Average Error: 41.0 → 0.5
Time: 18.8s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.9935258382624282402773019384767394512892:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.9935258382624282402773019384767394512892:\\
\;\;\;\;\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 r78600 = x;
        double r78601 = exp(r78600);
        double r78602 = 1.0;
        double r78603 = r78601 - r78602;
        double r78604 = r78601 / r78603;
        return r78604;
}
↓
double f(double x) {
        double r78605 = x;
        double r78606 = exp(r78605);
        double r78607 = 0.9935258382624282;
        bool r78608 = r78606 <= r78607;
        double r78609 = 3.0;
        double r78610 = pow(r78606, r78609);
        double r78611 = 1.0;
        double r78612 = pow(r78611, r78609);
        double r78613 = r78610 - r78612;
        double r78614 = r78606 / r78613;
        double r78615 = r78606 * r78606;
        double r78616 = r78611 * r78611;
        double r78617 = r78606 * r78611;
        double r78618 = r78616 + r78617;
        double r78619 = r78615 + r78618;
        double r78620 = r78614 * r78619;
        double r78621 = 0.5;
        double r78622 = 0.08333333333333333;
        double r78623 = r78622 * r78605;
        double r78624 = 1.0;
        double r78625 = r78624 / r78605;
        double r78626 = r78623 + r78625;
        double r78627 = r78621 + r78626;
        double r78628 = r78608 ? r78620 : r78627;
        return r78628;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original41.0
Target40.7
Herbie0.5
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.9935258382624282

    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 0.9935258382624282 < (exp x)

    1. Initial program 61.9

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

    2. Taylor expanded around 0 0.8

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

  4. Final simplification0.5

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.9935258382624282402773019384767394512892:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019303 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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