?

Average Accuracy: 96.7% → 99.9%
Time: 13.7s
Precision: binary64
Cost: 45956

?

\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
\[\begin{array}{l} t_0 := \mathsf{hypot}\left(k, \sqrt{\mathsf{fma}\left(k, 10, 1\right)}\right)\\ \mathbf{if}\;k \leq 10^{+130}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{t_0} \cdot \frac{{k}^{m}}{\frac{t_0}{a}}\\ \end{array} \]
(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}

Error?

Derivation?

  1. Split input into 2 regimes
  2. if k < 1.0000000000000001e130

    1. Initial program 99.9%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]

    if 1.0000000000000001e130 < k

    1. Initial program 85.2%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
    2. Applied egg-rr99.8%

      \[\leadsto \color{blue}{\frac{1}{\mathsf{hypot}\left(k, \sqrt{\mathsf{fma}\left(k, 10, 1\right)}\right)} \cdot \frac{{k}^{m}}{\frac{\mathsf{hypot}\left(k, \sqrt{\mathsf{fma}\left(k, 10, 1\right)}\right)}{a}}} \]
      Proof

      [Start]85.2

      \[ \frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]

      *-un-lft-identity [=>]85.2

      \[ \frac{\color{blue}{1 \cdot \left(a \cdot {k}^{m}\right)}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]

      add-sqr-sqrt [=>]85.2

      \[ \frac{1 \cdot \left(a \cdot {k}^{m}\right)}{\color{blue}{\sqrt{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot \sqrt{\left(1 + 10 \cdot k\right) + k \cdot k}}} \]

      times-frac [=>]85.2

      \[ \color{blue}{\frac{1}{\sqrt{\left(1 + 10 \cdot k\right) + k \cdot k}} \cdot \frac{a \cdot {k}^{m}}{\sqrt{\left(1 + 10 \cdot k\right) + k \cdot k}}} \]

      +-commutative [=>]85.2

      \[ \frac{1}{\sqrt{\color{blue}{k \cdot k + \left(1 + 10 \cdot k\right)}}} \cdot \frac{a \cdot {k}^{m}}{\sqrt{\left(1 + 10 \cdot k\right) + k \cdot k}} \]

      add-sqr-sqrt [=>]85.2

      \[ \frac{1}{\sqrt{k \cdot k + \color{blue}{\sqrt{1 + 10 \cdot k} \cdot \sqrt{1 + 10 \cdot k}}}} \cdot \frac{a \cdot {k}^{m}}{\sqrt{\left(1 + 10 \cdot k\right) + k \cdot k}} \]

      hypot-def [=>]85.2

      \[ \frac{1}{\color{blue}{\mathsf{hypot}\left(k, \sqrt{1 + 10 \cdot k}\right)}} \cdot \frac{a \cdot {k}^{m}}{\sqrt{\left(1 + 10 \cdot k\right) + k \cdot k}} \]

      +-commutative [=>]85.2

      \[ \frac{1}{\mathsf{hypot}\left(k, \sqrt{\color{blue}{10 \cdot k + 1}}\right)} \cdot \frac{a \cdot {k}^{m}}{\sqrt{\left(1 + 10 \cdot k\right) + k \cdot k}} \]

      *-commutative [=>]85.2

      \[ \frac{1}{\mathsf{hypot}\left(k, \sqrt{\color{blue}{k \cdot 10} + 1}\right)} \cdot \frac{a \cdot {k}^{m}}{\sqrt{\left(1 + 10 \cdot k\right) + k \cdot k}} \]

      fma-def [=>]85.2

      \[ \frac{1}{\mathsf{hypot}\left(k, \sqrt{\color{blue}{\mathsf{fma}\left(k, 10, 1\right)}}\right)} \cdot \frac{a \cdot {k}^{m}}{\sqrt{\left(1 + 10 \cdot k\right) + k \cdot k}} \]

