Solutions of Equations in One Variable Secant & Regula Falsi - PowerPoint PPT Presentation

Solutions of Equations in One Variable Secant & Regula Falsi Methods 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

Secant Derivation Secant Example Regula Falsi Outline Secant Method: Derivation & Algorithm 1 Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 2 / 25

Secant Derivation Secant Example Regula Falsi Outline Secant Method: Derivation & Algorithm 1 Comparing the Secant & Newton’s Methods 2 Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 2 / 25

Secant Derivation Secant Example Regula Falsi Outline Secant Method: Derivation & Algorithm 1 Comparing the Secant & Newton’s Methods 2 The Method of False Position (Regula Falsi) 3 Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 2 / 25

Secant Derivation Secant Example Regula Falsi Outline Secant Method: Derivation & Algorithm 1 Comparing the Secant & Newton’s Methods 2 The Method of False Position (Regula Falsi) 3 Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 3 / 25

Secant Derivation Secant Example Regula Falsi Rationale for the Secant Method Problems with Newton’s Method Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 4 / 25

Secant Derivation Secant Example Regula Falsi Rationale for the Secant Method Problems with Newton’s Method Newton’s method is an extremely powerful technique, but it has a major weakness: the need to know the value of the derivative of f at each approximation. Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 4 / 25

Secant Derivation Secant Example Regula Falsi Rationale for the Secant Method Problems with Newton’s Method Newton’s method is an extremely powerful technique, but it has a major weakness: the need to know the value of the derivative of f at each approximation. Frequently, f ′ ( x ) is far more difficult and needs more arithmetic operations to calculate than f ( x ) . Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 4 / 25

Secant Derivation Secant Example Regula Falsi Derivation of the Secant Method f ( x ) − f ( p n − 1 ) lim f ′ ( p n − 1 ) = . x − p n − 1 x → p n − 1 Circumvent the Derivative Evaluation Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 5 / 25

Secant Derivation Secant Example Regula Falsi Derivation of the Secant Method f ( x ) − f ( p n − 1 ) lim f ′ ( p n − 1 ) = . x − p n − 1 x → p n − 1 Circumvent the Derivative Evaluation If p n − 2 is close to p n − 1 , then f ′ ( p n − 1 ) ≈ f ( p n − 2 ) − f ( p n − 1 ) = f ( p n − 1 ) − f ( p n − 2 ) . p n − 2 − p n − 1 p n − 1 − p n − 2 Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 5 / 25

Secant Derivation Secant Example Regula Falsi Derivation of the Secant Method f ( x ) − f ( p n − 1 ) lim f ′ ( p n − 1 ) = . x − p n − 1 x → p n − 1 Circumvent the Derivative Evaluation If p n − 2 is close to p n − 1 , then f ′ ( p n − 1 ) ≈ f ( p n − 2 ) − f ( p n − 1 ) = f ( p n − 1 ) − f ( p n − 2 ) . p n − 2 − p n − 1 p n − 1 − p n − 2 Using this approximation for f ′ ( p n − 1 ) in Newton’s formula gives p n = p n − 1 − f ( p n − 1 )( p n − 1 − p n − 2 ) f ( p n − 1 ) − f ( p n − 2 ) Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 5 / 25

Secant Derivation Secant Example Regula Falsi Derivation of the Secant Method f ( x ) − f ( p n − 1 ) lim f ′ ( p n − 1 ) = . x − p n − 1 x → p n − 1 Circumvent the Derivative Evaluation If p n − 2 is close to p n − 1 , then f ′ ( p n − 1 ) ≈ f ( p n − 2 ) − f ( p n − 1 ) = f ( p n − 1 ) − f ( p n − 2 ) . p n − 2 − p n − 1 p n − 1 − p n − 2 Using this approximation for f ′ ( p n − 1 ) in Newton’s formula gives p n = p n − 1 − f ( p n − 1 )( p n − 1 − p n − 2 ) f ( p n − 1 ) − f ( p n − 2 ) This technique is called the Secant method Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 5 / 25

Secant Derivation Secant Example Regula Falsi Secant Method: Using Successive Secants y y 5 f ( x ) p 3 p 0 p 2 p p 4 p 1 x Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 6 / 25

