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

Problem: Write a program to find the interpolated value for x =1. 93,
using Newton backward method, for these tabulated data.
X 1 1.2 1.4 1.6 1.8 2.0
f 2.317 2.425 2.522 2.609 2.689 2.762
program Newton_Backward_Iterpolation
dimension x(20),y(20),d(20,20)
n=6
xx=1.93
data (x(i), i=1,6) /1,1.2,1.4,1.6,1.8,2/
data (y(i), i=1,6)/2.317,2.425,2.522,2.609,2.689,2.762/
print*,(x(i), i=1,6)
print*,(y(i), i=1,6)
s=y(n)
do i=1,n-1
do j=1,n-i
y(j)=y(j+1)-y(j)
d(i,j)=y(j)
enddo
enddo
r=(xx-x(n))/(x(2)-x(1))
do i=1,n-1
p=1
do j=0,i-1
p=p*(r+j)
enddo
f=1
do j=1,i; f=f*j; enddo
s=s+d(i,n-i)*p/f
enddo
print*,'Interpolation of(',xx,')=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Interpolation of(1.93)=2.737522
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