Average Error: 2.1 → 2.1
Time: 4.9s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
\[\frac{a \cdot {k}^{m}}{\left(\sqrt[3]{1 + 10 \cdot k} \cdot \sqrt[3]{1 + 10 \cdot k}\right) \cdot \sqrt[3]{1 + 10 \cdot k} + k \cdot k}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
\frac{a \cdot {k}^{m}}{\left(\sqrt[3]{1 + 10 \cdot k} \cdot \sqrt[3]{1 + 10 \cdot k}\right) \cdot \sqrt[3]{1 + 10 \cdot k} + k \cdot k}
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) {
	return ((a * pow(k, m)) / (((cbrt((1.0 + (10.0 * k))) * cbrt((1.0 + (10.0 * k)))) * cbrt((1.0 + (10.0 * k)))) + (k * k)));
}

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. Initial program 2.1

    \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
  2. Using strategy rm
  3. Applied add-cube-cbrt2.1

    \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{\left(\sqrt[3]{1 + 10 \cdot k} \cdot \sqrt[3]{1 + 10 \cdot k}\right) \cdot \sqrt[3]{1 + 10 \cdot k}} + k \cdot k}\]
  4. Final simplification2.1

    \[\leadsto \frac{a \cdot {k}^{m}}{\left(\sqrt[3]{1 + 10 \cdot k} \cdot \sqrt[3]{1 + 10 \cdot k}\right) \cdot \sqrt[3]{1 + 10 \cdot k} + k \cdot k}\]

Reproduce

herbie shell --seed 2020103 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))