Average Error: 41.1 → 0.7
Time: 9.3s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.8591649740027726966928867113892920315266:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, x, \frac{1}{x}\right) + \frac{1}{2}\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.8591649740027726966928867113892920315266:\\
\;\;\;\;\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 r77854 = x;
        double r77855 = exp(r77854);
        double r77856 = 1.0;
        double r77857 = r77855 - r77856;
        double r77858 = r77855 / r77857;
        return r77858;
}
↓
double f(double x) {
        double r77859 = x;
        double r77860 = exp(r77859);
        double r77861 = 0.8591649740027727;
        bool r77862 = r77860 <= r77861;
        double r77863 = 3.0;
        double r77864 = pow(r77860, r77863);
        double r77865 = 1.0;
        double r77866 = pow(r77865, r77863);
        double r77867 = r77864 - r77866;
        double r77868 = r77860 / r77867;
        double r77869 = r77860 * r77860;
        double r77870 = r77865 * r77865;
        double r77871 = r77860 * r77865;
        double r77872 = r77870 + r77871;
        double r77873 = r77869 + r77872;
        double r77874 = r77868 * r77873;
        double r77875 = 0.08333333333333333;
        double r77876 = 1.0;
        double r77877 = r77876 / r77859;
        double r77878 = fma(r77875, r77859, r77877);
        double r77879 = 0.5;
        double r77880 = r77878 + r77879;
        double r77881 = r77862 ? r77874 : r77880;
        return r77881;
}

Error

Bits error versus x

Target

Original41.1
Target40.6
Herbie0.7
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.8591649740027727

    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 0.8591649740027727 < (exp x)

    1. Initial program 61.6

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

    2. Taylor expanded around 0 1.1

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

    3. Simplified1.1

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

  4. Final simplification0.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.8591649740027726966928867113892920315266:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, x, \frac{1}{x}\right) + \frac{1}{2}\\ \end{array}\]

Reproduce

herbie shell --seed 2019347 +o rules:numerics
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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