table of contents
std::sqrt(std::valarray)(3) | C++ Standard Libary | std::sqrt(std::valarray)(3) |
NAME¶
std::sqrt(std::valarray) - std::sqrt(std::valarray)
Synopsis¶
Defined in header <valarray>
template< class T >
valarray<T> sqrt( const valarray<T>& va );
For each element in va computes the square root of the value of the
element.
Parameters¶
va - value array to apply the operation to
Return value¶
Value array containing square roots of the values in va.
Notes¶
Unqualified function (sqrt) is used to perform the computation.
If such function is
not available, std::sqrt is used due to argument-dependent lookup.
The function can be implemented with the return type different from
std::valarray.
In this case, the replacement type has the following properties:
* All const member functions of std::valarray are provided.
* std::valarray, std::slice_array, std::gslice_array, std::mask_array and
std::indirect_array can be constructed from the replacement type.
* For every function taking a const std::valarray<T>&
except begin() and end()
(since C++11), identical functions taking the replacement types shall
be added;
* For every function taking two const std::valarray<T>& arguments,
identical
functions taking every combination of const std::valarray<T>& and
replacement
types shall be added.
* The return type does not add more than two levels of template nesting over
the
most deeply-nested argument type.
Possible implementation¶
template<class T>
valarray<T> sqrt(const valarray<T>& va)
{
valarray<T> other = va;
for (T& i : other)
i = sqrt(i);
return other; // proxy object may be returned
}
Example¶
Finds all three roots (two of which can be complex conjugates) of
several Cubic
equations at once.
// Run this code
#include <cassert>
#include <complex>
#include <cstddef>
#include <iostream>
#include <numbers>
#include <valarray>
using CD = std::complex<double>;
using VA = std::valarray<CD>;
// return all n complex roots out of a given complex number x
VA root(CD x, unsigned n)
{
const double mag = std::pow(std::abs(x), 1.0 / n);
const double step = 2.0 * std::numbers::pi / n;
double phase = std::arg(x) / n;
VA v(n);
for (std::size_t i{}; i != n; ++i, phase += step)
v[i] = std::polar(mag, phase);
return v;
}
// return n complex roots of each element in v; in the output valarray first
// goes the sequence of all n roots of v[0], then all n roots of v[1], etc.
VA root(VA v, unsigned n)
{
VA o(v.size() * n);
VA t(n);
for (std::size_t i = 0; i != v.size(); ++i)
{
t = root(v[i], n);
for (unsigned j = 0; j != n; ++j)
o[n * i + j] = t[j];
}
return o;
}
// floating-point numbers comparator that tolerates given rounding error
inline bool is_equ(CD x, CD y, double tolerance = 0.000'000'001)
{
return std::abs(std::abs(x) - std::abs(y)) < tolerance;
}
int main()
{
// input coefficients for polynomial x³ + p·x + q
const VA p{1, 2, 3, 4, 5, 6, 7, 8};
const VA q{1, 2, 3, 4, 5, 6, 7, 8};
// the solver
const VA d = std::sqrt(std::pow(q / 2, 2) + std::pow(p / 3, 3));
const VA u = root(-q / 2 + d, 3);
const VA n = root(-q / 2 - d, 3);
// allocate memory for roots: 3 * number of input cubic polynomials
VA x[3];
for (std::size_t t = 0; t != 3; ++t)
x[t].resize(p.size());
auto is_proper_root = [](CD a, CD b, CD p) { return is_equ(a * b + p / 3.0,
0.0); };
// sieve out 6 out of 9 generated roots, leaving only 3 proper roots (per
polynomial)
for (std::size_t i = 0; i != p.size(); ++i)
for (std::size_t j = 0, r = 0; j != 3; ++j)
for (std::size_t k = 0; k != 3; ++k)
if (is_proper_root(u[3 * i + j], n[3 * i + k], p[i]))
x[r++][i] = u[3 * i + j] + n[3 * i + k];
std::cout << "Depressed cubic equation: Root 1: \t\t Root 2: \t\t
Root 3:\n";
for (std::size_t i = 0; i != p.size(); ++i)
{
std::cout << "x³ + " << p[i] <<
"·x + " << q[i] << " = 0 "
<< std::fixed << x[0][i] << " " << x[1][i]
<< " " << x[2][i]
<< std::defaultfloat << '\n';
assert(is_equ(std::pow(x[0][i], 3) + x[0][i] * p[i] + q[i], 0.0));
assert(is_equ(std::pow(x[1][i], 3) + x[1][i] * p[i] + q[i], 0.0));
assert(is_equ(std::pow(x[2][i], 3) + x[2][i] * p[i] + q[i], 0.0));
}
}
Output:¶
Depressed cubic equation: Root 1: Root 2: Root 3:
x³ + (1,0)·x + (1,0) = 0 (-0.682328,0.000000)
(0.341164,1.161541) (0.341164,-1.161541)
x³ + (2,0)·x + (2,0) = 0 (-0.770917,0.000000)
(0.385458,1.563885) (0.385458,-1.563885)
x³ + (3,0)·x + (3,0) = 0 (-0.817732,0.000000)
(0.408866,1.871233) (0.408866,-1.871233)
x³ + (4,0)·x + (4,0) = 0 (-0.847708,0.000000)
(0.423854,2.130483) (0.423854,-2.130483)
x³ + (5,0)·x + (5,0) = 0 (-0.868830,0.000000)
(0.434415,2.359269) (0.434415,-2.359269)
x³ + (6,0)·x + (6,0) = 0 (-0.884622,0.000000)
(0.442311,2.566499) (0.442311,-2.566499)
x³ + (7,0)·x + (7,0) = 0 (-0.896922,0.000000)
(0.448461,2.757418) (0.448461,-2.757418)
x³ + (8,0)·x + (8,0) = 0 (-0.906795,0.000000)
(0.453398,2.935423) (0.453398,-2.935423)
See also¶
applies the function std::pow to two valarrays or a valarray and
pow(std::valarray) a value
(function template)
sqrt computes square root (\(\small{\sqrt{x}}\)
sqrtf √
sqrtl x)
(C++11) (function)
(C++11)
sqrt(std::complex) complex square root in the range of the right half-plane
(function template)
Hidden category:¶
* Pages with unreviewed LWG DR marker
2024.06.10 | http://cppreference.com |