First Semester in Numerical Analysis with Julia, 2020a

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CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 157 Example 78. Approximate ∫ 1 cos xdx using Gauss-Legendre quadrature with n =3nodes. −1 Solution. From Table 4.1, and using two-digit rounding, we have ∫ 1 −1 cos xdx ≈ 0.56 cos(−0.77) + 0.89 cos 0 + 0.56 cos(0.77) = 1.69 and the true solution is sin(1) − sin(−1) = 1.68. So far we discussed integrating functions over the interval (−1, 1). What if we have a different integration domain? The answer is simple: change of variables! To compute ∫ b f(x)dx for any a
CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 158 In [1]: function gauss(f::Function) 0.2369268851*f(-0.9061798459)+ 0.2369268851*f(0.9061798459)+ 0.5688888889*f(0)+ 0.4786286705*f(0.5384693101)+ 0.4786286705*f(-0.5384693101) end Out[1]: gauss (generic function with 1 method) Now we compute 1 4 ∫ 1 −1 ( t+3 ) t+3 4 dt using the code: 4 In [2]: 0.25*gauss(t->(t/4+3/4)^(t/4+3/4)) Out[2]: 0.41081564812239885 The next theorem is about the error of the Gauss-Legendre rule. Its proof can be found in Atkinson [3]. The theorem shows, in particular, that the degree of accuracy of the quadrature rule, using n nodes, is 2n − 1. Theorem 80. Let f ∈ C 2n [−1, 1]. The error of Gauss-Legendre rule satisfies ∫ b a f(x)dx − n∑ w i f(x i )= i=1 2 2n+1 (n!) 4 f (2n) (ξ) (2n +1)[(2n)!] 2 (2n)! for some ξ ∈ (−1, 1). Using Stirling’s formula n! ∼ e −n n n (2πn) 1/2 , where the symbol ∼ means the ratio of the two sides converges to 1 as n →∞, it can be shown that 2 2n+1 (n!) 4 (2n +1)[(2n)!] 2 ∼ π 4 n . This means the error of Gauss-Legendre rule decays at an exponential rate of 1/4 n as opposed to, for example, the polynomial rate of 1/n 4 for composite Simpson’s rule. Exercise 4.3-1: Prove that the sum of the weights in Gauss-Legendre quadrature is 2, for any n.
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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 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 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 3. INTERPOLATION 98 and thu
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CHAPTER 3. INTERPOLATION 100 With t
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CHAPTER 3. INTERPOLATION 102 Exampl
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CHAPTER 3. INTERPOLATION 104 Exerci
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CHAPTER 5. APPROXIMATION THEORY 208
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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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