Math FPCore C Julia Wolfram TeX \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
\]
↓
\[\begin{array}{l}
t_0 := \mathsf{hypot}\left(k, \sqrt{\mathsf{fma}\left(k, 10, 1\right)}\right)\\
\mathbf{if}\;k \leq 10^{+130}:\\
\;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\
\mathbf{else}:\\
\;\;\;\;\frac{1}{t_0} \cdot \frac{{k}^{m}}{\frac{t_0}{a}}\\
\end{array}
\]
(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 1e+130)
(/ (* a (pow k m)) (+ (+ 1.0 (* k 10.0)) (* k k)))
(* (/ 1.0 t_0) (/ (pow k m) (/ t_0 a)))))) 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 <= 1e+130) {
tmp = (a * pow(k, m)) / ((1.0 + (k * 10.0)) + (k * k));
} else {
tmp = (1.0 / t_0) * (pow(k, m) / (t_0 / a));
}
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 <= 1e+130)
tmp = Float64(Float64(a * (k ^ m)) / Float64(Float64(1.0 + Float64(k * 10.0)) + Float64(k * k)));
else
tmp = Float64(Float64(1.0 / t_0) * Float64((k ^ m) / Float64(t_0 / a)));
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, 1e+130], 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[(1.0 / t$95$0), $MachinePrecision] * N[(N[Power[k, m], $MachinePrecision] / N[(t$95$0 / a), $MachinePrecision]), $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 10^{+130}:\\
\;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\
\mathbf{else}:\\
\;\;\;\;\frac{1}{t_0} \cdot \frac{{k}^{m}}{\frac{t_0}{a}}\\
\end{array}
Alternatives Alternative 1 Accuracy 99.9% Cost 7428
\[\begin{array}{l}
\mathbf{if}\;k \leq 1.2 \cdot 10^{+153}:\\
\;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\
\mathbf{else}:\\
\;\;\;\;{k}^{m} \cdot \frac{\frac{-a}{k}}{-k}\\
\end{array}
\]
Alternative 2 Accuracy 96.2% Cost 7176
\[\begin{array}{l}
t_0 := a \cdot {k}^{m}\\
\mathbf{if}\;k \leq 4 \cdot 10^{-17}:\\
\;\;\;\;t_0\\
\mathbf{elif}\;k \leq 1.36 \cdot 10^{+137}:\\
\;\;\;\;\frac{t_0}{k \cdot k}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{1}{k}}{\frac{k}{a}}\\
\end{array}
\]
Alternative 3 Accuracy 98.1% Cost 7172
\[\begin{array}{l}
\mathbf{if}\;k \leq 4 \cdot 10^{-17}:\\
\;\;\;\;a \cdot {k}^{m}\\
\mathbf{else}:\\
\;\;\;\;{k}^{m} \cdot \frac{\frac{-a}{k}}{-k}\\
\end{array}
\]
Alternative 4 Accuracy 97.8% Cost 7044
\[\begin{array}{l}
\mathbf{if}\;k \leq 4 \cdot 10^{-17}:\\
\;\;\;\;a \cdot {k}^{m}\\
\mathbf{else}:\\
\;\;\;\;\frac{{k}^{m}}{\frac{k}{\frac{a}{k}}}\\
\end{array}
\]
Alternative 5 Accuracy 92.7% Cost 6788
\[\begin{array}{l}
\mathbf{if}\;k \leq 63000000:\\
\;\;\;\;a \cdot {k}^{m}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{1}{k}}{\frac{k}{a}}\\
\end{array}
\]
Alternative 6 Accuracy 69.1% Cost 841
\[\begin{array}{l}
\mathbf{if}\;m \leq -420 \lor \neg \left(m \leq 2.9 \cdot 10^{+16}\right):\\
\;\;\;\;\left(1 + \frac{a}{k \cdot k}\right) + -1\\
\mathbf{else}:\\
\;\;\;\;\frac{a}{1 + k \cdot k}\\
\end{array}
\]
Alternative 7 Accuracy 70.2% Cost 841
\[\begin{array}{l}
\mathbf{if}\;m \leq -420 \lor \neg \left(m \leq 3.5 \cdot 10^{+16}\right):\\
\;\;\;\;\left(1 + \frac{a}{k \cdot k}\right) + -1\\
\mathbf{else}:\\
\;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\
\end{array}
\]
Alternative 8 Accuracy 63.6% Cost 712
\[\begin{array}{l}
\mathbf{if}\;k \leq -0.44:\\
\;\;\;\;\frac{a}{k \cdot k}\\
\mathbf{elif}\;k \leq 0.1:\\
\;\;\;\;a + -10 \cdot \left(k \cdot a\right)\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{a}{k}}{k}\\
\end{array}
\]
Alternative 9 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 10 Accuracy 63.3% 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 11 Accuracy 62.8% Cost 580
\[\begin{array}{l}
\mathbf{if}\;k \leq 1.25 \cdot 10^{+176}:\\
\;\;\;\;\frac{a}{1 + k \cdot k}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{a}{k}}{k}\\
\end{array}
\]
Alternative 12 Accuracy 63.3% Cost 580
\[\begin{array}{l}
\mathbf{if}\;k \leq 5 \cdot 10^{+152}:\\
\;\;\;\;\frac{a}{1 + k \cdot k}\\
\mathbf{else}:\\
\;\;\;\;\frac{a \cdot \frac{1}{k}}{k}\\
\end{array}
\]
Alternative 13 Accuracy 63.3% Cost 580
\[\begin{array}{l}
\mathbf{if}\;k \leq 10^{+77}:\\
\;\;\;\;\frac{a}{1 + k \cdot k}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{1}{k}}{\frac{k}{a}}\\
\end{array}
\]
Alternative 14 Accuracy 27.2% Cost 64
\[a
\]