Average Error: 39.6 → 0.6
Time: 12.1s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.9698251240266461:\\ \;\;\;\;\frac{e^{x}}{\left(\sqrt[3]{\log \left(e^{e^{x} - 1}\right)} \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}\right) \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{12} \cdot x + \left(\frac{1}{x} + \frac{1}{2}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.9698251240266461:\\
\;\;\;\;\frac{e^{x}}{\left(\sqrt[3]{\log \left(e^{e^{x} - 1}\right)} \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}\right) \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}}\\

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

\end{array}
double f(double x) {
        double r1753664 = x;
        double r1753665 = exp(r1753664);
        double r1753666 = 1.0;
        double r1753667 = r1753665 - r1753666;
        double r1753668 = r1753665 / r1753667;
        return r1753668;
}
↓
double f(double x) {
        double r1753669 = x;
        double r1753670 = exp(r1753669);
        double r1753671 = 0.9698251240266461;
        bool r1753672 = r1753670 <= r1753671;
        double r1753673 = 1.0;
        double r1753674 = r1753670 - r1753673;
        double r1753675 = exp(r1753674);
        double r1753676 = log(r1753675);
        double r1753677 = cbrt(r1753676);
        double r1753678 = r1753677 * r1753677;
        double r1753679 = r1753678 * r1753677;
        double r1753680 = r1753670 / r1753679;
        double r1753681 = 0.08333333333333333;
        double r1753682 = r1753681 * r1753669;
        double r1753683 = r1753673 / r1753669;
        double r1753684 = 0.5;
        double r1753685 = r1753683 + r1753684;
        double r1753686 = r1753682 + r1753685;
        double r1753687 = r1753672 ? r1753680 : r1753686;
        return r1753687;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.6
Target39.2
Herbie0.6
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.9698251240266461

    1. Initial program 0.0

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

    3. Applied add-log-exp0.0

      \[\leadsto \frac{e^{x}}{\color{blue}{\log \left(e^{e^{x} - 1}\right)}}\]
    4. Using strategy rm

    5. Applied add-cube-cbrt0.0

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

    if 0.9698251240266461 < (exp x)

    1. Initial program 60.1

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

    2. Taylor expanded around 0 0.9

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

    3. Taylor expanded around -inf 0.9

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

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.9698251240266461:\\ \;\;\;\;\frac{e^{x}}{\left(\sqrt[3]{\log \left(e^{e^{x} - 1}\right)} \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}\right) \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{12} \cdot x + \left(\frac{1}{x} + \frac{1}{2}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019152 
(FPCore (x)
  :name "expq2 (section 3.11)"

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

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