Average Error: 39.8 → 0.4
Time: 2.2s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -2.285697559519662 \cdot 10^{-4}:\\ \;\;\;\;e^{x} \cdot \frac{1}{x} - \frac{1}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right) + 1\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -2.285697559519662 \cdot 10^{-4}:\\
\;\;\;\;e^{x} \cdot \frac{1}{x} - \frac{1}{x}\\

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

\end{array}
double f(double x) {
        double r82828 = x;
        double r82829 = exp(r82828);
        double r82830 = 1.0;
        double r82831 = r82829 - r82830;
        double r82832 = r82831 / r82828;
        return r82832;
}
↓
double f(double x) {
        double r82833 = x;
        double r82834 = -0.0002285697559519662;
        bool r82835 = r82833 <= r82834;
        double r82836 = exp(r82833);
        double r82837 = 1.0;
        double r82838 = r82837 / r82833;
        double r82839 = r82836 * r82838;
        double r82840 = 1.0;
        double r82841 = r82840 / r82833;
        double r82842 = r82839 - r82841;
        double r82843 = 0.5;
        double r82844 = 0.16666666666666666;
        double r82845 = r82833 * r82844;
        double r82846 = r82843 + r82845;
        double r82847 = r82833 * r82846;
        double r82848 = r82847 + r82837;
        double r82849 = r82835 ? r82842 : r82848;
        return r82849;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
Target40.2
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.0002285697559519662

    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 -0.0002285697559519662 < x

    1. Initial program 60.0

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

    2. Taylor expanded around 0 0.5

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

    4. Applied associate-+r+0.5

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

    5. Simplified0.5

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -2.285697559519662 \cdot 10^{-4}:\\ \;\;\;\;e^{x} \cdot \frac{1}{x} - \frac{1}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right) + 1\\ \end{array}\]

Reproduce

herbie shell --seed 2020057 
(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))