NUmeric :LINEAR INTERPOLATION–QUADRATIC -LAGRANCE

NUmeric :LINEAR INTERPOLATION–QUADRATIC -LAGRANCE

sample input

T	700	720	740	760
V	0.0977	0.12184	0.1406	0.15509

 

//LINEAR INTERPOLATION

//QUADRATIC

//LAGRANCE

#include<stdio.h>

int main()

{

    double y[50],x[50],X,f1,f11,b2,f2,sum=0,G;

    int n,op;

printf("Enter the Range :");

scanf("%d",&n);

//....................................

printf("ENTER x =\n");

for(int i=0;i<n;i++)

{

    scanf("%lf",&x[i]);

}

//....................................

printf("ENTER y =\n");

for(int i=0;i<n;i++)

{

    scanf("%lf",&y[i]);

    //((y[2]-y[1])/(x[2]-x[1]));

}

printf("ENTER the value of X =\n");

scanf("%lf",&X);

//.....................................

    for(int i=0;i<n;i++)

    {    G=1;

        for(int j=0;j<n;j++)

        {

            if(i==j)

            {

                continue;

            }

            G=G*((X-x[j])/(x[i]-x[j]));

        }

        sum=sum+G*y[i];

    }

//.....................................

printf("\n\nEnter Your Option\n__________________________\n");

printf("1.lINEAR INTERPOLATION\n\n2.QUADRATIC\n\n3.LAGRANCE\n__________________________\n");

while(scanf("%d",&op)==1)

{

switch(op)

{

case 1:

f1=y[0]+(((y[1]-y[0])*(X-x[0]))/(x[1]-x[0]));

printf("LINEAR INTERPOLATION RESULT is = %lf\n\n",f1);

    break;

case 2:

    f11=y[0]+(((y[1]-y[0])*(X-x[0]))/(x[1]-x[0]));

    b2=((((y[2]-y[1])/(x[2]-x[1]))-((y[1]-y[0])/(x[1]-x[0])))/(x[2]-x[0]));

    f2=(f11+(b2*((X-x[0])*(X-x[1]))));

    printf("\nQUADRATIC RESULT is = %lf\n\n",f2);

    break;

case 3:

    printf("Lagrance RESULT is = %lf \n",sum);

    break;

default:

    printf("ERORR");

    break;

}

}

//.....................................

return 0;

}