Average Error: 2.2 → 0.1
Time: 4.7s
Precision: binary64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \leq 9.999499964503198 \cdot 10^{+95}:\\ \;\;\;\;\frac{\left(a \cdot {\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m}\right) \cdot {\left(\sqrt[3]{k}\right)}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k} \cdot \frac{{k}^{m}}{k} + \left(a \cdot \frac{{k}^{m}}{{k}^{3}}\right) \cdot \left(\frac{99}{k} - 10\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \leq 9.999499964503198 \cdot 10^{+95}:\\
\;\;\;\;\frac{\left(a \cdot {\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m}\right) \cdot {\left(\sqrt[3]{k}\right)}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\

\mathbf{else}:\\
\;\;\;\;\frac{a}{k} \cdot \frac{{k}^{m}}{k} + \left(a \cdot \frac{{k}^{m}}{{k}^{3}}\right) \cdot \left(\frac{99}{k} - 10\right)\\

\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
 (if (<= k 9.999499964503198e+95)
   (/
    (* (* a (pow (* (cbrt k) (cbrt k)) m)) (pow (cbrt k) m))
    (+ (+ 1.0 (* k 10.0)) (* k k)))
   (+
    (* (/ a k) (/ (pow k m) k))
    (* (* a (/ (pow k m) (pow k 3.0))) (- (/ 99.0 k) 10.0)))))
double code(double a, double k, double m) {
	return (((double) (a * ((double) pow(k, m)))) / ((double) (((double) (1.0 + ((double) (10.0 * k)))) + ((double) (k * k)))));
}
↓
double code(double a, double k, double m) {
	double tmp;
	if ((k <= 9.999499964503198e+95)) {
		tmp = (((double) (((double) (a * ((double) pow(((double) (((double) cbrt(k)) * ((double) cbrt(k)))), m)))) * ((double) pow(((double) cbrt(k)), m)))) / ((double) (((double) (1.0 + ((double) (k * 10.0)))) + ((double) (k * k)))));
	} else {
		tmp = ((double) (((double) ((a / k) * (((double) pow(k, m)) / k))) + ((double) (((double) (a * (((double) pow(k, m)) / ((double) pow(k, 3.0))))) * ((double) ((99.0 / k) - 10.0))))));
	}
	return tmp;
}

Error

Bits error versus a

Bits error versus k

Bits error versus m

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 2 regimes

  2. if k < 9.9994999645031976e95

    1. Initial program Error: 0.1 bits

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
    2. Using strategy rm

    3. Applied add-cube-cbrtError: 0.1 bits

      \[\leadsto \frac{a \cdot {\color{blue}{\left(\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right) \cdot \sqrt[3]{k}\right)}}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]

    4. Applied unpow-prod-downError: 0.1 bits

      \[\leadsto \frac{a \cdot \color{blue}{\left({\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m} \cdot {\left(\sqrt[3]{k}\right)}^{m}\right)}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]

    5. Applied associate-*r*Error: 0.1 bits

      \[\leadsto \frac{\color{blue}{\left(a \cdot {\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m}\right) \cdot {\left(\sqrt[3]{k}\right)}^{m}}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]

    if 9.9994999645031976e95 < k

    1. Initial program Error: 8.2 bits

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

    2. Taylor expanded around inf Error: 8.2 bits

      \[\leadsto \color{blue}{\left(99 \cdot \frac{e^{m \cdot \left(\log 1 - \log \left(\frac{1}{k}\right)\right)} \cdot a}{{k}^{4}} + \frac{e^{m \cdot \left(\log 1 - \log \left(\frac{1}{k}\right)\right)} \cdot a}{{k}^{2}}\right) - 10 \cdot \frac{e^{m \cdot \left(\log 1 - \log \left(\frac{1}{k}\right)\right)} \cdot a}{{k}^{3}}}\]

    3. SimplifiedError: 0.1 bits

      \[\leadsto \color{blue}{\frac{a}{k} \cdot \frac{{k}^{m}}{k} + \left(\frac{{k}^{m}}{{k}^{3}} \cdot a\right) \cdot \left(\frac{99}{k} - 10\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplificationError: 0.1 bits

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 9.999499964503198 \cdot 10^{+95}:\\ \;\;\;\;\frac{\left(a \cdot {\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m}\right) \cdot {\left(\sqrt[3]{k}\right)}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k} \cdot \frac{{k}^{m}}{k} + \left(a \cdot \frac{{k}^{m}}{{k}^{3}}\right) \cdot \left(\frac{99}{k} - 10\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020203 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))