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Wednesday, November 12, 2014

NUmeric : Newton Rapson code

 

Sample equation for Newton Rapson

where X=3;

F(x)=2x3-11.7x2+17.7x1-5

 

//NETON RAPSON
 
#include<stdio.h>
#include<math.h>
#include<iostream>
using namespace std;
int main()
{
    float xi,Es,a[40],xiii,old=0,Ea;
    int n,K=0;
    cout<< "HIghest power in the equation : " ;
    cin>>n;
    //.........................................
    printf("Enter Xi =");
    scanf("%f",&xi);
    printf("Enter Es =");
    scanf("%f",&Es);
    //..........................................
    printf("\n");
    printf("enter coefficient :\n");
    for (int i=n; i>=0; i--)
    {
        printf("X^%d =",i);
        scanf("%f",&a[i]);
    }
    //...........................................
    while(1)
    {
        float fxi=0,deff=0;
        printf("\n\n\nSTEP no: %d\n",K+1);
        for (int i=n; i>=0; i--)
        {
            fxi=fxi+a[i]*pow(xi,i);
        }
        for(int i=n; i>=0; i--)
        {
            deff=deff+i*a[i]*pow(xi,i-1);
        }
        printf("Xi   =  %.2f",xi);
        printf("\nfxi  =  %.2f",fxi);
        printf("\ndeff   =  %.2f",deff);
        xiii=(xi-(fxi/deff));
        printf("\nnew xi   =  %.2f",xiii);
 
        Ea=((xiii-old)/xiii)*100;
        if(Ea<0)
        {
            Ea=Ea*(-1);
        }
        printf("\nEa= %f",Ea);
            old=xiii;
            if(Ea<Es)
        {
            break;
        }
 
        xi=xiii;
 
        K++;
 
    }
    return 0;
}

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