Average Error: 40.8 → 0.6
Time: 16.9s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -0.0014285877116162612:\\ \;\;\;\;\frac{e^{x}}{\frac{e^{3 \cdot x} - 1}{1 + e^{x} \cdot \left(e^{x} + 1\right)}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{2} + \frac{1}{x}\right) + x \cdot \frac{1}{12}\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;x \le -0.0014285877116162612:\\
\;\;\;\;\frac{e^{x}}{\frac{e^{3 \cdot x} - 1}{1 + e^{x} \cdot \left(e^{x} + 1\right)}}\\

\mathbf{else}:\\
\;\;\;\;\left(\frac{1}{2} + \frac{1}{x}\right) + x \cdot \frac{1}{12}\\

\end{array}
double f(double x) {
        double r3305555 = x;
        double r3305556 = exp(r3305555);
        double r3305557 = 1.0;
        double r3305558 = r3305556 - r3305557;
        double r3305559 = r3305556 / r3305558;
        return r3305559;
}
↓
double f(double x) {
        double r3305560 = x;
        double r3305561 = -0.0014285877116162612;
        bool r3305562 = r3305560 <= r3305561;
        double r3305563 = exp(r3305560);
        double r3305564 = 3.0;
        double r3305565 = r3305564 * r3305560;
        double r3305566 = exp(r3305565);
        double r3305567 = 1.0;
        double r3305568 = r3305566 - r3305567;
        double r3305569 = r3305563 + r3305567;
        double r3305570 = r3305563 * r3305569;
        double r3305571 = r3305567 + r3305570;
        double r3305572 = r3305568 / r3305571;
        double r3305573 = r3305563 / r3305572;
        double r3305574 = 0.5;
        double r3305575 = r3305567 / r3305560;
        double r3305576 = r3305574 + r3305575;
        double r3305577 = 0.08333333333333333;
        double r3305578 = r3305560 * r3305577;
        double r3305579 = r3305576 + r3305578;
        double r3305580 = r3305562 ? r3305573 : r3305579;
        return r3305580;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.8
Target40.4
Herbie0.6
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.0014285877116162612

    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. Simplified0.0

      \[\leadsto \frac{e^{x}}{\frac{\color{blue}{e^{3 \cdot x} - 1}}{e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)}}\]

    5. Simplified0.0

      \[\leadsto \frac{e^{x}}{\frac{e^{3 \cdot x} - 1}{\color{blue}{e^{x} \cdot \left(1 + e^{x}\right) + 1}}}\]

    if -0.0014285877116162612 < x

    1. Initial program 60.1

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

    2. Taylor expanded around 0 0.9

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

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -0.0014285877116162612:\\ \;\;\;\;\frac{e^{x}}{\frac{e^{3 \cdot x} - 1}{1 + e^{x} \cdot \left(e^{x} + 1\right)}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{2} + \frac{1}{x}\right) + x \cdot \frac{1}{12}\\ \end{array}\]

Reproduce

herbie shell --seed 2019158 
(FPCore (x)
  :name "expq2 (section 3.11)"

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

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