Average Error: 2.1 → 2.1
Time: 4.6s
Precision: binary64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[a \cdot \frac{{k}^{m}}{1 + k \cdot \left(k + 10\right)}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
a \cdot \frac{{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 2.1

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

  2. Simplified2.1

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

  4. Applied *-un-lft-identity_binary642.1

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

  5. Applied times-frac_binary642.1

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

  6. Simplified2.1

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

  7. Final simplification2.1

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

Reproduce

herbie shell --seed 2020232 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))