C program for newton's backward interpolation?

Answer 1

#include<stdio.h>

#include<conio.h>

#include<process.h>

#include<math.h>

void main()

{

int n;

int i,j;

float ax[10];

float ay[10];

float x;

float y=0;

float h;

float p;

float diff[20][20];

float y1,y2,y3,y4;

clrscr();

printf("\t\t!! NEWTON GRAGORY FORWARD INTERPOLATION FORMULA!!\n");

printf("\t\t By KRISHAN \t\t\n");

printf("\t\t enter the no of terms ->");

scanf("%d",&n);

printf("\n\t\t enter the value in form of x->");

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

{

printf("\n\t\t enter the value of x%d->",i+1);

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

}

printf("\n\t\tenter the value in the form of y->");

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

{

printf("\n\t\tenter the value of y %d->",i+1);

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

}

printf("\n\t\tenter the value of x for");

printf("\n\t\t which u want the value of y->");

scanf("%f",&x);

h=ax[1]-ax[0];

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

diff[i][1]=ay[i+1]-ay[i];

for(j=2;j<=4;j++)

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

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

do

{

i++;

}

while(ax[i]<x);

i--;

p=(x-ax[i])/h;

y1=p*diff[i-1][1];

y2=p*(p+1)*diff[i-1][2]/2;

y3=p*(p+1)*(p-1)*diff[i-2][3]/6;

y4=(p+2)*(p+1)*p*(p-1)*diff[i-3][4]/24;

y=ay[i]+y1+y2+y3+y4;

printf("\n\t\t when x=%6.4f,y=%6.8f",x,y);

printf("\n\n\n\t\t\t!! PRESS ENTER TO EXIT!!");

getch();

}

Answer 2

/program for newton backward interpolation formula

#include <stdio.h>

#include <conio.h>

#include <math.h>

#define pf printf

#define sf scanf

#define maxn 100

#define order 4

void main()

{

float ax[maxn+1],ay[maxn+1],diff[maxn+1]

[order+1],nr=1,dr=1,x,p,h,yp;

int n,i,j,k;

clrscr();

pf("enter the value of n\n");

sf("%d",&n);

pf("enter the values in form x,y\n");

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

{sf("f",&ax[i],&ay[i]);}

pf("enter the value of x for which value of y is wanted\n");

sf("%f",&x);

h=ax[2]-ax[1];

for(i=n;i>=1;i--)

{diff[i][1]=ay[i]-ay[i-1];}

for(j=2;j<=order;j++)

{

for(i=n;i>j;i--)

{diff[i][j]=diff[i][j-1]-diff[i-1][j-1];}

}

i=n;

p=(x-ax[i])/h;

yp=ay[i];

for(k=1;k<=order;k++)

{

nr*=p+k-1;

dr*=k;

yp+=(nr/dr)*diff[i][k];

}

pf("when x =%f\t y= %f",x,yp);

getch();

}

Answer 3

//program for newton backward interpolation formula

#include <stdio.h>

#include <conio.h>

#include <math.h>

#define pf printf

#define sf scanf

#define maxn 100

#define order 4

void main()

{

float ax[maxn+1],ay[maxn+1],diff[maxn+1]

[order+1],nr=1,dr=1,x,p,h,yp;

int n,i,j,k;

clrscr();

pf("enter the value of n\n");

sf("%d",&n);

pf("enter the values in form x,y\n");

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

{sf("f",&ax[i],&ay[i]);}

pf("enter the value of x for which value of y is wanted\n");

sf("%f",&x);

h=ax[2]-ax[1];

for(i=n;i>=1;i--)

{diff[i][1]=ay[i]-ay[i-1];}

for(j=2;j<=order;j++)

{

for(i=n;i>j;i--)

{diff[i][j]=diff[i][j-1]-diff[i-1][j-1];}

}

i=n;

p=(x-ax[i])/h;

yp=ay[i];

for(k=1;k<=order;k++)

{

nr*=p+k-1;

dr*=k;

yp+=(nr/dr)*diff[i][k];

}

pf("when x =%f\t y= %f",x,yp);

getch();

}