Average Error: 40.9 → 0.6
Time: 2.8s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.9970535495123729052835415131994523108006:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.9970535495123729052835415131994523108006:\\
\;\;\;\;\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 r146121 = x;
        double r146122 = exp(r146121);
        double r146123 = 1.0;
        double r146124 = r146122 - r146123;
        double r146125 = r146122 / r146124;
        return r146125;
}
↓
double f(double x) {
        double r146126 = x;
        double r146127 = exp(r146126);
        double r146128 = 0.9970535495123729;
        bool r146129 = r146127 <= r146128;
        double r146130 = 3.0;
        double r146131 = pow(r146127, r146130);
        double r146132 = 1.0;
        double r146133 = pow(r146132, r146130);
        double r146134 = r146131 - r146133;
        double r146135 = r146127 / r146134;
        double r146136 = r146127 * r146127;
        double r146137 = r146132 * r146132;
        double r146138 = r146127 * r146132;
        double r146139 = r146137 + r146138;
        double r146140 = r146136 + r146139;
        double r146141 = r146135 * r146140;
        double r146142 = 0.5;
        double r146143 = 0.08333333333333333;
        double r146144 = r146143 * r146126;
        double r146145 = 1.0;
        double r146146 = r146145 / r146126;
        double r146147 = r146144 + r146146;
        double r146148 = r146142 + r146147;
        double r146149 = r146129 ? r146141 : r146148;
        return r146149;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.9
Target40.4
Herbie0.6
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.9970535495123729

    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 0.9970535495123729 < (exp x)

    1. Initial program 62.0

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

    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

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.9970535495123729052835415131994523108006:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019354 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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