Average Error: 41.1 → 0.7
Time: 3.0s
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 r84096 = x;
        double r84097 = exp(r84096);
        double r84098 = 1.0;
        double r84099 = r84097 - r84098;
        double r84100 = r84097 / r84099;
        return r84100;
}
↓
double f(double x) {
        double r84101 = x;
        double r84102 = exp(r84101);
        double r84103 = 0.8591649740027727;
        bool r84104 = r84102 <= r84103;
        double r84105 = 3.0;
        double r84106 = pow(r84102, r84105);
        double r84107 = 1.0;
        double r84108 = pow(r84107, r84105);
        double r84109 = r84106 - r84108;
        double r84110 = r84102 / r84109;
        double r84111 = r84102 * r84102;
        double r84112 = r84107 * r84107;
        double r84113 = r84102 * r84107;
        double r84114 = r84112 + r84113;
        double r84115 = r84111 + r84114;
        double r84116 = r84110 * r84115;
        double r84117 = 0.5;
        double r84118 = 0.08333333333333333;
        double r84119 = r84118 * r84101;
        double r84120 = 1.0;
        double r84121 = r84120 / r84101;
        double r84122 = r84119 + r84121;
        double r84123 = r84117 + r84122;
        double r84124 = r84104 ? r84116 : r84123;
        return r84124;
}

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)))