Average Error: 2.2 → 2.2
Time: 7.3s
Precision: binary64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
\[\frac{a \cdot {k}^{m}}{1 + k \cdot \left(k + 10\right)}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
\frac{a \cdot {k}^{m}}{1 + k \cdot \left(k + 10\right)}
(FPCore (a k m)
 :precision binary64
 (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))
(FPCore (a k m)
 :precision binary64
 (/ (* a (pow k m)) (+ 1.0 (* k (+ k 10.0)))))
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)) / (1.0 + (k * (k + 10.0)));
}

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 2.2

    \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
  2. Simplified2.2

    \[\leadsto \color{blue}{\frac{a \cdot {k}^{m}}{1 + k \cdot \left(k + 10\right)}}\]
  3. Final simplification2.2

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

Reproduce

herbie shell --seed 2020292 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))