Average Error: 2.0 → 1.9
Time: 7.9s
Precision: binary64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}
double code(double a, double k, double m) {
	return (((double) (a * ((double) pow(k, m)))) / ((double) (((double) (1.0 + ((double) (10.0 * k)))) + ((double) (k * k)))));
}
↓
double code(double a, double k, double m) {
	return ((double) ((a / ((double) (1.0 + ((double) (k * ((double) (10.0 + k))))))) * ((double) pow(k, m))));
}

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 2.0

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

  2. Simplified1.9

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

  3. Final simplification1.9

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

Reproduce

herbie shell --seed 2020182 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))