Average Error: 1.9 → 1.9
Time: 6.0s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
\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 f(double a, double k, double m) {
        double r269188 = a;
        double r269189 = k;
        double r269190 = m;
        double r269191 = pow(r269189, r269190);
        double r269192 = r269188 * r269191;
        double r269193 = 1.0;
        double r269194 = 10.0;
        double r269195 = r269194 * r269189;
        double r269196 = r269193 + r269195;
        double r269197 = r269189 * r269189;
        double r269198 = r269196 + r269197;
        double r269199 = r269192 / r269198;
        return r269199;
}
↓
double f(double a, double k, double m) {
        double r269200 = a;
        double r269201 = k;
        double r269202 = m;
        double r269203 = pow(r269201, r269202);
        double r269204 = r269200 * r269203;
        double r269205 = 1.0;
        double r269206 = 10.0;
        double r269207 = r269206 * r269201;
        double r269208 = r269205 + r269207;
        double r269209 = r269201 * r269201;
        double r269210 = r269208 + r269209;
        double r269211 = r269204 / r269210;
        return r269211;
}

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 1.9

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

  2. Final simplification1.9

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

Reproduce

herbie shell --seed 2019354 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))