programming-examples/c++/Computational_Geometry/C++ Program to Implement NEWTON'S BACKWARD DIFFERENCES INTERPOLATION METHOD.cpp
2019-11-15 12:59:38 +01:00

75 lines
2.3 KiB
C++

/*------ NEWTON'S BACKWARD DIFFERENCES INTERPOLATION METHOD -----------*/
/* THE PROGRAM GENERATES A BACKWARD DIFFERENCES TABLE FROM GIVEN
DATA, & CALCULATES THE VALUE OF f(x) AT GIVEN VALUE OF x.
INPUTS : 1) Number of entries of the data.
2) Values of 'x' & corresponding y = f(x).
3) Value of 'xr' at which y = f(x) to be calculated.
OUTPUTS : Interpolated value f(x) at x = xr. */
/*------------------------------ PROGRAM ---------------------------*/
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<conio.h>
void main()
{
double y[20][20],x[20],xr,yr,h,r,sum,fy,facto;
int i,j,k,n,m,t;
clrscr();
printf("\n\tNEWTON'S BACKWARD DIFFERENCES INTERPOLATION TECHNIQUE");
printf("\n\nEnter the number of entries (max 20) = ");
/* ENTER THE NUMBER OF ENTRIES IN THE TABLE */
scanf("%d",&n);
for(i = 0; i < n; i++)
{
/* LOOP TO GET x AND y = f(x) IN THE TABLE */
printf("x%d = ",i);
scanf("%lf",&x[i]);
printf(" y%d = ",i);
scanf("%lf",&y[i][0]);
}
printf("\nEnter the value of xr at which y = f(x)\nis to be "
"interpolated, xr = ");
scanf("%lf",&xr);
h = x[1] - x[0];
r = (xr - x[n-1])/h; /* CALCULATE VALUE OF 'r' */
printf("\nThe value of h = %lf and value of r = %lf\n",h,r);
k = 0;
for(j = 1; j < n; j++)
{
/* LOOP TO CALCULATE BACKWARD DIFFERENCES */
k++;
for(i = n-1; i >= k; i--)
{
y[i][j] = y[i][j-1] - y[i-1][j-1];
}
}
sum = 0;
for(t = 1; t < n; t++)
{
/* LOOP FOR NEWTON'S BACKWARD DIFFERENCE INTERPOLATION FORMULA */
fy = 1;
facto = 1;
for(m = 0; m < t; m++)
{
fy = fy * (r + m);
facto = facto * (m + 1);
}
fy = fy * (y[n-1][t]/facto);
sum = sum + fy;
}
yr = sum + y[n-1][0];
printf("\nThe value of y = f(x) at xr = %lf is yr = %lf", xr,yr);
}
/*--------------------------- END OF PROGRAM ---------------------------*/