Thread: Newton Raphson Method

  1. #1
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    Dec 2009
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    Newton Raphson Method

    Hello I am trying to programme c to use the newton raphson method to evaluate roots of an equation.

    My code is as follows:

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>
    double function(double );
    double function2(double );
    int main()
    {
        double x[100];
        double fx[100];
        double gradx[100];
        int n1=0;  
        x[0]=0;
        n1=-1;
    
        do  
        {
            n1++;
            fx[n1]=function(x[n1]);
            gradx[n1]=function2(x[n1]);
            x[n1+1]=x[n1]-(fx[n1]/gradx[n1]);
        }
        while(fx[n1+1]<1e-8||(fx[n1+1]>-1e-8));
    
        for(n1=0;n1<100;n1++)  
        {
            printf("%g\t %g\t %g\n",x[n1],fx[n1],gradx[n1]);
        }
    
        return 0;
    }
    
    double function(double argument1)
        {
            double result;
            double a;
            a=0.4+(8039.0/25000);
            result = sin(pow(argument1,a)-pow(argument1,(1/a))+(a*argument1));
            return result;
        }
    
    double function2(double argument1)
        {
            double result2;
            double aa;
            aa=0.4+(8039.0/25000);
            result2=((aa*pow(argument1, aa-1))-(1/aa*pow(argument1, (1/aa)-1))+aa)*sin(pow(argument1,aa)-pow(argument1,(1/aa))+(aa*argument1));
            return result2;
        }
    I'm getting an answer out but don't think it is the correct one. Any advice on where I am going wrong would be appreciated.

  2. #2
    DESTINY BEN10's Avatar
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    Quote Originally Posted by zeb1d1ah View Post
    Hello I am trying to programme c to use the newton raphson method to evaluate roots of an equation.

    My code is as follows:

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>
    double function(double );
    double function2(double );
    int main()
    {
        double x[100];
        double fx[100];
        double gradx[100];
        int n1=0;  
        x[0]=0;
        n1=-1;
    
        do  
        {
            n1++;
            fx[n1]=function(x[n1]);
            gradx[n1]=function2(x[n1]);
            x[n1+1]=x[n1]-(fx[n1]/gradx[n1]);
        }
        while(fx[n1+1]<1e-8&&(fx[n1+1]>-1e-8));
    
        for(n1=0;n1<100;n1++)  
        {
            printf("%g\t %g\t %g\n",x[n1],fx[n1],gradx[n1]);
        }
    
        return 0;
    }
    
    double function(double argument1)
        {
            double result;
            double a;
            a=0.4+(8039.0/25000);
            result = sin(pow(argument1,a)-pow(argument1,(1/a))+(a*argument1));
            return result;
        }
    
    double function2(double argument1)
        {
            double result2;
            double aa;
            aa=0.4+(8039.0/25000);
            result2=((aa*pow(argument1, aa-1))-(1/aa*pow(argument1, (1/aa)-1))+aa)*sin(pow(argument1,aa)-pow(argument1,(1/aa))+(aa*argument1));
            return result2;
        }
    I'm getting an answer out but don't think it is the correct one. Any advice on where I am going wrong would be appreciated.
    I think there is no need to explain it.
    HOPE YOU UNDERSTAND.......

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  3. #3
    Registered User
    Join Date
    Dec 2009
    Posts
    8
    Thanks for that. My main problem is that I want to take my calculated values of x, fx, and gradx and use it to calculate their next values. The value of x should converge using this method, however at the minute they are just all over the place. I think it's a problem with one of my loops, possibly the printf statement, just don't really have the experience to analyse it.

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