Average Error: 2.1 → 0.1
Time: 5.9s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 1.9661959056847304 \cdot 10^{150}:\\ \;\;\;\;\left(a \cdot {k}^{m}\right) \cdot \frac{2}{\mathsf{fma}\left(k, k, \mathsf{fma}\left(10, k, 1\right)\right) \cdot 2}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k} \cdot \frac{{\left({\left(\frac{1}{k}\right)}^{m}\right)}^{-1}}{k} + \left(99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 1.9661959056847304 \cdot 10^{150}:\\
\;\;\;\;\left(a \cdot {k}^{m}\right) \cdot \frac{2}{\mathsf{fma}\left(k, k, \mathsf{fma}\left(10, k, 1\right)\right) \cdot 2}\\

\mathbf{else}:\\
\;\;\;\;\frac{a}{k} \cdot \frac{{\left({\left(\frac{1}{k}\right)}^{m}\right)}^{-1}}{k} + \left(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 <= 1.9661959056847304e+150)) {
		temp = ((a * pow(k, m)) * (2.0 / (fma(k, k, fma(10.0, k, 1.0)) * 2.0)));
	} else {
		temp = (((a / k) * (pow(pow((1.0 / k), m), -1.0) / 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 < 1.9661959056847304e+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 div-inv0.1

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

    4. Simplified0.1

      \[\leadsto \left(a \cdot {k}^{m}\right) \cdot \color{blue}{\frac{2}{\mathsf{fma}\left(k, k, \mathsf{fma}\left(10, k, 1\right)\right) \cdot 2}}\]

    if 1.9661959056847304e+150 < k

    1. Initial program 10.0

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

    2. Taylor expanded around inf 10.0

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

    3. Simplified0.1

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)}\]
    4. Using strategy rm

    5. Applied fma-udef0.1

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

    6. Simplified0.1

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 1.9661959056847304 \cdot 10^{150}:\\ \;\;\;\;\left(a \cdot {k}^{m}\right) \cdot \frac{2}{\mathsf{fma}\left(k, k, \mathsf{fma}\left(10, k, 1\right)\right) \cdot 2}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k} \cdot \frac{{\left({\left(\frac{1}{k}\right)}^{m}\right)}^{-1}}{k} + \left(99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020058 +o rules:numerics
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))