Average Error: 2.1 → 0.1
Time: 9.7s
Precision: binary64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
↓
\[\begin{array}{l} \mathbf{if}\;k \leq 1.2939891137448822 \cdot 10^{+133}:\\ \;\;\;\;\begin{array}{l} t_0 := {k}^{\left(\frac{m}{2}\right)}\\ \frac{a \cdot \left(t_0 \cdot t_0\right)}{\mathsf{fma}\left(k, k + 10, 1\right)} \end{array}\\ \mathbf{else}:\\ \;\;\;\;\begin{array}{l} t_1 := a \cdot {k}^{m}\\ \mathsf{fma}\left(99, \frac{t_1}{{k}^{4}}, \mathsf{fma}\left(\frac{a}{k}, \frac{{k}^{m}}{k}, \mathsf{fma}\left(\frac{a}{{k}^{3}}, -10, \frac{-980}{\frac{{k}^{5}}{t_1}}\right)\right)\right) \end{array}\\ \end{array} \]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \leq 1.2939891137448822 \cdot 10^{+133}:\\
\;\;\;\;\begin{array}{l}
t_0 := {k}^{\left(\frac{m}{2}\right)}\\
\frac{a \cdot \left(t_0 \cdot t_0\right)}{\mathsf{fma}\left(k, k + 10, 1\right)}
\end{array}\\

\mathbf{else}:\\
\;\;\;\;\begin{array}{l}
t_1 := a \cdot {k}^{m}\\
\mathsf{fma}\left(99, \frac{t_1}{{k}^{4}}, \mathsf{fma}\left(\frac{a}{k}, \frac{{k}^{m}}{k}, \mathsf{fma}\left(\frac{a}{{k}^{3}}, -10, \frac{-980}{\frac{{k}^{5}}{t_1}}\right)\right)\right)
\end{array}\\


\end{array}

(FPCore (a k m)
 :precision binary64
 (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))
↓
(FPCore (a k m)
 :precision binary64
 (if (<= k 1.2939891137448822e+133)
   (let* ((t_0 (pow k (/ m 2.0))))
     (/ (* a (* t_0 t_0)) (fma k (+ k 10.0) 1.0)))
   (let* ((t_1 (* a (pow k m))))
     (fma
      99.0
      (/ t_1 (pow k 4.0))
      (fma
       (/ a k)
       (/ (pow k m) k)
       (fma (/ a (pow k 3.0)) -10.0 (/ -980.0 (/ (pow k 5.0) t_1))))))))
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 tmp;
	if (k <= 1.2939891137448822e+133) {
		double t_0_1 = pow(k, (m / 2.0));
		tmp = (a * (t_0_1 * t_0_1)) / fma(k, (k + 10.0), 1.0);
	} else {
		double t_1 = a * pow(k, m);
		tmp = fma(99.0, (t_1 / pow(k, 4.0)), fma((a / k), (pow(k, m) / k), fma((a / pow(k, 3.0)), -10.0, (-980.0 / (pow(k, 5.0) / t_1)))));
	}
	return tmp;
}

Error

Bits error versus a

Bits error versus k

Bits error versus m

Derivation

  1. Split input into 2 regimes

  2. if k < 1.29398911374488224e133

    1. Initial program 0.1

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

    2. Simplified0.0

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

    3. Applied sqr-pow_binary640.1

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

    4. Applied associate-*r*_binary640.1

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

    5. Applied associate-*l*_binary640.1

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

    if 1.29398911374488224e133 < k

    1. Initial program 9.3

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

    2. Simplified9.3

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

    3. Applied add-sqr-sqrt_binary649.3

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

    4. Applied associate-/r*_binary649.3

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

    5. Taylor expanded in k around inf 9.3

      \[\leadsto \color{blue}{\left(\frac{a \cdot e^{-1 \cdot \left(\log \left(\frac{1}{k}\right) \cdot m\right)}}{{k}^{2}} + 99 \cdot \frac{a \cdot e^{-1 \cdot \left(\log \left(\frac{1}{k}\right) \cdot m\right)}}{{k}^{4}}\right) - \left(10 \cdot \frac{a \cdot e^{-1 \cdot \left(\log \left(\frac{1}{k}\right) \cdot m\right)}}{{k}^{3}} + 980 \cdot \frac{a \cdot e^{-1 \cdot \left(\log \left(\frac{1}{k}\right) \cdot m\right)}}{{k}^{5}}\right)} \]

    6. Simplified0.1

      \[\leadsto \color{blue}{\mathsf{fma}\left(99, \frac{a \cdot {k}^{m}}{{k}^{4}}, \mathsf{fma}\left(\frac{a}{k}, \frac{{k}^{m}}{k}, \mathsf{fma}\left(\frac{a \cdot {k}^{m}}{{k}^{3}}, -10, \frac{-980}{\frac{{k}^{5}}{a \cdot {k}^{m}}}\right)\right)\right)} \]

    7. Taylor expanded in m around 0 0.1

      \[\leadsto \mathsf{fma}\left(99, \frac{a \cdot {k}^{m}}{{k}^{4}}, \mathsf{fma}\left(\frac{a}{k}, \frac{{k}^{m}}{k}, \mathsf{fma}\left(\color{blue}{\frac{a}{{k}^{3}}}, -10, \frac{-980}{\frac{{k}^{5}}{a \cdot {k}^{m}}}\right)\right)\right) \]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 1.2939891137448822 \cdot 10^{+133}:\\ \;\;\;\;\frac{a \cdot \left({k}^{\left(\frac{m}{2}\right)} \cdot {k}^{\left(\frac{m}{2}\right)}\right)}{\mathsf{fma}\left(k, k + 10, 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(99, \frac{a \cdot {k}^{m}}{{k}^{4}}, \mathsf{fma}\left(\frac{a}{k}, \frac{{k}^{m}}{k}, \mathsf{fma}\left(\frac{a}{{k}^{3}}, -10, \frac{-980}{\frac{{k}^{5}}{a \cdot {k}^{m}}}\right)\right)\right)\\ \end{array} \]

Reproduce

herbie shell --seed 2021329 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))