Math FPCore C Julia Wolfram TeX \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
\]
↓
\[\begin{array}{l}
t_0 := \mathsf{hypot}\left(k, \sqrt{\mathsf{fma}\left(k, 10, 1\right)}\right)\\
\mathbf{if}\;k \leq 6.8 \cdot 10^{+138}:\\
\;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\
\mathbf{else}:\\
\;\;\;\;\frac{{k}^{m}}{t_0} \cdot \frac{a}{t_0}\\
\end{array}
\]
(FPCore (a k m)
:precision binary64
(/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k)))) ↓
(FPCore (a k m)
:precision binary64
(let* ((t_0 (hypot k (sqrt (fma k 10.0 1.0)))))
(if (<= k 6.8e+138)
(/ (* a (pow k m)) (+ (+ 1.0 (* k 10.0)) (* k k)))
(* (/ (pow k m) t_0) (/ a t_0))))) double code(double a, double k, double m) {
return (a * pow(k, m)) / ((1.0 + (10.0 * k)) + (k * k));
}
↓
double code(double a, double k, double m) {
double t_0 = hypot(k, sqrt(fma(k, 10.0, 1.0)));
double tmp;
if (k <= 6.8e+138) {
tmp = (a * pow(k, m)) / ((1.0 + (k * 10.0)) + (k * k));
} else {
tmp = (pow(k, m) / t_0) * (a / t_0);
}
return tmp;
}
function code(a, k, m)
return Float64(Float64(a * (k ^ m)) / Float64(Float64(1.0 + Float64(10.0 * k)) + Float64(k * k)))
end
↓
function code(a, k, m)
t_0 = hypot(k, sqrt(fma(k, 10.0, 1.0)))
tmp = 0.0
if (k <= 6.8e+138)
tmp = Float64(Float64(a * (k ^ m)) / Float64(Float64(1.0 + Float64(k * 10.0)) + Float64(k * k)));
else
tmp = Float64(Float64((k ^ m) / t_0) * Float64(a / t_0));
end
return tmp
end
code[a_, k_, m_] := N[(N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision] / N[(N[(1.0 + N[(10.0 * k), $MachinePrecision]), $MachinePrecision] + N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
↓
code[a_, k_, m_] := Block[{t$95$0 = N[Sqrt[k ^ 2 + N[Sqrt[N[(k * 10.0 + 1.0), $MachinePrecision]], $MachinePrecision] ^ 2], $MachinePrecision]}, If[LessEqual[k, 6.8e+138], N[(N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision] / N[(N[(1.0 + N[(k * 10.0), $MachinePrecision]), $MachinePrecision] + N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Power[k, m], $MachinePrecision] / t$95$0), $MachinePrecision] * N[(a / t$95$0), $MachinePrecision]), $MachinePrecision]]]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
t_0 := \mathsf{hypot}\left(k, \sqrt{\mathsf{fma}\left(k, 10, 1\right)}\right)\\
\mathbf{if}\;k \leq 6.8 \cdot 10^{+138}:\\
\;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\
\mathbf{else}:\\
\;\;\;\;\frac{{k}^{m}}{t_0} \cdot \frac{a}{t_0}\\
\end{array}
Alternatives Alternative 1 Accuracy 99.6% Cost 39368
\[\begin{array}{l}
t_0 := \mathsf{hypot}\left(k, \sqrt{\mathsf{fma}\left(k, 10, 1\right)}\right)\\
t_1 := \sqrt{{k}^{m}}\\
\mathbf{if}\;m \leq -5 \cdot 10^{-20}:\\
\;\;\;\;\frac{a \cdot t_1}{\frac{\mathsf{fma}\left(k, k + 10, 1\right)}{t_1}}\\
\mathbf{elif}\;m \leq 10^{-26}:\\
\;\;\;\;\frac{\frac{a}{t_0}}{t_0}\\
\mathbf{else}:\\
\;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\
\end{array}
\]
Alternative 2 Accuracy 99.7% Cost 7428
\[\begin{array}{l}
\mathbf{if}\;k \leq 8.5 \cdot 10^{+138}:\\
\;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\
\mathbf{else}:\\
\;\;\;\;\frac{{k}^{m}}{\frac{k}{\frac{a}{k}}}\\
\end{array}
\]
Alternative 3 Accuracy 99.5% Cost 7300
\[\begin{array}{l}
\mathbf{if}\;k \leq 1.6 \cdot 10^{+123}:\\
\;\;\;\;\frac{{k}^{m}}{\frac{1 + k \cdot \left(k + 10\right)}{a}}\\
\mathbf{else}:\\
\;\;\;\;\frac{{k}^{m}}{\frac{k}{\frac{a}{k}}}\\
\end{array}
\]
Alternative 4 Accuracy 98.7% Cost 7172
\[\begin{array}{l}
\mathbf{if}\;k \leq 10:\\
