Average Error: 1.9 → 1.9
Time: 5.2s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\frac{1}{\frac{k \cdot \left(10 + k\right) + 1}{{k}^{m}}} \cdot a\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\frac{1}{\frac{k \cdot \left(10 + k\right) + 1}{{k}^{m}}} \cdot a
double code(double a, double k, double m) {
	return ((double) (((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) (((double) (1.0 / ((double) (((double) (((double) (k * ((double) (10.0 + k)))) + 1.0)) / ((double) pow(k, m)))))) * a));
}

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 1.9

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

  2. Simplified1.9

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

  4. Applied clear-num1.9

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

  5. Final simplification1.9

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

Reproduce

herbie shell --seed 2020128 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))