double legendre ( unsigned int n, double x );
/* floating-point-type */ legendre( unsigned int n,
/* floating-point-type */ x );
float legendref( unsigned int n, float x );
(2) (since C++17)long double legendrel( unsigned int n, long double x );
(3) (since C++17)template< class Integer >
double legendre ( unsigned int n, Integer x );
Computes the unassociated
Legendre polynomialsof the degree
nand argument
x.
The library provides overloads ofstd::legendre
for all cv-unqualified floating-point types as the type of the parameter x.(since C++23)
A) Additional overloads are provided for all integer types, which are treated as double.
[edit] Parameters n - the degree of the polynomial x - the argument, a floating-point or integer value [edit] Return valueIf no errors occur, value of the order-
nunassociated Legendre polynomial of
x, that is
\(\mathsf{P}_n(x) = \frac{1}{2^n n!} \frac{\mathsf{d}^n}{\mathsf{d}x^n} (x^2-1)^n \)(x2, is returned.
[edit] Error handlingErrors may be reported as specified in math_errhandling.
Implementations that do not support C++17, but support ISO 29124:2010, provide this function if __STDCPP_MATH_SPEC_FUNCS__
is defined by the implementation to a value at least 201003L and if the user defines __STDCPP_WANT_MATH_SPEC_FUNCS__
before including any standard library headers.
Implementations that do not support ISO 29124:2010 but support TR 19768:2007 (TR1), provide this function in the header tr1/cmath
and namespace std::tr1
.
An implementation of this function is also available in boost.math.
The first few Legendre polynomials are:
Function Polynomial legendre(0, x) 1 legendre(1, x) x legendre(2, x) 1 2 (3x2The additional overloads are not required to be provided exactly as (A). They only need to be sufficient to ensure that for their argument num of integer type, std::legendre(int_num, num) has the same effect as std::legendre(int_num, static_cast<double>(num)).
[edit] Example#include <cmath> #include <iostream> double P3(double x) { return 0.5 * (5 * std::pow(x, 3) - 3 * x); } double P4(double x) { return 0.125 * (35 * std::pow(x, 4) - 30 * x * x + 3); } int main() { // spot-checks std::cout << std::legendre(3, 0.25) << '=' << P3(0.25) << '\n' << std::legendre(4, 0.25) << '=' << P4(0.25) << '\n'; }
Output:
-0.335938=-0.335938 0.157715=0.157715[edit] See also [edit] External links
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