Average Error: 41.1 → 0.7
Time: 3.1s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.8591649740027726966928867113892920315266:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, x, \frac{1}{x}\right) + \frac{1}{2}\\ \end{array}\]
\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}:\\
\;\;\;\;\mathsf{fma}\left(\frac{1}{12}, x, \frac{1}{x}\right) + \frac{1}{2}\\

\end{array}
double f(double x) {
        double r78527 = x;
        double r78528 = exp(r78527);
        double r78529 = 1.0;
        double r78530 = r78528 - r78529;
        double r78531 = r78528 / r78530;
        return r78531;
}
↓
double f(double x) {
        double r78532 = x;
        double r78533 = exp(r78532);
        double r78534 = 0.8591649740027727;
        bool r78535 = r78533 <= r78534;
        double r78536 = 3.0;
        double r78537 = pow(r78533, r78536);
        double r78538 = 1.0;
        double r78539 = pow(r78538, r78536);
        double r78540 = r78537 - r78539;
        double r78541 = r78533 / r78540;
        double r78542 = r78533 * r78533;
        double r78543 = r78538 * r78538;
        double r78544 = r78533 * r78538;
        double r78545 = r78543 + r78544;
        double r78546 = r78542 + r78545;
        double r78547 = r78541 * r78546;
        double r78548 = 0.08333333333333333;
        double r78549 = 1.0;
        double r78550 = r78549 / r78532;
        double r78551 = fma(r78548, r78532, r78550);
        double r78552 = 0.5;
        double r78553 = r78551 + r78552;
        double r78554 = r78535 ? r78547 : r78553;
        return r78554;
}

Error

Bits error versus x

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. Simplified1.1

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

  4. Final simplification0.7

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

Reproduce

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

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

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