Average Error: 2.1 → 0.1
Time: 4.5s
Precision: binary64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 1.1892552418316243 \cdot 10^{152}:\\ \;\;\;\;a \cdot \frac{{k}^{m}}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{k \cdot \frac{10}{a \cdot {k}^{m}} + \left(\frac{1}{a \cdot {k}^{m}} + k \cdot \frac{k}{a \cdot {k}^{m}}\right)}\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 1.1892552418316243 \cdot 10^{152}:\\
\;\;\;\;a \cdot \frac{{k}^{m}}{1 + k \cdot \left(k + 10\right)}\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{k \cdot \frac{10}{a \cdot {k}^{m}} + \left(\frac{1}{a \cdot {k}^{m}} + k \cdot \frac{k}{a \cdot {k}^{m}}\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 <= 1.1892552418316243e+152)) {
		VAR = ((double) (a * ((double) (((double) pow(k, m)) / ((double) (1.0 + ((double) (k * ((double) (k + 10.0))))))))));
	} else {
		VAR = ((double) (1.0 / ((double) (((double) (k * ((double) (10.0 / ((double) (a * ((double) pow(k, m)))))))) + ((double) (((double) (1.0 / ((double) (a * ((double) pow(k, m)))))) + ((double) (k * ((double) (k / ((double) (a * ((double) pow(k, m))))))))))))));
	}
	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 < 1.1892552418316243e152

    1. Initial program 0.1

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

    2. Simplified0.1

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

    if 1.1892552418316243e152 < k

    1. Initial program 10.4

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

    3. Applied clear-num10.4

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

    4. Simplified10.4

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

    5. Taylor expanded around inf 10.4

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

    6. Simplified0.5

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 1.1892552418316243 \cdot 10^{152}:\\ \;\;\;\;a \cdot \frac{{k}^{m}}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{k \cdot \frac{10}{a \cdot {k}^{m}} + \left(\frac{1}{a \cdot {k}^{m}} + k \cdot \frac{k}{a \cdot {k}^{m}}\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2020184 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))