Initial-Value Problems for ODEs Eulers Method II: Error Bounds - PowerPoint PPT Presentation

Initial-Value Problems for ODEs Euler’s Method II: Error Bounds 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

Computational Lemmas Error Bound Example Outline Computational Lemmas 1 Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 2 / 25

Computational Lemmas Error Bound Example Outline Computational Lemmas 1 Error Bound for Euler’s Method 2 Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 2 / 25

Computational Lemmas Error Bound Example Outline Computational Lemmas 1 Error Bound for Euler’s Method 2 Error Bound Example 3 Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 2 / 25

Computational Lemmas Error Bound Example Outline Computational Lemmas 1 Error Bound for Euler’s Method 2 Error Bound Example 3 Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 3 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Lemma 1 For all x ≥ − 1 and any positive m , we have 0 ≤ ( 1 + x ) m ≤ e mx Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 4 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Proof of Lemma 1 Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 5 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Proof of Lemma 1 Applying Taylor’s Theorem with f ( x ) = e x , x 0 = 0, and n = 1 gives e x = 1 + x + 1 2 x 2 e ξ where ξ is between x and zero. Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 5 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Proof of Lemma 1 Applying Taylor’s Theorem with f ( x ) = e x , x 0 = 0, and n = 1 gives e x = 1 + x + 1 2 x 2 e ξ where ξ is between x and zero. Thus 0 ≤ 1 + x ≤ 1 + x + 1 2 x 2 e ξ = e x Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 5 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Proof of Lemma 1 Applying Taylor’s Theorem with f ( x ) = e x , x 0 = 0, and n = 1 gives e x = 1 + x + 1 2 x 2 e ξ where ξ is between x and zero. Thus 0 ≤ 1 + x ≤ 1 + x + 1 2 x 2 e ξ = e x and, because 1 + x ≥ 0, we have 0 ≤ ( 1 + x ) m ≤ ( e x ) m = e mx Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 5 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Lemma 2 If s and t are positive real numbers, { a i } k i = 0 is a sequence satisfying a 0 ≥ − t / s Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 6 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Lemma 2 If s and t are positive real numbers, { a i } k i = 0 is a sequence satisfying a 0 ≥ − t / s and a i + 1 ≤ ( 1 + s ) a i + t for each i = 0 , 1 , 2 , . . . , k − 1, Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 6 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Lemma 2 If s and t are positive real numbers, { a i } k i = 0 is a sequence satisfying a 0 ≥ − t / s and a i + 1 ≤ ( 1 + s ) a i + t for each i = 0 , 1 , 2 , . . . , k − 1, then a 0 + t − t � � a i + 1 ≤ e ( i + 1 ) s s s Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 6 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Proof of Lemma 2 (1/3) For a fixed integer i , the inequality a i + 1 ≤ ( 1 + s ) a i + t Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 7 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Proof of Lemma 2 (1/3) For a fixed integer i , the inequality a i + 1 ≤ ( 1 + s ) a i + t implies that a i + 1 ( 1 + s ) a i + t ≤ Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 7 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Proof of Lemma 2 (1/3) For a fixed integer i , the inequality a i + 1 ≤ ( 1 + s ) a i + t implies that a i + 1 ( 1 + s ) a i + t ≤ ( 1 + s )[( 1 + s ) a i − 1 + t ] + t ≤ Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 7 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Proof of Lemma 2 (1/3) For a fixed integer i , the inequality a i + 1 ≤ ( 1 + s ) a i + t implies that a i + 1 ( 1 + s ) a i + t ≤ ( 1 + s )[( 1 + s ) a i − 1 + t ] + t ≤ ( 1 + s ) 2 a i − 1 + [ 1 + ( 1 + s )] t = Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 7 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Proof of Lemma 2 (1/3) For a fixed integer i , the inequality a i + 1 ≤ ( 1 + s ) a i + t implies that a i + 1 ( 1 + s ) a i + t ≤ ( 1 + s )[( 1 + s ) a i − 1 + t ] + t ≤ ( 1 + s ) 2 a i − 1 + [ 1 + ( 1 + s )] t = � 1 + ( 1 + s ) + ( 1 + s ) 2 � ( 1 + s ) 3 a i − 2 + t ≤ Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 7 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas Proof of Lemma 2 (1/3) For a fixed integer i , the inequality a i + 1 ≤ ( 1 + s ) a i + t implies that a i + 1 ( 1 + s ) a i + t ≤ ( 1 + s )[( 1 + s ) a i − 1 + t ] + t ≤ ( 1 + s ) 2 a i − 1 + [ 1 + ( 1 + s )] t = � 1 + ( 1 + s ) + ( 1 + s ) 2 � ( 1 + s ) 3 a i − 2 + t ≤ . . . ( 1 + s ) i + 1 a 0 + 1 + ( 1 + s ) + ( 1 + s ) 2 + · · · + ( 1 + s ) i � � t ≤ Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 7 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas a i + 1 ≤ ( 1 + s ) i + 1 a 0 + 1 + ( 1 + s ) + ( 1 + s ) 2 + · · · + ( 1 + s ) i � � t Proof of Lemma 2 (2/3) Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 8 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas a i + 1 ≤ ( 1 + s ) i + 1 a 0 + 1 + ( 1 + s ) + ( 1 + s ) 2 + · · · + ( 1 + s ) i � � t Proof of Lemma 2 (2/3) But i 1 + ( 1 + s ) + ( 1 + s ) 2 + · · · + ( 1 + s ) i = ( 1 + s ) j � j = 0 is a geometric series with ratio ( 1 + s ) Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 8 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas a i + 1 ≤ ( 1 + s ) i + 1 a 0 + 1 + ( 1 + s ) + ( 1 + s ) 2 + · · · + ( 1 + s ) i � � t Proof of Lemma 2 (2/3) But i 1 + ( 1 + s ) + ( 1 + s ) 2 + · · · + ( 1 + s ) i = ( 1 + s ) j � j = 0 is a geometric series with ratio ( 1 + s ) that sums to 1 − ( 1 + s ) i + 1 = 1 s [( 1 + s ) i + 1 − 1 ] 1 − ( 1 + s ) Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 8 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas a i + 1 ≤ ( 1 + s ) i + 1 a 0 + � 1 + ( 1 + s ) + ( 1 + s ) 2 + · · · + ( 1 + s ) i � t Proof of Lemma 2 (3/3) Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 9 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas a i + 1 ≤ ( 1 + s ) i + 1 a 0 + � 1 + ( 1 + s ) + ( 1 + s ) 2 + · · · + ( 1 + s ) i � t Proof of Lemma 2 (3/3) Thus a i + 1 ≤ ( 1 + s ) i + 1 a 0 + ( 1 + s ) i + 1 − 1 a 0 + t − t � � t = ( 1 + s ) i + 1 s s s Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 9 / 25

Computational Lemmas Error Bound Example Euler’s Method: Computational Lemmas a i + 1 ≤ ( 1 + s ) i + 1 a 0 + � 1 + ( 1 + s ) + ( 1 + s ) 2 + · · · + ( 1 + s ) i � t Proof of Lemma 2 (3/3) Thus a i + 1 ≤ ( 1 + s ) i + 1 a 0 + ( 1 + s ) i + 1 − 1 a 0 + t − t � � t = ( 1 + s ) i + 1 s s s and using Lemma 1 with x = 1 + s gives a 0 + t − t � � a i + 1 ≤ e ( i + 1 ) s s . s Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 9 / 25

Computational Lemmas Error Bound Example Outline Computational Lemmas 1 Error Bound for Euler’s Method 2 Error Bound Example 3 Numerical Analysis (Chapter 5) Euler’s Method II: Error Bounds R L Burden & J D Faires 10 / 25