Numerical Analysis Programs Using Fortran 90 - University of ...

Problem: Write a program to find the interpolated value for x =1. 05 ,
using Stirling Interpolation method, for these tabulated data.
X 0.5 0.7 0.9 1.1 1.3 1.5
y 0.48 0.64 0.78 0.89 0.96 0.99
program Stirling_Formula_Iterpolation
dimension x(20),y(20),d(20,20)
n=6
xx=1.05
data (x(i), i=1,6) /.5,.7,.9,1.1,1.3,1.5/
data (y(i), i=1,6)/.48,.64,.78,.89,.96,.99/
print*,(x(i), i=1,6)
print*,(y(i), i=1,6)
s=y(4)
do i=1,n-1
do j=1,n-i
y(j)=y(j+1)-y(j)
d(i,j)=y(j)
print*,d(i,j)
enddo
enddo
r=(xx-x(4))/(x(2)-x(1))
p1=r*(d(1,3)+d(1,4))/2
p2=r**2/2*d(2,3)
p3=r*(r**2-1)/6*(d(3,2)+d(3,3))/2
p4=r**2*(r**2-1)/24*d(4,2)
s=s+p1+p2+p3+p4
print*,'Interpolation of(',xx,')=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Interpolation of(1.05)=8.660302 E-01
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