Average Error: 41.0 → 0.6
Time: 18.8s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.4621431440290150738370300587121164426208:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.4621431440290150738370300587121164426208:\\
\;\;\;\;\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}:\\
\;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\

\end{array}
double f(double x) {
        double r80777 = x;
        double r80778 = exp(r80777);
        double r80779 = 1.0;
        double r80780 = r80778 - r80779;
        double r80781 = r80778 / r80780;
        return r80781;
}
↓
double f(double x) {
        double r80782 = x;
        double r80783 = exp(r80782);
        double r80784 = 0.4621431440290151;
        bool r80785 = r80783 <= r80784;
        double r80786 = 3.0;
        double r80787 = pow(r80783, r80786);
        double r80788 = 1.0;
        double r80789 = pow(r80788, r80786);
        double r80790 = r80787 - r80789;
        double r80791 = r80783 / r80790;
        double r80792 = r80783 * r80783;
        double r80793 = r80788 * r80788;
        double r80794 = r80783 * r80788;
        double r80795 = r80793 + r80794;
        double r80796 = r80792 + r80795;
        double r80797 = r80791 * r80796;
        double r80798 = 0.5;
        double r80799 = 0.08333333333333333;
        double r80800 = r80799 * r80782;
        double r80801 = 1.0;
        double r80802 = r80801 / r80782;
        double r80803 = r80800 + r80802;
        double r80804 = r80798 + r80803;
        double r80805 = r80785 ? r80797 : r80804;
        return r80805;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

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. Recombined 2 regimes into one program.

  4. Final simplification0.6

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

Reproduce

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

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

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