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

75 lines
2.2 KiB
C++

/*------------ NEWTON'S DIVIDED DIFFERENCES INTERPOLATION METHOD --------*/
/* THE PROGRAM GENERATES A DIVIDED DIFFERENCES TABLE FROM GIVEN
DATA, AND IT CALCULATES THE VALUE OF f(x) AT GIVEN VALUE
OF xr.
INPUTS : 1) Number of entries of the data.
2) Values of 'x' & corresponding y = f(x).
3) Value of xr at which f(x) is to be interpolated.
VALUES OF x NEED NOT BE EQUALLY SPACED.
OUTPUTS : Interpolated value of 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],sum,fy,xr;
/* ARRAY OF y[n][n] ELEMENTS FOR DIVIDED DIFFERENCE TABLE */
int i,j,k,n,t,m;
clrscr();
printf("\n NEWTON'S DIVIDED DIFFERENCES INTERPOLATION METHOD");
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) is to be"
" calculated xr = ");
scanf("%lf",&xr);
k = n;
for(j = 1; j < n; j++)
{
/* LOOP TO CALCULATE DIVIDED DIFFERENCES IN THE TABLE */
k = k - 1;
for(i = 0; i < k; i++)
{
y[i][j] = (y[i+1][j-1] - y[i][j-1])/(x[i+j]-x[i]);
}
}
sum = 0;
for(t = 1; t < n; t++)
{
/* LOOP TO CALCULATE INTERPOLATED VALUE OF 'y' */
fy = 1;
for(m = 0; m < t; m++)
{
fy = fy * (xr - x[m]);
}
sum = sum + (fy * y[0][t]);
}
sum = sum + y[0][0];
printf("\nThe interpolated value of y at xr = %lf"
" is yr = %lf\n",xr,sum);
}
/*-------------------- END OF PROGRAM ---------------------------------*/