使用boost::math中的高斯-克朗罗德求积对复杂函数进行积分
使用boost::math中的高斯-克朗罗德求积对复杂函数进行积分
提问于 2019-05-15 03:09:51
我正在尝试使用Boost提供的高斯-克朗罗德求积在C++中将一个具有实数参数和复数返回值的函数(如f(x) =1+i)在域0，5上进行数值积分。
我已经使用boost来解决实际的积分器问题，所以库安装正确。documentation说“高斯-克朗罗德求积支持在实数行上定义的被积函数，并返回复数值。”
#include <iostream>
#include <cmath>
#include <complex>
#include <boost/math/quadrature/gauss_kronrod.hpp>
using namespace boost::math::quadrature;
using complex = std::complex<double>;
complex f(double t){
return complex{1, 1};
};
int main() {
complex error;
complex a{0};
complex b{5};
unsigned int max_depth = 0;
complex tolerance = 0;
complex Q = gauss_kronrod<complex, 61>::integrate(f, a, b, max_depth, tolerance, &error);
std::cout << Q << ", " << error << "\n";
return 0;
}我的预期结果是5+ 5i。我得到了编译器错误
/usr/include/boost/math/quadrature/gauss_kronrod.hpp:1871:17: error: no match for 'operator<=' (operand types are 'std::complex<double>' and 'std::complex<double>')这意味着一些运算符没有为std::complex定义。我是不是误读了文档，无法进行复杂的集成？
编辑:将代码更改为user14717提供的版本会给我一个新的编译器错误：
/usr/include/boost/math/quadrature/gauss_kronrod.hpp: In instantiation of 'static boost::math::quadrature::gauss_kronrod<Real, N, Policy>::value_type boost::math::quadrature::gauss_kronrod<Real, N, Policy>::integrate(F, Real, Real, unsigned int, Real, Real*, Real*) [with F = std::complex<double> (*)(double); Real = double; unsigned int N = 61; Policy = boost::math::policies::policy<boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy, boost::math::policies::default_policy>; boost::math::quadrature::gauss_kronrod<Real, N, Policy>::value_type = double]':
/home/olaf/testing numerical integration/boost/main.cpp:21:91: required from here
/usr/include/boost/math/quadrature/gauss_kronrod.hpp:1877:47: error: cannot convert 'std::complex<double>' to 'double' in return
return f(t*inv)*(1 + t_sq)*inv*inv;
^~~
/usr/include/boost/math/quadrature/gauss_kronrod.hpp:1890:32: error: cannot convert ?std::complex<double>? to 'double' in return
return f(arg)*z*z;
^
/usr/include/boost/math/quadrature/gauss_kronrod.hpp:1907:40: error: cannot convert 'std::complex<double>' to 'double' in return
return f(b - arg) * z * z;
^回答 1
Stack Overflow用户
回答已采纳
发布于 2019-05-15 03:21:40
修复方法如下：
#include <iostream>
#include <cmath>
#include <complex>
#include <boost/math/quadrature/gauss_kronrod.hpp>
using namespace boost::math::quadrature;
using complex = std::complex<double>;
complex f(double t){
return complex{1, 1};
};
int main() {
double error;
double a{0};
double b{5};
unsigned int max_depth = 0;
double tolerance = 0;
complex Q = gauss_kronrod<double, 61>::integrate(f, a, b, max_depth, tolerance, &error);
std::cout << Q << ", " << error << "\n";
return 0;
}你的错误是你在复杂类型上模板化了。模板参数必须是真实类型。
我注意到文档没有在这里提供示例，所以您实际上没有办法知道这一点。我会为此提交一个补丁。。。
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原文链接：
https://stackoverflow.com/questions/56137060
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