Average Error: 39.4 → 0.2
Time: 13.1s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000014996160052:\\ \;\;\;\;x + \left(\frac{1}{3} \cdot x - \frac{1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.0000014996160052:\\
\;\;\;\;x + \left(\frac{1}{3} \cdot x - \frac{1}{2}\right) \cdot \left(x \cdot x\right)\\

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

\end{array}
double f(double x) {
        double r2653771 = 1.0;
        double r2653772 = x;
        double r2653773 = r2653771 + r2653772;
        double r2653774 = log(r2653773);
        return r2653774;
}
↓
double f(double x) {
        double r2653775 = x;
        double r2653776 = 1.0;
        double r2653777 = r2653775 + r2653776;
        double r2653778 = 1.0000014996160052;
        bool r2653779 = r2653777 <= r2653778;
        double r2653780 = 0.3333333333333333;
        double r2653781 = r2653780 * r2653775;
        double r2653782 = 0.5;
        double r2653783 = r2653781 - r2653782;
        double r2653784 = r2653775 * r2653775;
        double r2653785 = r2653783 * r2653784;
        double r2653786 = r2653775 + r2653785;
        double r2653787 = log(r2653777);
        double r2653788 = r2653779 ? r2653786 : r2653787;
        return r2653788;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.4
Target0.3
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 (+ 1 x) < 1.0000014996160052

    1. Initial program 59.2

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

    2. Taylor expanded around 0 0.2

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

    3. Simplified0.2

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

    if 1.0000014996160052 < (+ 1 x)

    1. Initial program 0.2

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000014996160052:\\ \;\;\;\;x + \left(\frac{1}{3} \cdot x - \frac{1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019142 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))