PROGRAM FOR NEWTON’S FORWARD DIFF. FORMULA

/*PROGRAM FOR NEWTON’S FORWARD DIFF. FORMULA*/

#include<stdio.h>

#include<conio.h>

#define n 4

void main()

{

float nr=1,dr=1,p,x,y;

float ax[n],ay[n],delta[n-1][n-1];

int i=0,j=0,k;

clrscr();

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

{

printf("\n Enter the value of ax and ay:\n");

scanf("%f%f",&ax,&ay);

}

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

scanf("%f",&x);

for(i=0;i<n-1;i++)

{

delta[i][0]=(ay[i+1]-ay[i]);

printf("\n %f",delta[i][0]);

}

for(j=1;j<n-1;j++)

{

for(i=0;i<(n-1-j);i++)

{

delta[i][j]=delta[i+1][j-1]-delta[i][j-1];

printf("\n\n%f",delta[i][j]);

}

}

for(i=0;i<n-1;i++)

{

printf("\n\n");

}

for(i=0;i<n-1-j;i++)

{

printf("%f",delta[j][i]);

}

p=(x-ax[0]/ax[1]-ax[0]);

y=ay[0];

for(k=0;k<n-1;k++)

{

nr=nr*(p-k);

dr=dr*(k+1);

y+=((nr/dr)*delta[0][k]);

}

printf("ANSWER: VALUE OF Y AT%f IS %f",x,y);

getch();

}

/*OUTPUT

Enter the value of ax and ay:

35

13229

Enter the value of ax and ay:

45

31368

Enter the value of ax and ay:

55

55593

Enter the value of ax and ay:

65

87089

Enter the value of x:50

18139.000000

24225.000000

31496.000000

6086.000000

7271.000000

1185.000000

ANSWER: VALUE OF y AT 50.000000 IS 42645.687500*/