Average Error: 40.1 → 0.3
Time: 12.1s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.365072845514119328266117170400661962049 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{e^{x + x} - 1 \cdot 1}{e^{x} \cdot e^{x} - 1 \cdot 1}}{\frac{x}{e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right) + 1\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.365072845514119328266117170400661962049 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{e^{x + x} - 1 \cdot 1}{e^{x} \cdot e^{x} - 1 \cdot 1}}{\frac{x}{e^{x} - 1}}\\

\mathbf{else}:\\
\;\;\;\;x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right) + 1\\

\end{array}
double f(double x) {
        double r52882 = x;
        double r52883 = exp(r52882);
        double r52884 = 1.0;
        double r52885 = r52883 - r52884;
        double r52886 = r52885 / r52882;
        return r52886;
}
↓
double f(double x) {
        double r52887 = x;
        double r52888 = -0.00013650728455141193;
        bool r52889 = r52887 <= r52888;
        double r52890 = r52887 + r52887;
        double r52891 = exp(r52890);
        double r52892 = 1.0;
        double r52893 = r52892 * r52892;
        double r52894 = r52891 - r52893;
        double r52895 = exp(r52887);
        double r52896 = r52895 * r52895;
        double r52897 = r52896 - r52893;
        double r52898 = r52894 / r52897;
        double r52899 = r52895 - r52892;
        double r52900 = r52887 / r52899;
        double r52901 = r52898 / r52900;
        double r52902 = 0.5;
        double r52903 = 0.16666666666666666;
        double r52904 = r52903 * r52887;
        double r52905 = r52902 + r52904;
        double r52906 = r52887 * r52905;
        double r52907 = 1.0;
        double r52908 = r52906 + r52907;
        double r52909 = r52889 ? r52901 : r52908;
        return r52909;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.1
Target40.5
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.00013650728455141193

    1. Initial program 0.1

      \[\frac{e^{x} - 1}{x}\]
    2. Using strategy rm

    3. Applied flip--0.1

      \[\leadsto \frac{\color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}}{x}\]

    4. Simplified0.1

      \[\leadsto \frac{\frac{\color{blue}{e^{x + x} - 1 \cdot 1}}{e^{x} + 1}}{x}\]
    5. Using strategy rm

    6. Applied flip-+0.1

      \[\leadsto \frac{\frac{e^{x + x} - 1 \cdot 1}{\color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} - 1}}}}{x}\]

    7. Applied associate-/r/0.1

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

    8. Applied associate-/l*0.1

      \[\leadsto \color{blue}{\frac{\frac{e^{x + x} - 1 \cdot 1}{e^{x} \cdot e^{x} - 1 \cdot 1}}{\frac{x}{e^{x} - 1}}}\]

    if -0.00013650728455141193 < x

    1. Initial program 60.2

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

    2. Taylor expanded around 0 0.4

      \[\leadsto \color{blue}{\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)}\]

    3. Simplified0.4

      \[\leadsto \color{blue}{x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right) + 1}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.365072845514119328266117170400661962049 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{e^{x + x} - 1 \cdot 1}{e^{x} \cdot e^{x} - 1 \cdot 1}}{\frac{x}{e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right) + 1\\ \end{array}\]

Reproduce

herbie shell --seed 2019235 
(FPCore (x)
  :name "Kahan's exp quotient"
  :precision binary64

  :herbie-target
  (if (and (< x 1) (> x -1)) (/ (- (exp x) 1) (log (exp x))) (/ (- (exp x) 1) x))

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