# Ch 3 3 interpolation newton gregory

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Numerical Analysis Interpolation Chap 3_3 1 Equally Spaced Data (Newton Gregory Interpolation) 2 Newton Gregory Interpolation β’ If the x values are evenly spaced, the interpolating polynomial of degree n can be written in terms of the ordinary differences with x evaluate at S π ππ π = ΰ· π=0 π π β π0 π π₯ β π₯0 π= β π π  β 1 π  β 2 β¦ (π β π + 1) π = π π! βπ ππ = βπβ1 ππ+1 β βπβ1 ππ 3 Recall the Newton forward difference polynomial 4 Example β’ compute the interpolating polynomial p3(x) that fits the data equally spaced, with spacing h = 1. 5 Solution β’ equally spacing h = 1 β’ s = (x - x0)/h = x + 1 6 βπ ππ = βπβ1 ππ+1 β βπβ1 ππ i 0 x -1 ππ 3 β1 ππ β2 ππ β3 ππ -4-3=-7 1 0 -4 9 - - 7 =16 5- - 4=9 2 1 5 -20-16=-36 -11-9= -20 -6 β 5=-11 3 2 -6 7 8 Example 3 β’ Consider the following data, find f(1.83) i 0 1 2 3 4 x 1 3 5 7 9 ππ 0 1.0986 1.6094 1.9459 2.1972 9 Example 3 Solution i 0 x 1 ππ 0 βπ ππ = βπβ1 ππ+1 β βπβ1 ππ β1 ππ β2 ππ β3 ππ β4 ππ 1.0986 1 3 1.0986 -0.5878 0.5108 2 5 1.6094 0.4135 -0.1747 0.3365 3 7 1.9459 -0.3244 0.0891 -0.0852 0.2513 4 9 2.1972 10 Example 3 Solutionβ¦cont 11 Exercise 2 Estimate f (3.17)from the data using Newt ...
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