Average Error: 39.8 → 0.2
Time: 9.8s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 6.904258977099596 \cdot 10^{-05}:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le 6.904258977099596 \cdot 10^{-05}:\\
\;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\\

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

\end{array}
double f(double x) {
        double r11421068 = 1.0;
        double r11421069 = x;
        double r11421070 = r11421068 + r11421069;
        double r11421071 = log(r11421070);
        return r11421071;
}
↓
double f(double x) {
        double r11421072 = x;
        double r11421073 = 6.904258977099596e-05;
        bool r11421074 = r11421072 <= r11421073;
        double r11421075 = r11421072 * r11421072;
        double r11421076 = -0.5;
        double r11421077 = 0.3333333333333333;
        double r11421078 = r11421077 * r11421072;
        double r11421079 = r11421076 + r11421078;
        double r11421080 = r11421075 * r11421079;
        double r11421081 = r11421072 + r11421080;
        double r11421082 = 1.0;
        double r11421083 = r11421072 + r11421082;
        double r11421084 = log(r11421083);
        double r11421085 = r11421074 ? r11421081 : r11421084;
        return r11421085;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
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 < 6.904258977099596e-05

    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

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

    if 6.904258977099596e-05 < x

    1. Initial program 0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 6.904258977099596 \cdot 10^{-05}:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019125 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))