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Chapter 3

Chapter 3. INTERPOLATION. WHAT IS INTERPOLATION?. Given (x 0 ,y 0 ), (x 1 ,y 1 ), …, (x n ,y n ), finding the value of ‘y’ at a value of ‘x’ in ( x 0, x n ) is called interpolation. INTERPOLANTS. Polynomials are the most common choice of interpolants because they are easy to: Evaluate,

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Chapter 3

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  1. Chapter 3 INTERPOLATION

  2. WHAT IS INTERPOLATION? Given (x0,y0), (x1,y1), …, (xn,yn), finding the value of ‘y’ at a value of ‘x’ in (x0, xn) is called interpolation.

  3. INTERPOLANTS Polynomials are the most common choice of interpolants because they are easy to: • Evaluate, • Differentiate, and • Integrate.

  4. LAGRANGIAN INTERPOLATION Lagrangian inter polating polynomial is given by n å = f ( x ) L ( x ) f ( x ) n i i = 0 i = th where ‘ n ’ in f ( x ) stands for the n order polynomial that approximates the function y f ( x ) n ( ( ) ( ) + ) ( ) given at ( n 1 ) data points as x , y , x , y ,..., x , y , x , y , and - - 0 0 1 1 1 1 n n n n - x x n Õ j = L ( x ) i - x x = 0 j i j ¹ j i - = L ( x ) is a weighting function that includes a product of ( n 1 ) terms with terms of j i i omitted.

  5. Example A manufacturer of thermistors makes the following observations on a thermistor. Determine the temperature corresponding to 754.8 ohms using the Lagrangian method for interpolation. R T ° Ohm C 1101.0 25.113 911.3 30.131 636.0 40.120 451.1 50.128

  6. Solution

  7. Solution

  8. Solution

  9. NEWTONS DIVIDED DIFFERENCE • What is divided difference? f[x1] – f[x0] f[x0,x1] = x1 – x0 f[x1,x2] – f[x0,x1] f[x0,x1,x2] = x2 – x1 f[x0, x1, …, xk-1, xk] = for k = 3, 4, ….. n. These Ist, IInd... and kth order differences are denoted by f, 2f, …, kf.

  10. INTERPOLATION USING DIVIDED DIFFERENCE • The divided difference interpolation polynomial is: P(x) = f(x0) + (x – x0) f [x0, x1] + L + (x – x0) L (x – xn-1) f[x0, x1, …, xn]

  11. Example • For the data x: –1 0 2 5 f(x) : 7 10 22 235 • Find the divided difference polynomial and estimate f(1).

  12. Solution

  13. INTERPOLATION FOR EQUALLY SPACED POINTS Let (X0,,Y0), (X1,,Y1), …, (Xn,,Yn) be the given points with Xi+1 = Xi +h,i = 0,1,2,…, (n-1). • Finite Difference Operators • Forward difference operator f(xi) = f (xi + h) – f (xi) • Backward difference operator f(xi) = f (xi) – f (xi – h) • Central difference operator • f(xi) = f - f

  14. INTERPOLATION FOR EQUALLY SPACED POINTS • Shift operators • Averaging operator

  15. RELATION BETWEEN OPERATORS • fi = fi+1 = dfi+ ½ •  = E – 1,  = 1 – E-1 •  = E½ – E-½ •  = ( E½ + E- ½)

  16. NEWTON GREGORY FORWARD INTERPOLATION and f0 = y0. Then we have For convenience we put p =

  17. Example Estimate f (3.17)from the data using Newton Forward Interpolation. x: 3.1 3.2 3.3 3.4 3.5 f(x): 0 0.6 1.0 1.2 1.3

  18. Solution

  19. Solution

  20. NEWTON GREGORY BACKWARD INTERPOLATION FORMULA

  21. Example Estimate f(42) from the following data using newtonbackward interpolation. x: 20 25 30 35 40 45 f(x): 354 332 291 260 231 204

  22. Solution Here xn = 45, h = 5, x = 42 and p = - 0.6

  23. Solution

  24. INTERPOLATION USING CENTRAL DIFFERENCES • Suppose the values of the function f (x) are known at the points a -3h, a – 2h, a – h, a, a +h, a + 2h, a + 3h, ... etc. Let these values be y-3, y-2, y-1, y0, y1, y2, y3 ..., and so on. Then we can form the central difference table as: We can relate the central difference operator  with  and E using the operator relation  = E½.

  25. GAUSS FORWARD INTERPOLATION FORMULA

  26. GAUSS FORWARD INTERPOLATION FORMULA • The value p is measured forwardly from the origin and 0<p<1. • The above formula involves odd differences below the central horizontal line and even differences on the line. This is explained in the following figure.

  27. Example Find f(30) from the following table values using Gauss forward difference formula:

  28. Solution f (30) = 16.9217

  29. + + + æ ö æ ö æ ö æ ö p p 1 p 1 p 2 ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ 1 2 3 4 è ø è ø è ø è ø GAUSS BACKWARD INTERPOLATION FORMULA This formula is used to interpolate the value of the function for a negative value of p and – 1<p<0. By 2 3 substituting for y , y , y , ....... in terms of central D D D 0 0 0 difference in the Newton Forward Difference formula, we get 2 3 4 y + y + y + y + y +… D D D D P (x) = 0 - 1 - 1 - 2 - 2

  30. GAUSS BACKWARD INTERPOLATION FORMULA • This formula involves odd differences above the central horizontal line and even differences on the central line.

  31. Example Estimate cos 51 42' by Gauss backward interpolation from the following data: x:50 51 52 53 54 cosx: 0.6428 0.6293 0.6157 0.6018 0.5878

  32. Solution

  33. Solution P(51.7) = 0.6198

  34. STIRLING’S FORMULA • This formula gives the average of the values obtained by Gauss forward and backward interpolation formulae. For using this formula we should have – ½ < p< ½. We can get very good estimates if - ¼ < p < ¼. The formula is:

  35. Example • Using Sterling Formula estimate f (1.63) from the following table: x: 1.50 1.60 1.70 1.80 1.90 f(x):17.609 20.412 23.045 25.527 27.875 Solution X0 = 1.60, x = 1.63, h = 0.1

  36. Solution • Difference table

  37. BESSELS’ INTERPOLATION FORMULA • This formula involves means of even difference on and below the central line and odd difference below the line. • The formula is

  38. BESSELS’ INTERPOLATION FORMULA Note • When p = 1/2 , the terms containing odd differences vanish. • Then we get the formula in a more simple form:

  39. Example Use Bessels’ Formula to find (46.24)1/3 from the following table of x1/3.

  40. Solution x0 = 45, x = 46.24, h = 4 p = 0.31 Difference table is: Applying Bessels’ formula, we get (46.24)1/3 = 3.5893

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