Average Error: 39.0 → 0.3
Time: 13.8s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000000000502987:\\ \;\;\;\;x + \left(\frac{1}{3} \cdot x - \frac{1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.0000000000502987:\\
\;\;\;\;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 r1552551 = 1.0;
        double r1552552 = x;
        double r1552553 = r1552551 + r1552552;
        double r1552554 = log(r1552553);
        return r1552554;
}
↓
double f(double x) {
        double r1552555 = x;
        double r1552556 = 1.0;
        double r1552557 = r1552555 + r1552556;
        double r1552558 = 1.0000000000502987;
        bool r1552559 = r1552557 <= r1552558;
        double r1552560 = 0.3333333333333333;
        double r1552561 = r1552560 * r1552555;
        double r1552562 = 0.5;
        double r1552563 = r1552561 - r1552562;
        double r1552564 = r1552555 * r1552555;
        double r1552565 = r1552563 * r1552564;
        double r1552566 = r1552555 + r1552565;
        double r1552567 = log(r1552557);
        double r1552568 = r1552559 ? r1552566 : r1552567;
        return r1552568;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.0
Target0.2
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1 x) < 1.0000000000502987

    1. Initial program 59.3

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

    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

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

    if 1.0000000000502987 < (+ 1 x)

    1. Initial program 0.4

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000000000502987:\\ \;\;\;\;x + \left(\frac{1}{3} \cdot x - \frac{1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019141 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))