Average Error: 39.8 → 0.3
Time: 12.2s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.23807481420980637978890293027234292822 \cdot 10^{-4}:\\ \;\;\;\;e^{x} \cdot \frac{1}{x} - \frac{1}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right) + 1\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.23807481420980637978890293027234292822 \cdot 10^{-4}:\\
\;\;\;\;e^{x} \cdot \frac{1}{x} - \frac{1}{x}\\

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

\end{array}
double f(double x) {
        double r52873 = x;
        double r52874 = exp(r52873);
        double r52875 = 1.0;
        double r52876 = r52874 - r52875;
        double r52877 = r52876 / r52873;
        return r52877;
}
↓
double f(double x) {
        double r52878 = x;
        double r52879 = -0.00012380748142098064;
        bool r52880 = r52878 <= r52879;
        double r52881 = exp(r52878);
        double r52882 = 1.0;
        double r52883 = r52882 / r52878;
        double r52884 = r52881 * r52883;
        double r52885 = 1.0;
        double r52886 = r52885 / r52878;
        double r52887 = r52884 - r52886;
        double r52888 = 0.5;
        double r52889 = 0.16666666666666666;
        double r52890 = r52889 * r52878;
        double r52891 = r52888 + r52890;
        double r52892 = r52878 * r52891;
        double r52893 = r52892 + r52882;
        double r52894 = r52880 ? r52887 : r52893;
        return r52894;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
Target40.3
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 < -0.00012380748142098064

    1. Initial program 0.1

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

    3. Applied div-sub0.1

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

    5. Applied div-inv0.1

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

    if -0.00012380748142098064 < 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

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.23807481420980637978890293027234292822 \cdot 10^{-4}:\\ \;\;\;\;e^{x} \cdot \frac{1}{x} - \frac{1}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right) + 1\\ \end{array}\]

Reproduce

herbie shell --seed 2019209 
(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))