Average Error: 38.5 → 0.2
Time: 9.4s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000710122837:\\ \;\;\;\;x + \left(x \cdot \frac{1}{3} + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.000000710122837:\\
\;\;\;\;x + \left(x \cdot \frac{1}{3} + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\

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

\end{array}
double f(double x) {
        double r1650864 = 1.0;
        double r1650865 = x;
        double r1650866 = r1650864 + r1650865;
        double r1650867 = log(r1650866);
        return r1650867;
}
↓
double f(double x) {
        double r1650868 = x;
        double r1650869 = 1.0;
        double r1650870 = r1650868 + r1650869;
        double r1650871 = 1.000000710122837;
        bool r1650872 = r1650870 <= r1650871;
        double r1650873 = 0.3333333333333333;
        double r1650874 = r1650868 * r1650873;
        double r1650875 = -0.5;
        double r1650876 = r1650874 + r1650875;
        double r1650877 = r1650868 * r1650868;
        double r1650878 = r1650876 * r1650877;
        double r1650879 = r1650868 + r1650878;
        double r1650880 = log(r1650870);
        double r1650881 = r1650872 ? r1650879 : r1650880;
        return r1650881;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.5
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 (+ 1 x) < 1.000000710122837

    1. Initial program 59.0

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

    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

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

    if 1.000000710122837 < (+ 1 x)

    1. Initial program 0.2

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000710122837:\\ \;\;\;\;x + \left(x \cdot \frac{1}{3} + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019143 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))