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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 Error (motivation) Floating point number system –difference to real number system –problem of roundoff Introduced/propagated error Focus on numerical methods –three bugs 2

3 Any Questions? 3 About the exercise?

4 In this slide Rootfinding –multiplicity Bisection method –Intermediate Value Theorem –convergence measures False position –yet another simple enclosure method –advantage and disadvantage in comparison with bisection method 4

5 Rootfinding 5 Given a function f, find a x such that f(x)=0

6 6 Is a rootfinding problem

7 7

8 8

9 9

10 Multiplicity 10

11 Definition 11

12 Multiplicity for polynomials For polynomials, multiplicity can be determined by factoring the polynomial That’s easy, but 12

13 For non-polynomials What about this f(x)=0, where Clearly, f(0)=0, so the f(x) has a root at x=0 But what is the multiplicity of this root? – f(0)=f’(0)=f’’(0)=0, but f’’’(0)=-4 –the equation has a root of multiplicity 3 at x=0 13 answer

14 14

15 15

16 For non-polynomials What about this f(x)=0, where Clearly, f(0)=0, so the f(x) has a root at x=0 But what is the multiplicity of this root? – f(0)=f’(0)=f’’(0)=0, but f’’’(0)=-4 –the equation has a root of multiplicity 3 at x=0 16

17 Rootfinding methods 2 categories –simple enclosure methods –fixed point iteration schemes Simple enclosure –bisection and false position –guaranteed to converge to a root, but slow Fixed point iteration –Newton’s method and secant method –fast, but require stronger conditions to guarantee convergence 17

18 18

19 A pathological example 19

20 2.1 20 The Bisection Method

21 Bisection method The most basic simple enclosure method All simple enclosure methods are based on Intermediate Value Theorem 21

22 22 Drawing proof for Intermediate Value Theorem

23 In Plain English Find an interval of that the endpoints are opposite sign Since one endpoint value is positive and the other negative, zero is somewhere between the values, that is, at least one root on that interval 23

24 Bisection method The objective is to systematically shrink the size of that root enclosing interval The simplest and most natural way is to cut the interval in half Next is to determine which half contains a root –Intermediate Value Theorem, again Repeat the process on that half 24

25 Bisection method 25

26 In action 26 f(x)=x 3 +2x 2 -3x-1, and (a 1,b 1 )=(1,2)

27 27

28 28

29 Any Questions? 29

30 30 You know what the bisection method is, but so far it is not an algorithm, why?

31 An Algorithm 31 Requires a stopping condition

32 32 Convergence of {p n }

33 33

34 Note The bisection method converges to a root of f, not the root of f –what’s the difference? f(a)f(b)<0 –guarantees the existence of a root, but not uniqueness, and the bisection method converge to one of these roots The bisection method cannot locate roots of even multiplicity (the sign does not change on either side of such roots) –is common to all simple enclosure techniques 34

35 35 Rate of convergence, O(1/2 n ) Order of convergence, α=1 and λ=1/2 http://www.dianadepasquale.com/ThinkingMonkey.jpg

36 36 We are now in position to select a stopping condition

37 Convergence measures For any rootfinding technique, we have 3 convergence measures to construct the stopping condition –absolute error –relative error –test 37

38 Which is the Best? 38 No one is always better than another answer

39 39

40 Which is the Best? 40 No one is always better than another

41 Algorithm Suppose that we decide to use the absolute error, but we don’t know the value of p With the theorem, we can now construct an algorithm 41

42 42

43 Note Performance measure –number of f evaluations rather than number of iterations ( f could involve many floating point operations) Underflow –both f(a) and f(p) will approaching zero –work with the signs rather than the sign of the product f(a)f(p) 43

44 Summary of bisection method Advantage –straightforward –inexpensive (1 evaluation per iteration) –guarantee to converge Disadvantage –error estimation can be overly pessimistic –(drawing for a extreme case of bisection method) 44

45 Any Questions? 45 2.1 The Bisection Method

46 2.2 46 The Method of False Position

47 False position Very similar to bisection method Only differ in selecting p n 47

48 48

49 False position uses more information –values of f(a n ) and f(b n ) –rather than just the signs Selecting p n 49

50 Which method is better? 50

51 Which method is better From another aspect to only the convergence rate –bisection method provides a theoretical bound of error, but no error estimate –false position provides computable error estimate –(the only one advantage of false position) Thus, we can have a more appropriate stopping condition for false position –(we will use this advantage in Section 2.6) 51

52 52 Since false position has no theoretical bound of error, it requires effort to prove the convergence

53 53

54 54

55 Convergence analysis One observation to proceed the convergence analysis –one of the endpoints remains fixed –the other endpoint is just the previous approximation Namely – a n =a n-1, b n =p n-1 or – b n =b n-1, a n =p n-1 55 observation

56 56 The first problem

57 57 The second problem

58 58 The third problem

59 59

60 Convergence analysis One observation to proceed the convergence analysis –one of the endpoints remains fixed –the other endpoint is just the previous approximation Namely – a n =a n-1, b n =p n-1 or – b n =b n-1, a n =p n-1 60

61 Go back to the equation (4) 61 (b n -p)=(p n-1 -p)=e n-1

62 62

63 Any Questions? 63

64 Guarantee to convergence Now we know e n ≒ λ e n-1 One question that remains is –whether |λ| is less than 1 64 answer

65 The first condition 65 The remaining three conditions can be proved in a similar fashion

66 66 Now it’s time to select a stopping condition

67 Stopping condition Suppose the absolute error is used We have e n ≒ λ e n-1 We have to estimate e n 67

68 68

69 69 The first problem

70 70 The second problem

71 71 The third problem

72 Any Questions? 72 2.2 The Method of False Position


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