) 2 (x x 1) We have p (x) = b + 2c(x x 0) + 2d(x x 0)(x x 1) + d(x x 0) 2 then we have f(x 0) = a, f (x 0) = b f(x 1) = a + bh + ch 2, f (x 1) = b + 2ch + dh 2 (h = x 1 x 0) a, b, c, d could be solved.

19 Error estimates Theorem Suppose a = x 0 < x 1 < < x n = b, f(x) C n+1 [a, b], L n(x) is the Lagrange interpolation polynomial, then where E(f; x) = f(x) L n(x) ωn(x) (n + 1)! Mn+1 ω n(x) = (x x 0)(x x 1) (x x n), M n+1 = max x [a,b] f (n+1) (x). Remark: This theorem doesn t imply the uniform convergence when n.

20 Runge phenomenon Suppose take the equi-partitioned nodes f(x) = Lagrange interpolation (n = 10) x 2 x i = 1 + 2i, i = 0, 1,..., n n

21 Remark on polynomial interpolation Runge phenomenon tells us Lagrange interpolation could NOT guarantee the uniform convergence when n. Another note: high order polynomial interpolation is unstable! This drives us to investigate the piecewise interpolation.

22 Outline Examples Polynomial interpolation Piecewise polynomial interpolation

23 Piecewise linear interpolation Suppose we have n + 1 discrete points (x 0, y 0), (x 1, y 1),..., (x n, y n) Piecewise linear interpolation is to connect the discrete data points as

24 Tent basis functions Define the piecewise linear basis functions as x x 1, x [x 0, x 1], l n,0(x) = x 0 x 1 0, x [x 1, x n], x x i 1, x [x i 1, x i], x i x i 1 x x l n,i(x) = i+1, x [x i, x i+1], i = 1, 2,..., n 1, x i x i+1 0, x / [x i 1, x i+1], x x n 1, x [x n 1, x n], l n,n(x) = x n x n 1 0, x [x 0, x n 1].

25 Tent basis functions The sketch of tent basis function l i (x) 1 l 0 (x) l i (x)(0 <i<n) l n (x) x 0 x 1 x i 1 x i x i+1 x n 1 x n x Ξ 2.2: #}P s@aλ: l i (x) VΞ_

26 Piecewise linear interpolation function With the above tent basis function l n,i(x), we have l n,i(x j) = δ ij = { 1 i = j 0 i j The functions l n,i(x) form a basis in piecewise linear function space with nodes x i (i = 0, 1,..., n). Piecewise linear interpolation function p(x) = y 0l n,0(x) + y 1l n,1(x) + + y nl n,n(x) then p(x) satisfies the interpolation condition.

27 Cubic spline In order to make the interpolation curve more smooth, cubic spline is introduced. Formulation: Given discrete points (x 0, y 0), (x 1, y 1),..., (x n, y n), find function S h (x) such that (1) S h (x) is a cubic polynomial in each interval [x i, x i+1]; (2) S h (x i) = y i, i = 0, 1,..., n; (3) S h (x) C 2 [a, b].

28 Cubic spline Suppose we have n cubic polynomials in each interval, we have 4n unknowns totally. The interpolation condition gives 2n equations, S h (x) C 1 gives n 1 equations, S h (x) C 2 gives n 1 equations, so we have 4n 2 equations totally, we need some boundary conditions. Supplementary boundary conditions: (1) Fixed boundary: S h(x 0) = f (x 0),S h(x n) = f (x n); (2) Natural boundary: S h(x 0) = 0,S h(x n) = 0; (3) Periodic boundary: S h (x 0) = S h (x n), S h(x 0) = S h(x n), S h(x 0) = S h(x n). Each type of boundary condition gives 2 equations, thus we have 4n equations and 4n unknowns. The system could be solved theoretically. Problem: Why are piecewise cubic polynomials needed?)

29 Homework assignment Take interpolation points x k = 1 + 2k, k = 0, 1,..., n n for Runge function, plot the Lagrange polynomial of degree n (n = 1, 2,..., 15). Take interpolation points x k = cos kπ, k = 0, 1,..., n n for Runge function, plot the Lagrange polynomial of degree n (n = 1, 2,..., 15).