\[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)
\]
↓
\[\begin{array}{l}
t_0 := \sin^{-1} \left(\sqrt{0.5 + x \cdot -0.5}\right)\\
\frac{1}{\mathsf{fma}\left(\pi, 0.5, 2 \cdot t_0\right)} \cdot \left({\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{{\left({t_0}^{2}\right)}^{3}}\right)
\end{array}
\]
(FPCore (x)
:precision binary64
(- (/ PI 2.0) (* 2.0 (asin (sqrt (/ (- 1.0 x) 2.0))))))
↓
(FPCore (x)
:precision binary64
(let* ((t_0 (asin (sqrt (+ 0.5 (* x -0.5))))))
(*
(/ 1.0 (fma PI 0.5 (* 2.0 t_0)))
(- (* (pow PI 2.0) 0.25) (* 4.0 (cbrt (pow (pow t_0 2.0) 3.0)))))))double code(double x) {
return (((double) M_PI) / 2.0) - (2.0 * asin(sqrt(((1.0 - x) / 2.0))));
}
↓
double code(double x) {
double t_0 = asin(sqrt((0.5 + (x * -0.5))));
return (1.0 / fma(((double) M_PI), 0.5, (2.0 * t_0))) * ((pow(((double) M_PI), 2.0) * 0.25) - (4.0 * cbrt(pow(pow(t_0, 2.0), 3.0))));
}
function code(x)
return Float64(Float64(pi / 2.0) - Float64(2.0 * asin(sqrt(Float64(Float64(1.0 - x) / 2.0)))))
end
↓
function code(x)
t_0 = asin(sqrt(Float64(0.5 + Float64(x * -0.5))))
return Float64(Float64(1.0 / fma(pi, 0.5, Float64(2.0 * t_0))) * Float64(Float64((pi ^ 2.0) * 0.25) - Float64(4.0 * cbrt(((t_0 ^ 2.0) ^ 3.0)))))
end
code[x_] := N[(N[(Pi / 2.0), $MachinePrecision] - N[(2.0 * N[ArcSin[N[Sqrt[N[(N[(1.0 - x), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
↓
code[x_] := Block[{t$95$0 = N[ArcSin[N[Sqrt[N[(0.5 + N[(x * -0.5), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]}, N[(N[(1.0 / N[(Pi * 0.5 + N[(2.0 * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(N[Power[Pi, 2.0], $MachinePrecision] * 0.25), $MachinePrecision] - N[(4.0 * N[Power[N[Power[N[Power[t$95$0, 2.0], $MachinePrecision], 3.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)
↓
\begin{array}{l}
t_0 := \sin^{-1} \left(\sqrt{0.5 + x \cdot -0.5}\right)\\
\frac{1}{\mathsf{fma}\left(\pi, 0.5, 2 \cdot t_0\right)} \cdot \left({\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{{\left({t_0}^{2}\right)}^{3}}\right)
\end{array}