INTERPOLATION AND POLYNOMIAL APPROXIMATION

Transcription

1 ITERPOLATIO AD POLYOMIAL APPROXIMATIO Interpolation Methods Why would we be interested in interpolation methods? Interpolation method are the basis for other procedures that we will deal with: umerical differentiation umerical integration Solution of ODE (ordinary differential equations) and PDE (partial differential equations) Why would we be interested in interpolation methods? These methods demonstrate some important theory about polynomials and the accuracy of numerical methods. The interpolation of polynomials serve as an excellent introduction to some techniques for drawing smooth curves. Interpolation uses the data to approximate a function, which will fit all of the data points. All of the data is used to approximate the values of the function inside the bounds of the data. Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x, y). Interpolation: Problem Specification Input: A set S of data f( x, y 0 0 )( x 1, y 1 )( xn, y n)g( xi, y i) Output: A function () which interpolates the given set of data, i.e., ( x ) = y, for 0 in. i i ote. In general, the interpolating function () is OT unique. 1

2 LAGRAGE ITERPOLATIO Linear interpolation uses a line segment that passes two points ( x, y ) 1 1 ( x, y ) 0 0 The slope between ( x0, y0) and ( x1, y1 ) is y y m x x from the point slope formula y y m( x x ) 0 0 can be rearranged as x x y y0 y1 y0 x1 x0 0...( 1 ) When formula (1) is expanded, the result is a polynomial of degree 1. Evaluating Px ( ) at x and x 0 1 produces y and y 0 1 respectively. P( x ) y 0 0 P( x ) y 1 1.() Lagrange used slightly different method to find this polynomial as follows ( x x1) ( x x0) y P1 ( x) y0 y1 ( x x ) ( x x ) () Each term on the right hand side of () involves linear factor. Hence the sum is a polynomial of degree 1. The quotients in () are denoted by ( x x ) ( x x ) L x L x ( ) ( ) 1 0 1, 0( ), 1, 1( ) x0 x1 x1 x0 (4)

3 which are called the Lagrange coefficients polynomials based on the nodes x notation, () can be written in summation form ( ) 1 1 k 1, k 0 1, 0 1 1, 1 k0 P x y L y L y L..(5) and x 0 1. Using this The generalization of (5) is the construction of a polynomial P ( x ) of degree at most that passes through the 1 points ( x0, y0),( x1, y1 ),...,( x, y) and has the form P ( x) y L ( x) k, k k0..(6) where L, ( x) k j0 jk j0 jk x x x k j x j..(7) Using (6) and (7), find 1 st, nd, and rd Lagrange Interpolation polynomial. 1 st order Lagrange interpolation polynomial where ( ) 1 1 k 1, k 0 1, 0 1 1, 1 k0 P x y L y L y L ( x x ) ( x x ) L x L x ( ) ( ) 1 0 1, 0( ), 1, 1( ) x0 x1 x1 x0

4 nd order Lagrange interpolation polynomial where ( ) k, k 0, 0 1, 1, k0 P x y L y L y L y L ( x x )( x x ) ( x x )( x x ) ( x x )( x x L x L x L x ) ( )( ) ( )( ) ( )( ) , 0( ),, 1( ),, ( ) x0 x1 x0 x x1 x0 x1 x x x0 x x1 rd order Lagrange interpolation polynomial where ( ) k, k 0, 0 1, 1,, k0 P x y L y L y L y L y L ( x x )( x x )( x x ) ( x x )( x x )( x x ) L x L x ( )( )( ) ( )( )( ) 1 0, 0( ),, 1( ) x0 x1 x0 x x0 x x1 x0 x1 x x1 x ( x x )( x x )( x x ) ( x x )( x x )( x x ) L x L x ( )( )( ) ( )( )( ) , ( ),, ( ) x x0 x x1 x x x x0 x x1 x x ERROR TERMS AD ERROR BOUDS Theorem : Assume that 1 x a, b then, f C a, b and that x0, x1,..., x a, b are 1 nodes. If f ( x) P ( x) E ( x)..(1) where P ( x ) polynomial that can be approximate f( x ). f ( x) P ( x) y L ( x) () k, k k0 4

