1. 1 Numerical Differentiation

2. 2  First order derivatives  High order derivatives  Examples

3. 3 Motivation  How do you evaluate the derivative of a tabulated function.  How do we determine the velocity and acceleration from tabulated measurements. Time (second) Displacement (meters) 030.1 548.2 1050.0 1540.2

4. 4 Recall

5. For small values of h, the difference quotient [f (x 0 + h) − f (x 0 )]/h can be used to approximate f’(x 0 ) with an error bounded by M|h|/2, where M is a bound on |f’’(x)| for x between x0 and x0 +h. CISE301_Topic65

6. 6 Three Formula

7. 7 The Three Formulas

8. 8 Forward/Backward Difference Formula

9. 9 Central Difference Formula

10. 10 The Three Formula (Revisited)

11. 11 Higher Order Formulas

12. 12 Other Higher Order Formulas

13. Example  Use forward, backward and centered difference approximations to estimate the first derivate of: f(x) = –0.1x 4 – 0.15x 3 – 0.5x 2 – 0.25x + 1.2 at x = 0.5 using step size h = 0.5 and h = 0.25  Note that the derivate can be obtained directly: f’(x) = –0.4x 3 – 0.45x 2 – 1.0x – 0.25 The true value of f’(0.5) = -0.9125  In this example, the function and its derivate are known. However, in general, only tabulated data might be given. 13

14. Solution with Step Size = 0.5  f(0.5) = 0.925, f(0) = 1.2, f(1.0) = 0.2  Forward Divided Difference: f’(0.5)  (0.2 – 0.925)/0.5 = -1.45 | t | = |(-0.9125+1.45)/-0.9125| = 58.9%  Backward Divided Difference: f’(0.5)  (0.925 – 1.2)/0.5 = -0.55 | t | = |(-0.9125+0.55)/-0.9125| = 39.7%  Centered Divided Difference: f’(0.5)  (0.2 – 1.2)/1.0 = -1.0 | t | = |(-0.9125+1.0)/-0.9125| = 9.6% 14

15. Solution with Step Size = 0.25  f(0.5)=0.925, f(0.25)=1.1035, f(0.75)=0.6363  Forward Divided Difference: f’(0.5)  (0.6363 – 0.925)/0.25 = -1.155 | t | = |(-0.9125+1.155)/-0.9125| = 26.5%  Backward Divided Difference: f’(0.5)  (0.925 – 1.1035)/0.25 = -0.714 | t | = |(-0.9125+0.714)/-0.9125| = 21.7%  Centered Divided Difference: f’(0.5)  (0.6363 – 1.1035)/0.5 = -0.934 | t | = |(-0.9125+0.934)/-0.9125| = 2.4% 15

16. Discussion  For both the Forward and Backward difference, the error is O(h)  Halving the step size h approximately halves the error of the Forward and Backward differences  The Centered difference approximation is more accurate than the Forward and Backward differences because the error is O(h 2 )  Halving the step size h approximately quarters the error of the Centered difference. 16

17. Use the forward-difference formula to approximate the derivative of f (x) = ln x at x0 = 1.8 using h = 0.1, h = 0.05, and h = 0.01, and determine bounds for the approximation errors. 17 Example 2

18. Solution The forward-difference formula f (1.8 + h) − f (1.8) h with h = 0.1 gives (ln 1.9 − ln 1.8)/0.1 = (0.64185389 − 0.58778667)/0.1 = 0.5406722. Because f’’(x) = −1/x 2 and 1.8 < ξ < 1.9, a bound for this approximation error is |hf’’(ξ )|/2 =|h|/2ξ 2 < 0.1/2(1.8) 2 = 0.0154321. 18

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