Average Error: 39.8 → 0.2
Time: 18.8s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000001787741193:\\ \;\;\;\;x + \left(\frac{1}{3} \cdot x - \frac{1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.0000001787741193:\\
\;\;\;\;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 r3011563 = 1.0;
        double r3011564 = x;
        double r3011565 = r3011563 + r3011564;
        double r3011566 = log(r3011565);
        return r3011566;
}
↓
double f(double x) {
        double r3011567 = x;
        double r3011568 = 1.0;
        double r3011569 = r3011567 + r3011568;
        double r3011570 = 1.0000001787741193;
        bool r3011571 = r3011569 <= r3011570;
        double r3011572 = 0.3333333333333333;
        double r3011573 = r3011572 * r3011567;
        double r3011574 = 0.5;
        double r3011575 = r3011573 - r3011574;
        double r3011576 = r3011567 * r3011567;
        double r3011577 = r3011575 * r3011576;
        double r3011578 = r3011567 + r3011577;
        double r3011579 = log(r3011569);
        double r3011580 = r3011571 ? r3011578 : r3011579;
        return r3011580;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
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) < 1.0000001787741193

    1. Initial program 59.1

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

    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

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

    if 1.0000001787741193 < (+ 1 x)

    1. Initial program 0.2

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000001787741193:\\ \;\;\;\;x + \left(\frac{1}{3} \cdot x - \frac{1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019164 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))