Average Error: 39.2 → 0.4
Time: 8.5s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000000000006892:\\ \;\;\;\;x + \left(x \cdot \frac{1}{3} + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.0000000000006892:\\
\;\;\;\;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 r1404678 = 1.0;
        double r1404679 = x;
        double r1404680 = r1404678 + r1404679;
        double r1404681 = log(r1404680);
        return r1404681;
}
↓
double f(double x) {
        double r1404682 = x;
        double r1404683 = 1.0;
        double r1404684 = r1404682 + r1404683;
        double r1404685 = 1.0000000000006892;
        bool r1404686 = r1404684 <= r1404685;
        double r1404687 = 0.3333333333333333;
        double r1404688 = r1404682 * r1404687;
        double r1404689 = -0.5;
        double r1404690 = r1404688 + r1404689;
        double r1404691 = r1404682 * r1404682;
        double r1404692 = r1404690 * r1404691;
        double r1404693 = r1404682 + r1404692;
        double r1404694 = log(r1404684);
        double r1404695 = r1404686 ? r1404693 : r1404694;
        return r1404695;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.2
Target0.3
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1 x) < 1.0000000000006892

    1. Initial program 59.6

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

    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

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

    if 1.0000000000006892 < (+ 1 x)

    1. Initial program 0.7

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000000000006892:\\ \;\;\;\;x + \left(x \cdot \frac{1}{3} + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019156 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))