Average Error: 41.0 → 0.6
Time: 17.9s
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 r66796 = x;
        double r66797 = exp(r66796);
        double r66798 = 1.0;
        double r66799 = r66797 - r66798;
        double r66800 = r66797 / r66799;
        return r66800;
}
↓
double f(double x) {
        double r66801 = x;
        double r66802 = exp(r66801);
        double r66803 = 0.4621431440290151;
        bool r66804 = r66802 <= r66803;
        double r66805 = 3.0;
        double r66806 = pow(r66802, r66805);
        double r66807 = 1.0;
        double r66808 = pow(r66807, r66805);
        double r66809 = r66806 - r66808;
        double r66810 = r66802 / r66809;
        double r66811 = r66802 * r66802;
        double r66812 = r66807 * r66807;
        double r66813 = r66802 * r66807;
        double r66814 = r66812 + r66813;
        double r66815 = r66811 + r66814;
        double r66816 = r66810 * r66815;
        double r66817 = 0.5;
        double r66818 = 0.08333333333333333;
        double r66819 = r66818 * r66801;
        double r66820 = 1.0;
        double r66821 = r66820 / r66801;
        double r66822 = r66819 + r66821;
        double r66823 = r66817 + r66822;
        double r66824 = r66804 ? r66816 : r66823;
        return r66824;
}

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)))