I am investigating the equations which say that the error in the trapezium rule is proportional to (b-a)^3, where a and b are the lower and upper limits respectively, and that the error in the simpson's rule is proportional to (b-a)^5.

[img]http://img809.imageshack.us/i/eqns.png/[/img]

(First equation is for trapezium, second equation is for simpson)(Taken from Wikipedia)

I am considering the integral of cos(x) between a=0 and b.

I compute the corresponding error for different (b-a) values, keeping the lower limit 'a' constant and the number of strips 'n' constant. Then I use these values to plot a graph of log(error) vs log(b-a), which should produce a straight line graph of gradient 3 for the trapezium rule and of gradient 5 for the simpson's rule. (See image below)

[img]http://img695.imageshack.us/i/graphqm.png/[/img]

As you can see, the graphs behave as expected (the gradients are indeed 3 and 5), but only up to b≈2, and then they become a mess. Does this mean there is something wrong with my code, or is this related to the precision limit of a floating point? Either way, I don't really understand why the graphs only behave as expected for such a small range.

Any help would be gratefully appreciated. Thanks.

```
// Finding the relationship between error and range of limits
#include<cmath>
#include<fstream>
using namespace std;
double integrand(double);
double trapezium(int, double, double);
double simpson(int, double, double);
int main()
{
ofstream outfile("cosba.xls");
double a=0, b, Itrap, Isimp, errortrap, errorsimp;
const int n=10;
for(b=0.1;b<100.0;b+=0.1)
{
Itrap=trapezium(n,a,b);
Isimp=simpson(n,a,b);
errortrap=Itrap-(sin(b)-sin(a));
errorsimp=Isimp-(sin(b)-sin(a));
outfile<<log((b-a))<<"\t"<<log(abs(errortrap))<<"\t"<<log(abs(errorsimp))<<endl;
}
return 0;
}
double integrand(double x)
{
return cos(x);
}
double trapezium(int n, double a, double b)
{
double x=a, sumpar=0.0, totalsumpar, delta=(b-a)/n;
for(int i=1;i<=n-1;i++)
{
x+=delta;
sumpar+=integrand(x);
}
totalsumpar=(delta*0.5)*((2.0*(sumpar)+integrand(a)+integrand(b)));
return totalsumpar;
}
double simpson(int n, double a, double b)
{
double x, sumpar1=0.0, sumpar2=0.0, totalsumpar, delta=(b-a)/n;
x=a-delta;
for(int i=1;i<=n-1;i+=2)
{
x+=2.0*delta;
sumpar1+=integrand(x);
}
x=a;
for(int i=2;i<=n-2;i+=2)
{
x+=2.0*delta;
sumpar2+=integrand(x);
}
totalsumpar=(delta/3.0)*((4.0*sumpar1)+(2.0*sumpar2)+integrand(a)+integrand(b));
return totalsumpar;
}
```

*Edited 5 Years Ago by Xtian9*: n/a