/*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
Press any
key to continue
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