Average Error: 2.1 → 2.1
Time: 4.3s
Precision: binary64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
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) (a * ((double) pow(k, m)))) / ((double) (((double) (1.0 + ((double) (10.0 * k)))) + ((double) (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. Final simplification2.1

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

Reproduce

herbie shell --seed 2020155 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))