Average Error: 39.2 → 0.3
Time: 17.6s
Precision: 64
\[\log \left(1 + x\right)\]
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000098543586624089130054926499724388:\\ \;\;\;\;\mathsf{fma}\left(x, \mathsf{fma}\left(\frac{\frac{-1}{2}}{1 \cdot 1}, x, 1\right), \log 1\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000000098543586624089130054926499724388:\\
\;\;\;\;\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 r110161 = 1.0;
        double r110162 = x;
        double r110163 = r110161 + r110162;
        double r110164 = log(r110163);
        return r110164;
}

double f(double x) {
        double r110165 = 1.0;
        double r110166 = x;
        double r110167 = r110165 + r110166;
        double r110168 = 1.0000000985435866;
        bool r110169 = r110167 <= r110168;
        double r110170 = -0.5;
        double r110171 = r110165 * r110165;
        double r110172 = r110170 / r110171;
        double r110173 = fma(r110172, r110166, r110165);
        double r110174 = log(r110165);
        double r110175 = fma(r110166, r110173, r110174);
        double r110176 = log(r110167);
        double r110177 = r110169 ? r110175 : r110176;
        return r110177;
}

Error

Bits error versus x

Target

Original39.2
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) < 1.0000000985435866

    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

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

    if 1.0000000985435866 < (+ 1.0 x)

    1. Initial program 0.2

      \[\log \left(1 + x\right)\]
  3. Recombined 2 regimes into one program.
  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000098543586624089130054926499724388:\\ \;\;\;\;\mathsf{fma}\left(x, \mathsf{fma}\left(\frac{\frac{-1}{2}}{1 \cdot 1}, x, 1\right), \log 1\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019347 +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)))