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

59 lines
1.8 KiB
C++

/*----------------- LAGRANGE'S INTERPOLATION METHOD --------------------*/
/* THE PROGRAM CALCULATES THE VALUE OF f(x) AT GIVEN VALUE OF x
USING LAGRANGE'S INTERPOLATION METHOD.
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],x[20],xr,fy,num,den;
int i,j,n;
clrscr();
printf("\n LAGRANGE'S INTERPOLATION TECHNIQUE");
printf("\n\nEnter the number of entries (max 20) = ");
/* ENTER THE NUMBER OF ENTRIES */
scanf("%d",&n);
for(i = 0; i < n; i++)
{
/* LOOP TO GET x AND y = f(x) IN THE ARRAY */
printf("x%d = ",i);
scanf("%lf",&x[i]);
printf(" y%d = ",i);
scanf("%lf",&y[i]);
}
printf("\nEnter the value of xr at which y = f(x)\n\t\t\t"
"is to be interpolated, xr = ");
scanf("%lf",&xr);
fy = 0;
for(j = 0; j < n; j++)
{
/* LOOP TO CALCULATE LAGRANGE'S INTERPOLATION */
num = den = 1;
for(i = 0; i < n; i++)
{
if(i == j) continue;
num = num * (xr - x[i]);
den = den * (x[j] - x[i]);
}
fy = fy + ((num/den) * y[j]);
}
printf("\nThe value of y = f(x) at xr = %lf is yr = %lf", xr,fy);
}
/*--------------------------- END OF PROGRAM ----------------------------*/