lect3_1-1 - Interpolation and Lagrange Polynomials(3.1 1...

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Interpolation and Lagrange Polynomials - (3.1) 1. Polynomial Interpolation: Problem: Given n ± 1 pairs of data points x i , y i , i ² 0,1,..., n , we want to find a polynomial P k x   of lowest possible degree for which P k x i   ² y i , i ² 0,1,..., n . The polynomial P k x   is said to interpolate the data x i , y i , i ² 0,1,..., n and is called an interpolating polynomial. Graphically, P k x   is an approximation to f x   and satisfies the conditions : P k x i   ² f x i   , i ² 0,1,..., n . -10 -5 0 5 10 -3 -2 -1 1 2 3 x y ² f x   , -.-.- y ² P k x   Obviously, for this example P k x   is not a good approximation to f x   though P k x   satisfies the conditions: P k x i   ² y i for i ² 0,1,2. Note that the differences between a k th degree Taylor polynomial and a k th degree interpolating polynomial are: a. P Taylor x   ² f x   at only x ² x 0 and P interpolating x   ² f x   , at x ² x 0 , x 1 , ..., x n . b. P Taylor requires knowledge of f U , f UU , ... but P interpolating requires f x 0   , f x 1   , ..., f x n   . 2. Lagrange Interpolating Polynomials: a. Lagrange Polynomials: For k ² 0,1,..., n , define L n , k x   ² ± i ² 0, i p k n x " x i   x k " x i   ² x " x 0   ... x " x k " 1   x " x k ± 1   ... x " x n   x k " x 0   ... x k " x k " 1   x k " x k ± 1   ... x k " x n   . L n , k x   are called Lagrange polynomials . For example, let x i ² i , i ² 0,1,2,3. L 3,1 x   ² x x " 2   x " 3   1 " 0   1 " 2   1 " 3   ² 1 2 x x " 2   x " 3   Observe that Lagrange Polynomials have the following properties: i. L n , k x k   ² 1 L n , k x k   ² x k " x 0   ... x k " x k " 1   x k " x k ± 1   ... x k " x n   x k " x 0   ... x k " x k " 1   x k " x k ± 1   ... x k " x n   ² 1 ii. L n , k x i   ² 0 for i p k . L n , k x i   ² x i " x 0   ... x i " x i   ... x i " x k " 1   x i " x k ± 1   ... x i " x n   x k " x 0   ... x k " x i   ... x k " x k " 1   x k " x k ± 1   ... x k " x n   ² 0 1
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b. Lagrange Interpolating Polynomials: The polynomial P n x   ² ! k ² 0 n y k L n , k x   ² y 0 L n ,0 x   ± y 1 L n ,1 x   ± ... ± y n L n , n x   is called the n th Lagrange interpolating polynomial . Observe that for i ² 0,1,..., n P n x i   ² ! k ² 0 n y k L n , k x i   ² y i L n , i x i   ² y i . P n x   is an n th polynomial that agrees with f x   at x 0 , x 1 , ..., x n . Theorem 3 . 3 Suppose that x 0 , ..., x n are distinct numbers in the interval a , b and f n ± 1   is continuous in a , b . For each x in a , b , there exists a number c x   in a , b such that f x   ² P n x   ± f n ± 1   c x    n ± 1   ! x " x 0   x " x 1   ... x " x n   . Proof For x p x i and x in a , b , define g t   ² f t   " P n t   " f x   " P n x    ± k ² 0 n t " x k   x " x k . Clearly, g x i   ² 0, i ² 0,1,..., n . Observe that g x   ² 0. Since x p x i , g t   ² 0 at t ² x 0 , x 1 , ..., x n , x .
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