First Semester in Numerical Analysis with Julia, 2020a

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CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 149 Exercise 4.2-1: Show that the quadrature rule in Example 74 corresponds to taking n =4in the composite Simpson’s formula (4.4). Exercise 4.2-2: Show that the absolute error for the composite trapezoidal rule decays at the rate of 1/n 2 , and the absolute error for the composite Simpson’s rule decays at the rate of 1/n 4 , where n is the number of subintervals. Example 75. Determine n that ensures the composite Simpson’s rule approximates ∫ 2 1 x log xdx with an absolute error of at most 10−6 . Solution. The error term for the composite Simpson’s rule is b−a 180 h4 f (4) (ξ) where ξ is some number between a =1and b =2,andh =(b − a)/n. Differentiate to get f (4) (x) = 2 x 3 . Then b − a 180 h4 f (4) (ξ) = 1 180 h4 2 ξ 3 ≤ h4 90 where we used the fact that 2 ξ 3 than 10 −6 , that is, ≤ 2 1 =2when ξ ∈ (1, 2). Now make the upper bound less h 4 90 ≤ 10−6 ⇒ 1 n 4 (90) ≤ 10−6 ⇒ n 4 ≥ 106 90 ≈ 11111.11 which implies n ≥ 10.27. Since n must be even for Simpson’s rule, this means the smallest value of n to ensure an error of at most 10 −6 is 12. Using the Julia code for the composite Simpson’s rule that will be introduced next, we get 0.6362945608 as the estimate, using 10 digits. The correct integral to 10 digits is 0.6362943611, which means an absolute error of 2 × 10 −7 , better than the expected 10 −6 . Julia codes for Newton-Cotes formulas We write codes for the trapezoidal and Simpson’s rules, and the composite Simpson’s rule. Coding trapezoidal and Simpson’s rule is straightforward.
CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 150 Trapezoidal rule In [1]: function trap(f::Function,a,b) (f(a)+f(b))*(b-a)/2 end Out[1]: trap (generic function with 1 method) Let’s verify the calculations of Example 72: In [2]: trap(x->x^x,0.5,1) Out[2]: 0.42677669529663687 Simpson’s rule In [3]: function simpson(f::Function,a,b) (f(a)+4f((a+b)/2)+f(b))*(b-a)/6 end Out[3]: simpson (generic function with 1 method) In [4]: simpson(x->x^x,0.5,1) Out[4]: 0.4109013813880978 Recall that the degree of accuracy of Simpson’s rule is 3. This means the rule integrates polynomials 1,x,x 2 ,x 3 exactly, but not x 4 . Wecanusethisasawayto verify our code: In [5]: simpson(x->x,0,1) Out[5]: 0.5 In [6]: simpson(x->x^2,0,1) Out[6]: 0.3333333333333333
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First Semester in Numerical Analysi
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First Semester in Numerical Analysi
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Contents 1 Introduction 5 1.1 Revie
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Preface This book is based on my le
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Chapter 1 Introduction 1.1 Review o
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CHAPTER 1. INTRODUCTION 8 Solution.
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CHAPTER 1. INTRODUCTION 10 Let’s
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CHAPTER 1. INTRODUCTION 12 Out[8]:
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CHAPTER 1. INTRODUCTION 14 Press ]
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CHAPTER 1. INTRODUCTION 16 Matrix o
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CHAPTER 1. INTRODUCTION 18 stdlib/v
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CHAPTER 1. INTRODUCTION 20 In [37]:
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CHAPTER 1. INTRODUCTION 24 -0.62988
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CHAPTER 1. INTRODUCTION 26 5 7 9 He
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CHAPTER 1. INTRODUCTION 28 Sometime
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CHAPTER 1. INTRODUCTION 30 where s
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CHAPTER 1. INTRODUCTION 32 Notice t
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CHAPTER 1. INTRODUCTION 34 and real
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CHAPTER 1. INTRODUCTION 38 Chopping
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CHAPTER 1. INTRODUCTION 40 Machine
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CHAPTER 1. INTRODUCTION 42 The abso
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CHAPTER 1. INTRODUCTION 44 Next wil
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CHAPTER 1. INTRODUCTION 48 Exercise
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CHAPTER 1. INTRODUCTION 50 Sources
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CHAPTER 2. SOLUTIONS OF EQUATIONS:
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Chapter 3 Interpolation In this cha
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CHAPTER 3. INTERPOLATION 90 Here is
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CHAPTER 3. INTERPOLATION 92 basis f
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CHAPTER 3. INTERPOLATION 94 therefo
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CHAPTER 3. INTERPOLATION 96 Summary
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CHAPTER 5. APPROXIMATION THEORY 200
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Index Absolute error, 38 Beasley-Sp
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INDEX 220 Chebyshev, 203 Julia code
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First Semester in Numerical Analysis with Julia, 2020a

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