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Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
Interpolation functions
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Interpolation functions

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Interpolation functions

Interpolation functions

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  • 1. Interpolation produces a function that matches the given data exactly. The function then can be utilized to approximate the data values at intermediate points.
  • 2. Interpolation may also be used to produce a smooth graph of a function for which values are known only at discrete points, either from measurements or calculations.
  • 3. Given data points Obtain a function, P(x) P(x) goes through the data points Use P(x) To estimate values at intermediate points
  • 4. Given data points: At x0 = 2, y0 = 3 and at x1 = 5, y1 = 8 Find the following: At x = 4, y = ?
  • 5. P(x) should satisfy the following conditions: P(x = 2) = 3 and P(x = 5) = 8 P( x ) = 3 L0 ( x ) + 8 L1 ( x ) P(x) can satisfy the above conditions if at x = x0 = 2, L0(x) = 1 and L1(x) = 0 and at x = x1= 5, L0(x) = 0 and L1(x) = 1
  • 6. At x = x0 = 2, L0(x) = 1 and L1(x) = 0 and at x = x1= 5, L0(x) = 0 and L1(x) = 1 The conditions can be satisfied if L0(x) and L1(x) are defined in the following way. x−5 L0 ( x ) = and 2−5 x − x1 L0 ( x ) = x0 − x1 x−2 L1 ( x ) = 5−2 x − x0 and L1 ( x ) = x1 − x0
  • 7. P( x ) = 3 L0 ( x ) + 8 L1 ( x ) P( x ) = L0 ( x ) y0 + L1 ( x ) y1 Lagrange Interpolating Polynomial P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 )
  • 8. P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 )  x−5  x−2 P( x ) =  ( 3 ) +  ( 8 )  2−5 5−2 5x − 1 P( x ) = 3 5× 4 −1 P( 4) = = 6.333 3
  • 9. The Lagrange interpolating polynomial passing through three given points; (x0, y0), (x1, y1) and (x2, y2) is: P( x ) = L0 ( x ) y0 + L1 ( x ) y1 + L2 ( x ) y2 P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 ) + L2 ( x ) f ( x2 )
  • 10. ( x − x1 )( x − x2 ) L0 ( x ) = ( x0 − x1 )( x0 − x2 ) At x0, L0(x) becomes 1. At all other given data points L0(x) is 0.
  • 11. ( x − x0 )( x − x2 ) L1 ( x ) = ( x1 − x0 )( x1 − x2 ) At x1, L1(x) becomes 1. At all other given data points L1(x) is 0.
  • 12. ( x − x0 )( x − x1 ) L2 ( x ) = ( x2 − x0 )( x2 − x1 ) At x2, L2(x) becomes 1. At all other given data points L2(x) is 0.
  • 13. General form of the Lagrange Interpolating Polynomial P( x ) = L0 ( x ) y0 + L1 ( x ) y1 + L2 ( x ) y2 + ........... + Ln ( x ) yn P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 ) + L2 ( x ) f ( x2 ) + ........... + Ln ( x ) f ( xn )
  • 14. ( x − x0 )( x − x1 )...( x − xk −1 )( x − xk +1 )...( x − xn ) Lk ( x ) = ( xk − x0 )( xk − x1 )...( xk − xk −1 )( x k − x k +1 )...( xk − xn ) ( x − xi ) Lk ( x ) = ∏ i =0 ( x k − xi ) n i ≠k
  • 15. Numerator of Lk ( x ) ( x − x0 ) ( x − x1 ) ( x − x2 ) ×LL × ( x − xk −1 ) ( x − xk +1 ) × LL × ( x − xn −1 ) ( x − xn )
  • 16. Denominator of Lk ( x ) ( xk − x0 ) ( xk − x1 ) ( xk − x2 ) ×LL × ( xk − xk −1 ) ( xk − xk +1 ) × LL × ( xk − xn −1 ) ( xk − xn )
  • 17. Find the Lagrange Interpolating Polynomial using the three given points. ( x0 , y0 ) = ( 2, 0.5) ( x1 , y1 ) = ( 2.5, 0.4) ( x2 , y2 ) = ( 4, 0.25)
  • 18. ( x − x1 )( x − x2 ) L0 ( x ) = ( x0 − x1 )( x0 − x2 ) ( x − 2.5)( x − 4) L0 ( x ) = ( 2 − 2.5)( 2 − 4) = x − 6.5 x + 10 2
  • 19. ( x − x0 )( x − x2 ) L1 ( x ) = ( x1 − x0 )( x1 − x2 ) ( x − 2)( x − 4) L1 ( x ) = ( 2.5 − 2)( 2.5 − 4) − x + 6x − 8 = 0.75 2
  • 20. ( x − x0 )( x − x1 ) L2 ( x ) = ( x2 − x0 )( x2 − x1 ) ( x − 2)( x − 2.5) L2 ( x ) = ( 4 − 2)( 4 − 2.5) x − 4. 5 x + 5 = 3 2
  • 21. P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 ) + L2 ( x ) f ( x2 )
  • 22. ( ) P( x ) = x − 6.5 x + 10 ( 0.5 ) 2  − x + 6x − 8  ( 0.4 ) +   0.75   2  x − 4 .5 x + 5  ( 0.25 ) +   3   2
