Average Error: 39.1 → 0.4
Time: 4.1s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000000000413:\\ \;\;\;\;\mathsf{fma}\left(x, 1, \log 1 - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.0000000000000413:\\
\;\;\;\;\mathsf{fma}\left(x, 1, \log 1 - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\right)\\

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

\end{array}
double f(double x) {
        double r83092 = 1.0;
        double r83093 = x;
        double r83094 = r83092 + r83093;
        double r83095 = log(r83094);
        return r83095;
}
↓
double f(double x) {
        double r83096 = 1.0;
        double r83097 = x;
        double r83098 = r83096 + r83097;
        double r83099 = 1.0000000000000413;
        bool r83100 = r83098 <= r83099;
        double r83101 = log(r83096);
        double r83102 = 0.5;
        double r83103 = 2.0;
        double r83104 = pow(r83097, r83103);
        double r83105 = pow(r83096, r83103);
        double r83106 = r83104 / r83105;
        double r83107 = r83102 * r83106;
        double r83108 = r83101 - r83107;
        double r83109 = fma(r83097, r83096, r83108);
        double r83110 = log(r83098);
        double r83111 = r83100 ? r83109 : r83110;
        return r83111;
}

Error

Bits error versus x

Target

Original39.1
Target0.2
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000000000413

    1. Initial program 59.4

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

    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

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

    if 1.0000000000000413 < (+ 1.0 x)

    1. Initial program 0.7

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000000000413:\\ \;\;\;\;\mathsf{fma}\left(x, 1, \log 1 - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020024 +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)))