How to use boost bisection?

Question

Yesterday I had problems with another boost functions but luckily you guys helped me to solve them. Today I would need to know how to use bisection function properly.

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Answer 1

This is an example use of bisect. Consider solving the equation x^2 - 3x + 1 = 0:

struct TerminationCondition  {
  bool operator() (double min, double max)  {
    return abs(min - max) <= 0.000001;
  }
};

struct FunctionToApproximate  {
  double operator() (double x)  {
    return x*x - 3*x + 1;  // Replace with your function
  }
};

// ...
using boost::math::tools::bisect;
double from = 0;  // The solution must lie in the interval [from, to], additionally f(from) <= 0 && f(to) >= 0
double to = 1;
std::pair<double, double> result = bisect(FunctionToApproximate(), from, to, TerminationCondition());
double root = (result.first + result.second) / 2;  // = 0.381966...

EDIT: Alternatively, this is how you can use it with custom functions:

double myF(double x)  {
  return x*x*x;
}

double otherF(double x)  {
  return log(abs(x));
}

// ...
std::pair<double, double> result1 = bisect(&myF, from, to, TerminationCondition());
std::pair<double, double> result2 = bisect(&otherF, 0.1, 1.1, TerminationCondition());

Answer 2

Note that bisect also supports lambdas:

using boost::math::tools::bisect;
auto root = bisect
(
    [](double x){return x;},
    -1.0, 1.0,
    [](double l, double r){return abs(l-r) < 1e-8;}
);