CPNM LAB PROGRAM-11

11. Given a table of x and corresponding f(x) values, write a program which will determine f(x)   value at an intermediate x value using Lagrange’s interpolation.

FLOW CHART :-

 

Algorithm: 

Step – 1 : Start 

Step – 2 : Declare arrays x[10], f[10], variables, sum, product, y, i, j, n 

Step – 3 : Initialize sum to zero 

Step – 4 : Print enter value of n 

Step – 5 : Print enter x[i] value s 

Step – 6 : Initialize i to zero 

Step – 7 : Check whether i < n if true go to next Step else go to Step–9 

Step – 8 : Read x[i] values, increment i by 1 and go to Step–7  

Step – 9 : Print function values 

Step – 10 : Initialize i to zero 

Step – 11 : Check whether i < n if true go to next Step else go to Step–13 

Step – 12 : Read f[i] values, increment i by 1 and go to Step–11 

Step – 13 : Print x[i], f[i] values 

Step – 14 : Initialize i to zero 

Step – 15 : Check whether i < n if true execute Step–16 else go to Step–17 

Step – 16 : Print x[i], f[i] values, increment i by 1 and go to Step–15 

Step – 17 : Print enter intermediate value 

Step – 18 : Initialize i to zero 

Step – 19 : Check whether i < n if true execute from Step–19 to Step–25 otherwise go to Step–26 Step – 20 : Initialize product = 1 

Step – 21 : Initialize j to zero 

Step – 22 : Check whether j < n if true execute from Step–23 to Step–24 otherwise go to Step–25 Step – 23 : Check whether i!=j if true go to Step–24 otherwise increment by 1 and go to Step–22 

Step – 24 : Calculate product *=((y-x[j])/(x[i] - x[j])) 

Step – 25 : Calculate sum += f[i] * product 

Step – 26 : Print interpolated values sum 

Step – 27 :    Stop 

Program:

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#include<math.h>

  main()

{

int i, j, n;

float x[10], f[10], sum = 0, product = 1, y;

clrscr();

printf("Enter the value of n \n");

scanf("%d",&n);

printf("Enter %d values of x, f(x) \n",n);

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

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

printf("Enter the intermediate value \n");

scanf("%f",&y);

printf("\n x f(x) \n");

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

printf("\n %f \t %f\n",x[i],f[i]);

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

{

product = 1;

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

{

if(j != i)

product *=((y-x[j])/(x[i] - x[j]));

}

sum += f[i] * product;

}

printf("y = %f \t sum = %f \n",y,sum);

getch();

return 0;

}

 THIS CODE IS CREATED BY ECE THE WONDERS , HEMU THE SMART

OUTPUT :-

Enter the value of n

3

Enter 3 values of x, f(x)

5.0,7.0

6.0,6.0

7.0,5.0

Enter the intermediate value

6.5

 x f(x)

5.000000 7.000000

6.000000 6.000000

7.000000 5.000000

y = 6.500000 sum = 5.500000

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