Solutions of Equations in One Variable. Fixed-Point Iteration II

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1 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 c 2011 Brooks/Cole, Cengage Learning

2 Outline 1 Functional (Fixed-Point) Iteration Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 2 / 54

3 Outline 1 Functional (Fixed-Point) Iteration 2 Convergence Criteria for the Fixed-Point Method Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 2 / 54

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

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

6 Functional (Fixed-Point) Iteration Now that we have established a condition for which g(x) has a unique fixed point in I, there remains the problem of how to find it. The technique employed is known as fixed-point iteration. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 4 / 54

7 Functional (Fixed-Point) Iteration Now that we have established a condition for which g(x) has a unique fixed point in I, there remains the problem of how to find it. The technique employed is known as fixed-point iteration. Basic Approach To approximate the fixed point of a function g, we choose an initial approximation p 0 and generate the sequence {p n } n=0 by letting p n = g(p n 1 ), for each n 1. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 4 / 54

8 Functional (Fixed-Point) Iteration Now that we have established a condition for which g(x) has a unique fixed point in I, there remains the problem of how to find it. The technique employed is known as fixed-point iteration. Basic Approach To approximate the fixed point of a function g, we choose an initial approximation p 0 and generate the sequence {p n } n=0 by letting p n = g(p n 1 ), for each n 1. If the sequence converges to p and g is continuous, then ( ) p = lim p n = lim g(p n 1 ) = g lim p n 1 = g(p), n n n and a solution to x = g(x) is obtained. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 4 / 54

9 Functional (Fixed-Point) Iteration Now that we have established a condition for which g(x) has a unique fixed point in I, there remains the problem of how to find it. The technique employed is known as fixed-point iteration. Basic Approach To approximate the fixed point of a function g, we choose an initial approximation p 0 and generate the sequence {p n } n=0 by letting p n = g(p n 1 ), for each n 1. If the sequence converges to p and g is continuous, then ( ) p = lim p n = lim g(p n 1 ) = g lim p n 1 = g(p), n n n and a solution to x = g(x) is obtained. This technique is called fixed-point, or functional iteration. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 4 / 54

10 Functional (Fixed-Point) Iteration y p 2 5 g(p 1 ) p 3 5 g(p 2 ) p 1 5 g(p 0 ) (p 1, p 1 ) (p 1, p 2 ) (p 2, p 2 ) y 5 x (p 0, p 1 ) y p 3 5 g(p 2 ) p 2 5 g(p 1 ) p 1 5 g(p 0 ) (p 0, p 1 ) y 5 x (p 2, p 3 ) y 5 g(x) (p 2, p 2 ) (p 1, p 1 ) y 5 g(x) p 1 p 3 p 2 p 0 x p 0 p 1 p 2 x (a) (b) Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 5 / 54

11 Functional (Fixed-Point) Iteration Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 6 / 54

12 Functional (Fixed-Point) Iteration Fixed-Point Algorithm To find the fixed point of g in an interval [a, b], given the equation x = g(x) with an initial guess p 0 [a, b]: Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 6 / 54

13 Functional (Fixed-Point) Iteration Fixed-Point Algorithm To find the fixed point of g in an interval [a, b], given the equation x = g(x) with an initial guess p 0 [a, b]: 1. n = 1; Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 6 / 54

14 Functional (Fixed-Point) Iteration Fixed-Point Algorithm To find the fixed point of g in an interval [a, b], given the equation x = g(x) with an initial guess p 0 [a, b]: 1. n = 1; 2. p n = g(p n 1 ); Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 6 / 54

15 Functional (Fixed-Point) Iteration Fixed-Point Algorithm To find the fixed point of g in an interval [a, b], given the equation x = g(x) with an initial guess p 0 [a, b]: 1. n = 1; 2. p n = g(p n 1 ); 3. If p n p n 1 < ɛ then 5; Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 6 / 54

