Average Error: 39.3 → 0.6
Time: 11.6s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1:\\ \;\;\;\;1 \cdot x - \left(\left(\frac{x}{1} \cdot \frac{x}{1}\right) \cdot \frac{1}{2} - \log 1\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1:\\
\;\;\;\;1 \cdot x - \left(\left(\frac{x}{1} \cdot \frac{x}{1}\right) \cdot \frac{1}{2} - \log 1\right)\\

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

\end{array}
double f(double x) {
        double r3497721 = 1.0;
        double r3497722 = x;
        double r3497723 = r3497721 + r3497722;
        double r3497724 = log(r3497723);
        return r3497724;
}
↓
double f(double x) {
        double r3497725 = x;
        double r3497726 = 1.0;
        double r3497727 = r3497725 + r3497726;
        bool r3497728 = r3497727 <= r3497726;
        double r3497729 = r3497726 * r3497725;
        double r3497730 = r3497725 / r3497726;
        double r3497731 = r3497730 * r3497730;
        double r3497732 = 0.5;
        double r3497733 = r3497731 * r3497732;
        double r3497734 = log(r3497726);
        double r3497735 = r3497733 - r3497734;
        double r3497736 = r3497729 - r3497735;
        double r3497737 = log(r3497727);
        double r3497738 = r3497728 ? r3497736 : r3497737;
        return r3497738;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.3
Target0.2
Herbie0.6
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0

    1. Initial program 59.6

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

    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 1.0 < (+ 1.0 x)

    1. Initial program 1.3

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

  4. Final simplification0.6

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

Reproduce

herbie shell --seed 2019172 
(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)))