Average Error: 2.3 → 0.1
Time: 5.0s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
\[\begin{array}{l} \mathbf{if}\;k \le 8.31726067928948193 \cdot 10^{150}:\\ \;\;\;\;\frac{1}{\mathsf{fma}\left(k, k, \mathsf{fma}\left(10, k, 1\right)\right)} \cdot \left(a \cdot {k}^{m}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
\begin{array}{l}
\mathbf{if}\;k \le 8.31726067928948193 \cdot 10^{150}:\\
\;\;\;\;\frac{1}{\mathsf{fma}\left(k, k, \mathsf{fma}\left(10, k, 1\right)\right)} \cdot \left(a \cdot {k}^{m}\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 temp;
	if ((k <= 8.317260679289482e+150)) {
		temp = ((1.0 / fma(k, k, fma(10.0, k, 1.0))) * (a * pow(k, m)));
	} else {
		temp = 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 temp;
}

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 < 8.317260679289482e+150

    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

      \[\leadsto \color{blue}{\frac{a}{\frac{\left(1 + 10 \cdot k\right) + k \cdot k}{{k}^{m}}}}\]
    4. Using strategy rm
    5. Applied div-inv0.1

      \[\leadsto \frac{a}{\color{blue}{\left(\left(1 + 10 \cdot k\right) + k \cdot k\right) \cdot \frac{1}{{k}^{m}}}}\]
    6. Applied *-un-lft-identity0.1

      \[\leadsto \frac{\color{blue}{1 \cdot a}}{\left(\left(1 + 10 \cdot k\right) + k \cdot k\right) \cdot \frac{1}{{k}^{m}}}\]
    7. Applied times-frac0.1

      \[\leadsto \color{blue}{\frac{1}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot \frac{a}{\frac{1}{{k}^{m}}}}\]
    8. Simplified0.1

      \[\leadsto \color{blue}{\frac{1}{\mathsf{fma}\left(k, k, \mathsf{fma}\left(10, k, 1\right)\right)}} \cdot \frac{a}{\frac{1}{{k}^{m}}}\]
    9. Simplified0.1

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

    if 8.317260679289482e+150 < k

    1. Initial program 11.3

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
    2. Taylor expanded around inf 11.3

      \[\leadsto \color{blue}{\left(\frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{2}} + 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}}\right) - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}}\]
    3. Simplified0.2

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)}\]
  3. Recombined 2 regimes into one program.
  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 8.31726067928948193 \cdot 10^{150}:\\ \;\;\;\;\frac{1}{\mathsf{fma}\left(k, k, \mathsf{fma}\left(10, k, 1\right)\right)} \cdot \left(a \cdot {k}^{m}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020057 +o rules:numerics
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))