我正在寻找一种健壮的算法（或描述算法的论文），它可以使用封闭形式的解决方案找到多项式的根（理想情况下可以找到第 4 个 debree，但任何事情都可以）。我只对真正的根源感兴趣。
我第一次解决二次方程涉及到这个（我也有类似风格的三次/四次代码，但现在让我们关注二次方程）：
/**
* @brief a simple quadratic equation solver
*
* With double-precision floating-point, this reaches 1e-12 worst-case and 1e-15 average
* precision of the roots (the value of the function in the roots). The roots can be however
* quite far from the true roots, up to 1e-10 worst-case and 1e-18 average absolute difference
* for cases when two roots exist. If only a single root exists, the worst-case precision is
* 1e-13 and average-case precision is 1e-18.
*
* With single-precision floating-point, this reaches 1e-3 worst-case and 1e-7 average
* precision of the roots (the value of the function in the roots). The roots can be however
* quite far from the true roots, up to 1e-1 worst-case and 1e-10 average absolute difference
* for cases when two roots exist. If only a single root exists, the worst-case precision is
* 1e+2 (!) and average-case precision is 1e-2. Do not use single-precision floating point,
* except if pressed by time.
*
* All the precision measurements are scaled by the maximum absolute coefficient value.
*
* @tparam T is data type of the arguments (default double)
* @tparam b_sort_roots is root sorting flag (if set, the roots are
* given in ascending (not absolute) value; default true)
* @tparam n_2nd_order_coeff_log10_thresh is base 10 logarithm of threshold
* on the first coefficient (if below threshold, the equation is a linear one; default -6)
* @tparam n_zero_discriminant_log10_thresh is base 10 logarithm of threshold
* on the discriminant (if below negative threshold, the equation does not
* have a real root, if below threshold, the equation has just a single solution; default -6)
*/
template <class T = double, const bool b_sort_roots = true,
const int n_2nd_order_coeff_log10_thresh = -6,
const int n_zero_discriminant_log10_thresh = -6>
class CQuadraticEq {
protected:
T a; /**< @brief the 2nd order coefficient */
T b; /**< @brief the 1st order coefficient */
T c; /**< @brief 0th order coefficient */
T p_real_root[2]; /**< @brief list of the roots (real parts) */
//T p_im_root[2]; // imaginary part of the roots
size_t n_real_root_num; /**< @brief number of real roots */
public:
/**
* @brief default constructor; solves for roots of \f$ax^2 + bx + c = 0\f$
*
* This finds roots of the given equation. It tends to find two identical roots instead of one, rather
* than missing one of two different roots - the number of roots found is therefore orientational,
* as the roots might have the same value.
*
* @param[in] _a is the 2nd order coefficient
* @param[in] _b is the 1st order coefficient
* @param[in] _c is 0th order coefficient
*/
CQuadraticEq(T _a, T _b, T _c) // ax2 + bx + c = 0
:a(_a), b(_b), c(_c)
{
T _aa = fabs(_a);
if(_aa < f_Power_Static(10, n_2nd_order_coeff_log10_thresh)) { // otherwise division by a yields large numbers, this is then more precise
p_real_root[0] = -_c / _b;
//p_im_root[0] = 0;
n_real_root_num = 1;
return;
}
// a simple linear equation
if(_aa < 1) { // do not divide always, that makes it worse
_b /= _a;
_c /= _a;
_a = 1;
// could copy the code here and optimize away division by _a (optimizing compiler might do it for us)
}
// improve numerical stability if the coeffs are very small
const double f_thresh = f_Power_Static(10, n_zero_discriminant_log10_thresh);
double f_disc = _b * _b - 4 * _a * _c;
if(f_disc < -f_thresh) // only really negative
n_real_root_num = 0; // only two complex roots
else if(/*fabs(f_disc) < f_thresh*/f_disc <= f_thresh) { // otherwise gives problems for double root situations
p_real_root[0] = T(-_b / (2 * _a));
n_real_root_num = 1;
} else {
f_disc = sqrt(f_disc);
int i = (b_sort_roots)? ((_a > 0)? 0 : 1) : 0; // produce sorted roots, if required
p_real_root[i] = T((-_b - f_disc) / (2 * _a));
p_real_root[1 - i] = T((-_b + f_disc) / (2 * _a));
//p_im_root[0] = 0;
//p_im_root[1] = 0;
n_real_root_num = 2;
}
}
/**
* @brief gets number of real roots
* @return Returns number of real roots (0 to 2).
*/
size_t n_RealRoot_Num() const
{
_ASSERTE(n_real_root_num >= 0);
return n_real_root_num;
}
/**
* @brief gets value of a real root
* @param[in] n_index is zero-based index of the root
* @return Returns value of the specified root.
*/
T f_RealRoot(size_t n_index) const
{
_ASSERTE(n_index < 2 && n_index < n_real_root_num);
return p_real_root[n_index];
}
/**
* @brief evaluates the equation for a given argument
* @param[in] f_x is value of the argument \f$x\f$
* @return Returns value of \f$ax^2 + bx + c\f$.
*/
T operator ()(T f_x) const
{
T f_x2 = f_x * f_x;
return f_x2 * a + f_x * b + c;
}
};
代码很糟糕，我讨厌所有的门槛。但是对于在区间中有根的随机方程[-100, 100]，这还不错：
root response precision 1e-100: 6315 cases
root response precision 1e-19: 2 cases
root response precision 1e-17: 2 cases
root response precision 1e-16: 6 cases
root response precision 1e-15: 6333 cases
root response precision 1e-14: 3765 cases
root response precision 1e-13: 241 cases
root response precision 1e-12: 3 cases
2-root solution precision 1e-100: 5353 cases
2-root solution precision 1e-19: 656 cases
2-root solution precision 1e-18: 4481 cases
2-root solution precision 1e-17: 2312 cases
2-root solution precision 1e-16: 455 cases
2-root solution precision 1e-15: 68 cases
2-root solution precision 1e-14: 7 cases
2-root solution precision 1e-13: 2 cases
1-root solution precision 1e-100: 3022 cases
1-root solution precision 1e-19: 38 cases
1-root solution precision 1e-18: 197 cases
1-root solution precision 1e-17: 68 cases
1-root solution precision 1e-16: 7 cases
1-root solution precision 1e-15: 1 cases
请注意，此精度与系数的大小有关，通常在 10^6 范围内（因此最终精度远非完美，但可能大部分可用）。然而，如果没有阈值，它几乎是无用的。
我尝试过使用多精度算术，这通常效果很好，但往往会拒绝许多根，因为多项式的系数不是多精度的，并且某些多项式不能精确表示（如果在 2 度中有双根多项式，它通常要么将其分成两个根（我不介意），要么说根本没有根）。如果我想恢复甚至稍微不精确的根，我的代码就会变得复杂并且充满阈值。
到目前为止，我已经尝试过使用 CCmath，但要么我不能正确使用它，要么精度真的很差。此外，它在plrt().
我曾尝试使用 GNU 科学库gsl_poly_solve_quadratic()，但这似乎是一种幼稚的方法，并且在数值上不是很稳定。
天真地使用std::complex数字也被证明是一个非常糟糕的主意，因为精度和速度都可能很差（尤其是在代码中包含大量超越函数的三次/四次方程）。
将根恢复为复数是唯一的方法吗？然后不会遗漏任何根，用户可以选择根需要的精确度（从而忽略不太精确的根中的小假想分量）。