Composite Simpson's Rule in C++ - c++

Question

I've been trying to write a function to approximate an the value of an integral using the Composite Simpson's Rule.
template <typename func_type>
double simp_rule(double a, double b, int n, func_type f){
int i = 1; double area = 0;
double n2 = n;
double h = (b-a)/(n2-1), x=a;
while(i <= n){
area = area + f(x)*pow(2,i%2 + 1)*h/3;
x+=h;
i++;
}
area -= (f(a) * h/3);
area -= (f(b) * h/3);
return area;
}
What I do is multiply each value of the function by either 2 or 4 (and h/3) with pow(2,i%2 + 1) and subtract off the edges as these should only have a weight of 1.
At first, I thought it worked just fine, however, when I compared it to my Trapezoidal Method function it was way more inaccurate which shouldn't be the case.
This is a simpler version of a code I previously wrote which had the same problem, I thought that if I cleaned it up a little the problem would go away, but alas. From another post, I get the idea that there's something going on with the types and the operations I'm doing on them which results in loss of precision, but I just don't see it.
Edit:
For completeness, I was running it for e^x from 1 to zero
\\function to be approximated
double f(double x){ double a = exp(x); return a; }
int main() {
int n = 11; //this method works best for odd values of n
double e = exp(1);
double exact = e-1; //value of integral of e^x from 0 to 1
cout << simp_rule(0,1,n,f) - exact;

Answer 1

The Simpson's Rule uses this approximation to estimate a definite integral:
Where
and
So that there are n + 1 equally spaced sample points xi.
In the posted code, the parameter n passed to the function appears to be the number of points where the function is sampled (while in the previous formula n is the number of intervals, that's not a problem).
The (constant) distance between the points is calculated correctly
double h = (b - a) / (n - 1);
The while loop used to sum the weighted contributes of all the points iterates from x = a up to a point with an ascissa close to b, but probably not exactly b, due to rounding errors. This implies that the last calculated value of f, f(x_n), may be slightly different from the expected f(b).
This is nothing, though, compared to the error caused by the fact that those end points are summed inside the loop with the starting weight of 4 and then subtracted after the loop with weight 1, while all the inner points have their weight switched. As a matter of fact, this is what the code calculates:
Also, using
pow(2, i%2 + 1)
To generate the sequence 4, 2, 4, 2, ..., 4 is a waste, in terms of efficency, and may add (depending on the implementation) other unnecessary rounding errors.
The following algorithm shows how to obtain the same (fixed) result, without a call to that library function.
template <typename func_type>
double simpson_rule(double a, double b,
int n, // Number of intervals
func_type f)
{
double h = (b - a) / n;
// Internal sample points, there should be n - 1 of them
double sum_odds = 0.0;
for (int i = 1; i < n; i += 2)
{
sum_odds += f(a + i * h);
}
double sum_evens = 0.0;
for (int i = 2; i < n; i += 2)
{
sum_evens += f(a + i * h);
}
return (f(a) + f(b) + 2 * sum_evens + 4 * sum_odds) * h / 3;
}
Note that this function requires the number of intervals (e.g. use 10 instead of 11 to obtain the same results of OP's function) to be passed, not the number of points.
Testable here.

Answer 2

The above excellent and accepted solution could benefit from liberal use of std::fma() and templatize on the floating point type.
https://en.cppreference.com/w/cpp/numeric/math/fma
#include <cmath>
template <typename fptype, typename func_type>
double simpson_rule(fptype a, fptype b,
int n, // Number of intervals
func_type f)
{
fptype h = (b - a) / n;
// Internal sample points, there should be n - 1 of them
fptype sum_odds = 0.0;
for (int i = 1; i < n; i += 2)
{
sum_odds += f(std::fma(i,h,a));
}
fptype sum_evens = 0.0;
for (int i = 2; i < n; i += 2)
{
sum_evens += f(std::fma(i,h,a);
}
return (std::fma(2,sum_evens,f(a)) +
std::fma(4,sum_odds,f(b))) * h / 3;
}