Thread: Simpson's Rule (not composite simpson's rule)

Perhaps post your main() as well, so we can see what parameters you pass to this function.

We could also run the same test as you as well.

Originally Posted by Salem

Perhaps post your main() as well, so we can see what parameters you pass to this function.

We could also run the same test as you as well.

Code:

double simpsons ( double a, int n, double b, double ans);
double Anal_function (double a, double b);
int main ()
{
    double a, b, h, ans, Anal_ans;           ////variables to the function 
    int n;                                     ///variable to the function (n should be an integer)
    
    
    printf ("Please enter Start value (a):");
    scanf ("%lf", &a);
    
    printf ("Please enter End value (b);");
    scanf ("%lf", &b);
    
    printf("Please enter number of segments (n);");
    scanf ("lf", &n);
    
    h= (b-a)/2.0;                          ///defining segment size
    
    ans= simpsons (a, b, h, n);                        //// ans  is equal to the simpsons function, that is defined later)
    Anal_ans= Anal_function (a, b);                    //// here we linked the two variables that are defined differently.
    
    printf ("The value of Integral according to Simpson's Rule= %0.8f\n",ans);
    printf ("The value from the Analytical method = %0.8f\n", Anal_ans);
    
    system ("pause");
    return 0;
    
}


double simpsons ( double a, int n, double b, double h)
{
         int i;
         double ans[10000], x=a, final_ans, ans1, holding=0;
         
         for ( i = 0; i <= n-1; i++ )
     
              {  
                 ans[i] = exp (2*x);             ///function to be integrated e^2x ans[] is an array of all x values.
                 x = x + h; 
              }
     
         for ( i = 1; i <= n-2; i+=2 )
             {  
                ans1 = 4*ans[i];
                holding = holding + ans1; 
             }             
         
         h= (b+a)/2.0;
         
         final_ans = (h/3)*(ans[0] + holding + ans[n-1]);
         return final_ans;
}


double Anal_function (double a, double b)
{      
       double Anal_ans;
       Anal_ans = (0.5*(exp(2*b)))-(0.5*(exp(2*a)));
       return Anal_ans;
       }

what im trying here is to do the simpson's rule with 3 points only, the start value, end value and the midpoint.
where f(x) dx= ((b-a)/6)* [f(a) + 4f*(a+b/2) + f(b)].

$ gcc -W -Wall -Wextra foo.c -lm
foo.c: In function ‘main’:
foo.c:20:5: warning: too many arguments for format [-Wformat-extra-args]

Your scanf for n isn't all that it seems to be.
For a start, it lacks a %
Then it's completely wrong for reading an int anyway.

Additionally, telling us the values you would like to type in will help us check results.

> does anyone know where the problem is?
Well until you've shown us that you've fixed your scanf for n, we all know you're just passing garbage values into your function, so any further discussion is moot.

Nor have you told us what values you typed in for a, b (and n - perhaps).
Without this information, all we can do is look at the code and find semantic errors.

We can't run the code to examine in detail runtime behaviour to see where you're screwing up.