Average Error: 38.9 → 0.2
Time: 20.9s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.0001168771733402:\\ \;\;\;\;x + \left(\frac{1}{3} \cdot x - \frac{1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.0001168771733402:\\
\;\;\;\;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 r1101044 = 1.0;
        double r1101045 = x;
        double r1101046 = r1101044 + r1101045;
        double r1101047 = log(r1101046);
        return r1101047;
}
↓
double f(double x) {
        double r1101048 = x;
        double r1101049 = 1.0;
        double r1101050 = r1101048 + r1101049;
        double r1101051 = 1.0001168771733402;
        bool r1101052 = r1101050 <= r1101051;
        double r1101053 = 0.3333333333333333;
        double r1101054 = r1101053 * r1101048;
        double r1101055 = 0.5;
        double r1101056 = r1101054 - r1101055;
        double r1101057 = r1101048 * r1101048;
        double r1101058 = r1101056 * r1101057;
        double r1101059 = r1101048 + r1101058;
        double r1101060 = log(r1101050);
        double r1101061 = r1101052 ? r1101059 : r1101060;
        return r1101061;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.9
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) < 1.0001168771733402

    1. Initial program 58.9

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

    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 1.0001168771733402 < (+ 1 x)

    1. Initial program 0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.0001168771733402:\\ \;\;\;\;x + \left(\frac{1}{3} \cdot x - \frac{1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019151 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))