Newtons backward formula using C 



Sumit Kar, Numerical Methods, C, Timus Rak



/*

Newtons backward formula

x: 1 1.10 1.20 1.30
y: .4158 .8912 .9320 .9636
Compute::f(1.40)=?
*/
#include<conio.h>
#include<stdio.h>
#define n 4
void main()
{
 float x[n],y[n][n],xf,fx,u,h;
 int j,i,index,f=1;
 for(i=0;i<n;i++)
 {
 printf("\nEnter x[%d] & y[%d][0]",i+1,i+1);
 scanf("%f%f",&x[i],&y[i][0]);
 }
 for(j=1;j<n;j++)
 {
 for(i=0;i<n;i++)
 y[i][j]=y[i+1][j-1]-y[i][j-1];
 }
 for(i=0;i<n;i++)
 {
 for(j=0;j<n-1;j++)
  printf("%f\t",y[i][j]);
 printf("\n");
 }
 printf("\nEnter x");
 scanf("%f",&xf);
 index=n-1; 
 h=x[1]-x[0];
 u=(xf-x[index])/h;
 fx=y[index][0];
 for(i=1;i<n;i++)
 {
 if(i!=1)
    u=u*(u+i-1);
 f=f*i;
 fx=fx+(u*y[index][i])/f;
 }

 printf("\nResult %f",fx);
 getch();
}








































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