Average Error: 1.9 → 0.1
Time: 4.0s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 1.15382552312658998 \cdot 10^{60}:\\ \;\;\;\;a \cdot \frac{{k}^{m}}{1 + \mathsf{fma}\left(10, k, k \cdot k\right)}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 1.15382552312658998 \cdot 10^{60}:\\
\;\;\;\;a \cdot \frac{{k}^{m}}{1 + \mathsf{fma}\left(10, k, k \cdot k\right)}\\

\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\

\end{array}
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 VAR;
	if ((k <= 1.15382552312659e+60)) {
		VAR = (a * (pow(k, m) / (1.0 + fma(10.0, k, (k * k)))));
	} else {
		VAR = fma((exp((-1.0 * (m * log((1.0 / k))))) / k), (a / k), ((99.0 * ((a * exp((-1.0 * (m * log((1.0 / k)))))) / pow(k, 4.0))) - (10.0 * ((a * exp((-1.0 * (m * log((1.0 / k)))))) / pow(k, 3.0)))));
	}
	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 < 1.15382552312659e+60

    1. Initial program 0.1

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

    3. Applied associate-+l+0.1

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

    4. Simplified0.0

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

    6. Applied *-un-lft-identity0.0

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

    7. Applied times-frac0.1

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

    8. Simplified0.1

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

    if 1.15382552312659e+60 < k

    1. Initial program 6.3

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

    2. Taylor expanded around inf 6.3

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

    3. Simplified0.1

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 1.15382552312658998 \cdot 10^{60}:\\ \;\;\;\;a \cdot \frac{{k}^{m}}{1 + \mathsf{fma}\left(10, k, k \cdot k\right)}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020092 +o rules:numerics
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))