Average Error: 39.8 → 0.3
Time: 18.2s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.842286630228784469372194099179296244984 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{e^{x} \cdot \left(e^{x} \cdot e^{x}\right) - 1 \cdot \left(1 \cdot 1\right)}{e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{6} \cdot x + \frac{1}{2}\right) + 1\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.842286630228784469372194099179296244984 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{e^{x} \cdot \left(e^{x} \cdot e^{x}\right) - 1 \cdot \left(1 \cdot 1\right)}{e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)}}{x}\\

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

\end{array}
double f(double x) {
        double r8137908 = x;
        double r8137909 = exp(r8137908);
        double r8137910 = 1.0;
        double r8137911 = r8137909 - r8137910;
        double r8137912 = r8137911 / r8137908;
        return r8137912;
}
↓
double f(double x) {
        double r8137913 = x;
        double r8137914 = -0.00018422866302287845;
        bool r8137915 = r8137913 <= r8137914;
        double r8137916 = exp(r8137913);
        double r8137917 = r8137916 * r8137916;
        double r8137918 = r8137916 * r8137917;
        double r8137919 = 1.0;
        double r8137920 = r8137919 * r8137919;
        double r8137921 = r8137919 * r8137920;
        double r8137922 = r8137918 - r8137921;
        double r8137923 = r8137916 * r8137919;
        double r8137924 = r8137920 + r8137923;
        double r8137925 = r8137917 + r8137924;
        double r8137926 = r8137922 / r8137925;
        double r8137927 = r8137926 / r8137913;
        double r8137928 = 0.16666666666666666;
        double r8137929 = r8137928 * r8137913;
        double r8137930 = 0.5;
        double r8137931 = r8137929 + r8137930;
        double r8137932 = r8137913 * r8137931;
        double r8137933 = 1.0;
        double r8137934 = r8137932 + r8137933;
        double r8137935 = r8137915 ? r8137927 : r8137934;
        return r8137935;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
Target40.2
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.00018422866302287845

    1. Initial program 0.1

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

    3. Applied flip3--0.1

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

    4. Simplified0.1

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

    if -0.00018422866302287845 < x

    1. Initial program 60.1

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

    2. Taylor expanded around 0 0.5

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

    3. Simplified0.5

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.842286630228784469372194099179296244984 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{e^{x} \cdot \left(e^{x} \cdot e^{x}\right) - 1 \cdot \left(1 \cdot 1\right)}{e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{6} \cdot x + \frac{1}{2}\right) + 1\\ \end{array}\]

Reproduce

herbie shell --seed 2019173 
(FPCore (x)
  :name "Kahan's exp quotient"

  :herbie-target
  (if (and (< x 1.0) (> x -1.0)) (/ (- (exp x) 1.0) (log (exp x))) (/ (- (exp x) 1.0) x))

  (/ (- (exp x) 1.0) x))