\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
\]
↓
\[\frac{{k}^{-0.5}}{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\mathsf{fma}\left(k, 0.5, -0.5\right)\right)}}
\]
(FPCore (k n)
:precision binary64
(* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
↓
(FPCore (k n)
:precision binary64
(/ (pow k -0.5) (pow (* PI (* 2.0 n)) (fma k 0.5 -0.5))))
double code(double k, double n) {
return (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}
↓
double code(double k, double n) {
return pow(k, -0.5) / pow((((double) M_PI) * (2.0 * n)), fma(k, 0.5, -0.5));
}
function code(k, n)
return Float64(Float64(1.0 / sqrt(k)) * (Float64(Float64(2.0 * pi) * n) ^ Float64(Float64(1.0 - k) / 2.0)))
end
↓
function code(k, n)
return Float64((k ^ -0.5) / (Float64(pi * Float64(2.0 * n)) ^ fma(k, 0.5, -0.5)))
end
code[k_, n_] := N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
↓
code[k_, n_] := N[(N[Power[k, -0.5], $MachinePrecision] / N[Power[N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision], N[(k * 0.5 + -0.5), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\frac{{k}^{-0.5}}{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\mathsf{fma}\left(k, 0.5, -0.5\right)\right)}}
Alternatives
| Alternative 1 |
|---|
| Accuracy | 74.7% |
|---|
| Cost | 20036 |
|---|
\[\begin{array}{l}
\mathbf{if}\;k \leq 2 \cdot 10^{-33}:\\
\;\;\;\;\frac{\sqrt{n \cdot \left(\pi \cdot 2\right)}}{\sqrt{k}}\\
\mathbf{else}:\\
\;\;\;\;\frac{1}{\sqrt{\frac{k}{{\left(\pi \cdot \left(n + n\right)\right)}^{\left(1 - k\right)}}}}\\
\end{array}
\]
| Alternative 2 |
|---|
| Accuracy | 74.8% |
|---|
| Cost | 20032 |
|---|
\[\sqrt{\frac{1}{k}} \cdot {\left(n \cdot \left(\pi \cdot 2\right)\right)}^{\left(\frac{1 - k}{2}\right)}
\]
| Alternative 3 |
|---|
| Accuracy | 74.7% |
|---|
| Cost | 19908 |
|---|
\[\begin{array}{l}
\mathbf{if}\;k \leq 4 \cdot 10^{-28}:\\
\;\;\;\;\frac{\sqrt{n \cdot \left(\pi \cdot 2\right)}}{\sqrt{k}}\\
\mathbf{else}:\\
\;\;\;\;\sqrt{\frac{{\left(\pi \cdot \left(n + n\right)\right)}^{\left(1 - k\right)}}{k}}\\
\end{array}
\]
| Alternative 4 |
|---|
| Accuracy | 74.8% |
|---|
| Cost | 19904 |
|---|
\[\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}
\]
| Alternative 5 |
|---|
| Accuracy | 53.5% |
|---|
| Cost | 19844 |
|---|
\[\begin{array}{l}
\mathbf{if}\;k \leq 6 \cdot 10^{+170}:\\
\;\;\;\;\frac{\sqrt{n \cdot \left(\pi \cdot 2\right)}}{\sqrt{k}}\\
\mathbf{else}:\\
\;\;\;\;{\left({\left(\left(\pi \cdot 2\right) \cdot \frac{n}{k}\right)}^{3}\right)}^{0.16666666666666666}\\
\end{array}
\]
| Alternative 6 |
|---|
| Accuracy | 49.2% |
|---|
| Cost | 19584 |
|---|
\[\frac{\sqrt{n \cdot \left(\pi \cdot 2\right)}}{\sqrt{k}}
\]
| Alternative 7 |
|---|
| Accuracy | 37.0% |
|---|
| Cost | 13184 |
|---|
\[\sqrt{2 \cdot \left(\pi \cdot \frac{n}{k}\right)}
\]
| Alternative 8 |
|---|
| Accuracy | 37.0% |
|---|
| Cost | 13184 |
|---|
\[\sqrt{2 \cdot \frac{\pi}{\frac{k}{n}}}
\]
| Alternative 9 |
|---|
| Accuracy | 37.0% |
|---|
| Cost | 13184 |
|---|
\[\sqrt{n \cdot \frac{\pi \cdot 2}{k}}
\]
| Alternative 10 |
|---|
| Accuracy | 0.0% |
|---|
| Cost | 13056 |
|---|
\[\sqrt{\frac{\frac{\pi}{0}}{k}}
\]