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08 interpolation lagrange
This document contains lecture slides from an ENEM602 engineering mathematics course taught by Dr. Eng. Mohammad Tawfik in Spring 2007. It discusses Lagrange interpolation, which is a method for interpolating a polynomial that passes through a set of points. The document provides examples of using Lagrange interpolation for 2, 3 and 4 points. It derives the general Lagrange interpolation formula and shows an example of applying it to find a 3rd order polynomial to interpolate 4 given data points. Students are assigned homework problems applying Lagrange interpolation.
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08 interpolation lagrange
- 1. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Interpolation Lagrange Interpolation Polynomial
- 2. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Lagrange Method • First, we learned that a polynomial can pass by the points by using a simple polynomial with (n-1) terms. • Then, we learned a way that “looks like” the Taylor expansion (Newton’s method)
- 3. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Lagrange Method (cont’d) • Now, we will use polynomials that are zero at all points except the one we are evaluating at but in an easier form!
- 4. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik For the two points ( ) ( )2211 xxcxxcy −+−= 2 12 1 1 21 2 y xx xx y xx xx y − − + − − = ( ) ( )2121111 xxcxxcy −+−= ( )21 1 2 xx y c − = ( ) ( )2221212 xxcxxcy −+−= ( )12 2 1 xx y c − =
- 5. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Two lines added! 2 12 1 1 21 2 y xx xx y xx xx y − − + − − =
- 6. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Homework • Show that the polynomial obtained by solving a set of equations is equivalent to that obtained by Lagrange method
- 7. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik For the three points ( )( ) ( )( ) ( )( )133 322 211 xxxxc xxxxc xxxxcy −−+ −−+ −−= ( )( ) ( )( ) ( )( )11313 31212 211111 xxxxc xxxxc xxxxcy −−+ −−+ −−= ( )( )312121 xxxxcy −−= ( )( )3121 1 2 xxxx y c −− =
- 8. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Similarly ( )( )1323 3 1 xxxx y c −− = ( )( )3212 2 3 xxxx y c −− =
- 9. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Finally 3 23 2 13 1 2 32 3 12 1 1 31 3 21 2 y xx xx xx xx y xx xx xx xx y xx xx xx xx y − − − − + − − − − + − − − − =
- 10. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Three parabolas added!!! 3 23 2 13 1 2 32 3 12 1 1 31 3 21 2 y xx xx xx xx y xx xx xx xx y xx xx xx xx y − − − − + − − − − + − − − − =
- 11. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Example • Find a 3rd order polynomial to interpolate the function described by the given points using Lagrange’s method x Y -1 1 0 2 1 5 2 16
- 12. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Solution 4 34 3 24 2 14 1 3 43 4 23 2 13 1 2 42 4 32 3 12 1 1 41 4 31 3 21 2 y xx xx xx xx xx xx y xx xx xx xx xx xx y xx xx xx xx xx xx y xx xx xx xx xx xx y − − − − − − + − − − − − − + − − − − − − + − − − − − − =
- 13. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Solution 4 3 2 1 12 1 02 0 12 1 21 2 01 0 11 1 20 2 10 1 10 1 21 2 11 1 01 0 y xxx y xxx y xxx y xxx y − − − − + + + − − − − + + + − − − − + + + −− − −− − −− − =
- 14. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Solution ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( ) 4 3 2 1 6 11 2 21 2 211 6 21 y xxx y xxx y xxx y xxx y −+ + − −+ + −−+ + − −− =
- 15. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Solution ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )16 6 11 5 2 21 2 2 211 1 6 21 −+ + − −+ + −−+ + − −− = xxx xxx xxx xxx y
- 16. ENEM602 Spring 2007 Dr.Eng. Mohammad Tawfik Homework #6 • Solve the example presented in the previous lecture (Newton’s method) using Lagrange method • Chapter 18, pp. 505-506, numbers: 18.6, 18.7.