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Lecture -8 Newton’s divided difference formula: Note: The application of the Lagrange formula becomes quite lengthy and difficult if number of values in the data is more than four. It can be seen that inclusion of a point in the data list leads to add one more term in the formula and the numerator and denominator of each term get changed. Further, during calculations there is always a chance of committing some error due to a number of positive and negative signs in the coefficients of each term. Except this, there is no way to guess the degree of the resulted polynomial before the final result is obtained after lengthy calculations. e.g. suppose data corresponds to a polynomial of degree two and data points are five. So we have to simplify five terms with each coefficients as a polynomial of degree four. But, only after simplifications of these five terms we can obtain a polynomial of degree two. Hence, it is not advisable to use Lagrange formula for more than four points in the data set. To overcome this problem, we prefer to use the Newton’s divided difference formula derived as below: Let ( ), ( ) … , ( ) be the values of f(x) corresponding to the argument values as , , … , , not necessarily equally spaced. By definition of divided differences ( , )= ( ) ( ) ( )= ( )+( − Also ( , , )= or ( , ) ( , ) ( ). , ) or ( , ) = ( , ) + ( − ) ( , , Similarly ( , , )= ( , , )+( − ) ( , ................. .................. ( , , , )= , , ,…, +( − Multiplying the equation (2) by ( − ), (3) by ( finally the equation (4) by ( − )( − ) … ( equation (1), we have ( ) = ( ) + ( − ) ( , ) + ( − )( − ) ( +( − )( − ) … . ( − ….(1) ). , ...(2) , ). ...(3) ) ( , , , … , ). ... (4) − )( − ) and so on and − ) and adding to the ) ( , , ,…, , ) ...... )+ where the remainder is given by = ( − )( − ) ...... ( − ) ( , , , … ). Assuming that ( )is a polynomial of degree n, ( , , , … , ) vanishes being (n+1) th -order divided difference, so that ( ) = ( ) + ( − ) ( , ) + ( − )( − ) ( , , ) ...... ) ( , , , … , ). +( − )( − ) … ( − ......(5) This is called Newton’s divided difference Formula of Interpolation. Note: As Newton’s divided difference formula requires the divided difference table, it will be better if we rearrange the values of argument before preparing the table either in ascending or descending order. If will be helpful in preparing the table. Formula shows that it uses only the leading differences of various orders. So, once the table is prepared, it is very easy to use. Also, if any order differences are zeros, then one can predict the degree of interpolated polynomial from the table without any calculation. Example : Find the polynomial of the lowest possible degree which assumes the values 3, 12, 15, -21 when x has the values 3, 2, 1, -1, respectively. Hence find f(0). Solution : Divided difference table ( ) ∆ ( ) -1 -21 +18 1 15 -3 2 12 -9 3 3 ∆ ( ) ∆ -7 1 -3 Using Newton’s divided difference formula, we get ( ) = −21 + − (−1) 18 + ( + 1)( − 1) (−7) + ( + 1)( − 1)( − 2) 1 = −21 + 18( + 1) + ( − 1)(−7) + ( − 2 − + 2) f (x) = x3 - 9 x2 + 17 x + 6. Hence f (0) = 6. Example : The observed values of a function are respectively 168, 120, 72 and 63 at the four positions 3, 7, 9 and 10 of the independent variable. What is the best estimated value of the function at the position 6 of the independent variable? Solution : The divided difference table is given as follows : ( ) ∆ ( ) ∆ ( ) 3 168 7 120 9 72 10 63 Hence 120 − 168 = −12 7−3 72 − 120 = −24 9−7 −24 + 12 = −2 9−3 63 − 72 = −9 10 − 9 −9 + 24 =5 10 − 7 ∆ ( ) 5 − (−2) =1 10 − 3 f x f x0 x x0 f x0 , x1 x x0 x x1 f x0 , x1 , x2 x x0 x x1 x x2 f x0 , x1 , x2 , x3 = 168 + ( − 3)(−12) + ( − 3)( − 7)(−2) + ( − 3)( − 7)( − 9)(1) = − 21 + 119 − 27. When = 6, the estimated value of the function is given by (6) = 6 − 21 × 6 + 119 × 6 − 27 = 147