\;\;\;\;a \cdot \frac{{k}^{m}}{1 + k \cdot 10}\\
\mathbf{else}:\\
\;\;\;\;\frac{{k}^{m}}{\frac{k}{\frac{a}{k}}}\\
\end{array}
\]
Alternative 5 Accuracy 98.4% Cost 7044
\[\begin{array}{l}
\mathbf{if}\;k \leq 1:\\
\;\;\;\;a \cdot {k}^{m}\\
\mathbf{else}:\\
\;\;\;\;\frac{{k}^{m}}{\frac{k}{\frac{a}{k}}}\\
\end{array}
\]
Alternative 6 Accuracy 96.1% Cost 6921
\[\begin{array}{l}
\mathbf{if}\;m \leq -3.6 \cdot 10^{-9} \lor \neg \left(m \leq 7 \cdot 10^{-14}\right):\\
\;\;\;\;a \cdot {k}^{m}\\
\mathbf{else}:\\
\;\;\;\;\frac{a}{\left(1 + k \cdot 10\right) + k \cdot k}\\
\end{array}
\]
Alternative 7 Accuracy 74.8% Cost 968
\[\begin{array}{l}
\mathbf{if}\;m \leq -22000000000000:\\
\;\;\;\;a \cdot \left(\left(1 + \frac{0.1}{k}\right) + -1\right)\\
\mathbf{elif}\;m \leq 1.5 \cdot 10^{+24}:\\
\;\;\;\;\frac{a}{\left(1 + k \cdot 10\right) + k \cdot k}\\
\mathbf{else}:\\
\;\;\;\;\left(1 + a \cdot \frac{0.1}{k}\right) + -1\\
\end{array}
\]
Alternative 8 Accuracy 69.2% Cost 844
\[\begin{array}{l}
\mathbf{if}\;k \leq -1500000000:\\
\;\;\;\;a \cdot \left(\left(1 + \frac{0.1}{k}\right) + -1\right)\\
\mathbf{elif}\;k \leq -1.02 \cdot 10^{-206}:\\
\;\;\;\;\left(1 + a \cdot \frac{0.1}{k}\right) + -1\\
\mathbf{elif}\;k \leq 10:\\
\;\;\;\;\frac{a}{1 + k \cdot 10}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{a}{k}}{k}\\
\end{array}
\]
Alternative 9 Accuracy 74.8% Cost 840
\[\begin{array}{l}
\mathbf{if}\;m \leq -22000000000000:\\
\;\;\;\;a \cdot \left(\left(1 + \frac{0.1}{k}\right) + -1\right)\\
\mathbf{elif}\;m \leq 7 \cdot 10^{+23}:\\
\;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\
\mathbf{else}:\\
\;\;\;\;\left(1 + a \cdot \frac{0.1}{k}\right) + -1\\
\end{array}
\]
Alternative 10 Accuracy 63.3% Cost 712
\[\begin{array}{l}
\mathbf{if}\;k \leq -0.43:\\
\;\;\;\;\frac{a}{k \cdot k}\\
\mathbf{elif}\;k \leq 0.1:\\
\;\;\;\;a + -10 \cdot \left(k \cdot a\right)\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{a}{k}}{k}\\
\end{array}
\]
Alternative 11 Accuracy 63.3% Cost 712
\[\begin{array}{l}
\mathbf{if}\;k \leq -0.43:\\
\;\;\;\;\frac{1}{\frac{k \cdot k}{a}}\\
\mathbf{elif}\;k \leq 0.1:\\
\;\;\;\;a + -10 \cdot \left(k \cdot a\right)\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{a}{k}}{k}\\
\end{array}
\]
Alternative 12 Accuracy 63.4% Cost 712
\[\begin{array}{l}
\mathbf{if}\;k \leq -10:\\
\;\;\;\;\frac{1}{\frac{k \cdot k}{a}}\\
\mathbf{elif}\;k \leq 10:\\
\;\;\;\;\frac{a}{1 + k \cdot 10}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{a}{k}}{k}\\
\end{array}
\]
Alternative 13 Accuracy 66.9% Cost 712
\[\begin{array}{l}
\mathbf{if}\;k \leq -0.05:\\
\;\;\;\;a \cdot \left(\left(1 + \frac{0.1}{k}\right) + -1\right)\\
\mathbf{elif}\;k \leq 10:\\
\;\;\;\;\frac{a}{1 + k \cdot 10}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{a}{k}}{k}\\
\end{array}
\]
Alternative 14 Accuracy 39.1% Cost 585
\[\begin{array}{l}
\mathbf{if}\;k \leq -0.1 \lor \neg \left(k \leq 0.1\right):\\
\;\;\;\;a \cdot \frac{0.1}{k}\\
\mathbf{else}:\\
\;\;\;\;a\\
\end{array}
\]
Alternative 15 Accuracy 61.6% Cost 585
\[\begin{array}{l}
\mathbf{if}\;k \leq -1 \lor \neg \left(k \leq 1\right):\\
\;\;\;\;\frac{a}{k \cdot k}\\
\mathbf{else}:\\
\;\;\;\;a\\
\end{array}
\]
Alternative 16 Accuracy 63.0% Cost 584
\[\begin{array}{l}
\mathbf{if}\;k \leq -1:\\
\;\;\;\;\frac{a}{k \cdot k}\\
\mathbf{elif}\;k \leq 1:\\
\;\;\;\;a\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{a}{k}}{k}\\
\end{array}
\]
Alternative 17 Accuracy 27.0% Cost 64
\[a
\]