Function primitive_float_acot_pi 

pub fn primitive_float_acot_pi<T>(x: T) -> T
where
    Float: From<T> + PartialOrd<T>,
    for<'a> T: ExactFrom<&'a Float> + RoundingFrom<&'a Float> + PrimitiveFloat,

Computes $\operatorname{acot}(x)/\pi$, the arccotangent of a primitive float measured in half-turns, returning the result as a primitive float.

This is primitive_float_acot_with_period with a period of 2: see primitive_float_acot_with_period for the error bounds, the special cases, and the complexity, with $u = 2$. Either infinity gives a zero of its sign, $\pm0.0$ give $\pm1/2$, and $\pm1$ give $\pm1/4$; NaN gives NaN. Overflow is not possible, since $|\operatorname{acot}(x)/\pi| \leq 1/2$.

Worst-case complexity

$T(m) = O(m \log m \log\log m)$

$M(m) = O(m \log m)$

where $T$ is time, $M$ is additional memory, and $m$ is x.significant_bits().

Examples

use malachite_base::num::float::NiceFloat;
use malachite_float::float::arithmetic::acot::primitive_float_acot_pi;

assert!(primitive_float_acot_pi(f32::NAN).is_nan());
// a zero is half a half-turn, and the arccotangent is defined inside (-1, 1) too
assert_eq!(NiceFloat(primitive_float_acot_pi(0.0f32)), NiceFloat(0.5));
assert_eq!(
    NiceFloat(primitive_float_acot_pi(0.5f32)),
    NiceFloat(0.3524164)
);
assert_eq!(
    NiceFloat(primitive_float_acot_pi(f32::INFINITY)),
    NiceFloat(0.0)
);
assert_eq!(NiceFloat(primitive_float_acot_pi(1.0f32)), NiceFloat(0.25));
assert_eq!(
    NiceFloat(primitive_float_acot_pi(-1.0f32)),
    NiceFloat(-0.25)
);
assert_eq!(
    NiceFloat(primitive_float_acot_pi(2.5f32)),
    NiceFloat(0.12111894)
);
assert_eq!(
    NiceFloat(primitive_float_acot_pi(2.5f64)),
    NiceFloat(0.1211189415908434)
);