Average Error: 2.1 → 0.1
Time: 4.6s
Precision: binary64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \leq 1.0690098631054175 \cdot 10^{+152}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k} \cdot \frac{{k}^{m}}{k} + \left(a \cdot \frac{{k}^{m}}{{k}^{3}}\right) \cdot \left(\frac{99}{k} - 10\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \leq 1.0690098631054175 \cdot 10^{+152}:\\
\;\;\;\;\frac{a \cdot {k}^{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}
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 VAR;
	if ((k <= 1.0690098631054175e+152)) {
		VAR = (((double) (a * ((double) pow(k, m)))) / ((double) (((double) (1.0 + ((double) (k * 10.0)))) + ((double) (k * k)))));
	} else {
		VAR = ((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 VAR;
}

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 < 1.06900986310541749e152

    1. Initial program 0.1

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

    if 1.06900986310541749e152 < k

    1. Initial program 11.0

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

    2. Taylor expanded around inf 11.0

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

    3. Simplified0.1

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 1.0690098631054175 \cdot 10^{+152}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k} \cdot \frac{{k}^{m}}{k} + \left(a \cdot \frac{{k}^{m}}{{k}^{3}}\right) \cdot \left(\frac{99}{k} - 10\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020199 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))