Average Error: 39.1 → 0.6
Time: 14.1s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1:\\ \;\;\;\;\mathsf{fma}\left(x, \mathsf{fma}\left(\frac{\frac{-1}{2}}{1 \cdot 1}, x, 1\right), \log 1\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1:\\
\;\;\;\;\mathsf{fma}\left(x, \mathsf{fma}\left(\frac{\frac{-1}{2}}{1 \cdot 1}, x, 1\right), \log 1\right)\\

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

\end{array}
double f(double x) {
        double r53987 = 1.0;
        double r53988 = x;
        double r53989 = r53987 + r53988;
        double r53990 = log(r53989);
        return r53990;
}
↓
double f(double x) {
        double r53991 = 1.0;
        double r53992 = x;
        double r53993 = r53991 + r53992;
        bool r53994 = r53993 <= r53991;
        double r53995 = -0.5;
        double r53996 = r53991 * r53991;
        double r53997 = r53995 / r53996;
        double r53998 = fma(r53997, r53992, r53991);
        double r53999 = log(r53991);
        double r54000 = fma(r53992, r53998, r53999);
        double r54001 = log(r53993);
        double r54002 = r53994 ? r54000 : r54001;
        return r54002;
}

Error

Bits error versus x

Target

Original39.1
Target0.2
Herbie0.6
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0

    1. Initial program 59.6

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

    2. Taylor expanded around 0 0.3

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

    3. Simplified0.3

      \[\leadsto \color{blue}{\mathsf{fma}\left(x, \mathsf{fma}\left(\frac{\frac{-1}{2}}{1 \cdot 1}, x, 1\right), \log 1\right)}\]

    if 1.0 < (+ 1.0 x)

    1. Initial program 1.3

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

  4. Final simplification0.6

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

Reproduce

herbie shell --seed 2019326 +o rules:numerics
(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)))