Programming for Lagrange's interpolation method

/*lagrange's interpolation method*/

#include<stdio.h>

#include<math.h>

void

main()

{int i,j,n;

float x[20],y[20],x0,sum=0.0,num,deno;

printf("enter the total number of inputs:");

scanf("%d",&n);

printf("enter the value of x:");

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

{

                scanf("%f",& x[i]);

}

printf("enter the corresponding value of y:");

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

{

                scanf("%f",&y[i]);

}

printf("enter the value of x at which y is calculated:");

scanf("%f",& x0);

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

{

                num=1.0;

                deno=1.0;

                for(j=0;j<n;j++)

                {

                                if(j!=i)

                                {

                                                num=num*(x0-x[j]);

                                                deno=deno*(x[i]-x[j]);

                                }

                }

                sum=sum+(num/deno)*y[i];

}

printf("the required value of y is=%f\n",sum);

}

RESULT:

enter the total number of inputs:7

enter the value of x:2 3 4 5 6 7 8

enter the corresponding value of y:2 7 14 23 34 47 62

enter the value of x at which y is calculated:5.5

the required value of y is=28.250000

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