Interpolation and Curve Fitting - Numerical Methods in Engineering with MATLAB - page 114

All the P 's in the table are estimates of the root resulting fromdifferentorders
of interpolation involving different datapoints.For example, P 1 [ y 1 , y 2 ] is the root

obtained from linear interpolationbased on the first two points, and P 2 [ y 2 ,

y 4 ] is
the result from quadratic interpolationusing the last three points. The root obtained
from cubic interpolation overall four datapoints is x

y 3 ,

= P 3 [ y 1 , y 2 , y 3 , y 4 ] = 3 . 8317.

EXAMPLE 3.4
The datapoints in the table lieon the plot of f ( x ) = 4 . 8 cos π x / 20. Interpolate this

databyNewton's methodat x = 0 , 0 . 5 , 1 . 0 ,..., 8 . 0 and compare the results with the

“exact”values givenby y = f ( x ).

x 0 . 15 2 . 30 3 . 15 4 . 85 6 . 25 7 . 95

y 4 . 79867 4 . 49013 4 . 2243 3 . 47313 2 . 66674 1 . 51909

Solution
% Example 3.4 (Newton's interpolation)
xData = [0.15; 2.3; 3.15; 4.85; 6.25; 7.95];
yData = [4.79867; 4.49013; 4.22430; 3.47313;...
2.66674; 1.51909];
a = newtonCoeff(xData,yData);
' x yInterp yExact'

forx=0:0.5:8
y = newtonPoly(a,xData,x);
yExact = 4.8*cos(pi*x/20);
fprintf('%10.5f',x,y,yExact)
fprintf('\n')

end

The results are:

ans =

x yInterp yExact

0.00000 4.80003 4.80000

0.50000 4.78518 4.78520

1.00000 4.74088 4.74090

1.50000 4.66736 4.66738

2.00000 4.56507 4.56507

2.50000 4.43462 4.43462

3.00000 4.27683 4.27683

3.50000 4.09267 4.09267