  • 23. P( x ) = 0.05 x − 0.425 x + 1.15 2 The three given points were taken from the function 1 f ( x) = x
  • 24. 1 f ( 3) = = 0.333 3 An approximation can be obtained from the Lagrange Interpolating Polynomial as: P( 3) = 0.05( 3) − 0.425( 3) + 1.15 = 0.325 2
  • 25. Newton’s Interpolating Polynomials Newton’s equation of a function that passes through two points ( x0 , y 0 ) and ( x1 , y1 ) is P( x ) = a 0 + a1 ( x − x0 )
  • 26. P( x ) = a 0 + a1 ( x − x0 ) Set x = x 0 P ( x0 ) = y0 = a0 Set x = x1 P ( x1 ) = y1 = a0 + a1 ( x1 − x0 )
  • 27. y1 − y 0 a1 = x1 − x0 Newton’s equation of a function that passes through three points ( x0 , y 0 ) ( x1 , y1 ) is and ( x2 , y2 )
  • 28. P ( x ) = a0 + a1 ( x − x0 ) + a2 ( x − x0 ) ( x − x1 ) a 2 , set x = x 2 P ( x2 ) = a0 + a1 ( x2 − x0 ) + a2 ( x2 − x0 ) ( x2 − x1 ) To find
  • 29. y2 − y1 y1 − y0 − x2 − x1 x1 − x0 a2 = x2 − x0
  • 30. Newton’s equation of a function that passes through four points can be written by adding a fourth term . P ( x ) = a0 + a1 ( x − x0 ) + a2 ( x − x0 ) ( x − x1 ) + a3 ( x − x0 ) ( x − x1 ) ( x − x2 )
  • 31. P ( x ) = a0 + a1 ( x − x0 ) + a2 ( x − x0 ) ( x − x1 ) + a3 ( x − x0 ) ( x − x1 ) ( x − x2 ) The fourth term will vanish at all three previous points and, therefore, leaving all three previous coefficients intact.
  • 32. Divided differences and the coefficients The divided difference of a function, f with respect to xi is denoted as f [ xi ] It is called as zeroth divided difference and is simply the value of the function, f at xi f [ xi ] = f ( xi )
  • 33. The divided difference of a function, f with respect to xi and i +1 called as the first divided difference, is denoted x f [ xi , xi +1 ] f [ xi , xi +1 ] f [ xi +1 ] − f [ xi ] = xi +1 − xi
  • 34. The divided difference of a function, f with respect to xi , i +1 and i +2 called as the second divided difference, is denoted as x x f [ xi , xi +1 , xi + 2 ] f [ xi , xi +1 , xi +2 ] f [ xi +1 , xi +2 ] − f [ xi , xi +1 ] = xi + 2 − xi
  • 35. The third divided difference with respect to , , xi + 2 and i + 3 i i +1 x x f [ xi , xi +1 , xi +2 , xi +3 ] x f [ xi +1 , xi + 2 , xi +3 ] − f [ xi , xi +1 , xi + 2 ] = xi +3 − xi
  • 36. The coefficients of Newton’s interpolating polynomial are: a0 = f [ x0 ] a1 = f [ x0 , x1 ] a 2 = f [ x0 , x1 , x 2 ] a3 = f [ x0 , x1 , x 2 , x3 ] a 4 = f [ x0 , x1 , x 2 , x3 , x 4 ] and so on.
  • 37. First divided differences Second divided differences Third divided differences
  • 38. Example Find Newton’s interpolating polynomial to approximate a function whose 5 data points are given below. x f ( x) 2.0 0.85467 2.3 0.75682 2.6 0.43126 2.9 0.22364 3.2 0.08567
  • 39. i 0 xi f [ xi ] 2.0 f [ xi −1 , xi ] f [ xi − 2 , xi −1 , xi ] f [ xi −3 , , xi ] f [ xi −4 , , xi ] 0.85467 -0.32617 1 2.3 0.75682 -1.26505 -1.08520 2 2.6 0.43126 2.13363 0.65522 -0.69207 3 2.9 0.22364 4 3.2 0.08567 -0.29808 0.38695 -0.45990 -2.02642
  • 40. The 5 coefficients of the Newton’s interpolating polynomial are: a0 = f [ x0 ] = 0.85467 a1 = f [ x0 , x1 ] = −0.32617 a2 = f [ x0 , x1 , x2 ] = −1.26505 a3 = f [ x0 , x1 , x2 , x3 ] = 2.13363 a4 = f [ x0 , x1 , x2 , x3 , x4 ] = −2.02642
  • 41. P ( x ) = a0 + a1 ( x − x0 ) + a2 ( x − x0 ) ( x − x1 ) + a3 ( x − x0 ) ( x − x1 ) ( x − x2 ) + a4 ( x − x0 ) ( x − x1 ) ( x − x2 ) ( x − x3 )
  • 42. P ( x ) = 0.85467 − 0.32617 ( x − 2.0 ) -1.26505 ( x − 2.0 ) ( x − 2.3 ) + 2.13363 ( x − 2.0 ) ( x − 2.3) ( x − 2.6 ) −2.02642 ( x − 2.0 ) ( x − 2.3) ( x − 2.6 ) ( x − 2.9 ) P(x) can now be used to estimate the value of the function f(x) say at x = 2.8.
  • 43. P ( 2.8 ) = 0.85467 − 0.32617 ( 2.8 − 2.0 ) -1.26505 ( 2.8 − 2.0 ) ( 2.8 − 2.3 ) + 2.13363 ( 2.8 − 2.0 ) ( 2.8 − 2.3 ) ( 2.8 − 2.6 ) −2.02642 ( 2.8 − 2.0 ) ( 2.8 − 2.3) ( 2.8 − 2.6 ) ( 2.8 − 2.9 ) f ( 2.8 ) ≈ P ( 2.8 ) = 0.275

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