Average Error: 39.0 → 0.0
Time: 3.7s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\mathsf{log1p}\left(x\right)\]
\log \left(1 + x\right)
↓
\mathsf{log1p}\left(x\right)
double f(double x) {
        double r3906894 = 1.0;
        double r3906895 = x;
        double r3906896 = r3906894 + r3906895;
        double r3906897 = log(r3906896);
        return r3906897;
}
↓
double f(double x) {
        double r3906898 = x;
        double r3906899 = log1p(r3906898);
        return r3906899;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.0
Target0.3
Herbie0.0
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Initial program 39.0

    \[\log \left(1 + x\right)\]

  2. Simplified0.0

    \[\leadsto \color{blue}{\mathsf{log1p}\left(x\right)}\]

  3. Final simplification0.0

    \[\leadsto \mathsf{log1p}\left(x\right)\]

Reproduce

herbie shell --seed 2019120 +o rules:numerics
(FPCore (x)
  :name "ln(1 + x)"

  :herbie-target
  (if (== (+ 1 x) 1) x (/ (* x (log (+ 1 x))) (- (+ 1 x) 1)))

  (log (+ 1 x)))