Thread: Numerical Integration by Trap Rule.

accuracy>accuracyreq

for loop runs while condition is true

you set accuracy to 0, and I suppose enter some positive value to the accuracyreq

so the condition is never true and your loop should not run at all

Good point, it doesnt run the initial loop in which accuracy would be set to some other number (greater then 0 and greater then accuracyreq).

Having changed that, there appears to be an infinite loop (or else, my program takes 1+minute to run.)

Any ideas?

Ok, Update.

I have realized that when I input my accuracyreq, I am inputting it in the form of 0.999999 being very accurate and 0.000001 being very inaccurate. therefore, I have switched accuracy>accuracyreq into accuracyreq>accuracy meaning my loop will run while my accuracy is less then the accuracyreq.

It is now running, but when I printf my number of iterations and step number, the step number remains 3 (which is the initial step number, it isnt incrementing) and the iteration number comes out fairly large (over 20,000).

Not sure where to go next.

Current Code.

Code:

#include <stdio.h>
#include <math.h>

main()
{
  int n,i;
  double x1,xn,h,new,old,sum,y,x,accuracyreq,accuracy;

  printf("Input Accuracy:");
  scanf("%lf",&accuracyreq);

  printf("\nInput Upper Limit Value (Larger is More Accurate):");
  scanf("%lf",&xn);

  printf("The following is the answer to the integral cos(x)/sqrt(x): ");
  accuracy=100;
  new=2;
  sum=0;
  x=x1;
  old=0.5;
  
  
  for(n=3;accuracy<accuracyreq;n++)
  {
  h=xn-x1/(n-1);
  
 for(i=0;x<xn;i++)
  {
            
  y=2*cos(pow(x,2));
  if(x==x1||x==xn){
                   y=y/2;
                   }
  sum=sum+y;
  x=x+h;
   }
 
new=sum*h;
accuracy=new/old;

//printf("Accuracy is %lf",accuracy);
}
    
    printf("%lf\n",new);
    printf("%d\n",i);
    printf("%d\n",n);
    int end;
    scanf("%d",&end);
}