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Solutions of Equations in One Variable Fixed-Point Iteration 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

  1. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 5 Possible Transpositions to x = g ( x ) x − x 3 − 4 x 2 + 10 x = g 1 ( x ) = � 10 x = g 2 ( x ) = x − 4 x 1 � 10 − x 3 x = g 3 ( x ) = 2 � 10 x = g 4 ( x ) = 4 + x x − x 3 + 4 x 2 − 10 x = g 5 ( x ) = 3 x 2 + 8 x Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 10 / 54

  2. Fixed-Point Iteration Convergence Criteria Sample Problem Numerical Results for f ( x ) = x 3 + 4 x 2 − 10 = 0 n g 1 g 2 g 3 g 4 g 5 0 1.5 1.5 1.5 1.5 1.5 1 − 0 . 875 0.8165 1.286953768 1.348399725 1.373333333 2 6.732 2.9969 1.402540804 1.367376372 1.365262015 ( − 8 . 65 ) 1 / 2 3 − 469 . 7 1.345458374 1.364957015 1.365230014 1 . 03 × 10 8 4 1.375170253 1.365264748 1.365230013 5 1.360094193 1.365225594 10 1.365410062 1.365230014 15 1.365223680 1.365230013 20 1.365230236 25 1.365230006 30 1.365230013 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 11 / 54

  3. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 x = g ( x ) with x 0 = 1 . 5 x = g 1 ( x ) = x − x 3 − 4 x 2 + 10 Does not Converge � 10 x = g 2 ( x ) = x − 4 x Does not Converge x = g 3 ( x ) = 1 � 10 − x 3 Converges after 31 Iterations 2 � 10 x = g 4 ( x ) = Converges after 12 Iterations 4 + x x = g 5 ( x ) = x − x 3 + 4 x 2 − 10 Converges after 5 Iterations 3 x 2 + 8 x Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 12 / 54

  4. Fixed-Point Iteration Convergence Criteria Sample Problem Outline Functional (Fixed-Point) Iteration 1 Convergence Criteria for the Fixed-Point Method 2 Sample Problem: f ( x ) = x 3 + 4 x 2 − 10 = 0 3 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 13 / 54

  5. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration A Crucial Question How can we find a fixed-point problem that produces a sequence that reliably and rapidly converges to a solution to a given root-finding problem? Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 14 / 54

  6. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration A Crucial Question How can we find a fixed-point problem that produces a sequence that reliably and rapidly converges to a solution to a given root-finding problem? The following theorem and its corollary give us some clues concerning the paths we should pursue and, perhaps more importantly, some we should reject. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 14 / 54

  7. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Convergence Result Let g ∈ C [ a , b ] with g ( x ) ∈ [ a , b ] for all x ∈ [ a , b ] . Let g ′ ( x ) exist on ( a , b ) with | g ′ ( x ) | ≤ k < 1 , ∀ x ∈ [ a , b ] . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 15 / 54

  8. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Convergence Result Let g ∈ C [ a , b ] with g ( x ) ∈ [ a , b ] for all x ∈ [ a , b ] . Let g ′ ( x ) exist on ( a , b ) with | g ′ ( x ) | ≤ k < 1 , ∀ x ∈ [ a , b ] . If p 0 is any point in [ a , b ] then the sequence defined by p n = g ( p n − 1 ) , n ≥ 1 , will converge to the unique fixed point p in [ a , b ] . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 15 / 54

  9. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Proof of the Convergence Result Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 16 / 54

  10. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Proof of the Convergence Result By the Uniquenes Theorem, a unique fixed point exists in [ a , b ] . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 16 / 54

  11. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Proof of the Convergence Result By the Uniquenes Theorem, a unique fixed point exists in [ a , b ] . Since g maps [ a , b ] into itself, the sequence { p n } ∞ n = 0 is defined for all n ≥ 0 and p n ∈ [ a , b ] for all n . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 16 / 54

  12. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Proof of the Convergence Result By the Uniquenes Theorem, a unique fixed point exists in [ a , b ] . Since g maps [ a , b ] into itself, the sequence { p n } ∞ n = 0 is defined for all n ≥ 0 and p n ∈ [ a , b ] for all n . MVT and the assumption that Using the Mean Value Theorem | g ′ ( x ) | ≤ k < 1, ∀ x ∈ [ a , b ] , we write | p n − p | = | g ( p n − 1 ) − g ( p ) | Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 16 / 54

