1. CISE301_Topic6KFUPM1 SE301: Numerical Methods Topic 6 Numerical Differentiation Lecture 23 KFUPM Read Chapter 23, Sections 1-2

2. CISE301_Topic6KFUPM2 L ecture 23 Numerical Differentiation  First order derivatives  High order derivatives  Richardson Extrapolation  Examples

3. CISE301_Topic6KFUPM3 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. CISE301_Topic6KFUPM4 Recall

5. CISE301_Topic6KFUPM5 Three Formula

6. CISE301_Topic6KFUPM6 The Three Formulas

7. CISE301_Topic6KFUPM7 Forward/Backward Difference Formula

8. CISE301_Topic6KFUPM8 Central Difference Formula

9. CISE301_Topic6KFUPM9 The Three Formula (Revisited)

10. CISE301_Topic6KFUPM10 Higher Order Formulas

11. CISE301_Topic6KFUPM11 Other Higher Order Formulas

12. CISE301_Topic6KFUPM12 Richardson Extrapolation

13. CISE301_Topic6KFUPM13 Richardson Extrapolation

14. CISE301_Topic6KFUPM14 Richardson Extrapolation Table D(0,0)=Φ(h) D(1,0)=Φ(h/2)D(1,1) D(2,0)=Φ(h/4)D(2,1)D(2,2) D(3,0)=Φ(h/8)D(3,1)D(3,2)D(3,3)

15. CISE301_Topic6KFUPM15 Richardson Extrapolation Table

16. CISE301_Topic6KFUPM16 Example 1

17. CISE301_Topic6KFUPM17 Example First Column

18. CISE301_Topic6KFUPM18 Example Richardson Table

19. CISE301_Topic6KFUPM19 Example Richardson Table 1.08483 1.089881.09156 1.091151.09157 This is the best estimate of the derivative of the function. All entries of the Richardson table are estimates of the derivative of the function. The first column are estimates using the central difference formula with different h.

20. CISE301_Topic6KFUPM20 Example 2 The measured speed 'V m/sec.' of a car against time 'T, sec.' is as shown in the table below: a)Compute the car's acceleration at T=4 sec using the Richardson extrapolation with the best accuracy. b)What is the order of the error of the obtained result in part (a)? T(s)233.544.55678 V(m/s) 12101213.21415171612

21. CISE301_Topic6KFUPM21 Solution a)To find the acceleration, we have to differentiate V(4) From the table the max accuracy we can get is D(2,2) and h=2. First Column D(0,0) = [f(6)-f(2)]/2(2) = [17-12]/4 = 1.25 D(1,0) = [f(5)-f(3)]/2(1) = [15-10]/2 = 2.5 D(2,0) = [f(4.5)-f(3.5)]2/(0.5) = [14-12]/1 = 2 Second Column D(1,1) = (4/3)D(1,0)-(1/3)D(0,0) = (4/3)2.5 – (1/3)1.25 = 2.9167 D(2,1) = (4/3)D(2,0)-(1/3)D(1,0) = (4/3)2 – (1/3)2.5 = 1.8333 Third Column D(2,2) = (16/15)D(2,1)-(1/15)D(1,1) = (16/15)1.8333 – (1/15)2.9167 = 1.7611

22. CISE301_Topic6KFUPM22 Example Richardson Table 1.25 2.52.9167 21.83331.7611 This is the best estimate of the derivative of the function. b)The order is O(h 6 ). O(h 2 ) O(h 4 ) O(h 6 )