Average Error: 41.1 → 0.7
Time: 11.3s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.8591649740027726966928867113892920315266:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.8591649740027726966928867113892920315266:\\
\;\;\;\;\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 r54051 = x;
        double r54052 = exp(r54051);
        double r54053 = 1.0;
        double r54054 = r54052 - r54053;
        double r54055 = r54052 / r54054;
        return r54055;
}
↓
double f(double x) {
        double r54056 = x;
        double r54057 = exp(r54056);
        double r54058 = 0.8591649740027727;
        bool r54059 = r54057 <= r54058;
        double r54060 = 3.0;
        double r54061 = pow(r54057, r54060);
        double r54062 = 1.0;
        double r54063 = pow(r54062, r54060);
        double r54064 = r54061 - r54063;
        double r54065 = r54057 / r54064;
        double r54066 = r54057 * r54057;
        double r54067 = r54062 * r54062;
        double r54068 = r54057 * r54062;
        double r54069 = r54067 + r54068;
        double r54070 = r54066 + r54069;
        double r54071 = r54065 * r54070;
        double r54072 = 0.5;
        double r54073 = 0.08333333333333333;
        double r54074 = r54073 * r54056;
        double r54075 = 1.0;
        double r54076 = r54075 / r54056;
        double r54077 = r54074 + r54076;
        double r54078 = r54072 + r54077;
        double r54079 = r54059 ? r54071 : r54078;
        return r54079;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original41.1
Target40.6
Herbie0.7
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.8591649740027727

    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 0.8591649740027727 < (exp x)

    1. Initial program 61.6

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

    2. Taylor expanded around 0 1.1

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

  4. Final simplification0.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.8591649740027726966928867113892920315266:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019347 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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