  13. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Proof of the Convergence Result By the Uniquenes Theorem, a unique fixed point exists in [ a , b ] . Since g maps [ a , b ] into itself, the sequence { p n } ∞ n = 0 is defined for all n ≥ 0 and p n ∈ [ a , b ] for all n . MVT and the assumption that Using the Mean Value Theorem | g ′ ( x ) | ≤ k < 1, ∀ x ∈ [ a , b ] , we write | p n − p | = | g ( p n − 1 ) − g ( p ) | � | p n − 1 − p | � � ≤ � g ′ ( ξ ) Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 16 / 54

  14. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Proof of the Convergence Result By the Uniquenes Theorem, a unique fixed point exists in [ a , b ] . Since g maps [ a , b ] into itself, the sequence { p n } ∞ n = 0 is defined for all n ≥ 0 and p n ∈ [ a , b ] for all n . MVT and the assumption that Using the Mean Value Theorem | g ′ ( x ) | ≤ k < 1, ∀ x ∈ [ a , b ] , we write | p n − p | = | g ( p n − 1 ) − g ( p ) | � | p n − 1 − p | � � ≤ � g ′ ( ξ ) ≤ k | p n − 1 − p | where ξ ∈ ( a , b ) . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 16 / 54

  15. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Proof of the Convergence Result (Cont’d) Applying the inequality of the hypothesis inductively gives Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 17 / 54

  16. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Proof of the Convergence Result (Cont’d) Applying the inequality of the hypothesis inductively gives | p n − p | ≤ k | p n − 1 − p | Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 17 / 54

  17. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Proof of the Convergence Result (Cont’d) Applying the inequality of the hypothesis inductively gives | p n − p | ≤ k | p n − 1 − p | k 2 | p n − 2 − p | ≤ Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 17 / 54

  18. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Proof of the Convergence Result (Cont’d) Applying the inequality of the hypothesis inductively gives | p n − p | ≤ k | p n − 1 − p | k 2 | p n − 2 − p | ≤ k n | p 0 − p | ≤ Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 17 / 54

  19. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Proof of the Convergence Result (Cont’d) Applying the inequality of the hypothesis inductively gives | p n − p | ≤ k | p n − 1 − p | k 2 | p n − 2 − p | ≤ k n | p 0 − p | ≤ Since k < 1, n →∞ k n | p 0 − p | = 0 , n →∞ | p n − p | ≤ lim lim and { p n } ∞ n = 0 converges to p . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 17 / 54

  20. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then k n | p n − p | ≤ 1 − k | p 1 − p 0 | . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 18 / 54

  21. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then k n | p n − p | ≤ 1 − k | p 1 − p 0 | . Proof of Corollary (1 of 3) For n ≥ 1, the procedure used in the proof of the theorem implies that | p n + 1 − p n | = | g ( p n ) − g ( p n − 1 ) | Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 18 / 54

  22. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then k n | p n − p | ≤ 1 − k | p 1 − p 0 | . Proof of Corollary (1 of 3) For n ≥ 1, the procedure used in the proof of the theorem implies that | p n + 1 − p n | = | g ( p n ) − g ( p n − 1 ) | ≤ k | p n − p n − 1 | Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 18 / 54

  23. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then k n | p n − p | ≤ 1 − k | p 1 − p 0 | . Proof of Corollary (1 of 3) For n ≥ 1, the procedure used in the proof of the theorem implies that | p n + 1 − p n | = | g ( p n ) − g ( p n − 1 ) | ≤ k | p n − p n − 1 | ≤ · · · Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 18 / 54

  24. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then k n | p n − p | ≤ 1 − k | p 1 − p 0 | . Proof of Corollary (1 of 3) For n ≥ 1, the procedure used in the proof of the theorem implies that | p n + 1 − p n | = | g ( p n ) − g ( p n − 1 ) | ≤ k | p n − p n − 1 | ≤ · · · k n | p 1 − p 0 | ≤ Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 18 / 54