5 The error term E ( x ) has the form x x x x... x x 1! 0 1 ( 1) E ( x) f ( c)...( ) for some value of c c( x) that lies in the interval a, b x x0 x x1... x x 1! ( 1 max ) () E ( x) M...( 4) M f c x xx 0 Theorem : (Error bounds for Lagrange Interpolation, equally space nodes) Assume that f( x ) defined on ab,, which contains equally space nodes xk x0 hk. Additionally, assume that f( x ) and the derivatives of f( x ) up to the +1, are continuous and x0, x1, x1, x and x, x bounded on the special subinterval ( 1 M f ) ( c) M1 for x0 x x 1,, respectively ; that is The error terms corresponding the cases =1,, have the following useful bounds on their magnitude. hm E1( x) valid for x 0, x1 8 hm E( x) valid for x 0, x 9 4 hm4 E( x) valid for x 0, x 4 5

6 Example : The function y sin x is tabulated as x 0 4 y Using lagrange interpolation polynomial, find its value at x Solution Given x0 0, x1, x, y0 0, y , y 1 4 Find P ( x), P ( ) and E( x) 6 6 estimate the error. k, k 0, 0 1, 1, 1, 1, 0 0, 0 k0 P ( x) y L y L y L y L 0 y L y L ( where, y 0 then y L 0 ) x0 x ( x x )( x x ) 0 x L, ( x) 16 1 x ( x x )( x x ) x0 x ( x x )( x x ) 0 1 x L, ( x) 4 8 x ( x x )( x x ) x 8 x P ( x) x 1 x x 1. 8x 0. 81x 0. 66x 4 P ( x) 0. 6x x P P hm E( x) where h 9 4 ( Find M, f ) ( x) M in 0, f ( x) cos x, f ( x) sin x, f ( x) cos x f ( x) cos x maximum value of f ( x) in 0, M cos 0 1 then hm 1 E ( x)

7 Exercise 1: Find Lagrange interpolation polynomial which agrees with the following data and obtain a bound x for the errors in estimating the function f( x) sin. x y Exercise : Consider y cos x over,. P1 ( x), P ( x) and P ( x) Determine the error bounds for the Lagrange polynomials Exercise : a) Find a second degree polynomial approximating the function f ( x) x ln x, and passing from the points 1, f ( 1),, f ( ) and ( 4, f ( 4 )). b) Approximate f ( 5. ) using the polynomial in part (a) c) use the error formula ( 1 ) () c f E( x) x x0 x x1 x x... x x 1! To estimate the error at x=.5 Exercise 4: Use lagrange interpolation to approximate y(0) for the following data. x 4 y 5 4 7

8 EWTO POLYOMIALS If the Lagrange interpolation are used, there is no constructive relationship between P1 ( x) and P( x). Each polynomial has to be constructed individually and the work required to compute the higher degree polynomials involves many computations. We take the new approach and construct ewton s polynomials that have the recursive pattern. P ( x) a a ( x x ) ( 1) P ( x) a a ( x x ) a ( x x )( x x ) P ( x) a ( x x )( x x ) ( ) P ( x) a a ( x x ) a ( x x )( x x ) a ( x x )( x x )( x x ) P ( x) a ( x x )( x x )( x x ) ( ) 0 1 P ( x) P ( x) a ( x x )( x x )( x x )...( x 1) ( 4) The polynomial (4) is said to be ewton Polynomial with centers x 0, x 1,..., x 1. It involves sums of products of linear factors up to a( x x0)( x x1)( x x)...( x x 1 ) x So P ( x ) will simply to be an ordinary polynomials of degree. The ewton interpolation uses a divided difference method. The technique allows one to add additional points easily P x a a x x a x x x x For given set of data points: x, y, x, y and x, y If three data points are available, the estimate is improved by introducing some curvature into the line connecting the points. P ( x) a a ( x x ) a ( x x )( x x )

