Average Error: 39.1 → 0.2
Time: 12.7s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 0.00011600758426984505:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{3} \cdot x - \frac{1}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le 0.00011600758426984505:\\
\;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{3} \cdot x - \frac{1}{2}\right)\\

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

\end{array}
double f(double x) {
        double r2485728 = 1.0;
        double r2485729 = x;
        double r2485730 = r2485728 + r2485729;
        double r2485731 = log(r2485730);
        return r2485731;
}
↓
double f(double x) {
        double r2485732 = x;
        double r2485733 = 0.00011600758426984505;
        bool r2485734 = r2485732 <= r2485733;
        double r2485735 = r2485732 * r2485732;
        double r2485736 = 0.3333333333333333;
        double r2485737 = r2485736 * r2485732;
        double r2485738 = 0.5;
        double r2485739 = r2485737 - r2485738;
        double r2485740 = r2485735 * r2485739;
        double r2485741 = r2485732 + r2485740;
        double r2485742 = 1.0;
        double r2485743 = r2485732 + r2485742;
        double r2485744 = log(r2485743);
        double r2485745 = r2485734 ? r2485741 : r2485744;
        return r2485745;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

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

Derivation

  1. Split input into 2 regimes

  2. if x < 0.00011600758426984505

    1. Initial program 58.8

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

    2. Taylor expanded around 0 0.3

      \[\leadsto \color{blue}{\left(x + \frac{1}{3} \cdot {x}^{3}\right) - \frac{1}{2} \cdot {x}^{2}}\]

    3. Simplified0.3

      \[\leadsto \color{blue}{x + \left(x \cdot x\right) \cdot \left(\frac{1}{3} \cdot x - \frac{1}{2}\right)}\]

    if 0.00011600758426984505 < x

    1. Initial program 0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 0.00011600758426984505:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{3} \cdot x - \frac{1}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019139 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))