  25. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration | p n + 1 − p n | ≤ k n | p 1 − p 0 | Proof of Corollary (2 of 3) Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 19 / 54

  26. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration | p n + 1 − p n | ≤ k n | p 1 − p 0 | Proof of Corollary (2 of 3) Thus, for m > n ≥ 1, | p m − p n | = | p m − p m − 1 + p m − 1 − p m − 2 + · · · + p n + 1 − p n | Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 19 / 54

  27. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration | p n + 1 − p n | ≤ k n | p 1 − p 0 | Proof of Corollary (2 of 3) Thus, for m > n ≥ 1, | p m − p n | = | p m − p m − 1 + p m − 1 − p m − 2 + · · · + p n + 1 − p n | ≤ | p m − p m − 1 | + | p m − 1 − p m − 2 | + · · · + | p n + 1 − p n | Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 19 / 54

  28. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration | p n + 1 − p n | ≤ k n | p 1 − p 0 | Proof of Corollary (2 of 3) Thus, for m > n ≥ 1, | p m − p n | = | p m − p m − 1 + p m − 1 − p m − 2 + · · · + p n + 1 − p n | ≤ | p m − p m − 1 | + | p m − 1 − p m − 2 | + · · · + | p n + 1 − p n | k m − 1 | p 1 − p 0 | + k m − 2 | p 1 − p 0 | + · · · + k n | p 1 − p 0 | ≤ Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 19 / 54

  29. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration | p n + 1 − p n | ≤ k n | p 1 − p 0 | Proof of Corollary (2 of 3) Thus, for m > n ≥ 1, | p m − p n | = | p m − p m − 1 + p m − 1 − p m − 2 + · · · + p n + 1 − p n | ≤ | p m − p m − 1 | + | p m − 1 − p m − 2 | + · · · + | p n + 1 − p n | k m − 1 | p 1 − p 0 | + k m − 2 | p 1 − p 0 | + · · · + k n | p 1 − p 0 | ≤ k n � 1 + k + k 2 + · · · + k m − n − 1 � ≤ | p 1 − p 0 | . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 19 / 54

  30. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration 1 + k + k 2 + · · · + k m − n − 1 � | p m − p n | ≤ k n � | p 1 − p 0 | . Proof of Corollary (3 of 3) Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 20 / 54

  31. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration 1 + k + k 2 + · · · + k m − n − 1 � | p m − p n | ≤ k n � | p 1 − p 0 | . Proof of Corollary (3 of 3) However, since lim m →∞ p m = p , we obtain | p − p n | = m →∞ | p m − p n | lim Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 20 / 54

  32. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration 1 + k + k 2 + · · · + k m − n − 1 � | p m − p n | ≤ k n � | p 1 − p 0 | . Proof of Corollary (3 of 3) However, since lim m →∞ p m = p , we obtain | p − p n | = m →∞ | p m − p n | lim ∞ k n | p 1 − p 0 | � k i ≤ i = 1 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 20 / 54

  33. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration 1 + k + k 2 + · · · + k m − n − 1 � | p m − p n | ≤ k n � | p 1 − p 0 | . Proof of Corollary (3 of 3) However, since lim m →∞ p m = p , we obtain | p − p n | = m →∞ | p m − p n | lim ∞ k n | p 1 − p 0 | � k i ≤ i = 1 k n = 1 − k | p 1 − p 0 | . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 20 / 54

  34. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Example: g ( x ) = g ( x ) = 3 − x Consider the iteration function g ( x ) = 3 − x over the interval [ 1 3 , 1 ] starting with p 0 = 1 3 . Determine a lower bound for the number of iterations n required so that | p n − p | < 10 − 5 ? Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 21 / 54

  35. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Example: g ( x ) = g ( x ) = 3 − x Consider the iteration function g ( x ) = 3 − x over the interval [ 1 3 , 1 ] starting with p 0 = 1 3 . Determine a lower bound for the number of iterations n required so that | p n − p | < 10 − 5 ? Determine the Parameters of the Problem Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 21 / 54

