Average Error: 39.1 → 0.0
Time: 3.5s
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 r2053653 = 1.0;
        double r2053654 = x;
        double r2053655 = r2053653 + r2053654;
        double r2053656 = log(r2053655);
        return r2053656;
}
↓
double f(double x) {
        double r2053657 = x;
        double r2053658 = log1p(r2053657);
        return r2053658;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.1
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 39.1

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

  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 2019144 +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)))