Average Error: 2.2 → 0.1
Time: 4.8s
Precision: binary64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 7.924264108290504 \cdot 10^{146}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{{k}^{m} \cdot \frac{a}{k}}{k} + \left(a \cdot \frac{{k}^{m}}{{k}^{3}}\right) \cdot \left(\frac{99}{k} - 10\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 7.924264108290504 \cdot 10^{146}:\\
\;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\

\mathbf{else}:\\
\;\;\;\;\frac{{k}^{m} \cdot \frac{a}{k}}{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) (((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 <= 7.924264108290504e+146)) {
		VAR = ((double) (((double) (a * ((double) pow(k, m)))) / ((double) (((double) (1.0 + ((double) (k * 10.0)))) + ((double) (k * k))))));
	} else {
		VAR = ((double) (((double) (((double) (((double) pow(k, m)) * ((double) (a / k)))) / k)) + ((double) (((double) (a * ((double) (((double) pow(k, m)) / ((double) pow(k, 3.0)))))) * ((double) (((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 < 7.924264108290504e146

    1. Initial program 0.1

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

    if 7.924264108290504e146 < k

    1. Initial program 10.8

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

    2. Taylor expanded around inf 10.8

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

    3. Simplified0.1

      \[\leadsto \color{blue}{\frac{a}{k} \cdot \frac{{k}^{m}}{k} + \left(\frac{{k}^{m}}{{k}^{3}} \cdot a\right) \cdot \left(\frac{99}{k} - 10\right)}\]
    4. Using strategy rm

    5. Applied associate-*r/0.1

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

    6. Simplified0.1

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 7.924264108290504 \cdot 10^{146}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{{k}^{m} \cdot \frac{a}{k}}{k} + \left(a \cdot \frac{{k}^{m}}{{k}^{3}}\right) \cdot \left(\frac{99}{k} - 10\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020185 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))