  36. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Example: g ( x ) = g ( x ) = 3 − x Consider the iteration function g ( x ) = 3 − x over the interval [ 1 3 , 1 ] starting with p 0 = 1 3 . Determine a lower bound for the number of iterations n required so that | p n − p | < 10 − 5 ? Determine the Parameters of the Problem Note that p 1 = g ( p 0 ) = 3 − 1 3 = 0 . 6933612 and, since g ′ ( x ) = − 3 − x ln 3, we obtain the bound | g ′ ( x ) | ≤ 3 − 1 3 ln 3 ≤ . 7617362 ≈ . 762 = k . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 21 / 54

  37. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Use the Corollary Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 22 / 54

  38. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Use the Corollary Therefore, we have k n | p n − p | ≤ 1 − k | p 0 − p 1 | Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 22 / 54

  39. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Use the Corollary Therefore, we have k n | p n − p | ≤ 1 − k | p 0 − p 1 | . 762 n � � 1 � � ≤ 3 − . 6933612 � � 1 − . 762 � � Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 22 / 54

  40. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Use the Corollary Therefore, we have k n | p n − p | ≤ 1 − k | p 0 − p 1 | . 762 n � � 1 � � ≤ 3 − . 6933612 � � 1 − . 762 � � 1 . 513 × 0 . 762 n ≤ Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 22 / 54

  41. Fixed-Point Iteration Convergence Criteria Sample Problem Functional (Fixed-Point) Iteration Use the Corollary Therefore, we have k n | p n − p | ≤ 1 − k | p 0 − p 1 | . 762 n � � 1 � � ≤ 3 − . 6933612 � � 1 − . 762 � � 1 . 513 × 0 . 762 n ≤ We require that 1 . 513 × 0 . 762 n < 10 − 5 or n > 43 . 88 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 22 / 54

  42. Fixed-Point Iteration Convergence Criteria Sample Problem Footnote on the Estimate Obtained Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 23 / 54

  43. Fixed-Point Iteration Convergence Criteria Sample Problem Footnote on the Estimate Obtained It is important to realise that the estimate for the number of iterations required given by the theorem is an upper bound. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 23 / 54

  44. Fixed-Point Iteration Convergence Criteria Sample Problem Footnote on the Estimate Obtained It is important to realise that the estimate for the number of iterations required given by the theorem is an upper bound. In the previous example, only 21 iterations are required in practice, i.e. p 21 = 0 . 54781 is accurate to 10 − 5 . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 23 / 54

  45. Fixed-Point Iteration Convergence Criteria Sample Problem Footnote on the Estimate Obtained It is important to realise that the estimate for the number of iterations required given by the theorem is an upper bound. In the previous example, only 21 iterations are required in practice, i.e. p 21 = 0 . 54781 is accurate to 10 − 5 . The reason, in this case, is that we used g ′ ( 1 ) = 0 . 762 whereas g ′ ( 0 . 54781 ) = 0 . 602 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 23 / 54

  46. Fixed-Point Iteration Convergence Criteria Sample Problem Footnote on the Estimate Obtained It is important to realise that the estimate for the number of iterations required given by the theorem is an upper bound. In the previous example, only 21 iterations are required in practice, i.e. p 21 = 0 . 54781 is accurate to 10 − 5 . The reason, in this case, is that we used g ′ ( 1 ) = 0 . 762 whereas g ′ ( 0 . 54781 ) = 0 . 602 If one had used k = 0 . 602 (were it available) to compute the bound, one would obtain N = 23 which is a more accurate estimate. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 23 / 54

  47. Fixed-Point Iteration Convergence Criteria Sample Problem Outline Functional (Fixed-Point) Iteration 1 Convergence Criteria for the Fixed-Point Method 2 Sample Problem: f ( x ) = x 3 + 4 x 2 − 10 = 0 3 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 24 / 54

  48. Fixed-Point Iteration Convergence Criteria Sample Problem A Single Nonlinear Equation Example 2 We return to Example 1 and the equation x 3 + 4 x 2 − 10 = 0 which has a unique root in [ 1 , 2 ] . Its value is approximately 1 . 365230013. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 25 / 54

  49. Fixed-Point Iteration Convergence Criteria Sample Problem f ( x ) = x 3 + 4 x 2 − 10 = 0 on [ 1 , 2 ] y 14 y = f(x) = x 3 + 4x 2 −10 1 2 x −5 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 26 / 54

