First Semester in Numerical Analysis with Julia, 2020a

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CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 147 and p(x) =x 2 implies ∫ 1 0 p(x)dx = c 0 p(0) + c 1 p(x 1 ) ⇒ 1 3 = c 1x 2 1. We have three unknowns and three equations, so we have to stop here. Solving the three equations we get: c 0 =1/4,c 1 =3/4,x 1 =2/3. So the quadrature rule is of precision two and it is: ∫ 1 f(x)dx = 1 4 f(0) + 3 4 f(2 3 ). 0 Exercise 4.1-2: Find c 0 ,c 1 ,c 2 so that the quadrature rule ∫ 1 −1 f(x)dx = c 0f(−1) + c 1 f(0) + c 2 f(1) has degree of accuracy 2. 4.2 Composite Newton-Cotes formulas If the interval [a, b] in the quadrature is large, then the Newton-Cotes formulas will give poor approximations. The quadrature error depends on h =(b−a)/n (closed formulas), and if b−a is large, then so is h, hence error. If we raise n to compensate for large interval, then we face a problem discussed earlier: error due to the oscillatory behavior of high-degree interpolating polynomials that use equally-spaced nodes. A solution is to break up the domain into smaller intervals and use a Newton-Cotes rule with a smaller n on each subinterval: this is known as a composite rule. Example 74. Let’s compute ∫ 2 0 ex sin xdx. The antiderivative can be computed using integration by parts, and the true value of the integral to 6 digits is 5.39689. If we apply the Simpson’s rule we get: ∫ 2 0 e x sin xdx ≈ 1 3 (e0 sin0+4e sin 1 + e 2 sin 2) = 5.28942. If we partition the integration domain (0, 2) into (0, 1) and (1, 2), and apply Simpson’s rule
CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 148 to each domain separately, we get ∫ 2 e x sin xdx = ∫ 1 e x sin xdx + ∫ 2 0 0 1 e x sin xdx ≈ 1 6 (e0 sin0+4e 0.5 sin(0.5) + e sin 1) + 1 6 (e sin1+4e1.5 sin(1.5) + e 2 sin 2) =5.38953, improving the accuracy significantly. Note that we have used five nodes, 0, 0.5, 1, 1.5, 2, which splits the domain (0, 2) into four subintervals. The composite rules for midpoint, trapezoidal, and Simpson’s rule, with their error terms, are: • Composite Midpoint rule Let f ∈ C 2 [a, b], nbe even, h = b−a , and x n+2 j = a +(j +1)h for j = −1, 0, ..., n +1. The composite Midpoint rule for n +2subintervals is for some ξ ∈ (a, b). ∫ b a n/2 ∑ f(x)dx =2h f(x 2j )+ b − a 6 h2 f ′′ (ξ) (4.2) j=0 • Composite Trapezoidal rule Let f ∈ C 2 [a, b], h= b−a, and x n j = a+jh for j =0, 1, ..., n. The composite Trapezoidal rule for n subintervals is ∫ [ ] b f(x)dx = h ∑n−1 f(a)+2 f(x j )+f(b) − b − a 2 12 h2 f ′′ (ξ) (4.3) for some ξ ∈ (a, b). a j=1 • Composite Simpson’s rule Let f ∈ C 4 [a, b], nbe even, h = b−a, and x n j = a + jh for j =0, 1, ..., n. The composite Simpson’s rule for n subintervals is ∫ b a ⎡ ⎤ n f(x)dx = h 2 −1 n/2 ∑ ∑ ⎣f(a)+2 f(x 2j )+4 f(x 2j−1 )+f(b) ⎦ − b − a 3 180 h4 f (4) (ξ) (4.4) for some ξ ∈ (a, b). j=1 j=1
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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 5. APPROXIMATION THEORY 198
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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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