Average Error: 2.4 → 0.2
Time: 6.4s
Precision: binary64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \leq 5.95740836316349 \cdot 10^{+156}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\frac{{k}^{\left(-m\right)}}{a} + \frac{k}{e^{m \cdot \log k}} \cdot \left(\frac{k}{a} + \frac{10}{a}\right)}\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \leq 5.95740836316349 \cdot 10^{+156}:\\
\;\;\;\;\frac{a \cdot {k}^{m}}{1 + k \cdot \left(k + 10\right)}\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{\frac{{k}^{\left(-m\right)}}{a} + \frac{k}{e^{m \cdot \log k}} \cdot \left(\frac{k}{a} + \frac{10}{a}\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 5.95740836316349e+156)
   (/ (* a (pow k m)) (+ 1.0 (* k (+ k 10.0))))
   (/
    1.0
    (+
     (/ (pow k (- m)) a)
     (* (/ k (exp (* m (log k)))) (+ (/ k a) (/ 10.0 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 tmp;
	if (k <= 5.95740836316349e+156) {
		tmp = (a * pow(k, m)) / (1.0 + (k * (k + 10.0)));
	} else {
		tmp = 1.0 / ((pow(k, -m) / a) + ((k / exp(m * log(k))) * ((k / a) + (10.0 / a))));
	}
	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 < 5.95740836316349033e156

    1. Initial program 0.2

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

    2. Simplified0.1

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

    if 5.95740836316349033e156 < k

    1. Initial program 11.8

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

    2. Simplified11.8

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

    4. Applied clear-num_binary64_179811.8

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

    6. Applied *-un-lft-identity_binary64_179911.8

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

    7. Applied times-frac_binary64_180511.8

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

    8. Taylor expanded around inf 11.8

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

    9. Simplified0.4

      \[\leadsto \frac{1}{\color{blue}{\frac{{k}^{\left(-m\right)}}{a} + \frac{k}{e^{m \cdot \log k}} \cdot \left(\frac{k}{a} + \frac{10}{a}\right)}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 5.95740836316349 \cdot 10^{+156}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\frac{{k}^{\left(-m\right)}}{a} + \frac{k}{e^{m \cdot \log k}} \cdot \left(\frac{k}{a} + \frac{10}{a}\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2020288 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))