Average Error: 41.0 → 0.6
Time: 19.4s
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 r64710 = x;
        double r64711 = exp(r64710);
        double r64712 = 1.0;
        double r64713 = r64711 - r64712;
        double r64714 = r64711 / r64713;
        return r64714;
}
↓
double f(double x) {
        double r64715 = x;
        double r64716 = exp(r64715);
        double r64717 = 0.4621431440290151;
        bool r64718 = r64716 <= r64717;
        double r64719 = 3.0;
        double r64720 = pow(r64716, r64719);
        double r64721 = 1.0;
        double r64722 = pow(r64721, r64719);
        double r64723 = r64720 - r64722;
        double r64724 = r64716 / r64723;
        double r64725 = r64716 * r64716;
        double r64726 = r64721 * r64721;
        double r64727 = r64716 * r64721;
        double r64728 = r64726 + r64727;
        double r64729 = r64725 + r64728;
        double r64730 = r64724 * r64729;
        double r64731 = 0.08333333333333333;
        double r64732 = 1.0;
        double r64733 = r64732 / r64715;
        double r64734 = fma(r64731, r64715, r64733);
        double r64735 = 0.5;
        double r64736 = r64734 + r64735;
        double r64737 = r64718 ? r64730 : r64736;
        return r64737;
}

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)))