Average Error: 38.7 → 0.3
Time: 12.8s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000000000430527:\\ \;\;\;\;x + \left(\frac{1}{3} \cdot x + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.0000000000430527:\\
\;\;\;\;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 r3050546 = 1.0;
        double r3050547 = x;
        double r3050548 = r3050546 + r3050547;
        double r3050549 = log(r3050548);
        return r3050549;
}
↓
double f(double x) {
        double r3050550 = x;
        double r3050551 = 1.0;
        double r3050552 = r3050550 + r3050551;
        double r3050553 = 1.0000000000430527;
        bool r3050554 = r3050552 <= r3050553;
        double r3050555 = 0.3333333333333333;
        double r3050556 = r3050555 * r3050550;
        double r3050557 = -0.5;
        double r3050558 = r3050556 + r3050557;
        double r3050559 = r3050550 * r3050550;
        double r3050560 = r3050558 * r3050559;
        double r3050561 = r3050550 + r3050560;
        double r3050562 = log(r3050552);
        double r3050563 = r3050554 ? r3050561 : r3050562;
        return r3050563;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.7
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) < 1.0000000000430527

    1. Initial program 59.2

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

    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

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

    if 1.0000000000430527 < (+ 1 x)

    1. Initial program 0.5

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000000000430527:\\ \;\;\;\;x + \left(\frac{1}{3} \cdot x + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019165 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))