First Semester in Numerical Analysis with Julia, 2020a

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CHAPTER 1. INTRODUCTION 37 26.04166666666667 64.8 163.4013888888889 416.10158730158724 1067.6270089285715 2755.7319223985887 7147.658895778218 18613.92623376623 48638.84613847011 127463.00337621223 334864.62769059895 881657.9515664611 The previous version of the code gave overflow error when n =16. This version has no difficulty in computing n n /n! for n =16and 17. In fact, we can go as high as n = 700, In [3]: f(700) Out[3]: 1.5291379839716124e302 but not n = 750. In [4]: f(750) Out[4]: Inf Overflow in floating-point arithmetic yields the output Inf, which stands for infinity. We will discuss several features of computer arithmetic in the rest of this section. The discussion is easier to follow if we use the familiar base 10 representation of numbers instead of base 2. To this end, we introduce the normalized decimal floating-point representation: ±0.d 1 d 2 ...d k × 10 n where 1 ≤ d 1 ≤ 9, 0 ≤ d i ≤ 9 for all i =2, 3, ..., k. Informally, we call these numbers k−digit decimal machine numbers.
CHAPTER 1. INTRODUCTION 38 Chopping & Rounding Let x be a real number with more digits the computer can handle: x =0.d 1 d 2 ...d k d k+1 ...× 10 n . How will the computer represent x? Let’s use the notation fl(x) for the floating-point representation of x. There are two choices, chopping and rounding: • In chopping, we simply take the first k digits and ignore the rest: fl(x) =0.d 1 d 2 ...d k . • In rounding, if d k+1 ≥ 5 we add 1 to d k to obtain fl(x). If d k+1 < 5, then we simply do as in chopping. Example 11. Find 5-digit (k =5) chopping and rounding values of the numbers below: • π =0.314159265... × 10 1 Chopping gives fl(π) =0.31415 × 10 1 and rounding gives fl(π) =0.31416 × 10 1 . • 0.0001234567 We need to write the number in the normalized representation first as 0.1234567×10 −3 . Now chopping gives 0.12345 × 10 −3 and rounding gives 0.12346 × 10 −3 . Absolute and relative error Since computers only give approximations to real numbers, we need to be clear on how we measure the error of an approximation. Definition 12. Suppose x ∗ is an approximation to x. • |x ∗ − x| is called the absolute error • |x∗ −x| |x| is called the relative error (x ≠0) Relative error usually is a better choice of measure, and we need to understand why. Example 13. Find absolute and relative errors of 1. x =0.20 × 10 1 ,x ∗ =0.21 × 10 1 2. x =0.20 × 10 −2 ,x ∗ =0.21 × 10 −2 3. x =0.20 × 10 5 ,x ∗ =0.21 × 10 5 Notice how the only difference in the three cases is the exponent of the numbers. The absolute errors are: 0.01 × 10 , 0.01 × 10 −2 , 0.01 × 10 5 . The absolute errors are different since the exponents are different. However, the relative error in each case is the same: 0.05.
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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 96 Summary
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CHAPTER 3. INTERPOLATION 100 With t
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CHAPTER 3. INTERPOLATION 104 Exerci
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CHAPTER 3. INTERPOLATION 120 Exerci
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CHAPTER 3. INTERPOLATION 140 Recrea
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CHAPTER 4. NUMERICAL QUADRATURE AND
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Chapter 5 Approximation Theory 5.1
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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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