Secant Derivation Secant Example Regula Falsi The Secant Method p n = p n − 1 − f ( p n − 1 )( p n − 1 − p n − 2 ) f ( p n − 1 ) − f ( p n − 2 ) Procedure Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 7 / 25

Secant Derivation Secant Example Regula Falsi The Secant Method p n = p n − 1 − f ( p n − 1 )( p n − 1 − p n − 2 ) f ( p n − 1 ) − f ( p n − 2 ) Procedure Starting with the two initial approximations p 0 and p 1 , the approximation p 2 is the x -intercept of the line joining ( p 0 , f ( p 0 )) and ( p 1 , f ( p 1 )) . Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 7 / 25

Secant Derivation Secant Example Regula Falsi The Secant Method p n = p n − 1 − f ( p n − 1 )( p n − 1 − p n − 2 ) f ( p n − 1 ) − f ( p n − 2 ) Procedure Starting with the two initial approximations p 0 and p 1 , the approximation p 2 is the x -intercept of the line joining ( p 0 , f ( p 0 )) and ( p 1 , f ( p 1 )) . The approximation p 3 is the x -intercept of the line joining ( p 1 , f ( p 1 )) and ( p 2 , f ( p 2 )) , and so on. Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 7 / 25

Secant Derivation Secant Example Regula Falsi The Secant Method p n = p n − 1 − f ( p n − 1 )( p n − 1 − p n − 2 ) f ( p n − 1 ) − f ( p n − 2 ) Procedure Starting with the two initial approximations p 0 and p 1 , the approximation p 2 is the x -intercept of the line joining ( p 0 , f ( p 0 )) and ( p 1 , f ( p 1 )) . The approximation p 3 is the x -intercept of the line joining ( p 1 , f ( p 1 )) and ( p 2 , f ( p 2 )) , and so on. Note that only one function evaluation is needed per step for the Secant method after p 2 has been determined. Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 7 / 25

Secant Derivation Secant Example Regula Falsi The Secant Method p n = p n − 1 − f ( p n − 1 )( p n − 1 − p n − 2 ) f ( p n − 1 ) − f ( p n − 2 ) Procedure Starting with the two initial approximations p 0 and p 1 , the approximation p 2 is the x -intercept of the line joining ( p 0 , f ( p 0 )) and ( p 1 , f ( p 1 )) . The approximation p 3 is the x -intercept of the line joining ( p 1 , f ( p 1 )) and ( p 2 , f ( p 2 )) , and so on. Note that only one function evaluation is needed per step for the Secant method after p 2 has been determined. In contrast, each step of Newton’s method requires an evaluation of both the function and its derivative. Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 7 / 25

Secant Derivation Secant Example Regula Falsi The Secant Method: Algorithm To find a solution to f ( x ) = 0 given initial approximations p 0 and p 1 ; tolerance TOL ; maximum number of iterations N 0 . Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 8 / 25

Secant Derivation Secant Example Regula Falsi The Secant Method: Algorithm To find a solution to f ( x ) = 0 given initial approximations p 0 and p 1 ; tolerance TOL ; maximum number of iterations N 0 . 1 Set i = 2, q 0 = f ( p 0 ) , q 1 = f ( p 1 ) Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 8 / 25

Secant Derivation Secant Example Regula Falsi The Secant Method: Algorithm To find a solution to f ( x ) = 0 given initial approximations p 0 and p 1 ; tolerance TOL ; maximum number of iterations N 0 . 1 Set i = 2, q 0 = f ( p 0 ) , q 1 = f ( p 1 ) 2 While i ≤ N 0 do Steps 3–6: Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 8 / 25

Secant Derivation Secant Example Regula Falsi The Secant Method: Algorithm To find a solution to f ( x ) = 0 given initial approximations p 0 and p 1 ; tolerance TOL ; maximum number of iterations N 0 . 1 Set i = 2, q 0 = f ( p 0 ) , q 1 = f ( p 1 ) 2 While i ≤ N 0 do Steps 3–6: 3 Set p = p 1 − q 1 ( p 1 − p 0 ) / ( q 1 − q 0 ) . ( Compute p i ) Numerical Analysis (Chapter 2) Secant & Regula Falsi Methods R L Burden & J D Faires 8 / 25