Math FPCore C Julia Wolfram TeX \[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)
\]
↓
\[\begin{array}{l}
t_0 := \sqrt{0.5 - 0.5 \cdot x}\\
t_1 := \sin^{-1} t_0\\
\frac{1}{\frac{\mathsf{fma}\left(2 \cdot t_1, \mathsf{fma}\left(2, t_1, 0.5 \cdot \pi\right), {\pi}^{2} \cdot 0.25\right)}{{\pi}^{3} \cdot 0.125 - 8 \cdot {\left(0.5 \cdot \pi - \cos^{-1} t_0\right)}^{3}}}
\end{array}
\]
(FPCore (x)
:precision binary64
(- (/ PI 2.0) (* 2.0 (asin (sqrt (/ (- 1.0 x) 2.0)))))) ↓
(FPCore (x)
:precision binary64
(let* ((t_0 (sqrt (- 0.5 (* 0.5 x)))) (t_1 (asin t_0)))
(/
1.0
(/
(fma (* 2.0 t_1) (fma 2.0 t_1 (* 0.5 PI)) (* (pow PI 2.0) 0.25))
(- (* (pow PI 3.0) 0.125) (* 8.0 (pow (- (* 0.5 PI) (acos t_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 = sqrt((0.5 - (0.5 * x)));
double t_1 = asin(t_0);
return 1.0 / (fma((2.0 * t_1), fma(2.0, t_1, (0.5 * ((double) M_PI))), (pow(((double) M_PI), 2.0) * 0.25)) / ((pow(((double) M_PI), 3.0) * 0.125) - (8.0 * pow(((0.5 * ((double) M_PI)) - acos(t_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 = sqrt(Float64(0.5 - Float64(0.5 * x)))
t_1 = asin(t_0)
return Float64(1.0 / Float64(fma(Float64(2.0 * t_1), fma(2.0, t_1, Float64(0.5 * pi)), Float64((pi ^ 2.0) * 0.25)) / Float64(Float64((pi ^ 3.0) * 0.125) - Float64(8.0 * (Float64(Float64(0.5 * pi) - acos(t_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[Sqrt[N[(0.5 - N[(0.5 * x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[ArcSin[t$95$0], $MachinePrecision]}, N[(1.0 / N[(N[(N[(2.0 * t$95$1), $MachinePrecision] * N[(2.0 * t$95$1 + N[(0.5 * Pi), $MachinePrecision]), $MachinePrecision] + N[(N[Power[Pi, 2.0], $MachinePrecision] * 0.25), $MachinePrecision]), $MachinePrecision] / N[(N[(N[Power[Pi, 3.0], $MachinePrecision] * 0.125), $MachinePrecision] - N[(8.0 * N[Power[N[(N[(0.5 * Pi), $MachinePrecision] - N[ArcCos[t$95$0], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]
\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)
↓
\begin{array}{l}
t_0 := \sqrt{0.5 - 0.5 \cdot x}\\
t_1 := \sin^{-1} t_0\\
\frac{1}{\frac{\mathsf{fma}\left(2 \cdot t_1, \mathsf{fma}\left(2, t_1, 0.5 \cdot \pi\right), {\pi}^{2} \cdot 0.25\right)}{{\pi}^{3} \cdot 0.125 - 8 \cdot {\left(0.5 \cdot \pi - \cos^{-1} t_0\right)}^{3}}}
\end{array}
Alternatives Alternative 1 Accuracy 8.4% Cost 98304
\[\begin{array}{l}
t_0 := \sqrt{0.5 - 0.5 \cdot x}\\
t_1 := \sin^{-1} t_0\\
\frac{1}{\frac{\mathsf{fma}\left(2 \cdot t_1, \mathsf{fma}\left(2, t_1, 0.5 \cdot \pi\right), {\pi}^{2} \cdot 0.25\right)}{{\pi}^{3} \cdot 0.125 - 8 \cdot {\left(0.5 \cdot \pi - \cos^{-1} t_0\right)}^{3}}}
\end{array}
\]
Alternative 2 Accuracy 8.4% Cost 45568
\[{\left(\sqrt[3]{\mathsf{fma}\left(0.5 \cdot \pi - \cos^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right), -2, 0.5 \cdot \pi\right)}\right)}^{3}
\]
Alternative 3 Accuracy 8.4% Cost 26432
\[\frac{\pi}{2} + 2 \cdot \left(\cos^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right) - 0.5 \cdot \pi\right)
\]
Alternative 4 Accuracy 5.4% Cost 26176
\[\frac{\pi}{2} + 2 \cdot \left(\cos^{-1} \left(\sqrt{0.5}\right) - 0.5 \cdot \pi\right)
\]
Alternative 5 Accuracy 7.0% Cost 19968
\[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\frac{1}{\sqrt{\frac{2}{1 - x}}}\right)
\]
Alternative 6 Accuracy 7.0% Cost 19840
\[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)
\]
Alternative 7 Accuracy 4.1% Cost 19712
\[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\frac{1}{\sqrt{2}}\right)
\]
Alternative 8 Accuracy 4.1% Cost 19584
\[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{0.5}\right)
\]