Average Error: 39.2 → 0.3
Time: 2.6s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.4964858386511397 \cdot 10^{-4}:\\ \;\;\;\;\frac{1}{\frac{x}{e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, {x}^{2}, \mathsf{fma}\left(\frac{1}{3}, x, 1\right)\right) \cdot \mathsf{fma}\left(\frac{1}{36}, {x}^{2}, \mathsf{fma}\left(\frac{1}{6}, x, 1\right)\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.4964858386511397 \cdot 10^{-4}:\\
\;\;\;\;\frac{1}{\frac{x}{e^{x} - 1}}\\

\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(\frac{1}{12}, {x}^{2}, \mathsf{fma}\left(\frac{1}{3}, x, 1\right)\right) \cdot \mathsf{fma}\left(\frac{1}{36}, {x}^{2}, \mathsf{fma}\left(\frac{1}{6}, x, 1\right)\right)\\

\end{array}
double f(double x) {
        double r68850 = x;
        double r68851 = exp(r68850);
        double r68852 = 1.0;
        double r68853 = r68851 - r68852;
        double r68854 = r68853 / r68850;
        return r68854;
}
↓
double f(double x) {
        double r68855 = x;
        double r68856 = -0.00014964858386511397;
        bool r68857 = r68855 <= r68856;
        double r68858 = 1.0;
        double r68859 = exp(r68855);
        double r68860 = 1.0;
        double r68861 = r68859 - r68860;
        double r68862 = r68855 / r68861;
        double r68863 = r68858 / r68862;
        double r68864 = 0.08333333333333333;
        double r68865 = 2.0;
        double r68866 = pow(r68855, r68865);
        double r68867 = 0.3333333333333333;
        double r68868 = fma(r68867, r68855, r68858);
        double r68869 = fma(r68864, r68866, r68868);
        double r68870 = 0.027777777777777776;
        double r68871 = 0.16666666666666666;
        double r68872 = fma(r68871, r68855, r68858);
        double r68873 = fma(r68870, r68866, r68872);
        double r68874 = r68869 * r68873;
        double r68875 = r68857 ? r68863 : r68874;
        return r68875;
}

Error

Bits error versus x

Target

Original39.2
Target39.6
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 < -0.00014964858386511397

    1. Initial program 0.0

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

    3. Applied clear-num0.0

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

    if -0.00014964858386511397 < x

    1. Initial program 59.8

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

    2. Taylor expanded around 0 0.5

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

    3. Simplified0.5

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

    5. Applied add-cube-cbrt0.5

      \[\leadsto \color{blue}{\left(\sqrt[3]{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)} \cdot \sqrt[3]{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}\right) \cdot \sqrt[3]{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}}\]

    6. Taylor expanded around 0 0.5

      \[\leadsto \color{blue}{\left(\frac{1}{12} \cdot {x}^{2} + \left(\frac{1}{3} \cdot x + 1\right)\right)} \cdot \sqrt[3]{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}\]

    7. Simplified0.5

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{1}{12}, {x}^{2}, \mathsf{fma}\left(\frac{1}{3}, x, 1\right)\right)} \cdot \sqrt[3]{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}\]

    8. Taylor expanded around 0 0.5

      \[\leadsto \mathsf{fma}\left(\frac{1}{12}, {x}^{2}, \mathsf{fma}\left(\frac{1}{3}, x, 1\right)\right) \cdot \color{blue}{\left(\frac{1}{36} \cdot {x}^{2} + \left(\frac{1}{6} \cdot x + 1\right)\right)}\]

    9. Simplified0.5

      \[\leadsto \mathsf{fma}\left(\frac{1}{12}, {x}^{2}, \mathsf{fma}\left(\frac{1}{3}, x, 1\right)\right) \cdot \color{blue}{\mathsf{fma}\left(\frac{1}{36}, {x}^{2}, \mathsf{fma}\left(\frac{1}{6}, x, 1\right)\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.4964858386511397 \cdot 10^{-4}:\\ \;\;\;\;\frac{1}{\frac{x}{e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, {x}^{2}, \mathsf{fma}\left(\frac{1}{3}, x, 1\right)\right) \cdot \mathsf{fma}\left(\frac{1}{36}, {x}^{2}, \mathsf{fma}\left(\frac{1}{6}, x, 1\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020049 +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))