Average Error: 2.0 → 2.1
Time: 17.9s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\frac{{k}^{\left(2 \cdot \frac{m}{2}\right)}}{\frac{k \cdot \left(10 + k\right) + 1}{a}}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\frac{{k}^{\left(2 \cdot \frac{m}{2}\right)}}{\frac{k \cdot \left(10 + k\right) + 1}{a}}
double f(double a, double k, double m) {
        double r270777 = a;
        double r270778 = k;
        double r270779 = m;
        double r270780 = pow(r270778, r270779);
        double r270781 = r270777 * r270780;
        double r270782 = 1.0;
        double r270783 = 10.0;
        double r270784 = r270783 * r270778;
        double r270785 = r270782 + r270784;
        double r270786 = r270778 * r270778;
        double r270787 = r270785 + r270786;
        double r270788 = r270781 / r270787;
        return r270788;
}
↓
double f(double a, double k, double m) {
        double r270789 = k;
        double r270790 = 2.0;
        double r270791 = m;
        double r270792 = r270791 / r270790;
        double r270793 = r270790 * r270792;
        double r270794 = pow(r270789, r270793);
        double r270795 = 10.0;
        double r270796 = r270795 + r270789;
        double r270797 = r270789 * r270796;
        double r270798 = 1.0;
        double r270799 = r270797 + r270798;
        double r270800 = a;
        double r270801 = r270799 / r270800;
        double r270802 = r270794 / r270801;
        return r270802;
}

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. Using strategy rm

  3. Applied sqr-pow2.0

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

  4. Applied associate-*r*2.0

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

  5. Final simplification2.1

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

Reproduce

herbie shell --seed 2019291 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))