Average Error: 41.0 → 0.6
Time: 20.1s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.4621431440290150738370300587121164426208:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, x, \frac{1}{x}\right) + \frac{1}{2}\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.4621431440290150738370300587121164426208:\\
\;\;\;\;\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 r85960 = x;
        double r85961 = exp(r85960);
        double r85962 = 1.0;
        double r85963 = r85961 - r85962;
        double r85964 = r85961 / r85963;
        return r85964;
}
↓
double f(double x) {
        double r85965 = x;
        double r85966 = exp(r85965);
        double r85967 = 0.4621431440290151;
        bool r85968 = r85966 <= r85967;
        double r85969 = 3.0;
        double r85970 = pow(r85966, r85969);
        double r85971 = 1.0;
        double r85972 = pow(r85971, r85969);
        double r85973 = r85970 - r85972;
        double r85974 = r85966 / r85973;
        double r85975 = r85966 * r85966;
        double r85976 = r85971 * r85971;
        double r85977 = r85966 * r85971;
        double r85978 = r85976 + r85977;
        double r85979 = r85975 + r85978;
        double r85980 = r85974 * r85979;
        double r85981 = 0.08333333333333333;
        double r85982 = 1.0;
        double r85983 = r85982 / r85965;
        double r85984 = fma(r85981, r85965, r85983);
        double r85985 = 0.5;
        double r85986 = r85984 + r85985;
        double r85987 = r85968 ? r85980 : r85986;
        return r85987;
}

Error

Bits error versus x

Target

Original41.0
Target40.6
Herbie0.6
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.4621431440290151

    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 0.4621431440290151 < (exp x)

    1. Initial program 61.7

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

    2. Taylor expanded around 0 0.9

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

    3. Simplified0.9

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

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.4621431440290150738370300587121164426208:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, x, \frac{1}{x}\right) + \frac{1}{2}\\ \end{array}\]

Reproduce

herbie shell --seed 2019323 +o rules:numerics
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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