Average Error: 39.7 → 0.4
Time: 14.4s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -0.00017431816494063913:\\ \;\;\;\;\frac{e^{x}}{x} - \frac{1}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + \log \left(e^{\left(\frac{1}{2} + x \cdot \frac{1}{6}\right) \cdot x}\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -0.00017431816494063913:\\
\;\;\;\;\frac{e^{x}}{x} - \frac{1}{x}\\

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

\end{array}
double f(double x) {
        double r3056896 = x;
        double r3056897 = exp(r3056896);
        double r3056898 = 1.0;
        double r3056899 = r3056897 - r3056898;
        double r3056900 = r3056899 / r3056896;
        return r3056900;
}
↓
double f(double x) {
        double r3056901 = x;
        double r3056902 = -0.00017431816494063913;
        bool r3056903 = r3056901 <= r3056902;
        double r3056904 = exp(r3056901);
        double r3056905 = r3056904 / r3056901;
        double r3056906 = 1.0;
        double r3056907 = r3056906 / r3056901;
        double r3056908 = r3056905 - r3056907;
        double r3056909 = 0.5;
        double r3056910 = 0.16666666666666666;
        double r3056911 = r3056901 * r3056910;
        double r3056912 = r3056909 + r3056911;
        double r3056913 = r3056912 * r3056901;
        double r3056914 = exp(r3056913);
        double r3056915 = log(r3056914);
        double r3056916 = r3056906 + r3056915;
        double r3056917 = r3056903 ? r3056908 : r3056916;
        return r3056917;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.7
Target38.9
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 < -0.00017431816494063913

    1. Initial program 0.1

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

    3. Applied div-sub0.1

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

    if -0.00017431816494063913 < x

    1. Initial program 60.0

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

    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

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

    5. Applied add-log-exp0.5

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -0.00017431816494063913:\\ \;\;\;\;\frac{e^{x}}{x} - \frac{1}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + \log \left(e^{\left(\frac{1}{2} + x \cdot \frac{1}{6}\right) \cdot x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019141 
(FPCore (x)
  :name "Kahan's exp quotient"

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

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