Average Error: 41.4 → 0.6
Time: 2.9s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.94364060397973859029008281140704639256:\\ \;\;\;\;\frac{e^{x}}{\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.94364060397973859029008281140704639256:\\
\;\;\;\;\frac{e^{x}}{\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}\\

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

\end{array}
double f(double x) {
        double r95680 = x;
        double r95681 = exp(r95680);
        double r95682 = 1.0;
        double r95683 = r95681 - r95682;
        double r95684 = r95681 / r95683;
        return r95684;
}
↓
double f(double x) {
        double r95685 = x;
        double r95686 = exp(r95685);
        double r95687 = 0.9436406039797386;
        bool r95688 = r95686 <= r95687;
        double r95689 = 3.0;
        double r95690 = pow(r95686, r95689);
        double r95691 = 1.0;
        double r95692 = pow(r95691, r95689);
        double r95693 = r95690 - r95692;
        double r95694 = r95691 + r95686;
        double r95695 = r95691 * r95694;
        double r95696 = r95685 + r95685;
        double r95697 = exp(r95696);
        double r95698 = r95695 + r95697;
        double r95699 = r95693 / r95698;
        double r95700 = r95686 / r95699;
        double r95701 = 0.5;
        double r95702 = 0.08333333333333333;
        double r95703 = r95702 * r95685;
        double r95704 = 1.0;
        double r95705 = r95704 / r95685;
        double r95706 = r95703 + r95705;
        double r95707 = r95701 + r95706;
        double r95708 = r95688 ? r95700 : r95707;
        return r95708;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original41.4
Target41.0
Herbie0.6
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.9436406039797386

    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

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

    if 0.9436406039797386 < (exp x)

    1. Initial program 61.7

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

    2. Taylor expanded around 0 0.9

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

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.94364060397973859029008281140704639256:\\ \;\;\;\;\frac{e^{x}}{\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019352 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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