Average Error: 38.9 → 0.3
Time: 7.9s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.00000008116859207:\\ \;\;\;\;\mathsf{fma}\left(\frac{-1}{2}, \frac{{x}^{2}}{{1}^{2}}, \mathsf{fma}\left(1, x, \log 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.00000008116859207:\\
\;\;\;\;\mathsf{fma}\left(\frac{-1}{2}, \frac{{x}^{2}}{{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 r59235 = 1.0;
        double r59236 = x;
        double r59237 = r59235 + r59236;
        double r59238 = log(r59237);
        return r59238;
}
↓
double f(double x) {
        double r59239 = 1.0;
        double r59240 = x;
        double r59241 = r59239 + r59240;
        double r59242 = 1.000000081168592;
        bool r59243 = r59241 <= r59242;
        double r59244 = -0.5;
        double r59245 = 2.0;
        double r59246 = pow(r59240, r59245);
        double r59247 = pow(r59239, r59245);
        double r59248 = r59246 / r59247;
        double r59249 = log(r59239);
        double r59250 = fma(r59239, r59240, r59249);
        double r59251 = fma(r59244, r59248, r59250);
        double r59252 = log(r59241);
        double r59253 = r59243 ? r59251 : r59252;
        return r59253;
}

Error

Bits error versus x

Target

Original38.9
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) < 1.000000081168592

    1. Initial program 59.0

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

    2. Taylor expanded around 0 0.4

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

    3. Simplified0.4

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

    if 1.000000081168592 < (+ 1.0 x)

    1. Initial program 0.3

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.00000008116859207:\\ \;\;\;\;\mathsf{fma}\left(\frac{-1}{2}, \frac{{x}^{2}}{{1}^{2}}, \mathsf{fma}\left(1, x, \log 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020045 +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)))