Introduction
Lagranges interpolation formula
The code
Code: Cpp
#include <stdio.h>
#include <conio.h>
#define N 100
/*******************************************/
/***** Lagranges interpolation formula *****/
/*******************************************/
void main()
{
/*
y=(x-x2)(x-x3)...(x-xn)/(x1-x2)(x1-x3)...(x1-xn)*y[1]
+(x-x1)(x-x3)...(x-xn)/(x2-x2)(x2-x3)...(x2-xn)*y[2]
+...+
(x-x1)(x-x2)...(x-x(n-1))/(xn-x1)(xn-x2)...(xn-x(n-1))*y[n]
*/
float x[N]={1,3,13,21}; /* Values of x i's */
float y[N]={0,1,3,4}; /* Values of x i's */
int i,j; /* loop variable's */
float res; /* result of each sub term */
float ans=0; /* Final result */
float givx=2; /* Given x for which y is to be determined */
int n=4; /* No of points for interpolation */
char ch,newx; /* Choice inputing variables */
do
{
printf("\nEnter the total number of points for interpolation\n");
scanf("%d",&n);
printf("\nEnter the values of x\n");
for(i=0;i<n;i++)
scanf("%f",&x[i]);
printf("\nEnter the values of y\n");
for(i=0;i<n;i++)
scanf("%f",&y[i]);
do
{
printf("\nEnter the values of x to find y\n");
scanf("%f",&givx);
for(i=0;i<n;i++)
{
res=1;
for(j=0;j<n;j++)
{
if(i==j)
continue;
res=res*(givx-x[j])/(x[i]-x[j]);
}
ans=ans+res*y[i];
}
printf("\nanswer = %g\n",ans);
printf("\nDo you want y for new value of x[y/n]\n");
newx=(char)getche();
}while(newx=='Y' || newx=='y');
printf("\nDo you wish to continue[y/n]\n");
ch=(char)getche();
}while(ch=='Y' || ch=='y');
getch();
}
