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Numerical Analysis 1 EE, NCKU Tien-Hao Chang (Darby Chang)

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Presentation on theme: "Numerical Analysis 1 EE, NCKU Tien-Hao Chang (Darby Chang)"— Presentation transcript:

1 Numerical Analysis 1 EE, NCKU Tien-Hao Chang (Darby Chang)

2 In the previous slide Fixed point iteration scheme –what is a fixed point? –iteration function –convergence Newton’s method –tangent line approximation –convergence Secant method 2

3 In this slide Accelerating convergence –linearly convergent –Newton’s method on a root of multiplicity >1 –(exercises) Proceed to systems of equations –linear algebra review –pivoting strategies 3

4 2.6 4 Accelerating Convergence

5 Accelerating convergence Having spent so much time discussing convergence –is it possible to accelerate the convergence? How to speed up the convergence of a linearly convergent sequence? How to restore quadratic convergence to Newton’s method? –on a root of multiplicity > 1 5

6 Accelerating convergence Linearly convergence Thus far, the only truly linearly convergent sequence –false position –fixed point iteration Bisection method is not according to the definition 6

7 7

8 Aitken’s Δ 2 -method Substituting Eq. (2) into Eq. (1) Substituting Eq. (4) into Eq. (3) The above formulation should be a better approximation to p than p n 8

9 9

10 Aitken’s Δ 2 -method Accelerated? 10 which implies super- linearly convergence later answer

11 11

12 Any Questions? 12 About Aitken’s Δ 2 -method

13 Accelerating convergence Anything to further enhance? 13

14 14 Why not use p-head instead of p ?

15 Steffensen’s method 15

16 16 Restoring quadratic convergence to Newton’s method

17 17

18 18

19 Any Questions? 19

20 Two disadvantages Both the first and the second derivatives of f are needed Each iteration requires one more function evaluations 20 answer

21 Any Questions? 21 Chapter 2 Rootfinding (2.7 is skipped)

22 Exercise /4/21 9:00am to or hand over in class. You may arbitrarily pick one problem among the first three, which means this exercise contains only five

23 23

24 24

25 25

26 26

27 27 (Programming)

28 Chapter 3 28 Systems of Equations

29 Systems of Equations Definition 29

30 Linear Algebra Review (vectors and matrices)

31 Matrix Definitions 31

32 Any Questions? 32 m, n, m, i, j, E QUAL, S UM, S CALAR M ULTIPLICATION, P RODUCT …

33 The Inverse Matrix 33 (cannot be skipped)

34 34

35 Any questions? 35 answerquestion

36 The Determinant 36 (cannot be skipped, too)

37 37 cofactor

38 38

39 Link the concepts –All these theorems will be extremely important throughout this chapter Nonsingular matrices Determinants Solutions of linear systems of equations 39

40 40

41 41 (Hard to prove)

42 Any Questions? Linear Algebra Review

43 Gaussian Elimination (I suppose you have already known it)

44 An application problem 44

45 I 1 -I 2 -I 3 =0 I 2 -I 4 -I 5 =0 I 3 +I 4 -I 6 =0 2I 3 +I 6 =7 I 2 +2I 5 =13 -I 2 +2I 3 -3I 4 =0 45

46 Following Gaussian elimination 46

47 Any Questions? 47 Gaussian elimination

48 Gaussian elimination Operation Counts 48

49 Operation Counts Comparison Gaussian elimination –forward elimination –back substitution Gauss-Jordan elimination Compute the inverse matrix 49

50 Pivoting Strategy

51 51

52 52

53 53 Compare to x 1 =1, x 2 =7, x 3 =1

54 Pivoting strategy To avoid small pivot elements A scheme for interchanging the rows (interchanging the pivot element) Partial pivoting 54

55 55 In action

56 56

57 57 Compare to x 1 =1, x 2 =7, x 3 =1

58 Any Questions? 58

59 From the algorithm view How to implement the interchanging operation? –change implicitly Introduce a row vector r –each time a row interchange is required, we need only swap the corresponding elements of the vector –number of operations from 3n to 3 59 hint answer

60 60 In action

61 61 Without pivoting

62 62

63 63

64 x = [1.000, , , ] T –exact solution x = [1,-1,1,-1] T –no r x = [1.131, , , ] T 64

65 Scaled Partial Pivoting 65

66 Scaled partial pivoting An example 66

67 Any Questions? 67

68 Scaled partial pivoting A blind spot of partial pivoting 68 answer

69 Scaled partial pivoting 69

70 70

71 71 In action

72 72

73 73

74 x = [1.000, , 1.000, ] T –exact solution x = [1,-1,1,-1] T –no s x = [1.000, , , ] T –no r x = [1.131, , , ] T 74

75 Any Questions? Pivoting Strategy


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