Average Error: 39.4 → 0.3
Time: 4.3s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000000521483745075101978727616369724:\\ \;\;\;\;\mathsf{fma}\left(x, 1, \log 1 - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000000000521483745075101978727616369724:\\
\;\;\;\;\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 r73706 = 1.0;
        double r73707 = x;
        double r73708 = r73706 + r73707;
        double r73709 = log(r73708);
        return r73709;
}
↓
double f(double x) {
        double r73710 = 1.0;
        double r73711 = x;
        double r73712 = r73710 + r73711;
        double r73713 = 1.0000000005214837;
        bool r73714 = r73712 <= r73713;
        double r73715 = log(r73710);
        double r73716 = 0.5;
        double r73717 = 2.0;
        double r73718 = pow(r73711, r73717);
        double r73719 = pow(r73710, r73717);
        double r73720 = r73718 / r73719;
        double r73721 = r73716 * r73720;
        double r73722 = r73715 - r73721;
        double r73723 = fma(r73711, r73710, r73722);
        double r73724 = log(r73712);
        double r73725 = r73714 ? r73723 : r73724;
        return r73725;
}

Error

Bits error versus x

Target

Original39.4
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) < 1.0000000005214837

    1. Initial program 59.5

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

    2. Taylor expanded around 0 0.2

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

    3. Simplified0.2

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

    if 1.0000000005214837 < (+ 1.0 x)

    1. Initial program 0.3

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000000521483745075101978727616369724:\\ \;\;\;\;\mathsf{fma}\left(x, 1, \log 1 - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020001 +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)))