Average Error: 39.3 → 0.3
Time: 11.9s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.00000072941427853:\\ \;\;\;\;\left(\log 1 + 1 \cdot x\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.00000072941427853:\\
\;\;\;\;\left(\log 1 + 1 \cdot x\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\

\mathbf{else}:\\
\;\;\;\;\log \left(1 + x\right)\\

\end{array}
double f(double x) {
        double r47989 = 1.0;
        double r47990 = x;
        double r47991 = r47989 + r47990;
        double r47992 = log(r47991);
        return r47992;
}
↓
double f(double x) {
        double r47993 = 1.0;
        double r47994 = x;
        double r47995 = r47993 + r47994;
        double r47996 = 1.0000007294142785;
        bool r47997 = r47995 <= r47996;
        double r47998 = log(r47993);
        double r47999 = r47993 * r47994;
        double r48000 = r47998 + r47999;
        double r48001 = 0.5;
        double r48002 = 2.0;
        double r48003 = pow(r47994, r48002);
        double r48004 = pow(r47993, r48002);
        double r48005 = r48003 / r48004;
        double r48006 = r48001 * r48005;
        double r48007 = r48000 - r48006;
        double r48008 = log(r47995);
        double r48009 = r47997 ? r48007 : r48008;
        return r48009;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

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

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000007294142785

    1. Initial program 59.1

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

    2. Taylor expanded around 0 0.3

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

    if 1.0000007294142785 < (+ 1.0 x)

    1. Initial program 0.2

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.00000072941427853:\\ \;\;\;\;\left(\log 1 + 1 \cdot x\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019195 
(FPCore (x)
  :name "ln(1 + x)"

  :herbie-target
  (if (== (+ 1.0 x) 1.0) x (/ (* x (log (+ 1.0 x))) (- (+ 1.0 x) 1.0)))

  (log (+ 1.0 x)))