  50. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 Earlier, we listed 5 possible transpositions to x = g ( x ) x − x 3 − 4 x 2 + 10 x = g 1 ( x ) = � 10 x = g 2 ( x ) = x − 4 x 1 � 10 − x 3 x = g 3 ( x ) = 2 � 10 x = g 4 ( x ) = 4 + x x − x 3 + 4 x 2 − 10 x = g 5 ( x ) = 3 x 2 + 8 x Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 27 / 54

  51. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 Results Observed for x = g ( x ) with x 0 = 1 . 5 x = g 1 ( x ) = x − x 3 − 4 x 2 + 10 Does not Converge � 10 x = g 2 ( x ) = x − 4 x Does not Converge x = g 3 ( x ) = 1 � 10 − x 3 Converges after 31 Iterations 2 � 10 x = g 4 ( x ) = Converges after 12 Iterations 4 + x x = g 5 ( x ) = x − x 3 + 4 x 2 − 10 Converges after 5 Iterations 3 x 2 + 8 x Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 28 / 54

  52. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 x = g ( x ) with x 0 = 1 . 5 x = g 1 ( x ) = x − x 3 − 4 x 2 + 10 Does not Converge � 10 x = g 2 ( x ) = x − 4 x Does not Converge x = g 3 ( x ) = 1 � 10 − x 3 Converges after 31 Iterations 2 � 10 x = g 4 ( x ) = Converges after 12 Iterations 4 + x x = g 5 ( x ) = x − x 3 + 4 x 2 − 10 Converges after 5 Iterations 3 x 2 + 8 x Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 29 / 54

  53. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 x = g 1 ( x ) = x − x 3 − 4 x 2 + 10 Iteration for x = g 1 ( x ) Does Not Converge Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 30 / 54

  54. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 x = g 1 ( x ) = x − x 3 − 4 x 2 + 10 Iteration for x = g 1 ( x ) Does Not Converge Since 1 ( x ) = 1 − 3 x 2 − 8 x g ′ g ′ g ′ 1 ( 1 ) = − 10 1 ( 2 ) = − 27 � < 1. � � there is no interval [ a , b ] containing p for which � g ′ 1 ( x ) Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 30 / 54

  55. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 x = g 1 ( x ) = x − x 3 − 4 x 2 + 10 Iteration for x = g 1 ( x ) Does Not Converge Since 1 ( x ) = 1 − 3 x 2 − 8 x g ′ g ′ g ′ 1 ( 1 ) = − 10 1 ( 2 ) = − 27 � < 1. Also, note � � there is no interval [ a , b ] containing p for which � g ′ 1 ( x ) that g 1 ( 1 ) = 6 and g 2 ( 2 ) = − 12 so that g ( x ) / ∈ [ 1 , 2 ] for x ∈ [ 1 , 2 ] . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 30 / 54

  56. Fixed-Point Iteration Convergence Criteria Sample Problem Iteration Function: x = g 1 ( x ) = x − x 3 − 4 x 2 + 10 Iterations starting with p 0 = 1 . 5 n p n − 1 p n | p n − p n − 1 | 1 1.5000000 -0.8750000 2.3750000 2 -0.8750000 6.7324219 7.6074219 3 6.7324219 -469.7200120 476.4524339 p 4 ≈ 1 . 03 × 10 8 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 31 / 54

  57. Fixed-Point Iteration Convergence Criteria Sample Problem g 1 Does Not Map [ 1 , 2 ] into [ 1 , 2 ] g 1 ( x ) = x − x 3 − 4 x 2 + 10 y 6 2 x 1 2 −10 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 32 / 54

  58. Fixed-Point Iteration Convergence Criteria Sample Problem | g ′ 1 ( x ) | > 1 on [ 1 , 2 ] y 1 2 x −10 1 ( x ) = 1 − 3 x 2 − 8 x g ′ −27 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 33 / 54

