Average Error: 39.7 → 0.3
Time: 16.1s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.219218703283751786175040376924982865603 \cdot 10^{-4}:\\ \;\;\;\;\frac{e^{x}}{x} - \frac{1}{x}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{\mathsf{fma}\left(x, \mathsf{fma}\left(\frac{1}{6}, x, \frac{1}{2}\right), 1\right)}\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.219218703283751786175040376924982865603 \cdot 10^{-4}:\\
\;\;\;\;\frac{e^{x}}{x} - \frac{1}{x}\\

\mathbf{else}:\\
\;\;\;\;\log \left(e^{\mathsf{fma}\left(x, \mathsf{fma}\left(\frac{1}{6}, x, \frac{1}{2}\right), 1\right)}\right)\\

\end{array}
double f(double x) {
        double r50687 = x;
        double r50688 = exp(r50687);
        double r50689 = 1.0;
        double r50690 = r50688 - r50689;
        double r50691 = r50690 / r50687;
        return r50691;
}
↓
double f(double x) {
        double r50692 = x;
        double r50693 = -0.00012192187032837518;
        bool r50694 = r50692 <= r50693;
        double r50695 = exp(r50692);
        double r50696 = r50695 / r50692;
        double r50697 = 1.0;
        double r50698 = r50697 / r50692;
        double r50699 = r50696 - r50698;
        double r50700 = 0.16666666666666666;
        double r50701 = 0.5;
        double r50702 = fma(r50700, r50692, r50701);
        double r50703 = 1.0;
        double r50704 = fma(r50692, r50702, r50703);
        double r50705 = exp(r50704);
        double r50706 = log(r50705);
        double r50707 = r50694 ? r50699 : r50706;
        return r50707;
}

Error

Bits error versus x

Target

Original39.7
Target40.1
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.00012192187032837518

    1. Initial program 0.0

      \[\frac{e^{x} - 1}{x}\]
    2. Using strategy rm

    3. Applied div-sub0.1

      \[\leadsto \color{blue}{\frac{e^{x}}{x} - \frac{1}{x}}\]

    if -0.00012192187032837518 < x

    1. Initial program 60.1

      \[\frac{e^{x} - 1}{x}\]

    2. Taylor expanded around 0 0.4

      \[\leadsto \color{blue}{\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)}\]

    3. Simplified0.4

      \[\leadsto \color{blue}{\mathsf{fma}\left(x, \mathsf{fma}\left(\frac{1}{6}, x, \frac{1}{2}\right), 1\right)}\]
    4. Using strategy rm

    5. Applied add-log-exp0.4

      \[\leadsto \color{blue}{\log \left(e^{\mathsf{fma}\left(x, \mathsf{fma}\left(\frac{1}{6}, x, \frac{1}{2}\right), 1\right)}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.219218703283751786175040376924982865603 \cdot 10^{-4}:\\ \;\;\;\;\frac{e^{x}}{x} - \frac{1}{x}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{\mathsf{fma}\left(x, \mathsf{fma}\left(\frac{1}{6}, x, \frac{1}{2}\right), 1\right)}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019303 +o rules:numerics
(FPCore (x)
  :name "Kahan's exp quotient"
  :precision binary64

  :herbie-target
  (if (and (< x 1) (> x -1)) (/ (- (exp x) 1) (log (exp x))) (/ (- (exp x) 1) x))

  (/ (- (exp x) 1) x))