Average Error: 39.3 → 0.2
Time: 10.1s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000310505405878:\\ \;\;\;\;x + \left(\frac{1}{3} \cdot x - \frac{1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.0000310505405878:\\
\;\;\;\;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 r3143589 = 1.0;
        double r3143590 = x;
        double r3143591 = r3143589 + r3143590;
        double r3143592 = log(r3143591);
        return r3143592;
}
↓
double f(double x) {
        double r3143593 = x;
        double r3143594 = 1.0;
        double r3143595 = r3143593 + r3143594;
        double r3143596 = 1.0000310505405878;
        bool r3143597 = r3143595 <= r3143596;
        double r3143598 = 0.3333333333333333;
        double r3143599 = r3143598 * r3143593;
        double r3143600 = 0.5;
        double r3143601 = r3143599 - r3143600;
        double r3143602 = r3143593 * r3143593;
        double r3143603 = r3143601 * r3143602;
        double r3143604 = r3143593 + r3143603;
        double r3143605 = log(r3143595);
        double r3143606 = r3143597 ? r3143604 : r3143605;
        return r3143606;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.3
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 (+ 1 x) < 1.0000310505405878

    1. Initial program 58.9

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

    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 1.0000310505405878 < (+ 1 x)

    1. Initial program 0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000310505405878:\\ \;\;\;\;x + \left(\frac{1}{3} \cdot x - \frac{1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019162 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))