Average Error: 2.2 → 0.9
Time: 8.5s
Precision: binary64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq -3.3554869830284104 \cdot 10^{-254}:\\ \;\;\;\;\frac{{\left(\sqrt{k}\right)}^{m} \cdot \left(a \cdot {\left(\sqrt{k}\right)}^{m}\right)}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{elif}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 0:\\ \;\;\;\;\frac{1}{\frac{k}{e^{\log \left({k}^{m}\right)}} \cdot \left(\frac{k}{a} + \frac{10}{a}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{\frac{1 + k \cdot \left(k + 10\right)}{{k}^{m}}}\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq -3.3554869830284104 \cdot 10^{-254}:\\
\;\;\;\;\frac{{\left(\sqrt{k}\right)}^{m} \cdot \left(a \cdot {\left(\sqrt{k}\right)}^{m}\right)}{1 + k \cdot \left(k + 10\right)}\\

\mathbf{elif}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 0:\\
\;\;\;\;\frac{1}{\frac{k}{e^{\log \left({k}^{m}\right)}} \cdot \left(\frac{k}{a} + \frac{10}{a}\right)}\\

\mathbf{else}:\\
\;\;\;\;\frac{a}{\frac{1 + k \cdot \left(k + 10\right)}{{k}^{m}}}\\

\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 (<=
      (/ (* a (pow k m)) (+ (+ 1.0 (* k 10.0)) (* k k)))
      -3.3554869830284104e-254)
   (/ (* (pow (sqrt k) m) (* a (pow (sqrt k) m))) (+ 1.0 (* k (+ k 10.0))))
   (if (<= (/ (* a (pow k m)) (+ (+ 1.0 (* k 10.0)) (* k k))) 0.0)
     (/ 1.0 (* (/ k (exp (log (pow k m)))) (+ (/ k a) (/ 10.0 a))))
     (/ a (/ (+ 1.0 (* k (+ k 10.0))) (pow k m))))))
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 (((a * pow(k, m)) / ((1.0 + (k * 10.0)) + (k * k))) <= -3.3554869830284104e-254) {
		tmp = (pow(sqrt(k), m) * (a * pow(sqrt(k), m))) / (1.0 + (k * (k + 10.0)));
	} else if (((a * pow(k, m)) / ((1.0 + (k * 10.0)) + (k * k))) <= 0.0) {
		tmp = 1.0 / ((k / exp(log(pow(k, m)))) * ((k / a) + (10.0 / a)));
	} else {
		tmp = a / ((1.0 + (k * (k + 10.0))) / pow(k, m));
	}
	return tmp;
}

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 3 regimes

  2. if (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k))) < -3.3554869830284104e-254

    1. Initial program 0.1

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

    2. Simplified0.1

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

    4. Applied add-sqr-sqrt_binary64_14640.1

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

    5. Applied unpow-prod-down_binary64_15210.1

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

    6. Applied associate-*r*_binary64_13820.1

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

    if -3.3554869830284104e-254 < (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k))) < -0.0

    1. Initial program 3.1

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

    2. Simplified3.1

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

    4. Applied clear-num_binary64_14413.2

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

    5. Taylor expanded around inf 33.2

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

    6. Simplified1.3

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

    if -0.0 < (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k)))

    1. Initial program 0.4

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

    2. Simplified0.2

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

    4. Applied associate-/l*_binary64_13870.2

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

  4. Final simplification0.9

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq -3.3554869830284104 \cdot 10^{-254}:\\ \;\;\;\;\frac{{\left(\sqrt{k}\right)}^{m} \cdot \left(a \cdot {\left(\sqrt{k}\right)}^{m}\right)}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{elif}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 0:\\ \;\;\;\;\frac{1}{\frac{k}{e^{\log \left({k}^{m}\right)}} \cdot \left(\frac{k}{a} + \frac{10}{a}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{\frac{1 + k \cdot \left(k + 10\right)}{{k}^{m}}}\\ \end{array}\]

Reproduce

herbie shell --seed 2021024 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))