Interpolation & Polynomial Approximation Lagrange Interpolating - PowerPoint PPT Presentation

Interpolation & Polynomial Approximation Lagrange Interpolating Polynomials II Numerical Analysis (9th Edition) R L Burden & J D Faires Beamer Presentation Slides prepared by John Carroll Dublin City University � 2011 Brooks/Cole, Cengage Learning c

Error Bound Error Example 1 Error Example 2 Outline Interpolating Polynomial Error Bound 1 Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 2 / 25

Error Bound Error Example 1 Error Example 2 Outline Interpolating Polynomial Error Bound 1 Example: 2nd Lagrange Interpolating Polynomial Error Bound 2 Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 2 / 25

Error Bound Error Example 1 Error Example 2 Outline Interpolating Polynomial Error Bound 1 Example: 2nd Lagrange Interpolating Polynomial Error Bound 2 Example: Interpolating Polynomial Error for Tabulated Data 3 Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 2 / 25

Error Bound Error Example 1 Error Example 2 Outline Interpolating Polynomial Error Bound 1 Example: 2nd Lagrange Interpolating Polynomial Error Bound 2 Example: Interpolating Polynomial Error for Tabulated Data 3 Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 3 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound Theorem Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 4 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound Theorem Suppose x 0 , x 1 , . . . , x n are distinct numbers in the interval [ a , b ] and f ∈ C n + 1 [ a , b ] . Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 4 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound Theorem Suppose x 0 , x 1 , . . . , x n are distinct numbers in the interval [ a , b ] and f ∈ C n + 1 [ a , b ] . Then, for each x in [ a , b ] , a number ξ ( x ) (generally unknown) between x 0 , x 1 , . . . , x n , and hence in ( a , b ) , exists with f ( x ) = P ( x ) + f ( n + 1 ) ( ξ ( x )) ( x − x 0 )( x − x 1 ) · · · ( x − x n ) ( n + 1 )! Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 4 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound Theorem Suppose x 0 , x 1 , . . . , x n are distinct numbers in the interval [ a , b ] and f ∈ C n + 1 [ a , b ] . Then, for each x in [ a , b ] , a number ξ ( x ) (generally unknown) between x 0 , x 1 , . . . , x n , and hence in ( a , b ) , exists with f ( x ) = P ( x ) + f ( n + 1 ) ( ξ ( x )) ( x − x 0 )( x − x 1 ) · · · ( x − x n ) ( n + 1 )! where P ( x ) is the interpolating polynomial given by n � P ( x ) = f ( x 0 ) L n , 0 ( x ) + · · · + f ( x n ) L n , n ( x ) = f ( x k ) L n , k ( x ) k = 0 Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 4 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound Error Bound: Proof (1/6) Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 5 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound Error Bound: Proof (1/6) Note first that if x = x k , for any k = 0 , 1 , . . . , n , then f ( x k ) = P ( x k ) , and choosing ξ ( x k ) arbitrarily in ( a , b ) yields the result: f ( x ) = P ( x ) + f ( n + 1 ) ( ξ ( x )) ( x − x 0 )( x − x 1 ) · · · ( x − x n ) ( n + 1 )! Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 5 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound Error Bound: Proof (1/6) Note first that if x = x k , for any k = 0 , 1 , . . . , n , then f ( x k ) = P ( x k ) , and choosing ξ ( x k ) arbitrarily in ( a , b ) yields the result: f ( x ) = P ( x ) + f ( n + 1 ) ( ξ ( x )) ( x − x 0 )( x − x 1 ) · · · ( x − x n ) ( n + 1 )! If x � = x k , for all k = 0 , 1 , . . . , n , define the function g for t in [ a , b ] by g ( t ) = f ( t ) − P ( t ) − [ f ( x ) − P ( x )] ( t − x 0 )( t − x 1 ) · · · ( t − x n ) ( x − x 0 )( x − x 1 ) · · · ( x − x n ) n ( t − x i ) � = f ( t ) − P ( t ) − [ f ( x ) − P ( x )] ( x − x i ) i = 0 Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 5 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound n ( t − x i ) � g ( t ) = f ( t ) − P ( t ) − [ f ( x ) − P ( x )] ( x − x i ) i = 0 Error Bound: Proof (2/6) Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 6 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound n ( t − x i ) � g ( t ) = f ( t ) − P ( t ) − [ f ( x ) − P ( x )] ( x − x i ) i = 0 Error Bound: Proof (2/6) Since f ∈ C n + 1 [ a , b ] , and P ∈ C ∞ [ a , b ] , it follows that g ∈ C n + 1 [ a , b ] . For t = x k , we have n ( x k − x i ) � g ( x k ) = f ( x k ) − P ( x k ) − [ f ( x ) − P ( x )] ( x − x i ) = 0 − [ f ( x ) − P ( x )] · 0 = 0 i = 0 Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 6 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound n ( t − x i ) � g ( t ) = f ( t ) − P ( t ) − [ f ( x ) − P ( x )] ( x − x i ) i = 0 Error Bound: Proof (3/6) Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 7 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound n ( t − x i ) � g ( t ) = f ( t ) − P ( t ) − [ f ( x ) − P ( x )] ( x − x i ) i = 0 Error Bound: Proof (3/6) We have seen that g ( x k ) = 0. Furthermore, n ( x − x i ) � g ( x ) = f ( x ) − P ( x ) − [ f ( x ) − P ( x )] ( x − x i ) i = 0 = f ( x ) − P ( x ) − [ f ( x ) − P ( x )] = 0 Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 7 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound n ( t − x i ) � g ( t ) = f ( t ) − P ( t ) − [ f ( x ) − P ( x )] ( x − x i ) i = 0 Error Bound: Proof (3/6) We have seen that g ( x k ) = 0. Furthermore, n ( x − x i ) � g ( x ) = f ( x ) − P ( x ) − [ f ( x ) − P ( x )] ( x − x i ) i = 0 = f ( x ) − P ( x ) − [ f ( x ) − P ( x )] = 0 Thus g ∈ C n + 1 [ a , b ] , and g is zero at the n + 2 distinct numbers x , x 0 , x 1 , . . . , x n . Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 7 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound Error Bound: Proof (4/6) Since g ∈ C n + 1 [ a , b ] , and g is zero at the n + 2 distinct numbers Theorem there exists a x , x 0 , x 1 , . . . , x n , by Generalized Rolle’s Theorem number ξ in ( a , b ) for which g ( n + 1 ) ( ξ ) = 0. Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 8 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound Error Bound: Proof (4/6) Since g ∈ C n + 1 [ a , b ] , and g is zero at the n + 2 distinct numbers Theorem there exists a x , x 0 , x 1 , . . . , x n , by Generalized Rolle’s Theorem number ξ in ( a , b ) for which g ( n + 1 ) ( ξ ) = 0. So g ( n + 1 ) ( ξ ) 0 = � n � f ( n + 1 ) ( ξ ) − P ( n + 1 ) ( ξ ) − [ f ( x ) − P ( x )] d n + 1 ( t − x i ) � = dt n + 1 ( x − x i ) i = 0 t = ξ Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 8 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound Error Bound: Proof (4/6) Since g ∈ C n + 1 [ a , b ] , and g is zero at the n + 2 distinct numbers Theorem there exists a x , x 0 , x 1 , . . . , x n , by Generalized Rolle’s Theorem number ξ in ( a , b ) for which g ( n + 1 ) ( ξ ) = 0. So g ( n + 1 ) ( ξ ) 0 = � n � f ( n + 1 ) ( ξ ) − P ( n + 1 ) ( ξ ) − [ f ( x ) − P ( x )] d n + 1 ( t − x i ) � = dt n + 1 ( x − x i ) i = 0 t = ξ However, P ( x ) is a polynomial of degree at most n , so the ( n + 1 ) st derivative, P ( n + 1 ) ( x ) , is identically zero. Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 8 / 25

Error Bound Error Example 1 Error Example 2 The Lagrange Polynomial: Theoretical Error Bound Error Bound: Proof (5/6) Numerical Analysis (Chapter 3) Lagrange Interpolating Polynomials II R L Burden & J D Faires 9 / 25