To implements the Lagrange interpolation and Newton Gregory forward interpolation

       Algorithm:

Step 1. Read x,n

Step2. for i=1 to (n+1) is steps of 1 do Read xi,fi end for {the above 

           statements reads x,s and the corresponding values of  f is }

Step 3. Sum=0

Step 4. for i=1 to (n+1) in steps of 1 do

Step 5. Profvnc=1

Step 6. for J=1 to (n+1) in steps of 1 do

Step 7. If (j≠i) then prodfunc=prodfunc X(x-xj) / (xi-xj) endfor

Step 8. Sum=Sum+fi x Prodfunc {sum is the value of f at x} end for

Step 9. Write x, sum

Step 10. STOP

Program:

#include<stdio.h>

#include<math.h>

Main()

{

  Float y, x[20],f[20],sum,pf;

  Int I,j,n;

  Printf(“enter the value of n”);

  Scanf(“%d”,&n);

  Printf(“enter the value to be found”);

  Scanf(“%f”,&y);

  Printf(“enter the values of xi’s & fi’s”);

  For(i=0;i<n;i++)

   {

        Pf=1;

         For(j=0;j<n;j++)

         {

                     If(j!=i)

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

         }

         Sum += f[i] *pf;

 }

Printf(“\nx = %f ”,y);

Printf(“\n sum =%f ”,sum);

}

Input/Output:

Enter the value of n 4

Enter the value to be found 2.5

Enter the values for xi’s & fi’s

1                    1

2                    8

3                    27

4                    64

         X = 2.500000

         Sum = 15.625000

Conclusion: The program is error free

VIVA QUESATIONS

1) Define storage class ?

Ans: Storage class specifiers inform the complier how to store the variable; the storage clas specifiers in the c language are : auto, register, static,extern, typedef

           Newton  gregory  forward interpolation.

          Algorithm:

Step1: START

Step2: Read n

Step3: for i=0 to (n-1) do read xi,yi

Step4: read x

Step5: h←xi-x0

Step6: p←(x-xo)/n

Step7: for j=0 to n-2 do

   ∆1yj←yj+1-∆i-1

Step8: k←n-2

Step9: for i=2 to (n-1)do

    Step9.1: k←k-1

    Step9.2:for j=0 to  k do

      ∆iyj←∆i-1 yj+1-∆i-1yj

Step10:   Sumy←y0

Step11: Pvalue←1

Step12: Fact value←1

Step13: for l=1 to (n-1) do

   Step13.1: Pvalue←pvalue x (p-(l-1))

Step13.2: factvalue←factvaluex1

Step13.3: term←(pvalue x ∆ly) / factvalue

Step13.4: Sumy←Sumy+term

Step14: Print x,SUMY

Step15: STOP

 Program:

#include<stdio.h>

#include<math.h>

Main()

{

    Int i, j, n, k, l;

    Float sumy, h, term, p, z, pvalue;

    Float x[25], y[25], d[25][25], factvalue;

    Printf(“enter the value of n”);

    Scanf(“%d”,&n);

    Printf(“enter %d values for x, y \n”,n);

    For(i=0;i<n;i++)

     Scanf(“%f %f”, &x[i], &y[i]);

     Printf(“\n enter z”);

     Scanf(“%f”,&z);

     h = x[1] – x[0];

     p = (z - x[0] )/ h;

     for(j=0; j<n-2; j++)

         d[i][j] =y[j+1] – y[j];

         k=n-2;

    for(i=2; i<n; i++)

    {

         k++;

         for(j=0; j<=k; j++)

           d[i][j] =d[i-1][j+1] – d[i-1][j];

   }

   For(l=1; l<n; l++)

   {

         Pvalue *= (p-(l  - 1));

         Factvalue *= 1;

         Term = pvalue* d[l][0] / factvalue;

         Sumy += term;

 }

 Printf(“\n y value at z = %f is %f”, z, sumy);

}

Input/Output:

Enter n 7

Enter 7 data values for x, y

                  1921                35

1931                42

1941                58

1951                84

1961               120

1971              165

1981              220

Enter z 1925

Y value at z = 1925.000000 is 36.756710

Conclusion: The program is error free

VIVA QUESATIONS

1) What is the use of goto statement ?

Ans: The goto statement is used to alter the normal sequence of the program execution by unconditionally transferring control to some other part of the program.

 2) What is the use of continue statement ?

Ans: The continue statement is used to bypass the remainder of the current pass through a loop