  59. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 x = g ( x ) with x 0 = 1 . 5 x = g 1 ( x ) = x − x 3 − 4 x 2 + 10 Does not Converge � 10 x = g 2 ( x ) = x − 4 x Does not Converge x = g 3 ( x ) = 1 � 10 − x 3 Converges after 31 Iterations 2 � 10 x = g 4 ( x ) = Converges after 12 Iterations 4 + x x = g 5 ( x ) = x − x 3 + 4 x 2 − 10 Converges after 5 Iterations 3 x 2 + 8 x Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 34 / 54

  60. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 � 10 x = g 2 ( x ) = x − 4 x Iteration for x = g 2 ( x ) is Not Defined Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 35 / 54

  61. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 � 10 x = g 2 ( x ) = x − 4 x Iteration for x = g 2 ( x ) is Not Defined It is clear that g 2 ( x ) does not map [ 1 , 2 ] onto [ 1 , 2 ] and the sequence { p n } ∞ n = 0 is not defined for p 0 = 1 . 5. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 35 / 54

  62. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 � 10 x = g 2 ( x ) = x − 4 x Iteration for x = g 2 ( x ) is Not Defined It is clear that g 2 ( x ) does not map [ 1 , 2 ] onto [ 1 , 2 ] and the sequence { p n } ∞ n = 0 is not defined for p 0 = 1 . 5. Also, there is no interval containing p such that � < 1 � � � g ′ 2 ( x ) since g ′ ( 1 ) ≈ − 2 . 86 g ′ ( p ) ≈ − 3 . 43 and g ′ ( x ) is not defined for x > 1 . 58. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 35 / 54

  63. Fixed-Point Iteration Convergence Criteria Sample Problem � 10 Iteration Function: x = g 2 ( x ) = x − 4 x Iterations starting with p 0 = 1 . 5 n p n − 1 p n | p n − p n − 1 | 1 1.5000000 0.8164966 0.6835034 2 0.8164966 2.9969088 2.1804122 √ 3 2.9969088 − 8 . 6509 — Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 36 / 54

  64. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 x = g ( x ) with x 0 = 1 . 5 x = g 1 ( x ) = x − x 3 − 4 x 2 + 10 Does not Converge � 10 x = g 2 ( x ) = x − 4 x Does not Converge x = g 3 ( x ) = 1 � 10 − x 3 Converges after 31 Iterations 2 � 10 x = g 4 ( x ) = Converges after 12 Iterations 4 + x x = g 5 ( x ) = x − x 3 + 4 x 2 − 10 Converges after 5 Iterations 3 x 2 + 8 x Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 37 / 54

  65. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 x = g 3 ( x ) = 1 � 10 − x 3 2 Iteration for x = g 3 ( x ) Converges (Slowly) Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 38 / 54

  66. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 x = g 3 ( x ) = 1 � 10 − x 3 2 Iteration for x = g 3 ( x ) Converges (Slowly) By differentiation, 3 x 2 g ′ 3 ( x ) = − √ 10 − x 3 < 0 for x ∈ [ 1 , 2 ] 4 and so g= g 3 is strictly decreasing on [ 1 , 2 ] . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 38 / 54

  67. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 x = g 3 ( x ) = 1 � 10 − x 3 2 Iteration for x = g 3 ( x ) Converges (Slowly) By differentiation, 3 x 2 g ′ 3 ( x ) = − √ 10 − x 3 < 0 for x ∈ [ 1 , 2 ] 4 � > 1 for � � and so g= g 3 is strictly decreasing on [ 1 , 2 ] . However, � g ′ 3 ( x ) � ≈ − 2 . 12. � � x > 1 . 71 and � g ′ 3 ( 2 ) Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 38 / 54

  68. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 x = g 3 ( x ) = 1 � 10 − x 3 2 Iteration for x = g 3 ( x ) Converges (Slowly) By differentiation, 3 x 2 g ′ 3 ( x ) = − √ 10 − x 3 < 0 for x ∈ [ 1 , 2 ] 4 � > 1 for � � and so g= g 3 is strictly decreasing on [ 1 , 2 ] . However, � g ′ 3 ( x ) � ≈ − 2 . 12. A closer examination of { p n } ∞ � � x > 1 . 71 and � g ′ 3 ( 2 ) n = 0 will � < 1 � � show that it suffices to consider the interval [ 1 , 1 . 7 ] where � g ′ 3 ( x ) and g ( x ) ∈ [ 1 , 1 . 7 ] for x ∈ [ 1 , 1 . 7 ] . Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 38 / 54

