09 numerical differentiation
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This document outlines a course on interpolation and curve fitting taught by Dr. Eng. Mohammad Tawfik. It discusses the objectives of understanding the differences between regression and interpolation and how to fit polynomials to data. It provides examples of using linear, polynomial, and Newton's interpolation methods to find equations that fit sets of measured data points, including solving systems of equations and using tables. The document emphasizes that Newton's method has the advantage of not requiring matrix inversion.
09 numerical differentiation
- 1. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Interpolation/Curve Fitting
- 2. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Objectives • Understanding the difference between regression and interpolation • Knowing how to “best fit” a polynomial into a set of data • Knowing how to use a polynomial to interpolate data
- 3. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Measured Data
- 4. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Polynomial Fit!
- 5. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Line Fit!
- 6. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Which is better?
- 7. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Curve Fitting • If the data measured is of high accuracy and it is required to estimate the values of the function between the given points, then, polynomial interpolation is the best choice. • If the measurements are expected to be of low accuracy, or the number of measured points is too large, regression would be the best choice.
- 8. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Interpolation
- 9. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Why Interpolation? • When the accuracy of your measurements are ensured • When you have discrete values for a function (numerical solutions, digital systems, etc …)
- 10. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Acquired Data
- 11. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik But, how to get the equation of a function that passes by all the data you have!
- 12. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Equation of a Line: Revision xaay 21 += If you have two points 1211 xaay += 2212 xaay += = 2 1 2 1 2 1 1 1 y y a a x x
- 13. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Solving for the constants! 12 12 2 12 2112 1 & xx yy a xx yxyx a − − = − − =
- 14. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik What if I have more than two points? • We may fit a polynomial of order one less that the number of points we have. i.e. four points give third order polynomial.
- 15. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Third-Order Polynomial 3 4 2 321 xaxaxaay +++= For the four points 3 14 2 131211 xaxaxaay +++= 3 24 2 232212 xaxaxaay +++= 3 34 2 333213 xaxaxaay +++= 3 44 2 434214 xaxaxaay +++=
- 16. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik In Matrix Form = 4 3 2 1 4 3 2 1 3 4 2 24 3 3 2 23 3 2 2 22 3 1 2 11 1 1 1 1 y y y y a a a a xxx xxx xxx xxx Solve the above equation for the constants of the polynomial.
- 17. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Newton's Interpolation Polynomial
- 18. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Newton’s Method • In the previous procedure, we needed to solve a system of linear equations for the unknown constants. • This method suggests that we may just proceed with the values of x & y we have to get the constants without setting a set of equations • The method is similar to Taylor’s expansion without differentiation!
- 19. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Equation of a Line: Revision xaay 21 += If you have two points 1211 xaay += 2212 xaay += = 2 1 2 1 2 1 1 1 y y a a x x
- 20. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik For the two points 12 12 1 1 xx yy xx yy − − = − − ( ) 12 12 1 1 xx yy xx yxf − − = − − ( ) ( )1 12 12 1 xx xx yy yxf − − − +=
- 21. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik For the three points ( ) ( ) ( )( )213 121 xxxxa xxaaxf −−+ −+= 11 ya = 12 12 2 xx yy a − − = 13 12 12 23 23 3 xx xx yy xx yy a − − − − − − =
- 22. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Using a table xi yi x1 y1 x2 y2 x3 y3 13 12 12 23 23 xx xx yy xx yy − − − − − − 12 12 xx yy − − 23 23 xx yy − −
- 23. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik In General • Newton’s Interpolation is performed for an nth order polynomial as follows ( ) ( ) ( )( ) ( ) ( )nn xxxxa xxxxaxxaaxf −−++ −−+−+= + ...... 11 213121
- 24. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Example • Find a 3rd order polynomial to interpolate the function described by the given points x Y -1 1 0 2 1 5 2 16
- 25. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Solution: System of equations • A third order polynomial is given by: ( ) 3 4 2 321 xaxaxaaxf +++= ( ) 11 4321 =−+−=− aaaaf ( ) 20 1 == af ( ) 51 4321 =+++= aaaaf ( ) 168422 4321 =+++= aaaaf
- 26. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik In matrix form = −− 16 5 2 1 8421 1111 0001 1111 4 3 2 1 a a a a = 1 1 1 2 4 3 2 1 a a a a ( ) 32 2 xxxxf +++=
- 27. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Newton’s Method • Newton’s methods defines the polynomial in the form: ( ) ( ) ( )( ) ( )( )( )3214 213121 xxxxxxa xxxxaxxaaxf −−−+ −−+−+= ( ) ( ) ( )( ) ( )( )( )11 11 4 321 −++ ++++= xxxa xxaxaaxf
- 28. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Newton’s Method x Y -1 1 1 1 1 0 2 3 4 1 5 11 2 16
- 29. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Newton’s Method • Finally: ( ) ( ) ( )( ) ( )( )( )11 111 −++ ++++= xxx xxxxf ( ) ( ) ( ) ( )xxxxxxf −+++++= 32 11 ( ) 32 2 xxxxf +++=
- 30. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Advantage of Newton’s Method • The main advantage of Newton’s method is that you do not need to invert a matrix!
- 31. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Homework #6 • Chapter 18, pp. 505-506, numbers: 18.1, 18.2, 18.3, 18.5.