Average Error: 41.2 → 0.6
Time: 15.2s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.013991682066964498:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.013991682066964498:\\
\;\;\;\;\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}:\\
\;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\

\end{array}
double f(double x) {
        double r38827 = x;
        double r38828 = exp(r38827);
        double r38829 = 1.0;
        double r38830 = r38828 - r38829;
        double r38831 = r38828 / r38830;
        return r38831;
}
↓
double f(double x) {
        double r38832 = x;
        double r38833 = exp(r38832);
        double r38834 = 0.013991682066964498;
        bool r38835 = r38833 <= r38834;
        double r38836 = 3.0;
        double r38837 = pow(r38833, r38836);
        double r38838 = 1.0;
        double r38839 = pow(r38838, r38836);
        double r38840 = r38837 - r38839;
        double r38841 = r38833 / r38840;
        double r38842 = r38833 * r38833;
        double r38843 = r38838 * r38838;
        double r38844 = r38833 * r38838;
        double r38845 = r38843 + r38844;
        double r38846 = r38842 + r38845;
        double r38847 = r38841 * r38846;
        double r38848 = 0.5;
        double r38849 = 0.08333333333333333;
        double r38850 = r38849 * r38832;
        double r38851 = 1.0;
        double r38852 = r38851 / r38832;
        double r38853 = r38850 + r38852;
        double r38854 = r38848 + r38853;
        double r38855 = r38835 ? r38847 : r38854;
        return r38855;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original41.2
Target40.8
Herbie0.6
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.013991682066964498

    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 0.013991682066964498 < (exp x)

    1. Initial program 62.0

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

    2. Taylor expanded around 0 1.0

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

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.013991682066964498:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019199 
(FPCore (x)
  :name "expq2 (section 3.11)"

  :herbie-target
  (/ 1.0 (- 1.0 (exp (- x))))

  (/ (exp x) (- (exp x) 1.0)))