\[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)
\]
↓
\[\begin{array}{l}
t_0 := 0.25 \cdot {\pi}^{2}\\
t_1 := \cos^{-1} \left(\sqrt{\mathsf{fma}\left(-0.5, x, 0.5\right)}\right)\\
t_2 := {t_1}^{2} \cdot 4\\
\frac{\left(t_0 \cdot t_0 - t_2 \cdot t_2\right) \cdot \frac{1}{\mathsf{fma}\left(\pi, -0.5, t_1 \cdot -2\right)}}{t_0 + t_2}
\end{array}
\]
(FPCore (x)
:precision binary64
(- (/ PI 2.0) (* 2.0 (asin (sqrt (/ (- 1.0 x) 2.0))))))
↓
(FPCore (x)
:precision binary64
(let* ((t_0 (* 0.25 (pow PI 2.0)))
(t_1 (acos (sqrt (fma -0.5 x 0.5))))
(t_2 (* (pow t_1 2.0) 4.0)))
(/
(* (- (* t_0 t_0) (* t_2 t_2)) (/ 1.0 (fma PI -0.5 (* t_1 -2.0))))
(+ t_0 t_2))))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 = 0.25 * pow(((double) M_PI), 2.0);
double t_1 = acos(sqrt(fma(-0.5, x, 0.5)));
double t_2 = pow(t_1, 2.0) * 4.0;
return (((t_0 * t_0) - (t_2 * t_2)) * (1.0 / fma(((double) M_PI), -0.5, (t_1 * -2.0)))) / (t_0 + t_2);
}
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 = Float64(0.25 * (pi ^ 2.0))
t_1 = acos(sqrt(fma(-0.5, x, 0.5)))
t_2 = Float64((t_1 ^ 2.0) * 4.0)
return Float64(Float64(Float64(Float64(t_0 * t_0) - Float64(t_2 * t_2)) * Float64(1.0 / fma(pi, -0.5, Float64(t_1 * -2.0)))) / Float64(t_0 + t_2))
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[(0.25 * N[Power[Pi, 2.0], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[ArcCos[N[Sqrt[N[(-0.5 * x + 0.5), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]}, Block[{t$95$2 = N[(N[Power[t$95$1, 2.0], $MachinePrecision] * 4.0), $MachinePrecision]}, N[(N[(N[(N[(t$95$0 * t$95$0), $MachinePrecision] - N[(t$95$2 * t$95$2), $MachinePrecision]), $MachinePrecision] * N[(1.0 / N[(Pi * -0.5 + N[(t$95$1 * -2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(t$95$0 + t$95$2), $MachinePrecision]), $MachinePrecision]]]]
\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)
↓
\begin{array}{l}
t_0 := 0.25 \cdot {\pi}^{2}\\
t_1 := \cos^{-1} \left(\sqrt{\mathsf{fma}\left(-0.5, x, 0.5\right)}\right)\\
t_2 := {t_1}^{2} \cdot 4\\
\frac{\left(t_0 \cdot t_0 - t_2 \cdot t_2\right) \cdot \frac{1}{\mathsf{fma}\left(\pi, -0.5, t_1 \cdot -2\right)}}{t_0 + t_2}
\end{array}
Alternatives
| Alternative 1 |
|---|
| Accuracy | 8.4% |
|---|
| Cost | 149696 |
|---|
\[\begin{array}{l}
t_0 := 0.25 \cdot {\pi}^{2}\\
t_1 := \cos^{-1} \left(\sqrt{\mathsf{fma}\left(-0.5, x, 0.5\right)}\right)\\
t_2 := {t_1}^{2} \cdot 4\\
\frac{t_0 \cdot t_0 - t_2 \cdot t_2}{\mathsf{fma}\left(\pi, -0.5, t_1 \cdot -2\right) \cdot \left(t_0 + t_2\right)}
\end{array}
\]
| Alternative 2 |
|---|
| Accuracy | 8.4% |
|---|
| Cost | 71616 |
|---|
\[\begin{array}{l}
t_0 := \cos^{-1} \left(\sqrt{\mathsf{fma}\left(-0.5, x, 0.5\right)}\right)\\
\frac{1}{\mathsf{fma}\left(\pi, -0.5, t_0 \cdot -2\right)} \cdot \left(0.25 \cdot {\pi}^{2} + {t_0}^{2} \cdot -4\right)
\end{array}
\]
| Alternative 3 |
|---|
| Accuracy | 8.4% |
|---|
| Cost | 71488 |
|---|
\[\begin{array}{l}
t_0 := \cos^{-1} \left(\sqrt{\mathsf{fma}\left(-0.5, x, 0.5\right)}\right)\\
\frac{0.25 \cdot {\pi}^{2} + {t_0}^{2} \cdot -4}{\mathsf{fma}\left(\pi, -0.5, t_0 \cdot -2\right)}
\end{array}
\]
| Alternative 4 |
|---|
| Accuracy | 8.4% |
|---|
| Cost | 45248 |
|---|
\[{\left(\sqrt[3]{\mathsf{fma}\left(\pi, -0.5, 2 \cdot \cos^{-1} \left(\sqrt{\mathsf{fma}\left(-0.5, x, 0.5\right)}\right)\right)}\right)}^{3}
\]
| Alternative 5 |
|---|
| Accuracy | 8.4% |
|---|
| Cost | 19840 |
|---|
\[\pi \cdot -0.5 - -2 \cdot \cos^{-1} \left(\sqrt{0.5 + -0.5 \cdot x}\right)
\]
| Alternative 6 |
|---|
| Accuracy | 5.4% |
|---|
| Cost | 19584 |
|---|
\[\pi \cdot -0.5 + 2 \cdot \cos^{-1} \left(\sqrt{0.5}\right)
\]