9 A simple procedure can be used to determine the values of the coefficients x x a f ( x ) x x a 1 1 x x a f ( x1) f ( x0) x x 1 0 f ( x) f ( x1) f ( x1) f ( x0) x x1 x1 x0 x x 0 General Form of ewton s Interpolating Polynomials P ( x) f ( x ) ( x x ) f [ x, x ] ( x x )( x x ) f [ x, x, x ] a ( x x )( x x ) ( x x ) f [ x, x,, x ] f ( x ) 0 0 a f [ x, x ] a f [ x, x, x ] 1 0 a f [ x, x,, x, x ] f [ x, x ] i f [ x, x, x ] j i j k f ( x ) f ( x ) i x i x j j f [ x, x ] f [ x, x ] i j j k x i x k f [ x, x 1,, x1] f [ x 1, x,, x0] f [ x, x1,, x1, x0] x x0 and use to construct the divided differences in the following table x k f xk, f f,, f,,, f,,,, x 0 f x 0 x 1 f x 1 f x 1, x0 x f x f x, x x x 4 1 f x, x 1, x 0 f x f x, x f x, x, x 1 f x, x, x 1, x 0 f x 4 f x 4, x f x 4, x, x f x 4, x, x, x 1 f x 4, x, x, x 1, x 0 9

10 Theorem : ewton Polynomial Suppose that x, x,..., x ab,. There exist a unique polynomial P ( x ) of degree at most with the property that are +1 distinct numbers in The ewton form polynomial is j j 01,,..., f x P x for j P ( x) P ( x) a ( x x )( x x )( x x )...( x x ) where ak f x k, xk1,..., x1, x0 for k 01,,..., and the ewton approximation then f ( x) P ( x) E ( x) x x x x... x x 1! 0 1 ( 1) E ( x) f ( c) 10

11 EXAMPLE 1: Let f ( x) x 4x Construct the divided difference table based on the nodes x0 1, x1, x, x 4, x4 5, x5 6. Find the ewton polynomials P ( x ) base on the above nodes. x k k f x f, f,, f,,, f,,,, x f (1) 0 1 x1 f () 0 0 ( ) 1 x f () x 4 f (4) x4 5 f (5) x5 6 f (6) a, a, a 6, a 1, a 0 Where P ( x) a a x x a x x x x a x x x x x x P ( x) x 1 6 x 1 x 1 x 1 x x then add like terms P ( x) x 4x 11

12 EXAMPLE : For the function y f ( x), given the following table, use the divided difference to determine highest possible order and use the ewton s polynomial to determine the valu of 11 x if f ( x ). x k k f x f, f,, x0 0 f (0) 1 a = 0 x1 1 (1) f = a 1 x f () = a Highest possible order nd order P ( x) a a x x a x x x x P ( x) 11 x 0 x 0 x 1 then add like terms 1 11 P ( x) x x 1 find x where f ( x ) x x 1 x x 0 x x 8 0 x 4 ( x ) 0 then x 4 and x x x 0,, 4 0, 1

13 Exercise 1: Use the ewton divided-difference formula to compute the cubic interpolating x polynomial P ( x) for the function f( x) sin( ) using interpolation points x 0, x 1, x and x. What is an upper bound for the error at x=0.5? 0 1 Exercise : Given the table of the function f( x) x X 0 1 f(x) a) Write down the ewton polynomials P 1 (x), P (x), P (x). b) Evaluate f(.5) by using P (x). Exercise :For the function y f ( x) given the following table, use the divided differences to determine highest possible order and use ewton s polynomial to determine the value of x if f( x ) 0 x 0 1 y