Average Error: 39.1 → 0.3
Time: 13.4s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.00000202420025896543620547163300216198:\\ \;\;\;\;\mathsf{fma}\left(1, x, \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.00000202420025896543620547163300216198:\\
\;\;\;\;\mathsf{fma}\left(1, x, \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\

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

\end{array}
double f(double x) {
        double r64819 = 1.0;
        double r64820 = x;
        double r64821 = r64819 + r64820;
        double r64822 = log(r64821);
        return r64822;
}
↓
double f(double x) {
        double r64823 = 1.0;
        double r64824 = x;
        double r64825 = r64823 + r64824;
        double r64826 = 1.000002024200259;
        bool r64827 = r64825 <= r64826;
        double r64828 = log(r64823);
        double r64829 = fma(r64823, r64824, r64828);
        double r64830 = 0.5;
        double r64831 = 2.0;
        double r64832 = pow(r64824, r64831);
        double r64833 = pow(r64823, r64831);
        double r64834 = r64832 / r64833;
        double r64835 = r64830 * r64834;
        double r64836 = r64829 - r64835;
        double r64837 = log(r64825);
        double r64838 = r64827 ? r64836 : r64837;
        return r64838;
}

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

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

    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

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

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)))