Average Error: 38.9 → 0.2
Time: 23.2s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 0.00012152448360312907:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le 0.00012152448360312907:\\
\;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\

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

\end{array}
double f(double x) {
        double r7184783 = 1.0;
        double r7184784 = x;
        double r7184785 = r7184783 + r7184784;
        double r7184786 = log(r7184785);
        return r7184786;
}
↓
double f(double x) {
        double r7184787 = x;
        double r7184788 = 0.00012152448360312907;
        bool r7184789 = r7184787 <= r7184788;
        double r7184790 = -0.5;
        double r7184791 = 0.3333333333333333;
        double r7184792 = r7184791 * r7184787;
        double r7184793 = r7184790 + r7184792;
        double r7184794 = r7184787 * r7184793;
        double r7184795 = r7184794 * r7184787;
        double r7184796 = r7184787 + r7184795;
        double r7184797 = 1.0;
        double r7184798 = r7184787 + r7184797;
        double r7184799 = log(r7184798);
        double r7184800 = r7184789 ? r7184796 : r7184799;
        return r7184800;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.9
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 < 0.00012152448360312907

    1. Initial program 59.0

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

    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

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

    if 0.00012152448360312907 < x

    1. Initial program 0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 0.00012152448360312907:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019107 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))