Average Error: 38.6 → 0.3
Time: 7.8s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000000966332569518613127002026885748:\\ \;\;\;\;1 \cdot x - \left(\frac{1}{2} \cdot \left(\frac{x}{1} \cdot \frac{x}{1}\right) - \log 1\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.000000000966332569518613127002026885748:\\
\;\;\;\;1 \cdot x - \left(\frac{1}{2} \cdot \left(\frac{x}{1} \cdot \frac{x}{1}\right) - \log 1\right)\\

\mathbf{else}:\\
\;\;\;\;\log \left(x + 1\right)\\

\end{array}
double f(double x) {
        double r4065755 = 1.0;
        double r4065756 = x;
        double r4065757 = r4065755 + r4065756;
        double r4065758 = log(r4065757);
        return r4065758;
}
↓
double f(double x) {
        double r4065759 = x;
        double r4065760 = 1.0;
        double r4065761 = r4065759 + r4065760;
        double r4065762 = 1.0000000009663326;
        bool r4065763 = r4065761 <= r4065762;
        double r4065764 = r4065760 * r4065759;
        double r4065765 = 0.5;
        double r4065766 = r4065759 / r4065760;
        double r4065767 = r4065766 * r4065766;
        double r4065768 = r4065765 * r4065767;
        double r4065769 = log(r4065760);
        double r4065770 = r4065768 - r4065769;
        double r4065771 = r4065764 - r4065770;
        double r4065772 = log(r4065761);
        double r4065773 = r4065763 ? r4065771 : r4065772;
        return r4065773;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.6
Target0.3
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000009663326

    1. Initial program 59.2

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

    2. Taylor expanded around 0 0.3

      \[\leadsto \color{blue}{\left(\log 1 + 1 \cdot x\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}}\]

    3. Simplified0.3

      \[\leadsto \color{blue}{1 \cdot x - \left(\frac{1}{2} \cdot \left(\frac{x}{1} \cdot \frac{x}{1}\right) - \log 1\right)}\]

    if 1.0000000009663326 < (+ 1.0 x)

    1. Initial program 0.3

      \[\log \left(1 + x\right)\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000000966332569518613127002026885748:\\ \;\;\;\;1 \cdot x - \left(\frac{1}{2} \cdot \left(\frac{x}{1} \cdot \frac{x}{1}\right) - \log 1\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019170 
(FPCore (x)
  :name "ln(1 + x)"

  :herbie-target
  (if (== (+ 1.0 x) 1.0) x (/ (* x (log (+ 1.0 x))) (- (+ 1.0 x) 1.0)))

  (log (+ 1.0 x)))