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Chapter 4 Numerical Differentiation and Integration

Given x 0 , approximate f ’ ( x 0 ). +. -. f. (. x. h. ). f. (. x. ). =. 0. 0. f. '. (. x. ). lim. 0. h. . 0. h. h. h. x 0. x 1. x 1. x 0. Chapter 4 Numerical Differentiation and Integration.  4.1 Numerical Differentiation. forward. backward. 1/16.

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Chapter 4 Numerical Differentiation and Integration

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  1. Given x0, approximatef ’(x0). + - f ( x h ) f ( x ) = 0 0 f ' ( x ) lim 0 h  0 h h h x0 x1 x1 x0 Chapter 4Numerical Differentiation and Integration  4.1 Numerical Differentiation forward backward 1/16

  2. Chapter 4 Numerical Differentiation and Integration -- Numerical Differentiation Approximate f(x) by its Lagrange polynomial with interpolating points x0 and x0 + h. O(h) f ’(x0) = f (x) = f ’(x) = 2/16

  3. Chapter 4 Numerical Differentiation and Integration -- Numerical Differentiation Approximate f(x) by its Lagrange polynomial with interpolating points { x0, x1, …, xn }. f ’(xj) = Note: In general, more evaluation points produce greater accuracy. On the other hand, the number of functional evaluations grows and the roundoff error increases. Hence the numerical differentiation is unstable! 3/16

  4. Chapter 4 Numerical Differentiation and Integration -- Numerical Differentiation x–1 x0 x1 Example: Given three points x0, x0 + h, and x0 + 2h, please derive the three-point formulae for each of the points. Symmetric to formula 1, with h < 0. Five-point formulae are given on p.171 4/16

  5. Chapter 4 Numerical Differentiation and Integration -- Numerical Differentiation Excuses for not doing homework I could only get arbitrarily close to my textbook. I couldn't actually reach it. 2 1 h [ ]  = - - + + - x ( 4 ) f ( x ) f ( x h ) 2 f ( x ) f ( x h ) f ( ) 0 0 0 0 2 h 12 Approximate f ”(x0) Consider Taylor expansions of f(x0 + h) and f(x0 – h) at x0: HW: p.176-177 #7, 13 5/16

  6. Chapter 4 Numerical Differentiation and Integration -- Elements of Numerical Integration Approximate Integrate the Lagrange interpolating polynomial of f (x) instead. Error Select a set of distinct nodes a  x0 < x1 <…< xn b from [a, b]. The Lagrange polynomial is Idea Ak  4.3 Elements of Numerical Integration -- Numerical Quadrature interpolatory quadrature 6/16

  7. Chapter 4 Numerical Differentiation and Integration -- Elements of Numerical Integration Example: Consider the linear interpolation on [a, b], we have f(x) a b f(b) f(a) Definition: The degree of accuracy, or precision, of a quadrature formula is the largest positive integer n such that the formula is exact for xk for each k = 0, 1, …, n. Please determine the precision of this formula. Solution: Consider xk for each k = 0, 1, … trapezoidal rule x0 =1： = x ： = Degree of Precision = 1 x2 ：  7/16

  8. Chapter 4 Numerical Differentiation and Integration -- Elements of Numerical Integration n = 1: For equally spaced nodes： Let Cotes coefficient Note: Cotes coefficients does not depend on either f(x) or [a, b], and can be determined by n and ionly. Hence we can find these coefficients from a table. The formulae are called Newton-Cotes formulae.  Trapezoidal Rule Precision = 1 8/16

  9. Chapter 4 Numerical Differentiation and Integration -- Elements of Numerical Integration n = 2: n = 3: Simpson’s 3/8-Rule. Precision = 3, and n = 4: Cotes Rule. Precision = 5, and Theorem: For the (n+1)-point closed Newton-Cotes formula, there exists   (a, b) for which if n is even and f Cn+2[a, b], and if n is odd and f Cn+1[a, b]. Simpson’s Rule Precision = 3 HW: p.195 #7, 9, 11, 13 9/16

  10. Chapter 4 Numerical Differentiation and Integration -- Composite Numerical Integration /*MVT*/  4.4 Composite Numerical Integration Due to the oscillatory nature of high-degree polynomials, piecewise interpolation is applied to approximate f(x)  a piecewise approach that uses the low-order Newton-Cotes formulae. Composite Trapezoidal Rule: Apply Trapezoidal Rule on each [xk – 1, xk]: Oh come on, you don’t seriously consider h=(ba)/2 acceptable, do you? Don’t you forget the oscillatory nature of high- degree polynomials! Haven’t we had enough formulae? What’s up now? Why can’t you simply refine the partition if you have to be so picky? Uh-oh =Tn 10/16

  11. Chapter 4 Numerical Differentiation and Integration -- Composite Numerical Integration 4 4 4 4 4 Note: To simplify the notation, we may let n’ = 2n. Then and Composite Simpson’s Rule: =Sn 11/16

  12. Chapter 4 Numerical Differentiation and Integration -- Composite Numerical Integration Composite integration techniques are all stable. Example: Consider the Simpson’s Rule with n subintervals on [a, b]. Assume that f (xi) is approximated by f *(xi) such that f (xi) = f *(xi) + i for each i = 0, 1, …, n. Then the accumulated error e(h) is If | i| <  for all i = 0, 1, …, n, then When we refine the partition to ensure accuracy, the increased computation will NOT increase the roundoff error.  12/16

  13. Chapter 4 Numerical Differentiation and Integration -- Composite Numerical Integration Example: Use Trapezoidal rule and Simpson’s rule with n = 8 to approximate where where Solution: = 3.138988494 = 3.141592502 When programming we usually keep dividing the subintervals into 2 equally spaced smaller subintervals. That is, take n = 2k for k = 0, 1, … HW: p.204 #7(a)(b) Whenk = 9, T512 = 3.14159202 = 3.141592502 = S4 13/16

  14. Chapter 4 Numerical Differentiation and Integration -- Romberg Integration ( ( ( ( ( ( ( ( ( ( 1 0 2 3 2 1 1 0 0 0 ) ) ) ) ) ) ) ) ) ) T T T T T T T T T T 2 2 1 0 1 0 0 1 0 3 Since the error of Trapezoidal rule is when we reduce the length of each subinterval into a half, T1 = T2 = S1 =  T4 = S2 = C1 = T8 = S4 = C2 = R1 =  4.5 Romberg Integration Romberg sequence Solve for I : = Sn In general: <  ? Romberg method: <  ? <  ? … … … … … … 14/16

  15. Lab 08. Shape Roof Time Limit: 2 seconds; Points: 4 The kind of roof shown in Figure 1 is shaped from plain flat rectangular plastic board in Figure 2. Figure 1 Figure 2 The transection of the roof is a sine curve with altitude l centimeters. Given the length of the roof, your task is to calculate the length of the flat board needed to shape the roof. 15/16

  16. Chapter 4 Numerical Differentiation and Integration -- Richardson’s Extrapolation Generate high-accuracy results while using low-order formulae - 2 T ( ) T ( h ) 1 3 h - = - a - a - 2 3 0 0 2 I h h ... 2 3 - 2 1 2 4  4.2 Richardson’s Extrapolation Suppose that for some h 0, we have a formula T0(h) that approximates an unknown I, and that the truncation error has the form: T0(h)  I = 1 h + 2 h2 + 3 h3 + … Replace hby half its value, we have T0(h/2)  I = 1 (h/2) + 2 (h/2)2 + 3 (h/2)3 + … Q：How to improve the accuracy from O(h) to O(h2) ？ 16/16

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