Average Error: 2.0 → 0.1
Time: 11.1s
Precision: binary64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \leq 6.06190966441004 \cdot 10^{+44}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(e^{m}\right)}^{\log k}}{k} \cdot \frac{a}{k}\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \leq 6.06190966441004 \cdot 10^{+44}:\\
\;\;\;\;\frac{a \cdot {k}^{m}}{1 + k \cdot \left(k + 10\right)}\\

\mathbf{else}:\\
\;\;\;\;\frac{{\left(e^{m}\right)}^{\log k}}{k} \cdot \frac{a}{k}\\

\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 6.06190966441004e+44)
   (/ (* a (pow k m)) (+ 1.0 (* k (+ k 10.0))))
   (* (/ (pow (exp m) (log k)) k) (/ a k))))
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 <= 6.06190966441004e+44) {
		tmp = (a * pow(k, m)) / (1.0 + (k * (k + 10.0)));
	} else {
		tmp = (pow(exp(m), log(k)) / k) * (a / k);
	}
	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 < 6.0619096644100396e44

    1. Initial program 0.1

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

    2. Simplified0.0

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

    if 6.0619096644100396e44 < k

    1. Initial program 6.1

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

    2. Simplified6.1

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

    4. Applied add-sqr-sqrt_binary64_14646.1

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

    5. Applied times-frac_binary64_14486.1

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

    6. Taylor expanded around -inf 64.0

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

    7. Simplified0.2

      \[\leadsto \color{blue}{\frac{{\left(e^{m}\right)}^{\left(0 + \log k\right)}}{k} \cdot \frac{a}{k}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 6.06190966441004 \cdot 10^{+44}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(e^{m}\right)}^{\log k}}{k} \cdot \frac{a}{k}\\ \end{array}\]

Reproduce

herbie shell --seed 2021027 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))