Average Error: 39.3 → 0.3
Time: 6.5s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000031820615:\\ \;\;\;\;\mathsf{fma}\left(\frac{{x}^{2}}{{1}^{2}}, \frac{-1}{2}, \mathsf{fma}\left(1, x, \log 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.0000000031820615:\\
\;\;\;\;\mathsf{fma}\left(\frac{{x}^{2}}{{1}^{2}}, \frac{-1}{2}, \mathsf{fma}\left(1, x, \log 1\right)\right)\\

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

\end{array}
double f(double x) {
        double r125788 = 1.0;
        double r125789 = x;
        double r125790 = r125788 + r125789;
        double r125791 = log(r125790);
        return r125791;
}
↓
double f(double x) {
        double r125792 = 1.0;
        double r125793 = x;
        double r125794 = r125792 + r125793;
        double r125795 = 1.0000000031820615;
        bool r125796 = r125794 <= r125795;
        double r125797 = 2.0;
        double r125798 = pow(r125793, r125797);
        double r125799 = pow(r125792, r125797);
        double r125800 = r125798 / r125799;
        double r125801 = -0.5;
        double r125802 = log(r125792);
        double r125803 = fma(r125792, r125793, r125802);
        double r125804 = fma(r125800, r125801, r125803);
        double r125805 = log(r125794);
        double r125806 = r125796 ? r125804 : r125805;
        return r125806;
}

Error

Bits error versus x

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) < 1.0000000031820615

    1. Initial program 59.3

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

    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

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

    if 1.0000000031820615 < (+ 1.0 x)

    1. Initial program 0.4

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000031820615:\\ \;\;\;\;\mathsf{fma}\left(\frac{{x}^{2}}{{1}^{2}}, \frac{-1}{2}, \mathsf{fma}\left(1, x, \log 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020042 +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)))