Reputation: 81
If I have a function similar to this:
c=atan2( a, 0 )
I am trying to figure out what the a
value should be limited to in order to ensure atan2() does not result in any domain errors. So the question is how close to zero can a get before a domain error is raised. Can this atan2 function compute a number as small as 0.0000001? At what point does it read as zero?
Upvotes: 2
Views: 564
Reputation: 25489
The best way to find out is to try it yourself!
#include <iostream>
#include <stdio.h>
#include <limits>
#include <math.h>
using namespace std;
int main()
{
int powers[12] = {10, 11, 12, 13, 14, 15, 20, 22, 24, 26, 28, 30};
for (auto i : powers)
{
try
{
double a = pow(10, -i);
printf("a = 10^-%d; atan2(a, 0) = %.10e\n", i, atan2(a, 0));
// cout << "a = 10^" << -i << "; atan2(a, 0) = " << atan2(a, 0) << endl;
}
catch (int e)
{
cout << "Failed to compute atan2(10^" << -i << ", 0)" << endl;
}
}
double minvalue = numeric_limits<double>::min();
try
{
printf("a = %.10e; atan2(a, 0) = %.10e\n", minvalue, atan2(minvalue, 0));
// cout << "a = " << minvalue << "; atan2(a, 0) = " << atan2(minvalue, 0) << endl;
}
catch (int e)
{
cout << "Failed to compute atan2(" << minvalue << ", 0)" << endl;
}
return 0;
}
Output:
a = 10^-10; atan2(a, 0) = 1.5707963268e+00
a = 10^-11; atan2(a, 0) = 1.5707963268e+00
a = 10^-12; atan2(a, 0) = 1.5707963268e+00
a = 10^-13; atan2(a, 0) = 1.5707963268e+00
a = 10^-14; atan2(a, 0) = 1.5707963268e+00
a = 10^-15; atan2(a, 0) = 1.5707963268e+00
a = 10^-20; atan2(a, 0) = 1.5707963268e+00
a = 10^-22; atan2(a, 0) = 1.5707963268e+00
a = 10^-24; atan2(a, 0) = 1.5707963268e+00
a = 10^-26; atan2(a, 0) = 1.5707963268e+00
a = 10^-28; atan2(a, 0) = 1.5707963268e+00
a = 10^-30; atan2(a, 0) = 1.5707963268e+00
a = 2.2250738585e-308; atan2(a, 0) = 1.5707963268e+00
Works perfectly fine for the smallest possible positive double.
It even works for 5E-324
, which is the smallest positive denormal double.
double minvalue = std::numeric_limits<double>::denorm_min();
printf("a = %.10e; atan2(a, 0) = %.10e\n", minvalue, atan2(minvalue, 0));
Output:
a = 4.9406564584e-324; atan2(a, 0) = 1.5707963268e+00
Upvotes: 3