Average Error: 2.1 → 2.1
Time: 17.1s
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 r139813 = a;
        double r139814 = k;
        double r139815 = m;
        double r139816 = pow(r139814, r139815);
        double r139817 = r139813 * r139816;
        double r139818 = 1.0;
        double r139819 = 10.0;
        double r139820 = r139819 * r139814;
        double r139821 = r139818 + r139820;
        double r139822 = r139814 * r139814;
        double r139823 = r139821 + r139822;
        double r139824 = r139817 / r139823;
        return r139824;
}
↓
double f(double a, double k, double m) {
        double r139825 = k;
        double r139826 = 2.0;
        double r139827 = m;
        double r139828 = r139827 / r139826;
        double r139829 = r139826 * r139828;
        double r139830 = pow(r139825, r139829);
        double r139831 = 10.0;
        double r139832 = r139831 + r139825;
        double r139833 = r139825 * r139832;
        double r139834 = 1.0;
        double r139835 = r139833 + r139834;
        double r139836 = a;
        double r139837 = r139835 / r139836;
        double r139838 = r139830 / r139837;
        return r139838;
}

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. Using strategy rm

  3. Applied sqr-pow2.1

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

  4. Applied associate-*r*2.1

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

  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 2019298 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))