Average Error: 39.4 → 0.3
Time: 25.7s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -0.000188050340254322:\\ \;\;\;\;\frac{\left(\sqrt[3]{\log \left(e^{e^{x} - 1}\right)} \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}\right) \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{\left(x \cdot x\right) \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}{x}\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -0.000188050340254322:\\
\;\;\;\;\frac{\left(\sqrt[3]{\log \left(e^{e^{x} - 1}\right)} \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}\right) \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}}{x}\\

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

\end{array}
double f(double x) {
        double r3836001 = x;
        double r3836002 = exp(r3836001);
        double r3836003 = 1.0;
        double r3836004 = r3836002 - r3836003;
        double r3836005 = r3836004 / r3836001;
        return r3836005;
}
↓
double f(double x) {
        double r3836006 = x;
        double r3836007 = -0.000188050340254322;
        bool r3836008 = r3836006 <= r3836007;
        double r3836009 = exp(r3836006);
        double r3836010 = 1.0;
        double r3836011 = r3836009 - r3836010;
        double r3836012 = exp(r3836011);
        double r3836013 = log(r3836012);
        double r3836014 = cbrt(r3836013);
        double r3836015 = r3836014 * r3836014;
        double r3836016 = r3836015 * r3836014;
        double r3836017 = r3836016 / r3836006;
        double r3836018 = r3836006 * r3836006;
        double r3836019 = 0.16666666666666666;
        double r3836020 = r3836006 * r3836019;
        double r3836021 = 0.5;
        double r3836022 = r3836020 + r3836021;
        double r3836023 = r3836018 * r3836022;
        double r3836024 = r3836023 + r3836006;
        double r3836025 = r3836024 / r3836006;
        double r3836026 = r3836008 ? r3836017 : r3836025;
        return r3836026;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.4
Target38.6
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 < -0.000188050340254322

    1. Initial program 0.1

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

    3. Applied add-log-exp0.1

      \[\leadsto \frac{\color{blue}{\log \left(e^{e^{x} - 1}\right)}}{x}\]
    4. Using strategy rm

    5. Applied add-cube-cbrt0.1

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

    if -0.000188050340254322 < x

    1. Initial program 60.2

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

    2. Taylor expanded around 0 0.4

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

    3. Simplified0.4

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -0.000188050340254322:\\ \;\;\;\;\frac{\left(\sqrt[3]{\log \left(e^{e^{x} - 1}\right)} \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}\right) \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{\left(x \cdot x\right) \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}{x}\\ \end{array}\]

Reproduce

herbie shell --seed 2019152 
(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))