Average Error: 41.0 → 0.6
Time: 2.9s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.991275043797484545:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.991275043797484545:\\
\;\;\;\;\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 r114323 = x;
        double r114324 = exp(r114323);
        double r114325 = 1.0;
        double r114326 = r114324 - r114325;
        double r114327 = r114324 / r114326;
        return r114327;
}
↓
double f(double x) {
        double r114328 = x;
        double r114329 = exp(r114328);
        double r114330 = 0.9912750437974845;
        bool r114331 = r114329 <= r114330;
        double r114332 = 3.0;
        double r114333 = pow(r114329, r114332);
        double r114334 = 1.0;
        double r114335 = pow(r114334, r114332);
        double r114336 = r114333 - r114335;
        double r114337 = r114329 / r114336;
        double r114338 = r114329 * r114329;
        double r114339 = r114334 * r114334;
        double r114340 = r114329 * r114334;
        double r114341 = r114339 + r114340;
        double r114342 = r114338 + r114341;
        double r114343 = r114337 * r114342;
        double r114344 = 0.5;
        double r114345 = 0.08333333333333333;
        double r114346 = r114345 * r114328;
        double r114347 = 1.0;
        double r114348 = r114347 / r114328;
        double r114349 = r114346 + r114348;
        double r114350 = r114344 + r114349;
        double r114351 = r114331 ? r114343 : r114350;
        return r114351;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original41.0
Target40.6
Herbie0.6
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.9912750437974845

    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 0.9912750437974845 < (exp x)

    1. Initial program 61.8

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

    2. Taylor expanded around 0 1.0

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

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.991275043797484545:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020056 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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