Average Error: 41.0 → 0.5
Time: 18.6s
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 r62411 = x;
        double r62412 = exp(r62411);
        double r62413 = 1.0;
        double r62414 = r62412 - r62413;
        double r62415 = r62412 / r62414;
        return r62415;
}
↓
double f(double x) {
        double r62416 = x;
        double r62417 = exp(r62416);
        double r62418 = 0.9935258382624282;
        bool r62419 = r62417 <= r62418;
        double r62420 = 3.0;
        double r62421 = pow(r62417, r62420);
        double r62422 = 1.0;
        double r62423 = pow(r62422, r62420);
        double r62424 = r62421 - r62423;
        double r62425 = r62417 / r62424;
        double r62426 = r62417 * r62417;
        double r62427 = r62422 * r62422;
        double r62428 = r62417 * r62422;
        double r62429 = r62427 + r62428;
        double r62430 = r62426 + r62429;
        double r62431 = r62425 * r62430;
        double r62432 = 0.5;
        double r62433 = 0.08333333333333333;
        double r62434 = r62433 * r62416;
        double r62435 = 1.0;
        double r62436 = r62435 / r62416;
        double r62437 = r62434 + r62436;
        double r62438 = r62432 + r62437;
        double r62439 = r62419 ? r62431 : r62438;
        return r62439;
}

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)))