16 Functional (Fixed-Point) Iteration Fixed-Point Algorithm To find the fixed point of g in an interval [a, b], given the equation x = g(x) with an initial guess p 0 [a, b]: 1. n = 1; 2. p n = g(p n 1 ); 3. If p n p n 1 < ɛ then 5; 4. n n + 1; go to 2. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 6 / 54

17 Functional (Fixed-Point) Iteration Fixed-Point Algorithm To find the fixed point of g in an interval [a, b], given the equation x = g(x) with an initial guess p 0 [a, b]: 1. n = 1; 2. p n = g(p n 1 ); 3. If p n p n 1 < ɛ then 5; 4. n n + 1; go to End of Procedure. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 6 / 54

18 A Single Nonlinear Equation Example 1 The equation x 3 + 4x 2 10 = 0 has a unique root in [1, 2]. Its value is approximately Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 7 / 54

19 f (x) = x 3 + 4x 2 10 = 0 on [1, 2] y 14 y = f(x) = x 3 + 4x x 5 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 8 / 54

20 f (x) = x 3 + 4x 2 10 = 0 on [1, 2] Possible Choices for g(x) Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 9 / 54

21 f (x) = x 3 + 4x 2 10 = 0 on [1, 2] Possible Choices for g(x) There are many ways to change the equation to the fixed-point form x = g(x) using simple algebraic manipulation. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 9 / 54

22 f (x) = x 3 + 4x 2 10 = 0 on [1, 2] Possible Choices for g(x) There are many ways to change the equation to the fixed-point form x = g(x) using simple algebraic manipulation. For example, to obtain the function g described in part (c), we can manipulate the equation x 3 + 4x 2 10 = 0 as follows: 4x 2 = 10 x 3, so x 2 = 1 4 (10 x 3 ), and x = ± 1 2 (10 x 3 ) 1/2. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 9 / 54

23 f (x) = x 3 + 4x 2 10 = 0 on [1, 2] Possible Choices for g(x) There are many ways to change the equation to the fixed-point form x = g(x) using simple algebraic manipulation. For example, to obtain the function g described in part (c), we can manipulate the equation x 3 + 4x 2 10 = 0 as follows: 4x 2 = 10 x 3, so x 2 = 1 4 (10 x 3 ), and x = ± 1 2 (10 x 3 ) 1/2. We will consider 5 such rearrangements and, later in this section, provide a brief analysis as to why some do and some not converge to p = Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 9 / 54

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

25 Numerical Results for f (x) = x 3 + 4x 2 10 = 0 n g 1 g 2 g 3 g 4 g ( 8.65) 1/ Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 11 / 54

26 Solving f (x) = x 3 + 4x 2 10 = 0 x = g(x) with x 0 = 1.5 x = g 1 (x) = x x 3 4x Does not Converge x = g 2 (x) = x = g 3 (x) = x 4x Does not Converge 10 x 3 Converges after 31 Iterations x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Converges after 12 Iterations Converges after 5 Iterations Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 12 / 54

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

28 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

29 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

30 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

31 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

32 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

33 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

34 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

35 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. Using the Mean Value Theorem MVT and the assumption that 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

36 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. Using the Mean Value Theorem MVT and the assumption that g (x) k < 1, x [a, b], we write p n p = g(p n 1 ) g(p) g (ξ) p n 1 p Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 16 / 54

37 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. Using the Mean Value Theorem MVT and the assumption that g (x) k < 1, x [a, b], we write where ξ (a, b). p n p = g(p n 1 ) g(p) g (ξ) p n 1 p k p n 1 p Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 16 / 54

38 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

39 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

40 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

41 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

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

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

44 Functional (Fixed-Point) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then p n p k n 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

45 Functional (Fixed-Point) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then p n p k n 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

46 Functional (Fixed-Point) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then p n p k n 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

47 Functional (Fixed-Point) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then p n p k n 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

48 Functional (Fixed-Point) Iteration Proof of Corollary (2 of 3) p n+1 p n k n p 1 p 0 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 19 / 54

49 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 p n+1 p n Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 19 / 54

50 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 p n+1 p n p m p m 1 + p m 1 p m p n+1 p n Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 19 / 54

