Average Error: 38.8 → 0.3
Time: 4.3s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000959807234:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.0000000959807234:\\
\;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\

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

\end{array}
double f(double x) {
        double r88771 = 1.0;
        double r88772 = x;
        double r88773 = r88771 + r88772;
        double r88774 = log(r88773);
        return r88774;
}
↓
double f(double x) {
        double r88775 = 1.0;
        double r88776 = x;
        double r88777 = r88775 + r88776;
        double r88778 = 1.0000000959807234;
        bool r88779 = r88777 <= r88778;
        double r88780 = r88775 * r88776;
        double r88781 = log(r88775);
        double r88782 = r88780 + r88781;
        double r88783 = 0.5;
        double r88784 = 2.0;
        double r88785 = pow(r88776, r88784);
        double r88786 = pow(r88775, r88784);
        double r88787 = r88785 / r88786;
        double r88788 = r88783 * r88787;
        double r88789 = r88782 - r88788;
        double r88790 = log(r88777);
        double r88791 = r88779 ? r88789 : r88790;
        return r88791;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.8
Target0.3
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) < 1.0000000959807234

    1. Initial program 59.2

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

    2. Taylor expanded around 0 0.3

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

    if 1.0000000959807234 < (+ 1.0 x)

    1. Initial program 0.2

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000959807234:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020064 
(FPCore (x)
  :name "ln(1 + x)"
  :precision binary64

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

  (log (+ 1 x)))