      *-commutative [=>]85.2

      \[ \frac{1}{\mathsf{hypot}\left(k, \sqrt{\mathsf{fma}\left(k, 10, 1\right)}\right)} \cdot \frac{\color{blue}{{k}^{m} \cdot a}}{\sqrt{\left(1 + 10 \cdot k\right) + k \cdot k}} \]

      associate-/l* [=>]85.2

      \[ \frac{1}{\mathsf{hypot}\left(k, \sqrt{\mathsf{fma}\left(k, 10, 1\right)}\right)} \cdot \color{blue}{\frac{{k}^{m}}{\frac{\sqrt{\left(1 + 10 \cdot k\right) + k \cdot k}}{a}}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification99.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 10^{+130}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\mathsf{hypot}\left(k, \sqrt{\mathsf{fma}\left(k, 10, 1\right)}\right)} \cdot \frac{{k}^{m}}{\frac{\mathsf{hypot}\left(k, \sqrt{\mathsf{fma}\left(k, 10, 1\right)}\right)}{a}}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy99.9%
Cost7428
\[\begin{array}{l} \mathbf{if}\;k \leq 1.2 \cdot 10^{+153}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;{k}^{m} \cdot \frac{\frac{-a}{k}}{-k}\\ \end{array} \]
Alternative 2
Accuracy96.2%
Cost7176
\[\begin{array}{l} t_0 := a \cdot {k}^{m}\\ \mathbf{if}\;k \leq 4 \cdot 10^{-17}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;k \leq 1.36 \cdot 10^{+137}:\\ \;\;\;\;\frac{t_0}{k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{k}}{\frac{k}{a}}\\ \end{array} \]
Alternative 3
Accuracy98.1%
Cost7172
\[\begin{array}{l} \mathbf{if}\;k \leq 4 \cdot 10^{-17}:\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{else}:\\ \;\;\;\;{k}^{m} \cdot \frac{\frac{-a}{k}}{-k}\\ \end{array} \]
Alternative 4
Accuracy97.8%
Cost7044
\[\begin{array}{l} \mathbf{if}\;k \leq 4 \cdot 10^{-17}:\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{else}:\\ \;\;\;\;\frac{{k}^{m}}{\frac{k}{\frac{a}{k}}}\\ \end{array} \]
Alternative 5
Accuracy92.7%
Cost6788
\[\begin{array}{l} \mathbf{if}\;k \leq 63000000:\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{k}}{\frac{k}{a}}\\ \end{array} \]
Alternative 6
Accuracy69.1%
Cost841
\[\begin{array}{l} \mathbf{if}\;m \leq -420 \lor \neg \left(m \leq 2.9 \cdot 10^{+16}\right):\\ \;\;\;\;\left(1 + \frac{a}{k \cdot k}\right) + -1\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \end{array} \]
Alternative 7
Accuracy70.2%
Cost841
\[\begin{array}{l} \mathbf{if}\;m \leq -420 \lor \neg \left(m \leq 3.5 \cdot 10^{+16}\right):\\ \;\;\;\;\left(1 + \frac{a}{k \cdot k}\right) + -1\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \end{array} \]
Alternative 8
Accuracy63.6%
Cost712
\[\begin{array}{l} \mathbf{if}\;k \leq -0.44:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;k \leq 0.1:\\ \;\;\;\;a + -10 \cdot \left(k \cdot a\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \end{array} \]
Alternative 9
Accuracy61.6%
Cost585
\[\begin{array}{l} \mathbf{if}\;k \leq -1 \lor \neg \left(k \leq 1\right):\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;a\\ \end{array} \]
Alternative 10
Accuracy63.3%
Cost584
\[\begin{array}{l} \mathbf{if}\;k \leq -1:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;k \leq 1:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \end{array} \]
Alternative 11
Accuracy62.8%
Cost580
\[\begin{array}{l} \mathbf{if}\;k \leq 1.25 \cdot 10^{+176}:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \end{array} \]
Alternative 12
Accuracy63.3%
Cost580
\[\begin{array}{l} \mathbf{if}\;k \leq 5 \cdot 10^{+152}:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{a \cdot \frac{1}{k}}{k}\\ \end{array} \]
Alternative 13
Accuracy63.3%
Cost580
\[\begin{array}{l} \mathbf{if}\;k \leq 10^{+77}:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{k}}{\frac{k}{a}}\\ \end{array} \]
Alternative 14
Accuracy27.2%
Cost64
\[a \]

Error

Reproduce?

herbie shell --seed 2023137 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))