Average Error: 39.9 → 0.3
Time: 4.6s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.696884717927048171620674210657853109296 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{2} \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}{x}\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.696884717927048171620674210657853109296 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}{x}\\

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

\end{array}
double f(double x) {
        double r68287 = x;
        double r68288 = exp(r68287);
        double r68289 = 1.0;
        double r68290 = r68288 - r68289;
        double r68291 = r68290 / r68287;
        return r68291;
}
↓
double f(double x) {
        double r68292 = x;
        double r68293 = -0.00016968847179270482;
        bool r68294 = r68292 <= r68293;
        double r68295 = exp(r68292);
        double r68296 = 3.0;
        double r68297 = pow(r68295, r68296);
        double r68298 = 1.0;
        double r68299 = pow(r68298, r68296);
        double r68300 = r68297 - r68299;
        double r68301 = r68298 + r68295;
        double r68302 = r68298 * r68301;
        double r68303 = r68292 + r68292;
        double r68304 = exp(r68303);
        double r68305 = r68302 + r68304;
        double r68306 = r68300 / r68305;
        double r68307 = r68306 / r68292;
        double r68308 = 2.0;
        double r68309 = pow(r68292, r68308);
        double r68310 = 0.16666666666666666;
        double r68311 = r68292 * r68310;
        double r68312 = 0.5;
        double r68313 = r68311 + r68312;
        double r68314 = r68309 * r68313;
        double r68315 = r68314 + r68292;
        double r68316 = r68315 / r68292;
        double r68317 = r68294 ? r68307 : r68316;
        return r68317;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.9
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 < -0.00016968847179270482

    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

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

    if -0.00016968847179270482 < x

    1. Initial program 60.2

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

    2. Taylor expanded around 0 0.4

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

    3. Simplified0.4

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.696884717927048171620674210657853109296 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{2} \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}{x}\\ \end{array}\]

Reproduce

herbie shell --seed 2019308 
(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))