Average Error: 41.0 → 0.5
Time: 21.0s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.9935258382624282402773019384767394512892:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, x, \frac{1}{x}\right) + \frac{1}{2}\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.9935258382624282402773019384767394512892:\\
\;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\

\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(\frac{1}{12}, x, \frac{1}{x}\right) + \frac{1}{2}\\

\end{array}
double f(double x) {
        double r69861 = x;
        double r69862 = exp(r69861);
        double r69863 = 1.0;
        double r69864 = r69862 - r69863;
        double r69865 = r69862 / r69864;
        return r69865;
}
↓
double f(double x) {
        double r69866 = x;
        double r69867 = exp(r69866);
        double r69868 = 0.9935258382624282;
        bool r69869 = r69867 <= r69868;
        double r69870 = 3.0;
        double r69871 = pow(r69867, r69870);
        double r69872 = 1.0;
        double r69873 = pow(r69872, r69870);
        double r69874 = r69871 - r69873;
        double r69875 = r69867 / r69874;
        double r69876 = r69867 * r69867;
        double r69877 = r69872 * r69872;
        double r69878 = r69867 * r69872;
        double r69879 = r69877 + r69878;
        double r69880 = r69876 + r69879;
        double r69881 = r69875 * r69880;
        double r69882 = 0.08333333333333333;
        double r69883 = 1.0;
        double r69884 = r69883 / r69866;
        double r69885 = fma(r69882, r69866, r69884);
        double r69886 = 0.5;
        double r69887 = r69885 + r69886;
        double r69888 = r69869 ? r69881 : r69887;
        return r69888;
}

Error

Bits error versus x

Target

Original41.0
Target40.7
Herbie0.5
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.9935258382624282

    1. Initial program 0.0

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

    3. Applied flip3--0.0

      \[\leadsto \frac{e^{x}}{\color{blue}{\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)}}}\]

    4. Applied associate-/r/0.0

      \[\leadsto \color{blue}{\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)}\]

    if 0.9935258382624282 < (exp x)

    1. Initial program 61.9

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

    2. Taylor expanded around 0 0.8

      \[\leadsto \color{blue}{\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)}\]

    3. Simplified0.8

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

  4. Final simplification0.5

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.9935258382624282402773019384767394512892:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, x, \frac{1}{x}\right) + \frac{1}{2}\\ \end{array}\]

Reproduce

herbie shell --seed 2019303 +o rules:numerics
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

  :herbie-target
  (/ 1 (- 1 (exp (- x))))

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