51 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 p n+1 p n p m p m 1 + p m 1 p m p n+1 p n k m 1 p 1 p 0 + k m 2 p 1 p k n p 1 p 0 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 19 / 54

52 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 p n+1 p n p m p m 1 + p m 1 p m p n+1 p n k m 1 p 1 p 0 + k m 2 p 1 p k n p 1 p 0 k n 1 + k + k k m n 1) p 1 p 0. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 19 / 54

53 Functional (Fixed-Point) Iteration p m p n k n ( 1 + k + k k m n 1) 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

54 Functional (Fixed-Point) Iteration p m p n k n ( 1 + k + k k m n 1) p 1 p 0. Proof of Corollary (3 of 3) However, since lim m p m = p, we obtain p p n = lim m p m p n Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 20 / 54

55 Functional (Fixed-Point) Iteration p m p n k n ( 1 + k + k k m n 1) p 1 p 0. Proof of Corollary (3 of 3) However, since lim m p m = p, we obtain p p n = lim m p n m k n p 1 p 0 i=1 k i Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 20 / 54

56 Functional (Fixed-Point) Iteration p m p n k n ( 1 + k + k k m n 1) p 1 p 0. Proof of Corollary (3 of 3) However, since lim m p m = p, we obtain p p n = lim m p n m k n p 1 p 0 = i=1 k i k n 1 k p 1 p 0. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 20 / 54

57 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

58 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

59 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 ) = = and, since g (x) = 3 x ln 3, we obtain the bound g (x) ln = k. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 21 / 54

60 Functional (Fixed-Point) Iteration Use the Corollary Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 22 / 54

61 Functional (Fixed-Point) Iteration Use the Corollary Therefore, we have p n p k n 1 k p 0 p 1 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 22 / 54

62 Functional (Fixed-Point) Iteration Use the Corollary Therefore, we have p n p k n 1 k p 0 p n Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 22 / 54

63 Functional (Fixed-Point) Iteration Use the Corollary Therefore, we have p n p k n 1 k p 0 p n n Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 22 / 54

64 Functional (Fixed-Point) Iteration Use the Corollary Therefore, we have p n p k n 1 k p 0 p n n We require that n < 10 5 or n > Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 22 / 54

65 Footnote on the Estimate Obtained Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 23 / 54

66 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

67 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 = is accurate to Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 23 / 54

68 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 = is accurate to The reason, in this case, is that we used g (1) = whereas g ( ) = Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 23 / 54

69 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 = is accurate to The reason, in this case, is that we used g (1) = whereas g ( ) = If one had used k = (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

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

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

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

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

74 Solving f (x) = x 3 + 4x 2 10 = 0 Results Observed for x = g(x) with x 0 = 1.5 x = g 1 (x) = x x 3 4x x = g 2 (x) = x = g 3 (x) = x 4x Does not Converge Does not Converge 10 x 3 Converges after 31 Iterations x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Converges after 12 Iterations Converges after 5 Iterations Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 28 / 54

75 Solving f (x) = x 3 + 4x 2 10 = 0 x = g(x) with x 0 = 1.5 x = g 1 (x) = x x 3 4x Does not Converge x = g 2 (x) = x = g 3 (x) = x 4x Does not Converge 10 x 3 Converges after 31 Iterations x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Converges after 12 Iterations Converges after 5 Iterations Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 29 / 54

76 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 1 (x) = x x 3 4x 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

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

78 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 1 (x) = x x 3 4x Iteration for x = g 1 (x) Does Not Converge Since g 1 (x) = 1 3x 2 8x g 1 (1) = 10 g 1 (2) = 27 there is no interval [a, b] containing p for which g 1 (x) < 1. Also, note 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

79 Iteration Function: x = g 1 (x) = x x 3 4x Iterations starting with p 0 = 1.5 n p n 1 p n p n p n p Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 31 / 54

80 g 1 Does Not Map [1, 2] into [1, 2] y g 1 (x) = x x 3 4x x 10 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 32 / 54

81 g 1 (x) > 1 on [1, 2] y 1 2 x 10 g 1 (x) = 1 3x2 8x 27 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 33 / 54

