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Newton’s Divided Difference Formula

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Computer Oriented Numerical Analysis

What is interpolation?
Many times, data is given only at discrete points such as .

So, how then does one find the value of y at any other value of x ?

Well, a continuous function f(x) may be used to represent the data values with f(x) passing through the points (Figure 1). Then one can find the value of y at any other value of x .

This is called interpolation

Newton’s Divided Difference Formula:

To illustrate this method, linear and quadratic interpolation is presented first.

Then, the general form of Newton’s divided difference polynomial method is presented.

Published in: Engineering

Newton’s Divided Difference Formula

  1. 1. Introduction What is interpolation? Many times, data is given only at discrete points such as . , So, how then does one find the value of y at any other value of x ? Well, a continuous function f(x) may be used to represent the data values with f(x) passing through the points (Figure 1). Then one can find the value of y at any other value of x . This is called interpolation  ,, 00 yx  11, yx , ......, 11,  nn yx Interpolation of discrete data. Figure 1  00 , yx  11, yx  22, yx  33, yx  xf x y
  2. 2. Introduction Newton’s Divided Difference Formula: To illustrate this method, linear and quadratic interpolation is presented first. Then, the general form of Newton’s divided difference polynomial method is presented. To illustrate the general form, cubic interpolation is shown in Figure 1. Interpolation of discrete data. Figure 1  00 , yx  11, yx  22, yx  33, yx  xf x y
  3. 3. Linear Interpolation (i) Linear Interpolation: Given pass a linear interpolant through the data Where ),,( 00 yx ),,( 11 yx )()( 0101 xxbbxf  )( 00 xfb  01 01 1 )()( xx xfxf b    Linear interpolation Figure 2  00, yx  11, yx  xf1 x y
  4. 4. Example Que. The upward velocity of a rocket is given as a function of time in Table 1 (Figure 3). Determine the value of the velocity at t=16 seconds using first order polynomial interpolation by Newton’s divided difference polynomial method. Ans. For linear interpolation, the velocity is given by Since we want to find the velocity at t=16, and we are using a first order polynomial, we need to choose the two data points that are closest to t=16 that also bracket t=16 to evaluate it. The two points are t=15 and t=20. Then Table 1 Figure 3 )s(t )m/s()(tv 0 0 10 227.04 15 362.78 20 517.35 22.5 602.97 30 901.67 ,150 t 78.362)( 0 tv ,201 t 35.517)( 1 tv
  5. 5. Example Contd. Gives Hence At t=16 Table 1 Figure 3 )s(t )m/s()(tv 0 0 10 227.04 15 362.78 20 517.35 22.5 602.97 30 901.67
  6. 6. Quadratic Interpolation Given and fit a quadratic interpolant through the data. Noting and assume the quadratic interpolant is given by At , At , Giving, At Giving, Hence the quadratic interpolant is given by
  7. 7. Example Que. The upward velocity of a rocket is given as a function of time in Table 1 (Figure 3). Determine the value of the velocity at t=16 seconds using second order polynomial interpolation by Newton’s divided difference polynomial method. Ans. For quadratic interpolation, the velocity is given by Since we want to find the velocity at t=16, and we are using a secoond order polynomial, we need to choose the three data points that are closest to t=16 that also bracket t=16 to evaluate it. The three points are , and . Then Table 1 Figure 3 )s(t )m/s()(tv 0 0 10 227.04 15 362.78 20 517.35 22.5 602.97 30 901.67
  8. 8. Example Contd. Then Gives Table 1 Figure 3 )s(t )m/s()(tv 0 0 10 227.04 15 362.78 20 517.35 22.5 602.97 30 901.67
  9. 9. Example Contd. Hence At t=16, We get Table 1 Figure 3 )s(t )m/s()(tv 0 0 10 227.04 15 362.78 20 517.35 22.5 602.97 30 901.67
  10. 10. General Form Newton’s Divided Difference Polynomial In the two previous cases, we found linear and quadratic interpolants for Newton’s divided difference method. Let us revisit the quadratic polynomial interpolant formula Note that and are finite divided differences. and are the first, second, and third finite divided differences, respectively. ))(()()( 1020102 xxxxbxxbbxf  where )( 00 xfb  01 01 1 )()( xx xfxf b    02 01 01 12 12 2 )()()()( xx xx xfxf xx xfxf b        Rewriting, We denote the first divided difference by )(][ 00 xfxf  the second divided difference by 01 01 01 )()( ],[ xx xfxf xxf    and the third divided difference by 02 0112 012 ],[],[ ],,[ xx xxfxxf xxxf    where ],[ 0xf ],,[ 01 xxf and ],,[ 012 xxxf are called bracketed functions of their variables enclosed in square brackets.
  11. 11. General Form Newton’s Divided Difference Polynomial , General form of the Newton’s divided difference polynomial for 1n data points,        nnnn yxyxyxyx ,,,,......,,,, 111100  , as ))...()(( ....)()( 110 010   nn n xxxxxxb xxbbxf where ][ 00 xfb  ],[ 011 xxfb  ],,[ 0122 xxxfb   ],....,,[ 0211 xxxfb nnn   ],....,,[ 01 xxxfb nnn  where the definition of the th m divided difference is ],........,[ 0xxfb mm  0 011 ],........,[],........,[ xx xxfxxf m mm    
  12. 12. Example Table of divided differences
  13. 13. Example Que. The upward velocity of a rocket is given as a function of time in Table 1 (Figure 3). Determine the value of the velocity at t=16 seconds using second order polynomial interpolation by Newton’s divided difference for cubical interpolation. Ans. For a third order polynomial, the velocity is given by Since we want to find the velocity at and we are using a third order polynomial, we need to choose the four data points that are closest to that also bracket to evaluate it. The four points are , and Table 1 Figure 3 )s(t )m/s()(tv 0 0 10 227.04 15 362.78 20 517.35 22.5 602.97 30 901.67
  14. 14. Example Contd. Gives Table 1 Figure 3 )s(t )m/s()(tv 0 0 10 227.04 15 362.78 20 517.35 22.5 602.97 30 901.67 03 012123 3 ],,[],,[ tt tttvtttv b    105.22 37660.044453.0    3 104347.5  
  15. 15. Example Table of divided differences 0b 100 t 04.227 1b 148.27 2b ,151 t 78.362 37660.0 3b 914.30 3 104347.5   ,202 t 35.517 44453.0 248.34 ,5.223 t 97.602 b0 = 227.04; b1 = 27.148; b2 = 0.37660; b3 = 5.4347×10−3

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