Average Error: 39.1 → 0.3
Time: 17.9s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.00000202420025896543620547163300216198:\\ \;\;\;\;\mathsf{fma}\left(x, \mathsf{fma}\left(\frac{\frac{-1}{2}}{1 \cdot 1}, x, 1\right), \log 1\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.00000202420025896543620547163300216198:\\
\;\;\;\;\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 r100079 = 1.0;
        double r100080 = x;
        double r100081 = r100079 + r100080;
        double r100082 = log(r100081);
        return r100082;
}
↓
double f(double x) {
        double r100083 = 1.0;
        double r100084 = x;
        double r100085 = r100083 + r100084;
        double r100086 = 1.000002024200259;
        bool r100087 = r100085 <= r100086;
        double r100088 = -0.5;
        double r100089 = r100083 * r100083;
        double r100090 = r100088 / r100089;
        double r100091 = fma(r100090, r100084, r100083);
        double r100092 = log(r100083);
        double r100093 = fma(r100084, r100091, r100092);
        double r100094 = log(r100085);
        double r100095 = r100087 ? r100093 : r100094;
        return r100095;
}

Error

Bits error versus x

Target

Original39.1
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) < 1.000002024200259

    1. Initial program 59.1

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

    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 1.000002024200259 < (+ 1.0 x)

    1. Initial program 0.1

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.00000202420025896543620547163300216198:\\ \;\;\;\;\mathsf{fma}\left(x, \mathsf{fma}\left(\frac{\frac{-1}{2}}{1 \cdot 1}, x, 1\right), \log 1\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019323 +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)))