82 Solving f (x) = x 3 + 4x 2 10 = 0 x = g(x) with x 0 = 1.5 x = g 1 (x) = x x 3 4x Does not Converge x = g 2 (x) = x = g 3 (x) = x 4x Does not Converge 10 x 3 Converges after 31 Iterations x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Converges after 12 Iterations Converges after 5 Iterations Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 34 / 54

83 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 2 (x) = Iteration for x = g 2 (x) is Not Defined 10 x 4x Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 35 / 54

84 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 2 (x) = Iteration for x = g 2 (x) is Not Defined 10 x 4x 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

85 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 2 (x) = Iteration for x = g 2 (x) is Not Defined 10 x 4x 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 g 2 (x) < 1 since g (1) 2.86 g (p) 3.43 and g (x) is not defined for x > Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 35 / 54

86 Iteration Function: x = g 2 (x) = 10 x 4x Iterations starting with p 0 = 1.5 n p n 1 p n p n p n Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 36 / 54

87 Solving f (x) = x 3 + 4x 2 10 = 0 x = g(x) with x 0 = 1.5 x = g 1 (x) = x x 3 4x Does not Converge x = g 2 (x) = x = g 3 (x) = x 4x Does not Converge 10 x 3 Converges after 31 Iterations x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Converges after 12 Iterations Converges after 5 Iterations Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 37 / 54

88 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 3 (x) = x 3 Iteration for x = g 3 (x) Converges (Slowly) Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 38 / 54

89 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 3 (x) = x 3 Iteration for x = g 3 (x) Converges (Slowly) By differentiation, g 3 (x) = 3x 2 4 < 0 for x [1, 2] 10 x 3 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

90 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 3 (x) = x 3 Iteration for x = g 3 (x) Converges (Slowly) By differentiation, g 3 (x) = 3x 2 4 < 0 for x [1, 2] 10 x 3 and so g=g 3 is strictly decreasing on [1, 2]. However, g 3 (x) > 1 for x > 1.71 and g 3 (2) Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 38 / 54

91 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 3 (x) = x 3 Iteration for x = g 3 (x) Converges (Slowly) By differentiation, g 3 (x) = 3x 2 4 < 0 for x [1, 2] 10 x 3 and so g=g 3 is strictly decreasing on [1, 2]. However, g 3 (x) > 1 for x > 1.71 and g 3 (2) A closer examination of {p n } n=0 will show that it suffices to consider the interval [1, 1.7] where g 3 (x) < 1 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

92 Iteration Function: x = g 3 (x) = x 3 Iterations starting with p 0 = 1.5 n p n 1 p n p n p n Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 39 / 54

93 g 3 Maps [1, 1.7] into [1, 1.7] y g 3 (x) = x x Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 40 / 54

94 g 3 (x) < 1 on [1, 1.7] y g 3 3x2 (x) = 4 10 x x 1 Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 41 / 54

95 Solving f (x) = x 3 + 4x 2 10 = 0 x = g(x) with x 0 = 1.5 x = g 1 (x) = x x 3 4x Does not Converge x = g 2 (x) = x = g 3 (x) = x 4x Does not Converge 10 x 3 Converges after 31 Iterations x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Converges after 12 Iterations Converges after 5 Iterations Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 42 / 54

96 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 4 (x) = 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

97 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 4 (x) = x Iteration for x = g 4 (x) Converges (Moderately) By differentiation, and it is easy to show that g 4 (x) = 10 4(4 + x) 3 < < g 4 (x) < 0.15 x [1, 2] Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 43 / 54

98 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 4 (x) = x Iteration for x = g 4 (x) Converges (Moderately) By differentiation, and it is easy to show that g 4 (x) = 10 4(4 + x) 3 < < g 4 (x) < 0.15 x [1, 2] The bound on the magnitude of g 4 (x) is much smaller than that for g 3 (x) and this explains the reason for the much faster convergence. Numerical Analysis (Chapter 2) Fixed-Point Iteration II R L Burden & J D Faires 43 / 54

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