  69. Fixed-Point Iteration Convergence Criteria Sample Problem √ Iteration Function: x = g 3 ( x ) = 1 10 − x 3 2 Iterations starting with p 0 = 1 . 5 n p n − 1 p n | p n − p n − 1 | 1 1.500000000 1.286953768 0.213046232 2 1.286953768 1.402540804 0.115587036 3 1.402540804 1.345458374 0.057082430 4 1.345458374 1.375170253 0.029711879 5 1.375170253 1.360094193 0.015076060 6 1.360094193 1.367846968 0.007752775 30 1.365230013 1.365230014 0.000000001 31 1.365230014 1.365230013 0.000000000 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 39 / 54

  70. Fixed-Point Iteration Convergence Criteria Sample Problem g 3 Maps [ 1 , 1 . 7 ] into [ 1 , 1 . 7 ] y √ g 3 ( x ) = 1 10 − x 3 2 2 1 x 1 2 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 40 / 54

  71. Fixed-Point Iteration Convergence Criteria Sample Problem � � g ′ � 3 ( x ) � < 1 on [ 1 , 1 . 7 ] y 3 x 2 g ′ 3 ( x ) = − √ 4 10 − x 3 1 x 1 2 −1 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 41 / 54

  72. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 x = g ( x ) with x 0 = 1 . 5 x = g 1 ( x ) = x − x 3 − 4 x 2 + 10 Does not Converge � 10 x = g 2 ( x ) = x − 4 x Does not Converge x = g 3 ( x ) = 1 � 10 − x 3 Converges after 31 Iterations 2 � 10 x = g 4 ( x ) = Converges after 12 Iterations 4 + x x = g 5 ( x ) = x − x 3 + 4 x 2 − 10 Converges after 5 Iterations 3 x 2 + 8 x Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 42 / 54

  73. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 � 10 x = g 4 ( x ) = 4 + x Iteration for x = g 4 ( x ) Converges (Moderately) Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 43 / 54

  74. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 � 10 x = g 4 ( x ) = 4 + x Iteration for x = g 4 ( x ) Converges (Moderately) By differentiation, � 10 g ′ 4 ( x ) = − 4 ( 4 + x ) 3 < 0 and it is easy to show that � < 0 . 15 � � 0 . 10 < � g ′ 4 ( x ) ∀ x ∈ [ 1 , 2 ] Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 43 / 54

  75. Fixed-Point Iteration Convergence Criteria Sample Problem Solving f ( x ) = x 3 + 4 x 2 − 10 = 0 � 10 x = g 4 ( x ) = 4 + x Iteration for x = g 4 ( x ) Converges (Moderately) By differentiation, � 10 g ′ 4 ( x ) = − 4 ( 4 + x ) 3 < 0 and it is easy to show that � < 0 . 15 � � 0 . 10 < � g ′ 4 ( x ) ∀ x ∈ [ 1 , 2 ] � is much smaller than that for � � The bound on the magnitude of � g ′ 4 ( x ) � and this explains the reason for the much faster convergence. � � � g ′ 3 ( x ) Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 43 / 54

  76. Fixed-Point Iteration Convergence Criteria Sample Problem � 10 Iteration Function: x = g 4 ( x ) = 4 + x Iterations starting with p 0 = 1 . 5 n p n − 1 p n | p n − p n − 1 | 1 1.500000000 1.348399725 0.151600275 2 1.348399725 1.367376372 0.018976647 3 1.367376372 1.364957015 0.002419357 4 1.364957015 1.365264748 0.000307733 5 1.365264748 1.365225594 0.000039154 6 1.365225594 1.365230576 0.000004982 11 1.365230014 1.365230013 0.000000000 12 1.365230013 1.365230013 0.000000000 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 44 / 54

  77. Fixed-Point Iteration Convergence Criteria Sample Problem g 4 Maps [ 1 , 2 ] into [ 1 , 2 ] y � 10 g 4 ( x ) = 4+ x 2 1 x 1 2 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 45 / 54

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