LECTURE 3: Polynomial interpolation and numerical differentiation

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Example. Find the Newton form of the interpolating polynomial for the following table of values: x 1/3 1/4 1 y 2 −1 7 We will construct three successive polynomials p 0 , p 1 and p 2 . The first one is p 0 (x)=2 The next one is given by p 1 (x)= p 0 (x)+c 1 (x−x 0 )=2+c 1 (x−1/3) Using the interpolation condition p 1 (1/4)=−1, we obtain c 1 = 36, and p 1 (x)=2+36(x−1/3) Finally, p 2 (x)= p 1 (x)+c 2 (x−x 0 )(x−x 1 ) = 2+36(x−1/3)+c 2 (x−1/3)(x−1/4) The interpolation condition gives p 2 (1)=7, and thus c 2 =−38. The Newton form of the interpolating polynomial is p 2 (x)=2+36(x−1/3)−38(x−1/3)(x−1/4) Comparison of Lagrange and Newton forms Lagrange form p 2 (x)=−36(x−1/4)(x−1) − 16(x−1/3)(x−1) + 14(x−1/3)(x−1/4) Newton form p 2 (x)=2+36(x−1/3)−38(x−1/3)(x−1/4) These two polynomials are equivalent, i.e: p 2 (x)=−38x 2 +(349/6)x−79/6
2.5 Nested multiplication For efficient evaluation, the Newton form can be rewritten in a so-called nested form. This form is obtained by systematic factorisation of the original polynomial. Consider a general polynomial in the Newton form p n (x)=a 0 + a 1 (x−x 0 )+a 2 (x−x 0 )(x−x 1 )...+a n (x−x 0 )...(x−x n−1 ) This can be written as p n (x)=a 0 + n ∑ i=1 [ ] i−1 a i (x−x j ) ∏ j=0 Let us now rewrite the Newton form of the interpolating polynomial in the nested form by using the common terms(x−x i ) for factoring. We obtain p n (x)=a 0 +(x−x 0 )(a 1 +(x−x 1 )(a 2 +(x−x 2 )(a 3 +...+(x−x n−1 )a n )...)) This is called the nested form and its evaluation is done by nested multiplication. When evaluating p(x) for a given numerical value of x = t, we naturally begin with the innermost parenthesis: The quantity v n is p(t). v 0 = a n v 1 = v 0 (t− x n−1 )+a n−1 v 2 = v 1 (t− x n−2 )+a n−2 . v n = v n−1 (t− x 0 )+a 0 In the actual implementation, we only need one variable v. The following pseudocode shows how the evaluation of p(x = t) is done efficiently once the coefficients a i of the Newton form are known. The array (x i ) 0:n contains the n+1 nodes x i . real array (a i ) 0:n ,(x i ) 0:n integer i,n real t,v v←a n for i=n-1 to 0 step -1 do v←v(t− x i )+a i end for
Page 1 and 2: LECTURE 3: Polynomial interpolation
Page 3 and 4: Note that when l i (x) is evaluated
Page 5: 2.4 Newton form The inductive proof
Page 9 and 10: Example. Determine the quantities f
Page 11 and 12: 2.7 Recursive property of divided d
Page 13 and 14: The table with a completed second c
Page 15 and 16: The following procedure Coefficient
Page 17 and 18: Figure 1 shows a plot of the data p
Page 19 and 20: 2.9 Inverse interpolation A process
Page 21 and 22: We can simplify the notation by S i
Page 23 and 24: 3.1 Choice of nodes Contrary to int
Page 25 and 26: Example. In a previous example, we
Page 27 and 28: 4 Numerical differentiation We now
Page 29 and 30: 4.3 Richardson extrapolation theore
Page 31 and 32: 4.5 First-derivative formulas via i
Page 33 and 34: This allows us to write out the for

LECTURE 3: Polynomial interpolation and numerical differentiation

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